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SPEED AND ACCURACY BRING SUCCESS IN MATHEMATICS
Higher Secondary – second Year Mathematics
BLUE PRINT
Chapter
No.
1
2
3
4
5
6
7
8
9
10
No. of Questions
1 Mark
6 Marks
10 Marks
Chapters
Application of Matrices and
Determinants
Vector Algebra
Complex Numbers
Analytical Geometry
Differential Calculus Application - 1
Differential Calculus Application - 2
Integral calculus and its applications
Differential Equations
Discrete Mathematics
Probability Distributions
Total Number of Questions
1
Total
Marks
4
2
1
26
6
4
4
4
2
4
4
4
4
40
2
2
1
2
1
1
1
2
2
16
2
1
3
2
1
2
2
1
1
16
38
26
40
36
18
30
30
26
26
296
Easy to score good marks for average students;
TABLE – I
Chapter
No.
2
4
No. of
Chapters
Questions
No. of
10 Mark
Questions
Total
Marks
-----
20(2)
28(3)
20
30
6 & 3 Mark
Vector Algebra
Analytical Geometry
50 marks can be scored if we practice 20 Ten mark questions in Lesson -2 and 28 Ten
mark questions in Lesson – 4
TABLE – II
Chapter
No.
9
6
3
No. of
Chapters
Questions
No. of
10 Mark
Questions
Total
Marks
33+12(2)
-----
15(1)
11(1)
16(1)
22
10
10
6 & 3 Mark
Discrete Mathematics
Differentials calculus Application – II
Complex Number
42 marks can be scored if we practice 15 Ten mark questions 31 Six mark questions
12 Three mark questions in Lesson 9 and 11 Ten mark questions in Lesson – 6 and 16 Ten
mark question in Lesson – 3
TABLE – III
Chapter
No.
1
10
Questions
No. of
10 Mark
Questions
Total
Marks
35+13(2)
---
12
---
---
---
No. of
Chapters
6 & 3 Mark
Application of Matrices and
Determinants
Probability Distributions
12 Marks can be scored if we practice 35 Six mark questions and 13 Three mark questions
in Lesson – 1.
TABLE – IV
Chapter
No.
5
7
8
No. of
Chapters
Questions
No. of
10 Mark
Questions
Total
Marks
-------
-------
-------
6 & 3 Mark
Differentials calculus Application – I
Integral calculus and its applications
Differential Equations
30 marks can be achieved if we study all the 271 one mark questions in the text book. If we practice
completely from table I to VI we can score 134 marks.
2
EASY WAY TO GET; 100%
 Slip test should be conducted repeatedly on the 271 one words questions in the text book and 380
questions in the COME book
 We can easily answer 9 out of the 10 six mark questions in the question paper if we practice all the
question in lesson, 1,2,3,9,10

For the remaining questions you have to concentrate on all the lessons
 We can easily answer 9 out of the 10 ten mark questions in the question paper if we practice all the 10
mark question in lesson 1,2,3,4,6,9,10

For the remaining 1 question you have to concentrate on all the lesson
 Created question can be answered easily if we have practice on all the lessons
 At least 5 full portions should be written before the public exam. It will bring the following results
o We can assess if we could answer all the questions with in the stipulated time (3hours)
o We can analyze whether we could answer all the questions the extent of scoring full marks
o We can identify the hurdles to score full marks and pass accordingly
 Avoid writing without reading the questions thoroughly
o (E.g) Without reading the questions Examples 4.35 and 5.7, if we just read ‘ladder’ we may give
a completely and observing thepictures promptly
 Avoid answering in a hurry without reading the questions completely and observing the pictures
promptly
o (E.g) It is possible to answer using the formula of “PARABOLA” instead of “ELLIPSE” if one
observes the figure of a semi ellipse as inverted parabola in Example 4.32 & Exercise 4.2 – 10.
 Use pen for writing the answers and pencil for drawing the diagrams
 Don’t waste your precious time on colouring the pages
 Those who aim at centum marks should give extra attention to one mark questions.
3
MATHEMATICS
VECTOR ALGEBRA
Hints for solving 10 marks questions.
1.
2. if
+ a3
a1 + a 2
a1 b1
3. If
+
;
b1 + b 2
+ b3
; then
a2 b2 + a3 b 3
are perpendicular, then
4.
5. If
a1 + a 2
=
6. If
+ a3
;
a1
b1
a1 + a 2
+ a3
+ b3
b1 + b 2
a2
b2
;
; then
a3
b3
b1 + b 2
+ b3
a1
b1
c1
=
7. Any point on the straight line
; and
a2
b2
c2
=
( l λ + x1 , mλ + y1 , nλ + z1 ) where λ
c1 + c2
a3
b3
c3
=
+ c3
= λ is given by
R
8. The conditions for two given straight lines
intersect each other are :
(i) and
(ii)[
are not parallel, and
]=0
=
9. A Plane is parallel to the straight line
i) The plane is parallel to the vector l + m
=
then
+n
ii) The plane need not pass through the point (x1, y1, z1)
10. A plane contains the straight line
=
i) The plane is parallel to the vector l + m
=
+n
then
ii) The plane passes through the point (x1, y1, z1)
4
CHAPTER – 4
ANALYTICAL GEOMETRY
TIPS TO SOLVE 10 MARK QUESTIONS
y
1.
a) The parabola is open upwards;
b) The equation of the parabola is X2 = 4aY
x
0
2.
a) The parabola is open downwards
b) The equation of the parabola is X2 = -4aY ;
y
x
0
3.
y
a) The parabola is open right wards
b) The equation of the parabola is Y2 = 4aX
0
x
y
4.
a) The parabola is open Leftwards
b) The equation of the parabola is Y2 = - 4aX
0
x
D
M
P
5. In a parabola,
=1
F(focus)
D’
5
6. tan Ѳ = value of
at p(x1, y1)
y = f(x)
p(x1, y1)
Ѳ
7.
Straight line (l)
A Straight parallel to (l)
A Straight line perpendicular to (l)
ax + by + c = 0
ax + by + k = 0
bx - ay + k = 0
8. Let the separate equations of asymptotes be ax + by + c = 0 and l x + my + n = 0.
Therefore (i) The Combined equation of the asymptotes is (ax + by + c) (l x + my + n) = 0 and
(ii) The equation of the hyperbola is (ax + by + c) (l x + my + n) + k = 0
9. The asymptotes of a hyperbola pass through the centre of the hyperbola
10. In a rectangular hyperbola, the asymptotes are perpendicular to each other.
11.
S.no
Curve
1
Ellipse :
Condition that y = mx + c is a
tangent
,
,a>b
2
The point of contact
)
Hyperbola :
,
,
6
)
CHAPTER 6
APPLICATIONS OF DIFFERENTIAL CALCULUS – II
Needed formulae for solving 10 mark questions
1.
2.
3.
4.
5. d( )=
6. d (sin(ax) ) = a cos(ax)
7. d (cos(ax) ) = -a sin(ax)
= cos u
8.
9.
= cos u
;
= sec2 u
;
= sec2 u
10. If y = f(x), then
(i) dx = x
(ii) dy
y
(iii) dy =
(iv) f (x +
x)
y+ dy
7
CHAPTER 9
DISCRETE MATHEMATICS
TIPS FOR SOLVING 10 MARK QUESTIONS
S.No
A non- empty Let, G
1
G = { a+b
}
2
/ a,b
n
G=Z
4
G=Q
Addition
Multiplication
G= {2 / n
3
Binary Operation
a x b =a+ b+ 2
+
a*b=
Identify
element
e=0+0
e=2
0
Inverse element
The inverse of
(a+b
where -a, -b
a
The inverse of2
-a
2
where -a
, -4-a
is the
inverse element
e = -2
for a
e=3
for a
is the
inverse element
5
G=Q–{1}
a * b = a+b-ab
6
G = Q – { -1 }
a*b= a+b+ab
e=0
e=0
is the
for a
inverse element
is the
for a
inverse element
7
G= {Z /
=1}
Multiplication
1+ 0 i
is the inverse
For Z
element
8
G = Zn
Addition Modulo n
e = [0]
for [a]
, [n-a]
inverse element
9
10
G = set of all nth roots Multiplication
of unity
G=
Matrix Multiplication
e=1=
E=
for
inverse element
is the
is the
For
is the inverse
element
11
G=
Matrix Multiplication
E=
For
is the inverse
element
8
CHAPTER – 3
COMPLEX NUMBERS
Hints for solving 10 marks questions.
1. If z1 = a + i b, z2 = c + i d then z1 . z2 = (ac – bd) + i (ad + bc)
=
2. If z1 = a + i b, z2 = c + i d then
3. Let z = a + ib then
4. arg
+i
= a – i b and z.
= (a + ib) (a -ib) = a2 + b2
= arg z1 - arg z2
5.
6.
7.
=
g(x) = 0
8.
=
=
9. If x2 – 2 px + ( p2 +q2) = 0 Then, x = p
10. i) 1 = 1 [ cos 0 + I sin 0 ]
ii) -1 = 1 [ cos π + i sin π]
iii) i = 1 [ cos
+ i sin
qi
]
iv) -i = 1 [ cos (- )+ i sin(- ) ]
11.
i) 1 + i
ii) 1 – i
cos
cos
iii)
iv) v)
12.
13.
cos
i = 2 [ cos i
= 1 [ cos (-
+ i sin
- i sin
+ i sin
+ i sin )+ i sin (-
Let x, y be positive
x + i y = r [ cos
ii)
x - i y = r [ cos
iii)
-x - iy = r [cos (
iv)
-x – iy = r [ cos (
]
]
], where r = +
and cos
]
)]
x4 – x3 + x2 -x + 1 = 0
≠
9
; sin
then
CHAPTER - 8
DIFFERENTIAL EQUATION ( APPLICATION )
(10 MARKS)
Note :
Compulsarily write down the following two steps in solving all the problems in
this chapter
In all the practical problems we apply the principle that the rate of change of population is
directly proportional to the initial population.
Step1:
or
Stp2: A=c
CHAPTER – 7
INTEGRAL CALCULUS ( APPLICATION )
(10 MARKS)
Length of the curve
1. Exercise 7.5
Question no 1 and 2
2. Example 7.37 and 7.38
Surface of area of a solid
1. Exercise 7.5
Question no 3 and 4
2. Example 7.39 and 7.40
10
MATHEMATICS
2. VECTOR ALGEBRA (10 MARK)
Two questions for full test
Total number of questions : 20
1) Prove that Cos(A-B) = CosACosB + SinAsinB
3) Prove that Cos(A+B) = CosACosB – SinASinB
Solution:
Solution:
Let P(CosA, SinA) and
Let P(CosA, SinA) and Q(CosB, -SinB) be any two
Q(CosB, SinB) be any two
points on the unit circle with centre at the origin O.
points on the unit circle with
Let and be the unit vectors along the cocentre at the origin O.
ordinate axes.
Let and be the unit
=CosA +SinA
vectors along the
=CosB –SinB
co-ordinate axes.
=CosA +SinA
=CosB +SinB
=
Cos(A–B)
= Cos(A–B)……...(1)
.
.
=
Cos (A+B)
= Cos(A+B)………….(1)
.
=CosACosB – SinASinB…….(2)
(1)=(2)⇒ Cos (A+B)=CosACosB – SinASinB
.
= CosACosB+SinASinB ……(2)
(1)=(2) Cos (A–B)=CosACosB+SinASinB
4) Prove thatSin (A+B) = SinACosB +CosASinB
Solutons:
Let P(CosA, SinA) and Q (CosB, -SinB) be any two
points on the unit circle
with centre at the origin
O.
Let and be the unit
vectors along the
co-ordinate axes.
2) Prove that Sin(A-B) = SinACosB – CosASinB
Solution:
Let P(CosA, SinA) and
Q(CosB,SinB) be any two
points on the unit circle
with centre at the origin O.
Let and be the unit
vectors along the
co-ordinate axes
=CosA +SinA
=CosA +SinA
=CosB –SinB
=CosB +SinB
×
x
=
Sin (A–B)
=
Sin (A+B)
= Sin (A+B) …………..(1)
= Sin (A–B) ….(1)
×
×
=
=
=(SinACosB+CosASinB) ……….(2)
(1)=(2) ⇒ Sin (A+B)=SinACosB+CosASinB
= (SinACosB –CosASinB)…(2)
(1)=(2) ⇒ Sin (A–B)=SinACosB –CosASinB
11
5) Altitides of a triangle are concurrent prove by
vector method.
Solution:
Let ABC be the given triangle.
Let the altitudes AD and BE intersecting at O and take
it as the origin.
To prove that CO is
perpendicular to AB.
7) If
and
×
=
+ ,
=
+ ,
=
+
+
= + + 2 , then verify that
)×( ×
)=
] -[
]
Solution:
×
=
= (1–0)– (1–2)+ (0–2)
=
,
AD
=
,
=
=0⇒
.
=
BC ⇒
/
.
=0…………. (1)
×
=
CA ⇒
BE
=0⇒
/
= (2–1)– (4–1)+ (2–1)
)=0..........(2)
.
=
(1) +(2)⇒
=0
×
(
) . =0⇒
.
=0
⇒ OC
AB
Hence the altitudes of a triangle are concurrent.
6) If
=
=
-
×
×
+
-
,
=
+
)× (
)=
×
= (1–6)– (1+2)+
=
,
. Verify that
….(1)
]=
[
)=
(–3–1)
=1(0–1)–1(2–2)+1(2–0)=1
Solution:
]=
=
=1(0–1)–1(4–1)+1(2–0)=–2
= (0 – 5) – (6 – 0)+
(–2 –0)
]
=
= –2(
×(
)=
(–12 + 15)
×
…………(1)
.
=
.
=
=6
)= –9
=6(
)+9(
=
…………(2)
(1) =(2)⇒
×
×
+
= –5 –3 –4
(1)=(2)⇒
= (–6 – 6) – (–4 – 5)+
=
+
)=
12
)×( ×
–[
) – 1( +
]
+2
)
…..(2)
)=
] –[
]
8) Show that the lines
= =
=
=
10) Find the vector and cartesian equations of
the plane through the point (2,–1,–3) and
and
intersect and hence find the point
parallel to the lines
of intersection.
Solution:
Let
=
Let
,
=
– –3
–4 ;
=2 –3 +2
Vector equation is
, ]=
=
=
=
(
=
=
=0
= µ ⇒ Any point on theline is
(x–2)(–8) – (y+1)(14)+(z+3)(–13) = 0
–8x+16 – 14y–14–13z–39 = 0
8x+14y+13z+37 = 0
(2
…………..(2)
From (1) and (2) –λ + 1 = 0 (or) 3µ – 1 = –1
,
µ=0
∴ Point of intersection is (4, 0, –1)
9) Show that the lines
=
=
=
=
11) Find the vector and cartesian equations
of the plane through the point (1,3,2) and
parallel to the lines
and
intersect and find their point of
=
intersection.
Solution:
Let
Let
=2 + –
= – +3 ;
[
=
= +2 –
,
, ]=
=0
=
=
(
Let
= ⇒ Any point on the line is
=
=
=
+2
+3 ;
+ 3 )+t ( + 2 +2 )
=0
………….(1)
=
= +
= + +2 +s(
Cartesian equation is
∴ The lines are intersecting
Let
and
= +2 +2
Vector equation is
= +
+t
= +2 –
–
=
Solution:
= – ;
–
+t
=0
= ⇒ Any point on theline is
………….(1)
Let
+
=2 – –3 + s(
– 4 )+t (2 –3 + 2 )
Cartesian equation is
=0
∴ The lines are intersecting.
Let
and
=
=
=2 +3
–
=3 –
–
=
Solution:
= + –
=4 –
=3 – ,
[
=
= µ ⇒ Any point on the line is
(
…………..(2)
From (1) and (2)
⇒ λ – µ = 1 ……….(3)
–λ –1= 2µ +1 ⇒ – λ –2µ =2 ………….(4)
(3)+(4)⇒ µ = – 1, λ = 0
∴ Point of intersection is (1, –1, 0)
=0
(x–1)(–8)–(y–3)(1)+(z–2)(5) = 0
8x+y–5z–1 = 0
13
12) Find the vector and cartesian equations of
the plane containing the lines
and parallel to the line
=
=
15) Find the vector and cartesian equations of
the plane through the points (–1,1,1) and
(1,–1,1) and perpendicular to the plane
x +2y +2z = 5
=
=
Solution: = +2 +
=
Solution:
+3 , = +2 +
+ + ;
+t
Vector equation is
+3 ) + t( +2 + )
Cartesian equation is
= (1 – s)( + + + s(
Cartesian equation is
=0
=
+
+2
=
+
=
+2
;
+ )+t( + +
=– +
+2 +s(
)
+t
+ 2 )+ t (
+ +
)
Cartesian equation is
=0
(x + 1) (–4) – (y – 1) (4) + (z – 1) (6) = 0
– 4x – 4 – 4y + 4 + 6z –6 = 0
⇒ 2x + 2y –z + 3 = 0
16) Find the vector and cartesian equations of
the plane through the points (1,2,3) and
(2,3,1) perpendicular to the plane
3x – 2y +4z –5 = 0
=3 + +2
Vector equation is
+t
=0
(x – 2) (–3) – (y – 2) (–7) + (z – 1) (–5) = 0
–3x + 6 + 7y – 14 –5z + 5 = 0
3x – 7y +5z + 3 = 0
13) Find the vector and cartesian equations to the
plane through the point (–1,3,2) and perpendicular
to the planes x + 2y + 2z = 5 and 3x + y + 2z = 8
Let
+ ;
= (1–s) +
=0
Solution:
=
= +2 +2
Vectors equation is = +
= 2 + 2 + + s(
=
=0
Solution :
=
=0
Let
+
= +
;
Vector equation is
=
+
;
–2 + 4
=(1–s)
+
+t
=(1–s)( + +3 )+s(2
+ )+t(3 – +
(x+1)(2) – (y–3) (–4) + (z–2) (–5) = 0
Cartesian equation is
2x+4y –5z = 0
14) Find the vector and cartesian equations of
the plane passing through the points A (1,–2,3) and
=0
A (1,–2,3) and B(–1,2,–1) and parallel to
the line
=
Solution:Let
=
= –2 + 3
=–
–
;
=0
= 2 +3 + 4
Vector equation is = (1–s)
=(1–s)( –2 +3 )+s(
+
+t
– )+t(
Cartesian equation is
+
+
(x–1) (0) –(y–2) (10) + (z–3) (–5) = 0
)
–10y –5z + 35 = 0
=0
÷ by –5
=0
⇒ 2y +z – 7 = 0
(x–1) (28) –(y+2) (0) + (z–3) (–14) = 0
28x – 14z + 14 = 0 ⇒ 2x – z + 1 = 0
14
)
17) Find the vector and cartesian equations of
the plane containing the line
=
19) Find the vector and Cartesian equations of
the plane passing through the points with
=
position vectors
and passing through the point (–1, 1, –1)
Solution:
=
+ –
=
;
Vector equation is
+ ;
= (1 – s) +
= (1–s)(– + – ) +s(2
=
+3 – 2
–
)
Cartesian equation
Let
=
+4 +2
= (1–s–t)
+4 +2 )+s(2
(x – 3) (–6) – (y – 4) (13) + (z – 2) (28) = 0
–6x – 13y + 28z + 14 = 0
6x + 13y – 28z – 14 = 0
20) Derive the equation of the plane in the
intercept form
Solution:
Let a, b, c be the x, y and z intercepts of the plane
–
+
+
= (1–s–t) (2 +
∴ =
=
A(a, 0, 0)
B(0, b, 0)
C(0, 0, c)
+t
–
– )+
=0
= +6
Vector equation is
= (1–s–t)
+t
=0
18) Find the vector and cartesian equations of
the plane passing through the points (2, 2,–1),
(3, 4, 2) and (7, 0, 6)
Solution :
=
+
t(7 + )
Cartesian equation is
(x + 1) (–8) – (y – 1) (–10) + (z + 1) (7) = 0
–8x + 10y + 7z – 11 = 0
⇒8x-10y-7z+11= 0
+
–
Vector equation is = (1 – s – t)
=0
=
=2
=7
=0
Let
,
7
Solution :
+t
+ )+t(2 +
+4 +2 ,2
+ s(3
+ t (7 +
+
=
)
)
(i) Vector equation is
Cartesian equation is
= (1–s–t)
+
+t
= (1–s–t) a +
+ tc
(ii) Cartesian equation is
=0
=0
=0
=0
(x–2) (20) – (y–2) (–8) + (z+1) (–12) = 0
20x – 40 + 8y –16 –12z –12 = 0
(x – a) (bc–0)–y(–ac–0)+z(0+ab) = 0
20x+8y –12z –68 = 0
xbc – abc + yac + zab = 0
xbc+ yac + zab = abc
Dividing by abc
5x + 2y –3z –17 = 0
15
4. ANALYTICAL GEOMETRY (10 MARK)
Three questions for full testTotal number of questions 28
1) Find the axis, vertex, focus, directrix, equation
2) Find the axis, vertex, focus, directrix,
of the latus rectum and length of the latus
equation of the latus rectum and length latus
rectum of the parabola
rectum of the parabola
and hence draw their
and hence draw their graph,
Solution:
graph,
Solution:
(y + 3)2 = 8x , [ Y2 = 4aX]
a=2
The type is open rightward
= 12( y +1) ); [
]
a=3
About (0, 0)
About (0, –3)
The type is open upward
Axis
Y=0
y = 0–3, y = –3
About (0, 0) About (3, –1)
Vertex
(0, 0)
V(0, 0)+ (0, –3)
Axis
X=0
x = 0+3, x = 3
V(0, –3)
Vertex
(0, 0)
V (0, 0)+(3, –1)
Focus
(a, 0) = (2, 0)
F(2, 0)+ (0, –3)
V(3, –1)
F(2, –3)
Focus
(0, a) = (0, 3) F(0, 3) +(3, –1)
Directrix
X = –a, X = –2
x = –2+ 0, x = –2
F(3, 2)
Latus Rectum X = a, X = 2
x = 2 + 0, x = 2
Directrix
Y = –a, Y = –3 y = –3–1, y = –4
Length L.R
4a
8
Latus Rectum Y = a, Y = 3
y = 3–1, y = 2
Length L.R
3) Find the axis, vertex, focus, directrix,
equation of the latus rectum and length
latus rectum of the parabola
and hence draw their
graph,
Solution:
, [
a=2
The type is open leftward
About (0, 0)
About (1, 3)
Axis
Y=0
y = 0+3,
y=3
Vertex
(0, 0)
V (0, 0)+(1,3)
V(1, 3)
Focus
(–a, 0)=(–2, 0) F(–2, 0) +(–1, 3)
F(–3, 3)
Directrix
X = a, X = 2
x = 2+1 , x = 3
Latus Rectum X = –a, X= –2
x = – 2+1 , x = – 1
Length L.R
4a
8
4a
12
(4) Find the axis, vertex, focus, directrix,
equation of the latus rectum and length latus
rectum of the parabola
and hence draw their
graph,
Solution:
= –8( y +2); [
The type is open downward
About (0, 0)
Axis
X=0
Vertex
(0, 0)
Focus
(0, –a) = (0, –2)
Directrix
Y = a, Y = 2
Y = –a, Y =–2
4a
Latus Rectum
Length L.R
16
]
a=2
About (1, –2)
x =0+1, x = 1
V (0, 0) +(1, –2)
V(1, –2)
F (0, –2) +(1, –2)
F(1, –4)
y = 2–2, y = 0
y = –2–2, y =– 4
8
5) Find the eccentricity, centre, foci, vertices of
the following ellipse and draw the diagram
About (0, 0)
Centre
Foci
Solution:
(
+ 4(
(68+16+16)
= 100 a = 10;
= 25
Major axis is parallel to x–axis
b=5
Eccentricity =
=
=
About (0, 0)
C (0, 0)
Centre
(ae,0)=(5
Foci
C (0,0)
C(1, –4)
(0,ae)=(0,
(0, –
)
(1,–4+
)
(1,–4–
)
ae)=(0,
)
A (1,2),
Vertices A (0, a)= A (0,6)
(0, –a)= (0, –6)
(1,–10)
, ae = 5
About (4, 2)
C (4,2)
,0)
(–ae,0)=(–5
About (1, –4)
,0),
A (a,0)= A (10,0)
Vertices
(–a,0), (–10,0)
(5
,0)+(4,2)
= (4+5
,2)
(–5
7) Find the eccentricity, centre, foci, vertices of
the following ellipse and draw the diagram
,0)+(4,2)
= (4 – 5 ,2)
A (10,0) +(4,2)
= (14,2),
(–10,0) +(4,2)
= (–6,2)
Solution:
16(
16
(92+16+36)
;
=9
b=3
The major axis is parallel to y–axis
eccentricity =
6) Find the eccentricity, centre, foci, vertices of
thefollowing ellipse and draw the diagram
ae = 4 ×
Centre
Foci
Solution:
36(
36(
+4
= 144
Vertices
= 36 ⇒ a = 6;
=4 ⇒ b=2
The major axis is parallel to y–axis
eccentricity =
=
=
ae = 4
17
=
=
About (0, 0)
C (0,0)
(0,ae) =(0,
(0, –ae)=(0,
(44+36+64)
;
=
About (–1,2)
C (–1,2)
)
)
A (0, a)= A (0,4)
(0, –a)= (0, –4)
(–1,2+
)
(–1,2–
)
A (–1,6)
(–1, –2)
8) Find the eccentricity centre, foci and vertices
of the hyperbola
= 0 and also
trace the curve
Solution:
9(
– 16(
9
– 16
9) Find the eccentricity, centre, foci and
vertices of the following hyperbola
and draw the
diagram
Solution:
= 199
= 144
–4
(199 + 96 – 4)
=4
(11+ 9 –16)
;
= 16
a=4;
=9
=4
b =3
a=2 ,
Transverse axis is parallel to x–axis
Transverse axis is parallel to x–axis
eccentricity =
eccentricity =
=
=
ae = 5
About (0, 0)
Centre
C (0, 0)
Foci
(ae, 0) = (5, 0)
(–ae, 0) = (–5, 0)
Vertices
A (a, 0) = A (4, 0)
(–a, 0)=
(–4, 0)
= 1 b =1
=
=
ae =
About (1, –2)
C (1, –2)
(5, 0) +(1, –2)
(6, –2)
(–5, 0)+ (1, –2)
(–4, –2)
A (4, 0) +(1, –2)
A (5, –2)
(–4, 0) +(1, –2)
(–3, –2)
Centre
Foci
About (0, 0)
C (0, 0)
F1 (ae, 0) =(
F2(–ae,0)=(
Vertices
18
About (–3, 2)
c (–3,2)
0)
0)
A
(a, 0)=(2, 0)
(Typequathere.0),
(–a, 0)=(–2, 0)
(
2)
(
, 2)
A (–1,2),
(–5, 2)
10) Find the eccentricity, centre, foci and vertices
of the hyperbola
and draw
their diagram.
Solution :
11) Find the eccentricity, centre, foci and
vertices of the hyperbola
and draw their
diagram.
Solution:
(
9(
– 16(
9
– 3(
= –164
– 16
–3
= –144
(–164+36–16)
= –12
–
=4
= –18
(–18+9–3)
=1
a = 2,
= 12
b =2
Transverse axis is parallel to y–axis
–
= 1,
eccentricity =
=9
a = 3,
= 16
=
ae = 3 ×
ae = 4
=
Foci
Vertices
Centre
Foci
=5
About (0, 0)
Centre
= 2
b =4
Transverse axis is parallel to y–axis
eccentricity =
=
C (0, 0)
(0, ae) = (0,5)
(0, –ae) = (0,
A (0, a) = A (0, 3)
(0, –a)= (0, –3)
Vertices
About (–2, 1)
c (–2, 1)
)
(–2, 6)
(–2, –4)
A (–2, 4),
(–2, –2)
19
About (0, 0)
About (–3, 1)
C (0, 0)
c (–3 1)
(0, ae) = (0,4)
(0, –ae)= (0
)
A (0, a) = A (0, 2)
(0, –a)= (0, –2)
(–3, 5)
(–3, –3)
A (–3, 3),
(–3, –1)
12) A comet is moving in a parabolic orbit
around the sun which is at the focus of a
parabola. When the comet is 80 million Kms
from the sun, the line segment from the sun
to the comet makes an angle of
radians
14) On lighting a rocket cracker it gets
projected in a parabolic path and reaches the
ground 12mts away from the starting point.
Find the angle of projection.
with the axis of the orbit. Find (i) the
equation of the comets orbit (ii) how close
does the comet come nearer to the sun?
Solution:
Equation of the parabola
…….(1)
From the right ∆FQP
Solution:The equation is
………….(1)
The point (–6, –4) lies on the parabola
= –4a (–4)
cos (
a=9
(1)
2x = –9
=
=
= –9y
=–
⇒ FQ = 40
=
= 1 ⇒ FP = PM
=
=
=
Angle of projection is
80 = 2a + 40
2a = 40⇒ a = 20
(i) The equation of the comets orbit is
(ii) The shortest distance between the sun and
The comet = 20 million kms.
15) Assume that water issuing from the end of a
horizontal pipe, 7.5m above the ground,
describes a parabolic path. The vertex of the
parabolic path is at the end of the pipe. At a
position 2.5m below the line of the pipe, the
flow of water has curved outward 3m
beyond the vertical line through the end of
the pipe, How far beyond this vertical line
will the water strike the ground?
13) The girder of a railway bridge is in the
parabolic form with span 100ft. and the
highest point on the arch is 10ft. above the
bridge. Find the height of the bridge at 10ft.
to the left or right from the midpoint of the
bridge.
Solution:
Solution:
The equation is
= –4ay ………..(1)
The point (3, –2.5) lies on the parabola
The equation is
………….(1)
The point (50, –10) lies on the parabola
= –4a (–10)
4a = 250
⇒
……..(2)
B (10, ) lies on the parabola
100
⇒
= –4a (–2.5) a =
=–
y …..(2)
The point ( , –7.5) lies on the parabola
=
ft
Height of the bridge at the required place
= 10 – = = 9 feet
Another Method :
The point (50, –10) lies on the parabola
= –4a (–10) ….(1)
B (10, – ) lies on the parabola (10)2 = –4a (– ) ….(2)
(2)÷(1) ⇒
/10=100/2500 ⇒ =
Height of the bridge at the required place
(–7.5)
= 27
The water strikes the ground 3 m beyond
the vertical line.
Another Method :
The point (3, –2.5) lies on the parabola
= –4a (–2.5) ……. (1)
The point ( , –7.5) lies on the parabola
= –4a (–7.5) …… (2)
(2)÷(1) ⇒
/9 =3
⇒
= 27
=3
m
The water strikes the ground 3
the vertical line.
= 10 – = = 9 feet
20
=3
m beyond
m
16) A cable of a suspension bridge hangs in the
form a parabola when the load is uniformly
distributed horizontally. The distance
between two towers is 1500 ft, the points of
support of the cable on the towers are 200ft
above the road way and the lowest point on
the cable is 70 ft above the roadway. Find the
vertical distance to the cable from a pole
whose height is 122ft.
Solution:
The equation is
= 4ay ………..(1)
The point (750, 130) lies
