CBSE Class 12

Mathematics
Previous Year Question Paper 2020
Series: HMJ/2 Set- 2
Code no.65/2/2

• Please check that this question paper contains 15 printed pages.

• Code number given on the right hand side of the question paper should be
written on the title page of the answer-book by the candidate.
• Please check that this question paper contains 36 questions.
• Please write down the Serial Number of the question before attempting it.
• 15 minutes of time has been allotted to read this question paper. The
question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30
a.m., the students will read the question paper only and will not write any
answer on the answer script during this period.

MATHEMATICS

Time Allowed: 3 hours Maximum Marks: 80

General Instructions:
Read the following instructions very carefully and strictly follow them:
(i) This question paper comprises four Sections A, B, C and D. This question
paper carries 36 questions. All questions are compulsory.
(ii) Section A – Questions no. 1 to 20 comprises of 20 questions of 1 mark
each.
(iii) Section B – Questions no. 21 to 26 comprises of 6 questions of 2 marks
each.
(iv) Section C – Questions no. 27 to 32 comprises of 6 questions of 4 marks
each.
(v) Section D – Questions no. 33 to 36 comprises of 4 questions of 6 marks each.
(vi) There is no overall choice in the question paper. However, an internal choice
has been provided in 3 questions of one mark, 2 questions of two marks,
2 questions of four marks and 2 questions of six marks. Only one of the choices
in such questions have to be attempted.
(vii) In addition to this, separate instructions are given with each section and
question, wherever necessary.
(viii) Use of calculators is not permitted.

SECTION - A
Question numbers 1 to 20 carry 1 mark each.
Question numbers 1 to 10 are multiple choice type questions. Select the
correct option.
1. The area of a triangle formed by vertices O, A and B, where
 
   and OB=-3i-2j+k
OA=i+2j+3k    is 1 Mark

(A) 3 5 sq. units

(B) 5 5 sq. units

(C) 6 5 sq. units

(D) 4 sq. units
 
 j+3k and OB=-3i-2
Ans: Given, OA=i+2  j+k .

1  
We know, are of a triangle if it’s vectors are given is, A×B .
2

 j+3k
Therefore, here, OA=i+2

 j+k
And, OB=-3i-2
i j k
1   1
Therefore, OA×OB = 1 2 3
2 2
-3 -2 1

1   1 
OA×OB = i {( 2 )(1) - ( -2 )( 3)} -j {(1)(1) - ( -3)( 3)} +k {(1)( -2 ) - ( -3)( 2 )}
2 2
   1
⇒ OA×OB = × i {2+6} -j{1+9} +k {-2+6}
2
  1
 j+4k
⇒ OA×OB = × 8i-10
2
  1
⇒ OA×OB = 64 + 100 + 16
2
  1
⇒ OA×OB = 180
2
  1
⇒ OA×OB = × 6 5
2
 
⇒ OA×OB =3 5

Therefore, the area is 3 5 sq. units.

Thus, the correct answer is A.

 2 
2. If cos  sin -1 +cos -1x  =0 , then x is equal to 1 Mark
 5 
1
(A)
5
2
(B) -
5
2
(C)
5
(D) 1

-1 2 
Ans: Given, cos  sin +cos -1x  =0
 5 
−1
Taking cos on both sides, we get,
2
⇒ sin -1 +cos -1x=cos -1 ( 0 )
5
 2 π
⇒ sin -1 +cos -1x=cos -1x=
5 2

π
We know, cos -1x= -sin -1x , hence,
2
2 π π
⇒ sin -1 + -sin -1x=
5 2 2

 2 
⇒ sin -1x=sin -1  
 5
2
⇒ x=
5
Thus, the correct answer is C.

3. The interval in which the function f given by f ( x ) =x 2e-x is strictly

increasing, is 1 Mark
(A) ( −∞, ∞ )

(B) ( −∞,0 )

(C) ( 2,∞ )

(D) (0,2)
Ans: Given, f ( x ) =x 2e-x

Now, differentiating both sides with respect to x, we get,

⇒ f' ( x ) =2xe − x -x 2e-x

⇒ f' ( x ) =xe-x ( 2-x )

For, the function to be increasing,

f' ( x ) >0

⇒ xe − x ( 2-x ) >0

⇒ x ( 2-x ) >0

⇒ x ( 2-x ) <0
 -x
[Since, e can never be zero]
Using, the method of intervals, we get,

Since, x ( x-2 ) <0 , we will take the negative region.

Therefore, x ∈ ( 0,2 ) .

The correct option is D.

x-1
4. The function f ( x ) = is discontinuous at 1 Mark
x ( x 2 -1)

(A) exactly one point

(B) exactly two points
(C) exactly three points
(D) no point
x-1
Ans: Given, f ( x ) =
x ( x 2 -1)

We can write the function as,

x-1
⇒ f (x)=
x ( x-1)( x-1)
Here, the function is discontinuous if,
x ( x-1)( x-1) = 0
x=0 or x-1=0 or x+1=0
x=0 or x=1 or x=-1
Therefore, the function is discontinuous exactly at three points.
The correct option is C.
5. The function f:R → [ -1,1] defined by f ( x ) =cosx is 1 Mark

(A) both one-one and onto

(B) not one-one, but onto
(C) one-one, but not onto
(D) neither one-one, nor onto
Ans: Given, f:R → [ -1,1] defined by f ( x ) =cosx .

Let, f ( x1 ) = f ( x 2 )

⇒ cos x1 =
cos x 2
⇒ x1 = 2nπ ± x 2 ,n ∈
Therefore, the above equations have infinitely many solutions.
Hence, it is not a one-one function.
Also, range of cos x is [-1, 1], which is a subset of co-domain .
Hence, the function is also not onto.
Therefore, the function is neither one-one nor onto.
Thus, the correct option is D.

6. The coordinates of the foot of the perpendicular drawn from the point (2,
-3, 4) on the y-axis is 1 Mark
(A) (2, 3, 4)
(B) (-2, -3, -4)
(C) (0, -3, 0)
(D) (2, 0, 4)
Ans: Given point is P(2, -3, 4).
Any point on y-axis is given by Q(0, k, 0), where k is any real number
So direction ratio of PQ are -2,-3,-k,4.
We know direction ratio of y-axis is given by 0,1,0.
Now since PQ ⊥ y-axis
⇒ (0)(2)+(1)(-3-k)+(0)(4)=0
⇒ k=-3
Hence, coordinate of foot of perpendicular is Q(0, -3, 0).
Thus, the correct option is C.

7. The relation R in the set {1,2,3} given by R={(1,2),(2,1),(1,1)} is 1 Mark

(A) symmetric and transitive, but not reflexive
(B) reflexive and symmetric, but not transitive
(C) symmetric, but neither reflexive nor transitive
(D) an equivalence relation
Ans: The relation is not reflexive because (2,2),(3,3) are not present.
It is symmetric because, (1,2) ∈ R and also (2,1) ∈ R, which satisfies the
condition for a relation to be symmetric perfectly.
And, also, it is transitive because, (1,2) ∈ R, (2,1) ∈ R and also (1,1) ∈ R,
which satisfies the condition for a relation to be transitive perfectly.
Hence, the relation is symmetric and transitive but not reflexive.
Thus, the correct option is A.

8. The angle between the vectors i − j and j-k is 1 Mark

π
(A) -
3
(B) 0
π
(C)
3
2π
(D)
3
Ans: Given vectors are

 j
a=i-


b=i-k

( )( )
So, a.b= i-j . i-k

a.b= (1 × 0 ) + ( −1 × 1) + ( 0 × ( −1) )

⇒ a.b= − 1

Also, a = 12 + ( −1) + 0 = 2
2


b = 0 + 12 =2
  
We also know, a.b= a . b .cosθ

⇒ -1= 2. 2.cosθ
−1
⇒ cosθ=
2
2π
⇒ θ=
3
2π
The angle between the vectors is .
3
Thus, the correct option is D.

9. If A is a non-singular square matrix of order 3 such that A2=3A , then

value of |A| is 1 Mark
(A) -3
(B) 3
(C) 9
(D) 27
Ans: Given, A2=3A
Taking determinant on both sides,
⇒ A 2 = 3A

⇒ A 2 =33 A

⇒ A 2 =27 A

⇒ A =27
Therefore, the correct option is D.

 
10. If a = 4 and -3 ≤ λ ≤ 2 , then λa lies in 1 Mark

(A) [0,12]
(B) [2,3]
(C) [8,12]
(D) [-12,8]
Ans: The maximum value of λ is 2.
 
So, λa = λ . a

⇒ λa = 2.4 = 8

The minimum value of λ is -3.

 
So, λa = λ . a

⇒ λa =
−3 .4

⇒ λa = 3.4 =12

So, there are no value of λ which is negative.

For, λ =0, we get,
 
λa = λ . a

⇒ λa = 0.4 = 0

Therefore, the smallest value of λa is 0 .

Therefore, λa lies in [0, 12].

Thus, the correct option is A.

Fill in the blanks in question numbers 11 to 15.

11. If the radius of the circle is increasing at the rate of 0.5 cm/s, then the
rate of increase of its circumference is _____________ . 1 Mark
Ans: Let r be the radius and C the circumference of the circle.
Then, C=2πr
dr
It is given that =0.5 cm/s
dt
Now, C=2πr
Differentiating both sides w.r.t t, we get,
dr dr
⇒ =2π.
dt dt
dC
⇒ =2π.05
dt
dC
⇒ =π cm/s.
dt
Therefore, the rate in increase of the circumference is π cm/s.

2x -9 -4 8
12. If = , then value of x is ____________ . 1 Mark
-2 x 1 -2

2x -9 -4 8
Ans: =
-2 x 1 -2

(2x)(x)-(-9)(-2)=(-4)(-2)-(8)(1)
2x2-18=8-8
2x2-18=0
2x2-18=0
2x2=18
x2=9
x= ± 3
Therefore, the value of x is ± 3.
13. The corner points of the feasible region of an LPP are (0,0), (0,8), (2,7),
(5,4) and (6,0). The maximum profit P=3x+2y occurs at the point ______.
1 Mark
Ans: P( 0,0 ) =3(0)+2=0

P( 0,8) =3(0) +2(8)=16

P( 2,7 ) =3(2)+2(7)=20

P( 5,4 ) =3(5)+2(4)=23

P( 6,0 ) =3(6)+2(0)=18

The maximum value is at (5,4).

14. The range of the principal value branch of the function y-sec-1x is
____________ . 1 Mark

Ans: We know, sec-1x ∈ [ 0,π ] −  π

 .
2

Therefore, the range of the principal value branch of the function y-sec-1x is
π .
[ 0,π ] − 
2
Or

The principal value of cos -1  - 1  is ___________ . 1 Mark

 2

Ans: cos -1  - 1 
 2
2π 1
We know, cos -1 =-
3 2
 2π 
= cos −1  cos 
 3 

2π
=
3
 2π
Thus, the principle value of cos −1  − 1  is = .
 2 3

15. The distance between parallel planes 2x+y-2z-6=0 and 4x+2y-4z is

___________ units. 1 Mark
Ans: Given, 2x+y-2z-6=0 --------(1)
4x+2y-4z -------(2)
Multiplying (1) by 2, we get,
4x+2y-4z-12=0
4x+2y-4z=12
Therefore, we can write,
c1=12, c2=0
And, a=4, b=2, c=-4.
Therefore, the distance between the parallel lines is,
c 2 -c1
a 2 +b 2 +c 2

0-12
=
( 4 ) + ( 2 ) + ( -4 )
2 2 2

12
=
16 + 4 + 16

12
=
36
12
=
6

=2 units
Or
If P(1,0,-3) is the foot of the perpendicular from the origin to the plane,
then the Cartesian equation of the plane is _____________ . 1 Mark
Ans: The given foot of the perpendicular is P(1,0,-3).
The direction coefficients of the perpendicular are (1-0,0-0,-3,-0)
=(1,0,-3) .
Therefore, the equation of the plane is
a ( x-x1 ) +b ( y-y1 ) +c ( z-z1 ) =0

⇒ 1( x-1) +b ( y-0 ) -3 ( z- ( -3) ) =0

⇒ x-1-3 ( z+3) =0
⇒ x-1-3z-9=0
⇒ x-3z-10=0
Therefore, the equation of the plane is x-3z-10=0 .

Question numbers 16 to 20 are very short answer type questions.

π
2
16. Evaluate : ∫ xcos 2 xdx 1 Mark
π
2

Ans: Here, f ( x ) =xcos 2 x

Now, f ( -x ) = ( -x ) cos 2 ( -x )

⇒ f ( -x ) =-xcos 2 x =
-f ( -x )

Therefore, it is an odd function.

π
2

∫ xcos xdx=0 .
2
So,
π
-
2

x-1 y+4 z+4

17. Find the coordinates of the point where the line = = cuts the
3 7 2
xy-plane. 1 Mark
x-1 y+4 z+4
Ans: If the line = = cuts the XY plane.
3 7 2
Then, z=0.
So, let coordinates of point be (x,y,0).
x-1 y+4 z+4
Now, = = =k
3 7 2
Thus, x=3k+1, y=7k-4, z=2k-4
Since, z=0
2k-4=0
2k=4
k=2
Now, x=3(2)+1=7
y=7(2)-4=10
Therefore, the point is (7,10,0).

kx 2 +5 if x ≤ 1
18. Find the value of k, so that the function f ( x ) =  is
2 if x > 1
continuous at x=1 . 1 Mark
Ans: For, x>1, the value of f(x), such that,

f ( x ) x →1+ =2

For, , the value of f(x), such that,

f ( x ) x →1+ =k (1) + 5
2

f ( x ) x →1+ =k+5

For the function to be continuous

f ( x ) x →1+ =f ( x ) x →1−

2=k+5
k=-3

19. Find the integrating factor of the differential equation

 dy
x =2x 2 +y
dx
1 Mark
dy
Ans: Given, x =2x 2 +y
dx
Dividing both sides by x, we get,
dy y
=2x+
dx x
dy y
- =2x
dx x
1
Therefore, P ( x ) =
x
1
Thus, integrating factor, IF = e ∫ x
= dx

1
=e ∫
- = dx
x

=e-logx
-1
=elogx
=x -1
1
=
x
1
Thus, the integrating factor is .
x

20. Differentiate =sec 2 ( x 2 ) with respect to x 2 . 1

Mark
Ans: We need to find

(
d sec 2 ( x 2 ) )
dx 2
Let x2=t
 d ( sec 2 ( t ) )
So,
dt
=2.sec t.( sec t ) '
=2.sec t.sec t.tan t

=2.sec 2 t.tan t
Putting t=x2
=2.sec 2 x 2 .tan x 2
Or
dy
If y=f ( x 2 ) and f' ( x ) =e x , then find . 1 Mark
dx
Ans: Given, y=f ( x 2 )

Differentiating both sides w.r.t x, we get,

dy
= y=f' ( x 2 ) .2x
dx

Also, given, f' ( x ) =e x .

f' ( x 2 ) =e x2
= ex

dy
∴ 2xe x .
=
dx

SECTION - B
Question numbers 21 to 26 carry 2 marks each.
21. Find a vector r equally inclined to the three axes and whose magnitude
is 3 3 units. 2 Marks

Ans: We have r =3 3

Since,
 r is equally inclined to the three axes, direction cosines of the unit vector
r will be same.
i.e., l=m=n
Now, we know that,
l2 +m 2 +n 2 =1
⇒ l2 +l2 +l2 =1
⇒ 3l2 =1
1
⇒ l2 =
3
1
⇒ l= ±
3
1  1  1 
So, r=± i± j± k
3 3 3
 
∴ r =r .r

 1  1  1 
=2 3  ± i± j± k
 3 3 3 

= ± 2 i + j + k 
 
Or

Find the angle between unit vectors a and b so that 3a-b is also a unit
vector. 2 Marks

Ans: a and b are unit vectors and 3a-b is also unit vector
To find: Angle between a and b
Suppose angle between a and b is θ .
  
a.b= a . b .cosθ (Dot product of two vectors)

a.b=cosθ
   
As a and b are unit vector so, a = b =1 .
  
3a-b is also unit vector i.e. 3a-b = 1

Squaring both sides, we get,

  2
( 3a-b = 1)
 2 2 
( 3)
2
a + b -2. 3. a.b =1

⇒ 3.1 + 1 − 2. 3.cosθ=1

[Since, a.b=cosθ ]
⇒ 4-2 3.cosθ=1
⇒ 2 3.cosθ=3

3
⇒ cosθ=
2
π
⇒ θ=
6
π
Therefore, the angle between the two unit vectors is .
6

-3 2  1 0 
22. If A=   and I=   , find scalar k so that A2+I=kA. 2 Marks
 1 -1   0 1 
-3 2  1 0 
Ans: A=   , I= 0 1 
 1 -1  
A2+I=kA
A2=A×A

( −3)( −3) + ( 2 )(1) ( −3)( 2 ) + ( 2 )(1)

⇒ A2 =
 
 1( −3) + 1( −1) 1( 2 ) + ( −1)( −1) 

 9 + 2 −6 − 2 
⇒ A2 =
 −3 − 1 2 + 1 
 
 11 −8
⇒ A2 =
 −4 3 
 
A2+I=kA
  11 −8 1 0   −3 2 
⇒ +
   k
=  1 −1
 −4 3  0 1   
 11 + 1 −8 + 0   −3k 2k 
⇒ = 
 −4 + 0 3 + 1   1k −1k 
12 −8  −3k 2k 
⇒ = 
 −4 4   1k −1k 
Therefore, comparing the terms on both sides, we get,
k=-4.

secx-1 π
23. If f ( x ) = , find f'   . 2 Marks
secx+1 3
secx-1
Ans: Given, f ( x ) =
secx+1
1
Using, secx= .
cosx
1
−1
f (x)= cosx
1
+1
cosx

1-cosx
f (x)=
1+cosx

x
2sin 2
f (x)= 2
x
2sin 2
2
x x
Since, 1-cosx=2sin 2 and 1+cosx=2cos 2 . So, we get,
2 2
x
sin
f (x)= 2 =tan x
x 2
cos
2
Differentiating both sides w.r.t x, we get,
d  x
f' ( x ) =  tan 
dx  2
x 1
f' ( x ) =sec 2 .
2 2
1 x
f' ( x ) = sec 2
2 2
π 1  π 
Therefore, f'   = sec 2  
3 2  2.3 
π 1 π
⇒ f'   = sec 2  
3 2 6
2
 π 1 2 
⇒ f'   =  
 3  2 3 
π 2
⇒ f'   =
3 3
Or

Find f' ( x ) if f ( x ) = ( tanx )

tanx
. 2 Marks

Ans: Given, f ( x ) = ( tanx )

tanx

Let, f ( x ) = ( tanx )
tanx
=y
Taking log on both sides, we get,
logy=log ( tanx tanx )

⇒ logy=tanx log ( tanx )

Differentiating both sides w.r.t x , we get,

1 dy 1
⇒ . =tanx. .sec 2 x+log ( tanx ) .sec 2 x
y dx tanx
1 dy
⇒ . =sec 2 x+sec 2 xlog ( tanx )
y dx
 dy
⇒ =y sec 2 x + sec 2 x log ( tan x ) 
dx
dy
⇒ =ytanx tan x sec 2 x 1 + log ( tan x ) 
dx
Since, y=tanx tanx .

Therefore, f' ( x ) =ytanx tanx sec 2 x 1+log ( tanx )  .

tan 3 x
24. Find the value of integral: ∫ dx 2
cos 3 x
Marks
tan 3 x
Ans: I = ∫ dx
cos3 x
sin 3 x
⇒ I= ∫ dx
cos3 x.cos3 x
sin 3 x.sinx
⇒ I= ∫ dx
cos 6 x

⇒ I= ∫
(1 − cos x ) sinx dx
2

cos 6 x
sinx sinx
⇒ I= ∫ 6
dx-∫ dx
cos x cos 4 x
Now, let, cosx=t
-sinxdx=dt
sinxdx=-dt
1 1
Therefore, I= ∫ - 6
dt+ ∫ 4 dt
t t
1 1
⇒ I=-∫ 6
dt+ ∫ 4 dt
t t
 t -6+1   t -4+1 
⇒ I=-  +  +c
 -6+1   -4+1 
  t −5   t −3 
⇒ I=-   +   +c
 -5   -3 
1 1
⇒ I= - +c
5t 5 3t 3
Substituting t=cosx, we get,
1 1
⇒ I= - +c
5cos x 3cos3 x
5

tan 3 x 1 1
Therefore, ∫ dx= - +c .
cos3 x 5cos5 x 3cos3 x

x-5
25. Show that the plane x-5y-2z=1 contains the line =y=2-z . 2 Marks
3

Ans: Given:

Plane: x-5y-2z=1

In vector form, we can write the equation of plane as,


(
r. i-5j-2k
  =1
)

(
Direction ratio of the plane P= i-5j-2k
 
)
x-5
Line: =y=2-z
3

In vector form, we can write the equation of line as,


(
  +λ 3i+
r= 5i+2k ) (
 j-k
)
Direction ratio of the plane

(
 j-k
L= 3i+ )
 
Now, p.L= i-5j-2k(
  . 3i+
)(
 j-k
)
 
p.L= (1)( 3) + ( -5 )(1) + ( -2 )( -1)
 
⇒ p.L = 3 − 5 + 2 = 0

Hence this given plane contain the given line.

