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SAMPLE PAPER- 1
CLASS: XII
MATHEMATICS
Time Allowed: 3 hours
Max.Marks:100
General Instructions:
All questions are compulsory
1. This question paper consists of 26 questions divided into three sections A, B and C. Section A comprises
of 6 questions of one mark each, section B comprises of 13 questions of four marks each and section C
comprises of 07 questions of six marks each
2. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement
of the question
3. There is no overall choice. However, internal choice has been given.
4. Use of calculators is not permitted. You may ask for logarithmic tables, if required.
SECTION - A (1 mark questions)
2
 d2y 
 dy 
1. Write the order and degree of the differential equation  2   cos 
0
 dx 
 dx 
k 1 
 is singular
 2 4 
2. Find the value of k if the matrix 
3. Write integrating factor of -
𝑥





𝑑𝑦
− 𝑦 = 𝑠𝑖𝑛𝑥
𝑑𝑥



4. Find the projection of a  2 i  j  k on b  i  2 j  k
5. If
𝑎⃗ is a unit vector and (𝑥⃗ − 𝑎⃗). (𝑥⃗ + 𝑎⃗) = 24, find |𝑥⃗|.
6. Write the direction cosines of a line equally inclined to the three co-ordinate axes.
SECTION - B (4 marks questions)
7. Let A be a nonsingular square matrix of order 3 × 3. Then prove that |adj A|=|A|2.
OR
Using properties of determinants prove that
1
1
1  a 2  b2
2ab
2ab
1 a  b
2b
2a
2
2b
2
2a
1  a 2  b2
2

 2x  1
1  1  y 

cos


2 
2 
 1 x  2
 1  y 
8. Write in the simplest form tan  sin 
2
OR
Simplify
 1 3 2 


9. Find the inverse of the matrix  3 0 5  using elementary row transformation.
 2 5 0 

 1  a 2  b2

3






10. For any two vectors a and b prove the triangle inequality a  b  a  b
11. Differentiate: y 
12. Find
x  3x 2  4
3x 2  4 x  5
with respect to x
d2y
, when x  a(cos    sin  ), and , y  a(sin    cos  )
dx 2
13. Examine the following function for continuity.
 3x
14. Evaluate:
2
1
dx
 13 x  10
8
15. Evaluate:
 x  5 dx, by using properties of definite integrals
2

16. Prove that:
x
0 1  sin x dx  
OR
xe x
 ( x  1)2 dx
Evaluate
17. Find the shortest distance between the lines:
18. Using matrices solve the system 2 x  3 y  5 z  11; 3x  2 y  4 z  5; x  y  2 z  3
19. A company has two plants to manufacture machines. Plant A manufactures 70%and plant B
manufactures 30% machines. At plant A, 80% machines are rated of standard quality and at plant B,
90% machines are rated of standard quality. A machine is chosen at random and is found to be of
standard quality. What is the probability that it was manufactured by plant A ?
SECTION - C (6 mark questions)
20. Consider f : R   5,   given by f ( x)  9 x  6 x  5 . Show that f is invertible .find the inverse
2
of f.
OR
Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and P2 have same number
of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle
triangle T with sides 3, 4 and 5?
21. An open box with a square base is to be made out of a given quantity of cardboard of area c2 .Show that
the maximum volume of the box is
c3
6 3
22. Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x 
a
2
23. Two go downs A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to
3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of
transportation per quintal from the godowns to the shops are given in the following table:
Transportation cost per quintal (in Rs)
From/To
A
B
D
6
4
E
3
2
F
2.50
3
How should the supplies be transported in order that the transportation cost is minimum?
What is the minimum cost?
24. Solve the following differential equation: (x2 − y2) dx + 2 xydy = 0.
25. Find the equation of the plane which contains the line of intersection of the planes
 






r.( i  2 j  3 k )  4  0 and r.(2 i  j  k )  5  0 and which is perpendicular to the plane




r.(5 i  3 j  6 k )  8  0
26. An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there
will be at least 4 successes.
********
ANSWER KEY WITH MARKING SCHEME
To SAMPLE PAPER
CLASS XII
MATHEMATICS
Q.No
1
2
Value points
The highest order derivative present in the given differential equation is 2nd order derivative.
Therefore, its order is 2.
The given differential equation is not a polynomial equation in its derivatives. Hence, its degree
is not defined.
For singular matrix detA=0
k
1
2 4
3
0k 
1
2