on the parabola
= 4a (130)
(1) ⇒
17) A cable of a suspension bridge is in the form
of a parabola whose span is 40 mts. The road
way is 5 mts below the lowest point of the
cable. If an extra support is provided across
the cable 30 mts above the ground level, find
the length of the support if the height of the
pillars are 55 mts.
Solution:
The equation is
= 4ay ………..(1)
The point (20, 50) lies on the parabola
= 4a (50)
a=2
(1)
= 8y ……(2)
The point ( , 25) lies on the parabola
(2) ⇒
= 8 × 25
= 200
4a =
y…….(2)
The point P( , 52) lies on the parabola
(2) ⇒
=
× 52
= 150
ft
=
Vertical distance to the cable from a pole =
= 300
ft
Another Method :
The point (750, 130) lies on the parabola
(750)2 = 4a (130) ….(1)
B (x1, 52) lies on the parabola ( x1)2 = 4a (52) ….(2)
(2)÷(1) ⇒ ( x1)2 = (750)2(52)/130 ⇒ = 150
Vertical distance to the cable from a pole =
= 10
ft
Length of the support =
= 20
ft
ft
= 300
ft
18) A Kho–kho player in a practice session while
running realizes that the sum of the distances
from the two kho–kho poles from him is
always 8 m. Find the equation of the path
traced by him if the distance between the
poles is 6m.
Solution:
The equation is
………..(1)
19)The orbit of the planet mercury around the
sun is in elliptical shape with sun at a focus.
The semi–major axis is of length 36 million
miles and the eccentricity of the orbit is
0.206. Find (i) How close the mercury gets
to sun? (ii) The greatest possible distance
between mercury and the sun.
Solution:
The equation is
…..(1)
2a = 8
Semi major axis = a
= 36 million miles
e = 0.206 ⇒ ae = 7.416 A = a – ae = 36 – 7.416
= 28.584 milion miles
= a +ae = 36 + 7.416
= 43.416 million miles
(i) The closest distances of the mercury from
the sun = 28.584 million miles
(ii) The greatest distance of the mercury from
the sun = 43.416 million miles.
a=4
ae = 3
= a2 – a2 e2
= 16 – 9 = 7
(1) ⇒
Which is the equation of the path traced by the
kho–kho player.
21
20) A satellite is travelling around the earth in an
elliptical orbit having the earth at a focus and
22) The arch of a bridge is in the shape of a
semi–ellipse having horizontal span of 40 ft
and 16 ft high at the centre. How high is the
arch, 9 ft from the right or left of the centre.
of eccentricity . The shortest distance that
the satellite gets to the earth is 400 kms. Find
the longest distance that the satellite gets
from the earth.
Solution:
The equation is
………..(1)
Solution:
2a = 40
a = 20
b = 16
CA – C =400⇒ a – ae=400
………..(2)
Point (9,
) lies on the ellipse
ae=800 × =400
Longest distance between the satellite from
the earth = a + ae = 800 + 400 = 1200 km
= 1–
=
=
ft
The required height =
21) An arch is in the form of a semi–ellipe whose
span is 48 feet wide. The height of the arch is
20 feet. How wide is the arch at the height of
10 feet above the base?
ft
Solution:
23) The ceiling in a hallway 20 ft wide is in the
shape of a semi ellipse and 18 ft high at the
centre. Find the height of the ceiling 4 feet
from either wall if the height of the side
walls is 12
ft.Solution:
The equation is
The equation is
………..(1)
………..(1)
2a = 48 ⇒ a = 24, ∴ b = 20
2a = 20 ⇒ a = 10,
b=6
………..(2)
Point (
(1) ⇒
lies on the ellipse
………..(2)
Point (6,
= × 576
=
= 1–
=
× 24
=
= 4.8
Required height of the ceiling = 12 + 4.8
= 16.8 ft
= 12
The required width = 24
lies on the ellipse
ft.
22
24) A ladder of length 15m moves with its ends
always touching the vertical wall and the
horizontal floor. Determine the equation of
the locus of a point P on the ladder, which is
6m from the end of the ladder in contact with
the floor.
Solution:
Let P(
be any point on the line AB such that AP=6
and BP=9
Assum:
___PAO = ___BPQ =
From the right ∆ PQB
Cos
=
From the right ∆ ARP;
Sin
=
Cos 2
+ Sin2
+
Locus P(
=1
=1
is
26) Find the equation of the rectangular
hyperbola which has for one of its
asymptotes the line x + 2y – 5 = 0 and passes
through the points (6, 0)and (–3, 0)
Solution:
Equation of the asymptote is x + 2y –5 = 0
The other asymptote is of the form 2x – y + l = 0
Combined equation of the asymptote is
(x + 2y – 5) (2x – y + l ) = 0
Equation of the rectangular hyperbola is of the form
(x + 2y – 5) (2x – y + l ) + k = 0
It passes through (6, 0)
(1) ⇒ (6 + 0 – 5) (12 – 0 + l ) + k = 0
l + k = –12 …………(2)
Again it passes through (–3, 0)
(1) ⇒ (–3 + 0 –5) (–6 – 0 + l ) + K = 0
(–8) (–6 + l ) + k = 0
–8 + k = –48 ………….(3)
Solving (2)&(3) l = 4, k = –16
(1)⇒ (x + 2y – 5 ) (2x – y + 4) – 16 = 0
This is the required equation
27) Prove that the line 5x + 12y = 9 touches the
hyperbola
and find its point
of contact.
Solution:
25) Find the equation of the hyperbola if the
asymptotes are parallel to x + 2y – 12 = 0,
5x + 12y = 9⇒
,⇒m=
x – 2y + 8 = 0, (2, 4) is the centre of the
⇒
⇒
hyperbola and it passes through (2, 0)
Solution:
=
=9×
–1=
;
The asymptotes are parallel to
⇒
x + 2y – 12 = 0, x – 2y + 8 = 0
∴ The line touches the hyperbola
The asymptotes are of the form
x + 2y + = 0 …....(1)
(I
Point of contact=
=
x – 2y + m = 0 ……(2)It passes through the centre(2, 4)
Point of contact = (5, )
(1) ⇒
∴ = –10
(2) ⇒
∴m=6
Equations of the asymptotes are
(1) ⇒ x + 2y – 10 = 0
(2) ⇒ x – 2y + 6 = 0
Combined equation of the asymptote is
(x + 2y – 10) (x –2y + 6) = 0
Equation of the hyperbola is of the form
(x + 2y – 10) (x – 2y + 6) + k = 0 ………(3)
It passes through (2, 0)
(–8) (8) + k = 0 ⇒ – 64 + k = 0 ⇒ k = 64
(3) ⇒ (x + 2y – 10)(x – 2y +6) +64 = 0
This is the required equation of the hyperbola.
, c=
28) Show that the line x – y + 4 =0 is a tangent to
the ellipse . Find the
co–ordinates of the point of contact
Solution:
x – y +4 = 0 ⇒ y = x + 4 ⇒ m = 1, c = 4
⇒
= 16;
⇒
⇒
= 12 × 1 + 4, = 16
⇒
,
∴The line touches the ellipse,
Point of contact =
=(-3, 1)
23
  12x1 4 
, 
4
 4
=
6. DIFFERENTIAL CALCULUS APPLICATIONS – II (10 Mark Questions)
One Question for full Test
Total number of questions: 11
3) Trace the curve
1) Trace the curve y =
2) Trace the curvey =
Domain: (Extent :
Horizontal extent : (Vertical extent : (Intercepts :
x intercept = -1
y intercept = 1
Origin :
Does not pass
through the origin
Domain : (Extent
Horizontal extent : (Vertical extent : (Intercepts :
x intercept = 0
y intercept = 0
Origin :
Passes through the
origin
Domain : [0,
Extent
The curve exists in
first and fourth quadrant
Intercepts :
x intercept = 0
y intercept = 0
Origin :
Passes through the origin
Symmetry
Not symmetrical
about any axis
Asymptotes
No asymptote
Monotonicity
The curve is increasing
throughout (
Symmetry
Symmetrical about
the origin
Asymptotes
No asymptote
Monotonicity
The curve is increasing
throughout (
Symmetry
Symmetrical about
the x axis
Asymptotes
No asymptote
Monotonicity
Special points
Concave downward in (Concave upward in (0,
Point of inflection (0, 1)
Special points
Concave downward in (Concave upward in (0,
Point of inflection (0, 0)
4) Using Eulers theorem, prove that
The branch y = , is decreasing
Special points
(0, 0) is not a point of inflection.
x
, if u =
Solution:
Solution;
=
⇒
=
= f (x, y)
u=
⇒ f is a homogeneous function in x and y of
degree
⇒ tan u =
= 2f
f
x
= 2 tan u
sin u
+ ycos u
= f (x, y)
⇒ f is a homogeneous function in x and y of
degree 2
∴ By Euler’s theorem,
∴ By Euler’s theorem,
x
is increasing
5) Using Euler’s theorem, prove that
tanu, if
xcosu
The branch y =
+
= sin u
tan u
24
= 2 tan u
6) Verify Euler’s theorem for f(x, y) =
9) Verify
Solution;
Solution:
f(tx, ty) =
=
=
f (x, y)
=
⇒ f is a homogeneous function in x and y of degree –1
∴ To verify Euler’s theorem, we have to prove that
;
=
= –f
=
for the functionu =
=
=–
=
=
……(1)
=
=
……(2)
(1)=(2)
=–
=
=–f
10) If w =
∴ Euler’s theorem is verified
7. Verify
and v = y log x, find
and
Solution:
for the function
w=
u=
Solution:
=
v = y log x ⇒
. =
=
. =
=
=
=
=
=
……(1)
=
=
=
=
……. (2)
=
=
From (1) & (2)
8. Verify
11. Use differentials to find an approximate
for the function
value for the given number y =
u = sin 3x cos 4y
Solution:
u = sin 3xcos 4y
+
Solution: Consider
= 1,
= 3 cos 3xcos 4y
= –4sin3x sin 4y
=
= –4 cos 3x 3 sin 4y
∴f(x+dx) = y+dy
= –12 cos 3x sin 4y ………. (1)
=
∴
1+ 0.0066 = 1.0066
Consider
x = 1,
= 3 cos 3x (–sin 4y) 4
= –12 cos 3x sin 4y ………. (2)
y=
(1) = (2)
dy =
∴
(1) + (2) ⇒
25
1+ 0.005 = 1.005
+
= 2.0166
1.0066 + 1.005
9. Discrete Mathematics (10 Mark)
One question for full test
Total number of questions : 15
1) Prove that the set of four functions
2) Show that
on the set of non zero complex numbers C –
defined by
and
forms an abelian group
where
w ≠ 1 form a group with respect
to matrix multiplication.
with respect to the composition of functions.
Solution: Let G =
o – Composition of functions
o
Solution:I =
C=
1)Closure axiom:
From the table closure axiom is true.
2)Associative axiom
Composition of functions is always associative
3) Existence of identity
G is the identity element.
4) Existence of inverse
Inverses of
are
respectively.
5) Commutative axiom
From the table,(symmetrical about main diagonal)
commutative axiom is true.
∴ (G, o is an abelian group.
3)Show that (
forms a group
Solution:Let G =
∙7– Multiplication modulo 7
∙7 [1] [2] [3]
[1] [1] [2] [3]
[2] [2] [4] [6]
[3] [3] [6] [2]
[4] [4] [1] [5]
[5] [5] [3] [1]
[6] [6] [5] [4]
[4]
[4]
[1]
[5]
[2]
[6]
[3]
[5]
[5]
[3]
[1]
[6]
[4]
[2]
,B=
are
respectively. ∴ (G,∙7) is a group.
26
,
E=
Let G =
.
I
A
B
C
D
E
I
I
A
B
C
D
E
A
A
B
I
E
C
D
B
B
I
A
D
E
C
C
C
D
E
I
A
B
D
D
E
C
B
I
A
E
E
C
D
A
B
I
1) Closure axiom
Form the table closure axiom is true
2) Associative axiom
Matrix multiplication is always associative
3) Existence of identity
I=
G is the identity element.
4) Existence of inverse: Inverses of
I, A, B, C, D, E are I, B, A, C, D, E respectively
∴ G is a group under multiplication of matrices.
(4) Show that the set
forms an
abelian group under multiplication modulo 11.
Solution: Let G =
∙11 – Multiplication modulo 11
∙11
[1] [3] [4] [5] [9]
[1]
[1] [3] [4] [5] [9]
[3]
[3] [9] [1] [4] [5]
[4]
[4] [1] [5] [9] [3]
[5]
[5] [4] [9] [3] [1]
[9]
[9] [5] [3] [1] [4]
1) Closure axiom:
From the table closure axiom is true.
2) Associative axiom
Multiplication modulo 11 is always associative
3) Existence of identity
[1] G is the identity element.
4) Existence of inverse
Inverses of
respectively.
5) Commutative axiom
From the table commutative axiom is true.
∴ (G,∙11 is an abelian group.
[6]
[6]
[5]
[4]
[3]
[2]
[1]
1) Closure axiom:
From the table closure axiom is true.
2) Associative axiom
Multiplication modulo 7 is always associative
3) Existence of identity
[1] G is the identity element.
4) Existence of inverse
Inverses of
,D=
, A=
5) Show that the set G of all matrices of the form
where x
6) Show that the set of all matrices of the form
, is a group under matrix
,a
multiplication.
Solution:
R–
, forms an abelian group under
matrix multiplication
Solution:
Let G =
Let G =
1) Closure axiom :
1) Closure axiom:
X=
Y=
G, x ≠ 0, y ≠ 0
XY =
G, [∵2xy ≠ 0]
where a, b
R–
G,
.
G[∵ ab ≠ 0]
∴ Closure axiom is true.
2) Associative axiom:
Matrix multiplication is always associative
3) Identity axiom:
Let E =
be the identity element
⇒ XE = X ⇒ 2xe = x, e =
E=
,B=
AB =
∴ Closure axiom is true.
2) Associative axiom:
Matrix multiplication is always associative
3) Identity axiom:
Let E =
A=
AE =
be the identity element
⇒ ae = a ⇒ e = 1
G is the identity element.
G is the identity element.
4) Inverse axiom :
4) Inverse axiom :
Let
=
Let
be the inverse of X
⇒
=
,
=
=E ⇒
be the inverse of
=1 ⇒
=
G is the inverse of
G is the inverse of
5) Commutative axiom:
AB =
∴ G is a group under matrix multiplication.
=
= BA
∴ Commutative axiom is true
∴ G is an abelian group under matrix multiplication.
27
7) Show that (Z,
ia an infinite abelian group where
8) Show that the set G of the positive rationals
forms a group under the composition defined
is defined as a b = a + b + 2
Solution:
Z = The set of all integers
a b = a+ b+ 2
1) Closure axiom
⩝ a, b Z, a b = a + b + 2 Z
∴ Closure axiom is true.
2) Associative axiom:
⩝ a, b, c Z
a (b c) = a (b+ c+ 2)
= a + (b + c + 2) + 2
=a+b+c+4
(a b c = (a + b + 2) c
=a+b+c+4
⇒ a (b c) = (a b c
∴ Associative axiom is true.
3) Existence of Identity
Let e be the identity element
= ⇒ a + e + 2 = a ⇒ e = –2
–2 Z is the identity element.
4) Existence of Inverse
Let
be the inverse of a
= –2 ⇒
⇒
Z is the inverse of a
5) Commutative axiom:
⩝ a, b Z
a b = a+b+2 = b + a + 2 =
Commutative axiom is true.
Z is an infinite set.
∴ (Z, is an infinite abelian group
by a
=
for all a, b G
Solution:
G = The set of all positive rationals
a
=
1) Closure axiom: ⩝ a, b G , a
=
G
Closure axiom is true.
2) Associative axiom: ⩝ a, b, c G
=
=
=
=
⇒ a (b c) = (a b c
∴ Associative axiom is true
3) Existence of Identity
Let e be the identity element
⇒
⇒ = a, e = 3
G is the identity element.
4) Existence of Inverse
Let
be the inverse of a
∴
=3⇒
=3⇒
G is the inverse of a
∴ (G,
28
is a group
=
9) Let G be the set of all rational numbers except 1
and be defined on G by
for all a, b G. Show that (G, is an infinite
abelian group.
Solution:
G = The set of all rational numbers except 1
10) Show that the set G of all rational numbers
except –1 forms an abelian group with respect
to the operation given by
=
for all a, b G.
Solution:
G = The set of all rational numbers except –1
=
1) Closure axiom:
a, b, G, a ≠ –1 and b ≠ –1
Suppose
1) Closure axiom:
a, b, G, a ≠ 1, b ≠ 1
Suppose
a = 1 (or) b = 1 ⇒⇐ to a, b, G,
a = –1 (or) b = –1 ⇒⇐ to a, b,
∴
G
Closure axiom is true
2) Associative axiom
⩝ a, b, c G
=
=
∴
G
Closure axiom is true
2) Associative axiom
⩝ a, b, c G
=
=
=
=
=(
=(
=
=
=(
=
=
⇒
=
∴ Associative axiom it true
3) Existence of Identity
Let e be the identity element
=
e(1+ a)⇒ e = 0, [ ∴ a ≠ –1]
0 G is the identity element
4) Existence of Inverse
Let
be the inverse of a ⇒
=
∴ Associative axiom is true.
3) Existence of Identity
Let e be the identity element
⇒
e(1 – a) = 0 ⇒ e = 0, since a ≠ 1
0
is the identity element
4) Existence of Inverse
Let
be the inverse of a ⇒
a ≠ 1,
is the inverse of a
G is the inverse of a
Inverse axiom is true
5) Commutative axiom
a, b, G
Inverse axiom is true
5) Commutative axiom
a, b, G
a*b=a+b+ab=b+a+ba=b*a
Commutative axiom is true.
∴ (G, ) is an abelian group.
G contains infinite number of elements.
∴ (G, ) is an infinite abelian group.
Commutative axiom is true.
∴ (G, ) is an abelian group.
G contains infinite number of elements.
∴ (G, ) is an infinite abelian group
29
G,
11) Show that the set G =
is an
infiniteabelion group with respect to addition.
Solution:
1) Closure axiom
⩝ a+
,c+d
G Where a, b, c, d Q
Since a + c, b + d G.
∴ Closure axiom is true.
2) Associative axiom
Addition is always associative
3) Identity axiom
⩝a+
G, there exist an element
0=0+0
=
. =
∴ Associative axiom is true.
3) Identity axiom
⩝
G, there exists an element
= 1 G such that
.1 =
∴ 1 G is the identity element.
4) Inverse axiom
⩝
G, there existy an element
G such that
=
=
∴
is the inverse of
Inverse axiom is true
5) Commutative axiom
⩝
G
=
=
= .
Commutative axiom is true
∴ (G, .) is an abelian group
G such that
0 G is the identity element
4) Inverse axiom
⩝a+
G, there exist an element
G such that
=
+
is the inverse of
5) Commutative axiom
⩝a+
, c+
(a +
) + (c +
12) Show that the set G =
is an abelian
group under multiplication.
Solution:
Given G =
1) Closure axiom
⩝
G, Where a, b z
=
G, since a+b z
∴ Closure axiom is true.
2) Associative axiom
⩝
G
)= .
=
G
)= (a + c) + (b + d)
= (c + a) + (d + b)
= (c +
) + (a +
)
Commutative axiom is true.
G Contains infinite number of elements.
∴(G,+) is an infinite abelian group
30
=1
13) Show that the set M of complex numbers z with
the condition =1 forms a group with respect
to the operation of multiplication of complex
numbers.
Solution:
M = Set of all complex numbers having
modulus value 1.
1) Closure axiom
⩝
M⇒
M
since
=
=1
∴ Closure axiom is true.
2) Associative axiom
Multiplication of complex numbers is always
associative
3) Identity axiom
⩝
M there exists an element 1 M such that
z . 1 = 1 . z =z
1 G is the identity element.
4) Inverse axiom
⩝ z M, there exists an element
z. =
.z=1
4) Inverse axiom
⩝
G, there exists an element
G
such that .
=
=
=l
is the invese of
Inverse axiom is true
5) Commuative axiom
⩝
G
.
=
=
=
Commutative axiom is true.
G. contains finite number of elements.
∴ (G, .)is a finite abelian group.
15) Show that
Solution:
Let
=
1) Closure axiom
[]
M such that
forms group
=
⩝ [ ], [m]
,
0
,m<n
[]
∴ Closure axiom is true.
2) Associative axiom
Addition modulo n is always associative
3) Identity axiom
[0]
is the identity element.
4) Inverse axiom :
⩝[ ]
, there exist an element
[∴
∴ is the inverse of z
(M, .) is a group
14) Show that the
roots of unity form an abelian
group of finite order with usual multiplication.
Solution:
Let G =
,
1) Closure axiom
Let ,
G, 0 , m
To prove
,
=
G
Case (i) if + m n then
G
Case (ii) if +m n
By division algoritham
+ m =(q.n) + r where 0 r < n
=
=
.
= .
G
∴ Closure axiom is true.
2) Associative axiom
Multiplication is always associative for the set
of complex numbers.
3) Identity axiom
⩝
G, there exists an element l G such that
. 1=1.
=
∴ 1 G is the identity element.
31
[n– ]
[]
such that
=
is the inverse of [ ]
Inverse axiom is true.
∴(
,
) is a group.
= [n] = [0]
3. COMPLEX NUMBERS. (10 MARK)
One question for full testTotal number of questions : 16
1) P represents the variable complex number z.
4) P represents the variable complex number z.
Find the locus of P, if Im
= –2
Find the locus of P if arg
Solution:
=
Solution : Let z = x + iy
arg
Let z = x + iy
=
arg (z – 1) – arg(z + 1) =
⇒ arg (x + iy – 1) – arg(x + iy + 1) =
=
=
=
⇒ arg [(x – 1) + iy] – arg[(x + 1)+ iy] =
×
–
=
Im
=
= –2
–x(2x + 1) +2y(1 – y) = – 2[(1 – y)2 + x2 ]
–2x2 – x + 2y – 2y2 = –2(1 + y2 – 2y + x2)
–x + 2y = –2 + 4y
x + 2y – 2 = 0
∴Locus of P is a straight line
= tan
=
2y =
(
)
=0
∴ Locus of P is a circle
5) P represents the variable complex number z.
2) P represents the variable complex number z.
Find the locus of P, if Re
=1
Find the locus of P if arg
Solution :
Let z = x+iy
Solution :
Let z = x + iy
arg
=
=
=
⇒ arg (z – 1) – arg(z + 3) =
=
⇒ arg (x + iy – 1) – arg(x + iy + 3) =
×
Re
arg[(x – 1 + iy] – arg[(x + 3) + iy] =
=1
–
(x – 1)x + y(y + 1) = x2 + (y + 1)2
x2 – x + y2 + y = x2 + y2 + 1 + 2y
⇒ x+y+1=0
Locus of P is a straight line
=
=
3) P represents the variable complex number z.
Find the locus of P if Re
= tan
=1
Solution :
Let z = x + iy
=
Re
=
=
=
=
×
0=
=1
=0
Locus of P is a circle
(x + 1)x + y(y + 1) = x2 + (y + 1)2
x2 + x + y2 + y = x2 + y2 + 1 + 2y
x–y–1=0
Locus of P is a straight line
32
tan = ∞ =
6) If α and β are the roots of x2 – 2x + 2 = 0 and
cot θ = y + 1, show that
9) Find all the values of
=
Solution :
Solution :x2 – 2x + 2 = 0
x = 1 ± i, Let α = 1 + i and β = l–i ⇒ α – β = 2i
Given cot θ = y + 1 ⇒ y = cot θ – 1 =
r=2,   /6
consider
2 ( cos
–1
+ i sin )
=
(y + α)n =
=
=
(y + α)n =
=
(y + β)n =
=
=
[ cos(
7) If α and β are the roots of the equation
x2 – 2px + (p2 + q2) = 0 and tan θ =
show that
Solution :
⇒y+p=
(y + α)n =
⇒y=
+ i sin
=
=
(
(
=
=
=
= qn–1
=
8) If α and β are the roots of the equation
x2– 2x + 4 = 0. Prove that
αn – βn = i2n+1sin and deduct α9 – β9
11) Find all the values of
Solution :
x2 – 2x + 4 =0
⇒ x=1±i
r=2,   /3
α = 1+ i
= 2 ( cos + i sin )
 = 1– i
αn
=
αn –
n
Solution: –i
+ isin
– i sin
=
)
, k = 0, 1, 2, 3
When k = 0,
)
When k = 1,
When k = 2,
When k = 3,
n=9
Product
=i
) + i sin (
=
= 2n 2i sin
β9
= cos (
=
= i 2n+1sin
α9 –
and hence
=
(cos
n = 2n (cos
;k = 0, 1, 2
prove that the product of the values is 1
= 2 ( cos – i sin )
2n
)
r=2,   5/6
–p
= qn
(y + β)n =
consider
2 ( cos
Solution :x2 –2px + (p2 + q2) = 0 ⇒ x = p ± qi
Let α = p + qi and β = p – qi ⇒ α – β = 2iq
(y + )n =
)], k = 0, 1, 2
10) Find all the values of
= qn–1
tanθ =
) + i sin(
=
29+1 sin
=
α9 – β9 = 0
=
33
=1
) ( r=2, 
 /3
12) Solve the equation x9 + x5 – x4 – 1 =0
Solution :
x9 + x5 – x4 – 1 = 0
x5(x4 + 1) – 1(x4 + 1) = 0
(x5 – 1)(x4 + 1) = 0
x5 – 1 = 0 ⇒ x =
15) If
=2 cos and
(i)
– nф)
(ii)
– nф)
Solution:
Let x = cos
=
=
= 2 cos , show that
; y = cosф + isinф
=
, k = 0, 1, 2, 3, 4
= cos(m –nф) + i sin (m – nф)….(1)
x4 + 1 = 0 ⇒ x =
∴
=
= cos(m –nф) – i sin (m – nф)…..(2)
(1) + (2) ⇒
=
=
– nф)
(1) – (2) ⇒
, k = 0, 1, 2, 3
– nф)
13) Solve the equation x7 + x4 + x3 + 1 = 0
Solution :
Given x7 + x4 + x3 + 1 = 0
x4(x3 + 1)+1(x3 + 1) = 0
(x4 + 1)(x3 + 1) = 0
16) If a = cos 2 + i sin 2 , b = cos 2 + isin 2
x4 + 1= 0 ⇒ x =
Solution:
i)abc=(cos2 +isin2
and c = cos 2 + i sin 2
(i)
(ii)
=
x=
1= 0 ⇒ x =
∴
=
= 2 cos (
= 2 cos (
(cos2 +isin2
(cos2 +isin2
+i sin 2(
=cos2(
, k = 0, 1, 2, 3
x3 +
+
prove that
=cos(
+i sin (
….(1)
=cos(
–i sin (
…(2)
∴(1)+(2)⇒
+
x=[
=2cos(
, k = 0, 1, 2
(ii)
14) Solve the equation x4 – x3 + x2 – x + 1 = 0
Solution:
x4 – x3 + x2 – x +1 = 0
=
+
=
=cos2(
=0
=cos2(
= –1, x ≠–1
∴(3)+(4)⇒
x=
+ =2cos2(
x=
=2 cos2(
=
x = cos
+ isin
When k = 0, x =
, k = 0, 1, 3, 4 as x ≠ –1
+i
When k = 1, x =
+i
When k = 3, x =
+i
When k = 4, x =
+i
(The root formed by k = 2 should be excluded,
since x ≠ –1)
34
+ i sin 2(
i sin2(
….(3)
…(4)
1. Matrices and Determinants- 6 and 3 mark Questions
1)IfA=
4) If A =
,verify that;A(adjA)=(adjA)A=
B=
verify that
Solution:
;
=
AB =
|A|=
=2
|A| I2 =
…………(1)
=
=
=1
0
Adj(AB)=
adj A =
A (adjA)=
=
……(2)
(adj A) A =
=
……(3)
=
From (1),(2),(3), A(adjA)(adjA) A=
....(1)
Adj(A) =
=
=1
0
Adj B =
= -11
Adj(B) =
=
=
=
Adj A =
B=
Solution:
=
=1 0
=
verify thatA(adj A)=(adj A) A =
2)IfA=
Adj(AB)=
=
=
…………(1)
=
adj A =
=
From (1),(2)
A (adjA)=
=
……(2)
(adj A) A =
=
...…(3)
5) If A =
Solution:
3) If A=
AB =
verify that
and B
=
and B =
verify that (
From (1),(2),(3), A(adjA)(adjA) A=
.......(2)
=
;.
=
=
=
=2–3= –1 ≠ 0
Solution:
AB =
=
=
=
=
_____(1)
………..(1)
=
=
From (1), (2),
=
adj (AB)
=
= –1 ≠ 0
…(2)
=
=
=
35
=
=1 ≠ 0
=
=
=
=
adj A =
=
11) If A=
Show that the adjoint of A
______(2)
(1)=(2)⇒(
=
6)Find the adjoint matrix of A =
is A itself
adj A =
A=
7) Find the adjoint matrix of A =
[Aij] =
adj A =
adj A = [Aij]T
adj A =
=
=A
8) Find the adjoint matrix of A =
12) If A =
[Aij] =
Show that A=
adj A = [Aij]T
adj A =
A=
(A.
A*A=
X
=
=
= I
=
.A =I)
∴A=
9) Find the adjoint matrix of A =
13)Find the inverse of
[Aij] =
Solution:
adj A = [Aij]T
A=
adj A =
=
=1(1–1) – 0+3(–2–1)= –9
adj A =
=
10) Find the adjoint matrix of
A=
[Aij] =
adj A = [Aij]T
adj A =
36
=
adj A
16. Find the inverse of
14 . Find the inverse of
A=
A =
=
==
adj A =
=1
= 35
adj A =
=
=
=
=
=
adj A
=
adj A
=
adj A
17 . Find the inverse of
=
A =
15 . Find the inverse of
=
=5
A =
adj A =
=
=
= -1
=
adj A =
=
18. Find the inverse of
=
=
adj A
A =
=
=
=2
=
adj A =
37
=
=
=
19. Find the inverse of
=
22) Solve by matrix inversion method
adj A
7x+3 y = -1, 2x + y = 0
A=
=
inverse exists.
Inverse does not exists
 1 - 3
adj A = 