26. A fair dice is thrown two times. Find the probability distribution of the
number of sixes. Also determine the mean of the number of sixes. 2 Marks

Ans: The dice is thrown twice.

Therefore, the sample space is

Therefore, no. of sample with 0 sixes =25

No. of sample with 1 sixes =10

no. of sample with 2 sixes =1

X 0 Sixes 1 Six 2 Sixes
p(X) 25 10 1

Now, Mean = ∑ X.P ( X )

25 10 1
=0× +1× +2×
36 36 36

10 2
= +
36 36

12
=
36
 1
=
3

SECTION - C

Question numbers 27 to 32 carry 4 marks each.

27. Solve the following differential equation:

 yx  y
x
y
 1-e  dy+e  1-  dx=0 ( x ≠ 0 ) . 4 Marks
   x

 xy  y
x 
y
Ans: Given, 1-e  dy+e 1-  dx=0
   x

 xy  y
x 
y
⇒ 1-e  dy=-e 1-  dx
   x

y
 y
-e 1- 
x
dy  x −−− 1
⇒ = ()
dx  xy 
1-e 
 

dy
Now, let, =F ( x,y )
dx
y
 y
-e x 1- 
dy  x
∴ =F ( x,y ) =
dx  xy 
1-e 
 
λy
 λy 
-e 1- 
λx

Now, F ( λx,λy ) =  λx 
 λxλy

1-e 
 
 λy
 λy 
-e 1- 
λx

F ( λx,λy ) =  λx  F x,y
= ( )
 λxλy

1-e 
 

So, F ( λx,λy ) =F ( x,y ) =λ o F ( x,y )

Thus, F(x, y) is a homogeneous function.

Therefore, the given differential equation is a homogeneous differential

equation.

Now, let, y=xv

dy dv
⇒ =v+x.
dx dx

Now, substituting these values in (1), we get,

y
 y
-e 1-  x
dy  x
⇒ =
dx  xy 
1-e 
 

dy -e (1-v )
v
⇒ v+x. =
dx (1+e v )

dy -e (1-v )
v
⇒ x. = −v
dx (1+e v )

dy -e v +ve v
⇒ x. = -v
dx (1+e v )

dy -e +ve -v (1+e )
v v v

x. = -v
dx (1+ev )
 dy -e v +ve v -v-ve v
⇒ x. = -v
dx (1+ev )
dy -e v -v
⇒ x. =
dx (1+e v )

Now, by method of substitution of differential equation, we get,

1+e v dx
⇒ v
dv=-
v+e x

Now, integrating both sides,

1+e v dx
⇒∫ dv=- ∫x
v+e v

1+e v
⇒∫ dv=-log x +logc
v+e v

Now, putting v+e v = t

⇒ (1+e v ) dv=dt

Thus, our equation becomes,

dt
⇒∫ =-log x +logc
t

⇒ log t =-log x +logc

Putting back, t = v+e v , we get,

⇒ log v+e v =-log x +logc

⇒ log v+e v +log x =logc

⇒ log ( v+e v ) . x =logc

⇒ log ( v+e v ) .x =logc

⇒ log vx+e v x =logc

⇒ vx+e v x=c

y
Putting back, ⇒ y=vx ⇒ v= , we get,
x
y
⇒ y+e x=c
x

28. A cottage industry manufactures pedestal lamps and wooden shades.

Both the products require machine time as well as craftsman time in the
making. The number of hour(s) required for producing unit of each and
the corresponding profit is given in the following table:
Craftsman
Item Machine Time Profit(in ₹)
Time
Pedestal lamp 1.5 hours 3 hours 30
Wooden shades 3hours 1 hours 20

In a day, the factory has availability of not more than 42 hours of machine
time and 24 hours of craftsman time. Assuming that all items
manufactured are sold, how should the manufacturer schedule his daily
production in order to maximise the profit? Formulate it as an LPP and
solve it graphically. 4 Marks

Ans: Let number of pedestal lamps =x

Number of wooden shades =y
Maximize Profit: P=30x+20y
According to the question:
1.5x+3y ≤ 42
x y
⇒ + ≤1
28 14
Therefore, the intercepts will be (28, 0), (0, 14).
3x+y ≤ 24
x y
⇒ + ≤1
8 24
Therefore, the intercepts will be ( 8,0 ) , ( 0,24 ) .
x ≥ 0, y ≥ 0

Check profit at Corner points

At C(0,0),
P30(0)+20(0)=0
At B(0. 14),
P=30(0)+20(14)=280
At F(4,12),
P=30(4)+20(12)=360 [Max]
At D(8,0),
P=30(8)+20(0)=240
Maximum profit = Rs 360 at (number of pedestal lamps) x=4 and (Number of
wooden shades) y=12.
π
2
29. Evaluate the value of integral: ∫ sin2x tan -1 ( sinx ) dx 4 Marks
0
 π
2
Ans: Given, ∫ sin2x tan -1 ( sinx ) dx
0

π
2
= ∫ sin2x cosx tan -1 ( sinx ) dx
0

Let, sin x= t
Differentiating both sides w.r.t x
Cosx dx=dt ,
x 0 π
2
t = sinx Sin 0 π
sin   = 1
=0 2
Substituting x and dx, we get,
1
= ∫ 2t tan -1 ( t ) dt
0

1
=2 ∫ t tan -1 ( t ) dt
0

Now, using integration by parts, with function 1 as tan-1t and function 2 as t,

we get,
  d ( tan -1t )  
=2 tan t ∫ tdt-∫ 
 -1
∫ tdt  dt 
 dt
  

 -1  t 2   1   t 2   
=2  tan t ∫ t  -∫  2     dt 
  2   t +1   2   

 t2 -1 1  t
2
 
=2  tan t- ∫  2  dt 
2 2  t +1  

 t2 
=t tan t-∫  2 td
2 -1

 t +1 
t2
Let I1 = ∫ 2 dt .
t +1
 t2 + 1 −1
∫ t 2 +1 dt
⇒ I1 =

1
⇒ I1 = ∫ dt-∫ dt
1+t 2
⇒ I1 =t-tan -1
Thus, our equation becomes,
=t 2 tan -1t  t-tan -1t 

=t 2 tan -1t-t+tan -1t

1
1
Now, 2 ∫ t tan =1 ( t ) dt=  t 2 tan -1t-t+tan -1t 
0
0

1
2 ∫ t tan =1 ( t ) dt= 12 tan −1 1 − 1 − + tan −1 1 − 0 − 0 + tan −1 0 
0

1
π π
⇒ 2 ∫ t tan =1 ( t ) dt=  -1+  -0
0 4 4

1
π
⇒ 2 ∫ t tan =1 ( t ) dt= -1
0
2

π
2
π
Therefore, ∫ sin2xtan -1 ( sinx ) dx= -1.
0
2

30. Check whether the relation R in the set N of natural numbers given by
R={(a,b):a is a divisor of b} is reflexive, symmetric or transitive. Also
determine whether R is an equivalence relation. 4 Marks

Ans: Reflexivity:

Let there be a natural number n ,

We know that n divides n, which implies nRn.

So, Every natural number is related to itself in relation R.

Thus, relation R is reflexive .

Transitivity:

Let there be three natural numbers a,b,c and let aRb, bRc

aRb implies a divides a and bRc implies b divides c, which as combined implies
that a divides c i.e. aRc.

So, Relation R is also transitive.

Symmetry:

Let there be two natural numbers a,b and let aRb,

aRb implies a divides b but it can't be assured that b necessarily divides a.

For ex, 2R4 as 2 divides 4 but 4 does not divide 2 .

Thus Relation R is not symmetric.

Hence, the relation is not an equivalence relation.

Or

1 2 1 4
Prove that tan -1 +tan -1 = sin -1 . 4 Marks
4 9 2 5

1 2 1 4
Ans: To Prove, tan -1 +tan -1 = sin -1
4 9 2 5

1 2
LHS= tan -1 +tan -1
4 9

 1 2 
-1
 4+9 
=tan 
1 2
 1- × 
 4 9
 -1 -1 -1  x+y  
∴ tan x+tan y=tan  
  1-xy  

 17 
 
= tan −1  36 
34
 
 36 

1
= tan −1  
2

 1 
1 −1  2. 2 
= sin 
2 1
1+ 
 4

 -1 1 -1  2x  
∴ tan x= sin  2 
 2  1+x  

 
1 1
= sin −1  
2 5
 
4

1 4
= sin −1  
2 5

1 4
RHS= = sin −1
2 5

Hence, LHS = RHS.

31. Find the equation of the plane passing through the points (1,0,-2), (3,-
1,0) and perpendicular to the plane 2x-y+z=8 . Also find the distance of the
plane thus obtained from the origin. 4 Marks

Ans: Given points, P(1,0,-2,), Q(3,-1,0)

Given plane, 2x-y+z=8.

 j+k
Normal vector of given plane, n1 =2i-

 j+2k
Now, PQ=2i-

Normal vector of required plane,

  
(
n 2 = n1×PQ )
i j k
= 2 -1 1
2 -1 2

 j+0k
=-1.i-2

 j
=-i-2

Required equation of plane,

-1(x-1)+(-2)(y-0)+0(z-2)=0

-x+1-2y+0=0

x+2y=1

Therefore, the required equation of the plane is, x+2y=1.

Now, distance from origin (0,0,0) is,

a1.a+b1.b+c1.c+d
=
a 2 +b 2 +c 2

0.1+0.2+0.0+ ( -1)
=
12 22 +02

-1
=
1+4
 1
=
5

1
Therefore, distance from origin is units.
5

y dy x+y
32. If tan -1   =log x 2 +y 2 , prove that = . 4 Marks
x dx x-y

y
Ans: Given, tan -1   =log x 2 +y 2
x

y
⇒ tan -1   =log ( x 2 +y 2 )
x

Now, differentiating both sides w.r.t x,

1 d y 1 1 d 2 2
⇒ 2   = . 2 2
. ( x +y )
 y  dx  x  2 x +y dx
1+  
x

 d 
x -y
1  dx  1 1 d  dy 
⇒ 2  2  = . 2 2 .  2x+2y 
y x  2 x +y dx  dx 
1+    
x

 dy   dy 
x -y 2  x+y 
x  2
 1 dx
⇒ 2 2 . dx2  = .  2 2 
x +y  x  2 x +y
 

dy dy
⇒ x+y =x -y
dx dx

dy dy
⇒x -y =x+y
dx dx
 dy
⇒ ( x-y ) = ( x+y )
dx

dy x+y
⇒ =
dx x-y

Hence, proved.

Or
=1
If y=eacos x , -1<x<1, then show that

d 2 y dy
(1-x ) dx2 -x dx -ay 2 =0 .
2
4 Marks

=1
Ans: Given curve is y=eacos x

Differentiating given curve,

d
( acos -1x )
-1
y'=eacos x .
dx

-1
y'=eacos x .
( -a )
1-x 2

-ay
⇒ y'= -- (1)
1-x 2

∴ y=eacos x 
-1

 

On differentiating above equation again w.r.t x, we get

 
-1
-1 x.eacos x
-a  -aeacos x + 
 1-x 2
n
⇒y =  
(1-x 2 )
 acos-1x x.eacos x 
-1

⇒ (1-x ) y =-a  -ae

2 n
+ 
 1-x 2
 
 -1
a.eacos x
⇒ (1-x ) y =a e
2 n 2 acos-1x
-
1-x 2

⇒ (1-x 2 ) y n =a 2 y+xy'

[from (1)]

⇒ (1-x 2 ) y n -xy' − a 2 y=0

d 2 y dy 2
⇒ (1-x ) 2 -x -a y=0
2

dx dx

Hence, proved.

SECTION - D

Question numbers 33 to 36 carry 6 marks each.

33. Amongst all open (from the top) right circular cylindrical boxes of
volume 125π cm3, find the dimensions of the box which has the least
surface area. 6 Marks

Ans: Given that the volume of the right circular cylindrical box V=125π cm³.

Let the radius of the cylinder be h and the height be equal to h.

Volume, V=πr 2 h

⇒ 125π=πr 2 h

⇒ r 2 h=125

125
⇒ h=
r2

Surface area of the box, S=πrh+π r 2

 125 
S=πr  2  +πr 2
 r 
  250  2
S=   +πr
 r 

Differentiating S w.r.t r to find the point of minima,

dS -250π
⇒ = 2 +2πr
dr r

Therefore, for the point of minima,

dS
⇒ =0
dr

-250π
⇒ +2πr=0
r2

250π
⇒ 2πr=
r2

⇒ r 3 =125

⇒ r = 5cm

[Only positive value will be considered, as length can’t have negative value]

125 125
Now, h= = =5 cm
r2 25

The dimension of the cylindrical box is radius, r=5 cm and height, h=5cm.

34. Using integration, find the area lying above x-axis and included
between the circle x 2 +y 2 =8x and inside the parabola y 2 =4x . 6 Marks

Ans:
The given equations are

x 2 +y 2 =8x -------(1)

y 2 =4x -----(2)

From ,

x 2 -8x+y 2 =0

⇒ x 2 -2.5x+16+y 2 =16

⇒ ( x-4 ) +y 2 = ( 4 ) ------------(3)
2 2

Therefore, the equation (1) is a circle with centre (4,0) and has a radius 4.

Also, y 2 =4x is a parabola with vertex at origin and the axis along the x-axis
opening in the positive direction.

To find the intersection points of the curves, we solve both the equation.

∴ x 2 +4x=8x

x 2 -4x=0

X(x-4)=0

x=0 and x=4

When, x=4,y=±4 .

But since, it is given above the x-axis.

So, y=4.
4
Therefore, area, A= ∫ y 2 -y1 dx
0

4
= ∫ ( y 2 -y1 )dx
0

[∴ y2 >y1 ]
4

0
( 2
)
= ∫  16- ( x-4 ) -2 x  dx


[from (2) and (3)]

4 4

0
( 2
)
= ∫  16- ( x-4 )  dx-∫ -2 xdx
 0

4
4  3
 ( x-4 ) 2 16  x-4    x 2 
 16- ( x-4 ) + sin -1   -  
 2 2  4 0  3 
 2 0
4 4
 ( x-4 ) 2 16 -1  x-4   4  32 
 16- ( x-4 ) + sin   -  x 
 2 2  4 0 3  0

 ( 4-4 ) 2 16 -1  4-4    ( 0-4 ) 2 16 -1  0-4  

 4  32 32 
=  16- ( 4-4 ) + sin    - 16- ( 0-4 ) + sin     + 3 -0 
  2 2  4    2 2  2   3  

 ( 0 )   4  32 
=  (
16-0+8sin -1 ( 0 )  -2 16-16+8sin -1 ( -1) )  - 4 
 2   3 

4  32 
= ( 0+0 ) - ( 0-8sin (1) )  -  4 
 -1

3 
 π 4
=8 - .23
2 3

32
=4π-
3

32
Hence, the required area of the region is 4π- sq. units.
3

Or

Using the method of integration, find the area of the triangle ABC ,
coordinates of whose vertices are A(2,0), B(4,5) and C(6,3). 6 Marks

Ans: The vertices of Δ ABC are A(2,0), B(4,5) and C(6,3).

Equation of line segment AB is

 5-0 
y-0=   ( x-2 )
 4-2 

5
⇒ y=   ( x-2 ) --- (1)
2

Equation of line segment BC is

  3-5 
y-5=   ( x-4 )
 6-4 

 −2 
⇒ y-5=   ( x-4 )
 2 

⇒ y-5=-x+4

⇒ y=-x+4------- ( 2 )

Equation of line segment CA is

 0-3 
⇒ y-3=   ( x-6 )
 2-6 

 -3 
⇒ y-3=   ( x-6 )
4

⇒ 4 ( y-3) =3 ( x-6 )

⇒ 4y-12=3x-18

⇒ 4y=3x-6

3
⇒ y= ( x-2 ) --- ( 3)
4

Area( Δ ABC)=Area(ABDA)+Area(BDECB)-Area(AECA)
4 6 6
5 3
= ∫ ( x-2 ) dx+ ∫ ( -x+9 ) dx-∫ ( x-2 )dx
22 4 2
4

4 6 6
5 3
= ∫ ( x-2 ) dx+ ∫ ( -x+9 ) dx- ∫ ( x-2 )dx
22 4
42

4 6 6
5  x2   -x 2  3  x2 
=  -2x  +  +9x  -  -2x 
2 2 2  2 4 4  2 2
 5 3
= [8-8-2+4] + [-18+54+8-36] - [18-12-2+4]
2 4

5 3
= [ 2 ] + [ 8] - [ 8]
2 4

= 5+8-6

= 7 sq. units

Therefore, the area of the triangle is 7 sq. units.

 5 -1 4 
35. If A=  2 3 5  , find A −1 and use it to solve the following system of
 
 5 -2 6 
equations:

5x-y+4z=5

2x+3y+5z=2

5x-2y+6z=-1 6 Marks

 5 -1 4 
Ans: Given, A=  2 3 5  .
 
 5 -2 6 

 5 -1 4 
Now, A =  2 3 5 
 
 5 -2 6 

⇒ A =5 (18+10 ) +1(12-25 ) +4 ( -4-15 )

⇒ A =140-13-76

⇒ A =51

Now, we have to find the cofactor matrix.

, where, A ij = ( -1) M ji
i+j
=  A ij 
3×3
 3 5
A11 = ( -1)
1+1
M11 = =18+10=82
-2 6

2 5
A12 = ( -1) =- (12 − 25 ) =
1+2
M12 = 13
5 6

2 3
A13 = ( -1)
1+3
M13 = =-4-15=-19
5 -2

−1 4
A 21 = ( -1) =- ( 6 + 8 ) =
2+1
M 21 = −2
−2 6

5 4
A 22 = ( -1)
2+ 2
M 22 = =30-20 = 10
5 6

5 −1
A 23 = ( -1) =- ( −10 + 5 ) =
2+3
M 23 = 5
5 −2

5 4
A 31 = ( -1)
3+1
M 31 = =-5-12 = −17
2 5

5 4
A 32 = ( -1) =- ( 25 − 8 ) =
3+ 2
M 32 =- −17
2 5

5 −1
A 33 = ( -1)
3+3
M 33 = =15+2 = 17
2 3

Therefore, the cofactor matrix is,

 A11 A12 A13   28 13 -19 

A A 22 A 23  =  -2 10 5 
 21   
 A 31 A 32 A 33  -17 -17 17 

 28 13 -19 
∴ adjA=  -2 10 5 
 
-17 -17 17 
 1
∴ A -1 = .adjA
A

 28 13 -19 
1 
= -2 10 5 
51  
-17 -17 17 

Now, given set of equations is,

5x-y+4z=5

2x+3y+5z=2

5x-2y+6z=-1

The equations can be written in matrix form as,

 5 -1 4   x   5 
2 3 5  y  =  2 
    
 5 -2 6   z  -1

This is of the form AX=B, where

 5 -1 4  x  5
A= 2 3 5 ,X= y ,B=  2 
   
     
 5 -2 6   z  -1

Now, multiplying AX=B by A-1, we get,

A-1(AX)=A-1B

A-1(AX)=A-1B

IX=A-1B

Now, substituting the values, we get,

 28 13 -19   5 
1 
⇒ X= -2 10 5 2
51   
-17 -17 17  -1
  140 − 4 + 17 
1
⇒ X=  65 + 20 + 17 
51  
 −95 + 10 − 17 

 153 
1 
⇒ X= 102 
51  
 −102 

x   3 
⇒  y =  2 
   
 z  -2 

Therefore, by equality of matrices.

x=3, y=2, z=-2

This is the required solution.