Projection of a on b =
½M
½M
½M
Integrating factor = 1/x
4
Marks
½M
1M
 
a.b

b

2  2 1
5

1 4 1
6
1M
5
|𝑥⃗|.=5 units
6
1M
Here l=m=n
l2+m2+n2=1
1M.
So 3l2=1.
So l=m=n=±1/√3.
7
We know that,
1M
1M
1M
1M
OR
1  a 2  b2
2ab
2b
2ab
1 a  b
2b
2a
2
2
2a
1  a 2  b2
1  a 2  b2
2ab
0
1 a  b

2b
2
b(1  a  b )
2
2
1 a  b
2
1  a 2  b2
0
0
1 a  b

(C1  C1  bC3 )
2a
2 a
2
1M
2b
2
2
2
2
2
1
0
2b
 (1  a  b ) 0
1
2a
2
2 2
(C2  C2  aC3 )
2a
b(1  a  b ) a (1  a  b ) 1  a  b
2
1M
2
2
½M
2
taking (1  a 2  b 2 )
b a 1  a 2  b 2
as common factor from firs t an d sec ond columns
1
0
0
 (1  a  b ) 0
1
2a
2
2 2
b a 1  a  b
2
1
0
0
 (1  a  b ) 0
1
0
2
2 2
(C3  C3  2bC1 )
2
½M
(C3  C3  2aC2 )
b a 1  a 2  b 2
1
0
0
 (1  a  b ) 0
1
0  (1  a 2  b 2 )3
2
2 3
½M
½M
b a 1
8
2
1

 2x  1
1  1  y 
Take y  tan  sin 1 

cos

2 
2 
 1 x  2
 1  y 
2
 2x 
sin 1 
 2 tan 1 x
2 
1

x


 1 y2 
cos 1 
 2 tan 1 y
2 
 1 y 
1
1

then, y  tan  2 tan 1 x  2 tan 1 y 
2
2





 x  y   x  y 
 tan  tan 1 
  

 1  xy    1  xy 

OR

1M
1M
1M
1M
2M
2M
9
We write A=IA
 1 3 2  1 0 0 
 3 0 5   0 1 0  A

 

 2 5 0  0 0 1 
1 3 2  1 0 0 
  0 9 11   3 1 0  A by R2  R2  3R1
 
 2 5 0  0 0 1 
1 3 2   1 0 0 
 0 9 11   3 1 0  Aby R3  R3  2 R1
0 1 4   2 0 1 
1M
=
1M
1 0 10   5 0 3 
 0 9 11   3 1 0  A by R1  R1  3R3
0 1 4   2 0 1 
1 0 10   5 0 3
 0 1 21   13 1 8 A by R2  R2  8 R3
0 1 4   2 0 1
0 10
−5 0 3
1 21] = [−13 1 8] 𝐴 𝑅3 → 𝑅3 + 𝑅2
0 25
−15 1 9
−5
0
3
10
−13 1
8
21] = [ 3 1
9 ] 𝐴 𝑏𝑦 𝑅3 → 𝑅3 /25
−
1
5 25 25
−2 −3
1
5
5
1 0 0
−2 4 11
⇒ [0 1 0] =
𝐴 𝑏𝑦 𝑅1 → 𝑅1 − 10𝑅3 , 𝑎𝑛𝑑 𝑅2 → 𝑅2 − 21𝑅3
5
25 25
0 0 1
3 1
9
−
[ 5 25 25 ]
−2 −3
1
5
5
−2
4
11
⇒ 𝐴−1 =
5
25 25
3 1
9
[− 5 25 25 ]
1M
1
⇒ [0
0
1 0
⇒ [0 1
0 0
1M
10. The inequality holds trivially in case either
1M
1M
1M
1M
11
Taking log on both sides we get


