Download Poornima University, For any query, contact us at :8875666617,18

AIEEE/2012/MATHS
S.
No
Q.1
Questions
sinx
The equation e
–e
−sinx
Solutions
−4 = 0 has
Sol:1 (b)
(a) infinite number of real roots (b) no real roots
esinx – e−sinx =4
1
t– t =4
(c) exactly one real root
t2 – 4t – 1 = 0
(d) exactly four real roots
⇒ Let esinx = t
4+2 5
⇒t=
esinx = 2± 5
Let â and b be two unit vectors. If the vectors c = â + 2b and d = 5 â - 4b are
perpendicular to each other, then the angle between â and b is
(a)
Q.3
π
(b)
6
π
(c)
2
π
3
(d)
9
(b)9
2
9
c∙ d= 0
1
⇒a∙b=
2
4
7
(a)7
Q.4
∴ hence no solution
⇒ 5|a|2 + 6a ∙ b − 8|b|2 = 0
π
⇒ a∙ b =
3
4
Rate of decrease of volume v = 3 πr 3
dv
dt
dr
(d)2
Statement1:The sum of the series 1+ (1+2+4) + (4+6+9) + (9+12+16) +…….
dt
4
dr
= 3 π3r 2 dt
72
72π = 4πr 2
2
dr
dt
= 4.9.9 = 9
Sol:4 (b)
+ (361+380+400) is 8000.
Tn = (n – 1)2 + (n – 1) n + n2
Statement 2:
n
(k 3 − k − 1)3 = n3 for any natural number n.
k=1
=
(n−1)3 − n 3
n−1 −n
= n3 − (n − 1)3
T1 = 13 − 03
(a) Statement 1 is false, statement 2 is true
T2 = 23 − 13 ……….
(b) Statement 1 is true, statement 2 is true; statement 2 is
T20 = 203 − 193
a correct explanation for statement 1
(c) Statement 1 is true, statement 2 is true; statement 2 is
≤ esinx ≤ e
Sol:3 (c)
4
v = πr 2
3
After 49 minutes volume = 4500π −49(72π) = 972π
4
πr 3 = 972π
⇒ r 3 = 729
⇒r=9
3
⇒
(c)9
e
Sol:2 (C)
π
A spherical balloon is filled with 4500π cubic meters of helium gas. If a leak
in the balloon causes the gas to escape at the rate of 72π cubic meters per
minute, then the rate (in meters per minute) at which the radius of the balloon
decreases 49 minutes after the leakage began is
1
−1≤sin x ≤ 1
esinx =2+ 5 not possible
esinx =2− 5 not possible
Q.2
⇒ t=2+ 5
2
S20 = (20)3 − 03 = 8000
∴ kth bracket is (k – 1)2 + k(k – 1) + k2 = 3k2 – 3k + 1.
not a correct explanation for statement 1
(d) Statement 1 is true, statement 2 is false
Poornima University, For any query, contact us at :8875666617,18
1
AIEEE/2012/MATHS
Q. 5
The negation of the statement “If I become a teacher, then I will open a
Sol:5 (a)
~(~p ⋁ q) = p ∧ ~q
school” is
(a) I will become a teacher and I will not open a school
(b) Either I will not become a teacher or I will not open a
school
(c) Neither I will become a teacher nor I will open a school
(d) I will not become a teacher or I will open a school
Q.6
If the integral ʃ
5 tanx
tanx −2
dx = x + a ln |sin x – 2 cos x| + k, then a is equal to
Sol: 6 (d)
ʃ
(a) –1
(b) –2
(c) 1
(d) 2
5 tanx
tanx −2
⇒ʃ
= 2ʃ
dx = ʃ
5 sinx
sinx −2cosx
dx
2(cos x+2 sin x )+sin x−2cosx )
sinx −2cosx
cos x+2 sin x
sinx −2cosx
dx
dx + ʃ dx + k Let sin x – 2 cos x = t and after solving
= 2 log |sin x – 2 cos x| + x + k ∴ a = 2
Q.7
Statement 1: An equation of a common tangent to the parabola y2 = 16√3X
and the ellipse 2x2 + y2 = 4 is y = 2x + 2 3.