Or

x x 2 1+x 3
If x,y,z are different and y y 2 1+y 3 = 0 , then using properties of
z z 2 1+z 3
determinants show that 1+xyz=0. 6 Marks

x x 2 1+x 3
Ans: Given, y y 2 1+y3 = 0 .
z z 2 1+z3

x x 2 1+x 3
Let, ∆ = y y 2 1+y3
z z 2 1+z3

Now, expanding elements of C3 into two determinants,

 x x2 1 x x2 x3
= y y2 1 + y y2 y3
z z2 1 z z2 z3

Taking x,y,z common from R1,R2,R3 in 2nd determinant,

x x2 1 1 x x2
=y y 2 1 + xyz 1 y y2
z z2 1 1 z z2

Replacing C3 ↔ C2 in 1st determinant,

x 1 x2 1 x x2
= ( −1) y 1 y 2 + xyz 1 y y 2
z 1 z2 1 z z2

Replacing C1 ↔ C2 in 1st determinant,

1 x x2 1 x x2
= ( -1)( -1) 1 y y 2 +xyz 1 y y 2
1 z z2 1 z z2

1 x x2 1 x x2
= 1 y y 2 +xyz 1 y y 2
1 z z2 1 z z2

1 x x2
= 1 y y 2 (1+xyz )
1 z z2

Using R 2 → R 2 − R1 and R 3 → R 3 − R1 ,

1 x x2
= 1-1 y-x y 2 -x 2 (1+xyz )
1-1 z-x z 2 -x 2
 1 x x2
= 1-1 y-x ( y-x )( y+x ) (1+xyz )
1-1 z-x ( z-x )( z+x )
Taking common factor (y-x) from R2 and (z-x) from R3,

1 x x2
=0 1 ( y+x ) (1+xyz )( y-x )( z-x )
0 1 ( z+x )
Expanding determinant through C1, we get,

1{( z+x ) - ( y-x )} (1+xyz )( y-x )( z-x )

= [ z-y ] (1+xyz )( y-x )( z-x )

= (1+xyz )( y-x )( z-x )( z-y )

Given, x,y,z are different.

Therefore, (x-y) ≠ (z-x) ≠ (z-y) ≠ 0

x x 2 1+x 3
Given, y y 2 1+y3 = 0
z z 2 1+z3

(1+xyz )( y-x )( z-x )( z-y ) =0

Since, (x-y) ≠ (z-x) ≠ (z-y) ≠ 0

Therefore, 1+xyz=0

Hence, proved.

36. A card from a pack of 52 cards is lost. From the remaining cards of the
pack, two cards are drawn randomly one-by-one without replacement and
are found to be both kings. Find the probability of the lost card being a king.
6 Marks

Ans: Let E1 be the event that the card is a king.

And, E2 be the event that the card is not a king.

Let A denote the lost card.

Out of 52 cards 4 are king and 48 are non-king.

4 1
Probability that the card is a king, P ( E1 ) = =
52 13

48 12
Probability that the card is not a king, P ( E 2 ) = =
52 13

Two cards can be drawn out of 4 king in 4 C2 ways and 2 kings can be drawn
out of 51 cards in 51 C2 ways.

Probability of getting two kings out of the remaining cards if the lost card is a
king,

 A  3 C2
P   = 51
 E1  C1

A 3
⇒ P =
 E1  51.50
2

A 1
⇒ P =
 E1  425

Probability of getting two kings out of the remaining cards if the lost card is not
a king,

 A  4C
⇒ P   = 51 2
 E2  C2
 4.3
A
⇒ P = 2
 E 2  51.50
2

A 2
⇒ P =
 E 2  425

Therefore, probability of getting two cards when on lost card is king, is,

1 1
.
 E1  13 425
P =
 A  1 . 1 + 12 . 2
13 425 13 425

1 1
.
 E1  13 425
P =
 A  1 . 1 + 12 . 2
13 425 13 425

E  1
P 1 =
 A  25

1
Probability that the lost card is a king is .
25
 CBSE Question Paper 2019
Class 12 Mathematics

Time allowed: 3 hours

Maximum Marks: 100

General Instructions:

(i) All questions are compulsory.

(ii) This question paper contains 29 questions divided into four sections A, B, C and D.
Section A comprises of 4 questions of one mark each, Section B comprises of 8
questions of two marks each, Section C comprises of 11 questions of four marks
each and Section D comprises of 6 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per
the exact requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 1
question of Section A, 3 questions of Section B, 3 questions of Section C and 3
questions of Section D. You have to attempt only one of the alternatives in all
such questions.
(v) Use of calculators is not permitted. You may ask logarithmic tables, if required.

SECTION-A

1. If A is a square matrix of order 3 with A  4 then write the value of 2A

Solution. Since, order of the matrix,

A 4
2A   2  A
n

2A   2   4
3

2A  32
Therefore, the value of 2A is  32

dy
2. If y = sin-1x + cos-1x ,find
dx

Solution.

1
y  sin 1 x  cos 1 x


dy d

dx dx
sin 1 x  cos 1 x 

d
dx
 
sin 1 x 
d
dx
cos 1 x 
1 1
 
1 x 2
1  x2
0
dy
Therefore, 0
dx

3
 d4 y    dy  
2 2

3. Write the order and degree of the differential equation  4  =  x +   

 dx    dx  

Solution. Since,
3
 d 4 y    dy  
2 2


 4    
x 
 dx    dx  
2
 d4y 
6 2 4
 dy  2  dy   dy 
 4   x     3x    3x  
3

 dx   dx   dx   dx 

d4y
The highest power raised to is 2and degree of the differential equation is 2
dx 4

4. If the line has the direction ratios -18,12,-4, then what are its direction cosines?

OR

Find the Cartesian equation of the line which passes through the point (-2,4,-5) is parallel
x+3 4-y z +8
to the line = =
3 5 6

Solution.

2
The direction ratios of the lines are  18,12, 4
18 12 4
Direction cosines of the lines are  , ,
18  12  4
2 2 2
18  12  4
2 2 2
18  122  42
2

9 6 2
Hence, direction cosine of line are  , , 
11 11 11

OR
The cartesion equation of the line which passes through the point  2, 4, 5  and is parallel to the line
x3 y 4 z 8 x 2 y 4 z 5
  is  
3 5 6 3 5 6

SECTION - B

5.If * is defined on the set R of all real number by * : a*b = a2 + b 2 find the identity element
if exist in R with respect to *

Solution. As per the question

Let b be the identity element then
a *b  b * a  a

 a   b 
2 2
a *b  a
  a   b    a 
2 2 2

b0
Similarly,

b    a 
2 2
ba  a
 b    a    a 
2 2 2

b0
Therefore, 0 is the identity element

0 2   0 3a 
6. If A =   and kA =   then find the value of k,a and b
 3 -4  2b 24
Solution. Given,

3
  0 3a 
kA    i 
 2b 24 
0 2  0 2k 
A  ,implies kA    ii 
 3 4  3k 4k 
 0 2k   0 3a 
3k 4k    2b 24 
   
4k  24  k  6
3a  2k  a  4
2b  3k  b  9

sin x  cos x
7. Find  dx, 0  x  π / 2
1  sin 2 x

Solution. According to question,

sin x  cos x π
let I   dx, 0  x 
1  sin 2 x 2
sin x  cos x
I  dx
sin 2 x  cos 2 x  2sin x·cos x
sin x  cos x
 dx
 sin x  cos x 
2

sin x  cos x
 dx
sin x  cos x
let sin x  cos x  t
  cos x  sin x  dx  dt
1
I  dt
t
  ln t  C
1
 ln    C
t
 1 
 I  ln  C
 sin x  cos x 

sin  x  a 
8.Find  dx
sin  x  a 
OR

  log x 
2
Find dx

Solution

4
 sin  x  a 
Let I   dx
sin  x  a 
sin  x  a   2a 
I  dx
sin  x  a 
sin  x  a ·cos  2a   cos  x  a ·sin  2a 
 dx
sin  x  a 
  cos  2a  dx   cot  x  a ·sin  2a  dx
 x·cos  2a   log sin  x  a  ·sin  2a   C

OR

Let I    log x  dx
2

 I  1· log x  dx
2

2 x log x
 I  x· log x   
2
dx
x
 I  x· log x   I1  c1 .....  i 
2

I1   2·log xdx
x
 I1  2 x·log x  2  dx
x
 I1  2 x·log x  2 x  c2 .....  ii 
I  x· log x   2 x·log x  2 x  c1  c2
2

I  x· log x   2 x·log x  2 x  C  whereC  c1  c2 

2

9. From the differential equation representing the family of curves y 2  m  a 2  x 2  by

eliminating the arbitrary constant m and a

5
Solution
 
The equation y 2  m a 2  x 2 where m and a are arbitrary constants


y 2  m a2  x2  ..... i 
dy
2y  2mx .....  ii 
dx
y dy
 2m  2
x dx
 d y  dy 2 
2
2  y 2      2m .....  iii 
 dx  dx  
 d 2 y  dy 2  y dy
2y 2     2
 dx  dx   x dx
2
d 2 y  dy   y  dy
y 2     0
dx  dx   x  dx
2
d 2 y  dy   y  dy
therefore the required differential equation is y 2       0
dx  dx   x  dx

10.Find the unit vector perpendicular to both the vectors a and b, where a = ˆi - 7jˆ + 7kˆ andb = 3iˆ - 2jˆ + 2kˆ
OR
Show that the vectors ˆi - 2jˆ + 3k,
ˆ -2iˆ + 3j-
ˆ 4kˆ andiˆ - 3jˆ + 5kˆ are coplanner

Solution
a  iˆ  7 ˆj  7 kˆ and b  3iˆ  2 ˆj  2kˆ
let n be the vector perpendicular to a and b
n  a b
iˆ ˆj kˆ
n  1 7 7  19 ˆj  19kˆ
3 2 2

19 ˆj  19kˆ
nˆ 
19  19 2 2
 
1 ˆ ˆ
2
jk 
OR

6
let a  iˆ  2 ˆj  3kˆ
b  2iˆ  3 ˆj  4kˆ
c  iˆ  3 ˆj  5kˆ
1 2 3
 a b c   2 3 4
 
1 3 5
 115  12   2  10  4   3  6  3
 3  12  9
0
therefore, a,b,c are coplanar

11. Mother, father and son line up at random for a family photo. If A and B are two events
given by A = Son on one end, B = Father in the middle, find P(B/A).

Solution

If mother (M), father (F), and son (S) line up for the family picture, then the sample space
will be
S = {MFS, MSF, FMS, FSM, SMF, SFM} = A= {MFS, FMS, SMF, SFM}
2 1
P  A  B  
6 3
2 1
P  B  
6 3
4 2
P  A  
6 3
1
P  A  B 3 1
P  B | A   
P  A 2 2
3

12. Let X be a random variable which assumes values x1, x2, x3, x4 such that 2P(X = x1) =
3P(X = x2) = P(X = x3) = 5P(X = x4). Find the probability distribution of X.

OR

A coin is tossed 5 times. Find the probability of getting (i) at least 4 heads, and (ii) at most
4 heads.

Solution

7
Let P  X  x3   x
x
P  X  x1  
2
x
P  X  x2  
3
x
P  X  x4  
5
4

P  x   1
i 1
i

P  x1   P  x2   P  x3   P  x4   1
x x x
  x  1
2 3 5
30
x
61
15 10 30 6
P  X  x1   ; P  X  x2   ; P  X  x3   ; P  X  x4  
61 61 61 61
So, the probability distribution function will be
X 1 2 3 4
15 10 30 6
P  X  xi 
61 61 61 61

OR

Total number of probability of tossing a coin 5 times is 32

(i) Probability of getting atleast 4 heads

8
P  X  4   P  X  5
1 4 0 5
1 1 1 1
5
C4      5C5    
2 2 2 2
5 5
1 1
 5C4    5C5  
2 2
6 3
 
32 16
 ii  probability of getting at most 4 head
P  X  1  P  X  2   P  X  3  P  X  4 
5 5 5 5
1 1 1 1
5
C1    5C2    5C3    5C4  
2 2  2  2
5
1
   5  10  10  5
2
15

16
SECTION – C

 1 
14. If tan 1 x  cot 1 x  tan 1   , x  0 then find the value of x and hence find the value
 3
2
of sec1  
 x

Solution

9
  1 
tan 1 x  cot 1 x  tan 1  , x  0
 3
1  1   1 
 tan 1 x  tan 1    tan 1    cot 1 x  tan 1   , x  0 
x  3   x 
 1 
 x 
 tan 1  x  tan 1  1 
  
 1  x·1   3
 x
x 1 1
2
 
2x 3
 3x 2  2 x  3  0
 3x 2  3x  x  3  0
  
 3x x  3  1 x  3  0 

 x 3  
3x  1  0
1
x , 3
3
x  0, x  3
2  2 
 sec 1    sec 1  
x  3
2  
 sec 1    sec 1  sec 
x  6
2 
 sec 1   
x 6

15. Using properties of determinant prove that

bc a a
b ca b  4abc
c c a b

Solution

10
 bc a a
Let   b ca b
c c ab
R1  R1  R2  R3
0 2c 2b
  b ca b
c c ab
Expending R1
ca b b b b ca
0   2c    2b 
c ab c ab c c
 
 2c ab  b 2  bc  2b bc  c 2  ac  
 2abc  2cb 2  2bc 2  2b 2 c  2bc 2  2abc
 4abc

dy
16. If  sin x   x  y, find
y

dx

Solution

 sin x   x  y
y

log  sin x   log  x  y 

y

 y log  sin x   log  x  y  .....  i 

dy d d
log  sin x ·  y· log  sin x    log  x  y  
dx dx dx 
dy cos x 1  dy 
 log  sin x ·  y·  ·1  
dx sin x  x  y   dx 
dy  1  1
 log  sin x     y·cot x
dx   x  y    x  y 


dy


1  xy  y 2 ·cot x 
dx  x  y ·log  sin x   1

d2y
      dy
2
17. If y  sec1 x , x  0 showthat x2 x2  1 2
 2 x3  x 2  0
dx dx

Solution

11
  
2
y  sec 1 x , x  0


dy 1
 2sec x·

d sec 1 x 
dx dx
dy 1
  2sec 1 x· .....  i 
dx x x2 1
  2x  
  x 2
 1  x  
d2y   2 x2 1  
 2  2 2 2
1 1 
  2sec x 
dx 
 x x  1   
 
x2 x2 1  

 

d2y
 2  2 2 2
 1 
  2sec 1 x·
1
 2 2

 x 1  2x2 

 .....  ii 
dx 
 x x  1   x x 2  1  x x  1  
d2y 
 2 2 2
1  dy  x 1  2 x 2 
 


dx 2  
 x x  1  dx  x 2 x 2  1   
d2y

 x2 x2 1  dx 2  dy

 2 x3  x ·  2  0
dx

18. Find the equation of a tangent and the normal to the curve y 
 x  7  at the
 x  2  x  3
point where it cuts the x-axis

Solution

12
Equation of the curve is

y
 x  7
 x  2  x  3
put y=0 in the above equation we get x=7
dy  x  2 · x  3   x  7 · 2 x  5 

 x  2  · x  3
2 2
dx
The slope of the tangent at point  7, 0  is
dy 20 1
mt   
dx  7,0 400 20
1
 y  0   x  7   x  20 y  7  0
20
mt ·mn  1
1
 mn   20
1
20
Equation of the normal is
 y  0   20  x  7   20 x  y  140  0

sin 2 x
19. Find   sin 2
 
x  1 sin 2 x  3
dx

Solution

13
 sin 2 x
  sin 2

x  1 sin 2 x  3
dx

2sin x·cos x
I  dx
 
sin x  1 sin 2 x  3
2

let sin 2 x  3  t  2sin x·cos xdx  dt
Therefore,
dt
I 
t  2 t
1  1 1
I  
2 t2 t 
  dt

1
 I  ln  t  2   ln t   c
2
1 t2
 I  ln  c
2  t 
t2
 I  ln c
t
sin 2 x  1
 I  ln c
sin 2 x  3

π
b b 3
dx
20. Prove that  f  x  dx   f  a  b  x  dx and hence evaluate  1
a a π tan x
6

Solution

14
let a  b  x  t
 dx   dt
when x  a, t  b and x  b, t  a
b a

 f  x  dx   f  a  b  t  dt
a b
b
 b a

  f  a  b  t  dt  a f  x  dx   b f  x  dx 
a 
b
 b b

  f  a  b  x  dx  a f  x  dx  a f  t  dt 
a 
π π
3 3
dx cos xdx
let I   
π 1 tan x π cos x  sin x
6 6

   
cos    x dx
3 6 
3
I
      
6 cos    x   sin    x 
3 6  3 6 

3
sin xdx
 .....  iii 
 sin x  cos x
6


3
  
2 I  dx   x 3   
 6 3 6 6
6


I
12

21. Show that 1  x 2  dy  2 xydx  cot xdx

Solution

15
1  x  dy  2 xydx  cot xdx
2

dy 2 xy cot x
 
dx 1  x 2
1  x2
The Linear differential equation is
2x

IF=e 
pdx
e 1 x 2
 1  x2
the general solution is
 cot x 

y 1  x2    
1  x
2
1  x 2  dx  C

 
 
 y 1  x 2  log sin x   C

22. let a,b, c be the three vectors such that a = 1, b = 2, c = 3. If the projection of a and b

is equal to the projection of c along a and b, c are perpendicular to each other then find 3a - 2b + 2c

Solution

a  1, b  2, c  3
b·a
the projection of b along a 
a
c·a
the projection of c along a 
a
b·a c·a
 
a a
 b·a  c·a .....  i 

 3a  2b  2c · 3a  2b  2c   9 a
2
 6a·b  6a·c  6b·a  4 b  4b·c  6c·a  4c·b  4 c
2 2

2 2
3a  2b  2c  9 a  4 b  4 c  12a·b  12a·c  8b·c
2 2

2 2
3a  2b  2c  9 a  4 b  4 c
2 2

2
 3a  2b  2c  9 1  4  4  4  9  61

 3a  2b  2c  61

SECTION – D

16
23.Find the value of λ for which the following lines are perpendicular to each other
1
y+
x-5 2- y 1- z x 2 = z -1
= = ; =
5λ + 2 5 -1 1 2λ 3
hence, find whether the lines intersect or not

Solution

x 5 y  2 z 1
  ...1
5  2 5 1
and
1
y
x
 2  z 1 ... 2 
1 2 3
a1  5  2, b1  5, c1  1 and
a2  1, b2  2 , c2  3
a1a2  b1b2  c1c2  0
 5  2   5  2   1 3  0
5  5  0
   1

1 1 1 
24.If A = 0 1 3  , find A -1
1 -2 1 
hence, solve the following system of equations
x+y+z = 6
y + 3z = 11
x - 2y + z = 0

Solution

17
1 1 1
 0 1 3
 
1 2 1
Cofactors
A11  7, A12  3, A13  1
A21  3, A22  0, A23  3
A31  2, A32  3, A33  1
Adj  A 
A1 
A
 7 3 1  7 3 2 
T

Adj  A    3 0 3    3 0 3
 
 2 3 1   1 3 1 
A 9
 7 3 2 
A   3 0 3
1 1
9
 1 3 1 
For system of equations
AX  B
X  A1 B
 x  7 3 2   6 
 y   1  3 0 3 11
  9  
 z   1 3 1   0 
 x 9
 y   1 18 
  9 
 z   27 
x  1, y  2, z  3

25. Show that the height of a cylinder, which is open at the top, having a given surface
area and greatest volume, is equal to the radius of its base.

Solution

18
Let R be the radius
H be the height
V be the volume
S be the total surface area
V   R2 H
S   R 2  2 RH
S   R2
H 
2 R
Substituting value of H in V

V
1
2

SR   R 3 
dV 1
dR 2

 S  3 R 2 
dV
0
dR
1

 S  3 R 2  0
2

S
R
3
2
dV 1
  0  6 R 
dR 2 2
 3 R
S
V is greatest when R 
3
S
S  
H 3
S
2
3
2S
H
3
S
2
3
S
H
3

26. Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using
integration.

Solution

19
Let A  1,1 , B  0,5  and C  3, 2 
The equation of line AB is
5 1
y 1   x  1
0 1
y  4x  5
The equation of line BC is
25
y 5   x  0
30
y  x  5
The equation of line CA is
1 2
y2  x  3
1  3
x 5
y 
4 4
Required area  Area of ABC
The equation of line CA is
1 2
y2  x  3
1  3
x 5
y 
4 4

20
27. Find the equation of the plane passing through the intersection of the planes
and and parallel to x-axis. Hence, find the
distance of the plane from x-axis.