- 2 7 
20. Find the inverse of A=
=
=
X=
adj A =
=
⇒ x = -1 and y = 2
adj A
23) Solve by matrix inversion method
=
2x - y = 7, 3x – 2y = 11
=
inverse exists.
21. Find the inverse of A=
=
- 2 1
adj A = 

 - 3 2
=2
adj A =
=
adj A
=
X=
=
⇒ x = 3 and y = -1
38
24) Solve by matrix inversion method
~
x+y = 3, 2x +3 y = 8
This matrix is in the echelon form.
Number of non-zero rows is 2
inverse exists.
(A)=2
 3 - 1
adj A = 

- 2 1 
27 Find the rank of the matrix
Solution:
A=
X=
1 1  1 0 


~ 2  3 0 1


0 1
2 1 
⇒ x = 1 and y = 2
R1  R3
25)Find the rank of the matrix
~
Solution:
A=
~
This matrix is in the echelon form.
~
,
Number of non-zero rows is 3
(A)=3
This matrix is in the echelon form.
Number of non-zero rows is 1
28) Find the rank of the matrix
(A)=1
Solution:
A=
~
26) Find the rank of the matrix
A=
~
This matrix is in the echelon form.
non-zero rows is 2.
(A)=2
39
Number of
29) Find the rank of the matrix
31) Find the rank of the matrix
Solution:
Solution:
A=
A=
~
~
~
~
~
~
This matrix is in the echelon form.
Number of non-zero rows is 3
This matrix is in the echelon form.
(A)=3
Number of non-zero rows is 2
(A)=2
32) Find the rank of the matrix
30) Find the rank of the matrix
Solution:
Solution:
A=
A=
~
,
~
~
~
This matrix is in the echelon form.
Number of non-zero rows is 1
This matrix is in the echelon form.
(A)=1
Number of non-zero rows is 2
(A)=2
40
33) Find the rank of the matrix
35) Find the rank of the matrix
Solution:
Solution:
A=
A=
~
~
~
~
This matrix is in the echelon form.
This matrix is in the echelon form.
Number of non-zero rows is 2
Number of non-zero rows is 2
(A)=2
(A)=2
34) Find the rank of the matrix
36) Find the rank of the matrix
Solution:
Solution:
A=
A=
~
~
~
~
~
~
~
~
This matrix is in the echelon form.
Number of non-zero rows is 2
(A)=2
This matrix is in the echelon form.
Number of non-zero rows is 2
37) Solve by determinant method
(A)=2
2x +3y =5,4x +6y =12
Solution:
i)
∴The system is inconsistant and it has no solution.
41
38) Solve by determinant method
40) Solve by determinant method
3x +2y =5 , x +3y =4 .
x -y = 2 , 3y =3x-7
Solution:
Solution:
x -y = 2 , 3x -3y= 7
The given equations are
,
∴ It has no solution.
∴ It has unique solution.
41) Solve by determinant method
=
By Cramer’s rule x
=
x +y +2z =0, 2x+ y- z =0,2x + 2y +z =0
= 1
Solution:
=
y
≠0
=
= 1
∴ It has unique solution.
The given system is homogenious
. ∴ it is consistant and has trivial solution
∴ Solution x = 1, y=1
∴ Solution (x,y,z) == (0,0,0)
39) Solve by determinant method
x +y =3 , 2 x + 3y =7
42) Solve by determinant method
Solution:
4x +5y =9,8x +10y =18
Solution:
∴ It has unique solution.
(i)
(ii)
=
By Cramer’s rule x
(iii) Atleast one
=
y
∴ the system is consistant and has infinitely many
solutions.
= 2
=
=
of
To find the solution Let y = t ,
⇒4x = 9-5t
= 1
∴(x, y)=
∴ Solution x = 2, y=1
42
x=
t
R
t
R
43) Solve by determinant method 2x +3y =8,4x +6y
=16
45) Solve x + y +2z =4,2x +2y +4z =8,
3x +3y +6z =10
Solution:
Solution:
(i)
(ii)
(iii) Atleast one
of
∴ the system is consistant and has infinitely many
solutions.
To find the solution, Let y = t t
⇒2x +3t =8
∴(x, y)=
R
(i)
(ii)
x=
t
(iii) All the 2 x 2 minors of –is 0
(iv) But not all the 2 x 2 minors of
R
–are
zero
∴The system is inconsistant and it has no
solution.
46) Examine the consistency of the system of
equations. If it is consistent then solve the same.
x + y +z =7, x +2y +3z =18, y +2z =6
44) ) Solve by determinant method
2x +2y +z =5, x– y + z =1,3x + y +2z =4
The matrix equation is
Solution:
=
AX = B
= -6 ≠ 0
=
(i)
(ii)
∴The system is inconsistant and it has no solution.
This matrix is in the echelon form.
(A)=2
also
(A,B)=3
(A) ≠
(A,B)
∴The system is inconsistant and it has no solution.
43
47) Examine the consistency of the system of
equations. If it is consistent then solve the same.
x -4 y +7z =14, 3x +8y -2z =13, 7x-8 y +26z =5
The matrix equation is
=
AX = B
=
This matrix is in the echelon form.
(A)=2
also
(A,B)=3
(A) ≠
(A,B)
∴The system is inconsistant and it has no solution.
48) Reversal law of inverses
law: If A and B are any two non-singular matrices
of the same order, then
Proof: A and B are non singular matrices.
∴ AB is a non singular matrix
(AB)
=
= I ….(1)
(AB) = I ………….(2)
Similarlly
From (1), (2)
=
(AB)
=
(AB) = I
is the inverse of (AB)
∴
Hence proved.
44
9.Discrete Mathematics - 6 marks questions and answers
1) Example
(iii) construct the truth table for (pq)(~q) .
q
(pq)
~q
(pq)(~q)
p
T
T
F
F
T
F
T
F
T
T
T
F
F
T
F
T
F
T
F
F
(iv) construct the truth table for ~ [(~ p)(~ )]
2) Example
p
q
~p
~q
T
T
F
F
F
T
T
F
F
T
F
T
F
T
T
F
F
T
F
F
T
T
T
F
3) Example 9.5: construct the truth table for (pq)(~ r )
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
(pq)
(pq)(~ r )
T
T
F
F
F
F
F
F
F
T
F
T
F
T
F
T
T
T
F
T
F
T
F
T
4) Example 9.6: construct the truth table for (pq)  r
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
(pq)
(pq)  r
T
T
T
T
T
T
F
F
T
F
T
F
T
F
F
F
5) Exercise 9.2 – 7 construct the truth table for (p q )  [~ (pq )]
p
q
(p q
~ (pq )]
(p q )  [~ (pq )]
T
T
T
F
T
T
F
F
T
T
F
T
F
T
T
F
F
F
T
T
6) Exercise 9.2 - 9 construct the truth table for (p q )  r
p
q
r
(p q )
(p q )  r
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
T
T
T
T
F
F
T
T
T
T
T
T
T
F
45
7) Exercise 9.2-10 construct the truth table for (pq )  r
p
q
r
(pq )
(pq )  r
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
F
F
F
F
F
F
T
T
T
F
T
F
T
F
8) Example 9.7: Show that ~ (p  q )
(~ p) ( ~q )
p  q
~ (p  q )
p
T
T
F
F
q
T
F
T
F
p
q
T
T
F
F
F
T
F
F
T
F
F
T
T
F
F
F
F
T
T
T
T
T
T
F
F
F
F
T
The last columns are identical /
9) Example 9.10: (i)Show that the statement [ (~ p )  (~ q ) ]  p is a tautology
p
q
T
T
F
F
F
T
T
F
F
T
T
T
F
T
T
F
T
T
F
F
T
T
T
T
The last column contains only T /
is a tautology.
10) Example 9.10: (ii) show that [(~ q)p) ]q is a contradiction
p
q
T
T
F
F
F
T
F
T
T
F
F
T
F
F
F
F
F
T
F
F
The last column contains only F /
is a contradiction.
11) Example 9.11: use the truth table to determine whether the statement [(~ p)q] [ p (~q )]is a tautology
p
q
T
T
F
F
T
F
T
T
F
F
T
F
T
T
F
T
T
F
T
F
T
F
F
T
T
T
F
T
46
The last column contains only T / the given statement is a tautology.
12) EXERCISE 9.3 (i) Use the truth table to establish the following statement is a tautology or a contradiction
[(~ p) q ) ]  p.
p
(~ p) q )
(~ p) q ) ]  p
q
T
T
F
F
F
T
F
F
F
F
F
T
T
T
F
F
F
T
F
F
The last column contains only F /
is a contradiction.
13) EXERCISE 9.3 (ii) Use the truth table to establish the following statement is a tautology or a contradiction
(p  q )  [~ (p  q ) ]
p
q
T
T
T
F
T
T
F
T
F
T
F
T
T
F
T
F
F
F
T
T
The last column contains only T /
is a tautology.
14) EXERCISE 9.3 (iii) Use the truth table to establish the following statement is a tautology or a contradiction
[ p  (~q ) ]  [ ( ~ p )  q ]
p
q
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
F
T
F
F
The last column contains only T /
T
F
T
T
T
T
T
T
is a tautology.
15) EXERCISE 9.3 ( iv) Use the truth table to establish the following statement is a tautology or a contradiction
q [p  (~q ) 
p
q
T
T
F
F
T
F
T
F
F
T
F
T
T
T
F
T
is a tautology.
The last column contains only T /
T
T
T
T
16) EXERCISE 9.3 (v) Use the truth table to establish the following statement is a tautology or a contradiction
p  ( ~p ) ]  [ (~ q) p ]
p
q
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
F
F
F
F
The last column contains only F /
47
(~ q) p
[ p  ( ~p ) ] [ (~ q) p ]
F
T
F
F
F
F
F
F
is a contradiction .
[
17) EXERCISE 9.3 - 2. Show that p  q  (~ p )  q.
P
q
T
T
F
F
T
F
T
F
P
T
F
T
T
q
T
T
T
F
F
T
F
F
The last columns are identical / p  q
F
F
T
T
 (~ p )  q .
T
F
T
T
18) EXERCISE 9.3 - 3. Show that p  q  ( p  q )  ( q  p ).
p
q
p q
T
T
F
F
T
F
T
F
p
q
T
F
F
T
p  q
q  p
(p  q)(q  p)
T
T
T
T
T
F
F
T
F
T
T
F
F
F
T
T
The last columns are identical / p  q  ( p  q )  ( q  p ) .
T
F
F
T
19) EXERCISE 9.3 - 4. Show that p q[ (~ p )  q ) [ (~q )  p ) 
p
q
p q
T
T
F
F
p
T
T
F
F

T
F
T
F
T
F
F
T
q
T
F
F
T
T
F
F
T
F
T
T
T
F
T
F
F
T
T
T
T
The last columns are identical / p q [ (~ p )  q ) [ (~q )  p ) ;.
20) EXERCISE 9.3 - 5. Show that ~ (pq )  (~p)(~q ).
p
q
(pq
~ (pq )
T
T
T
F
T
F
F
T
F
T
F
T
F
F
F
T
48
T
F
F
T
p
q
T
T
F
F
F
T
F
F
T
T
F
T
T
F
T
F
F
T
T
T
The last columns are identical / ~ (pq ) (~p)(~q ) .
21) EXERCISE 9.3 - 6.Show that
and
p
q
T
T
F
F
T
F
T
F
are not equivalent.
T
F
T
T
T
T
F
T
/ p  q and q  p are not equivalent.
22). EXERCISE 9.3 - 7. Show that (pq)
( p q) is a tautology.
pq
p q
p
q
T
T
T
T
T
T
F
F
T
T
F
T
F
T
T
F
F
F
F
T
The last column contains only T / (pq)
( p q)
(pq)
( p q) is a tautology .
23).Group:
Definition :
A non-empty set G, together with an operation * i.e., (G, *) is said to be a
group if it satisfies the following axioms
(1) Closure axiom : a, b ∈ G ⇒ a * b ∈ G
(2) Associative axiom : ∀a, b, c ∈ G, (a * b) * c = a * (b * c)
(3) Identity axiom : There exists an element e ∈ G such that a * e = e * a = a, ∀a ∈ G.
( 4) Inverse axiom : ∀a ∈ G there exists an element a−1∈G such that a−1 * a = a * a−1 = e.
e is called the identity element of G and a−1 is called the inverse of a in G.
24) State and prove cancellation laws of a group:
Proof
Let G be a group. Then for all a, b, c G.
(i)
(Left Cancellation Law)
(ii)
(Right Cancellation Law)
Proof:
(i)
(ii)
(
b
49
25) State and prove reversal law of inverses of a group.
Proof:
Let G be a group a, b G. Then
It is enough to prove
=
is the inverse of
To prove (i)
(ii)
=e
=
=
=
=
=
is the inverse of
=
i.e.,
26) Prove that (Z,+) is an infinite abelian group.
Solution :
Z = set of all integers
Closure axiom
Sum of 2 integers is also an integer
Associative axiom
Usual addition is always associative
Identity axiom
0 Z is the identity element
Inverse axiom
is the inverse of
Commutative axiom Addition is always commutative
Z contain infinite number of elements.
(Z,+) is an infinite abelian group.
27)Example 9.13: (R – {0}, . ) is an infinite abelian group.
Solution:
Closure axiom
Product of two non zero real numbers is also a non zero real number.
Associative axiom
Usual multiplication is always associative
Identity axiom
1 R – {0} is the identity element
Inverse axiom
1/a R – {0} is the inverse of a R– {0}
Commutative axiom Multiplication is always commutative
R– {0} contain infinite number of elements.
(R– {0},.) is an infinite abelian group.
28)Show that the cube roots of unity forms a finite abelian group under multiplication.
Solution :
1
1
1
1
1
Closure axiom
Associative axiom
Identity axiom
Inverse axiom
from the table closure axiom is true
Usual multiplication is always associative
1 is the identity element
The inverse of 1 is 1
The inverse of is
The inverse of
is
50
Multiplication is always commutative
G is finite group. (G,.) is a finite abelian group
Commutative axiom
29)Show that the fourth roots of unity forms a finite abelian group under multiplication.
Solution :
1
-1
i
-i
1
1
-1
i
-i
Closure axiom
Associative axiom
Identity axiom
Inverse axiom
Commutative axiom
-1
-1
1
-i
i
i
i
-i
-1
1
-i
-i
i
1
-1
from the table closure axiom is true
Usual multiplication is always associative
1 is the identity element
The inverse of 1 is 1
The inverse of -1is -1
The inverse of i is –i
The inverse of -i is i
Multiplication is always commutative
G is finite group. (G,.) is a finite abelian group
30) Example 9.16 : Prove that (C, +) is an infinite abelian group.
Solution:
(i) Closure axiom : Sum of two complex numbers is always a complex number.
Closure axiom is true.
(ii) Associative axiom : Addition is always associative in C
(iii) Identity axiom : o ∈ C is the identity element .
(iv) Inverse axiom : − z ∈ C is the inverse of z ∈ C
(v) Commutative property : Addition of complex numbers is always commutative
Since C is an infinite set (C, +) is an infinite abelian group.
31) Example 9.17 : Show that the set of all non-zero complex numbers is an abelian
group under the usual multiplication of complex numbers.
Solution:
(i) Closure axiom : Let G = C − {0} Product of two non-zero complex
numbers is again a non-zero complex number.
(ii) Associative axiom :Multiplication is always associative.
(iii) Identity axiom : 1 ∈ G is the identity element .
(iv) Inverse axiom : 1/z ∈ G is the inverse of z ∈ G.
(v) Commutative property : Multiplication of complex numbers is always commutative.
∴ (C-{0} , .) is an abelian.
32) Example 9.19 : Show that the set of all 2 × 2 non-singular matrices forms a non-abelian infinite group under matrix
multiplication, (where the entries belong to R).
Solution:Let G be the set of all 2 × 2 non-singular matrices, where the entries belong to R.
(i) Closure axiom : Since product of two non-singular matrices is again non-singular and the order is 2 × 2,
the closure axiom is satisfied.
(ii) Associative axiom : Matrix multiplication is always associative
and hence associative axiom is true.
(iii) Identity axiom :
is the identity element .
(iv) Inverse axiom : the inverse of A ∈ G, exists i.e. A−1 exists and is of order 2 × 2 .
51
Thus the inverse axiom is satisfied.
Hence the set of all 2 × 2 non-singular matrices forms a group under matrix multiplication.
Further, matrix multiplication is non-commutative and the set contain infinitely many elements.
The group is an infinite non-abelian group.

1 0    1 0  1 0    1 0  

, 

 ,  0  1  ,  0  1 
0
1
0
1

 
 

 
 form an abelian group, under
33) Example 9.21 : Show that the set G= 
multiplication of matrices.
Solution :
Let
.
I
A
B
C
I
I
A
B
C
A
A
I
C
B
B
B
C
I
A
C
C
B
A
I
(i) Closure axiom : All the entries in the multiplication table are members of G.
∴ Closure axiom is true.
(ii) associative axiom : Matrix multiplication is always associative
(iii) identity axiom : I is the identity element in G.
(iv) inverse axiom : I is the inverse of I
A is the inverse of A
B is the inverse of B
C is the inverse of C
From the table it is clear that “.” is commutative. ∴ G is an abelian group under matrix
multiplication.
3 Mark Questions:34. Theorem : In a group G, (a−1)−1 = a for every a ∈ G.
Proof :
We know that a−1 ∈ G and hence (a−1)−1 ∈ G. Clearly a * a−1 = a−1 * a = e
a−1*(a−1)−1= (a−1)−1* a−1 = e
⇒ a * a−1 = (a−1)−1* a−1
⇒ a = (a−1)−1 (by Right Cancellation Law)
35. Theorem :
The identity element of a group is unique.
Proof : Let G be a group. If possible let e1 and e2 be identity elements in G.
Treating e1 as an identity element we have e1 * e2 = e2
… (1)
Treating e2 as an identity element, we have e1 * e2 = e1
… (2)
From (1) and (2), e1 = e2
∴ Identity element of a group is unique.
52
36. Theorem :
The inverse of each element of a group is unique.
Proof :
Let G be a group and let a ∈ G.
If possible, let a1 and a2 be two inverses of a.
Treating a1 as an inverse of ‘a’ we have a * a1 = a1 * a = e.
Treating a2 as an inverse of ‘a’, we have a * a2 = a2 * a = e
Now a1 = a1 * e = a1 * (a * a2) = (a1 * a) * a2 = e * a2 = a2
⇒ Inverse of an element is unique.
Do it your self:Example 9.4:
Contruct the truth table for the following statement:
(i)  ~ p   ~ q  
EXERCISE 9.2
(ii) ~  ~ p   q 
Contruct the truth table for the following statement:
(1) p  ~ q 
(2) ~ p   ~ q 
(3) ~  p  q 
(4)  p  q   ~ p 
(5)  p  q   ~ q 
(6) ~  p  ~ q  
(8)  p  q   ~ q 
53
CHAPTER – I (APPLICATION OF MATRICES AND DETERMINANTS)
BOOK BACK QUESTIONS
1.
is
The rank of the matrix
1) 1
2) 2
2.
The rank of the diagonal matrix
3.
1) 0
2) 2
If A = [2 0 1], then the rank of the A AT
1) 1
2) 2
4.
If A =
3) 3
4) 4
is
3) 3
4) 5
3) 3
4) 0
3) 1
4) 2
then the rank of the A AT
1) 3
2) 0
5.
If the rank of the matrix
6.
1) 1
2) 2
3) 3
If A is a scalar matrix with scalar k ≠ 0, of order 3, then A–1 is
1) I
2) I
3) I
is 2, then λ is
4) any real number
4) kI
7.
If the matrix
3) k ≠ – 4
4) k ≠ 4
8.
1) k is any real number 2) k = – 4
If A=
, then (adj A)A =
1)
3)
4)
9.
has an inverse then the values of k
2)
If A is a square matrix of order n, then | adjA | is
1) | A |2
2) | A |n
3) | A |n–1
10. The inverse of the matrix
1)
4) | A |
is
2)
3)
4)
11. If A is a matrix of order 3, then det (k A)
1) k3(detA)
2) k2(detA)
3) k(detA)
4) (detA)
12. If I is the unit matrix of order n, where k ≠ 0 is a constant, then adj(kI) =
1) kn(adjI)
2) k(adjI)
3) k2 (adjI)
4) kn - 1(adjI)
13. If A and B are any two matrices such that AB = O and A is non-singular, then
1) B = O
2) B is singular
3) B is non-singular 4) B = A
12
14. If A =
then A is
1)
15. The inverse of
1)
2)
3)
4)
is
2)
3)
4)
16. In a system of 3 linear non-homogenous equation with three unknowns, if Δ = 0 and Δx = 0, Δy≠0 and Δz = 0
then the system has
1) unique solution
2) two solutions
3) infinitely many solutions 4) no solution
54
17. The system of equations ax + y + z = 0, x + by + z = 0, x + y + cz = 0 has a non-trivial solution, then
=
1) 1
2) 2
3) – 1
18. If aex + bey= c, pex + qey= d and Δ1 =
1)
2)
, Δ2 =
4) 0
, Δ3 =
, then the value of (x, y) is
3)
4)
19. If the equation – 2x + y + z = l, x – 2y + z = m, x + y – 2z = nsuch that l + m + n = 0, then the system has
1) a non-zero unique solution
2) trivial solution 3) infinitely many solution
4) No solution
CREATED QUESTIONS
2  4
1. The rank of the matrix 
 is
 1 2 
(1) 1
(2) 2
(3) 0
(4) 8
7  1
 is
2 1 
2. The rank of the matrix 
(1) 9
(2) 2
(3) 1
(4) 5
3. If A and B are matrices conformable to multiplication then (AB)T is
(1) ATBT
(2) BTAT
(3) AB
(4) BA
4. (AT)-1 is equal to
(1) A-1
(2) AT
(3) A
(4) (A-1)T
5. If (A) = r, then which of the following is correct?
(1) All the minors of order r which do not vanish.
(2) A has atleast one minor of r which does not vanish and all higher order minors vanish.
(3) A has atleast one (r + 1) order minor which vanishes.
(4) All (r + 1) and higher order minors should not vanish.
6. Which of the following is not elementary transformation?
(1)
Ri  R j
(2) Ri  2 Ri  R j
(3)
Ci  C j  Ci
(4)
Ri  Ri  C j
7. Equivalent matrices are obtained by
(1) taking inverses
(2) taking transposes
(3) taking adjoints
(4) taking finite number of elementary transformations
8. In echelon form, which of the following is incorrect?
(1) Every row of A which has all its entries 0 occur below every row which has a non-zero entry.
(2) The first non-zero entry in each non-zero row is 1.
(3) The number of zeros before the first non-zero element in a row is less than the number of such
zeros in the next row.
(4) Two rows can have same number of zeros before the first non-zero entry.
9. If   0 then the system is
(1) Consistent and has unique solution
(2) Consistent and has infinitely many solutions
(3) Inconsistent
(4) Either consistent or inconsistent.
10) In the system of 3 linear equations with three unknowns, If  = 0 and one of  x,  y,  z is non-zero,
then the system is
(1) consistent
(2) inconsistent
(3) consistent and the system reduces to two equations
(4) consistent and the system reduces to a single equation.
(11) In the system of 3 linear equations with three unknowns, if  = 0,  x = 0,  y = 0,  z = 0 and atleast
one 2 x 2 minor of   0, then the system is
(1) consistent
(2) inconsistent
(3) consistent and the system reduces to two equations
(4) consistent and the system reduces to a single equation.
(12) In the system of 3 linear equations with three unknowns, if  = 0 and all 2 x 2 minors of  = 0 and
atleast one 2 x 2 minor of  x or  y or  z is non-zero, then the system is
(1) consistent
(2) inconsistent
(3) consistent and the system reduces to two equations.
(4) consistent and the system reduces to a single equation
55
(13) In the system of 3 liner equations with three unknowns, if  = 0 and all 2 x 2 minors of  ,  x,  y,  z
are zeros and atleast one non-zero element is in  , then the system is
(1) consistent
(2) inconsistent
(3) consistent and the system reduces to two equations
(4) consistent and the system reduces to a single equation.
(14) Every homogeneous system (linear)
(1) is always consistent
(2) has only trivial solution
(3) has infinitely many solution
(4) need not be consistent
(15) If  (A) =  [A B], then the system is
(1) consistent and has infinitely many solution.
(2) consistent and has a unique solution.
(3) consistent
(4) inconsistent
(16) If  (A) =  [A B] = the number of unknowns, then the system is
(1) consistent and has infinitely many solution.
(3) consistent.
(2) consistent and has a unique solution.
(4) inconsistent.
(17)  (A)   [A B], then the system is
(1) consistent and has infinitely many solution.
(3) consistent
(2) consistent and has a unique solution.
(4) inconsistent
(18) In the system of 3 linear equations with three unknowns,  (A) =  [A, B] = 1, then the system
(1) has unique solution.
(2) reduces to 2 equations and has infinitely many solution.
(3) reduces to a single equation and has infinitely many solution. (4) is inconsistent.
(19) In the homogenous system with three unknowns,  (A) = number of unknowns, then the system has
(1) only trivial solution.
(2) reduces to 2 equations and has infinitely many solution.
(3) reduces to a single equation and has infinitely many solution.
(4) no solution
(20) In the system of 3 linear equations with three unknowns, in the non homogenous system
 (A) =  [A B] = 2, then the system
(1) has unique solution.
(2) reduces to 2 equations and has infinitely many solution.
(3) reduces to a single equation and has infinitely many solution
(4) is inconsistent
(21) In the homogenous system,  (A) < the number of unknowns, then the system has
(1) only trivial solution.
(2) trivial solution and infinitely many non- trivial solutions.
(3) only non-trivial solutions. (4) no solution.
(22) Cramer’s rule is applicable only (with three unknowns), when
(1)   0
(2)  = 0
(3)  = 0,  x  0
(4)  x =  y =  z = 0.
(23) Which of the following statement is correct regarding homogenous system?
(1) always inconsistent.
(2) has only trivial solution. (3) has only non- trivial solutions.
(4) has only trivial solution only if rank of the coefficient matrix is equal to the number of
unknowns.
CHAPTER – II (VECTOR ALGEBRA)
BOOK BACK QUESTIONS
20. If is a non-zero vector and m is a non-zero scalar then m is a unit vector, if
1) m = ± 1
2) a = | m |
3) a =
4) a =1
21. If
1) θ =
22. If
and
are two unit vectors and θ is the angle between them, then (
2) θ =
3) θ =
include an angle 120° and their magnitude are 2 and
1)
2) –
25. The vectors 2 + 3 + 4
1) a = 2, b = 3, c = – 4
then . is equal to
3) 2
23. If = × ( × ) + × ( × ) + × ( × ) then
1) u is an unit vector
2) = + +
3) =
24. If + + = 0, | | = 3, | | = 4, | | = 5 then the angle between
1)
2)
3)
) is a unit vector if
4) θ =
4) –
4)
and
≠
is
4)
and a + b + c are perpendicular when
2) a = 4, b = 4, c = 5
3) a = 4, b = 4, c = – 5 4) a = –2, b = 3, c = 4
56
26. The area of the parallelogram having a diagonal 3 + – and a side – 3 + 4 is
1) 10
2) 6
3)
4) 3
27. If
then
2) is perpendicular to
3) | | = | | 4) and are unit vectors
1) is parallel to
28. If , and + are vectors of magnitude λ then the magnitude of
is
3) λ
4) 1
1) 2 λ
2) λ
29. If × ( × ) + × ( × ) + × ( × ) = × then
1) =
2) =
3)
and are parallel
4) = or = or and are parallel
30. If
=2 + + ,
= – + 3 + 2 then the area of the quadrilateral PQRS is
1) 5
2) 10
31. The projection of
on a unit vector
1) tan–1
32. If the projection of
1)
3)
equals thrice the area of parallelogram OPRQ. Then
2) cos–1
on
4)
3) sin–1
and the projection of on
2)
3)
is
4) sin–1
are equal then the angle between
4)
+ and
–
is
33. If × ( × ) = ( × ) × for non-coplanar vectors , , then
2) parallel to
3) parallel to
4) + + =
1) parallel to
34. If a line makes 45°, 60° with positive direction of axes x and y then the angle it makes with the z axis is
1) 30°
2) 90°
3) 45°
4) 60°
35. If [ × , × , × ] = 64 then [ , , ] is
1) 32
2) 8
3) 128
4) 0
36. If [ + , + , + ] = 8 then [ , , ] is
1) 4
2) 16
3) 32
4) – 4
37. The value of [ + , + , + ] is equal to
1) 0
2) 1
3) 2
4) 4
38. The shortest distance of the point (2, 10, 1) from the plane . (3 – + 4 ) = 2
1) 2
2)
3) 2
4)
39. The vector ( × ) × ( × ) is
1) perpendicular to , , and
2) parallel to the vectors ( × ) and ( × )
3) parallel to the line of intersection of the plane containing and and the plane containing and
4) perpendicular to the line of intersection of the plane containing and and the plane containing and
40. If , , are a right handed triad of mutually perpendicular vectors of magnitude a, b, c then the value of
[ , , ] is
4) abc
1) a2b2c2
2) 0
3) abc
41. If , , are non-coplanar and [ × , × ,
1) 2
2) 3
42.
= s + t is the equation of
1) a straight line joining the points and
43. If the magnitude of the moment about the point
then the value of a is
1) 1
2) 2
44. The equation of the line parallel to
1)
3)
= ( + 5 + 3 ) + t( + 3 + 5 )
=( +5 +
) + t( + 3 + 5 )
× ]=[
3) 1
+
,
+ ,
+ ] then [
4) 0
, ,
] is
2) xoy plane 3) yoz plane 4) zox plane
+ of a force + a – ) acting through the point + is
3) 3
4) 4
and passing through the point (1, 3. 5) in vector form is
2)
4)
= ( + 3 + 5 ) + t( + 5 + 3 )
= ( + 3 + 5 ) + t( + 5 +
)
45. The point of intersection of the line = ( – ) + t (3 + 2 + 7 ) and the plane . ( + – ) = 8 is
1) (8, 6, 22)
2) (–8, –6, – 22)
3) (4, 3, 11)
4) (– 4, – 3, – 11)
57
46. The equation of the plane passing through the point (2, 1, –1) and the line of intersection of the planes
. ( + 3 – ) = 0 and . ( + 2 ) = 0 is
3) 2x + y – z + 5 = 0
4) 2x – y + z = 0
1) x + 4y – z = 0
2) x + 9y + 11z = 0
47. The work done by the force = + + acting on a particle, if the particle is displaced from A(3, 3, 3) to the
point B(4, 4, 4) is
1) 2 units
2) 3 units
3) 4 units
4) 7 units
48. If = – 2 + 3 and = 3 + + 2 then a unit vector perpendicular to and is
1)
2)
3)
4)
49. The point of intersection of the lines
and
1) (0, 0, –4)
2) (1, 0, 0)
50. The point of intersection of the lines
= (– + 2 + 3 ) + t (–2 + + ) and
1) (2, 1, 1)
2) (1, 2, 1)
51. The shortest distance between the lines
1)
is
3) (0, 2, 0)
4) (1, 2, 0)
= (2 + 3 + 5 ) + 5( + 2 + 3 )
3) (1, 1, 2)
4) (1, 1, 1)
and
is
2)
3)
4)
and
52. The shortest distance between the parallel lines
1) 3
2) 2
53. The following two lines are
3) 1
is
4) 0
and
1) parallel
2) intersecting
3) skew
4) perpendicular
54. The centre and radius of the sphere given by x2 + y2 + z2 – 6x + 8y –10z +1 = 0 is
1) (– 3, 4, – 5), 49
2) (–6, 8, – 10) 1
3) (3, – 4, 5), 7
4) (6, – 8, 10), 7
CREATED QUESTIONS
 





 





 








(1) The value of a . b when a = i – 2 j + k and b = 4 i – 4 j + 7 k is
(1) 19
(2) 3
(3) – 19
(4) 14

(2) The value of a . b when a = j + 2 k and b = 2 i + k is
(1) 2
(2) – 2
(3) 3
(4) 4
(3) The value of a . b when a = j – 2 k and b = 2 i +3 j – 2 k is
(1) 7
(2) – 7
(3) 5
(4) 6








(4) If m i +2 j + k and 4 i – 9 j + 2 k are perpendicular, then m is
(1) – 4
(2) 8
(3) 4





(4) 12

5) If 5 i – 9 j + 2 k and m i + 2 j + k are perpendicular, then m is
(1)

5
16
5
16
(2)

(3)

16
5
(4)
 

 16
5


(6) If a and b are two vectors such that a = 4, b =3 and a . b = 6, then the angle between a and b is
(1)

6

6
(2)

(3)




3
(4)

3


(7) The angle between the vectors 3 i – 2 j – 6 k and 4 i – j + 8 k is
 34 

 63 
(1) cos-1 
 34 

 63 
 34 

 63 
(2) sin-1  



(3) sin-1 
 34 

 63 
(4) cos-1  

(8) The angle between the vectors i – j and i – k is
(1)

3
(2)
 2
3
(3)
58

3
(4)
2
3






(9) The projection of the vector 7 i + j – 4 k on 2 i + 6 j + 3 k is
(1)
7
8
8
(2)
(3)
66
 






8
7


(10) a . b , when a = 2 i + 2 j – k and b = 6 i – 3 j + 2 k is
(1) 4
(2) – 4
(3) 3




66
8
(4)
(4) 5


(11) If the vectors 2 i +  j + k and i – 2 j + k are perpendicular to each other, then  is
(1)
2
3
2
3
(2)
(3)
















3
2
(4)
(12) If the vectors a = 3 i + 2 j + 9 k and b = i + m j + 3 k are perpendicular, then m is
(1) – 15
(2) 15
(3) 30
3
2
(4) – 30
(13) If the vectors a = 3 i + 2 j + 9 k and b = i + m j + 3 k are parallel, then m is
(1)
3
2
2
3
(2)
(3)
3
2
  


(4)
2
3
(4)
3

(14) If a , b , c are three mutually perpendicular unit vectors, then a  b  c =
(1) 3

(2) 9



(3) 3 3


(15) If a  b = 60, a  b = 40 and b = 46, then a is
(1) 22
(2) 21

 

(3) 18




(4) 11


(16) Let u , v and w be vectors such that u + v + w = 0 . If u = 3, v = 4 and w = 5,
 






then u . v + v . w + w . u is
(2) – 25
(1) 25
(17) The projection of i – j on z- axis is
(1) 0
(2) 1




(3) 5
(4)
(3) – 1
(4) 2
5


(18) the projection of i + 2 j – 2 k on 2 i – j + 5 k is
(1)
 10
(2)
10
30
(3)
30




1
3
(4)
10
30


(19) The projection of 3 i + j – k on 4 i – j + 2 k is
(1)
9
(2)
9
21
(3)
21
81
(4)
21
 81
21



(20) The work done in moving a particle from the point A with position vector 2 i – 6 j + 7 k to the







point B, with position vector 3 i – j – 5 k by a force F = i + 3 j – k is
(1) 25
(2) 26
(3) 27



(4) 28

(21) The work done by the force F = a i + j + k is moving the point of application from (1, 1, 1) to
(2, 2, 2) along a straight line is given to be 5 units. The value of a is
(1) – 3
(2) 3
(3) 8
(4) – 8


 


(22) If a = 3, b = 4 and a . b = 9, then a x b is
(1) 3 7
(2) 63
(3) 69
59
(4)
69




 
(23) The angle between two vectors a and b is, if a x b = a . b
(1)

4
(2)






3
(3)

6



(4)

2
(4)

2

(24) If a = 2, b = 7 and a x b = 3 i – 2 j + 6 k , then the angle between a and b is
(1)

4
(2)