Sol: 7 (b)
y2 = mx + 16 3x
4 3
Statement 2: If the line y = mx + m , (m≠0) is a common tangent to the
parabola y2 = 16 3x and the ellipse 2x2 +y2 = 4, then m satisfies m4 +2m2 =24.
y = mx +
1 0 0
1
Let A = 2 1 0 . If u1 and u2 are column matrices such that Au1 = 0 and
3 2 1
0
0
Au2 1 then u1 +u2 is equal to
0
(a)
−1
1
0
(b)
−1
1
−1
−1
(c) −1
0
m
48
x2
2
+
y2
4
=1
is tangent to parabola
which is tangent to ellipse
(a) Statement 1 is false, statement 2 is true
(b) Statement 1 is true, statement 2 is true; statement 2 is a
correct explanation for statement 1
(c) Statement 1 is true, statement 2 is true; statement 2 is not a
correct explanation for statement 1
(d) Statement 1 is true, statement 2 is false
Q.8
4 3
⇒
⇒ m 2 = 2m2 + 4
if c2 = a2m2 +b2
⇒ m4 + 2m2 = 24
⇒m2 = 4
Sol: 8 (d)
1
(d) −1
−1
Poornima University, For any query, contact us at :8875666617,18
1 0 0
A= 2 1 0
Let u1 =
3 2 1
1
1
Au1 = 0
⇒u1 = −2
0
1
0
0
Au2 = 1
⇒ u2 = 1
0
−2
a
d
b ;u2 e
c
f
1
⇒ u1+u2 = −1
−1
2
AIEEE/2012/MATHS
Q.9
If n is a positive integer, then ( 3 + 1)2n − ( 3 − 1)2n is
Sol: 9 (a)
( 3 + 1)2n − ( 3 − 1)2n = ( 3 + 1)2
− ( 3 − 1)2
n
= 2n (2 + 3)n − (2 − 3)n
(b) an odd positive integer
=2n [n C0 2n +n C1 2n−1 3+n C2 2n−2 3 + ⋯ . . ] − [n C0 2n −n C1 2n−1 3+n C2 2n−2 3 − ⋯ ]
(c) an even positive integer
= 2n+1[n C1 2n−1 3+n C3 2n−3 3 3 + ⋯]=2n-1 3 (some integer)
(d) a rational number other than positive integers
Which is irrational
If 100 times the 100th term of an AP with non zero common difference equals
the 50 times its 50th term, then the 150th term of this AP is
Sol: 10 (d)
100(T100) = 50(T50)
⇒ 2[a + 99d] = a + 49d
th
(a) –150
(c) 150
n
= (4 + 2 3)n − (4 − 2 3)n
(a) an irrational number
Q.10
(b) 150 times its 50 term
(d) zero
⇒ a + 149d = 0
⇒ T150 = 0
Q.11
In a ∆PQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R
is equal to
(a)
5π
6
π
π
(b) 6
(c) 4
(d)
Sol: 11 (b)
3 sin P + 4 cos Q = 6
4 sin Q + 3 cos P = 1
From squaring and adding
⇒ 9 + 16 + 24 (sin P cos Q + cos P sin Q) = 37
⇒ 24 sin (P + Q) = 12
3π
4
1
sin(π - r) = 2 (if R =
π
Q.13
Sol: 12 (a)
(a) x – 2y + 2z – 3 = 0
(b) x – 2y + 2z + 1 = 0
(c) x – 2y + 2z – 1 = 0
(d) x – 2y + 2z + 5 = 0
perpendicular distance from
O(0, 0, 0) to (1) is 1
If the line 2x + y = k passes through the point which divides the line segment
joining the points (1, 1) and (2, 4) in the ratio 3 : 2, then k equals
Sol:13 (c)
|k|
1+4+4
= 1 ⇒ |k| = 3
5
(b) 5
(c) 6
(d)
11
Poornima University, For any query, contact us at :8875666617,18
11
2
⇒ k = ±3 ∴ x – 2y + 2z – 3 = 0
6+2 12+2
5
,
5
8 14
P=
5
5
P=
π
then O < P Q < 6
2x+y = K
Point p =
29
6
cos Q < 1 and sinp < 2 then 3 sinp + 4 cos Q <
An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit
distance from the origin is
(a)
5π
1
R=6
Q.12
3
,
8 14
5
,
5
A(1,2)
5
lies on 2x + y = k
⇒
16
5
+
14
5
=k
P
⇒k=
B(2,4)
30
5
=6
AIEEE/2012/MATHS
Q.