Solution

a  2iˆ  5 ˆj  3kˆ, b  2iˆ  3 ˆj  5kˆ, c  5iˆ  3 ˆj  3kˆ

 r  a · b  a    c  a   0
  ˆ ˆ ˆ
   ˆ ˆ ˆ
  ˆ ˆ

  r   2 i  5 j  3 k  · 4 i  8 j  8 k    3 i  2 j    0
       
  ˆ ˆ ˆ
  ˆ ˆ ˆ

  r   2 i  5 j  3 k  · 2 i  3 j  4 k   0
    
x2 y 5 z 3
2  2 3  5 5  3  0
5  2 3  5 3  3
x2 y 5 z 3
 4 8 8 0
3 2 0
  x  2 16    y  5  24    z  3 32   0
 2x  3y  4z  7
2  2  3  3  2  1  4  3  5   7
 22  22
   1
Therefore, point of intersection is 1, 1, 2 

28. There are two boxes I and II. Box I contains 3 red and 6 Black balls. Box II contains 5
red and black balls. One of the two boxes, box I and box II is selected at random and a ball
is drawn at random. The ball drawn is found to be red. If the probability that this red ball
comes out from box II is ' a find the value of n

Solution

21
E1  selecting box I
E2  selecting box II
A  getting a red ball from selected box
1 1
P  E1   , P  E1  
2 2
 A 3 1
P   
 E1  9 3
 A 5
P  
 E2  n  5

 A
P  E2  P  
E   E2 
P 2  
 A  A  A
P  E1  P    P  E2  P  
 E1   E2 
1 5

3
 2 n5
5 11 1 5
2 3 2 n5
3 15

5 n  20

 n  20  3  75
3n  15
n5

22
 XII CBSE - BOARD - MARCH - 2018
CODE ( 65/2 )
Date: 21.03.2018 Mathematics - Solutions
Section- A

1. If a * b denotes the larger of ‘a’ and ‘b’ and if a  b   a * b   3 , then write the value of  5  10  , where
* and  are binary operations.

Sol:  5  10    5*10   3  10  3  13

 
2. Find the magnitude of each of the two vectors a and b , having the same magnitude such that the angle
9
between them is 600 and their scalar product is
2
Sol: Given :
9
a  b and   60o and a  b 
2
 a  b  a b cos 

9
 a a cos 60o
2
9 2 1
 a 
2 2
2
a 9

a =3 = b

 0 a 3
A   2 0 1
3. If the matrix is skew symmetric, find the values of ‘a’ and ‘b’.
b 1 0 

Sol:  A is skew symmetric matrix

a12  a 21  a  2
and a 31  a13  b  3
4. Find the value of tan
1
3  cot 1  3  
Sol: tan 1  3   cot   3   k say 
1

as cot 1   x     cot 1 x

 k  tan 1  3      cot  3 
1

  
  
3  6
 
 
3 6

 
2


2

Section- B

5. The total cost C  x  associated with the production of x units of an item is given by

C  x   0.005 x 3  0.02 x 2  30 x  5000 . Find the marginal cost when 3 units are produced, where by
marginal cost we mean the instantaneous rate of total cost at any level of output.

Sol: C  x   0.005x 3  0.02x 2  30x  5000

d
Marginal cost  CM  
dx
 C  x    0.005  3x 2 0.02  2x  30

x 3
CM  0.005  3  9  0.02  2  3  30
 0.135  0.12  30
 30.135  0.12
 30.015

 1  cos x 
6. Differentiate tan 1   with respect to x.
 sin x 

 1  cos x 
Sol: Let y  tan 1  
 sin x 
  2cos2 x 2 
 y  tan 1  
 2sin x 2 cos x 2 
 tan 1  cot x 2 

   x 
 tan 1  tan    
  2 2 
 x
y  
2 2
dy d   x  1
   
dx dx  2 2  2

 2 3 
7. Given A    compute A1 and show that 2 A  9 I  A .
  4 7 

 2 3
Sol: A 
 4 7 

A  14  12  2
 A11  7 A12  4 A31  3 A 22  2
T T
A A 22  7 4 7 3
adj  A    11     
 A 21 A 22  3 2  4 2

1 1 7 3
 A 1  adj  A   
A 2  4 2 

7 3
L.H.S.  2A 1   
 4 2
9 0   2 3 7 3 
R.H.S.  9I  A     
0 9   4 7   4 2 
L.H.S. =R.H.S.

8. Prove that : 3sin 1 x  sin 1  3 x  4 x 3  , X    1 , 1 

 2 2 

1 1
Sol: When  x
2 2
We have,
1 1    
  x          3 
2 2 6 6 2 2
 1 1
Also,   x    1  3x  4x 3  1
2 2
 sin 3  3x  4x 3
 3  sin 1  3x  4x 3 

⇒3sin −1 x =sin −1 (3x −4x 3 )

9. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that
the red die resulted in a number less than 4.

Sol: S  1,1 , 1, 2  ..... 6,6 

 n  s   36
A = Red die resulted in a number less than 4.
 {1,1 , 1, 2  , 1,3 ,  2,1 ,  2, 2  ,  2,3 ,  3,1 ,  3, 2  ,  3,3 ,  4,1 ,  4, 2  ,  4,3 ,

 5,1 ,  5, 2  ,  5,3 ,  6,1 ,  6, 2  ,  6,3}

n  A   18C1  18
B = sum of number is 8
B   4,4 ,  6,2 ,  2,6 ,  5,3 ,  3,5

n  B   5 C1  5

A  B   5,3 ,  6, 2 

n  A  B   2C1  2

B
 P    Probability of sum of number 8 when Red die resulted in a number less than
A
P  A  B n  A  B 2 1
4   
P A n A 18 9

10. If  is the angle between two vectors i  2 j  3k and 3i  2 j  k find sin  .

Sol: a  ˆi  2 j  3k,
ˆ b  3iˆ  2ˆj  kˆ

ˆi ˆj kˆ
a  b  1 2 3   4  ˆi  8jˆ  4kˆ
3 2 1

2 2 2
ab   4  8   4  16  64  16  96
 4 6
a  1   4   9  14

b  9  4  1  14

96 4 16 2 6
sin    
14  14 14 7

11. Find the differential equation representing the family of curves y  aebx 5 , where a and b are arbitary
constants.
Sol: y  ae bx  e5
y  ae bx  e5
y   e bx where e5a  
Differentiate w.r.t. ‘x’
dy
 be bx
dx
dy
  by
dx
dy
dx  b
y
Again differentiate w.r.t. ‘x’
d 2 y dy dy
y  
dx 2 dx dx  0
y2
2
d 2 y  dy 
y 2    0
dx  dx 

cos 2 x  2sin 2 x
12. Evaluate :  cos2 x dx
cos 2x  2sin 2 x
Sol: I dx
cos 2 x
1  2sin 2 x  2sin 2 x
I dx
cos 2 x
I   sec 2 x dx
I = tan x + C
 Section- C

d2y dy 2
13. If y  sin  sin x  , prove that 2  tan x  y cos x  0
dx dx

Sol: y  sin  sin x  .....1

dy
dy
 cos  sin x   cos x  dx  cos  sin x  .....  2 
dx cos x
d2 y
  cos  sin x   sin x  cos x sin  sin x  cos x ....  3 
dx 2
Put (1) and (2) in (3)

 dy 
2
d y  dx  2
2
     sin x  y cos x
dx  cos x 
 

d2 y dy
2
  tan x  y cos 2 x
dx dx
d 2 y dy
2
 tan x  y cos 2 x  0
dx dx


14. Find the particular solution of the differential equation e x tan ydx   2  e x  sec 2 ydy  0 , given that y 
4
when x  0

Sol: e x tan y dx  (2  e x ) sec 2 y dy  0

e x tan y dx  (e x  2) sec 2 y dy

e x dx sec2 y dy

 e x  2  tan y

ln e x  2  ln tan y  ln C

ln e x  2  ln (C tan y )

e x  2  C tan y


Given: x  0, y 
4

 
eo  2  C tan  
4
  
eo  2  C tan  
4

1  2  C 1  C   1

 e x  2   tan y

e x  2  tan y  0

(OR)

dy 
Find the particular solution of the differential equation  2 y tan x  sin x , given that y  0 when x 
dx 3

dy
Sol:  (2 tan x) y  sin x
dx

dy
 py  Q
dx
P  2 tan x and Q  sin x

I .F  e  e 
Pdx 2 tan x dx
 e 2lnsec x
2
 elnsec x  sec2 x

Soln. y ( I .F )   Q ( I . F )dx

y  sec 2 x   sin x  sec 2 x dx

y sec 2 x   tan x sec x dx

y sec2 x  sec x  C


Given y  0 x
3


sec C 0
3


C   sec 2
3

 y sec 2 x  sec x  2

y sec2 x −sec x +2 =0
15. Find the shortest distance between the lines.
 
r  4i  j   i  2 j  3k and r  i  j  2k   2i  4 j  5k
       
Sol: r  4i  j   i  2 j  3k  a   b (say)
   
r  i  j  2k   2i  4 j  5k  c   d (say)
   
 c  a  i  j  2k  4i  j   3i  0 j  2k
   
i j k
 b  d  1 2 3  2i  j  0 k
2 4 5

| b d |  4  1  5

(c  a)  (b  d )
Shortest distance 
| b d |

6 6
  units
5 5

16. Two numbers are selected at random (without replacement) from the first five positive integers. Let X
denote the larger of the two numbers obtained. Find the mean and variance of X.
Sol: X can take values as 2,3,4,5 such that

P  X  2  = probability that the larger of two number 2.

= prob. of getting 1 in first selection and 2 in second selection getting 2 in first selection and 1 in second
selection.

1 1 1 1 2
 P  X  2     
5 4 5 4 20
similarly,

2 1 1 2 4
 P  X  3     
5 4 5 4 20

3 1 1 3 6
 P  X  4     
5 4 5 4 20

4 1 1 4 8
 P  X  5     
5 4 5 4 20
 X 2 3 4 5
P X  2 4 6 8
20 20 20 20

2 4 6 8
E  X   2  3  4   5
20 20 20 20

80
 4
20

2 4 6 8
E  X 2   4  9   16   25 
20 20 20 20

340
  17
20
2
V  X   E  X 2    E  X 

=17 −16
=1
17. Using propeties of determinants, prove that

1 1 1  3x
1 3y 1 1  9  3xyz  xy  yz  zx 
1 1  3z 1

Sol:

1 1 1  3x
L.H .S .  1  3 y 1 1
1 1  3z 1

C1  C1  C2 ; C3  C3  C2

0 1 3x
 3y 1 0
3z 1  3z 3z

0 1 x
  3  3 y 1 0
 z 1  3z  z

 9  1  yz  0   x  y  3 zy  z  

 9  yz  xy  3xyz  xz 
  9  3xyz  xy  yz  zx   R.H .S .
Hence proved.

18. Find the equations of the tangent and the normal, to the curve 16 x 2  9 y 2  145 at the point  x1 , y1  where
x1  2 and y1  0

Sol:  P  x1 , y1    2, y1  lies on 16 x 2  9 y 2  145

2
16  2   9 y12  145

9 y12  145  64
9 y12  81
y12  9
y1   3
But y1  0  y1  3
 P   2,3
16 x 2  9 y 2  145 ...(i)
dy
32 x  18 y 0
dx
dy 32 x 16 x
 
dx 18 y 9y
16  2 32
Slope of tangent  m 2,3  
93 27
' 27
Slope of normal  m 2,3 
32
Equation of tangent is,
32
 y  3   x  2
27
27 y  81  32 x  64
32 x  27 y  145  0
Equation of normal is,
27
 y  3   x  2
32
32 y  96  27 x  54
27 x  32 y  54  96  0
27 x  32 y  42  0

(OR)
 x4
Find the intervals in which the function f  x    x3  5 x 2  24 x  12 is
4
(a) strictly increasing, (b) strictly decreasing.

Sol: f '  x   x3  3x 2  10 x  24

f '  x    x  3 x  2  x  4 

f  x  is strictly increasing

if f '  x   0

+ +
-3 2 4

 x   3, 2    4,  

f  x  is strictly decreasing if f '  x   0

 x   , 3   2, 4 

2 cos x
19. Find :  1  sin x  1  sin x dx2

2 cos x
Sol: Let I   dx
(1  sin x) (1  sin 2 x)

Let sin x  t
cos dx  dt

2
I   dt
(1  t ) (1  t 2 )
Consider

2 A Bt  C
 
(1  t )(1  t ) 1  t t 2  1
2

A(t 2  1)  ( Bt  C ) (1  t )

(1  t ) (t 2  1)

2  At 2  A  Bt  C  Bt 2  Ct

 ( A  B) t 2  ( B  C )t  ( A  C )
 A  B  0, B  C  0 AC 2
 A  1, B  1, C 1

 1 2t 1 
 I   2  2 dt
 1  t 2(t  1) t  1 

1
  log 1  t  log t 2  1  tan 1 (t )  C
2

1 t 2 1
 log 2
 tan 1 (t )  C
2 (1  t )

1 sin 2 x  1
 log  tan 1  sin x   C
2 (1  sin x) 2

20. Suppose a girl throws a die. If she gets 1 or 2 she tosses a coin three times and notes the number of tails. If
she gets 3,4,5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained
exactly one ‘tail’, what is the probability that she threw 3,4,5 or 6 with the ride ?
Sol: Let A be the event that girl will get 1 or 2
2 1
P  A  
6 3
Let B be the event that girl will get 3, 4, 5 or 6
4 2
P  B  
6 3

 A = Probability of exactly one till given she will get 1 or 2  83

P T

 B  = Probability of exactly one till given she will get 3, 4, 5 or 6  12

P T

 
P  B P T
 T   P  A  P T  P  BB  P T
P B
 A  B
2 1

 3 2
1 3 2 1
  
3 8 3 2

1
 3
1 1

8 3
 1
 3
11
83
8

11

    
21. Let a  4i  5 j  k , b  i  4 j  5k and c  3i  j  k . Find a vector d which is perpendicular to both c
  
and b and d  a  21

Sol: Since d is perpendicular to both c and b , therefore, if is parallel to c  b

 d   (c  b)

i j k
  3 1 1
1 4 5

  (5  4)i  (15  1) j  (12 1) k

 
  i 16 j 13k
 
Given that

d  a  21

 i 16 j 13k  4i  5 j  k  21

  
  4  80  13  21

21 1
 
63 3

1 
 d  i  16 j  13k
 
3

 1   16   13 
    i    j    k
 3  3   3 

22. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a
given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width
. If the cost is to be least when depth of the tank is half of its width. If the cost is to be borne by nearby
settled lower income families, for whom water will be provided, what kind of value is hidden in this question
?
Sol: Let the length, width and height of the open tank be x, x and y units respectively. Then, its volume is x 2 y
and the total suface area is x 2  4xy.
It is given that the tank can hold a given quantity of water. This means that its volume is constant. Let it be V.
Then,
V  x2y
The cost of the material will be least if the total surface area is least. Let S denote the total surface area.
Then,
S  x 2  4xy
We have to minimize S object to the condition that the volume V is constant.
Now,
S  x 2  4xy
4V
 S  x2 
x
dS 4V d 2S 8V
  2x  2 and 2  2  3
dx x dx x
dS
The critical numbers of S are given by  0. y
dx y
y
dS x
Now, 0 y
dx x
x
4V
 2x  2  0 x
x
 2x 3  4V  0
 2x 3  4x 2 y
 x  2y
d 2S 8V
Clearly, 2
 2  3  0 for all x.
dx x
Hence, S is minimum when x = 2y i.e. the depth (height) of the tank is half of its width.
Comment : Base is directly proportional to height.

2 dy
If  x  y   xy , find
2 2
23. .
dx
Sol: Given :
2
x 2
 y 2   xy

x 4  y 4  2 x 2 y 2  xy
diff. w.r.t. x .
 dy  dy   dy 
4 x3  4 y 3  2  2 x 2 y  2 xy 2    x  y 
dx  dx   dx 

dy dy dy
4 y3  4 x2 y  x  y  4 x 3  4 xy 2
dx dx dx

dy
dx
 4 y 3  4 x 2 y  x   y  4 x3  4 xy 2
dy y  4 x 3  4 xy 2

dx 4 y 3  4 x 2 y  x
(OR)

dy 
If x  a  2  sin 2  and y  a 1  cos 2  , find when   .
dx 3

dy
Sol: y  a 1  cos 2   ,  a  0  2sin 2  
d

dy
  2a sin 2
d

dx
x  a  2  sin 2  ,  a  2  2 cos 2 
d

dy
dy d 2 a sin 2   dx 
     0
dx dx 2a 1  cos 2    d 
d

2 sin  cos 
   cot 
2 sin 2 

dy   1
   cot    
dx 3 3
Section- D

 /4
sin x  cos x
24. Evaluate :  16  9 sin 2 x dx
0

 /4
sin x  cos x
Sol: Let I   16  9sin 2x dx
0

Here, we express the denominator in terms sin x - cos x which is integration of numerator.
 2
Clearly,  sin x  cos x   sin 2 x  cos 2 x  2sin x cos x  1  sin 2x
2
 sin 2x  1   sin x  cos x 
/4
sin x  cos x
 I  16  9 1   sin x  cos x  dx
0  2

 /4
sin x  cos x
 I  25  9  sin x  cos x  2
dx
0

Let sin x  cos x  t.Then,d  sin x  cos x   dt   cos x  sin x  dx  dt.

  
Also, x  0  t  sin 0  cos 0  1 and x   t  sin  cos  0
4 4 4
0 0 0
dt 1 dt 1 dt
 I 1 25  9t 2  9 1 25 2  9  5 2
t 1 2
9   t
3
0
1   5/3 t 
 I   log
9 2 5 / 3  5 / 3  t  1

1   2 / 3  1   1  1 1 1
 I  log1  log     log1  log      log1  log 4   log 4  log 2
30   8 / 3   30   4   30 30 15

(OR)

x  3x  e x dx
2
Evaluate :
1

3 b

Sol: I   ( x 2  3 x  e x )dx   f ( x )dx (say)

1 a

when f ( x)  x 2  3 x  e x ; a 1, b  3

b  a 3 1 2
h  
n n n

f (a  rh)  f (1  rh)  (1  rh) 2  3(1  rh)  e1 rh

 4  5rh  r 2 h 2  e  e rh

 r 2 h 2  5rh  4  e  e rh
b n

 f ( x)dx  lim  h f (a  rh)

n 
a r 1
 n
 lim  h (r 2 h 2  5rh  4  e  e rh )
n 
r 1

 n n n n

 lim   r 2 h3  5  rh 2   4h  e e rh  h 
n 
 r 1 r 1 r 1 r 1 

 8 n(n  1) (2n  1) 4 n(n  1) 2   e nh 1   

 lim  3   5 2   4   n  e  eh  h    h 
n   n 6 n 2 n e  1
     

 
 
  8 n  n 1   2n 1  20 n  n 1  e n 1 (e 2  1)  
 lim             8 
n    6 n n n  n n 2 n n n  eh 1   
    
  h   

as n    h  0

b
4  1  1  1 e h 1 (e2  1)
 f ( x)dx  1 lim 1    lim  2    lim10 1 1    8  lim
h  0  eh 1 
a
3 n 
 n  n   n  n   n
 
 h 

4 e(e2  1)
 1 2  10 11  8 
3 1

8
  10  8  e3  e
3

8
  18  e3  e
3

62 3
 e e
3

25. A factory manufactures two types of screws A and B, each type requiring the use of two machines, an
automatic and a hand - operated. It takes 4 minutes on the automatic and 6 minutes on the hand oper-
ated machines to manufacture a packet of screws ‘B’. Each machine is availble for at most 4 hours on
any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a
profit of Rs. 1. Assuming that he can sell all the screws he manufactrures, how many packets of each type
should the factory owner produce in a day in order to maximize his profit ? Formulate the above LPP and
solve it graphically and find the maximum profit.
Sol: Let the factory manufactures x screws of type A and y screws of type B on each day.
 x  0, y  0
Given that
 Screw A Screw B Availibility
Automatic machine 4 6 4  60  240 minutes
Hand operate machine 6 3 4  60  240 minutes
Profit 70 paise 1 rupee

The constraints are

4 x  6 y  240

6 x  3 y  240
Total profit
z  0.70 x  1 y

 L.P.P. is

maximise z  0.7 x  y
subject to ,
2 x  3 y  120

2 x  y  80

x  0, y  0

80 E (0,80)

70

60

50

40 C (0,40)

30

20 B (30,20)

10
O
10 20 30 40 50 60 70
A (40,0) D (60,0)

 common feasible region is OCBAO

 Correct point Z  0.7 x  y
A  40,0  Z  A   28
B 30, 20 Z  B   41 maximum
C 0, 40 Z  C   40
O  0, 0 Z O  0

The maximum value of ‘Z’ is 41 at  30, 20  . Thus the factory showed produce 30 packages at screw A
and 20 packages of screw B to get the maximum profit of Rs.41

26. Let A   x  Z : 0  x  12 show that R   a, b  : a, b  A, a  b  is divisible by 4} is an equivalence

relation. Find the set of all elements related to 1. Also write the equivalence class [2].
Sol: We have,
R   a, b  : a  b is a multiple of 4, where a, b  A  x  Z:0  x  12  0,1, 2,...,12.
We observe the following properties of relation R.
Reflexivity : For any a  A, we have
a  a  0, which is a multiple of 4.

  a,a   R

Thus,  a,a   R for all a  A.