3
(3)
(25) The d.c.s of a vector whose direction ratios are 2, 3, – 6 are
2 3 6
2 3 6
(1)  , ,
(2)  , ,


7 7 7 
 49 49 49 
(26) The unit normal vectors to the plane 2x – y + 2z = 5 are



(1) 2 i – j + 2 k
(2)

1  
(2 i – j + 2 k )
3

6
(3)  2 , 3 ,  6 


 7
7

1  
(2 i – j + 2 k )
3
(3) –


(4)  2 , 3 , 6 
7 7 7
7 
(4) 

1  
(2 i – j + 2 k )
3


(27) The length of the perpendicular from the origin to the plane r . (3 i + 4 j + 12 k ) = 26 is
(1) 26
(2)
26
169
(3) 2


(4)
1
2
(4)
7
30


(28) The distance from the origin to the plane r . (2 i – j + 5 k ) = 7 is
7
30
(1)
30
7
(2)

(3)


30
7

(29) Chord AB is a diameter of the sphere r  (2 i  j  6 k ) = 18 with coordinate of A as (3, 2, – 2).
The coordinates of B is
(1) (1, 0, 10)
(2) (– 1, 0, – 10)



(3) (– 1, 0, 10)
(4) (1, 0, – 10)

(30) The centre and radius of the sphere r  (2 i  j  4 k ) = 5 are
(1) (2, – 1, 4) and 5
(2) (2, 1, 4) and 5


(3) (– 2, 1, 4) and 6 (4) (2, 1, – 4) and 5


(31) The centre and radius of the sphere 2 r  (3 i  j  4 k ) = 4 are
3 1

, ,2  ,4
 2 2

3 1

3 1

, ,2  and 2 (3) 
, ,2  , 6
 2 2

 2 2

(1) 
3 1

, ,2  and 5
 2 2

(2) 
(4) 

(32) The vector equation of a plane passing through a point where P.V. is a and perpendicular to a

vector n is
 
 














(1) r . n = a . n
(2) r x n = a x n
(3) r + n = a + n
(4) r – n = a – n
(33) The vector equation of a plane whose distance from the origin is p and perpendicular to a unit

vector n is


 
(1) r . n = p

(2) r . n = q

(3) r x n = p
(4) r . n = p

(34) The non- parametric vector equation of a plane passing through a point whose P.V. is a and


parallel to u and v is


  


(1)  r  a , u , v  = 0
   
(2)  r u v  = 0


60
 



(3)  r a u x v  = 0


   
(4)  a u v  = 0


 
(35) The non parametric vector equation of a plane passing through the points whose P.V.s are a , b

and parallel to v is

 
 


(1)  r  a b a v  = 0



 

(2)  r b a v  = 0





b v = 0



(3)  a




(4)  r a b  = 0


(36) The non-parametric vector equation of plane passing through three non- collinear points whose
  
P.V.s are a , b , c is


 
 




(1)  r  a b a c  a  = 0


a b = 0


(2)  r


(3)  r



   
b c  = 0 (4)  a b c  = 0







(37) The vector equation of a plane passing through the line of intersection of the planes r . n 1 = q 1 and



r . n 2 = q 2 is







(1) ( r . n 1 – q 1 ) +  ( r . n 2 – q 2 ) = 0





(3) r x n 1 + r x n 2 = q1 + q2






(2) r . n 1 – r . n 2 = q 1 +  q 2



(4) r x n 1 – r x n 2 = q1 + q2


 
(38) The angle between the line r = a + t b and plane r . n = q is connected by the relation
 
 
 
a .n
(1) cos  
q
b.n
a .b
(2) cos  


b
n
(3) sin  

 
(4) sin  
n
b.n


b
n
(39) The vector equation of a sphere whose centre is origin and radius ‘a’ is



(1) r = a



(2) r – c = a

(3) r = a

(4) r = a
CHAPTER – III (COMPLEX NUMBERS)
BOOK BACK QUESTIONS
55. The value of
+
is
1) 2
2) 0
3) – 1
4) 1
56. The modulus and amplitude of the complex number [e3 - iπ/4 ]3 are respectively
1) e9 ,
2) e9 , –
3) e6 , –
4) e9 , –
57. If (m – 5) + i(n + 4) is the complex conjugate of (2m + 3) + i(3n– 2) then (n, m) are
1)
2)
3)
4)
58. If x2 + y2 = 1 then the value of
59.
60.
61.
62.
63.
64.
is
1) x – iy
2) 2x
3) – 2 iy
4) x + iy
The modulus of the complex number 2 + i is
2)
3)
4) 7
1)
2
2
If A + iB = (a1 + ib1) (a2 + ib2) (a3 + ib3) then A + B is
1) a12 + b12 + a22 + b22 + a32 + b32 2) (a1 + a2 + a3 )2 + (b1 + b2 + b3 )2
3) (a12 + b12) (a22 + b22) (a32 + b32)
4) (a12 + a22 + a32) (b12 + b22 + b32)
If a = 3 + i and z = 2 –3i then the points on the Argand diagram representing az, 3az and – az are
1) Vertices of a right angled triangle
2) Vertices of an equilateral triangle
3) Vertices of an isosceles triangle
4) Collinear
The points z1, z2, z3, z4 in the complex plane are the vertices of a parallelogram taken in order if
and only if
1) z1 + z4 = z2 + z3 2) z1 + z3 = z2 + z4
3) z1 + z2 = z3 + z4
4) z1 – z2 = z3 – z4
If z represents a complex number then arg(z) + arg( ) is
1) π/3
2) π/2
3) 0
4) π/4
If the amplitude of a complex number is π/2 then the number is
1) purely imaginary
2) purely real
3) 0
4) neither real nor imaginary
61
65. If the point represented by the complex number iz is rotated about the origin through the angle π/2 in the
counter clockwise direction, then the complex number representing the new position is
1) iz
2) – iz
3) – z
4) z
66. The polar form of the complex number ( i25)3 is --------1) cos + i sin
2) cos + isin
3) cos – isin 4) cos – isin
67. If P represents the variable complex number z and if | 2z –1| = 2| z | then the locus of P is
1) the straight line x = 1/4
2) the straight line y = 1/4
3) the straight line z = 1/2
3) the circle x2 + y2 – 4x –1 =0
68.
=
1) cos+i sin
69. If zn = cos + isin
2) cos – i sin
then z1, z2 ….z6 is
3) sin – icos 4) sinθ + icos
1) 1
2) – 1
3) i
70. If – lies in the third quadrant, then z lies in the --------1) first quadrant
2) second quadrant 3) third quadrant
71. If x = cosθ + isinθ, the value of xn +
4) – i
4) fourth quadrant
1) 2 cosnθ
2) 2 i sin nθ
3) 2 sin nθ
4) 2 icosnθ
72. If a = cos α – i sin α, b = cos β – i sin β, c = cos γ – i sin γ then (a2 c2 – b2) / abc is
1) cos 2(α – β + γ) + i sin 2(α – β + γ)
2) – 2 cos (α – β + γ) 3) – 2i sin (α – β + γ) 4) 2 cos (α – β + γ)
73. z1 = 4 + 5i, z2 = –3 + 2i, then is
1)
2)
3)
4)
74. The value of i + i22 + i23 + i24 + i25 is
1) i
2) – i
3) 1
4) – 1
75. The conjugate of i13 + i14 + i15 + i16 is
1) 1
2) – 1
3) 0
4) – i
76. If –i + 2 is one root of the equation ax2 – bx+ c = 0, then the other root is
1) – i – 2
2) i – 2
3) 2 + i
4) 2i + i
77. The quadratic equation whose roots are ± i is
1) x2 + 7 = 0
2) x2 – 7
=0
3) x2 + x + 7 = 0
4) x2 – x – 7 = 0
78. The equation having 4 – 3i and 4 + 3i as roots is
1) x2 + 8x + 25 = 0
2) x2 + 8x – 25 = 0
3) x2 – 8x + 25 = 0
4) x2 – 8x – 25 = 0
is the root of the equation ax2 + bx + 1 = 0, where a, b are real then (a, b) is
79. If
1) (1, 1)
2) (1, – 1)
3) (0, 1)
4) (1, 0)
80. If –i + 3 is a root of x2 – 6x + k = 0, then the value of k is
3)
4) 10
1) 5
2)
81. If ω is a cube root of unity then the value of (1– ω + ω2)4 + (1+ ω – ω2)4 is
4) – 32
1) 0
2) 32
3) – 16
82. If ω is the nth root of unity then
1) 1+ ω2 + ω4 + … =
ω + ω3 + ω5 + …
2) ωn = 0
3) ωn = 1
4) ω = ωn –1
2
4
8
83. If ω is the cube root of unity then the value of (1– ω) (1 – ω ) (1– ω ) (1 – ω ) is
1) 9
2) – 9
3) 16
4) 32
CREATED QUESTIONS
 35 is
(1) i 35
(2) – i 35
(2) The complex number form of 3 –  7 is
(1) – 3 + i 7
(2) 3 – i 7
(3) Real and imaginary parts of 4 – i 3 are
(1) 4, 3
(2) 4, – 3
(1) The complex number form of
(3) i  35
(4) 35 i
(3) 3 – i7
(4) 3 + i7
(3) – 3 , 4
(4)
62
3,4
3
i are
2
3
3
(1) 0,
(2) , 0
2
2
5) The complex conjugate of 2 + i 7 is
(1) – 2 + i 7
(2) – 2 – i 7
4) Real and imaginary parts of
(6)The complex conjugate of – 4 – i9 is
(1) – 4 + i9
(2) 4 + i9
(7) The complex conjugate of 5 is
(3) 2, 3
(4) 3, 2
(3) 2 – i 7
(4) 2 + i 7
(3) 4 – i9
(4) – 4 – i9
(1) 5
(2) – 5
(3) i 5
(4) – i 5
8) The standard form (a + ib) of 3 + 2i + (– 7 – i) is
(1) 4 – i
(2) – 4 + i
(3) 4 + i
(4) 4 + 4i
(9) If a + ib = (8 – 6i) – (2i – 7), then the values of a and b are
(1) 8, – 15
(2) 8, 15
(3) 15, 9
(4) 15, – 8
(10) If p + iq = (2 – 3i)(4 + 2i), then q is
(1) 14
(2) – 14
(3) – 8
(4) 8
(11) The conjugate of (2 + i)(3 – 2i) is
(3) – 8 + i
(4) 8 + i
(1) 8 – i
(2) – 8 – i
(12) The real and imaginary parts of (2 + i)(3 – 2i) are
(1) – 1, 8
(2) – 8, 1
(3) 8, – 1
(4) – 8, – 1
(13) The modulus values of – 2 + 2i and 2 – 3i are
(1) 5 , 5
(2) 2 5 , 13
(3) 2 2 , 13
(4) – 4, 1
(14) The modulus values of – 3 – 2i and 4 + 3i are
(1) 5, 5
(2) 5 , 7
(3) 6 , 7
(4) 13 , 5
(15) The cube roots of unity are
(1) in G.P. with common ratio 
(2) in G.P. with common difference  2
(3) in A.P. with common difference  (4) in A.P. with common difference with  2
(16) The arguments of nth roots of a complex number differ by
(1)
2
n
(2)

n
(3)
3
n
(4)
4
n
(17) Which of the following statements is correct?
(1) negative complex numbers exist
(2) order relation does not exist in real numbers
(3) order relation exist in complex numbers (4) (1+i) > (3-2i) is meaningless
(18) Which of the following are correct?
(a) Re (z)  z
(1) (a), (b)
(19) The values of z  z is
(1) 2 Re (z)
(20) The value of z – z is
(1) 2Im (z)
(21) The value of z z is
(1) z
(b) lm (z)  z
(d) ( z n )  ( z ) n
(c) z = z
(2) (b), (c)
(3) (b), (c) and (d)
(2) Re (z)
(3) Im (z)
(4) 2Im (z)
(2) 2iIm (z)
(3) Im (z)
(4) i Im (z)
(2) z
(3) 2 z
(4) 2 z
2
(4) (a),(c) and (d)
2
(22) If z  z 1 = z  z 2 , then the locus of z is
(1) a circle with centre at the origin
(2) a circle with centre at z1
(3) a straight line passing through the origin (4) is a perpendicular bisector of the line joining z1 and z2
(23) If  is a cube roots of unity, then
(1)  2 = 1
(2) 1 +  = 0
(3) 1 +  +  2 = 0
(4) 1 –  +  2 = 0
63
(24) The principal value of arg z lies in the interval
(1) 0,  
(2) (–  ,  ]
(3) [0,  ]
(4) (–  , 0]
 2


(25) If z1 and z2 are any two complex numbers, then which one of the following is false?
(1) Re(z1 + z2) = Re(z1) + Re(z2)
(2) Im(z1 + z2) = Im(z1) + Im(z2)
(3) arg(z1 + z2) = arg z1 + arg z2
(4) z 1  z 2 = z 1 z 2
(26) The fourth roots of unity are
(1) 1  i, – 1  i
(2)  i, 1  i
(3)  1,  i
(27) The fourth roots of unity form the vertices of
(1) an equilateral triangle
(2) a square (3) a hexagon
(28) Cube roots of unity are
(1) 1,
 1 i 3
2
(2) i, – 1 
i 3
2
(3) 1,
1 i 3
2
(4) 1, – 1
(4) a rectangle
(4) i,
1 i 3
2
p
q
(29) The number of values of (cos   i sin ) where p and q are non-zero integers prime to each other, is
(1) p
(2) q
(3) p + q
(4) (p – q)
(30) The value of e i  +e - i  is
(1) 2cos 
(2) cos 
(3) 2 sin 
(4) sin 
(31) The value of e i  – e - i  is
(1) sin 
(2) 2 sin 
(3) i sin 
(4) 2i sin 
(32) Geometrical interpretation of z is
(1) reflection of z on real axis
(2) reflection of z on imaginary axis
(3) rotation of z about origin
(4) rotation of z about origin through  /2 in clockwise direction
(33) If z1 = a + ib, z2 = – a + ib, then z1 – z2 lies on
(1) real axis
(2) imaginary axis
(3) the line y = x
(4) the line y = – x
(34) Which one of the following is incorrect?
(1) (cos  + isin  )n = cos n  + isinn 
(2) (cos  – isin  )n = cosn  – isinn 
(3) (sin  + i cos  )n = sinn  + icos n 
(4)
1
= cos  – isin 
cos   i sin 
(35) Polynomial equation P(x) = 0 admits conjugate pairs of imaginary roots only, if the coefficients are
(1) imaginary
( 2) complex
(3) real
(4) either real of complex
(36) Identify the correct statement
(1) Sum of the moduli of two complex numbers is equal to their modulus of the sum
(2) Modulus of the product of the complex numbers is equal to sum of their moduli
(3) Arguments of the product of two complex numbers is the product of their arguments.
(4) Argument of the product of two complex numbers is equal to sum of their arguments.
(37) Which of the following is not true?
(1) z 1  z 2 = z 1 + z 2
(2) z 1 z 2 = z 1 z 2
(3) Re(z) =
zz
2
(4) Im (z) =
zz
2i
(38) If z1 and z2 are complex numbers, then which of the following is meaningful?
(1) z1 < z2
(2) z1 > z2
(3) z1  z2
(4) z1  z2
(39) Which of the following is incorrect ?
(1) Re (z)  z
(2) Im (z)  z
(3) z z = z 2
(4) Re (z)  z
(40) Which of the following is incorrect ?
(1) z 1  z 2  z 1 + z 2
(2) z 1  z 2  z 1 + z 2
(3) z 1  z 2  z 1 – z 2
(4) z 1  z 2  z 1 + z 2
(41) Which of the following is incorrect?
(1) z is the mirror image of z on the real axis
(2) The polar form of z is (r, –  )
(3) – z is the point symmetrical to z about the origin
(4) The polar form of – z is ( – r, –  )
64
(42) Which of the following is incorrect ?
(1) Multiplying a complex number by i is equivalent to rotating the number counter clock wise
about the origin through an angle 900
(2) Multiplying a complex number by – i is equivalent to rotating the number clockwise about
the origin through an angle 900
(3) Dividing a complex number by i is equivalent to rotating the number counter clockwise about
the origin through an angle 900
(4) Dividing a complex number by i is equivalent to rotating then number clockwise about the
origin through an angle 900
(43) Which of the following is incorrect regarding nth roots of unity?
(1) the number of distinct roots is n
(2) the roots are in G.P. with common ratio cis
(3) the arguments are in A.P. with common difference
2
n
2
n
(4) Product of the roots is 0 and the sum of the root is  1
(44) Which of the following are true?
(i) If n is a positive integer, then (cos  + isin  )n = cosn  + isin n 
(ii) If n is a negative integer, then (cos  + isin  )n = cos n  – i sin n 
(iii) If n is a fraction, then cosn  + isinn  is one of the values of (cos  + isin  )n
(iv) If in is a negative integer, then (cos  + i sin  )n = cosn  + i sinn 
(1) (i), (ii), (iii), (iv)
(2) (i), (iii), (iv)
(3) (i), (iv)
(4) (i) only
(45) If O(0, 0), A(z1), B(z2), B’(–z2) are the complex numbers in a Argand plane, then which of the following
are correct ?
(i) In the parallelogram OACB, C represents z1 + z2
(ii) In the argand plane, E represents z1z2 where OE = OA.OB and OE makes an angle
arg(z1) + arg( z2) with positive real axis.
(iii) In the parallelogram OB’DA, D represents z1 – z2
(iv) In the argand plane, F represents
z1
OA
where OF =
and OF makes an angle
z2
OB
arg(z1) – arg(z2) with positive real axis.
(1) (i), (ii), (iii), (iv)
(2) (i), (iii) , (iv)
(46) If Z = 0, then the arg (Z) is
(1) 0
(2) 
(3) (i), (iv)
(3)

2
(4) (i) only
(4) indeterminate
CHAPTER – IV (ANALYTICAL GEOMETRY)
BOOK BACK QUESTIONS
84. The axis of the parabola y2 – 2y + 8x – 23 = 0 is
2) x = –3
3) x = 3
4) y = 1
1) y = –1
85. 16x2 – 3y2 – 32x – 12y – 44 = 0 represents
1) an ellipse
2) a circle
3) a parabola
4) a hyperbola
86. The line 4x + 2y = c is a tangent to the parabola y2 = 16x then c is
1) –1
2) –2
3) 4
4) – 4
87. The point of intersection of the tangents at t1 = tand t2 = 3tto the parabola y2 = 8x is
1) (6t2, 8t)
2) (8t, 6t2)
3) (t2, 4t)
4) (4t, t2)
2
88. The length of the latus rectum of the parabola y – 4x + 4y +8 = 0
1) 8
2) 6
3) 4
4) 2
2
89. The diretrix of the parabola y = x + 4 is
1) x =
2) x = –
3) x = –
4) x =
90. The length of the latus rectum of the parabola whose vertex is (2, –3) and the diretrix is x = 4 is
1) 2
2) 4
3) 6
4) 8
91. The focus of the parabola x2 = 16y is
4) (0, –4)
1) (4, 0)
2) (0, 4)
3) (– 4, 0)
65
92. The vertex of the parabola x2 = 8y – 1 is
1)
2)
93. The line 2x + 3y + 9 = 0 touches the parabola
1) (0, – 3)
2) (2, 4)
3)
y2
4)
= 8x at the point
3)
4)
y2
94. The tangents at the end of any focal chord to the parabola = 12x is intersect on the line
1) x – 3 = 0
2) x + 3 = 0
3) y + 3 = 0
4) y – 3 = 0
95. The angle between the two tangents drawn from the point (– 4, 4) to y2 = 16x is
1) 45°
2) 30°
3) 60°
4) 90°
96. The eccentricity of the conic 9x2 + 5y2 – 54x – 40y + 116 = 0 is
2)
3)
4)
1)
97. The length of the semi-major and the length of semi-minor axis of the ellipse
1) 26, 12
2) 13, 24
3) 12, 26
4) 13, 12
98. The distance between the foci of the ellipse 9x2 + 5y2 =180
1) 4
2) 6
3) 8
4) 2
99. If the length of major and semi-minor axes of an ellipse are 8, 2 and their corresponding equations are
y – 6 = 0 and x + 4 = 0 then the equations of the ellipse is
1)
2)
3)
4)
100. The straight line 2x – y + c = 0 is a tangent to the ellipse 4x2 + 8y2 = 32, if c is
1) ±
2) ± 6
3)36
4) ± 4
101. The sum of the distance of any point on the ellipse 4x2 + 9y2 = 36 from ( , 0) and (– , 0) is
1) 4
2) 8
3) 6
4) 18
102. The radius of the director circle of the conic 9x2 + 16y2 = 144 is
1)
2) 4
3) 3
4) 5
103. The locus foot of the perpendicular from the focus to a tangent of the curve 16x2 + 25y2 = 400 is
1) x2 + y2 = 4
2) x2 + y2 = 25
3) x2 + y2 = 16
4) x2 + y2 = 9
104. The eccentricity of the hyperbola 12y2 – 4x 2 – 24x + 48y – 127 = 0
1) 4
2) 3
3) 2
4) 6
105. The eccentricity of the hyperbola whose latus rectum is equal to half of its conjugate axis is
1)
2)
3)
4)
106. The difference between the focal distance of any point on the hyperbola
is 24
and the eccentricity is 2. Then the equation of the hyperbola is
1)
2)
3)
4)
107. The directrices of the hyperbola x2 – 4(y – 3)2 = 16 are
1) y =
2) x =
3) y =
4) x =
108. The line 5x – 2y + 4k = 0 is a tangent to 4x2– y2 = 36 then k is
1)
2)
3)
4)
109. The equation of the chord of contact of tangents from (2, 1) to the hyperbola
1) 9x – 8y – 72 = 0
2) 9x + 8y + 72 = 0
3) 8x – 9y – 72 = 0
110. The angle between the asymptotes to the hyperbola
1)
2)
is
4) 8x + 9y + 72 = 0
is
3)
4)
111. The asymptotes to the hyperbola 36y2 – 25x2 + 900 = 0 are
1) y =
2) y =
3) y =
4) y =
112. The product of the perpendiculars drawn from the point (8, 0) on the hyperbola
1)
2)
3)
4)
113. The locus of the point of intersection of perpendicular tangents to the hyperbola
1) x2 + y2 = 25
2) x2 + y2 = 4
to its asymptotes is
3) x2 + y2 = 3
66
4) x2 + y2 = 7
is
114. The eccentricity of the hyperbola with asymptotes x + 2y – 5 = 0, 2x – y + 5 = 0
1) 3
2)
3)
4) 2
115. Length of the semi-trasverse axis of the rectangular hyperbola xy = 8 is
1) 2
2) 4
3) 16
4) 8
116. The asymptotes of the rectangular hyperbola xy = c2 are
1) x = c, y = c
2) x = 0, y = c
3) x = c, y = 0
4) x = 0, y = 0
117. The co-ordinate of the vertices of the rectangular hyperbola xy = 16 are
1) (4, 4), (–4, –4)
2) (2, 8), (–2, –8)
3) (4, 0), (–4, 0)
4) (8, 0), (–8, 0)
118. One of the foci of the rectangular hyperbola xy = 18 is
1) (6, 6)
2) (3, 3)
3) (4, 4)
4) (5, 5)
119. The length of the latus rectum of the rectangular hyperbola xy= 32 is
2) 32
3) 8
4) 16
1) 1)
120. The area of the triangle formed by the tangent at any point on the rectangular hyperbola xy = 72 and its
asymptotes is
1) 36
2) 18
3) 72
4) 144
meets the curve again at
121. The normal to the rectangular hyperbola xy = 9 at
1)
2)
3)
4)
CREATED QUESTIONS
y2 =
(1) The axis of the parabola
4x is
(1) x = 0
(2) y = 0
(3) x = 1
(2) The vertex of the parabola y2 = 4x is
(1) (1, 0)
(2) (0, 1)
(3) (0, 0)
2
(3) The focus of the parabola y = 4x is
(1) (0, 1)
(2) (1, 1)
(3) (0, 0)
(4) The directrix of the parabola y2 = 4x is
(2) x = – 1
(3) y = 1
(1) y = – 1
2
(5) The equation of the latus rectum of y = 4x is
(1) x = 1
(2) y = 1
(3) x = 4
(6) The length of the latus rectum of y2 = 4x is
(1) 2
(2) 3
(3) 1
(7) The axis of the parabola x2 = – 4y is
(1) y = 1
(2) x = 0
(3) y = 0
(8) The vertex of the parabola x2 = – 4y is
(1) (0, 1)
(2) (0, – 1)
(3) (1, 0)
(9) The focus of the parabola x2 = – 4y is
(3) (0, 1)
(1) (0, 0)
(2) (0, – 1)
(10) The directrix of the parabola x2 = – 4y is
(1) x = 1
(2) x = 0
(3) y = 1
(11) The equation of the latus rectum of x2 = – 4y is
(1) x = – 1
(2) y = – 1
(3) x = 1
2
(12) The length of the latus rectum of x = – 4y is
(1) 1
(2) 2
(3) 3
(13) The axis of the parabola y2 = – 8x is
(1) x = 0
(2) x = 2
(3) y = 2
2
(14) The vertex of the parabola y = – 8x is
(1) (0, 0)
(2) (2, 0)
(3) (0, – 2)
(15) The focus of the parabola y2 = – 8x is
(1) (0, – 2)
(2) (0, 2)
(3) (– 2, 0)
(16) The equation of the directrix of the parabola y2 = – 8x is
(1) y + 2 = 0
(2) x – 2 = 0
(3) y – 2 = 0
(17) The equation of the latus rectum of y2 = – 8x is
(1) y – 2 = 0
(2) y + 2 = 0
(3) x – 2 = 0
(18) The length of the latus rectum y2 = – 8x is
(1) 8
(2) 6
(3) 4
67
(4) y = 1
(4) (0, – 1)
(4) (1, 0)
(4) x = 1
(4) y = – 1
(4) 4
(4) x = 1
(4) (0, 0)
(4) (1, 0)
(4) y = 0
(4) y = 1
(4) 4
(4) y = 0
(4) (2, – 2)
(4) (2, 0)
(4) x + 2 = 0
(4) x + 2 = 0
(4) – 8
(19) The axis of the parabola x2 = 20y is
(1) y = 5
(2) x = 5
(3) x = 0
(4) y = 0
2
(20) The vertex of the parabola x = 20y is
(1) (0, 5)
(2) (0, 0)
(3) (5, 0)
(4) (0, – 5)
(21) The focus of the parabola x2 = 20y is
(1) (0, 0)
(2) (5, 0)
(3) (0, 5)
(4) (– 5, 0)
2
(22) The equation of the directrix of the parabola x = 20y
(1) y – 5 = 0
(2) x + 5 = 0
(3) x – 5 = 0
(4) y + 5 = 0
(23) The equation of the latus rectum of the parabola x2 = 20y is
(3) y + 5 = 0
(4) x + 5 = 0
(1) x – 5 = 0
(2) y – 5 = 0
(24) The length of the latus rectum of the parabola x2 = 20y is
(1) 20
(2) 10
(3) 5
(4) 4
(25) If the centre of the ellipse is (2, 3) one of the foci is (3, 3), then the other focus is
(1) (1, 3)
(2) (– 1, 3)
(3) (1, – 3)
(4) (– 1, – 3)
(26) The equations of the major and minor axes of
x 2 y2

 1 are
9
4
(3) x = 0, y = 0 (4) y = 0, x = 0
(1) x = 3, y = 2 (3) x = – 3, y = – 2
(27) The equations of the major and minor axes of 4x2 + 3y2 = 12 are
(1) x = 3 , y = 2
(2) x = 0, y = 0 (3) x = 3 , y = – 2
(4) y = 0, x = 0
(28) The lengths of minor and major axes of
x 2 y2

 1 are
9
4
(1) 6, 4
(2) 3, 2
(3) 4, 6
(29) The lengths of major and minor axes of 4x2 + 3y2 = 12 are
(1) 4, 2 3
(2) 2, 3
(3) 2 3 , 4
4
(4)
3 ,2
2
2
(30) The equations of the directrices of
(4) 2, 3
y
x

 1 are
16
9
16
16
7
(4) y = 
25
4
(3) y  
4
25
2
2
x
y

 1 are
(32) The equations of the latus rectum of
16
9
(1) y =  7
(2) x   7
(3) x =  7
(4) y = 
(1) y  
(2) x = 
(3) x = 
7
7
16
7
(31) The equations of the directrices of 25x2 + 9y2 = 225 are
(1) x = 
4
25
(2) x  
25x2
(33) The equations of the latus rectum of
(1) y =  5
(2) x =  5
2
(34) The length of the latus rectum of
(1)
9
2
(2)
9
5
(2)
+
9y2
= 225 are
(3) y =  4
(4)  =  4
y
x

 1 is
16
9
9
16
(4)
16
9
(3)
25
9
(4)
5
18
(3)
2
5
(4)
4
5
(3)
(35) The length of the latus rectum of 25x2 + 9y2 = 225 is
18
5
x 2 y2

 1 is
(36) The eccentricity of the ellipse
25 9
1
3
(1)
(2)
5
5
(1)
(4) y =  7
2
2
9
68
25
4
x 2 y2

 1 is
4
9
3
(2)
5
(37) The eccentricity of the ellipse
(1)
5
3
(3)
(38) The eccentricity of the ellipse 16x2 + 25y2 = 400 is
(1)
4
5
(2)
3
5
(3)
x 2 y2
(39) Centre of the ellipse

 1 is
25 9
(1) (0, 0)
(2) (5, 0)
(1) (0, 3)
y
x

 1 is
4
9
(2) (2, 3)
2
(4)
2
3
3
4
(4)
2
5
(3) (3, 5)
(4) (0, 5)
(3) (0, 0)
(4) (3, 0)
2
2
(40) The centre of the ellipse
3
5
2
y
x

 1 are
25 9
(1) (0,  5)
(2) (0,  4)
2
y2
x

 1 are
(42) The foci of the ellipse
4
9
(2) (0,  5 )
(1) (  5, 0)
(3) (  5, 0)
(4) (  4, 0)
(3) (0,  5)
(4) ( 
(43) The foci of the ellipse 16x2 + 25y2 = 400 are
(1) (  3, 0)
(2) (0,  3)
(3) (0,  5)
(4) (  5, 0)
(3) (  5, 0)
(4) (  3, 0)
(3) (  3, 0)
(4) (0,  2)
(41) The foci the ellipse
2
5 , 0)
2
y
x

 1 are
25 9
(1) (0,  5)
(2) (0,  3)
x2 y 2

 1 are
(45) The vertices of the ellipse
4
9
(1) (0,  3)
(2) (  2, 0)
(44) The vertices of the ellipse
16x2
25y2
(46) The vertices of the ellipse
+
= 400 are
(1) (0,  4)
(2) (  5, 0)
(3) (  4, 0)
(4) (0,  5)
(47) If the centre of the ellipse is (4, – 2) and one of the foci is (4, 2), then the other focus is
(1) (4, 6)
(2) (6, – 4)
(3) (4, – 6)
(4) (6, 4)
(48) The equations of transverse and conjugate axes of the hyperbola
x 2 y2

 1 are
9
4
(1) x = 2 ; y = 3 (2) y = 0 ; x = 0
(3) x = 3 ; y = 2 (4) x = 0 ; y = 0
(49) The equations of transverse and conjugate axes of the hyperbola 16y2 – 9x2 = 144 are
(1) y = 0 ; x = 0 (2) x = 3 ; y = 4 (3) x = 0 ; y = 0
(4) y = 3 ; x = 4
(50) The equations of transverse and conjugate axes of the hyperbola 144x2 – 25y2 = 3600 are
(1) y = 0 ; x = 0
(2) x = 12 ; y = 5
(3) x = 0 ; y = 0 (4) x = 5 ; y = 12
(51) The equations are transverse and conjugate axes of the hyperbola 8y2 – 2x2 = 16 are
(1) x = 2 2 ; y = 2
(2) x  2 ; y = 2 2
(3) x = 0 ; y = 0
(4) y  0; x  0
x2
y2
(52) The equations of the directrices of the hyperbola

 1 are
9
4
13
9
13
(1) y = 
(2) x = 
(3) y = 
9
9
13
9
(4) x  
13
(53) The equations of the directrices of the hyperbola 16y2 – 9x2 = 144 are
(1) x = 
5
9
(2) y = 
9
5
(3) x = 
69
9
5
(4) y  
5
9
(54) The equations of the latus rectum of the hyperbola
x 2 y2