1
log( x  3)  log( x 2  4)  log(3 x 2  4 x  5)
2
differentiating , w.r.t.x, we get
log y 
1 dy 1  1   2 x   6 x  4  
 



y dx 2  x  3   x 2  4   3x 2  4 x  5  
dy 1
hence, 
dx 2
12
 x  3  x 2  4  
1   2 x   6 x  4 




2
3 x  4 x  5  x  3   x 2  4   3x 2  4 x  5  
dx
 a cos 
d
dy
 a sin 
d
dy dy dx


dx d d
 tan 
d 2 y d  dy  d
     tan  
dx 2 dx  dx  dx
13
1½M
1½M
1M
1M
1M
1½ M
½M
d
d sec3 

 tan   
d
dx
a
The given function is
½M
This function f is defined at all points of the real line.
Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5
Case I: c < 5
Then, f (c) = 5 − c
½M
1M
Therefore, f is continuous at all real numbers less than 5.
Case II : c = 5
Then,
1M
Therefore, f is continuous at x = 5
Case III: c > 5
1M
Therefore, f is continuous at all real numbers greater than 5.
Hence, f is continuous at every real number and therefore, it is a continuous
function.
14
1
dx
 13 x  10
1
 3x 2  13x  10dx
1
1
 
dx
3 x 2  13 x  10
3
3
I=
 3x
2
½M
1½ M

1
1
dx
2
2

3 
13   17 
 x    
3  6


13 17
x 

1
1
6 6
 
log
13 17
3  2  17
x 
6
6 6


15
8






1M
1
3x  2
log
c
17
3 x  15
5
8
2
5
x  5 dx   x  5 dx   x  5 dx
2
1M
5
1½ M
8
    x  5 dx    x  5 dx
2
1½ M
5
5
8
 x2
  x2

   5x     5x 
 2
2  2
5
9
1M

16
I=
x
 1  sin x dx
0


 x
 1  sin   x  dx
1M
0


 x
 1  sin x dx
1M
0

II 
=

 1  sin x dx
0

1  sin x
dx
cos 2 x
0



1M

   sec x  sec x tan x dx
2
0
1M
 2
 I 
Hence proved
OR
 x  1  1 e x
xe x
dx
 ( x  1)2 dx =  ( x  1)2
1½ M
 1
1 
  ex 

dx which is of the form
x  1  x  12 


1½ M
x
 e ( f  f ')dx
1M
 1 
c
 x 1 
Hence given integral = e x f ( x)  c  e x 
17
Sol. The equations of the given lines are
It is known that the shortest distance between the lines,
and
given
1½M
by,
...(1)
1½M
Comparing the given equations, we obtain
1M
Substituting all the values in equation (1), we obtain
18
1M
1M
Therefore, the shortest distance between the two lines is
units.
 1 3 2   x   11 
 3 0 5   y    5
   
Corresponding matrix equation is 
 2 5 0   z   3
1M
AX  B
 X  A1B
1
Where A 
1
adjA
A
A  1
0 1 2 
adjA   2 9 23
1 5 13 
1M
 0 1 2 
1 
A 
2 9 23

1
 1 5 13 
 x
0 1 2  1  1 
 y   1  2 9 23  2   2
  1 
   
 z 
1 5 13   3  3
x  1; y  2; z  3 is the solution
1
19
Let E: machine is of standard quality.
7
3
; P( B) 
10
10
8
9
P( E / A)  ; P( E / B) 
10
10
P( A).P( E / A)
P( A / E ) 