So, R is reflexive.
Symmetry : Let  a, b   R. Then,

 a,b   R
 a  b is a multiple of 4

 a  b  4 for some   N

 b  a  4 for some   N  a  b  b  a 

  b,a   R
So, R is symmetric.
Transitivity : Let  a, b  R and  b,c   R.Then,

 a, b   R and  b,c   R
 a  b is a multipleof 4 and b  c is a multipleof 4

 a  b  4 and b  c  4 for some ,   N

 a  b  4 and b  c  4
 a  c  4  4
  a  c is a multiple of 4
 a  c is a multipleof 4

  a,c   R
Thus,  a, b   R and  b,c   R   a,c   R
So, R is transitive.
Hence, R is an equivalence relation.
Let x be an element of A such that  x,1  R. Then,

x  1 is a multipleof 4

 x  1  0,4,8,12
 x  1  0, 4,8,12
 x  1,5,9  13 A
Hence, the set of all elements of A which are related to 1 is {1, 5, 9} i.e. [1] = [1, 5, 9].
&  2   2, 6,10

(OR)

x
Show that the function f : R  R defined by f  x   2
, x  R is neither one – one nor onto.
x 1
Also , if g : R  R is defined as g  x   2 x  1 find fog  x 

x
Sol: f : R  R, f ( x)  2
,  x R
x 1

x1
f ( x1 )  2
x 1
2

f ( x1 )  f ( x2 )

x1 x
2
 22
x  1 x2  1
1

x1 x22  x1  x2 x12  x2

x1 x22  x2 x12  x1  x2  0

x1 x2 ( x2  x1 ) 1( x2  x1 )  0

( x1 x2 1) ( x2  x1 )  0

x1 x2 1 or x1  x2
 f ( x) is not one-one
 x
also y  x 2  1

x2 y  x  y  0
  0 if x is real

 B 2  4 AC  0

(1) 2  4  y  y  0

1 4 y2  0
(1  2 y ) (1  2 y )  0
(2 y  1) (2 y  1)  0

1 1
  y
2 2
Codomain  R

 1 1
But range    , 
 2 2
 Function is not onto
x
f ( x)  2
as f : R  R
x 1

g ( x )  2 x  1as g : R  R

g ( x)
( fog ) ( x)  f ( g ( x )) 
( g ( x) 2  1)

2x 1

(2 x  1) 2  1

2 x 1
 2
4 x  4 x 11

2 x 1
 2
4x  4x  2

27. Using integration, find the area of the region in the first quadrant enclosed by the x – axis, the line y  x
and the circle x 2  y 2  32

Sol: Put y  x in x 2  y 2  32

 x 2  x 2  32
 2 x 2  32

x 2  16
x4

4 32 45°
A   yline dx   ycircle dx (4, 0) ( 32, 0)
0 4

4 32
A   xdx  ( 32  x 2 )dx
0 4

4 32
 x2 
    ( 32)2  x 2 dx
 2 0 4

32
x 32  x 
 8   32  x 2  sin 1  
2 2  32  

    4 
  8    0  16    2 16  16sin 1  
 2   32   

 1 
 8  8  8  16sin 1  
 2


 8  16   8  4  4 sq units
4

2 3 5
28. If A  3 2 4 , find A1 . Use it solve the system of equations.
1 1 2

2 x  3 y  5 z  11

3 x  2 y  4 z  5

x  y  2 z  3

 2 3 5 
A   3 2 4 
Sol:
 1 1 2 

 A  2  4  4   3  6  4   5  3  2   0  6  5  1  0
 Now, A11  0, A12  2, A12  1
A 21  1, A 22  9, A 23  5
A31  2, A 32  23, A 33  13

 0 1 2   0 1 2 
1
A  1
 adjA     2 9 23   2 9 23 .... 1
A
1 5 13   1 5 13

Now, the given system of equations can be written in the form of AX  B, where

 2 3 5  x  11 
A   3 2 4  , X   y  and B   5
   
 1 1 2   x   3

The solution of the systemof equations is given by X  A 1B,

X  A 1B
 x   0 1 2  11 
  y    2 9 23  5  Using 1 
 z   1 5 13  3

0  5  6  1 
  22  45  39   2 

 11  25  39  3 

Hence, x =1, y =2, and z =3

(OR)

1 2 3

Using elementary row transformations, find the inverse of the matrix A   2 5 7 

 2 4 5

1 2 3
Sol: A   2 5 7 
 2 4 5

A  1 25  28  2  10  14   3  8  10 

 3  2  4  3 2  9  8  1  0

A1 exists.
A  A1  I
 1 2 3 1 0 0 
 2 5 7  A1  0 1 0
   
 2 4 5 0 0 1 

R2  R2  2 R1 ; R3  R3  2 R1

1 2 3  1 0 0
0 1 1 A1   2 1 0 
   
0 0 1  2 0 1 

R1  R1  2 R2

1 0 1  5 2 0
0 1 1 A1   2 1 0
   
0 0 1  2 0 1 

R1  R1  R3 ; R2  R2  R3

1 0 0   3 2 1
0 1 0 A1   4 1 1
   
0 0 1   2 0 1 

 3 2 1
I  A  A   4 1 1
1 1

 2 0 1 

29. Find the distance of the point (−1, −5, −10) from the point of intersection of the line
 
r  2i  j  2k   3i  4 j  2k and the plane r  i  j  k  5
   
Sol: Cartesian equation of line and plane,
x  2 y 1 z  2
  :  Line 
3 4 2
x y  z 5  0 :  Plane 

Let Q  ,  ,   be point of intersection of line and plane which will satisfy both equation.
  2  1   2
    (say)
3 4 2
  3  2,   4  1,   2  2
also       5  0
3  2  4  1  2  2  5  0
 0
   2,   1,   2  Q   2, 1, 2 
2 2 2
  PQ    1  2    5  1   10  2 
 9  16  144
 169
= 13 units
CBSE-XII-2017 EXAMINATION CBSE-XII-2017 EXAMINATION

SET-1 MATHEMATICS
Series GBM Paper & Solution Code: 65/1
GM 3 Hrs.
Time: Max. Marks: 100

General Instruction:
(i) All questions are compulsory
(ii) The question paper consists of 29 questions divided into four section A, B, C and D. Sections A comprises of questions of
one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each
and Section D comprises of 6 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 3 questions of
six marks each. You have to attempt only one of the alternatives in all such questions.
(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION – A
Question numbers 1 to 4 carry 1 mark each
8 0 
1. If for any 2 × 2 square matrix A, A (adj A)    , then write the value of |A|.
0 8 
Solution:
8 0 
A ( adj A)   ,
0 8 
by using property
A (adj A) = |A| In
8 0 
| A | I n   
0 8 
1 0 
| A | I n  8   | A | 8
0 1 

2. Determine the value of 'k' for which the following function is continuous at x = 3:
 ( x  3)2  36
 ,x  3
f ( x)   x 3
 ,x  3
 k
Solution:
(x + 3)2 36
lim f(x) = lim
x 3 x 3 x 3
( x  3  6)( x  3  6)
 lim
x 3 ( x  3)
= 12
given that f(x) is continuous at x = 3
 lim f ( x)  f (3)
x 3

 k  12

www.vedantu.com 1 / 22
 CBSE-XII-2017 EXAMINATION

sin 2 x  cos2 x
3. Find:  dx
sin x cos x
Solution:
sin 2 x  cos2 x
 sin x cos x dx
 cos 2 x
 2 dx
sin x
 2 cot 2 x dx
2log | sin 2 x |
 C
2
  log | sin 2 x | C

4. Find the distance between the planes 2x – y + 2z = 5 and 5x – 2⋅5y + 5z = 20.

Solution:

2x – y + 2z = 5 …(1)

5x – 2⋅5 y + 5z = 20

or 2x – y + 2z = 8 …(2)

Distance between plane (1) & (2)

d1  d2 3
  1
a 2  b2  c 2 9

SECTION - B
Question numbers 5 to 12 carry 2 marks each

5. If A is a skew-symmetric matrix of order 3, then prove that det A = 0.

Solution:
 0 a b
Let A = a 0 c  be a skew symmetric matrix of order 3
 
 b c 0
 0 a b
 | A |  a 0 c 
 b c 0 
| A |  a(0  bc)  b(ac  0)
= – abc + abc = 0 Proved

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6. Find the value of c in Rolle's theorem for the function f(x) = x3 – 3x in [  3 , 0].
Solution:
f(x) = x3 – 3x
(i) f(x) being a polynomial is continuous on [  3 , 0]
(ii) f(  3 ) = f (0) = 0
(iii) f′(x) = 3x2 – 3 and this exist uniquely on [  3 , 0]
∴ f(x) is derivable on (  3 , 0)
∴ f(x) satisfies all condition of Rolle's theorem
∴ There exist at least one c ∈ (  3 , 0) where f′(c) = 0
⇒ 3c2 – 3 = 0
⇒ c = ±1 ⇒ c = – 1

7. The volume of a cub is increasing at the rate of 9 cm 3 . How fast is it surface area increasing when the
s
length of an edge is 10 cm?
Solution:
Assumed volume of cube = V
dV
Given that,  9cm3/sec
dt
dA
?
dt
l  10 cm
dV d 3 dl
 (l )  9  3l 2  9
dt dt dt
dl 3
 . ..................(1)
dt l 2
dA d dl 3
Now  (6l 2 )  12l  12l  2 (form (1))
dt dt dt l
36 36
   3.6 cm2/sec
l 10

8. Show that the function f ( x)  x3  3x2  6x  100 is increasing on R.

Solution: f(x) = x2 – 3x2 + 6x - 100

f′(x) = 3x2 – 6x + 6
f′(x) = 3 (x2 – 2x + 2)
f′(x) = 3 [(x – 1)2 + 1]
f′(x) > 0 for all x ∈ R
So, f(x) is increasing on R.

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 CBSE-XII-2017 EXAMINATION

9. The x- coordinate of a point on the joining the points P(2, 2, 1) and Q(5, 1,-2) is 4. Find its z-coordinate.
Solution:

Let R divides PQ in the ratio k: 1

 5k  2 k  2 2k  1 
R , , 
 k 1 k 1 k 1 
Given x co-ordinate of R = 4
5k  2
 4
k 1
k 2
(2)  1
 z co- ordinate =  1
2 1

10. A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number
obtained is even" and B be the event "number obtained is red". Find if A and B are independent events.

Solution:
3 1
A = {2, 4, 6} P( A)  
6 2
B = {1, 2, 3}
3 1
A ∩ B = {2} P( B)  
6 2
1
P(A ∩ B) =
6
1 1 1
Here, P(A) P(B)   
2 2 4
Since, P(A ∩ B) ≠ P(A) P(B), so events A and B are not independent events.

11. Tow tailors, A and B, earn ₹ 300 and ₹ 400 per day respectively A can stitch 6 shirts and 4 pairs of
trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of
them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labor cost,
formulate this as an LPP.
Solution:

Let tailor A and B works for X days and Y days respectively

 x  0, y  0
Minimum number of shirts = 60

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 CBSE-XII-2017 EXAMINATION

 6 x  10 y  60
3x  5 y  8
Minimum no of trouser = 32
 4 x  4 y  32
 x y 8
Let z be the total labor cost
 z = 300 x + 400 y
 The given L. P. problem reducers to: z = 300x + 400y
x  0, y  0, 3x + 5y  30 and x + y  8

dx
12. Find:  5  8x  x 2

Solution:
dx
 
{( x  4) 2  21}
dx

( 21)  ( x  4) 2
2

1 21  ( x  4)
 log C
2 21 21  ( x  4)

SECTION – C
Question numbers 13 to 23 carry 4 marks each

x 3 x3 
13. If tan 1  tan 1  , then find the value of x.
x4 x4 4
Solution:
 x3 x3
  
tan 1  x  4 x  4 
  x 9  4
2

1   2 
  x  16  
( x  4)( x  3)  ( x  3)( x  4)
1
( x 2  16)  ( x 2  9)
2 x 2  24  7
2 x 2  7  24
17
x2 
2
17
x
2

14. Using properties of determinants, prove that

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 CBSE-XII-2017 EXAMINATION

a 2  2a 2a  1 1
2a  1 a  2 1  (a  1)3
3 3 1
OR
Find matrix A such that
 2 1  1 8 
 1 0  A   1 2 
   
 3 4   9 22 
   
Solution:
Use R1 = R1 – R2; R2 = R2 – R3; R3 = R3
L. H. S.
a2  1 a  1 0
 2a  2 a  1 0
3 3 1
(a  1)(a  1) (a  1) 0
 2(a  1) (a  1) 0
3 3 1
Taking common (a  1)2
(a  1) 1 0
 (a  1)2
2 1 0
3 3 1
 (a  1)2 [(a  1)(1  0)  1(2  0)]
 (a  1)2 [(a  1)  2]
 (a  1)3
= R. H. S.
OR
Let matrix A is
a b 
A 
c d 
 2 1  1 8
1 0   a b    1 2
   c d   
 3 4  9 22 
 2a  c 2b  d   1 8
 a b    1 2
   
 3a  4c 3b  4d   9 22 
Comparing both the sides
2a ⎼ c = ⎼1,
2b ⎼ d = ⎼ 8

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 CBSE-XII-2017 EXAMINATION

After solving we get

C = 3, d = ⎼ 4
1 2
So, A   
3 4

dy
15. If x y  y x  ab , then find .
dx
OR
2
d 2 y  dy 
If e y (x + 1) = 1, then show that   .
dx2  dx 

Solution:

We have x y  y x  ab

d y d
Differentiating W. r. t . x, we get ( x )  ( y x )  0. ……(1)
dx dx

Let u = x y log u = y log x

1 du 1 dy du y dy 
  y.  log x. ;   u   log x 
u dx x dx dx x dx 

d y y dy 
or ( x )  x y   log x  ……(2)
dx x dx 

Let v = y x logv xlogy

1 dv 1 dy dy  x dy 
  x.  log y.1;   v  log y 
v dx y dx dx  y dx 

d x  x dy 
or ( y )  yx   log y  ……(3)
dx  y dx 

Using (2) and (3) in (1),

y dy   x dy 
We get. x y   log x   y x   log y   0. …….(4)
x dx   y dx 

x 1 dy y 1 dy y x log y  yx y 1
 ( x log x  xy
y
)  ( y log y  yx ) or
x
 y
dx dx x log x  xy x 1

OR
Let ey (x + 1) = 1

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 CBSE-XII-2017 EXAMINATION

dy
ey(1) + (x + 1)ey 0
dx
dy
 ( x  1)  1  0 ……(1)
dx
Again differentiating W. r. t . x,

d 2 y  dy 
 ( x  1)    .1  0
dx 2  dx 
dy
d2y
  dx
dx 2
( x  1)

d2y dy dy
2
 . [equation (1)]
dx dx dx
2
d 2 y  dy 
 
dx2  dx 

cos 
16. Find:  (4  sin 2
)(5  4cos2 )
d

Solution:

cos 
 (4  sin 2
)(5  4cos2 )
d

cos 

(4  sin )(5  4(1  sin 2 ))
2

cos  d 
 (sin 2   )(4sin 2   1)
Put sin θ = t

Cos θ dθ = dt

1
 I  dt
(4  t )(1  4t 2 )
2

Consider

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 CBSE-XII-2017 EXAMINATION

1 At  B Ct  D
 
(4  t )(1  4t ) 4  t 2 1  4t 2
2 2

1  ( At  B)(1  4t 2 )  (Ct  D)(4  t 2 )

 At  B  4 At 3  4 Bt 2  4Ct  Ct 3  4 D  Dt 2
 (4 A  C )t 3  (4 B  D)t 2  ( A  4C )t  ( B  4 D)
4 A  C  0  C  4 A
4 B  D  0  D  4 B
A  4C  0  A  4C
B  4D  1
1 4
By solving we get A = 0, B =  , C 
15 15
1 4
1 15
= + 15
(4 + t 2 )(1+ 4t 2 ) 4 + t 2 1+ 4t 2
1 1 4 1 1
I   
15 4  t 2
dt   
15 4 1  t 2
dt
4
1 1 1  t  1 1  t 
   tan     tan  1
C
15 2  2  15 1 1 
 2
2
1 t 2
  tan 1    tan 1 (2t )  C
30  2  15
2 1  sin  
 tan 1 (2sin )  tan 1  C
15 30  2 


x tan x
17. Evaluate:  sec x  tan x dx
0

OR
4
Evaluate:  {| x  1|  | x  2 |  | x  4 |}dx
1

Solution:

x tan x
I  dx ...(1)
0 sec x  tan x

x(  x)( tan x)
I  dx
0  sec x  tan x

(  x) tan x
I  dx ...(2)
0 sec x  tan x
Adding (1) & (2)

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 CBSE-XII-2017 EXAMINATION

 tan x

2I   dx
0 sec x  tan x

/ 2 tan x
 2 I  2  dx
0 sec x  tan x

 0
2a
f ( x)dx  2 f ( x)dx whenever f (2a  x)  f ( x)
0
a

/ 2tan x
I   dx
0 sec x  tan x
 / 2 tan x(sec x  tan x)
I   dx
0 sec2 x  tan 2 x
/ 2
I   (sec x tan x  tan 2 x)dx
0
/ 2
  (sec x tan x  sec2 x  1)dx
0

I   sec x  tan x  x 0
/ 2

  
   lim  sec x  tan x    sec  
 x  2 
 2 
1  sin x 2
  lim  
 cos x 2
x
2

1  sin 2 x 2
  lim  
 cos x(1  sin x) 2
x
2


2
 
2
OR
Let f (x) = | x – 1| +| x – 2 | + |x – 4|
We have three critical points x = 1, 2, 4
(i) when x < 1
(ii) when 1  x < 2
(iii) when 2  x < 4
(iv) when x  4

F (x) = – (x – 1) – (x – 2) – (x – 4) if x<1
= (x – 1) – (x – 2) – (x – 4) if 1  x<2
= (x – 1) + (x – 2) – (x – 4) if 2  x<4
= (x – 1) + (x – 2) + (x – 4) if x 4

∴ f(x) = –3x + 7 if x<1

= –x + 5 if 1  x<2
=x+1 if 2  x<4
= 3x –7 if x 4

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 CBSE-XII-2017 EXAMINATION

4
 I   f ( x)dx
1
2 4
 I   f ( x)dx   f ( x  1)dx
1 2
2 4
 I   ( x  5)dx   ( x  1)dx
1 2
2 4
 x2   x2 
   5x     x 
 2 1  2 2
 4   1   16  4 
    10      5     4     2 
 2   2   2  2 
9 23
 8   12  4 
2 2

18. Solve the differential equation (tan–1 x – y)dx = (1 + x2) dy.

Solution:

We have

dy tan 1 x  y

dx 1  x2
dy y tan 1 x
 
dx 1  x 2 1  x 2
1
 1 x2 dx 1
I .F .  e  e tan x

tan 1 x tan 1
y.e tan  
1
 e dx
1  x2

Put t = tan1

1.dx
dt 
1  x2

 t.et  1.et dt
1
y.etan x  tet  et  c
1 1
y.etan x  (tan 1 x  1)etan x
c
1
y  tan 1 x  1  cetan x

19. Show that the points A, B, C with position vectors 2iˆ  ˆj  kˆ, iˆ  3 ˆj  5kˆ and 3iˆ  4 ˆj  4kˆ respectively,
are the vertices of a right-angled triangle, Hence find the area of the triangle.