 1 are
9
4
(1) y=  13
(2) y =  13
(3) x =  13
2
2
(55) The equations of the latus rectum of the hyperbola 16y – 9x = 144 are
(1) y =  5
(2) x =  5
(3) y =  5
(4) x = 
13
(4) x = 
5
2
2
y
x

 1 is
9
4
4
8
(1)
(2)
3
3
2
y
x2
(57) The eccentricity of the hyperbola

1
9
25
34
5
(2)
(1)
3
3
(56) The length of the latus rectum of
(3)
(59) The foci of the hyperbola
(4)
34
3
(3)
(58) The centre of the hyperbola 25x2 – 16y2 = 400 is
(1) (0, 4)
(2) (0, 5)
2
3
2
9
4
34
5
(4)
(3) (4, 5)
(4) (0, 0)
2
y
x

 1 are
9
25
(2) (34,0)
(1) (0,  34 )
(3) (0,  34)
(4) (  34 , 0)
2
2
(60) The vertices of the hyperbola 25x – 16y = 400 are
(2) (  4, 0)
(3) (0,  5)
(4) (  5, 0)
(1) (0,4)
2
(61) The equation of the tangent at (3, – 6) to the parabola y = 12x is
(1) x – y – 3 = 0
(2) x + y – 3 = 0
(3) x – y + 3 = 0
(4) x + y + 3 = 0
2
(62) The equation of the tangent at (– 3, 1) to the parabola x = 9y is
(1) 3x – 2y – 3 = 0
(2) 2x – 3y + 3 = 0
(3) 2x + 3y + 3 = 0
(4) 3x + 2y + 3 = 0
(63) The equation of chord of contact of tangents from the point (– 3, 1) to the parabola y2 = 8x is
(2) 4x + y + 12 = 0
(3) 4y – x – 12 = 0
(4) 4y – x + 12 = 0
(1) 4x – y – 12 = 0
(64) The equation of chord of contact of tangents from (2, 4) to the ellipse 2x2 + 5y2 = 20 is
(1) x – 5y + 5 = 0
(2) 5x – y + 5 = 0
(3) x + 5y – 5 = 0
(4) 5x – y – 5 = 0
(65) The equation of chord of contact of tangents from (5, 3) to the hyperbola 4x2 – 6y2 = 24 is
(1) 9x + 10y + 12 = 0
(2) 10x + 9y – 12 = 0
(3) 9x – 10y + 12 = 0 (4) 10x – 9y – 12 = 0
(66) The combined equation of the asymptotes to the hyperbola 36x2 – 25y2 = 900 is
(1) 25x2 + 36y2 = 0
(2) 36x2 – 25y2 = 0
(3) 36x2 + 25y2 = 0
(4) 25x2 – 36y2 = 0
2
2
(67) The angle between the asymptotes of the hyperbola 24x – 8y = 27 is
(1)

3
(2)

2
or
3
3
 a 2a 
, 
2
m m 
(2) 
(3)
2
3
(4)
(68) The point of contact of the tangent y = mx + c and the parabola y2 = 4ax is
 a 2a 
, 2
m m 
x 2 y2
(69) The point of contact of the tangent y = mx + c and the ellipse 2  2  1 is
a
b
2
2
2
2
b a m
a m b 
 a 2m  b2 


,
, 
,
(1) 
(2) 
(3) 
c 
c 
c 
 c
 c
 c
x 2 y2
(70) The point of contact of the tangent y = mx + c and the hyperbola 2  2  1 is
a
b
2
2
2
2
 am b 
a m b 
  a 2m  b2
, 
, 
,
(2) 
(3) 
(1) 
c 
c 
c
 c
 c
 c
(1) 
 2a a 
, 
2
m m
  a  2a 
,

2
m 
m
(3) 
70
 2
3
(4) 
  a 2m  b2
,
c
 c
(4) 






  am 2  b 2 

,
c 
 c
(4) 
(71) The true statements of the following are
(a) Two tangents and 3 normals can be drawn to a parabola from a point.
(b) Two tangents and 4 normals can be drawn to an ellipse from a point.
(c) Two tangents and 4 normals can be drawn to an hyperbola from a point.
(d) Two tangents and 4 normals can be drawn to an R.H. from a point.
(1) (a), (b), (c) and (d) (2) (a), (b) only
(3) (c), (d) only
(4) (a), (b) and (c)
2
(72) If ‘t1’, ‘t2’ are the extremities of any focal chord of a parabola y = 4ax, then t1 t2 is
(1) – 1
(3)  1
(2) 0

(73) The normal at ‘t1’ on the parabola y2 = 4ax meets the parabola at ‘t2’, then  t 1 

(1) – t2
(2) t2
(1)
al3
+
2alm2
+
m2n
=0
(3)
a 2 b 2 (a 2  b 2 ) 2


l 2 m2
n2
2alm2 +
m2n
(1)
=0
(3)
a 2 b 2 (a 2  b 2 ) 2


l 2 m2
n2
(4)
1
t2
x 2 y2

 1 is
a 2 b2
a 2 b 2 (a 2  b 2 ) 2
(2) 2  2 
l
m
n2
a 2 b 2 (a 2  b 2 ) 2
(4) 2  2 
l
m
n2
(75) The condition that the line lx + my + n = 0 may be a normal to the hyperbola
al3 +
1
2
2
 is
t 1 
(3) t1 + t2
(74) The condition that the line lx + my + n = 0 may be a normal to the ellipse
(4)
x 2 y2

 1 is
a 2 b2
a 2 b 2 (a 2  b 2 ) 2
(2) 2  2 
l
m
n2
a 2 b 2 (a 2  b 2 ) 2
(4) 2  2 
l
m
n2
(76) The condition that the line lx + my + n = 0 may be a normal to the parabola y2 = 4ax is
(1) al3 + 2alm2 + m2n = 0
a 2 b 2 (a 2  b 2 ) 2
(3) 2  2 
l
m
n2
a 2 b 2 (a 2  b 2 ) 2


l 2 m2
n2
a 2 b 2 (a 2  b 2 ) 2
(4) 2  2 
l
m
n2
(2)
(77) The chord of contact of tangents from any point on the directrix of the parabola y2 = 4ax passes through its ….
(1) vertex
(2) focus
(3) directrix
(4) latus rectum
(78) The chord of contact of tangents from any point on the directrix of the ellipse
through its….
(1) vertex
(2) focus
(3) directrix
x 2 y2

 1 passes
a 2 b2
(4) latus rectum
(79) The chord of contact of tangents from any point on the directrix of the hyperbola
x 2 y2

 1 passes
a 2 b2
through its…..
(a) vertex
(2) focus
(3) directrix
(4) latus rectum
2
(80) The point of intersection of tangents at ‘t1’ and ‘t2’ to the parabola y = 4ax is
(1) (a(t1 + t2), at1t2)
(2) (at1t2, a(t1 + t2)) (3) (at2, 2at)
(4) (at1t2, a(t1 – t2))
(81) If the normal to the R.H. xy = c2 at ‘t1’ meets the curve again at ‘t2’, then t13 t2 =
(1) 1
(2) 0
(3) – 1
(4) – 2
(82) The locus of the point of intersection of perpendicular tangents to the parabola y2 = 4ax is
(1) latus rectum
(2) directrix
(3) tangent at the vertex
(4) axis of the parabola
71
(83) The locus of the foot of perpendicular from the focus on any tangent to the ellipse
(1) x2 + y2 = a2 – b2
(2) x2 + y2 = a2 (3) x2 + y2 = a2 + b2
x 2 y2

 1 is
a 2 b2
(4) x = 0
x 2 y2
(84) The locus of the foot of perpendicular from the focus on any tangent on the hyperbola 2  2  1 is
a
b
(1) x2 + y2 = a2 – b2
(2) x2 + y2 = a2
(3) x2 + y2 = a2 + b2
(4) x = 0
2
(85) The locus of the foot of perpendicular from the focus on any tangent to the parabola y = 4ax is
(1) x2 + y2 = a2 – b2
(2) x2 + y2 = a2
(3) x2 + y2 = a2 + b2
(4) x = 0
x 2 y2
(86) The locus of point of intersection of perpendicular tangents to the ellipse 2  2  1 is
a
b
(1) x2 + y2 = a2 – b2
(2) x2 + y2 = a2
(3) x2 + y2 = a2 + b2
(4) x = 0
2
2
(87) The locus of point of intersection of perpendicular tangents to the hyperbola
y
x
 2  1 is
2
a
b
(1) x2 + y2 = a2 – b2
(2) x2 + y2 = a2
(3) x2 + y2 = a2 + b2
(88) The condition that line lx + my + n = 0 may be a tangent to the parabola y2 = 4ax is
(1) a2l2 + b2m2 = n2
(2) am2 = ln
(3) a2l2 – b2m2 = n2
(1) a2l2 + b2m2 = n2
(2) am2 = ln
(4) 4c2lm = n2
2
2
(89) The condition that the line lx + my + n = 0 may be a tangent to the ellipse
(4) x = 0
y
x
 2  1 is
2
a
b
(3) a2l2 – b2m2 = n2
(4) 4c2lm = n2
2
(90) The condition that the line lx + my + n = 0 may be a tangent to the hyperbola
2
y
x
 2  1 is
2
a
b
(1) a2l2 + b2m2 = n2
(2) am2 = ln
(3) a2l2 – b2m2= n2
(4) 4c2lm = n2
(91) The condition that the line lx + my + n = 0 may be a tangent to the rectangular hyperbola xy = c2 is
(1) a2l2 + b2m2 = n2
(2) am2 = ln
(3) a2l2 – b2m2= n2
(4) 4c2lm = n2
(92) The foot of a perpendicular from a focus of the hyperbola on an asymptote lies on the ……
(1) centre
(2) corresponding directrix
(3) vertex
(4) L.R.
CHAPTER – V (DIFFERENTIAL CALCULUS APPLICATIONS I)
BOOK BACK QUESTIONS
1.
2.
The gradient of the curve y = – 2x3 +3x + 5 at x = 2 is
1) –20
2) 27
3) – 16
The rate of change of area A of a circle of radius r is
1) 2πr
2) 2πr
3) πr2
4) – 21
4) π
3.
The velocity v of a particle moving along a straight line when at a distance x from the origin is given by a + bv2 =
x2 where a andb are constants. Then the acceleration is
1)
2)
3)
4)
4.
A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1 cm3 / min. The rate at
which the diameter is decreasing when the diameter is 10 cm is
1)
cm / min
2)
cm / min
3)
cm / min
4)
cm / min
5.
The slope of the tangent to the curve y = 3x2 + 3sin x at x = 0
1) 3
2) 2
3) 1
4) –1
The slope of the normal to the curve y = 3x2 at the point whose x coordinate is 2 is
2)
3)
4)
1)
6.
7.
The point on the curve y = 2x2 – 6x – 4 at which the tangent is parallel to the x- axis is
1)
2)
3)
4)
8.
The equation of the tangent to the curve y =
1) 5y + 3x = 2
2) 5y – 3x = 2
at the point (–1, –1/5) is
3) 3x – 5y = 2
72
4) 3x +3y = 2
9.
The equation of the normal to the curve θ = at the point (–3, –1/3) is
1) 3θ = 27t – 80
2) 5θ = 27t – 80
10. The angle between the curves
1)
3) 3θ = 27t + 80
and
2)
is
3)
emx
11. The angle between the curve y =
1) tan–1
2) tan–1
and y =
4) θ =
4)
e–mxfor
m> 1 is
3) tan–1
4) tan–1
12. The parametric equations of the curve x2/3 + y2/3 = a2/3 are
1) x = a sin3 θ; y = a cos3 θ
2) x = a cos3 θ; y = a sin3 θ
3
3
3) x = a sin θ; y = a cos θ
4) x = a3cos θ; y = a3 sin θ
13. If the normal to the curve x2/3 + y2/3 = a2/3 makes an angle θ with the x- axis then the slope of the normal is
2) tan θ
3) – tan θ
4) cot θ
1) – cot θ
14. If the length of the diagonal of a square is increasing at the rate of 0.1 cm /sec. What is the rate of increase of its
area when the side is cm?
4) 0.15 cm2/sec
1) 1.5 cm2/sec
2) 3 cm2/sec
3) 3 cm2/sec
15. What is the surface area of a sphere when the volume is increasing at the same rate as its radius
1) 1
2)
3) 4π
4)
16. For what values of x is the rate of increase of x3 – 2x2 +3x +8 is twice the rate of increase of x
2)
3)
4)
1)
17. The radius of a cylinder is increasing at the rate of 2 cm /sec and its altitude is decreasing at the rate of 3 cm
/sec. The rate of change of volume when the radius is 3 cm and the altitude is 5 cm is
1) 23 π
2) 33 π
3) 43 π
4) 53 π
18. If y = 6x – x3 and x increases at the rate of 5 units per second, the rate of change of slope when x = 3 is
1) – 90 units / sec
2) 90 units / sec
3) 180 units / sec
4) – 180 units / sec
19. If the volume of an expanding cube is increasing at the rate of 4 cm3 /sec then the rate of change of surface area
when the volume of the cube is 8 cubic cm is
1) 8 cm2/sec
2) 16 cm2/sec
3) 2 cm2/sec
4) 4 cm2/sec
2
20. The gradient of the tangent to the curve y = 8 + 4x - 2x at the point where the curve cuts the y-axis is
1) 8
2) 4
3) 0
4) – 4
21. The angle between the parabolas y2 = x and x2 = y at the origin is
1) 2tan–1
2) tan–1
3)
4)
22. For the curve x = etcost; y = et sin t the tangent line is parallel to the x-axis when t is equal to
1) –
2)
3) 0
4)
23. If the normal makes an angle θ with positive x-axis then the slope of the curve at the point where the normal is
drawn is
1) – cot θ
2) tan θ
3) – tan θ
4) cot θ
24. The value of ‘a’ so that the curves y = 3ex and y = e–x intersect orthogonally is
1) – 1
2) 1
3)
4) 3
25. If s = t3 – 4t2 + 7, the velocity when the acceleration is zero is
1) m/sec
2) – m/sec
3) m/sec
4) – m/sec
26. If the velocity of a particle moving along a straight line is directly proportional to the square of its distance from
a fixed point on the line. Then its acceleration is proportional to
1) s
2) s2
3) s3
4) s4
2
27. The Rolle’s constant for the function y = x on [– 2, 2] is
1)
2) 0
3) 2
4) – 2
28. The ‘c’ of Lagranges Mean Value Theorem for the function f(x) = x2 + 2x – 1 ;a = 0, b = 1 is
1) – 1
2) 1
3) 0
4)
73
29. The value of ‘c’inRolle’s Theorem for the function f(x) = cos on [π, 3π] is
1) 0
2) 2 π
3)
4)
30. The value ‘c’ of Lagranges Mean Value Theorem for the function f(x) =
2)
3)
1)
when a = 1and b = 4 is
4)
31.
1) 2
2) 0
3) ∞
4) 1
1) ∞
2) 0
3) log
4)
32.
33. If f(a) = 2; f′(a) = 1; g(a) = –1; g′(a) = 2 then the value of
is
1) 5
2) – 5
3) 3
34. Which of the following function is increasing in (0, ∞)
1) ex
2)
3) – x2
35. The function of f(x) = x2 – 5x + 4 is increasing in
1) (– ∞, 1)
2) (1, 4)
36. The function of f(x) = x2 is decreasing in
1) (– ∞, ∞)
2) (–∞, 0)
37. The function y = tan x – x is
1) an increasing function in
3) increasing in
and decreasing in
4) – 3
4) x –2
3) (4, ∞)
4) everywhere
3) (0, ∞)
4) (– 2, ∞)
2) a decreasing function in
4) decreasing in
and increasing in
38. In a given semi circle of diameter 4 cm a rectangle is to be inscribed. The maximum area of the rectangle is
1) 2
2) 4
3) 8
4) 16
39. The least possible perimeter of a rectangle of area 100 m2 is
1) 10
2) 20
3) 40
4) 60
2
40. If f(x) = x – 4x + 5 on [0, 3] then the absolute maximum value is
1) 2
2) 3
3) 4
4) 5
41. The curve y = – e–x is
1) concave upward for x>0
2) concave downward for x>0
3) everywhere concave upward
4) everywhere concave downward
42. Which of the following curves is concave downward?
1) y = – x2
2) y = x2
3) y = ex
4) y = x2 + 2x – 3
43. The point of inflexion of the curve y = x4 is at
1) x = 0
2) x = 3
3) x = 12
4) nowhere
3
2
44. The curve y = ax + bx + cx + d has a point of inflexion at x = 1 then
1) a + b = 0
2) a + 3b = 0
3) 3a + b = 0
4) 3a + b = 1
CREATED QUESTIONS
(1) Let “h” be the height of the tank. Then the rate of change of pressure “p” of the tank with respect to the
height is
dp
dh
(2) If the temperature  oC of the certain metal rod of “l” metres is given by l = 1 + 0.0005  +0.0000004  2 ,
(1)
dh
dt
(2)
dp
dt
(3)
dh
dp
(4)
Then the rate of change of l in m/Co when the temperature is 100oC is
(1) 0.00013 m/Co
(2) 0.00023 m/Co
(3) 0.00026 m/Co
(4) 0.00033 m/Co
(3) The following graph gives the functional relationship between distance and time of a
moving car in m/sec. The speed of the car is
(1)
x
m/s
t
(2)
t
m/s
x
(3)
dx
m/s
dt
74
(4)
dt
m/s
dx
(4) The distance – time relationship of a moving body is given by y = F(t), then the acceleration of the body is the
(1) gradient of the velocity / time graph (2) gradient of the distance / time graph
(3) gradient of the acceleration / time graph (4) gradient of the velocity / distance graph
(5) The distance travelled by a car in “t” seconds is given by x = 3t3 – 2t2 + 4t – 1. Then the initial velocity
and initial acceleration respectively are
(1) (– 4m/s, 4m/s2)
(2) (4m/s, – 4m/s2) (3) (0,0)
(4) (18.25 m/s, 23 m/s2)
(6) The angular displacement of a fly wheel in radians is given by  = 9t2 – 2t3. The time when the angr
acceleration zero is
(1) 2.5 s
(2) 3.5s
(3) 1.5s
(4) 4.5s
(7) Food pockets were dropped from an helicopter during the flood and distance fallen in “t” seconds is given
by y =
1 2
gt ( g  9.8m / s 2 ) . Then the speed of the food pocket after it has fallen for “2” seconds is
2
(1) 19.6 m/sec
(2) 9.8 m/sec
(3) – 19.6 m/sec
(8) An object dropped from the sky follows the law of motion x =
(4) – 9.8 m/sec
1 2
gt (g = 9.8m/sec2). The acceleration of the
2
Object, when t = 2 is
(1) – 9.8 m/sec2
(2) 9.8 m/sec2
(3) 19.6 m/sec2
(4) – 19.6 m/sec2
(9) A missile fired from ground level rises x metres vertically upwards in “t” seconds and x = t (100 – 12.5t).
Then the maximum height reached by missile is
(1) 100 m
(2) 150m
(3) 250 m
(4) 200m
(10) A continuous graph x = f(x) is such that f’(x)   as x x1 at (x1, y1). Then y = f(x) has a
(1) vertical tangent y = x1
(2) horizontal tangent x = x1
(3) vertical tangent x = x1
(4) horizontal tangent y = y1
(11) The curve y = f(x) and y = g(x) cut orthogonally at the point of intersection. If
(1) slope of f(x) = slope of g(x)
(2) slope of f(x) + slope of g(x) = 0
(3) slope of f(x) /slope of g(x) = – 1
(4) [slope of f(x)] [ slope of g(x)] = – 1
(12) The law of the mean can also be put in the form
(1) f(a + h) = f(a) – hf’(a +  h) ; 0 <  < 1
(2) f(a + h) = f(a) + hf’(a +  h) ; 0 <  < 1
(4) f(a + h) = f(a) – hf’(a –  h) ; 0 <  < 1
(3) f(a + h) = f(a) + hf’(a –  h) ; 0 <  < 1
(13) l ‘Hopital’s rule cannot be applied to
x 1
as x  0 because f(x) = x + 1 and g(x) = x + 3 are
x3
(1) not continuous
(3) not in the indeterminate form as x 0
(14) If lim g ( x) = b and f is continuous at x = b, then
(2) not differentiable
(4) in the indeterminate form as x  0
lim g ( x)
lim f ( x)
lim f ( g ( x))
xa
(1) lim g ( x) = f
xa
(3)
(15)
lim
x 0
=g
xa
x
tan x
(2) lim f ( g ( x)) = f
x a
xa
(4) lim f ( g ( x))  f
x a
xa
lim g ( x)
lim g ( x)
x a
x a
is
(1) 1
(2) – 1
(3) 0
(4) 
(16) f is a real valued function defined on an interval I  R ( R being the set of real numbers ) increases on L.
Then
(1) f(x1)  f(x2) whenever x1< x2 x1, x2  I
(2) f(x1)  f(x2) whenever x1 < x2 x1, x2  I
(3) f(x1)  f(x2) whenever x1 > x2
x1, x2  I (4) f(x1) > f(x2) whenever x1 > x2, x1, x2  I
(17) If a real valued differentiable function y = f(x) defined on an open interval I is increasing, then
(1)
dy
0
dx
(2)
dy
0
dx
(3)
dy
0
dx
(4)
dy
0
dx
(18) f is a differentiable function defined on an interval I with positive derivative, then f is
(1) increasing on I
(2) decreasing on I
75
(3) strictly increasing on I
(4) strictly decreasing on I
(19) The function f(x) = x3 is
(1) increasing
(2) decreasing
(3) strictly decreasing
(4) strictly increasing
(20) If the gradient of a curve changes from positive just before P to negative just after, then “P” is a
(1) minimum point
(2) maximum point (3) inflexion point
(4) discontinuous point
(21) The function f(x) = x2 has
(1) a maximum value at x = 0
(2) minimum value at x = 0
(3) finite no. of maximum values
(4) infinite no. of maximum values.
22) The function f(x) = x3 has
(1) absolute maximum
(2) absolute minimum
(3) local maximum
(4) no extrema
(23) If f has a local extremum at a and if f’ (a) exists, then
a) f’(a) < 0
(2) f’(a) > 0
(3) f’(a) = 0
(4) f”(a) = 0
(24) In the following figure, the curve y = f(x) is
(1) concave upward
(2) convex upward
(3) changes from concavity to convexity
(4) changes from convexity and concavity
(25) The point that separates the convex part of a continuous curve from the concave part is
(1) the maximum point
(2) the maximum point (3) the inflextion point
(4) critical point
(26) f is a twice differentiable function on a interval I and if f”(x) > 0 for all x in the domain I of f, then f is
(1) concave upward
(2) convex upward
(3) increasing
(4) decreasing
(27) x = x0 is a root of even order for the equation f’(x) = 0, then x = x0 is a
(1) maximum point
(2) minimum point
(3) inflexion point (4) critical point
(28) If x0 is the x-coordinate of the point of inflection of a curve y = f(x), then (Second derivative exists)
(1) f(x0) = 0
(2) f’(x0) = 0
(3) f”(x0) = 0
(4) f”(x0)  0
(29) The statement “If f is continuous on a closed interval [a, b] then f attains an absolute maximum value f(c)
and an absolute minimum value f(d) at some number c and d in [a, b]” is
(1) The extreme value theorem
(2) Fermat’s theorem
(3) Law of Mean
(4) Rolle’s theorem
(30) The statement : “If f has a local extremum (minimum or maximum) at c and if f’(c ) exists, then f’(x) = 0” is
(1) the extreme value theorem
(2) Fermat’s theorem
(3) Law of Mean
(4) Rolle’s theorem
(31) Identify the false statement
(1) all the stationary numbers are critical numbers. (2) at the stationary point the first derivative zero
(3) at critical numbers the first derivative need not exist
(4) all the critical numbers are stationary numbers
(32) Identify the correct statement
(a) a continuous function has local maximum then it has absolute maximum
(b) a continuous function has local minimum then it has absolute minimum
(c) a continuous function has absolute maximum then it has local maximum
(d) a continuous function has absolute minimum then it has local minimum
(1) (a) and (b)
(2) (a) and (c)
(3) (c) and (d)
(4) (a), (c) and (d)
(33) Identify the correct statements
(a) Every constant function is an increasing function
(b) Every constant function is a decreasing function
(c) Every identify function is an increasing function
(d) Every identify function is a decreasing function
(1) (a), (b) and (c)
(2) (a) and (c)
(3) (c) and (d)
(4) (a), (c) and (d)
(34) Which of the following statement is incorrect?
(1) Initial velocity means velocity at t = 0
(2) Initial acceleration means acceleration at t = 0
(3) If the motion is upward, at the maximum height, the velocity is not zero
(4) If the motion is horizontal, v = 0 when the particle comes to rest
(35) Which of the following statements are correct (m1 and m2 are slopes of two lines)
(a) If the two lines are perpendicular, then m1m2 = – 1
76
(b) If m1 m2 = – 1, then the two lines are perpendicular
(c) If m1 = m2, then the two lines are parallel
(d) If m1 = –
1
, then the two lines are perpendicular
m2
(1) (b),(c) and (d)
(2) (a),(b) and (d)
(3) (c) and (b)
(4) (a) and (b)
36) One of the conditions of Rolle’s theorem is
(1) f is defined and continuous on (a, b)
(2) f is differentiable on [a, b]
(3) f(a) = f(b)
(4) f is differentiable on (a, b]
(37) If a and b are two roots of a polynomial f(x) = 0, then Rolle’s theorem says that there exists atleast
(1) one root between a and b for f’(x) = 0 (2) two roots between a and b for f’(x) = 0
(3) one root between a and b for f”(x) = 0
(4) two roots between a and b for f”(x) = 0
38) A real valued function which is continuous on [a,b] and differentiable on (a,b) then there exists
atleast one ‘c’ in
(1) [a, b] such that f’(c) = 0
(2) (a, b) such that f’(c) = 0
f ( b )  f (a )
f ( b )  f (a )
=0
(4) (a, b) such that
= f’(c)
ba
ba
(39) In the law of mean, the value of ‘  ’ satisfies the condition
(1)  > 0
(2)  < 0
(3)  < 1
(4) 0 <  < 1
(3) (a, b) such that
(40) Which of the following statements are correct?
(a) Rolle’s theorem is particular case of Lagranges law of mean
(b) Lagranges law of mean is a particular case of generalized law of mean (Cauchy)
(c) Lagranges law of mean is a particular case of Rolle’s theorem
(d) Generalised law of mean is a particular case of Lagranges law of mean (Cauchy)
(1) (b), (c)
(2) (c), (d)
(3) (a), (b)
(4) (a), (d)
CHAPTER – VI (DIFFERENTIAL CALCULUS APPLICATIONS II)
BOOK BACK QUESTIONS
45. If u = xy then
is equal to
1) yxy–1
2) u log x
46. If
1) 0
47. If
1) u
3) u log y
4) xyx–1
and f = sin u then f is a homogenous function of degree
then
+
2) 1
is equal to
3) 2
4) 4
2) u
3) u
4) –u
48. The curve y2 (x – 2) = x2 (1 + x) has
1) an asymptote parallel to x-axis 2) an asymptote parallel to y-axis
3) asymptotes parallel to both axes
4) no asymptote
is equal to
49. If x = rcosθ ;y = rsin θ, then
1) sec θ
2) sin θ
3) cos θ
4) cosec θ
50. Identify the true statements in the following
(i) If a curve is symmetrical about the origin, then it is symmetrical about both axes
(ii) If a curve is symmetrical about both the axes, then it is symmetrical about the origin
(iii) A curve f(x, y) = 0 is symmetrical about the line y = x if f(x, y) = f(y, x)
(iv) For the curve f(x, y) = 0, if f(x, y) = f(–y, –x), then it is symmetrical about the origin
1) (ii), (iii)
2) (i), (iv)
3) (i), (iii)
4) (ii), (iv)
then
+
is
51. If
1) 0
2) u
3) 2u
4) u–1
52. The percentage error in the 11th root of the number 28 is approximately ______ times the percentage error in 28
1)
2)
3) 11
4) 28
53. The curve a2y2 = x2 (a2 – x2) has
1) only one loop between x = 0 and x = a
3) two loops between x = – a and x = a
2) two loops between x = 0 and x = a
4) no loop
77
54. An asymptote to the curve y2 (a + 2x) = x2 (3a – x) is
3) x = a/2
1) x = 3a
2) x = – a/2
2
2
55. In which region the curve y (a + x) = x (3a – x) does not lie
1) x> 0
2) 0 <x < 3a
3) x ≤ – a and x> 3a
56. If u = y sin x then
1) cosx
57. If
then
4) – a<x < 3a
is equal to
+
2) cosy
is equal to
3) sin x
4) 0
1) 0
2) 1
3) 2u
2
2
2
58. The curve 9y = x (4 – x ) is symmetrical about
1) y-axis
2) x-axis
3) y = x
59. The curve ay2 = x2 (3a – x) cuts the y-axis at
2) x = 0,x = 3a
3) x = 0,x = a
1) x = – 3a,x = 0
CREATED QUESTIONS
(1) For the function y =
(1) 1
4) x = 0
x3 +
2x2,
the value of dy when x = 2 and dx = 0.1 is
(2) 2
(3) 3
u
(2) If u = x4 + y3 + 3x2y2 + 3x2y, then
is
x
4) u
4) both the axes
4) x = 0
(4) 4
(1) 4x3 + 6xy2 + 6xy
(2) 3x2 + 6xy2 + 3xy2
(3) 4x3 – 6x2y + 6xy2
(4) 4x3 + 6x2y2 + 3xy
(3) If u = f(x, y), then with usual notations, uxy = uyx if
(1) u is continuous
(2) ux is continuous
(3) uy is continuous (4) u, ux, uy are continuous
(4) If u = f(x, y) is a differentiable function of x and y; x and y are differentiable functions of t, then
du f dx f y
=
.  .
 x dt y  t
dt
u f x f y
(4)
= . 
.
t x t y  t
f
f
(5) If f(x, y) is a homogenous functions of degree n, then x
=
y
x
y
du f x f y
=
.  .
x t y  t
dt
du f dx f dy
(3)
=
.  .
x dt y d t
dt
(1)
(1) f
(2)
(2) nf
(3) n (n – 1) f
(4) n (n + 1) f
(3) 12x2y – 6x
(4) 12xy2 – 6x
(3) 12x2y – 6x
(4) 12x2 – 6x
(3) 12x2y – 6x
(4) 12x2 + 6y2 + 6y
(3) 12x2y – 6x
(4) 3y2 + 6x2y + 3x2
(3) x-3/4dx
(4) 0
(3) 5x5dx
(4) 5x5
 u
is
x y
2
(6) If u (x, y) = x4 + y3 + 3x2y2 + 3x2y, then
(1) 12xy + 6x
(2) 12xy – 6x
 u
=
y x
2
(7) If u (x, y) = x4 + y3+ 3x2y2 + 3x2y, then
(1) 12xy + 6x
(2) 12xy – 6x
 u
=
x 2
2
(8) If u(x, y) = x4 + y3 + 3x2y2 + 3x2y, then
(1) 3y2 + 6x2y + 3x2
(2) 6y + 6x2
(9) If u (x, y) = x4 + y3 + 3x2y2 + 3x2y, then
 2u
=
y 2
(1) 6y + 6x2
(2) 12xy – 6x
(10) The differential on y of the function y =
(1) 1 x 3 / 4
(2) 1 x 3 / 4 dx
4
4
(11) The differential of y, if y = x5 is,
(1) 5x4
(2) 5x4dx
(12) The differential of y, if y =
4
x is
x 4  x 2  1 is
78
1

1
(4 x3  2 x) 2 dx
2
1

1
(3) (4 x 3  2 x) 2
2
(1)
(13) The differential of y, if y =
(1)
7
dx
(2 x  3) 2
1