14
Let x1, x2, ......, xn be n observations, and let X be their arithematic mean and
σ2 be their variance.
Statement 1: Variance of 2x1, 2x2, ......, 2xn is 4σ2 .
Statement 2: Arithmetic mean of 2x1, 2x2, ......, 2xn is 4x .
Sol:14 (d)
Xi 2
σ2 =
Xi 2
−
n
n
Variance of 2x1, 2x2,……., 2xn =
(a) Statement 1 is false, statement 2 is true
=4
(b) Statement 1 is true, statement 2 is true; statement 2 is a
correct explanation for statement 1
Xi2
n
−
=
2x 1 + 2x 2 +⋯+2x n
The population p(t) at time t of a certain mouse species satisfies the differential
dp (t)
equation
= 0.5 p(t) – 450. If p(o) = 850, then the time at which the
dt
population becomes zero is
d(p t )
dt
1
= p(t) – 450
p t −ʃ900
=
2
d (p t )
2ʃ p
(d) In 18
t −900
= ʃ dt
2 In Ip(t) – 900| = t + K
⇒ 2In 50 = 0 + K
t=0
t = 2 (In 900 – In 50) = 2In
Q.
16
Q.17
Let a, b ∈ R be such that the function f given by f(x) = In |x| + bx2 + ax, x ≠ 0
has extreme values at x = –1 and x = 2.
Statement 1: f has local maximum at x = –1 and at x = 2.
1
−1
Statement 2: a = 2 = and b = 4
(a) Statement 1 is false, statement 2 is true
(b) Statement 1 is true, statement 2 is true; statement 2 is a
correct explanation for statement 1
(c) Statement 1 is true, statement 2 is true; statement 2 is not a
correct explanation for statement 1
(d) Statement 1 is true, statement 2 is false
y
The area bounded between the parabolas x2 = 4 and x2 = 9y, and the straight
line y = 2 is
(a) 20 2
(b)
10 2
3
(c)
20 2
3
=2x
2
dt
(b) In 9
2
n
Sol: 15 (a)
d(p t )
(c) In 18
x 1 + x 2 +⋯+x n
Statement 2 is false.
(d) Statement 1 is true, statement 2 is false
1
n
= 4σ2
=2
n
correct explanation for statement 1
(a) 2 In 18
n
n
Xi 2
−
Statement 1 is true. A. M. of 2x1,2x2,……., 2xn
(c) Statement 1 is true, statement 2 is true; statement 2 is not a
Q.15
Xi 2
(2x i )2
(d) 10 2
900
50
⇒K = 2 In 50
= 2 In 18.