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 CBSE-XII-2017 EXAMINATION

Solution:

AB  iˆ  2 ˆj  6kˆ
BC  2iˆ  ˆj  kˆ
CA  iˆ  3 ˆj  5kˆ
BC.CA  0
BC  CA

ABC is a right angled triangle

1
 | BC || AC |
2
1
 4  1  1 1  9  25
2
1
= 6 35
2
1
 210
2

20. Find the value of λ, if four points with position vectors 3iˆ  6 ˆj  9kˆ, iˆ  2 ˆj  3kˆ, 2iˆ  3 ˆj  kˆ and
4iˆ  6 ˆj  kˆ, are coplanar.

Solution:
We have

P.V. of A  3iˆ  6 ˆj  9kˆ

P.V. of B  iˆ  2 ˆj  3kˆ

AB  2iˆ  4 ˆj  6kˆ
AD  iˆ  3 ˆj  8kˆ
AD  iˆ  (  9)kˆ

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 CBSE-XII-2017 EXAMINATION

Now,

2 4 6
AB.( AC  AD)  1 3 8  0
1 0 (  9)

 –2(–3λ + 27) + 4(–λ + 9 +8) –6(0 + 3) = 0

 6λ – 54 – 4λ + 68 – 18 = 0
2λ – 4 = 0
λ=2

AB, AC, AD are coplanar and so the points A, B, C and D are coplanar.

21. There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random
without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and
variance of X.

Solution:
X denote sum of the numbers so, X can be 4, 6, 8, 10, 12

Mean = XP( x)  8

212 20
Variance  X2P( x)  (XP( x))2   64 
3 3

22. Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular.
Previous year results report that 70% of all students who have 100% attendance attain A grade and 10%
irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at

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 CBSE-XII-2017 EXAMINATION

random from the school and he was found to have an A grade. What is the probability that the student has
100% attendance? Is regularity required only in school? Justify your answer.
Solution:
Let E1 be students having 100% attendance
E2 be students having irregular attendance
E be students having A grade

30 70
P( E1 )  P( E2 ) 
100 100
 E  70 30
P     21%
 E1  100 100
 E  10 70
P     7%
 E2  100 100

By Baye's theorem,

E 30 21
P( E1 ) P   
 E1   E1  100 100 63 63
So, P      
E  E  E 30

21

70

7 63  49 112
P( E1 ) P    P( E2 ) P  
 E2   E2  100 100 100 100

23. Maximize Z = x + 2y
Subject to the constraints

x + 2y ≥ 100
2x – y ≤ 0
2x + y ≤ 200
x, y  0
Solve the above LPP graphically.

Solution:
x + 2y = 100
2x – y = 0 ……(1)
2x + y = 200 ……(2)
x = 0, y = 0 ……(3)

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 CBSE-XII-2017 EXAMINATION

Corner points are A (100, 0), B(50, 100), C(20, 40)

Maximum at point B and maximum value 250

SECTION – D
Question numbers 24 to 29 carry 6 marks each
 4 4 4  1 1 1 
24. Determine the product 7 1 3  1 2 2 and use it to solve the system of equations
  
 5 3 1 2 1 3 
x – y + z = 4, x – 2y – 2z = 9, 2x + y +3z = 1.
Solution:
Product of the matrices
 4 4 4  1 1 1 
7 1 3  1 2 2
  
 5 3 1 2 1 3 
4  4  8 4  8  4 4  8  12
 7  1  6 7  2  3 7  2  9 
 5  3  2 5  6  1 5  6  3 
8 0 0
 0 8 0  8I3
0 0 8
1
1 1 1   4 4 4 
Hence 1 2 2  7 1 3 
  1
  8 
2 1 3   5 3 1
Now, given system of equations can be written in matrix form, as follows

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 CBSE-XII-2017 EXAMINATION

1
1 1 1   x   4
1 2 2   y   9 
     
 2 1 3   z  1 
1
 x  1 1 1   4
 y   1 2 2  9 
     
 z   2 1 3  1 
x   4 4 4   4
 y   1  7 1 3   9 
  8   
 z   5 3 1 1 
x   16 36 4 
 y   1  28 9 3
  8 
 z   20 27 1
x  24 
 y   1 16
  8 
 z  8 
24 16 8
x , y , z
8 8 8
x = 3, y = –2, z = –1

 4 4 4x  3
25. Consider f : R     R    given by f (x)  . Show that f is bijective. Find the inverse of f
 3 3 3x  4
and hence find f-1 (0) and x such that f-1 (x) = 2.
OR
Let A = Q × Q and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) ∈
A. Determine, whether * is commutative and associative. Then, with respect to * on A
(i) Find the identity element in A.
(ii) Find the invertible elements of A.
Solution:
4x  3  4
f ( x)  , x  R   
3x  4  3
F is one – one 

Let x1, x2  R  
4
  and f (x1) = f (x2)
 3
4 x  3 4 x2  3
 1 
3x1  4 3x2  4
 12 x1 x2  16 x1  9 x2  12  12 x1 x2  9 x1  16 x2  12
 7 x1  7 x2  x1  x2
 f is one – one
F is onto 

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 CBSE-XII-2017 EXAMINATION

4
Let k  R    be any number
3
4x  3
f ( x)  k 
3x  4
 4 x  3  3kx  4k
4k  3
x
4  3k
4k  3 4
Also 
4  3k 3
implies – 9 = – 16 (which is impossible)
 4k  3 
f  = k i.e. f is onto
 4  3k 
 The function f is invertible i.e. f-1 exist inverse of f
Let f-1 (x) = k
f (k) = x
4k  3
 x
3k  4
4x  3
k 
4  3x
4x  3  4
 f 1 ( x)  , x  R   
4  3x  3
3
f 1 (0)  
4
and when
f-1 (x) = 2
4x  3
 2
4  3x
 4x  3  8  6x
 10 x  11
11
x
10
OR
(i) Let (e, f) be the identify element for *
 for (a, b)  Q × Q, we have
(a, b) * (e, f) = (a, b) = (e, f) * (a, b)
 (ae, af + b) = (a, b) = (ea, eb + f)
 ae = a, af + b = b, a = ea, b = eb + f
 e = 1, af = 0, e = 1, b = (1) b + f
( a need not be '0')
 e = 1, f = 0, e = 1, f = 0
 (e, f) = (1, 0)  Q × Q
 (1, 0) is the identify element of A

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 CBSE-XII-2017 EXAMINATION

(ii) Let (a, b)  Q × Q

Let (c, d)  Q × Q
such that
(a, b) * (c, d) = (1, 0) = (c, d) * (a, b)
⇒ (ac, ad + b) = (1, 0) = (ca, cb + d)
⇒ ac = 1, ad + b = 0, ca = 1, cb + d = 0
1 b 1
 c  , d   ,   b  d  0(a  0)
a a a
 1 b 
 ( c, d )   ,  (a  0)
a a 
 1 b 
 for a  0, (a, b) 1   , 
a a 

26. Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is
a cube.
Solution: If each side of square base is x and height is h then volume
V
V = x2h  h  2
x
S is surface area then
V 
S  4hx  2 x 2  4  2  x  2 x 2
x 
4V
S  2x2
x
Diff. w. r. to x
dS 4V d 2S 8V
  2  4 x and 2
 3 4
dx x dx x
dS 4V
Now  0  4x  2
dx x
 x V  x V
3 1/3

d 2S
at x = V1/3, 0
dx 2
 S is minimum when x = V1/3
V V
and h  2  2/3  V 1/3  x  h
x V
 x  h means it is a cube

27. Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A
(4, 1), B (6, 6) and C (8, 4).
OR
Find the area enclosed between the parabola 4y = 3x and the straight line 3x – 2y + 12 = 0.
2

Solution:

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 CBSE-XII-2017 EXAMINATION

6 1
Equation of AB is y – 1 = ( x  4)
64
 2y – 2 = 5x – 20
5x
 y  9
2
Equation of BC is
46
 y 6  ( x  6)
86
 y = – x + 12
Equation of AC is
4 1
 y 1  ( x  4)
84
 4y – 4 = 3x – 12
3x
 y  2
4
Area of  ABC = area ABED + area BEFC – area ADFC
6  5x  6  3x 
    9  dx     x  12  dx     2  dx
8

4
 2  6 4
 4 
6 8 6
 5x2    x2   3x 2 
  9x     12 x     2 x   7 sq units
 4 4  2 6  8 4
OR
Parabola 4y = 3x …(1)
2

line 3x – 2y + 12 = 0 …(2)
3x  12
from (2) y 
2
putting this value of y is (1) we get
6x + 24 = 3x2
 x = 4, – 2
when x = 4 then y = 12
x = – 2 then y = 3
Required area
4 4
   y of line  dx   ( y of parabola) dx
2 2

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 CBSE-XII-2017 EXAMINATION

 3x  12 3x 2 
 
4
  dx
2
 2 4 
3 4
  (8  2 x  x 2 ) dx
4 2
4
3 x2 
 8 x  x 2    27 sq. units
4 3  2

dy
28. Find the particular solution of the differential equation (x – y)  ( x  2 y), given that y = 0 when x =
dx
1.
Solution:
dy
( x  y)  ( x  2 y)
dx
dy x  2 y

dx x y
Let y = Vx
dy dV
V  x
dx dx
dV x  2(Vx)
V  x 
dx x  Vx
dV 1  2V
V  x 
dx 1  V
dV 1  2V  V  V 2
x 
dx 1V
1V dx
 dV  
1V V 2
x
1  (2V  1)  3  dx
   2 
dV  
2  1V V  x
1  2V  1 dV  dx
   dV  3 2

2  1V V 2
1V V  x

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 CBSE-XII-2017 EXAMINATION

1 3 dV
  log |1  V  V 2 |   2
 log | x |  C
2 2  1  3
2

V     
 2  2 
 1
1 3 2  V  
  log |1  V  V 2 |  tan 1  2   log | x |  C
2 2 3  3 
 
 2 
 y 
 2 1
1 y y2
  log 1   2  3 tan 1  x   log | x |  C
2 x x  3 
 
we have y = 0 when x = 1
 1 
 0  3 tan 1  0C
 3
1
 C  3 tan 1
3
 Solution
 y 
 2 1
1 y y2 1 1
  log 1   2  3 tan 1  x   log | x |  3 tan
2 x x  3  3
 

29. Find the coordinates of the point where the line through the points (3, –4, –5) and (2, –3, 1), crosses the
plane determined by the points (1, 2, 3), (4, 2, –3) and (0, 4, 3).
OR
A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C.
1 1 1 1
Show that the locus of the centroid of triangle ABC is 2  2  2  2 .
x y z p
Solution:
Equation of line passing through
(3, – 4, – 5) and (2, – 3, 1)
x 3 y  4 z 5
  ...(1)
1 1 6
Equation of plane passing through
(1, 2, 3) (4, 2, – 3) and (0, 4, 3)
x 1 y  2 z  3
3 0 6  0
1 2 0
 (x – 1) (12) – (y – 2) (– 6) + (z – 3) (6) = 0
 2x + y + z – 7 = 0 …(2)
Let any point on line (1)

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 CBSE-XII-2017 EXAMINATION

is P (– k + 3, k – 4, 6k – 5)
it lies on plane
 2(– k + 3) + k – 4 + 6k – 5 – 7 = 0
5k = 10
 k=2
 P (1, –2, 7)
OR
Let the equation of plane
x y z
   1 ...(1)
a b c
It cut the co-ordinate axes at A, B and C
 A (a, 0, 0), B (0, b, 0), C (0, 0, c)
Let the centroid of  ABC be (x, y, z)
 a b c
  x  , y  , z   ...(2)
 3 3 3
given that distance of plane (1) from origin is 3p

1
  3p
1 1 1
 
a 2 b2 c2
1 1 1 1
 2
 2 2 2
a b c 9p
from (2)
1 1 1 1
 2 2 2 2
9x 9 y 9z 9p
1 1 1 1
 2  2  2  2 Proved
x y z p
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CBSE-XII-2017 EXAMINATION CBSE-XII-2015 EXAMINATION

SET-1 MATHEMATICS
Series SSO/1 Paper & Solution Code: 65/1/1/D
Time:
GM 3 Hrs. Max. Marks: 100

General Instructions:
(i) All questions are compulsory.
(ii) Please check that this Question Paper contains 26 Questions.
(iii) Marks for each question are indicated against it.
(iv) Questions 1 to 6 in Section-A are Very Short Answer Type Questions carrying one mark each.
(v) Questions 7 to 19 in Section-B are Long Answer I Type Questions carrying 4 marks each.
(vi) Questions 20 to 26 in Section-C are Long Answer II Type Questions carrying 6 marks each
(vii) Please write down the serial number of the Question before attempting it.

SECTION – A
Question numbers 1 to 6 carry 1 mark each.
1. If a = 7i + j 4k and b = 2i +6j+3k, then find the projection of a on b .
Solution:
a.b 8
p 
b 7

2. Find λ , if the vectors a  iˆ  3 ˆj  kˆ, b  2iˆ  ˆj  kˆ and c  ˆj  3kˆ are coplanar.
Solution:
1 3 1
2 1 1  0    7
0  3

3. If a line makes angles 900 ,600 and θ with x, y and z- axis respectively, where θ is Acute, then find θ .
Solution:

cos2   cos2   cos2   1   
2 3 6

i j
4. Write the element a 23 of a 3 × 3 matrix A = (aij ) whose elements a ij are given a ij = .
2
Solution:
23 1
a23  
2 2

A
5. Find the differential equation representing the family of curves v = + B, where A and B are arbitrary
r
constants.
Solution:
dv A d 2v dv
  2 ,  r 2 2  2r 0
dr r dr dr

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 CBSE-XII-2015 EXAMINATION

6. Find the integrating factor of the differential equation

e2 x y dx
=1.
x x dy
Solution:
1
I .F   dx  e2 x

e x

SECTION – B
Question numbers 7 to 19 carry 4 marks each.
 2 0 1
7. If A   2 1 3  find A2 5A + 4I and hence find a matrix X such that A2 5A + 4I + X = O
 1 1 0 
 
OR
 1 2 3
If A   0 1 4 , find ( A)1 .
 
2 2 1 
Solution:
5 1 2 

Getting A   9
2
2 5 
0 1 2 

5 1 2   10 0 5   4 0 0 
A2  5 A  4I   9 2 5    10 5 15    0 0 0 
0 1 2   5 5 0   0 0 4 

 1 1 3 
  1 3 10 
 5 4 2 

1 1 3 
 X   1 3 10 
 5 4 2 
 
OR
 1 0 2 
A '   2 1 2 
3 4 1
 
A'  1(9)  2(5)  9  10  1  0
 9 8 2 
 
Adj A '   8 7 2 
 5 4 1 
 

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 CBSE-XII-2015 EXAMINATION

 9 8 2 
  A '   8 7 2 
1

 5 4 1 
 

 a 1 0 
8. if f(x )   ax a 1 , using properties of determinants find the value of F(2x) - f(x).

ax 2 ax a 
Solution:
a 1 0
f  x   ax a 1
ax 2 ax a
R2  R2  xR1 and R3  R3  x2 R1
a 1 0
f  x   0 a  x 1 (For bringing 2 zeroes in any row/column
0 ax  x2 a
 f ( x)  a(a2  2ax  x2 )  a( x  a)2
 f (2x)  f ( x)  a 2x  a  a  x  a 
2 2

 a x (3x  2a)

dx
9. Find :  sin x  sin 2x
OR
Integrate the following w. r. t. x
x 2 3x +1
1 x2
Solution:
dx dx sin x  dx
 sin x  sin 2x   sin x (1  2 cos x)   1  cos x 1  cos x 1  2cos x 
dt
  where cos x = t
1  t 1  t 1  2t 
 1 1 4 
  6  2  3  dt
 1  t 1  t 1  2t 
 
1 1 2
  log 1  t  log 1  t  log 1  2 cos x  c
6 2 3
OR
x  3x  1
2 2  3x  1  x 2
 dx
 1  x2
dx  
1  x2

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 CBSE-XII-2015 EXAMINATION

1 x
 2 dx  3 dx   1  x2 dx
1 x 2
1 x 2

x 1
 2sin 1 x  3 1  x2  1  x2  sin 1 x  c
2 2
3 1
or  sin 1 x  (6  x) 1  x2  c
2 2

 (cos ax  sin bx) dx

2
10. Evaluate :


Solution:
  
I   cos ax  sin bx  dx    cos ax  sin 2 bx  dx   2cos ax sin bx dx
2 2

  

 I1  I 2

I1  2  cos2 ax  sin 2 bx  dx (being an even fun.)
0

I2 = 0 (being an odd fun.)


 I  I1   1  cos 2ax  1  cos 2bx  dx
0

 sin 2ax sin 2bx 
 2 x  
 2a 2b  0
 1 sin 2b 
 2   sin 2a  or 2
 2a 2b 

11. A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1
or 2 appears on it, then bag A is chosen, otherwise bag b. If two balls are drawn at random (without
replacement ) from the select bag, find the probability of one of them being red and another black.
OR
An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained.
Solution:
. Let E1 : selecting bag A, and E2 : selecting bag B.
 P( E1 )  1 , P( E2 )  2
3 3
Let A : Getting one Red and one balck ball
4
C 6 C1 8 7
C 3 C 7
 P  A E1   101  , P  A E2   101 1 
C2 15 C2 15
P( A)  P(E1)  P( A E1)  P(E2 )  P  A E2 
1 8 2 7 22
    
3 15 3 15 45
OR

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 CBSE-XII-2015 EXAMINATION

32
Mean   x P( x)  2
16
Variance   x2 P( x)    xP( x)  
80
  2  1
2 2

16

12. If r x i yj zk, find (r i) . (r j) xy

Solution:
 
r  i  xiˆ  yjˆ  zkˆ xiˆ   ykˆ  zjˆ

r  j   xiˆ  yjˆ  zkˆ  ˆj  xkˆ  ziˆ

 r  iˆ,  r  j   oiˆ  zjˆ  ykˆ  ziˆ  ojˆ  xkˆ  xy

 r  iˆ   r  j   xy  xy  xy  0
13. Find the distance between the point (-1 , -5, -10) and the point of intersection of the line
x 2 y +1 z 2
= = and the plane x – y + z = 5.
3 4 12
Solution:
x  2 y 1 z  2
. Any point on the line   is  3  2,4  1,12  2
3 4 12
If this is the point of intersection with plane x – y + z = 5
Then 3  2  4  1  12  2  5  0    0
 Point of intersection is (2, –1, 2)
2
Required distance  (2  1)2  (1  5)2   2  10  13

14. If sin cot 1 (x +1) = cos (tan -1x), then find x.

OR
5π2
If (tan 1x)2 + (cot 1x)2 = , then find x.
8
Solution:

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 CBSE-XII-2015 EXAMINATION

1
Writing cot 1 ( x  1)  sin 1
1  ( x  1)2
1
and tan 1 x  cos1
1  x2
 1   1 
sin  sin 1   cos  cos 1 
 1  ( x  1)2   1  x2 

1
1  x2  2 x  1  1  x2  x  
2
OR
5 2   5
2 2

 tan x    cot x    tan 1 x     tan 1 x  

1 2 1 2 2

8 2  8
3 2
 2  tan 1 x    tan 1 x 
2
0
8
   2  3 2
tan 1 x   3 
4 4, 4
 x  1

1+ x 2 + 1 x 2 dy
15. If y = tan 1
, x 2  1, then find .
1+ x 2 1 x2 dx
Solution:
Putting x2  cos , we get
 1  cos  1  cos 
y  tan 1  
 1  cos  1  cos 
 cos  sin   1  tan 
1  2 2  1  2
 tan  tan
 cos  sin    1  tan  
 2 2  2
  1
y      cos1 x2
4 2 4 2
dy 1 1 x
  2x  
dx 2 1 x 4
1  x4

d2 y dy
16. If x = a cos  + b sin  , y = a sin  - b cos  , show that y2 2
x + y = 0.
dx dx
Solution:
dx
 a sin   b cos 
d
dy
 a cos  b sin 
d

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 CBSE-XII-2015 EXAMINATION

dy a cos  b sin  x
  
dx a sin   b cos  y
dy
Or y  x  0
dx
d 2 y dy dy
 y 2   1  0
dx dx dx

17. The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing
when the side of the triangle is cm ?
Solution:
Let x be the side of an equilateral triangle
dx
  2cm/s.
dt
3x 2
Area (A) 
4
dA 3 dx
  x
dt 2 dt
dA 3
    20   2  20 3 cm 2 s
dt 2