1 4
( x  x 2  1) 2 (4 x3  2 x)dx
2
1

1
(4) ( x 4  x 2  1) 2 (4 x3  2 x)
2
(2)
x2
is
2x  3
1
(2)
dx
(2 x  3) 2
(3)
7
dx
(2 x  3) 2
(4)
7
(2 x  3) 2
(14) The differential of y, if y = sin2x is
(1) 2cos2x
(2) 2cos2x.dx
(3) – 2cos2x.dx (4) cos2x.dx
(15) The differential of xtan x is
(1) (xsec2x + tan2x)
(2) (xsec2x – tanx)dx (3) xsec2xdx
(4) (xsec2x + tanx)dx
(16) If u (x, y) = x4 + y3 + 3x2y2 + 3x2y, then
u
is
y
(1) 3y2 + 6xy + 3x2
(2) 3y2 + 6xy2 + 3x2
(3) 3y2 + 6x2y + 3x2 (4) 3y2 + 6x2y2 + 3x2
(17) The curve y2 = x2(1 – x2) is defined only for
(1) x  2 and x  – 2
(2) x  1 and x  – 1
(3) x  – 1 and x  1
(4) x < 1 and x > – 1
(18) The curve y2 = x2 (1 – x2) is symmetrical about
(1) x-axis only (2) y-axis only (3) x and y axes only
(4) x, y axes and the origin
(19) The curve y2 = x2 (1 – x2) has
(1) only the loop between x = 0 and x = 1
(2) two loops between x = – 1 and x = 0
(3) two loops between x = – 1 and 0; 0 and 1
(4) no loop
(20) The curve y2 = x2(1 – x2) has
(1) an asymptote x = – 1
(2) an asymptote x = 1
(3) two asymptotes x = 1 and x = – 1
(4) no asymptote
2
2
(21) The curve y (2 + x) = x (6 – x) exists for
(1) – 2 < x  6
(2) – 2  x  6
(3) – 2 < x < 6
(4) – 2  x < 6
(22) The x-intercept of the curve y2 (2 + x) = x2(6 – x) is
(1) 0
(2) 0, 6
(3) 2
(4) – 2
2
2
(23) The asymptote to the curve y (2 + x) = x (6 – x) is
(1) x = 2
(2) x = – 2
(3) x = 6
(4) x = – 6
(24) The curve y2(2 + x) = x2 (6 – x) has
(1) only one loop between x = 0 and x = 6
(2) two loops between x = 0 and x = 6
(3) only one loop between x = – 2 and x = 6
(4) two loops between x = – 2 and x = 6
(25) The curve y2 = x2(1 – x) is defined only for
(1) x  1
(2) x  1
(3) x < 1
(4) x > 1
(26) The curve y2 = x2(1 – x) is symmetrical about
(1) y-axis only (2) x-axis only
(3) both the axes
(4) origin only
2
2
(27) The curve y = x (1 – x) has
(1) an asymptote y = 0
(2) an asymptote x = 1
(3) an asymptote y = 1
(4) no asymptote
(28) The curve y2 = x2(1 – x) has
(1) only one loop between x = –1 and x = 0
(2) only one loop between x = 0 and x = 1
(3) two loops between x = – 1 and x = 1
(4) no loop
(29) The curve y2 = (x – a)(x – b)2, a, b > 0 and a > b does not exist for
(1) x  a
(2) x = b
(3) b < x < a
(4) x = a
(30) The curve y2 = (x – a)(x – b)2 is symmetrical about
(1) origin only (2) y-axis only
(3) x-axis only
(4) both x and y axes
2
2
(31) The curve y = (x – a)(x – b) has a, b > 0 and a > b
(1) an asymptote x = a
(2) an asymptote x = b (3) an asymptote y = a
(4) no asymptote
(32) The curve y2 = (x – a)(x – b)2, a, b > 0 and a > b has
(1) a loop between x = a and x = b
(2) two lops between x = a and x = b
(3) two loops between x = 0 and x = a (4) no loop
(33) The curve y2(1 + x) = x2(1 – x) is defined for
(1) – 1  x  1
(2) – 1 < x  1
(3) – 1  x < 1
(4) – 1 < x < 1
79
(34) The curve y2 (1 + x) = x2(1 – x) is symmetrical about
(1) both the axes
(2) origin only (3) y-axis only
(4) x-axis only
2
2
(35) The asymptote to the curve y (1 + x) = x (1 – x) is
(4) x = – 1
(1) x = 1
(2) y = 1
(3) y = – 1
(36) The curve y2 (1 + x) = x2 (1 – x) has
(1) a loop between x = – 1 and x = 1
(2) a loop between x = – 1 and x = 0
(3) a loop between x = 0 and x = 1
(4) no loop
(37) The curve a2 y2 = x2 (a2 – x2) is defined for
(2) x < a and x > – a
(1) x  a and x  – a
(3) x  – a and x  a
(4) x  a and x > – a
2
2
2
2
2
(38) The curve a y = x (a – x ) is symmetrical about
(1) x-axis only (2) y-axis only (3) both the axes
(4) both the axes and origin
(39) The curve a2y2 = x2(a2 – x2) has
(1) an asymptote x = a (2) an asymptote x = – a
(3) an asymptote x = 0
(4) no asymptote
2
2
2
2
2
(40) The curve a y = x (a – x ) has
(1) a loop between x = a and x = – a
(2) two loops between x = – a and x = 0; x = 0 and x = a
(3) two loops between x = 0 and x = a (4) no loop
(41) The curve y2 = (x – 1)(x – 2)2 is not defined for
(1) x  1
(2) x  2
(3) x < 2
(4) x < 1
2
2
(42) The curve y = (x – 1)(x – 2) is symmetrical about
(1) both x and y axes
(2) x-axis only
(3) y-axis only
(4) both the axes and origin
(43) The curve y2 = (x – 1)(x – 2)2 has
(1) an asymptote x =1
(2) an asymptote x = 2
(3) two asymptotes x = 1 and x = 2
(4) no asymptote
(44) The curve y2 = (x – 1)(x – 2)2 has
(1) two loops between x = 0 and x = 2
(2) one loop between x = 0 and x = 1
(3) one loop between x = 1 and x = 2
(4) no loop
CHAPTER – VII (INTEGRAL CALCULUS)
BOOK BACK QUESTIONS
60. The value of
is
1)
2)
1)
2) 0
62. The value of
2)
65. The value of
1)
4) π
3)
4)
3) log 2
4) log 4
is
1) 0
1) 3π/16
3)
is
63. The value of
64. The value of
4) π
is
61. The value of
1)
3) 0
2) 2
is
2) 3/16
3) 0
4) 3π/8
is
2)
3) 0
80
4)
66. The value of
is
1) π
2) π/2
3) π/4
4) 0
67. The area bounded by the line y = x, the x-axis, the ordinates x =1, x = 2 is
1)
2)
3)
4)
68. The area of the region bounded by the graph of y = sin x and y = cosx between x = 0 and x = is
1)
+1
2)
–1
3) 2
69. The area between the ellipse
–2
4) 2
+2
and its auxillary circle is
1) πb(a – b)
2) 2πa(a – b)
3) πa(a – b)
70. The area bounded by the parabola y2 = x and its latus rectum is
1)
2)
3)
71. The volume of the solid obtained by revolving
1) 48π
2) 64π
72. The volume, when the curve y =
1) 100π
2)
4) 2πb(a – b)
4)
about the minor axis is
3) 32π
4) 128π
from x = 0 to x = 4 is rotated about x-axis
3)
4)
73. The volume generated when the region bounded by y = x, y = 1, x = 0 is rotated about y-axis
1)
2)
3)
4)
74. Volume of solid obtained by revolving the area of the ellipse
about major and minor axes are in
the ratio
1) b2:a2
2) a2:b2
3) a:b
4) b : a
The volume generated by rotating the triangle with vertices at (0, 0), (3, 0) and (3, 3) about x-axis is
1) 18π
2) 2π
3) 36π
4) 9π
The length of the arc of the curve x2/3 + y 2/3 = 4 is
1) 48
2) 24
3) 12
4) 96
The surface area of the solid of revolution of the region bounded by y = 2x, x = 0 and x = 2 about x-axis is
1) 8 π
2) 2 π
3) π
4) 4 π
The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance 2 and 4 from
the centre is
1) 20π
2) 40π
3) 10π
4) 30π
75.
76.
77.
78.
CREATED QUESTIONS

(1) If In = sin xdx , then In =
n
1 n 1
n 1
In - 2
sin x cosx +
n
n
1
n 1
(3)  sin n 1 x cosx –
In - 2
n
n
1 n 1
n 1
In - 2
sin x cosx +
n
n
1
n 1
(4)  sin n 1 x cosx +
In
n
n
(1) 
2a

(2)
(2)
a
f ( x) dx  2 f ( x) dx, if
0
0
(1) f(2a – x) = f(x)
(2) f(a – x) = f(x)
(3) f(x) = – f(x)
(4) f(– x) = f(x)
(2) f(2a – x) = – f(x)
(3) f(x) = – f(x)
(4) f(– x) = f(x)
2a
3)
 f (x) dx = 0 if
0
(1) f(2a – x) = f(x)
a
4) If f(x) is an odd function, then
 f (x) dx is
a
81
a
a
(2) f ( x ) dx

(4) f (a  x ) dx
(3) 0
0
0
a
a


(1) 2 f ( x ) dx
0
a


(5) f ( x ) dx + f (2a  x ) dx =
0
0
a
a


0
2a
2a
(2) 2 f ( x ) dx
(1) f ( x ) dx
(3)
0
 f (x) dx
(4)
 f (a  x) dx
0
0
a
 f (x) dx is
(6) If f(x) is even, then
a
a
a

(1) 0
(2) 2 f ( x ) dx
a

(4)  2 f ( x ) dx
0
0
(3) f ( x ) dx
0

a
(7)
 f ( x) dx is
0
a
(1)

a
f ( x  a)dx
(2)
0

a
f (a  x)dx
(3)
0

a
f (2a  x) dx
(4)
 f ( x  2a)dx
0
0
b
(8)
 f ( x) dx is
a
a
b

(1) 2 f ( x) dx
(2)

b
f (a  x)dx
a
0
(9) If n is a positive integer, then
n!
an
x
n
(4)
 f (a  b  x)dx
a
e ax dx
0
(2)

b
f (b  x)dx
a

(1)
(3)
(n  1)!
an
=
(3)
(n  1)!
a n 1
(4)
n!
a n 1
/ 2
(10) If n is odd, then
 cos
n
x dx is
0
n 1 n  3 n  5 1 
.
.
...
n n2 n4 2 2
n 1 n  3 n  5 2
(4)
.
.
... .1
n n2 n4 3
n n2 n4 
(1)
.
.
...
n 1 n  3 n  5 2
n n2 n4 3
(3)
.
.
... .1
n 1 n  3 n  5 2
(2)
/ 2
(11) If n is even, then
 sin
n
xdx is
0
n 1 n  3 n  5 1 
.
.
...
n n2 n4 2 2
n 1 n  3 n  5 2
.
.
... .1
(4)
n n2 n4 3
n n2 n4 
.
.
...
n 1 n  3 n  5 2
n n2 n4 3
.
.
... .1
(3)
n 1 n  3 n  5 2
(1)
(2)
/ 2
(12) If n is even, then
 cos
n
xdx is
0
n 1 n  3 n  5 1 
.
.
...
n n2 n4 2 2
n 1 n  3 n  5 2
(4)
.
.
... .1
n n2 n4 3
n n2 n4 
.
.
...
(1)
n 1 n  3 n  5 2
n n2 n4 3
(3)
.
.
... .1
n 1 n  3 n  5 2
(2)
82
/ 2
(13) If n is odd, then
 sin
n
xdx is
0
n 1 n  3 n  5 1 
.
.
...
n n2 n4 2 2
n 1 n  3 n  5 2
(4)
.
.
... .1
n n2 n4 3
n n2 n4 
.
.
...
n 1 n  3 n  5 2
n n2 n4 3
(3)
.
.
... .1
n 1 n  3 n  5 2
(1)
(2)
b

(14) f ( x )dx 
a
b
a

a

(1)  f ( x )dx
a
b

(2)  f ( x )dx

(3)  f ( x )dx
b
(4) 2 f ( x)dx
0
0
(15) The area bounded by the curve x = g(y) to the right of y-axis and the two lines y= c and y = d is given by
d
a
 x dx
(1)
(2)
c
d
d
 x dy
(3)
c
 y dy
(4)
c
 x dy
c
(16) The area bounded by the curve x = f(y), y-axis and the lines y = c and y = d is rotated about y-axis.
Then the volume of the solid is
d

d

d

d

(1)  x dy
(2)  x dx
(3)  y dx
(4)  y 2 dy
c
c
c
c
2
2
2
(17) The area bounded by the curve x = f(y) to the left of y-axis between the lines y = c and y = d is
d
d
 x dy
(1)
d

(2)  x dy
c
(3)
c
d
 y dx

(4)  y dx
c
c
(18) The arc length of the curve y = f(x) from x = a to x = b is
b
(1)

a
2
 dy 
1    dx
 dx 
d
(2)

c
2
 dx 
1    dx
 dy 
2
 dy 
(3) 2 y 1  
 dx
 dx 
a
b
2
 dx 
 dx
(4) 2 y 1  
 dy 
a
b
(19) The surface area obtained by revolving the area bounded by the curve y = f(x), the two ordinates x = a,
x = b and x-axis, about x-axis is
2
 dy 
1    dx
 dx 
b
(1)

a

(20)
x
d
(2)

c
2
 dx 
1    dx
 dy 
2
 dy 
(3) 2 y 1  
 dx
dx


a
b
e  4 x dx is
5
0
(1)

(21)
e
6!
46
 mx
(2)
6!
45
(3)
5!
46
(4)
5!
45
(2)
7!
m7
(3)
m!
7 m 1
(4)
7!
m8
(2)
6!
26
(3) 2 6 6!
x 7 dx is
0
(1)

(22)

m!
7m
x 6 e  x / 2 dx is
0
(1)
6!
27
83
2
 dx 
 dx
(4) 2 y 1  
dy


a
b
7
(4) 2 6!
(23) If In =
 cos
n
x dx , then In =
1
n 1
In-2
cos n 1 x sin x 
n
n
1
n 1
(3) cos n 1 x sin x 
In-2
n
n
n 1
In-2
n
1
n 1
(4) cos n 1 x sin x 
In-2
n
n
(2) cos n 1 x sin x 
(1) 
CHAPTER – VIII (DIFFERENTIAL EQUATIONS)
BOOK BACK QUESTIONS
79. The integrating factor of
+ 2 = e4x is
1) logx
2) x2
3) ex
80. If cosx is an integrating factor of the differential equation
4) x
+ Py = Q, then P =
1) – cot x
2) cot x
3) tan x
81. The integrating factor of dx + xdy = e– y sec2ydy is
1) ex
2) e– x
3) e y
is
82. The integrating factor of
1) ex
2) log x
83. Solution of
4) – tan x
4) e– y
3)
4) e– x
+ mx = 0, where m< 0 is
1) x = cemy
2) x = ce– my
3) x = my + c
2
84. y = cx – c is the general solution of the differential equation
1) (y′)2 – xy′ + y = 0
2) y′′ = 0
3) y′ = c
85. The differential equation
4) x = c
4) (y′)2 + xy′ + y = 0
+ 5y1/3 = x is
1) of order 2 and degree 1
2) of order 1 and degree 2
3) of order 1 and degree 6
4) of order 1 and degree 3
86. The differential equation of all non-vertical lines in a plane is
1)
=0
2)
=0
3)
=m
4)
=m
87. The differential equation of all circles with the centre at the origin is
1) xdy + ydx = 0
2) xdy – ydx = 0
3) xdx + ydy = 0
88. The integrating factor of the differential equation + py = Q
1) ∫pdx
2) ∫Q dx
3) e∫Qdx
2
2x
89. The complementary function of (D + 1)y = e is
1) (Ax + B) ex
2) Acosx + B sin x
3) (Ax + B) e2x
90. A particular integral of (D2 – 4D +4)y = e2x is
1)
e2x
2) xe2x
3) xe–2x
4) xdx – ydy = 0
4) e∫pdx
4) (Ax + B) e–x
4) e–2x
91. The differential equation of the family of lines y = mx is
1)
=m
2) ydx–xdy = 0
3)
=0
4) ydx+xdy = 0
92. The degree of the differential equation
1) 1
2) 2
3) 3
93. The degree of the differential equation
4) 6
where c is a constant is
1) 1
2) 3
3) – 2
4) 2
94. The amount present in a radio active element disintegrates at a rate proportional to its amount. The differential
equation corresponding to the above statement is (k is negative)
1)
2) = kt
3) = kp
4) = – kt
84
95. The differential equation satisfied by all the straight lines in xy plane is
1)
= a constant
keλxthen
96. If y =
1) = λy
2)
=0
3) y +
its differential equation is
2) = ky
3)
=0
4)
+ ky = 0
+y=0
4)
= eλx
97. The differential equation obtained by eliminating a and b from y = ae3x + be–3x is
1)
+ ay = 0
2)
– 9y = 0
3)
–9
=0
4)
+ 9x = 0
98. The differential equation formed by eliminating A and B from the relation y = ex (Acosx + B sin x) is
1) y2 + y1 = 0
2) y2 – y1 = 0
3) y2 – 2y1 + 2y = 0
4) y2 – 2y1 – 2y = 0
99. If
then
1) 2xy + y2 + x2 = c
2) x2 + y2 – x + y = c
100. If f ′(x) = and f (1) = 2 then f (x) is
2) (x +2)
1) – (x +2)
3) x2 + y2 – 2xy = c
4) x2 – y2 – 2xy = c
3) (x
4) x(
+2)
+2)
101. On putting y =vx, the homogenous differential equation x2dy + y(x + y)dx = 0 becomes
1) xdv + (2v + v2)dx = 0 2) vdx + (2x + x2)dv = 0 3) v2dx – (x + x2)dv = 0 4) vdv + (2x + x2)dx = 0
102. The integrating factor of the differential equation - y tan x = cosx is
1) secx
2) cosx
3) etanx
4) cot x
2
2x
103. The P.I. of (3D + D –14)y = 13e is
1) 26xe2x
2) 13xe2x
3) xe2x
4) x2 / 2 e2x
104. The particular integral of the differential equation f(D)y = eax where f(D) = (D – a) g(D), g(a) ≠ 0 is
1) meax
2)
3) g(a) eax
4)
CREATED QUESTIONS
3
d 3 y  d 2 y   dy 
    y7
(1) The order and degree of the differential equation are

dx 3  dx 2   dx 
(1) 3, 1
(2) 1, 3
(3) 3, 5
(2) The order and degree of the differential equation y = 4
(1) 2, 1
(4) 2, 3
dy
dx
are
 3x
dx
dy
(2) 1, 2
(3) The order and degree of the differential equation are
5
(3) 1, 1
(4) 2, 2
2
d2y 
 dy  


4
  

dx 2 
 dx  
3
4
(1) 2, 1
(2) 1, 2
(3) 2, 4
(4) The order and degree of the differential equation are (1 + y’)2 = y’2
(1) 2, 1
(2) 1, 2
(3) 2, 2
(4) 4, 2
(4) 1, 1
dy
(5) The order and degree of the differential equation are
+ y = x2
dx
(1) 1, 1
(2) 1, 2
(3) 2, 1
(6) The order and degree of the differential equation are y’ + y2 = x
(1) 2, 1
(2) 1, 1
(3) 1, 0
(7) The order and degree of the differential equation are y” + 3y’2 + y3 = 0
(1) 2, 2
(2) 2, 1
(3) 1, 2
2
(8) The order and degree of the differential equation are
(1) 2, 1
d y
+x=
dx 2
(2) 1, 2
y
(4) 0, 1
(4) 0, 1
(4) 3, 1
dy
dx
(3) 2, 1 /2
d2y
(9) The order and degree of the differential equation are
–y+
dx 2
85
(4) 2, 2
3
2
 dy d 3 y 

 3  = 0
 dx dx 
(1) 2, 3
(2) 3, 3
(3) 3, 2
(10) The order and degree of the differential equation are y” = (y – y’3)2/ 3
(1) 2, 3
(2) 3, 3
(3) 3, 2
(11) The order an degree of the differential equation are y’ + (y”)2 = (x + y’)2
(1) 1, 1
(2) 1, 2
(3) 2, 1
(12) The order and degree of the differential equation are y’+ (y”)2 = x (x + y”)2
(1) 2, 2
(2) 2, 1
(3) 1, 2
(4) 2, 2
(4) 2, 2
(4) 2, 2
(4) 1, 1
2
dx
 dy 
 x2
 +x=
dy
dx
 
(13) The order and degree of the differential equation are 
(1) 2, 2
(2) 2, 1
(3) 1, 2
(4) 1, 3
(14) The order and degree of the differential are sinx (dx + dy) = cosx (dx – dy)
(1) 1, 1
(2) 0, 0
(3) 1, 2
(4) 2, 1
(15) The differential equation corresponding to xy = c2, where c is an arbitary constant, is
(1) xy” + x = 0
(2) y” = 0
(3) xy’ + y = 0
(4) xy” – x = 0
(16) In finding the differential equation corresponding to y = emx, where m is the arbitrary constant, then m is
(1)
y
y'
(2)
y'
y
(17) The solution of a linear differential equation

(3) y (I.F.) = 
(1) y (I.F.) =
(3) y’
dx
+ Px = Q, where P and Q are functions of y, is
dy
(I.F.) Q dx + c
(2) x (I.F.) =
(I.F.) Q dy + c
(4) x (I.F.) =
(18) The solution of the linear differential equation
(1) y (I.F.) =
(3) y (I.F.) =


(I.F.) Q dx + c
(I.F.) Q dy + c
(4) y


(I.F.) Q dy + c
(I.F.) Q dx + c
dy
+ Py = Q where p and Q are functions of x is
dx

(4) x (I.F.) = 
(2) x (I.F.) =
(I.F.) Q dy + c
(I.F.) Q dx + c
(19) Identify the incorrect statement
(1) The order of a differential equation is the order of the highest order derivative occurring in it.
(2) The degree of the differential equation is the degree of the highest order derivative which occurs in
it (the derivatives are free from radicals and fractions)
dy f 1 ( x , y)
=
is the first order first degree homogenous differential equation
dx f 2 ( x , y)
dy
(4)
+ xy = ex is a linear differential equation in x.
dx
(3)
CHAPTER – IX (DISCRETE MATHEMATICS)
BOOK BACK QUESTIONS
105. Which of the following are statements?
(i) May God bless you
(ii) Rose is a flower (iii) Milk is white
(iv) 1 is a prime number
1) (i), (ii), (iii)
2) (i), (ii), (iv)
3) (i), (iii), (iv)
4) (ii), (iii), (iv)
106. If a compound statement is made up of three simple statements, then the number of rows in the truth table is
1) 8
2) 6
3) 4
4) 2
107. If p is T and q is F, then which of the following have the truth value T ?
(i) p q
(ii) ~ p q
(iii) p ~ q
(iv) p ~ q
1) (i), (ii), (iii)
2) (i), (ii), (iv)
3) (i), (iii), (iv)
4) (ii), (iii), (iv)
108. The number of rows in the truth table of ~ [ p (~ q)] is
1) 2
2) 4
3) 6
4) 8
109. The conditional statement p→q is equivalent to
1) p q
2) p ~ q
3) ~ p q
4) p q
110. Which of the following is tautology?
1) p q
2) p q
3) p ~ p
4) p ~ p
86
111. Which of the following is contradiction?
1) p q
2) p q
3) p ~ p
4) p ~ p
112. p↔ q is equivalent to
4) (p→q) (q→p)
1) p→ q
2) q→p
3) (p→q) (q→p)
113. Which of the following is not a binary operation on R?
1) a * b = ab
2) a * b = a – b
3) a * b =
4) a * b =
114. A monoid becomes a group if it also satisfies the
1) closure axiom
2) associative axiom 3) identity axiom
4) inverse axiom
115. Which of the following is not a group?
1) (Zn , +n)
2) (Z, +)
3) (Z, .)
4) (R, +)
116. In the set of integers with operation * defined by a * b = a + b – ab, the value of 3*(4*5) is is
1) 25
2) 15
3) 10
4) 5
117. The order of [7] in (Z9 , +9) is
1) 9
2) 6
3) 3
4)1
118. In the mulplicative group of cube root of unity, the order of w2 is
1) 4
2) 3
3) 2
4)1
119. The value of [3] +11 ([5] +11 [6]) is
1) [0]
2) [1]
3) [2]
4) [3]
. Then the value of (3 * 4) * 5 is
120. In the set of real numbers R, an operation * is defined by a * b =
3) 25
4) 50
1) 5
2)
121. Which of the following is correct
1) An element of a group can have more than one inverse.
2) If every element of a group is its own inverse, then the group is abelian.
3) The set of all 2 × 2 real matrices forms a group under matrix multiplication
4) (a * b) –1 = a–1 * b–1 for all a, b∈G
122. The order of – i in the mulplicative group of 4th roots of unity is
1) 4
2) 3
3) 2
4)1
123. In the mulplicative group of nth roots of unity, the inverse of ωk is (k<n)
1) ω1/k
2) ω –1
3) ωn–k
4) ωn/k
124. In the set of integers under the operation * defined by a * b = a + b – 1, the identity element is
1) 0
2) 1
3) a
4) b
CREATED QUESTIONS
(1) Which of the following are statements?
(i) Chennai is the capital of Tamilnadu.
(ii) The earth is a planet.
(iii) Rose is a flower.
(iv) Every triangle is an isosceles triangle.
(1) all
(2) (i) and (ii)
(3) (ii) and (iii)
(4) (iv) only
(2) Which of the following are not statements?
(i) Three plus four is eight
(ii) The sun is a planet.
(iii) Switch on the light.
(iv) Where are you going?
(1) (i), (ii)
(2) (ii), (iii)
(3) (iii) and (iv)
(4) (iv) only
(3) The truth values of the following statements are
(i) Ooty is in Tamilnadu and 3 + 4 = 8 (ii) Ooty is in Tamilnadu and 3 + 4 = 7
(iii) Ooty is in Kerala and 3 + 4 = 7
(iv) Ooty is in Kerala and 3 + 4 = 8
(1) FTFF
(2) FFFT
(3) TTFF
(4) TFTF
(4) The truth values of the following statements are
(ii) Chennai is in India or 2 is an irrational number.
(i) Chennai is in India or 2 is an integer.
(iii) Chennai is in China or 2 is an integer. (iv) Chennai is in China or 2 is an irrational number.
(1) TFTF
(2) TFFT
(3) FTFT
(4) TTFT
(5) Which of the following are not statements?
(i) All natural numbers are integers. (ii) A square has five sides.
(iii) The sky is blue
(iv) How are you?
(1) (iv) only
(2) (i) and (iv)
(3) (i), (ii), (iii)
(4) (iii) and (iv)
87
(6) Which of the following are statements?
(i) 7 + 2 < 10
(ii) The set of rational numbers is finite.
(iii) How beautiful you are?
(iv) Wish you all success.
(1) (iii) (iv)
(2) (i), (ii)
(3) (i), (iii)
(4) (ii), (iv)
(7) The truth values of the following statements are
(i) All the sides of a rhombus are equal in length.
(ii) 1+ 19 is an irrational number.
(iii) Milk is white.
(iv) The number 30 has four prime factors.
(1) TTTF
(2) TTTT
(3) TFTF
(4) FTTT
(8) The truth values of the following statements are
(i) Paris is in France.
(ii) Sinx is an even function.
(iii) Every square matrix is non-singular.
(iv) Jupiter is a planet.
(1) TFFT
(2) FTFT
(3) FTTF
(4) FFTT
(9) Let p be “Kamala is going to school” and q be “There are twenty students in the class”. “Kamala is not going
to school or there are twenty students in the class” stands for
(2) p  q
(3) ~ p
(4) ~ p  q
(1) p  q
(10) If p stands for the statement “Sita likes reading” and q for the statement “Sita likes playing’.
“Sita likes neither reading nor playing” stands for
(1) ~p  ~q
(2) P  ~q
(3) ~p  q
(4) p  q
(11) If p is true and q is unknown, then
(3) p  (~p) is true
(4) p  q is true
(1) ~P is true
(2) p  (~p) is false
(12) If p is true and q is false, then which of the following is not true?
(1) p  q is false
(2) p  q is true
(3) p  q is false
(4) p  q is true
(13) Which of the following is not true?
(1) Negation of a negation of a statement is the statement itself
(2) If the last column of its truth table contains only T then it is tautology
(3) If the last column of its truth table contains only F then it is contradiction
(4) If p and q are any two statements then p  q is a tautology.
(14) Which of the following are binary operation on R?
(a) a* b = min {a,b}
(b) a*b = max {a,b}
(c) a*b = a
(d) a*b = b
(1) all
(2) (a), (b) and (c)
(3) (b), (c) and (d)
(4) (c ), (d)
(15) ‘+’ is not a binary operation on
(1) N
(2) Z
(3) C
(4) Q – {0}
(16) ‘–‘ is a binary operation on
(1) N
(2) Q – {0}
(3) R – {0}
(4) Z
(17) ‘  ’ is a binary operation on
(1) N
(2) R
(3) Z
(4) C – {0}
(18) In congruence modulo 5, {x  z / x = 5k + 2, k  Z} represents
(1) [0]
(2) [5]
(3) [7]
(4) [2]
(19) [5] .12 [11] is
(1) [55]
(2) [12]
(3) [7]
(4) [11]
(20) [3] +8 [7] is
(1) [10]
(2) [8]
(3) [7]
(4) [2]
(21) In the group (G, .); G = {1, – 1, i, – i}, order of – 1 is
(1) – 1
(2) 1
(3) 2
(4) 0
(22) In the group (G, .); G = {1, – 1, i, – i}, order of – i is
(1) 2
(2) 0
(3) 4
(4) 3
(23) In the group (G, .); G = {1,  ,  2},  is cube root of unity, O(  2) is
(1) 2
(2) 1
(3) 4
(4) 3
(24) In the group (Z4, +4), order of [0] is
(1) 1
(2) 
(3) can’t be determined
(4) 0
(25) In the group (Z4 , +4) ; O([3]) is
(1) 4
(2) 3
(3) 2
(4) 1
(26) In (S, o), xoy = x, for all x, y  S, then ‘o’ is
(1) only associative
(2) only commutative
(3) associative and commutative
(4) neither associate nor commutative.
88
(27) In (N, *), x* y = max {x, y}, x, y  N, then (N, *) is
(1) only closed
(2) only semi group
(3) only monoid
(4) a group
(28) The set of positive even integers, with usual multiplication forms.
(1) a finite group
(2) only a semi group
(3) only a monoid
(4) an infinite group
(29) The set of positive even integers, with usual addition forms
(1) a finite group
(2) only a semi group
(3) only a monoid
(4) an infinite group
(30) In the group (Z5 – {[0]}, .5}, O([3]) is
(1) 5
(2) 3
(3) 4
(4) 2
(31) In the group (G, .); G = {1, – 1, i, – i } , then the order of 1 is
(1) 2
(2) 0
(3) 4
(4) 1
(32) In the group (G, .); G = {1, – 1, i, – i } , then the order of i is
(1) 2
(2) 0
(3) 4
(4) 3
(33) In the group (G, .); G = {1,  ,  2},  is cube root of unity then O(  ) is
(1) 2
(2) 1
(3) 4
(4) 3
2
(34) In the group (G, .); G = {1,  ,  },  is cube root of unity then O(1) is
(1) 2
(2) 1
(3) 4
(4) 3
(35) In the group (Z4, +4), order of [1] is
(1) 1
(2) 
(3) can’t be determined
(4) 4
(36) In the group (Z4, +4), order of [2] is
(1) 1
(2) 2
(3) can’t be determined
(4) 0
(37) In the group (Z5 , – {[0], .5), O([2]) is
(1) 5
(2) 3
(3) 4
(4) 2
(38) In the group (Z5 , – {[0], .5), O([4]) is
(1) 5
(2) 3
(3) 4
(4) 2
(39) In the group (Z5, – {[0]}, .5), O([1]) is
(1) 1
(2) 2
(3) 3
(4) 4
CHAPTER – X (PROBABILITY DISTRIBUTIONS)
BOOK BACK QUESTIONS
125. If
is a probability density function then the value of k is
1)
2)
126. If
3)
4)
, – ∞ <x< ∞ is a p.d.f. of a continuous random variable X, then the value of A is
1) 16
2) 8
3) 4
127. A random variable X has the following probability distribution
X
0
1
2
3
4
5
P(X = x) 1/4 2a
3a
4a
5a 1/4
thenP(1≤ x ≤ 4) is
1)
2)
3)
4) 1
4)
128. A random variable X has the following probability mass function as follows
X
–2
3
1
P(X = x)
then the value of λ is
1) 1
2) 2
3) 3
4) 4
129. X is a discrete random variable which takes the values 0, 1, 2 and P(X = 0) =
, P(X = 1) =
then the value of P(X = 2) is
1)
2)
3)
4)
130. A random variable X has the following probability distribution function
X
0
1
2
3
4
5
6
7
2
2
2
P(X = x)
0
k
2k
2k
3k
k
2k
7k + k
The value of k is
89
1)
2)
3)0
4) –1 or
131. Given E(X + c) = 8 and E(X – c) = 12 then the value of c is
2) 4
3) –4
4) 2
1) –2
132. X is a random variable which takes the values 3, 4 and 12 with probabilities ,
and
. Then E(X ) is
1) 5
2) 7
3) 6
4) 3
133. Variance of random variable X is 4. Its mean is 2. Then E(X 2) is
1) 2
2) 4
3) 6
4) 8
134. μ2 = 20, μ2′ = 276 for a discrete random variable X . Then the mean of the random variable X is
1) 16
2) 5
3) 2
4) 1
135. Var (4X + 3) is
1) 7
2) 16 Var (X)
3) 19
4) 0
136. In 5 throws of a die, getting 1 or 2 is a success. The mean number of success is
2)
3)
4)
1)
137. The mean of a binomial distribution is 5 and its standard deviation is 2. Then the value of n and p are
1)
2)
3)
4)
138. If the mean and standard deviation of a binomial distribution are 12 and 2 respectively. Then the value of
its parameter p is
2)
3)
4)
1)
139. In 16 throws of a die, getting an even number is considered a success. Then the variance of success is
1) 4
2) 6
3) 2
4) 256
140. A box contains 6 red and 4 white balls, if 3 balls are drawn at random, the probability of getting 2 white balls
without replacement, is
1)
2)
3)
4)
141. If 2 cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same colours
without replacement, is
1)
2)
3)
4)
142. If in a Poisson distribution P(X = 0) = k then the variance is
1) log
2) log k
3) eλ
4)
143. If a random variable X follows Poisson distribution such that E(X 2) = 30 then the variance of the distribution is
1) 6
2) 5
3) 30
4) 25
144. The distribution function F(X) of a random variable X is --------1) a decreasing function
2) a non-decreasing function
3) a constant function
4) increasing first and then decreasing
145. For a Poisson distribution with parameter λ = 0.25 the value of the 2nd moment about the origin is
1) 0.25
2) 0.3125
3) 0.0625
4) 0.025
146. In a Poisson distribution if P(X = 2) = P(X = 3) then the value of its parameter λ is
1) 6
2) 2
3) 3
4) 0
147. If f(x) is a p.d.f. of a normal distribution with mean μ then
1) 1
2) 0.5
3) 0
148. The random variable X follows normal distribution f(x) =
1)
2)
3)
is
4) 0.25
then the value of c is
4)
149. If f(x) is a probability distribution function of a normal variateX and X ~ N(μ, σ2) then
is
1) undefined
2) 1
3) 0.5
40 – 0.5
150. The marks scored by 400 students in a mathematics test were normally distributed with mean 65. If 120
students got more marks above 85, the number of students securing marks between 45 and 65 is
1) 120
2) 20
3) 80
4) 160
90
CREATED QUESTIONS
(1) A discrete random variable takes
(1) only a finite number of values
(2) all possible values between certain given limits.
(3) Infinite number of values
(4) A finite or countable number of values
(2) A continuous random variable takes
(1) only a finite number of values
(2) all possible values between certain given limits.
(3) infinite number of values
(4) a finite or countable number of values
(3) If X is a discrete random variable, then P (X  a) =
(3) 1 – P (X < a)
(4) 0
(1) P (X < a)
(2) 1 – P (X  a)
(4) If X is a continuous random variable then P (X  a) =
(1) P(X < a)
(2) 1 – P (X > a)
(3) P(X > a)
(4) 1 – P(X  a – 1)
(5) If X is a continuous random variable then P (a < X < b) =
(1) P (a  X  b)
(2) P (a < X  b)
(3) P (a  X < b)
(4) all the three above
(6) A continuous random variable X has p.d.f. f(x), then
(2) f(x)  0
(3) f(x)  1
(4) 0 < f(x) < 1
(1) 0  f(x)  1
(7) A discrete random variable X has probability, mass function p(x), then
(1) 0  p(x)  1
(2) p(x)  0
(3) p(x)  1
(4) 0 < p(x) < 1
(8) Mean and variance of binomial distribution are
(1) nq, npq
(2) np, npq
(3) np, np
(4) np, npq
(9) Which of the following is or are correct regarding normal distribution curve?
(a) symmetrical about the line X =  (mean) (b) Mean = median = mode
(c) Unimodal
(d) Points of inflection are at X =   
(1) (a), (b) only
(2) (b), (d) only
(3) (a), (b), (c) only
4) all
(10) For a standard normal distribution the mean and variance are
(1)  ,  2
(2)  , 
(3) 0, 1
(4) 1, 1
(11) The p.d.f. of the standard normal variate Z is  (z) =
1
(1)
 z2
1
e 2
2 
(2)
1 z2
e
2
(3)
1
2
e
1 2
z
2
(4)
1
2
e
1
 z2
2
(12) If X is a discrete random variable, then which of the following is correct?
(1) 0  F(x) < 1
(2) F (–  ) = 0 and F(  )  1
(3) P(X = xn) = F(xn) - F(xn-1)
(4) F(x) is a constant function
(13) If X is a continuous random variable, then which of the following is incorrect?
(1) F’(x) = f(x)
(2) F (  ) = 1 and F(–  ) = 0
(3) p [a  x  b] = F(b) – F(a)
(4) p [a  x < b]  F(b) – F(a)
(14) Which of the following are correct ?
(i) E(aX + b) = aE(X) + b
(ii)  2 =  2’– (  1’)2
(iii)  2 = variance
(iv) var (aX +b) = a2 var (X)
(1) all
(2) (i), (ii), (iii)
(3) (ii), (iii)
(4) (i), (iv)
(15) which of the following is not true regarding the normal distribution?
(1) skewness is zero
(2) mean = median = mode
(3) the points of inflection are at x =   
(4) maximum height of the curve is
********
91
1
2
HSC–10 Marks Questions not asked–(MARCH 06 – JUNE 16)
(1) APPLICATION OF MATRICES AND DETERMINANTS (15)
EXERCISE 1.1.
(3)
Find the adjoint of the matrix
3  3 4
A  2  3 4  and verify the result A ( adj A )  ( adj A ) A  A . 
0  1 1 
3  3 4 
A  2  3 4
0  1 1 
(6)
Find the inverse of the matrix
(7)
Show that the adjoint of A  2
(9)
 2 2 1
1 