Sol: 16 (b)
1
f´(x) =x +2b x + a
for extreme values f ′ (x) = 0
⇒ f´ (−1) = 0
⇒ a – 2b = 1
f´ (2) = 0
⇒ a + 4b = – 2 ⇒ a = 2, b = – 4
1
1
1
f´´(−1), f´´(2) are negative. f has local maxima at – 1, 2
Sol: 17 (c)
Required area
A=2
=5
Poornima University, For any query, contact us at :8875666617,18
2
0
3 y−
y 3/2 2 = 10
3/2
3
y
2
dy =2
23/2 − 0
=20 2
3
25 y
dy
0 2
4
AIEEE/2012/MATHS
Q.18
Q.19
Assuming the balls to be identical except for difference in colors’, the number
of ways in which one or more balls can be selected from 10 white, 9 green and
7 black balls is
(a) 880
(b) 629
(c) 630
(d) 879
If f: R → R is a function defined by f(x) = [x] cos
the greatest integer function, then f is
2X−1
2
π, where[X] denotes
Sol: 18 (d)
ways = (10 + 1)(9 + 1)(7 + 1) – 1 = 11 ×10 × 8 – 1 = 879.
Sol: 19 (a)
If the lines
(a) – 1
X−1
2
=
Y+1
(b)
3
2
9
=
Z−1
4
and
(c)
X−3
1
9
=
y−k
2
Z
= intersect, then k is equal to
1
(d) 0
2
2
π =[x]cos x −
1
2
π
=[x] sin π x is continuous for every real x.
(a) continuous for every real x
(b) discontinuous only at x = 0
(c) discontinuous only at non-zero integral values of x
(d) continuous only at x = 0
Q.20
2x−1
F(x) = [x]cos
Sol:20 (c)
x−1
2
=
x−3
1
y+1
3
=
y−k
=
2
z−1
4
=
z
1
a (1, -1, 1); r = a + λb
c (3, k, 0);
b (2, 3, 4)
r = c + μd
d (1, 2, 1)
These lines will intersect lines are coplanar
a - c, b , d are coplanar
∴ [a – c, b,d] = 0
⇒ 2(-5) – (k +1) (-2) – 1 (1) = 0
9
⇒k=2
Q.21
Three numbers are chosen at random without replacement from {1, 2, 3,
......8}. The probability that their minimum is 3, given that their maximum is 6,
is
(a)
3
(b)
8
1
Sol: 21 (b)
Let A be the event that maximum is 6.
B be event that minimum is 3
1
P(A) =
5
2
(c) 4
(d) 5
P(B) =
5C
2
8C
3
5C
2
8C
3
P(A∩B) =
Q.22
Z2
If z ≠ 1 and Z−1 is real, then the point represented by the complex number z
files
(the numbers < 6 are 5)
(the numbers < 3 are 5)
5C
2
8C
3
Required probability is P
B
A
=
P(A ∩B)
P(A )
=
Sol: 22 (a)
(a) either on the real axis or on a circle passing through the origin
(b) on a circle with centre at the origin
(c) either on the real axis or on a circle not passing through the
origin
Poornima University, For any query, contact us at :8875666617,18
Let z = x + iy (∵ x ≠ 1 as z ≠ 1)
z2 = (x2 – y2) + i(2xy)
z2
2−1
is real
⇒its imaginary part = 0
⇒ 2xy (x – 1) – y(x2 – y2) = 0
⇒ y(x2 + y2 – 2x) = 0
2C
1
5C
3
2
1
= 10 = 5.
5
AIEEE/2012/MATHS
⇒y = 0; x2 + y2 – 2x = 0
∴ z lies either on real axis or on a circle through origin.