18. Find :  ( x  3) 3  4 x  x2 dx.

Solution:
1
Writing x  3    4  2x   1
2
1
  ( x  3) 3  4x  x2 dx    (4  2x) 3  4 x  x2 dx   7  ( x  2)2 dx
2
1 3 x2 7  x2
  (3  4 x  x2 ) 2  3  4 x  x2  sin 1  c
3 2 2  7 

19. Three schools A, B and C organized a mela for collecting found for helping the rehabilitation of flood
victims. They sold hand made fans, mats and plates from recycled material at a cost of Rs. 25 , Rs. 100, Rs.
50, each. The number of articles sold are given below :

Find the found collected by each school separately by selling the above articles, Also Find the total founds
collected for the purpose. Write one value generated by the above situation.
Solution:

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 CBSE-XII-2015 EXAMINATION

HF . M P
A  40 50 20   25   7000 
B  25 40 30  100    6125 
C  35 50 40   50   7875 
Funds collected by school A : Rs. 7000,
School B : Rs. 6125, School C : Rs. 7875
Total collected : Rs. 21000
For writing one value

SECTION – C
Question numbers 20 to 26 carry 6 marks each.
20. Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b ) R (c, d ) if
ad (b + c) = bc (a + d). show that R is an equivalence relation.
Solution:
 a, b  N ,(a, b)R(a, b) as ab (b  a)  ba(a  b)
 R is reflexive .................... (i)
Let (a, b) R (c, d) for (a, b), (c, d)  N  N
 ad (b + c) = bc (a + d) .................... (ii)
Also (c, d) R (a, b) cb (d + a) = da (c + b) (using ii)
 R is symmetric .................... (iii)
Let (a, b) R (c, d) and (c, d) R (e, f), for a, b, c, d, e, f  N
 ad (b + c) = bc (a + d) and cf (d + e) = de (c + f)
bc ad d e c f
  and 
bc ad de cf
1 1 1 1 1 1 1 1
i.e    and   
c b d a e d f c
1 1 1 1 1 1 1 1
adding we get       
c b e d d a f c
 af (b + e) = be (a + f)
Hence (a, b) R (e, f)  R is transitive ................... (iv)
Form (i), (iii) and (iv) R is an equivalence relation

21. Using integration find the area of the triangle formed by positive x- zxis and tangent and normal to the
circle x 2 + y2 = 4 at (1, 3).
OR
3
Evaluate  (e2 3 x  x2  1) as a limit of a sum.
1
Solution:

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 CBSE-XII-2015 EXAMINATION

Eqn. of normal (OP) : y = 3x

Eqn. of tangent (PQ) is
1 1
y  3   ( x  1) i.e y  (4  x)
3 3
Coordinates of Q (4, 0)
1 4
1
 Req. area   3x dx    4  x  dx
0 1 3
1 4

x2  1  x2 
 3    
2 
4 x
2 3
0 1

3 1  1
  16  8  4    2 3 sq. units
2 3 2
OR
3

 e  x2  1 dx here h 
23 x 2
1
n
 lim h  f (1)  f (1  h)  f (1  2h)  .......  f (1  (n 1)h)
h0

 lim h  e1  2   e13h  2  2h  h2    e16h  2  4h  4h2   .......

h0

 
 e13 n 1 h  2  2  n  1 h   n  1 h2 
2


  
 lim h e1 1  e3h  e6h  .....  e3 n1h  2n  2h 1  2  ....   n  1 
h0 
2

 
 e3nh 1 nh(nh  h) nh  nh  h  2nh  h  
 lim h  e1  3n  h  2nh  2  
h0  e  1 2 6 

 e1
 e 1  4  4  8   e1 e 1  32
6 6

3 3 3 3

1
22. Solve the differential equation : (tan y x)dy = (1+ y2 )dx.
OR
dy xy
Find the particular solution of the differential equation = 2 2 given that y = 1, When x = 0 .
dx x + y
Solution:
Given differential equation can be written as
dx 1 tan 1 y
  x 
dy 1  y 2 1 y2

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 CBSE-XII-2015 EXAMINATION

 Integrating factor is etan 1 y

tan 1y
tan 1 y  e
 Solution is : x  e tan 1
y dy
1 y2
 x  etan 1 y   t et dt where tan 1 y  t
 t et  et  c  e
tan 1y
 tan 1
y 1  c
1
or x  tan 1 y 1  c e tan y

OR
y
dy x
Given differential equation is 
 
2
dx
1 y
x
y dv v
Putting  v to get v  x 
x dx 1  v2
dv v v3
x  v 
dx 1  v2 1  v2
v2  1 dx
  3 dv  
v x
1
 log v  2   log x  c
2v
x2
 log y  2  c
2y
x2
x  0, y  1  c  0  log y  0
2 y2

x 1 y +1 z 1 x 3 y+k z
23. If lines = = and = = intersect , then find the value of K and hence find the
2 3 4 1 2 1
equation of the plane containing these lines.
Solution:
x 1 y  1 z 1
. Any point on line   is  2  1,3 1, 4  1
2 3 4
2  1  3 3 1  k 4  1 3 9
       , hence k 
1 2 1 2 2
Eqn. of plane containing three lines is
x 1 y  1 z 1
2 3 4 0
1 2 1
  5  x 1  2  y  1  1 z 1  0
i.e. 5x  2 y  z  6  0

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 CBSE-XII-2015 EXAMINATION

2 1
24. If A and B are two independent events such that P( A  B)  and P( A  B)  , then find P (A) and
15 6
P (B).
Solution:


P A B   2
15
 
 P A  P  B 
2
15

  1
P A B   P A  P B 
6
    1
6
1  P  A  P  B   or P  B   P  A  P  B  
2 2
………………….(i)
15 15
P  A 1  P  B    or P  A  P  A  P  B   ………………………..(ii)
1 1
6 6
1 2 1
From (i) and (ii) P  A  P  B    
6 15 30
 1 
Let P (A) = x, P (B) = y  x    y 
 30 
 1  2
(i)  y    y  y   30 y  29 y  4  0
2

 30  15
Solving to get y  1 or y  4
6 5
 x  1 or x  5
5 6
Hence P  A  1
5, 
P B 1
6  OR P  A  5 6, P  B  4 5 
25. Find the local maxima and local minima, of the function f ( x)  sin x, 0  x  2. Also find the local
maximum and local minimum values.
Solution:
f  x   sin x  cos x, 0  x  2
f '  x   0  cos x  sin x  0 or tan x  1,
7
 x  3 ,
4 4

 4    12  12 i.e ve so, x  3 4 is LocalMaxima

f " 3

and f " 7     i.e ve so, x  7 is LocalMinima

1 1
4 2 2 4
1 1
Local Maximum value    2
2 2
1 1
Local Minimum value     2
2 2

26. Find graphically, the maximum value of z = 2x + 5y, subject to constraints given below :

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 CBSE-XII-2015 EXAMINATION

2 x  4 y  8.
3x  y  6
x y4
x  0, y  0
Solution:

Correct graphs of three lines

Correctly shading
feasible region
Vertices are
A (0, 2), B (1.6, 1.2), C (2, .0)
Z = 2x + 5y is maximum
at A (0, 2) and maximum value = 10

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CBSE-XII-2017 EXAMINATION CBSE-XII-2014 EXAMINATION

MATHEMATICS
Paper & Solution Code: 65/1
Time: 3 Hrs. Max. Marks: 100

General Instructions:
(i) All question are compulsory.
(ii) The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions
of one mark each, Section B comprises of 12 questions of four marks each and Section C comprises of 7 questions of six
marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2
questions of six marks each. You have to attempt only one of the alternatives in all such questions.
(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION A
Question numbers 1 to 10 carry 1 mark each.

1. If R   x, y  : x  2 y  8 is a relation on N, write the range of R.

Solution:

R   x, y  : x  2 y  8 is a relation on N

Then we can say 2y = 8 – x

x
y  4
2
so we can put the value of x, x = 2, 4, 6 only
we get y = 3 at x = 2
we get y = 2 at x = 4
we get y = 1 at x = 6
so range = {1, 2, 3} Ans.

2. If tan 1 x  tan 1 y  , xy  1, then write the value of x + y + xy.
4
Solution:

tan 1 x  tan 1 y 
4
x y 
 tan 1 
1  xy 4
x y 
  tan
1  xy 4
x y
  1 or, x  y  1  xy
1  xy

or, x  y  xy  1 Ans.

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 CBSE-XII-2014 EXAMINATION

3. If A is a square matrix such that A2 = A, then write the value of 7A – (I + A)3, where I is an identity
matrix.
Solution:
A2 = A
7A – (I + A)3
7A – [(I + A)2(I + A)] = 7A – [( I I + AA + 2AI) (I + A)]
= 7A – [I + A2 + 2AI] [I + A]
= 7A – [I + A + 2A] [I + A]
= 7A – [I + 3A] [I + A]
= 7A – [I I + IA + 3AI + 3A2]
= 7A – [I + A + 3A + 3A]
= 7A – [I + 7A]
= – I Ans.

 x  y z  1 4
4. If    , find the value of x + y.
2 x  y w  0 5
Solution:
 x  y z  1 4
If    then x + y = ?
2 x  y w  0 5
we can compare the element of 2 matrices. so
x – y = – 1 … (1)
2x – y = 0 … (2)
On solving both eqn we get → x = 1, y = 2
so x + y = 3 Ans.

3x 7 8 7
5. If  , find the value of x.
2 4 6 4
Solution:
3x 7 8 7

2 4 6 4
on expanding both determinants we get
12x + 14 = 32 – 42
12x + 14 = –10
12x = –24
x = –2 Ans.

x
6. If f (x) =  t sin t dt , then write the value of f ′ (x).
0

Solution:
x
f ( x)   t sin t dt
0

⇒ f′ (x) = 1⋅x sin x – 0

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 CBSE-XII-2014 EXAMINATION

= x sin x Ans.

7. Evaluate :
4
x
2 x2  1dx
Solution:
4
x
I   2 dx
2 x 1

at x  2
 2 xdx  dt t 5
1 at x  4
xdx  dt
Put x2+1=t 2 t  17

17
1/ 2
I  
4 t
dt

1 17
 log t 
2 4
1
 log 17  log 4
2
1
 log 17 / 4  Ans.
2

8. Find the value of 'p' for which the vectors 3iˆ  2 ˆj  9kˆ and iˆ  2 pjˆ  3kˆ are parallel.
Solution:
Let a  3iˆ  2 ˆj  9kˆ, b  iˆ  2 pjˆ  3kˆ
If a, b are parallel vector then their exist a, λ such that
ab
  
So 3iˆ  2 ˆj  9kˆ   iˆ  2 pjˆ  3kˆ 

put λ = 3 in 2= –2pλ
2 = –2p.3
1
p   Ans.
3
9.Find a .(b  c), if a  2iˆ  ˆj  3kˆ, b  iˆ  2 ˆj  kˆ and c  3iˆ  ˆj  2kˆ.

Solution:

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 CBSE-XII-2014 EXAMINATION

If a  2iˆ  ˆj  3kˆ, b  iˆ  2 ˆj  kˆ, c  3iˆ  ˆj  2kˆ

2 1 3
 
Then a . b  c  1 2 1
3 1 2
expand along R1 = 2[4 – 1] – 1 [–2 – 3] + 3 [– 1 – 6]
= 6 + 5 – 21 = – 10

3  x y  4 2z  6
10. If the Cartesian equations of a line are   , write the vector equation for the line.
5 7 4

Solution:
3  x y  4 2z  6
Cartesian eqn of line is   ,
5 7 4
x 3 y  4 z 3
we can write it as  
5 7 2
so vector eqn is r   3i  4 j  3k    (5iˆ  7 ˆj  2kˆ)
where λ is a constant

SECTION B
Question numbers 11 to 22 carry 4 marks each.
x
11. If the function f : R → R be given by f(x) = x2 + 2 and g : R → R be given by g(x) = , x ≠ 1, find
x 1
fog and gof and hence find fog (2) and gof (–3).
Solution:
f : R  R; f ( x)  x2  2
x
g : R  R; g ( x)  , x 1
x 1
fog  f ( g ( x))
2
 x   x 
 f   2
 x 1  x 1
x2
 2
 x  1
2

x 2  2( x  1) 2

( x  1) 2
x2  2 x2  4 x  2

 x  1
2

3x 2  4 x  2

 x  1
2

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gof  g ( f ( x))
 g ( x 2  2)
x 2
 2

x2  2 1
1 2
x 2
 2  1 x 1
2
x

3 2  4  2  2
2

 fog  2   6
 2  1
2

1 11 1
gof  3  1   1
(3)  1 10 10
2

 1 x  1 x   1 1
12. Prove that tan 1     cos x,
1
 x 1
 1 x  1 x  4 2 2
OR
 x2 1  x  2  
If tan 1    tan    , find the value of x.
 x4  x4 4
Solution:
 1  x  1  x   1 1 1
tan 1     cos x,  x 1
 1  x  1  x  4 2 2
In LHS
put x = cos 2θ
 1  cos2  1  cos2 
tan 1  
 1  cos2  1  cos2 
1
 1  2cos2   1  1  1  2sin 2  
 tan  
 1  2cos   1  1  1  2sin  
2 2

 cos   sin  
 tan 1  
 cos   sin  
1  tan  
 tan 1  
1  tan  
 tan( 4)  tan  
 tan 1  
1  tan( 4).tan  
 tan 1  tan( 4)  
 x  cos 2 
  
   as  cos 1 x 
4 so,  
 2 
 1
  cos 1 x  RHS proved
4 2
OR

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 x2 1  x  2  
tan 1    tan    (1)
 x4  x4 4

 x2 x2 
  
Use formula, tan 1  x4 x4   
1   x  2  .  x  2   4
  x  4   x  4  

  x  2 x  4   x  2.( x  4)  
 tan 1  
  x  4. x  4   x  2 .( x  2)  4


 x  2 x  4   x  2. x  4  1
 x  4  . x  4    x  2  .  x  2 
x2  8  2 x  x2  8  2 x
 1
x2  16  x2  4
2 x2  16
 1
12

 2 x2  12  16  4
 x2  2 x 2

13. Using properties of determinants, prove that

x y x x
5 x  4 y 4 x 2 x  x3
10 x  8 y 8x 3x

Solution:

x y x x
To prove, 5x  4 y 4 x 2 x  x3
10 x  8 y 8x 3x

x x x y x x
LHS  5x 4 x 2 x  4 y 4 x 2 x
10 x 8x 3x 8 y 8x 3x

1 1 1 1 1 1
 x 5 4 2  yx 4 4 2
3 2

10 8 3 8 8 3

Applying C1 → C1 – C2, C2 → C2 – C3 in the first determinant

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 CBSE-XII-2014 EXAMINATION

0 0 1
 x 1 2 2  yx2  0
3

2 5 3

As the first two columns of the 2nd determinant are same.

Expanding the first determinant through R1

1 2
 x3.1.  x3 (5  4)
2 5

 x3  RHS thus proved

dy 
14. Find the value of at   , if x  ae (sin   cos ) and y  ae (sin   cos ).
dx 4
Solution:

y  ae (sin   cos )

x  ae (sin   cos )

dy dy / d 
 (Applying parametric differentiation) … (1)
dx dx / d 
dy
Now,  ae (cos   sin )  ae (sin   cos )
d
 2ae (cos ) (Applying product Rule)

dx
 ae (cos   sin )  ae (sin   cos )
d
 2ae (sin )

dy dx
Substituting the values of and in (1)
d d

dy 2ae cos 
  cot 
dx 2ae sin 
dy 
Now at  
dx 4

[cot ]   cot  1.
4

15. If y = Peax + Qebx, show that

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 CBSE-XII-2014 EXAMINATION

d2y dy
2
 (a  b)  aby  0.
dx dx
Solution:
If y = Peax + Qebx …(1)
dy
 aPeax  bQebx ...(2)
dx
d2y
2
 a2 Peax  b2Qebx ...(3)
dx
multiplying … (1) by ab
we get, aby = abPeax + abQebx … (4)
multiplying (2) by (a + b)
dy
we get,, (a  b)  (a  b)(aPeax  bQebx )  (a 2 Peax  b2 Pebx )  (abPeax  abQebx )
dx
dy
or, (a2bPeax  b2Qebx )  (a  b)  (abPeax  abQebx )
dx
d2y dy
or, 2
 (a  b)  aby  0
dx dx

16. Find the value(s) of x for which y = [x (x – 2)]2 is an increasing function.

OR
x2 y 2
Find the equations of the tangent and normal to the curve 2  2  1 at the point
a b
 
2a , b .

Solution:
y = [x (x – 2)]2
we know, for increasing function we have f '( x)  0

d 
 f '( x)  2[ x( x  2)]  x( x  2) 
 dx 
d 2
Or, f '( x)  2[ x( x  2)] ( x  2 x)
dx
= 2x(x – 2) (2x – 2)
= 4x(x – 2) (x – 1)
For f '( x)  0

i.e., 4x(x – 1) (x – 2)  0
the values of x are :

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 CBSE-XII-2014 EXAMINATION

x [0,1]  [2, ]

OR

The slope of the tangent at  

2 a, b to the curve
x2 y 2

a 2 b2
1

2x 2 yy '
 2 0
a2 b

b2 x  b2 2a b 2
 y'    
a2 y ( 2a , b )
a 2b a

The equation of the tangent :

b 2
y b  ( x  2a) {using point-slope form : y –y1 = m(x – x1)}
a

ay  ab  b 2x  2ab

or b 2x  ay  ab  0

Normal :
1
The slope of the normal =
dy / dx

1 a
 
b 2 b 2
a
Equation of Normal :
a
y b  ( x  2a)
b 2

yb 2  b2 2  ax  2a2

or ax  b 2 y  2(a2  b2 )  0

17. Evaluate :

4 x sin x
 1  cos
0
2
x
dx

OR

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 CBSE-XII-2014 EXAMINATION

Evaluate :

x2
 x2  5x  6
dx

Solution:

4 x sin x
I  dx
0 1  cos 2
x

I 
4(  x)sin(  x)
0 1  cos (  x)
2
dx Applying  f (a  x)   f (x)
 
4 sin x 4 x sin x
I  dx   dx
0 1  cos x 0 1  cos x
2 2

Or,

4 sin x
I  dx  I
0 1  cos x
2


sin x
2I  4 dx
0 1  cos 2
x
/ 2
sin x  2a 2
2I  4.2  
0
dx
1  cos2 x 
Applying 0 f ( x) dx  2 0 f ( x)dx if f (2a – x)= f (x)
/ 2
sin x
I  4  dx
0 1  cos2 x

put cos x = t   sin xdx  dt

as well for x = 0, x  /2

t=1 t=0
0
dt
 I  4
1 1 t
2

1
dt b a
I  4  f ( x)dx   f ( x)dx
0 1 t
2
a b

I  4 tan 11  tan 1 0


 4  2 .
4
OR

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 CBSE-XII-2014 EXAMINATION

x2
 x  5x  6
2
dx

d 
put , x + 2    ( x2  5x  6)   
 dx 
x  2  2x  5  

comparing coefficients of x both sides

1  2    1/ 2
comparing constant terms both sides,
2  5  

1
or, 2  5   
5
5 1
or,   2  
2 2
1 1
x2 (2 x  5) 
 dx   2 2 dx {as x  2  (2 x  5)  }
x  5x  6
2
x  5x  6
2

1
(2 x  5) 1 dx
I   2 dx  
x2  5x  6 2 x2  5x  6
( I1 ) (I2 )

 I  I1  I 2

1 (2 x  5)
I1  
2 x  5x  6
2
dx, put x2  5x  6  t

 (2x  5)dx  dt

 

1 dt 1 t 1/ 2 1

      C  t  C  x  5x  6  C
1/ 2 2
1
2 t 2   1
 2 
1 dx
I2 
2 x  5x  6
2

1 dx 1 dx
2
   2
25 25 2  5  1
2
x  5x    6
2

4 4 x    
 2  2

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 CBSE-XII-2014 EXAMINATION

1  5  5 1 
2 2

.log  x     x        C

2  2  2 2 
 

1  5 
.log  x    x2  5x  6   C
2  2 
Substituting the values of I1 and I2 in (1)
we get,

1  5 
I  x2  5x  6  log  x    x2  5x  6   c
2  2 

dy
18. Find the particular solution of the differential equation  1  x  y  xy, given that y = 0 when x = 1.
dx
Solution:
dy
 (1  x)  y(1  x)
dx
dy
Or,  (1  y)(1  x)
dx
dy
Or,  (1  x)dx
1 y

dy
 1  y   (1  x)dx
x2
log |1  y | x   C
2
given y = 0 when x = 1
1
i.e., log | l 0 | 1  C
2
3
C 
2
 The particular solution is
x2 3
log |1  y | x .
2 2
or the answer can expressed as

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 CBSE-XII-2014 EXAMINATION

x2  2 x  3
log |1  y |
2
 2 x 3)/ 2
or 1  y  e( x
2

 2 x 3)
or, y  e( x  1.
2

dy
19. Solve the differential equation (1  x2 )  y  etan 1x .
dx
Solution:
dy
(1  x2 )  y  etan 1x
dx

dy y e tan 1
 
dx 1  x2 1  x2
It is a linear differential equation of 1st order.
comparing with standard LDE
dy
 P( x) y  Q( x)
dx
1 e tan 1x
P( x)  ; Q ( x ) 
1  x2 1  x2
1
 dx
Integrating factor IF  e Pdx  e 1 x  etan 1x
2

Solution of LDE

y.IF   IF Q( x)dx  C

etan 1x
 y.etan 1x   etan 1x . dx  C
1  x2
1
(etan x )2
y.e tan 1 x
 dx  C ....(1) y
1  x2
1
(etan x )2
To solving  dx
1  x2
1
Put etan x
t
1 1
or etan x .  dt
1  x2

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 CBSE-XII-2014 EXAMINATION

1 1
etan x .etan x
 dx   tdt
1  x2
1
t2 (etan x )2
 C  C
2 2
Substituting in (1)
1

tan 1 x (etan x )2
y.e  C
2

20. Show that the four points A, B, C and D with position vectors 4iˆ  5 ˆj  kˆ,  ˆj  kˆ, 3iˆ  9 ˆj  4kˆ and
4(iˆ  ˆj  kˆ) respectively are coplanar.