P( A).P( E / A)  P( B).P( E / B)
P( A) 
20
56
83
for x1 , x2  R
f ( x1 )  f ( x2 )  9( x12  x2 2 )  6( x1  x2 )  0
 ( x1  x2 ) 9( x1  x2 )  6  0
As 9( x1  x2 )  6  0 therefore ( x1  x2 )  0
1M
1M
1M
1M
1M
1M
 x1  x2  f is 1  1
y   5,   there exists x  R , such that
Let for
y  9 x2  6 x  5  x 
 x
2½ M
1  6  y
3
6  y 1
 R
hence onto.
3
1M
Therefore the function is one and onto showing that it is invertible
f 1 ( y) 
f 1 ( x) 
6  y 1
3
½M
6  x 1
3
OR
1M
Sol. R = {(P1, P2): P1 and P2 have same the number of sides}
R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with itself.
1M
1M
Let (P1, P2) ∈ R.
⇒ P1 and P2 have the same number of sides.
⇒ P2 and P1 have the same number of sides.
⇒ (P2, P1) ∈ R
1M
∴ R is symmetric.
Now,
1M
Let (P1, P2), (P2, P3) ∈ R.
⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.
⇒ P1 and P3 have the same number of sides.
⇒ (P1, P3) ∈ R
1M
∴ R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons
which have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
21
Let x be the side of the square base and y be the height of cuboid.
S = c2
1M
x 2  4 xy  c 2
y
V
c 2  4 xy
4x
1M
1M
1 2
c x  x 3 
4
dV 1 2
 c  3 x 2 
dx 4 
c
x
is the critical po int
3
d 2V
dx 2

x
c
3
Therefore,
1M
1
3c
0
 6 x  c 
4
2 3
x
3
V is maximum when x 
c
3
1 2 c
c3  c 2
c
.



4
3 3 3 6 3
The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line,
Vmax 
22
1M
, is the area ABCDA.
1M
2M
½M
It can be observed that the area ABCD is symmetrical about x-axis.
Area ABCD = 2 × Area ABC
1M
1M
1M
½M
Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line,
, is
units.
23
Let godown A supply x and y quintals of grain to the shops D and E respectively.
Then, (100 − x − y) will be supplied to shop F.
The requirement at shop D is 60 quintals since x quintals are transported from
godown A. Therefore, the remaining (60 −x) quintals will be transported from
godown B.
Similarly, (50 − y) quintals and 40 − (100 − x − y) = (x + y − 60) quintals will be
transported from godown B to shop E and F respectively.
The given problem can be represented diagrammatically as follows.
1½ M
1M
Total transportation cost z is given by,
½M
The given problem can be formulated as
Minimize z = 2.5x + 1.5y + 410 … (1)
subject to the constraints,
1½ M
The feasible region determined by the system of constraints is as follows.
The corner points are A (60, 0), B (60, 40), C (50, 50), and D (10, 50).The values
of z at these corner points are as follows.
24
Corner point
z = 2.5x + 1.5y + 410
A (60, 0)
560
B (60, 40)
620
C (50, 50)
610
D (10, 50)
510
1M
½M
→
Minimum
The minimum value of z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50
quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0
quintals, and 0 quintals respectively.
1M
The minimum cost is Rs 510.
Sol. Given that y =1 when x = 1 and
It is a homogeneous differential equation.
1M
1M
Substituting (2) and (3) in (1), we get:
1M
1M
Integrating both sides, we get:
1M
It is given that when x = 1, y = 1
(1)2+ (1)2 = C (1)
⇒C=2
Thus, the required equation is y2 + x2 = 2x.
25
Given lines are
x  3 y 1 z  5


;
3
1
5
x 1 y  2 z  5


1
2
5
If the lines are coplanar then
1M
3  1 1  2 5  5
3
1
5
1
2
5
1½ M
0
 10  10  0,true
1M
Hence the lines are coplanar.
Equation of the plane containing the lines is
1½ M
x  3 y 1 z  5
3
1
5
1
2
5
0
  x  3 5   y 1 10   z  5 5  0
1M
 x  2 y  z  0 is the required equation to the plane
26
The probability of success is twice the probability of failure.
Let the probability of failure be x.
1. ∴ Probability of success = 2x
1M
1M
Let p =
and q =
Let X be the random variable that represents the number of successes in six trials.
By binomial distribution, we obtain
P (X = x) =
Probability of at least 4 successes = P (X ≥ 4)
= P (X = 4) + P (X = 5) + P (X = 6)
1M
1M
1M
1M