(d) on the imaginary axis
Q.23
Q.24
Let P and Q be 3×3 matrices with P ≠Q. If P3 = Q3 and P2 Q = Q2 P, then
determinant of (P2 + Q2 ) is equal to
(a) –2
(b) 1
(c) 0
(d) –1
If g(x) =
g(x)
(a) g(π)
X
cos
0
Sol:23 (c)
P3 = Q3
P3 – P2Q = Q3 – Q2P
P2(P – Q) + Q2 (P – Q) = O
(P2 + Q2)(P – Q) = O ⇒ |P2 + Q2| = 0
Sol: 24 (b or d)
4t dt , then g(x + π) equals
g(x) =
(b) g(x) + g(π)
(c) g(x) – g(π)
(d) g(x) . g(π)
x
0
cos 4t dt
⇒g´(x) = cos 4x ⇒ g(x) =
⇒ g(x) =
sin 4×
4
sin 4×
4
+k
[∵g(0)=0]
g(x+π) = g(x) + g(π) = g(x) – g(π) (∵ g(π) = 0)
Q.25
The length of the diameter of the circle which touches the x-axis at the point
(1, 0) and passes through the point (2, 3) is
(a)
10
3
3
(b) 5
c
6
5
(d)
5
3
Sol: 25 (a)
Let (h, k) be centre.
(h – 1)2 + (k – 0)2 = k2 _ h = 1
5
(h – 2)2 + (k – 3)2 = k2 _ k =3
∴ diameter is 2k =
Q.26
Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can
be formed such that Y ⊆ X, Z⊆ X and Y ∩ Z is empty, is
10
3
Sol: 26 (b)
Every element has 3 options. Either set y, Set Z or None .
(a) 52
Ordered pairs = 35
(b) 35
(c) 25
(d) 53
Poornima University, For any query, contact us at :8875666617,18
6
AIEEE/2012/MATHS
Q.27
An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its
semiminor axis and a diameter of the circle x2 + (y – 2)2 = 4 as its semi-major
axis. If the centre of the ellipse is the origin and its axes are the coordinate
axes, then the equation of the ellipse is
(a) 4x2 + y2 = 4
(b) x2 + 4y2 = 8
(c) 4x2 + y2 = 8
(d) x2 +
Sol: 27 (d)
Semi minor axis b = 2
Semi major axis a = 4
x2
x2
⇒16 +
Y2
4
Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.
Statement 1: f´(4) = 0
Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and
f (2) = f (5).
Sol: 28 (b)
f(x) = 3, 2 ≤ x≤5
f’(x) = 0, 2 < x <5
f’(4) = 0
(a) Statement 1 is false, statement 2 is true
(b) Statement 1 is true, statement 2 is true; statement 2 is a
correct explanation for statement 1
(c) Statement 1 is true, statement 2 is true; statement 2 is not a
correct explanation for statement 1
(d) Statement 1 is true, statement 2 is false
Q.29
A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q
such that it forms a triangle OPQ, where O is the origin. If the area of the
triangle OPQ is least, then the slope of the line PQ is
1
(a) − 4
(b) −4
(c) −2
1
(d) − 2
continuous
2
Sol: 29 (c)
Equation of line passing through (1, 2) with slope m is y –
2 = m(x – 1)
Area of ∆OPQ =
∆=
m 2 +4−4m
2m
∆ is least if
Q.30
Let ABCD be a parallelogram such that AB = q , AD = p and ∠BAD be an
acute angle. If r is the vector that coincides with the altitude directed from the
vertex B to the side AD, then r is given by
(a) r = 3q –
3(p .q )
(p.p )
(b) r = – q +
(c) r = q +
=1
⇒ x2 + 4y2 = 16.
4y2 = 16
Q.28
Y2
Equation of ellipse = a 2 + b 2 = 1
p.p
p .q
p.p
(d) r = –3q –
p
p
3(p .q )
(p.p )
2
2|m |
m
∆=
2
=m
2
2
+m −2
⇒ m2 = 4 ⇒ m=+2
⇒ m=−2
Sol: 30 (b)
p
p .q
m
(m−2)2
p
Poornima University, For any query, contact us at :8875666617,18
AE = vector component of q on p
AE =
(p .q )
p
(p .q )
(p .q )p
⇒ q+r =
(p .q )
∴ From ∆ABE; AB +BE = AE
(p .q )
⇒ r=– q+(p.p ) p
5
7