If A 
 2 1 2 , prove that
3 
 1  2 2
  1  2  2


1  2  is
2  2 1 
and verify that
A3  A 1 .
3 AT .
A 1  AT .
EXERCISE 1.2.
(4)
Solve by matrix inversion method eachof the following system of linear equations:
2x  y  z  7, 3x  y  5z  13, x  y  z  5 .
EXERCISE 1.4
(4)
Solve the following non-homogeneous system of linear equations determinant method:
x  y  z  4 ; x  y  z  2 ; 2x  y  z  1
(6)
Solve the following non-homogeneous system of linear equations determinant method:
3 x  y  z  2 ; 2x  y  2z  6 ; 2x  y  2z  2
(7)
Solve the following non-homogeneous system of linear equations determinant method:
x  2y  z  6 ; 3x  3y  z  3 ; 2x  y  2z  3
1
1 1


Example 1.4 : If A  1 2  3 ,verify A adj A  adj A A  A I 3


2  1 3 
Example 1.18:
Solve the following non-homogeneous equations of three unknowns.
(2) x  y  2z  6 ; 3x  y  z  2 ; 4x  2y  z  8
(4) x  y  2z  4 ; 2x  2y  4z  8 ; 3x  3 y  6z  12
Example 1.21:
Solve : x  y  2z  0 ; 3x  2y  z  0 ; 2x  y  z  0
Example 1.23 :
Examine the consistency of the equations.
Example 1.25:
Verify whether the given system of equations is consistent. If it is consistent, solve them:
2x  3y  7z  5 , 3x  y  3z  13, 2x  19y  47z  32
x  y  z  5 ,  x  y  z  5 , 2x  2y  2z  10
Example 1.27:
Solve the following homogeneous linear equations.
x  2 y  5z  0 ,
3x  4 y  6z  0 ,
(2) VECTOR ALGEBRA (0)
92
x yz 0
(3) COMPLEX NUMBERS (1)
EXERCISE 3.2
(8)
 z 1  

 z 3 2
(v) If P represents the variable complex number z. Find the locus of P, if arg 
(4) ANALYTICAL GEOMETRY (1)
EXERCISE 4.2.
(6)
Find the eccentricity, centre, foci, vertices of the following ellipses and draw the diagram:
(ii) x 2  4 y 2  8x  16 y  68  0
(5) DIFFERENTIAL CALCULUS APPLICATION – (12)
EXERCISE 5.1
(9)
Gravel is being dumped from a conveyor belt at a rate of 30 ft3 / min and its coarsened such that it forms a pile in the shape of
a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft
high?
Example 5.14 :
Example 5.15 :
Find the equations of tangent and normal to the curve 16 x
2
2
 9 y  144 at
( x1, y1 )
where x  2 and y  0 .
1
Find the equations of the tangent and normal to the ellipse x  a cos  , y  b sin  at the point  

4
1
.
EXERCISE 5.2
(5)
Find the equations of those tangents to the circle x 2  y 2  52 ,which are parallel to the straight line 2 x  3 y  6 .
Example 5.35 :
Evaluate :
lim x
sin x
x  0
EXERCISE 5.9
(3)
Find the local maximum and minimum values of the following functions:
(iii)
x 4  6x 2
(iv)
x
2

1
3
(v)
sin 2   0, 

(vi)
t  cos t
Example 5.52 : A farmer has 2400 feet of fencing and want to fence of a rectangular field that borders a straight river. He needs no
fence along the river. What ar the dimensions of the field that has the largest area?
Example 5.63 : Discuss the curve
y  x 4  4x 3 with respect to concavity and points of inflection.
EXERCISE 5.11
(5) Find the intervals of concavity and the points of inflection of the function f    sin 2 in
0 ,  
(6) DIFFERENTIAL CALCULUS APPLICATION - II (1)
Example 6.18 : If w  u 2 e v where u 
w
x
and v  y log x , find
and
y
x
93
w
y
(7) INTEGRAL CALCULUS (4)
Example 7.26 : Find the area between the line y  x  1 and the curve
y  x 2  1.
Example 7.28 : Find the area of the region enclosed by y 2  x and y  x  2
Example 7.31 : Find the area of the region bounded by the ellipse
Example 7.32 : Find the area of the curve y  x  5
2
2
x2
a2

y2
b2
1
x  6 (i) between x = 5 and x = 6
(ii) between x = 6 and x = 7
(8) DIFFERENTIAL EQUATIONS (4)
Example 8.13 : Solve :
2


xy  x dy  y dx  0
Example 8.15 : Solve : 1  e x
EXERCISE 8.4 Solve : (5)
y
dx  e
x y
1  x y  dy
 0 given that
y  1 , where x  0
 
dy y
  sin x 2
dx x
Example 8.38 : A drug is excreted in a patients urine. The urine is monitored continuously using a catheter. A patient is administered
10 mg of drug at time t = 0 , which is excreted at a Rate of  3t 1 2 mg/h.
(i) What is the general equation for the amount of drug in the patient at time t > 0 ?
(ii) When will the patient be drug free?
(9) DISCRETE MATHEMATICS (1)


Example 9.22 : Show that the set G  a  b 2 / a, b  Q is an infinite abelian group with respect to addition.
(10) PBOBABILITY DISTRIBUTIONS (0)
************************************************************
94
MODEL TESTS – 10 MARK Questions
TEST - 1
1. Prove by vector method that Cos (A – B) = Cos A Cos B + Sin A Sin B.
2. Find the vector and cartesian equation of the plane through the point (1, 3, 2) and parallel to the lines
x 1 y  2 z  3
x  2 y 1 z  2


and


2
1
3
1
2
2 .
3. Find the vector and cartesian equation of the plane through the points (1, 2, 3) and (2, 3, 1) perpendicular
to the plane 3x – 2y + 4z – 5 = 0 .












4. If a = i + j + k , b = 2 i + k , c = 2 i + j + k ,




  




d = i + j + 2k ,
  

Verify that ( a x b ) x ( c x d ) = [ a b d ] c – [ a b c ] d .
5. Derive the equation of the plane in the intercept form.
TEST – 2
1. Prove by vector method that Sin (A – B) = Sin A Cos B – Cos A Sin B.
2. Find the vector and cartesian equation of the plane containing the line
and parallel to the line
x  1 y 1 z  1
.


3
2
1
3. Find the vector and cartesian equation of the plane containing the line
x  2 y  2 z 1


2
3
3
x  2 y  2 z 1


2
3
2
and passing through the point (–1, 1, – 1).
x  2 y 1  z 1
4. Show that the lines x  1  y  1  z and
intersect and find their


1
1
1
3
2
1
point of intersection.
5. Find the vector and cartesian equation of the plane passing through the points with position vectors








3 i + 4 j + 2k , 2i – 2 j – k and 7i + k .
TEST – 3
1. Prove by vector method that Cos (A + B) = Cos A Cos B – Sin A Sin B.
2. Find the vector and cartesian equations of the plane through the point (2, – 1, – 3) and parallel to
x  2 y 1 z  3
x 1
y 1
z2
the lines
and
.




3
2
4
2
3
2
3. Find the vector and cartesian equations of the plane passing through the points (– 1, 1, 1) and (1, – 1, 1)
and perpendicular to the plane x + 2y + 2z = 5.



















4. If a = 2i + 3 j – k , b =  2 i + 5 k , c = j – 3 k , Verify that a x ( b x c ) = ( a . c ) b – ( a . b ) c .
5. Prove by vector method that the Altitudes of a triangle are concurrent.
TEST – 4
1. Prove by vector method that Sin (A + B) = Sin A Cos B + Cos A Sin B.
2. Find the vector and cartesian equation to the plane through the point (– 1, 3, 2) and perpendicular to
the planes x + 2y + 2z = 5 and 3x + y + 2z = 8.
3. Find the vector and cartesian equations of the plane passing through the points A( 1, – 2, 3) and B(– 1, 2, –1)
x2
y 1 z 1
.


2
3
4
x 1 y 1 z 1
x4 y
z 1
4. Show that the lines
intersect and hence find the


and


3
1
0
2
0
3
and is parallel to the line
point of intersection.
5. Find the vector and cartesian equations of the plane passing through the points (2, 2, – 1), (3, 4, 2)
and (7, 0, 6).
95
TEST – 5
1. Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum of the
parabola y 2  8x  6 y  1 0 and hence draw the graph.
2. Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum of the
parabola x 2  2x  8y  17  0 and hence draw the graph.
3. Find the eccentricity, centre, foci, vertices of the ellipse 16x 2  9y 2  32x  36y  92 and draw the diagram.
4. Find the eccentricity, centre, foci and vertices of the hyperbola 9x 2  16y 2  18x  64y – 199 = 0 and draw
the diagram.
5) Show that the line x – y + 4 = 0 is a tangent to the ellipse x 2  3 y 2  12 . Find the co-ordinates of
the point of contact.
TEST – 6
1. Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum of the
parabola x 2  6x  12 y  3  0 and hence draw the graph.
2. Find the eccentricity, centre, foci, vertices of the ellipse x 2  4y 2  8x  16y  68  0 and draw the diagram.
3. Find the eccentricity, centre, foci and vertices of the hyperbola x 2  3y 2  6x  6y  18  0 and draw
the diagram.
4. Find the eccentricity, centre, foci and vertices of the hyperbola 9x 2  16y 2  36x  32y  164 = 0 and
Draw the diagram.
5. Find the equation of the hyperbola if its asymptotes are parallel to x + 2y – 12 = 0 and x – 2y + 8 = 0.
(2, 4) is the centre of the hyperbola and it passes through (2, 0).
TEST – 7
1. Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum of the
2
parabola y  8x  6y  9  0 and hence draw the graph.
2. Find the eccentricity, centre, foci, vertices of the ellipse 36x 2  4y 2  72x  32y  44 = 0. and draw the
diagram.
3. Find the eccentricity, centre, foci and vertices of the hyperbola x 2  4y 2  6x  16y  11 0 and draw
the diagram.
4. Prove that the line 5x + 12y = 9 touches the hyperbola x 2  9y 2  9 and find its point of contact.
5. Find the equation of the rectangular hyperbola which has for one of its asymptotes the line x + 2y – 5 = 0
and passes through the points (6, 0) and (– 3, 0).
TEST – 8
1.The girder of a railway bridge is in the parabolic form with span 100 ft. and the highest point on the arch is
10ft. above the bridge. Find the height of the bridge at 10ft. to the left or right from the midpoint of the
bridge.
2. An arch is in the form of a semi – ellipse whose span is 48 feet wide. The height of the arch is 20 feet.
How wide is the arch at a height of 10 feet above the base?
3. A satellite is traveling around the earth in an elliptical orbit having the earth at a focus and of
eccentricity ½ . The shortest distance that the satellite gets to the earth is 400 kms. Find the longest
distance that the satellite gets from the earth.
4. A kho – kho player in a practice session while running realizes that the sum of the distances from the two
kho – kho poles from him is always 8m. Find the equation of the path traced by him if the distance
between the poles is 6m.
5. A cable of a suspension bridge is in the form of a parabola whose span is 40 mts. The road way is 5 mts
below the lowest point of the cable. If an extra support is provided across the cable 30 mts above the
ground level, find the length of the support if the height of the pillars are 55 mts.
96
TEST – 9
1. On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4mts
when it is 6mts away from the point of projection. Finally it reaches the ground 12mts away from the
starting point. Find the angle of projection.
2. A comet is moving in a parabolic orbit around the sun which is at the focus of a parabola. When the comet

is 80 million kms from the sun. the line segment from the sun to the comet makes an angle of
radians
3
with the axis of the orbit. Find (i) the equation of the comet’s orbit (ii) how close does the comet come
nearer to the sun? (Take the orbit as open rightward).
3. The arch of a bridge is in the shape of a semi–ellipse having a horizontal span of 40ft and 16ft high
at the centre. How high is the arch, 9ft from the right or left of the centre.
4. The ceiling in a hallway 20ft wide is in the shape of a semi ellipse and 18ft high at the centre. Find the
height of the ceiling 4 feet from either wall if the height of the side walls is 12 ft.
TEST – 10
1. Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a
parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5m below the line
of the pipe, the flow of water has curved outward 3m beyond the vertical line through the end of the pipe.
How far beyond this vertical line will the water strike the ground?
2. A cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed
horizontally. The distance between two towers is 1500 ft. the points of support of the cable on the towers
are 200 ft above the road way and the lowest point the cable is 70 ft. above the roadway. Find the
vertical distance to the cable (parallel to the roadway) from a pole whose height is 122 ft.
3. The orbit of the planet mercury around the sun is in elliptical shape with sun at a focus. The semi-major
axis is of length 36 million miles and the eccentricity of the orbit is 0.206. Find
(i) how close the mercury gets to sun? (ii) the greatest possible distance between mercury and sun.
4. A ladder of length 15m moves with its ends always touching the vertical wall and the horizontal
floor. Determine the equation of the locus of a point P on the ladder, which is 6m from the end of the
ladder in contact with the floor.
TEST – 11
1. Use differentials to find an approximate value for y =
x
 2u
 2u
2. Verify
for the function u = tan– 1   .

x y y x
y
3
1.02  4 1.02 .
TEST – 12
w
x
w
and v = y log x , find
and
.
x
y
y
2. Trace the curve y = x3 + 1.
1
3. Verify Euler’s theorem for f(x, y) =
.
2
x  y2
1. If w = u2ey, where u =
TEST – 13
 u
 u
x
y

for the function u = 2  2 .
x y y x
y x
2
3
2. Trace the curve y = 2x .
2
2
1. Verify
3. Using Euler’s theorem prove that x
 xy 
u
u 1
.
+ y = tan u, if u = sin 1 
 x y
x
y 2


97
TEST – 14
 u
 u
for the function u = sin 3x cos 4y.

x y y x
2. Trace the curve y = x3 .
 x 3  y3 
u
u
 , Using Euler’s theorem, prove that x
3. If u = tan 1 
+y
= sin 2u.
y
x
 xy 
2
2
1. Verify
TEST – 15
1. Show that (Z, *) is an infinite abelian group where * is defined as a * b = a + b + 2.
2. Show that the set {[1], [3], [4], [5], [9] } forms an abelian group under multiplication modulo 11.
3. Show that (Zn, +n) forms a group.
TEST – 16
1. Show that the set G of all positive rationals forms a group under the composition * defined by
ab
a*b=
for all a, b  G.
3
 1 0    0    2 0   0 1   0  2   0  
, 
,  2
, 
, 
, 
 where 3 1,  1
2. Show that 
2 


0 
 0 1   0    0    1 0    0   
forms a group with respect to matrix multiplication.
3. Let G be the set of all rational numbers except 1 and * be defined on G by a * b = a + b – ab for all
a, b  G. Show that (G, *) is an infinite abelian group.
TEST – 17
1. Show that the set G = ( 2 / n Z } is a abelian group under multiplication.
2. Prove that the set of four functions f l , f 2 , f 3 , f 4 on the set of non – zero complex numbes C – {0} defined
1
1
by f1(z) = z, f2(z) = – z, f3(z) = and f4(z) = – ,  z C – {0} forms an abelian group with respect
z
z
to the composition of functions.
a o
, a  R – {0} forms an abelian group under
3. Show that the set of all matrices of the form 
o o
matrix multiplication.
n
TEST – 18
1. Show that the set M of complex numbers z with the condition |z| = 1 forms a group with respect
to the operation of multiplication of complex numbers.
2. Show that the set G = {a + b 2 / a, b Q } is an infinite abelian group with respect to addition.
3. Show that the set G of all rational numbers except – 1 forms an abelian group with respect to the
operation * given by a * b = a + b + ab for all a, b  G.
TEST – 19
1. Show that the nth roots of unity form an abelian group of finite order with usual multiplication.
x x
 , where x  R – {0}, is a group under matrix
2. Show the set G of all matrices of the form 
x x
multiplication.
3. Show that (Z7 – {[0]}, . 7) forms a group.
98
TEST – 20
 2 z  1
1. P represents the variable complex number z, Find the locus of P, if Im 
  2
 iz  1 
 z 1  
2. P represents the variable complex number z, Find the locus of P, if arg 
 .
 z 1 3
1
1
3. If x + = 2 cos  and y +
= 2 cos  , show that
x
y
(i)
xm
yn

 2 cos (m  n) .
yn
xm
4. Solve the equation
(ii)
xm
yn

 2 i sin (m  n) .
yn
xm
x 4  x 3  x 2  x  1 0 .
TEST – 21
 z  1
1. P represents the variable complex number z, Find the locus of P, if Re 
1.
 z  i 
q
2. If  and  are the roots of the equations x2 – 2px + (p2 + q2) = 0 and tan  =
yp
( y  ) n  ( y  ) n
n 1 sin n
Show that
=q
.
 
sin n 
2
3
(  3  i) .
x 9  x 5  x 4  1 0 .
3. Find all the values of
4. Solve the equation
TEST – 22
 z 1  
1. P represents the variable complex number z, Find the locus of P, if arg 
 .
 z  3  2
2. If  and  are the roots of x2 – 2x + 4 = 0, Prove that
 n   n = i2 n 1 sin n
3
3. If a = cos 2 + i sin 2, b = cos 2 + i sin 2 and c = cos 2 + i sin 2 , prove that
and deduct
1
a 2b 2  c 2
 2 cos (     ) . (ii)
 2 cos 2(     ) .
(i) abc 
abc
abc
4. Solve the equation
x7  x 4  x3 1  0 .
TEST – 23
 z 1
1. P represents the variable complex number z, Find the locus of P, if Re 
  1.
 z i
2. If  and  are the roots of the equations x2 – 2x + 2 = 0 and cot  = y + 1. Show that
( y  ) n  ( y  ) n
 
=
sin n
.
sin n 
3
1
3 4
 and hence prove that the product of the values is 1.
3. Find all the values of   i

2
2


4. Find all the values of

3 i

2
3
.
99
 9  9 .
MODEL TESTS – 6 MARK Questions
TEST – 1
1 2
0  1
1
1 1
and B  

 verify that  AB   B A .
1 1
1 2 
1. If A  
3
1
1
2
2. Find the inverse of the matrix A = 2  2

3. (i) Find the adjoint of the matrix A  a
c
4. If
 1 2 
A 
,
 1  4
b
d 
 1
0 
 1
(ii) Solve by determinant method: x  y  3 ; 2x  3y  7
verify the result A adj A  adj A A  A I 2
5. (i) Solve: x  y  2z  0 ; 2x  y  z  0 ; 2x  2y  z  0 (ii) Solve by determinant method : x  y  2 ; 3y  3x  7
6. State and prove reversal law for inverses of matrices.
1
 1

 4 2 
7. (i) Find the inverse of the matrix:  2
(ii) Find the rank of the matrix  2

 5
2 3 
4  6
1  1
1  5  1

  2 1  5
1 5  7 2 
3
8. Find the rank of the matrix 1
9. Solve the following non-homogeneous equations of three
unknowns: x  y  2z  4 ; 2x  2 y  4z  8 ; 3x  3 y  6z  10
10. Solve by matrix inversion method: x  y  3 , 2x  3 y  8
TEST – 2
1. Construct the truth table for the statement :  p  q    ~ q 
2. Construct the truth table for  p  q   r
3. Show that p  q    ~ p  q     ~ q  p 
4. Use the truth table to establish the statement  p  ~ q   ~ p  q is tautology or contradiction.
5. Use the truth table to establish the statement
 ~ p  q   p
is tautology or contradiction.
6. Show that the cube roots of unity forms a finite abelian group under multiplication.
7. Prove that  Z ,   is an infinite abelian group.
8. (i) Prove that identity element of a group is unique. (ii) Prove that inverse element of an element of a group
is unique.
9. State and prove reversal law on inverse of a group.
10. Find the order of each element in the Group of fourth roots of unity under multiplication.
100
TEST – 3
1 2 3
1. (i) Find the adjoint of the matrix: 0 5 0
(ii) Solve by determinant method: 3x  2y  5 ; x  3y  4
2 4 3
2. Find the rank of the matrix
6 12 6


 1 2 1
4 8 4
3. Find the inverse of the matrix A =
4. Find the adjoint of the matrix
 1 2  2


 1 3 0 
 0  2 1 
1 2 
A 

3  5
and verify the result A ( adj A )  ( adj A ) A  A . 
5. Examine the consistency of the system of equations: x  y  z  7 ; x  2y  3z  18 ; y  2z  6
6. Solve by matrix inversion method: 7x  3y  1, 2x  y  0
7. If A  5
2

7
 3
 2  1
B

 1 1 
and
verify that  AB T  B T AT
8. Solve the non-homogeneous system of linear equations by determinant method: 4x  5y  9 ; 8x  10y  18
9. If
5 2
A 

7 3 
10 For
and
 2  1
B

 1 1 
  1 2  2
A   4  3 4 
 4  4 5 
verify that  AB 1  B 1 A 1
show that A  A 1 .
TEST – 4
1. Construct the truth table for the statement :  p  q    ~ q 
2. Construct the truth table for  p  q   r
3. Show that p  q   p  q    q  p 
4. Use the truth table to determine whether the statement
 ~ p  q    p   ~ q 
is a tautology.
5. Use the truth table to establish the statement  p   ~ p    ~ q  p  is tautology or contradiction.
6. Prove that the set of all 4th roots of unity forms an abelian group under multiplication.
7. Show that R  0, . is an infinite abelian group. Here ‘ . ‘ denotes usual multiplication.
8. (i) Show that
a 
1 1
 a  a  G , a group.
(ii) Find the order of each element in the Group of cube roots of unity under multiplication.
9. State cancellation laws on groups and prove any one of them.
10. Find the order of each element in the Group Z 5  0,  5
101

MODEL TESTS – 1 MARK Questions
TEST – 1
0 1], then rank of AAT is
(1) 1
(2) 2
(3) 3
2. If A is a scalar matrix with scalar k  0, of order 3, then A– 1 is
1
1
1
(1) 2 I
(2) 3 I
(3) I
k
k
k
0 0 1 
3. The inverse of the matrix 0 1 0 is
1 0 0
1. If A = [2
1 0 0
(1) 0 1 0
0 0 1
1
4. AT  is equal to
1) A 1

(4) 0
(4) kI
 0 0 1
(2)  0 1 0
 1 0 0
0 0 1 
(3) 0 1 0
1 0 0
  1 0 0
(4)  0  1 0
 0 0 1
2) AT
3) A
4) A 1
 


T

5. If a and b are two unit vectors and  is the angle between them, then ( a + b ) is a unit vector, if



2
(1)  =
(2)  =
(3)  =
(4)  =
4
2
3
3






6. The area of the parallelogram having a diagonal 3 i  j  k and a side i  3 j  4 k is
3
30
(1) 10 3
(2) 6 30
(3)
(4) 3 30
2











7. If a x ( b x c ) + b x ( c x a ) + c x ( a x b ) = x x y , then



(1) x = 0

(2) y = 0



(3) x and y are parallel





(4) x = 0 or y = 0 or x and y are parallel.

    
   
8. If [ a x b , b x c , c x a ] = 64 ; then  a , b , c  is


(1) 32
(2) 8
(3) 128
9. The value of a  b when a  j  2 k and b  2 i  k is
1) 2
2) - 2
3) 3
(4) 0
4) 4
10. The value of a  b when a  j  2 k and b  2 i  3 j  2 k is
1) 7
2) - 7
3) 5


4) 6
3i ( / 4 ) 3
11. The modulus and amplitude (argument) of the complex number e
are respectively


 3
 3
(1) e9,
(2) e9,
(3) e6,
(4) e9,
2
2
4
4
12. If (m – 5) + i(n + 4) is the complex conjugate of (2m + 3) + i(3n – 2) , then (n, m) are
1

 1

 1 
1 
(1)   ,8 
(2)   ,8 
(3)  ,8 
(4)  ,8 
2

 2

 2 
2 
13. If z represents a complex number, then arg(z) + arg( z ) is
(1)

4
(2)

2
(3) 0
102
(4)

3
14.The complex number form of 3   7 is
1)  3  i 7
2) 3  i 7
3) 3  i 7
4) 3  i 7
2
15.The point of intersection of the tangents at t1 = t and t2 = 3t to the parabola y = 8x is
(1) (6t2, 8t)
(2) (8t, 6t2)
(3) (t2, 4t)
(4) (4t, t2)
2
16. The directrix of the parabola y = x + 4 is
15
15
17
17
(1) x =
(2) x = –
(3) x = –
(4) x =
4
4
4
4
17. The line 2x + 3y + 9 = 0 touches the parabola y2 = 8x at the point
9


9
(1) (0, – 3)
(2) (2, 4)
(3)   6, 
(4)  ,6 
2

2

2
18. The focus of the parabola y  4 x is
1) (0, 1)
2) (1, 1)
3) (0, 0)
4) (1, 0)
19. The slope of the tangent to the curve y = 3x2 + 3sin x at x = 0 is
(1) 3
(2) 2
(3) 1
(4) – 1
20. The point on the curve y = 2x2 – 6x – 4 at which the tangent is parallel to the x-axis is
 5  17 
  5  17 
  5 17 
 3  17 
(2) 
(3) 
(4)  ,
(1)  ,
,
, 



2 
2 2 
 2
 2 2
2 2 
21. If the normal to the curve x2/3 + y 2/3 = a 2/3 makes an angle  with the x-axis, then the slope of the
normal is
(1) – cot 
(2) tan 
(3) – tan 
(4) cot 
22. Let “ h “ be the height of the tank. Then the rate of change of pressure “ p “ of the tank with respect to
height is
1)
dh
dt
2)
dp
dt
3)
u
is equal to
x
(1) yx y – 1
(2) u log x
24. The curve y2 (x – 2) = x2 (1 + x) has
(1) an asymptote parallel to x-axis
(3) asymptotes parallel to both axes
/2
sin x  cos x
25. The value of 
dx is
1

sin
x
cos
x
0
dh
dp
4)
dp
dh
23. If u = x y , then
(1)

2
(3) u log y
(4) x yx – 1
(2) an asymptote parallel to y-axis
(4) no asymptotes
(2) 0
(3)

4
(4) 

26. The value of
 sin
4
x dx is
0
(1)
3
16
(2)
3
16
(3) 0

2
(3)
(4)
3
8

27. The value of
 sin
2
x cos 3 x dx is
0
(1) 
(2)

4
(4) 0
2a
28.  f x  dx  0 if
0
1) f 2a  x  f x
2) f 2a  x   f x
103
3) f x   f x
4) f  x  f x
29. The integrating factor of the differential equation dx + xdy = e – y sec2y dy is
(1) ex
(2) e – x
(3) ey
(4) e – y
2
 dx 
30. The differential equation   + 5y1/3 = x is
 dy 
(1) of order 2 and degree 1
(2) of order 1 and degree 2
(3) of order 1 and degree 6
(4) of order 1 and degree 3
31. The differential equation of all non-vertical lines in a plane is
d2y
d2y
dy
dy
(1)
=0
(2)
(3)
=m
(4)
0
m
dx
dx
dx 2
dx 2
32. The order and degree of the differential equation y  4
dy
dx
 3x
are
dx
dy
1) 2, 1
2) 1, 2
3) 1, 2
4) 2, 2
33. Which of the following are statements?
(i) May God bless you
(ii) Rose is a flower
(iii) Milk is white
(iv) 1 is a prime
number
(1) (i), (ii), (iii)
(2) (i), (ii), (iv)
(3) (i), (iii), (iv)
(4) (ii), (iii), (iv
34. Which of the following is a contradiction ?
(1) p  q
(2) p  q
(3) p  ~ p
(4) p  ~ p
35. p  q is equivalent to
(1) p  q
(2) q  p
(3) (p  q)  (q  p)
(4) (p  q)  (q  p)
36. The truth values of the following statements are
ii) Chennai is in India or 2 is an irrational number
i) Chennai is in India or 2 is an integer.
iii) Chennai is in China or 2 is an integer iv) Chennai is in China or 2 is an irrational number
1) T F T F
2) T F F T
3) F T F T
4) T T F T
 2
37. If f(x) = kx , 0  x  3 is a probability density function, then the value of k is
else where
 0,
1
(1)
(2) 1
(3) 1
(4) 1
3
6
9
12
38. X is a discrete random variable which takes the values 0, 1, 2 and P(X = 0) = 144 , P(X = 1) = 1 ,
169
169
then the value of P(X = 2) is
(1) 145
(2) 24
(3) 2
(4) 143
169
169
169
169
39. X is a random variable taking the values 3, 4 and 12 with probabilities 1 , 1 and 5 . Then E(x) is
3 4
12
(1) 5
(2) 7
(3) 6
(4) 3
40. A discrete random variable takes
1) only a finite number of values
2) all possible values between certain given limits
3) infinite number of values
4) a finite or countable number of values
---------------
104
TEST – 2
1 
1) If A =  2  , then the rank of AAT is
 3
(1) 3
(2) 0
(3) 1
 1 3 2 
2) If the matrix  1 k  3 has an inverse, then the values of k
 1 4 5 
(1) k is any real number
(2) k = – 4
3) k  – 4
3) If A is a square matrix of order n, then | adj A | is
(1) | A|2
(2) |A|n
4) If A and B are matrices conformable to multiplication then ABT is
2) BT AT
3) AB
1) AT BT


(4) 2
(4) k  4
(3) |A|n – 1
(4) |A|
4) BA


5) If a and b include an angle 120o and their magnitude are 2 and 3 , then a . b is equal to
(2) – 3
(1) 3





(3) 2
3
2
(4) 

6) The vectors 2 i  3 j  4 k and a i  b j  c k are perpendicular, when
(1) a = 2, b = 3, c = – 4
(2) a = 4, b = 4, c = 5
(3) a = 4, b = 4, c = – 5
(4) a = – 2, b = 3, c = 4








7) If PR = 2 i  j  k , QS =  i  3 j  2 k , then the area of the quadrilateral PQRS is
(1) 5 3

(2) 10 3


(3)
5 3
2
(4)

3
2




8) If the projection of a on b and projection of b on a are equal, then the angle between a + b and a – b is


2

(1)
(2)
(3)
(4)
3
4
3
2
9) If m i  2 j  k and 4 i  9 j  2 k are perpendicular then m is
1) - 4
2) 8
3) 4
4) 12
10) If 5 i  9 j  2 k and m i  2 j  k are perpendicular then m is
1)
5
16
2)
11) If x2 + y2 = 1, then the value of
(1) x – iy
5
16
3)
1  x  iy
is
1  x  iy
(2) 2x
16
5
(3) – 2iy
4)
 16
5
(4) x + iy
100
100
 1  i 3 
1  i 3 
12) The value of 
 is
 + 
2
2




(1) 2
(2) 0
(3) – 1
(4) 1
13) If a = 3 + i and z = 2 –3i, then the points on the Argand diagram representing az, 3az and – az are
(1) vertices of a right angled triangle
(2) vertices of an equilateral triangle
(3) vertices of an isosceles triangle
(4) collinear
14) Real and imaginary parts of
1) 0, 3/2
3
i
2
are
2) 3/2, 0
3) 2, 3
105
4) 3, 2
15) 16x2 – 3y2 – 32x – 12y – 44 = 0 represents
(1) an ellipse
(2) a circle
(3) a parabola
(4) a hyperbola
16) The length of the latus rectum of the parabola whose vertex is (2, – 3) and the directrix x = 4 is
(1) 2
(2) 4
(3) 6
(4) 8
2
17) The vertex of the parabola x = 8y –1 is
1
 1 
1 
 1

(1)   ,0 
(2)  ,0 
(3)  0, 
(4)  0, 
8
 8 
8 
 8

2
18) The length of the Latus Rectum of y  4 x is
1) 2
2) 3
3) 1
4) 4
19) The slope of the tangent to the curve y = 3x2 + 3sin x at x = 0 is
(1) 3
(2) 2
(3) 1
(4) – 1
1
20) The equation of the normal to the curve  = at the point (– 3, – 1/3) is
t
1
(1) 3  = 27 t – 80
(2) 5  = 27t – 80
(3) 3  = 27 t + 80
(4)  =
t
21) What is the surface area of a sphere, when the volume is increasing at the same rate as its radius?
1
4
(3) 4
(4)
(1) 1
(2)
2
3
22) If the temperature   C of the certain metal rod of “ l“ meters is given by l  1  0.00005  0.0000004  2 then
the rate
of change of lin m / C ° when the temperature is 100°C is
1) 0.00013 m C
2) 0.00023 m C
3) 0.00026 m C
4) 0.00033 m C
 x 4  y4 
 and f = sinu, then f is a homogeneous function of degree
23) If u = sin– 1  2
2 
x y 
(1) 0
(2) 1
(3) 2
1
u
u
24) If u =
, then x
+ y
is equal to
2
2
x
y
x y
1
3
(1) u
(2) u
(3) u
2
2
/2

25) The value of
0
(4) – u
cos 5 / 3 x
dx is
cos 5 / 3 x  sin 5 / 3 x

2
(1)
(4) 4
(2)

4
(3) 0
(2)
1
30
(3)
(2)
1
3
(3) 0
(4) 
1
26) The value of
 x(1  x)
4
dx is
0
(1)
1
12
1
24
(4)
1
20
(4)
2
3
 /4
27) The value of
 cos
3
2 x dx is
0
2
3
n
28) I n   sin x dx then I n 
(1)
1
n 1
I n 2
n
n
1
n 1
3)  sin n 1 x cos x 
I n 2
n
n
1
n
n 1
I n 2
n
1
n 1
4)  sin n 1 x cos x 
In
n
n
1)  sin n 1 x cos x 
2) sin n 1 x cos x 
106
29) If cosx is an integrating factor of the differential equation
(1) – cot x
dy
+ Py = Q, then P =
dx
(3) tan x
(2) cotx
dx
30) Solution of the differential equation
+ mx = 0, where m < 0 is
dy
(1) x = cemy
(2) x = ce – my
(3) x = my + c
31) The differential equation of all circles with centre at the origin is
(1) xdy + ydx = 0
(2) xdy – ydx = 0
(3) xdx + ydy = 0
32) The order and degree of the differential equation
dy
 y  x2
dx
(4) – tan x
(4) x = c
(4) xdx – ydy = 0
are
1) 1, 1
2) 1, 2
3) 2, 1
4) 0, 1
33) If a compound statement is made up of three simple statements, then the number of rows in the
truth table is
(1) 8
(2) 6
(3) 4
(4) 2
34) The conditional statement p  q is equivalent to
(2) p  ~ q
(3) ~ p  q
(4) p  q
(1) p  q
35) If p is T and q is F, then which of the following have the truth value T?
(i) p  q (ii) ~ p  q
(iii) p  ~ q
(iv) p  ~ q
(1) (i), (ii), (iii)
(2) (i), (ii), (iv)
(3) (i), (iii), (iv)
(4) (ii), (iii), (iv)
36) The truth values of the following statements are
i. Ooty is in Tamilnadu and 3 + 4 = 8
ii. Ooty is in Tamilnadu and 3 + 4 = 7
iii. Ooty is in Kerala and 3 + 4 = 7
iv. Ooty is in Kerala and 3 + 4 = 8
1) F,T,F,F
2) F,F,F,T
3) T,T,F,F
4) T,F,T,F
1
37) If f(x) = A
,    x   is a p.d.f of a continuous random variable X, then the value of A is
 16  x 2
(1) 16
(2) 8
(3) 4
(4) 1
38) A random variable X has the following probability mass function as follows:
X
P(X = x)
–2
3


1

6
4
12
Then the value of  is
(1) 1
(2) 2
(3) 3
39) A random variable X has the following probability distribution
Then P(1  x  4) is
(2) 2
21
7
40) A continuous random variable takes
1) only a finite number of values
3) infinite number of values
(1) 10
(4) 4
(3) 1
(4) 1
14
2
2) all possible values between certain given limits
4) a finite or countable number of values
---------------
107
TEST – 3
1  1 2 
1) The rank of the matrix 2  2 4 is


2  4 8 
(1) 1
(2) 2
(3) 3
(4) 4

1 0 
  1 is 2, then  is
 1 0  
2) If the rank of the matrix  0

(1) 1
(2) 2
(3) 3
3) If I is the unit matrix of order n, where k  0 is a constant, then adj(kI) =
(1) kn (adj I)
(2) k (adj I )
(3) k2 (adj I)
(4) any real number
(4) kn – 1 (adj I)
 2  4
 is
1 2 
4) The rank of the matrix 
1) 1
2) 2
3) 0
4) 8




5) The shortest distance of the point (2, 10, 1) from the plane r . (3 i – j + 4 k ) = 2 26 is
1
(1) 2 26
(2) 26
(3) 2
(4)
26


6) The projection of OP on a unit vector OQ equals thrice the area of parallelogram OPRQ. Then POQ is
1
 3
(2) cos– 1  
(3) sin– 1  3 
(4) sin– 1  1 
(1) tan– 1


3
 10 
 3
 10 



7) If a , b , c are a right handed triad of mutually perpendicular vectors of magnitude a, b, c. Then the



value of [ a , b , c ] is
1
abc
(4) abc
2
8) If a line makes 45o, 60o with positive direction of axes x and y, then the angle it makes with the z axis is
(1) a2 b2 c2
(2) 0
(3)
(1) 30o
(2) 90o
(3) 45o
(4) 60o
9) The angle between the vectors i  j and j  k is
1)

3
2)
 2
3
3)
10) If the vectors a  3 i  2 j  9 k and b  i  m j  3 k
1) -15
2) 15
11) The polar form of the complex number (i25)3 is


(1) cos + isin
(2) cos  + isin 
2
2
1
12) If x = cos  + isin  , then the value of xn + n is
x
(1) 2 cos n 
(2) 2i sin n 
z1
13) z1 = 4 + 5i, z2 = – 3 + 2i ; then
is
z2
2 22
2 22
(1)
(2)   i
 i
13 13
13 13
108

3
4)
2
3
are perpendicular then m is
3) 30
4) -30


– isin
2
2
(3) cos  – isin 
(4) cos
(3) 2 sin n 
(4) 2i cos n 
(3) 
2 23
 i
13 13
(4)
2 22
 i
13 13
14) The standard form a  ib  of 3  2 i  (7  i) is
1) 4  i
2)  4  i
3) 4  i
4) 4  4 i
2
2
15) The distance between the foci of the ellipse 9x + 5y = 180 is
(1) 4
(2) 6
(3) 8
(4) 2
16) The locus of foot of perpendicular from the focus to a tangent of the curve 16x2 + 25y2 = 400 is
(2) x2 + y2 = 25
(3) x2 + y2 =16
(4) x2 + y2 = 9
(1) x2 + y2 = 4
x 2 y2

 1 is
16 9
(3 ) 8x – 9y – 72 = 0 (4) 8x + 9y + 72 = 0
17) The equation of the chord of contact of tangents from (2, 1) to the hyperbola
(1) 9x – 8y – 72 = 0 (2) 9x + 8y + 72 = 0
18) The focus of the parabola x 2  20 y is
1) (0,0)
2) (5,0)
3) (0,5)
4) (-5,0)
3
19) If y = 6x – x and x increases at the rate of 5 units per second, the rate of change of slope, when x = 3 is
(1) – 90 units / sec
(2) 90 units / sec
(3) 180 units / sec
(4) – 180 units / sec
a
20) The value of ‘a’ so that the curves y = 3ex and y = e– x intersect orthogonally is
3
1
(1) – 1
(2) 1
(3)
(4) 3
3
21) The ‘c’ of Lagranges Mean Value Theorem for the function f (x) = x2 + 2x – 1; a = 0, b = 1 is
1
(1) – 1
(2) 1
(3) 0
(4)
2
22) The curve y = f ( x ) and y = g ( x ) cut orthogonally if at the point of intersection
2) slope of f ( x ) + slope of g ( x ) = 0
1) slope of f ( x ) = slope of g ( x )
3) slope of f ( x ) / slope of g ( x ) = -1
4) [ slope of f ( x ) ] [ slope of g ( x ) ] = -1
23) If u = y sin x, then
2u
is equal to
x y
(1) cosx
(2) cosy
(3) sinx
24) If f ( x , y ) is a homogeneous functions of degree n then x
(4) 0
f
f
y

x
y
2) n f
3) n ( n - 1 ) f
4) n ( n + 1 ) f
1) f
25) The surface area of the solid of revolution of the region bounded by y = 2x, x = 0 and x = 2 about x-axis is
(1) 8 5 
(2) 2 5 
(3) 5 
(4) 4 5 
26) The curved surface are of a sphere of radius 5, intercepted between two parallel planes of distance
2 and 4 from the centre is
(2) 40 
(3) 10 
(4) 30 
(1) 20 
2/3
2/3
27) The length of the arc of the curve x + y = 4 is
(1) 48
(2) 24
(3) 12
(4) 96
 2
28) If n is odd then
n
 cos x dx
0
1)
n n2 n4




n 1 n  3 n  5
2
2)
3)
n n2 n4
3


   1
n 1 n  3 n  5
2
4)
109
n 1 n  3 n  5
1 



n n2 n4
2 2
n 1 n  3 n  5
2


   1
n n2 n4
3
29) The amount present in a radio active element disintegrates at a rate proportional to its amount. The
differential equation corresponding to the above statement is ( k is negative)
dp
dp
dp
dp k
(1)
(2)
(3)
(4)
 kt
 kp
 kt

dt
dt
dt
dt p
30) The differential equation formed by eliminating A and B from the relation y = ex (Acos x + Bsin x) is
(1) y2 + y1 = 0
(2) y2 – y1 = 0
(3) y2 – 2y1 + 2y = 0 (4) y2 – 2y1 –2y = 0
dy
31) The integrating factor of the differential equation
– y tan x = cos x is
dx
(1) sec x
(2) cos x
(3) etanx
(4) cot x
32) The order and degree of the differential equation sin x dx  dy   cos x dx  dy  are
1) 1, 1
2) 0, 0
3) 1, 2
4) 2, 1
33) The number of rows in the truth table of ~ [p  (~ q)] is
(1) 2
(2) 4
(3) 6
(4) 8
34) A monoid becomes a group, if it also satisfies the
(1) closure axiom
(2) associative axiom
(3) identity axiom
(4) inverse axiom
35) The value of [3] +11 ( [5] +11 [6] ) is
(1) [0]
(2) [1]
(3) [2]
(4) [3]
36) Let p be “ Kamala is going to school “ and q be “ There are twenty students in the class “. “ Kamala is not
going
to school or there are twenty students in the class “ stands for
1) p  q
2) p  q
3) ~ p
4) ~ p  q
37) If the mean and standard deviation of a binomial distribution are 12 and 2 respectively. Then the
value of its parameter p is
(1) 1
(2) 1
(3) 2
(4) 1
2
3
3
4
38) A box contains 6 red and 4 white balls. If 3 balls are drawn at random, the probability of getting
2 white balls without replacement, is
(1) 1
(2) 18
(3) 4
(4) 3
25
10
20
125
39) In a Poisson distribution, if P(X = 2) = P(X = 3), then the value of its parameter  is
(1) 6
(2) 2
(3) 3
(4) 0
40) For a standard normal distribution the mean and variance are
1)  ,  2
2)  , 
3) 0,1
4) 1,1
---------------
110
TEST – 4
1) The rank of the diagonal matrix
 1



2




0


4 


0
(1) 0
1
, then (adjA)A =
4
2
2) If A = 
3
1

(1)  5
0


0
1

5
(2) 2
(3) 3
(4) 5
1 0 
(2) 

0 1 
5 0 
(3) 

0  5
5 0 
(4) 

0 5 
(3) k det (A)
(4) det (A)
3) 1
4) 5
3) If A is a matrix of order 3, then det(kA)
(2) k2 det (A)
(1) k3 det (A)
 7  1
 is
2 1 
4) The rank of the matrix 
1) 9
2) 2


5) If a is a non- zero vector and m is non-zero scalar, then m a is a unit vector, if
1
(1) m =  1
(2) a = | m|
(3) a =
(4) a = 1
|m|






6) If p , q and p + q are vectors of magnitude  , then the magnitude of p  q is
(1) 2 




3
(2)


7) If a + b + c = 0 ; a = 3 ;
2
(3)
(4) 1



b = 4 ; c = 5, then the angle between a and b is
5

2
(3)
(4)
3
3
2
x  3 y  3 2z  5
8) The equation of the line parallel to
and passing through the point (1, 3, 5)


1
5
3
in vector form is
(1)


6

(2)












(1) r = ( i + 5 j + 3 k ) + t ( i + 3 j + 5 k )
(2) r = ( i + 3 j + 5 k ) + t ( i + 5 j + 3 k )













3
3 
(3) r = ( i + 5 j + k ) + t ( i + 3 j + 5 k )
(4) r = ( i + 3 j + 5 k ) + t ( i + 5 j + k )
2
2
9) The projection of the vector 7 i  j  4 k on 2 i  6 j  3 k is
1)
7
8
2)
8
3)
66
8
7
4)
66
8
10) a  b , when a  2 i  2 j  k and b  6 i  3 j  2 k is
1) 4
2) -4
3) 3
4) 5
11) If the point represented by the complex number iz is rotated about the origin through the angle  /2
in the counter clockwise direction, then the complex number representing the new position is
(1) iz
(2) – iz
(3) – z
(4) z
n
n
12) If zn = cos
+ i sin
, then z1 . z2 . . . z6 is
3
3
(1) 1
(2) – 1
(3) i
(4) – i
111
13) If – z lies in the third quadrant, then z lies in the
(1) first quadrant
(2) second quadrant (3) third quadrant
(4) fourth quadrant
2  3i
then q is
14) If p  i q 
4  2i
1) 14
2) - 14
3) - 8
4) 8
2
2
15) The straight line 2x – y + c = 0 is a tangent to the ellipse 4x + 8y = 32, if c is
(1)  2 3
(2)  6
(4)  4
(3) 36
2
2
16) The sum of the distances of any point on the ellipse 4x + 9y = 36 from ( 5 , 0) and (– 5 , 0) is
(1) 4
(2) 8
(3) 6
(4) 18
2
2
17) The directrices of the hyperbola x – 4(y – 3) = 16 are
(1) y = 
8
(2) x = 
8
(3) y = 
5
8
5
5
2
18) The equation of the latus rectum of y   8x is
1) y – 2 = 0
2) y + 2 = 0
3) x – 2 = 0
mx
– mx
19) The angle between the curve y = e and y = e
for m > 1 is
(4) x = 
5
8
4) x + 2 = 0
 2m 
  2m 
 2m 

(2) tan– 1 
(3) tan– 1 
(4) tan– 1  2

2 
2 
1 m 
 m 1
 1 m 
20) The gradient of the tangent to the curve y = 8 + 4x – 2x2 at the point, where the curve cuts the y–axis is
(1) 8
(2) 4
(3) 0
(4) – 4
21) The radius of a cylinder is increasing at the rate of 2cm / sec and its altitude is decreasing at the rate of
3cm / sec. The rate of change of volume, when the radius is 3 cm and the altitude is 5 cm is
(1) 23 
(2) 33 
(3) 43 
(4) 53 
2
3
22) The angular displacement of a fly wheel in radius is given by   9t  2t .The time when the angular
 2m 
(1) tan– 1  2

 m 1 
acceleration zero is
1) 2.5 s
2) 3.5 s
2
23) An asymptote to the curve y (a + 2x) = x2 (3a – x) is
(1) x = 3a
(2) x = – a/2
24) If u = f ( x , y ) then with usual notations, u xy  u yx if
1) u is continuous
25) The volume, when the curve y =
(1) 100 
2) u x is continuous
3) 1.5 s
4) 4.5 s
(3) x = a/2
(4) x = 0
3) u y is continuous 4) u , u x , u y are continuous
3  x 2 from x = 0 to x = 4 is rotated about x-axis is
100
100
100


(2)
(3)
(4)
9
3
3
26) Volume of solid obtained by revolving the area of the ellipse
x 2 y2

 1 about major and minor
a 2 b2
axes are in the ratio
(1) b2 : a2
(2) a2 : b2
(3) a : b
(4) b : a
27) The volume generated when the region bounded by y = x, y = 1, x = 0 is rotated about y-axis is



2
(1)
(2)
(3)
(4)
4
2
3
3
112
a
28) If f ( x ) is even then  f x  dx is
a
a
a
2) 2  f x  dx
1) 0
3)  f x  dx
0
a
4)  2  f x  dx
0
0
29) A particular integal of the differential equation (D – 4D + 4) y = e is
2
(1)
x 2 2x
e
2
2x
(3) xe– 2x
(2) xe2x
(4)
x 2 x
e
2
30) If f ’(x) =
x and f(1) = 2, then f(x) is
3
2
3
2
(1) – ( x x  2)
(2) ( x x  2)
(3) ( x x  2)
(4)  x( x  2)
3
2
3
2
31) The particular integral of the differential equation f(D)y = eax , where f (D) = (D – a)g(D), g(a)  0 is
(1) meax
(2)
e ax
g (a )
32) The order and degree of the differential equation
(3) g(a) eax
d2y
dx
2
x
y
(4)
xe ax
g (a )
dy
are
dx
1) 2, 1
2) 1, 2
3) 2 , 1/2
4) 2, 2
33) In the set of integers with operation * defined by a * b = a + b – ab, the value of 3 * (4*5) is
(1) 25
(2) 15
(3) 10
(4) 5
34) The order of [7] in (Z 9 , + 9) is
(1) 9
(2) 6
(3) 3
(4) 1
35) Which of the following is not a group?
(1) ( Z n , + n )
(2) ( Z , + )
(3) ( Z , . )
(4) ( R , + )
36) The truth values of the following statements are
i. All the sides of a rhombus are equal in length
ii. 1 19 is an irrational number
iii. Milk is white
iv. The number 30 has four prime factors.
1) T T T F
2) T T T T
3) T F T F
4) F T T T
37) Var(4X + 3) is
(1) 7
(2) 16Var (X)
(3) 19
(4) 0
38) If in a Poisson distribution P(X = 0) = k, then the variance is
1
(1) log 1
(2) log k
(3) e 
(4)
k
k
39) If 2 cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same
colours without replacement, is
26
(1) 1
(2)
51
2
40) Mean and variance of binomial distribution are
1) nq, npq
2) np,
npq
(3) 25
51
(4) 25
102
3) np, np
4) np, npq
---------------
113
TEST – 5
a b
c b
a c
; 2=
, 3=
, then the value of (x, y) is
p q
d q
p d
1) If aex + bey = c; pex + qey = d and  1=


 



(2)  log 2 , log 3  (3)  log 1 , log 1
1
1 
3
2


  
(1)  2 , 3 
 1 1 





 
(4)  log 1 , log 1 
2
3 

2) In a system of 3 linear non-homogenous equation with three unknowns, if  = 0,  x = 0,  y  0
and  z = 0, then the system has
(1) unique solution (2) two solutions
(3) infinitely many solutions (4) no solutions.
3) If A and B are any two matrices such that AB = 0 and A is non- singular, then
(1) B = 0
(2) B is singular
(3) B is non-singular
(4) B = A
4) In the homogeneous system  A  the number of unknowns then the system has
1) only trivial solution
2) trivial solution and infinitely many non-trivial solutions
3) only non-trivial solutions
4) no solution
2
5) The centre and radius of the sphere given by x + y2 + z2 – 6x + 8y – 10z + 1 = 0 is
(1) (– 3, 4, – 5), 49 (2) (– 6, 8, – 10), 1 (3) (3, – 4, 5), 7
(4) (6, – 8, 10), 7
x 1 y  2 z  3
x6 y4 z 4
6) The point of intersection of the lines
and
is




6
4
8
2
4
2
(1) (0, 0, – 4)
(2) (1, 0, 0)
(3) (0, 2, 0)
(4) (1, 2, 0)
7) The equation of the plane passing through the point (2, 1, – 1) and the line of intersection of the







planes r . ( i +3 j – k ) = 0 and r . ( j + 2 k ) = 0 is
(1) x + 4y – z = 0
(2) x + 9y + 11z = 0


(3) 2x + y – z + 5 = 0 (4) 2x – y + z = 0

8) r = s i + t j is the equation of


(1) a straight line joining the points i and j (2) xoy plane (3) yoz plane (4) zox plane



9) If a , b , c are a right handed triad of mutually perpendicular vectors of magnitude a, b, c. Then the



value of [ a , b , c ] is
(1) a2 b2 c2
(2) 0
(3)
1
abc
2
(4) abc
10) The projection of 3 i  j  k on 4 i  j  2 k is
1)
9
2)
21
9
3)
21
22
23
24
81
21
25
4)
 81
21
11) The value of i + i + i + i + i is
(1) i
(2) – i
(3) 1
(4) –1
1 i
12) If
is a root of the equation ax2 + bx + 1 = 0, where a, b are real, then (a, b) is
1 i
(1) (1, 1)
(2) (1, – 1)
(3) (0, 1)
(4) (1, 0)
13) If  is the nth root of unity, then
(1) 1 +  2 +  4 + . . . =  +  3 +  5 + . . . (2)  n = 0
(3)  n = 1
(4)  =  n - 1
114
14) The conjugate of 2  i 3  2 i  is
1) 8  i
2)  8  i
3)  8  i
4) 8  i
2
15) The angle between the two tangents drawn from the point (– 4, 4) to y = 16x is
(1) 45o
(2) 30o
(3) 60o
(4) 90o
16) The radius of the director circle of the conic 9x2 + 16y2 = 144 is
7
(1)
(2) 4
(3) 3
(4) 5
x 2 y2

 1 is
16 9
(3 ) 8x – 9y – 72 = 0 (4) 8x + 9y + 72 = 0
17) The equation of the chord of contact of tangents from (2, 1) to the hyperbola
(1) 9x – 8y – 72 = 0 (2) 9x + 8y + 72 = 0
18) The directrix of the parabola x 2   4 y is
1) x = 1
2) x = 0
3) y = 1
4) y = 0
3
19) The gradient of the curve y =  2 x  3x  5 at x = 2 is
(2) 27
(3) – 16
(4) – 21
(1) – 20
2/3
2/3
2/3
20) The parametric equations of the curve x + y = a are
(1) x = a sin3  ; y = a cos3 
(2) x = a cos3  ; y = a sin3 
(3) x = a3 sin  ; y = a3 cos 
(4) x = a3 cos  ; y = a3 sin 
21) The angle between the parabolas y2 = x and x2 = y at the origin is


3
4
(1) 2 tan– 1  
(2) tan– 1  
(3)
(4)
2
4
4
3
22) The law of the mean can also be put in the form
1) f a  h  f a   hf ' a  h 0    1 2) f a  h  f a   hf ' a  h 0    1
3) f a  h  f a   hf ' a  h 0    1 4) f a  h  f a   hf ' a  h 0    1
23) Identify the true statements in the following:
(i)
If a curve is symmetrical about the origin, then it is symmetrical about both axes.
(ii)
If a curve is symmetrical about both axes, then it is symmetrical about the origin.
(iii)
A curve f(x, y) = 0 is symmetrical about the line y = x if f(x, y) = f(y, x).
(iv)
For the curve f(x, y) = 0, if f(x, y) = f(– y, – x), then it is symmetrical about the origin.
(1) (ii), (iii)
(2) (i), (iv)
(3) (i), (iii)
(4) (ii), (iv)
u
u
y
24) If u = f   , then x
+ y
is equal to
x
y
x
(1) 0
(2) 1
(3) 2u
(4) u
x 2 y2

 1 and its auxillary circle is
a 2 b2
(1)  b(a – b)
(2) 2  a (a – b)
(3)  a (a – b)
(4) 2  b(a – b)
2
26) The area bounded by the parabola y = x and its latus rectum is
4
1
2
8
(1)
(2)
(3)
(4)
3
6
3
3
27) The area of the region bounded by the graph of y = sin x and y = cos x between x = 0 and x =  /4 is
25) The area between the ellipse
2 +1
(1)
(2)
2 –1
(3) 2 2 – 2
(4) 2 2 + 2
b
28)  f x  dx is
a
a
1) 2  f x  dx
0
b
b
2)  f a  x  dx
3)  f b  x  dx
a
a
115
b
4)  f a  b  x  dx
a
29) Integrating factor of the differential equation
(1) ex
dy
2
1
+
y = 2 is
dx x log x
x
(2) log x
(3)
1
x
(4) e – x
30) The differential equation of the family of lines y = mx is
d2y
dy
=m
(2) ydx – xdy = 0
(3)
0
dx
dx 2
31) The complementary function of the differential equation (D2 + 1)y = e2x is
(1) (Ax + B)ex
(2) Acosx + Bsinx
(3) (Ax + B)e2x
(1)
(4) ydx + x dy = 0
(4) (Ax + B)e – x
3
32) The order and degree of the differential equation
 dy d 3 y  2
 
 0

y

 dx dx 3 
dx 2


d2y
1) 2, 3
2) 3, 3
33) Which of the following is a tautology?
(2) p  q
(1) p  q
34) which of the following is not a binary operation on R
are
3) 3, 2
4) 2, 2
(3) p  ~ p
(4) p  ~ p
(1) a * b = ab
(2) a * b = a – b
(3) a * b = ab
(4) a * b = a  b
2
35) In the multiplicative group of cube root of unity, the order of w is
(1) 4
(2) 3
(3) 2
(4) 1
36) If p is true and q is false then which of the following statements is not true ?
1) p  q is false
2) p  q is true
3) p  q is false
4) p  q is true
37) A random variable X has the following probability mass function
X
0
1
2
3
4
5
6
7
2
2
2
P(X = x)
0
k
2k
2k
3k
k
2k
7k + k
The value of k is
(2) 1
(3) 0
(4) – 1 or 1
(1) 1
8
10
10
38) The mean of a binomial distribution is 5 and its standard deviation is 2. Then the value of n and p are
2
4
(1)  ,25 
5

4
(2)  25, 

1
(3)  ,25 
5
5

1
(4)  25, 

5
2
39) If a random variable X follows Poisson distribution such that E(X ) = 30, then the variance
of the distribution is
(1) 6
(2) 5
(3) 30
(4) 25
40) The p.d.f of the standard normal variate Z is  z  
1)
1
2 

e
1 2
z
2
2)
1
2
e
 z2
3)
---------------
116
1
2
e
1 2
z
2
4)
1
2

e
1 2
z
2
2