OR

The scalar product of the vector a  iˆ  ˆj  kˆ with a unit vector along the sum of vectors b  2iˆ  4 ˆj  5kˆ
and c  iˆ  2 ˆj  3kˆ is equal to one. Find the value of  and hence find the unit vector along b  c.

Solution:

If P.V of A  4iˆ  5 ˆj  kˆ

B   ˆj  kˆ

C  3iˆ  9 ˆj  4kˆ

D  4(iˆ  ˆj  kˆ)

Points A, B, C, D all Coplanar if  AB AC AD  0  (1)

So, AB  PV . . of A   4iˆ  6 ˆj  2kˆ

. . of B  PV

AC  PV . . of A   iˆ  4 ˆj  3kˆ
. . of C  PV

AD  PV . . of A   8iˆ  ˆj  3kˆ
. . of D  PV

So, so for  AB AC AD

4 6 2
 1 4 3
8 1 3

expand along R1 
– 4[12 + 3] + 6[–3 + 24] – 2[1 +32]
= – 60 + 126 – 66

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 CBSE-XII-2014 EXAMINATION

=0
So, we can say that point A, B, C, D are Coplanar proved
OR

Given  a  iˆ  ˆj  kˆ

b  2iˆ  4 ˆj  5kˆ
c  iˆ  2 ˆj  3kˆ

So, b  c  (2  )iˆ  6 ˆj  2kˆ

(2  )iˆ  6 ˆj  2kˆ
Unit vector along (b  c ) 
(2  )2  36  4

(2  )iˆ  6 ˆj  2kˆ

(2  )2  40

given that dot product of a with the unit vector of b  c is equal to 1

So, apply given condition
(2  )  6  2
1
2    40
2

 2    4  (2  )2  40

Squaring 36  2  12  4  2  4  40
 8  8
   1.

21. A line passes through (2, –1, 3) and is perpendicular to the lines

r  (iˆ  ˆj  kˆ)  (2iˆ  2 ˆj  kˆ) and

r  (2iˆ  ˆj  3kˆ)  (iˆ  2 ˆj  2kˆ). Obtain its equation in vector and Cartesian form.

Solution:

Line L is passing through point  (2iˆ  ˆj  3kˆ)

If L1  r  (iˆ  ˆj  kˆ)  (2iˆ  2 ˆj  kˆ)

L2  r  (2iˆ  ˆj  3kˆ)  (iˆ  2 ˆj  2kˆ)

Let dr of line L = a1, a2, a3

The eqn of L in vector form 

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 CBSE-XII-2014 EXAMINATION

r  (2iˆ  ˆj  3kˆ)  k(a1iˆ  a2 ˆj  a3kˆ)

k is any constant.
so by condition that L1is perpendicular to L a1a2 + b1b2 + c1c2 = 0
2a1 – 2a2 + a3 = 0 … (1)
and also

L  L2

so, a1 + 2a2 + 2a3 = 0 …(2)

Solve (1), (2)
3a1 + 3a3 = 0
 a3  a1

put it in (2)
a1 + 2a2 – 2a1 = 0
a
a2  1 let
2

 a 
so dr of L =  a1 , 1 , a1 
 2 

 1 
so we can say dr of L  1, , 1
 2 
so eqn of L in vector form

 ˆj 
r  (2iˆ  ˆj  3kˆ)  k  iˆ   kˆ 
 2 
x  2 y 1 z  3
3-D form   
1 1/2 1

22. An experiment succeeds thrice as often as it fails. Find the probability that in the next five trials, there
will be at least 3 successes.
Solution:
In Binomial distribution

( p  q)n n C0. pn n C1. pn 1.q1 n C2. pn 2.q2  ....... n Cn .qn

if p = probability of success
q = prob. of fail
given that p = 3q …(1)
we know that p + q = 1
so, 3q + q = 1

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 CBSE-XII-2014 EXAMINATION

1
q
4
3
So, p 
4
Now given  n = 5 we required minimum 3 success
(p + q)5 = 5C0.p5 + 5C1.p4.q1 + 5C2.p3.q2
5 4 3 2
 3  3 1  3 1
 C0 .   5C1. 
5
.  5C2 .  . 
 4  4  4  4  4

35 5.34 10.33
   5
45 45 4

35  5.34  10.33 33[9  15  10] 34  27 459

    .
45 45 16  64 512

SECTION C
Question numbers 23 to 29 carry 6 marks each.
23. Two schools A and B want to award their selected students on the values of sincerity, truthfulness and
helpfulness. The school A wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to
3, 2 and 1 students respectively with a total award money of ₹ 1,600. School B wants to spend ₹ 2,300 to
award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values
as before). If the total amount of award for one prize on each value is ₹ 900, using matrices, find the award
money for each value. Apart from these three values, suggest one more value which should be considered
for award.
Solution:
Let Matrix D represents number of students receiving prize for the three categories :
D=
Number of students SINCERITY TRUTHFULNESS HELPFULNESS
of school
A 3 2 1
B 4 1 3
One student for each 1 1 1
value
x 
X   y  where x, y and z are rupees mentioned as it is the question, for sincerity, truthfulness and
 z 
helpfulness respectively.
1600 
E   2300 is a matrix representing total award money for school A, B and for one prize for each value.
900 

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 CBSE-XII-2014 EXAMINATION

We can represent the given question in matrix multiplication as :

DX = E
3 2 1  x  1600 
or 4 1 3  y   2300
    
1 1 1  z  900 
Solution of the matrix equation exist if | D | ≠ 0
3 2 1
i.e., 4 1 3  3[1  3]  2[4  3]  1[4  1]
1 1 1
=–6–2+3
=–5
therefore, the solution of the matrix equation is
X = D-1 E
1
To find D-1; D-1 = adj ( D)
| D|
Cofactor Matrix of D
2 1 3 
  1 2 1
 5 5 5
Adjoint of D = adj (D)
2 1 5 
  1 2 5
 3 1 5
{transpose of Cofactor Matrix}
2 1 5 
 D1 
1  1 2 5
5  
 3 1 5
Now, X = D-1E
2 1 5  1600 
1 
  1 2 5 2300
5
 3 1 5 900 
 x   200
 y   300 
   
 z   400
∴ x = 200, y = 300, z = 400. Ans.
Award can also be given for Punctuality.

24. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of
4r 8
radius r is . Also show that the maximum volume of the cone is of the volume of the sphere.
3 27

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 CBSE-XII-2014 EXAMINATION

Solution:
Let R and h be the radius and height of the cone.
r be the radius of sphere.

4r
To show h 
3
and Maximum Volume of Sphere
8
 Volume of Sphere
27
In  C  h  r
∴ (h – r)2+ R2 = r2 {Pythagorus Theorem}
 R 2  r 2  (h  r ) 2
1
Volume of cone : V  R 2h
3
1
or, V  (r 2  (h  r )2 )h
3
1
V  [r 2  h2  r 2  2hr ]h
3
1
V  [2h2r  h3 ]
3
dV
For maxima or minima, 0
dh
dV 1
Now,  [4hr  3h2 ]
dh 3
dV
Putting, 0
dh
We get 4hr = 3h2
4r
h
3
2
dV 1
 [4r  6h]
dh2 3
4r
Putting h =
3
dV 1 
2
6.4r 
   4r 
dh 2
3  3 
1
  [4r ]
3

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 CBSE-XII-2014 EXAMINATION

Which is less than zero, therefore

4r
h is a Maxima
3
4r
and the Volume of the cone at h 
3
will be maximum,
1
V  R 2h
3
1
 [r 2  (h  r )2 ]h
3
1  2  4r    4r 
2

  r    r    
3   3    3 
1  8r 2   4r 
 
3  9   3 
8  4r 3 
  
27  3 
8
= (Volume of the sphere)
27

25. Evaluate :
1
 cos4 x  sin4 xdx
Solution:
1
 cos4 x  sin4 xdx
1
4
dx
 cos x
1  tan 4 x
sec2 x sec 2 xdx

1  tan 4 x
(1  tan 2 x)sec2 xdx

1  tan 4 x
put tan x = t ⇒ sec2x dx = dt
(1  t 2 )dt

1 t4
( 1 2  1)dt
 t {dividing each by t 2 }
1 2
t
t2


1  1t  dt
2

t  1t 
2
2

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 CBSE-XII-2014 EXAMINATION

1  1
Put t   z  1  2  dt  dz
t  t 
dz 1
 2  tan 1 z C
z 2 2
1  1 
 tan 1  tan x  C
2  tan x 
1
 tan 1 (tan x  cot x)  C
2

26. Using integration, find the area of the region bounded by the triangle whose vertices are (– 1, 2), (1, 5)
and (3, 4).
Solution:
Let A = (–1, 2)
B = (1, 5)
C = (3, 4)

We have to find the area of  ABC

 25 
Find eqn of Line AB  y  5     ( x  1)
 1  1 
3
y  5  ( x  1)
2
2y – 10 = 3x – 3
3x – 2y + 7 = 0 …(1)
3x  7
y
2
54
Eqn of BC  y  4     ( x  3)
 1 3 
1
y  4  ( x  3)
2
2 y  8  x  3
x  2 y  11  0 .......(2)
11  x
y
2
 24 
Eqn of AC  y  4     ( x  3)
 1  3 
1
y  4  ( x  3)  2 y  8  x  3
2
x – 2y + 5 = 0 …(3)

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 CBSE-XII-2014 EXAMINATION

x5
y
2
 3x  7   11  x   x 5
1 3 3
So, required area =   dx    dx    dx
1 
2  1
2  1 
2 
1 3 3
1  3x 2  1 x2  1  x2 
   7 x   11x      5x 
2 2  1 2  2 1 2  2  1
1  3  3  1  9  1  1  9  1 
   7     7    33    11      15     5 
2  2  2  2  2  2  2  2   2 
1 1
 [14  22  4  24]  [36  28]  4 square unit
2 2

27. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y +
4z = 5 which is perpendicular to the plane x – y + z = 0. Also find the distance of the plane obtained above,
from the
origin.
OR
Find the distance of the point (2, 12, 5) from the point of intersection of the line
r  2iˆ  4 ˆj  2kˆ  (3iˆ  4 ˆj  2kˆ) and the plane r .(iˆ  2 ˆj  kˆ)  0.
Solution:
Eqn of given planes are
P1  x + y + z – 1 = 0
P2  2x + 3y + 4z – 5 = 0
Eqn of plane through the line of intersection of planes P 1, P2 is
P1 +  P2 = 0
(x + y + z – 1) +  (2x + 3y + 4z – 5) = 0
(1 + 2  ) x + (1 + 3  ) y + (1 + 4  ) z + (– 1 – 5  ) = 0 … (1)
given that plane represented by eqn (1) is perpendicular to plane
x–y+z=0
so we use formula a1a2 + b1b2 + c1c2 = 0
so (1 + 2  ).1 + (1 + 3  ). (– 1) + (1 + 4  ).1 = 0
1 + 2 – 1 – 3 + 1 + 4 = 0
3 + 1 = 0
1

3
1
Put    in eqn (1) so we get
3
 2  4 2
1   x  (1  1) y  1   z   0
 3  3 3
x z 2
  0
3 3 3
x – z + 2 = 0 Ans.
OR
General points on the line:

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 CBSE-XII-2014 EXAMINATION

x = 2 + 3, y = – 4 + 4, z = 2 + 2
The equation of the plane :
r .(iˆ  2 ˆj  kˆ)  0
The point of intersection of the line and the plane :
Substituting general point of the line in the equation of plane and finding the particular value of  .
[(2  3)iˆ  (4  4) ˆj  (2  2)kˆ].(iˆ  2 ˆj  kˆ)  0
(2  3).1  (4  4)(2)  (2  2).1  0
12  3  0 or,   4
∴ the point of intersection is :
(2 + 3 (4), –4 +4(4), 2 + 2(4)) = (14, 12, 10)
Distance of this point from (2, 12, 5) is
 (14  2)2  (12  12)2  (10  5)2 {Applying distance formula}
 122  52
= 13 Ans.

28. A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each
type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12
labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum
labour hours available per week are 180 and 30 respectively. The company makes a profit of ₹ 80 on each
piece of type A and ₹120 on each piece of type B. How many pieces of type A and type B should be
manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the
maximum profit per week?
Solution:
Let pieces of type A manufactured per week = x
Let pieces of type B manufactured per week = y
Companies profit function which is to be maximized : Z = 80x + 120y
Fabricating hours Finishing hours
A 9 1
B 12 3
Constraints : Maximum number of fabricating hours = 180
9x  12 y  180  3x  4 y  60 K
Where 9x is the fabricating hours spent by type A teaching aids, and 12y hours spent on type B.
and Maximum number of finishing hours = 30
 x  3y  30
where x is the number of hours spent on finishing aid A while 3y on aid B.
So, the LPP becomes :
Z (MAXIMISE) = 80x + 120 y
Subject to 3x  4 y  60
x  3 y  30
x0
y0
Solving it Graphically :

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 CBSE-XII-2014 EXAMINATION

Z = 80x + 120y at (0, 15)

= 1800
Z = 1200 at (0, 10)
Z = 1600 at (20, 0)
Z = 960 + 720 at (12, 6)
= 1680
Maximum profit is at (0, 15)
∴ Teaching aid A = 0
Teaching aid B = 15
Should be made

29. There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that
comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of
the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the
two-headed coin?
OR
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote
the larger of the two numbers obtained. Find the probability distribution of the random variable X, and
hence find the mean of the distribution.
Solution:
If there are 3 coins.
Let these are A, B, C respectively
For coin A  Prob. of getting Head P(H) = 1
3
For coin B  Prob. of getting Head P(H) =
4
For coin C  Prob. of getting Head P(H) = 0.6
 
we have to find P A
H
 Prob. of getting H by coin A
So, we can use formula
 
P H .P( A)
 
P A  A
H
 
P H
A  
.P( A)  P H
B  
.P( B)  P H .P(C )
C
1
Here P(A) = P(B) = P(C) = (Prob. of choosing any one coin)
3

   
P H  1, P H  , P H  0.6
A B 4
3
  C
Put value in formula so

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 CBSE-XII-2014 EXAMINATION

1
1.
 
P A  3 
1
1.  .  (0.6) 1  0.75  0.6
H 1 3 1 1
3 4 3 3
100

235
20
 Ans.
47
OR
First six numbers are 1, 2, 3, 4, 5, 6.
X is bigger number among 2 number so
Variable (X) 2 3 4 5 6
Probability
P(X)
if X = 2
for P(X) = Prob. of event that bigger of the 2 chosen number is 2
So, Cases = (1, 2)
1 1
So, P( X )  6  ....(1)
C2 15
if X = 3
So, favourable cases are = (1, 3), (2, 3)
2 2
P( x)  6  ....(2)
C2 15
if X = 4  favourable casec = (1, 4), (2, 4), (3, 4)
3
P( X )  ....(3)
15
if X = 5  favourable casec = (1, 5), (2, 5), (3, 5), (4, 5)
4
P( X )  ....(4)
15
if X = 6  favourable casec = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6)
5
P( X )  ....(5)
15
We can put all value of P(X) in chart, So
Variable (X) 2 3 4 5 6
Probability 1 2 3 4 5
P(X) 15 15 15 15 15
1 2 3 4 5
and required mean  2.   3.   4    5.   6. 
 15   15   15   15   15 
70 14
  Ans.
15 3

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CBSE-XII-2017 EXAMINATION CBSE-XII-2013 EXAMINATION

MATHEMATICS
Paper & Solution Code: 65/1
Time: 3 Hrs. Max. Marks: 70

General Instructions:
(i) All questions are compulsory.
(ii) The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises
of10 questions of one mark each, Section B comprises of 12 questions of four marks each and Section
Comprises of 7 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement
of the questions.
(iv) There is no overall choice, However, internal choice has been provided in 4 questions of four marks
each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such
questions.
(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

1. Write the principal value of tan–1 ( 3)  cot 1 ( 3

Solution:
tan 1 ( 3)   / 3
cot 1 ( 3)     / 6
Hence
 / 3  (   / 6)   / 2

  3 
2. Write the value of tan–1 2sin  2cos1  .
 2 
 
Solution:
3 
cos 1 
2 6
 tan 1 (2sin( 2. / 6))
 
 tan 1  2.sin 
 3
 3
 tan 1  2. 1
  tan 3   / 3
 2 

 0 1 2
3. For what value of x, is the matrix A = 1 0 3  a skew-symmetric matrix ?
 
 x 3 0 
Sol. The value of determinant of skew symmetric matrix of odd order is always equal to zero

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 CBSE-XII-2013 EXAMINATION

 0 1 2 
 1 0 3   0
 
 x 3 0 
 1(0  3x)  2(3  0)  0
 3x  6  0  x2

 1 1
4. If matrix A =   and A2 = kA, then write the value of k.
1 1 
Sol. Given A2 = kA
 1 1  1 1  1 1
 1 1   1 1   k  1 1 
     
 2 2   k k 
 2 2   k k   k  2
  

5. Write the differential equation representing the family of curves y = mx, where m is an arbitrary constant.
Sol. y = mx .....(1)
differentiating with respect to x, we get
dy/dx = m
∴ differential equation of curve
xdy
y
dx

2 3 5 
6. If Aij is the cofactor of the element aij of the determinant 6 0 4  , then write the value of a32 ⋅ A32.
 
1 5 7
2 3 5 
Sol. 6 0 4 
 
1 5 7 
A32  (1)3 2 M32 where M32is the min or of a 32 .
2 5
A32  (1)3 2
6 4
2 5
A32    A32  (8  30)
6 4
A32  22
 a32A32  5(22)  110

7. P and Q are two points with position vectors 3a  2b and a  b respectively. Write the position vector of
point R which divides the line segment PQ in the ratio 2 : 1externally. 1

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 CBSE-XII-2013 EXAMINATION

Sol. P.V. of P is 3a  2b

P.V. of Q is a  b
Point R divides segment PQ in ratio 2 : 1 externally.
(P.V.of p)1  (P.V.of Q)(2)
P.V.of R 
1 2
(3a  2b)(1)  (a  b)(2) a  4b
P.V.of R  
1 2 1
P.V.of R  4b  a

8. Find x , if for a unit vector a,(x  a).(x  a)  15.

Sol. Given a  1

(x  a).(x  a)  15
2 2
x  a  15
2
x  1  15
2
x  15  1
2
x  16

x 4

9. Find the length of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 21 = 0.
ax1  by1  cz1  d
Sol. p
a 2  b2  c2

0  0  0  21 21 21
p p p p3
2 3 6
2 2 2
49 7

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 CBSE-XII-2013 EXAMINATION

10. The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of
its total revenue (marginal revenue). If the total revenue (in rupees) received from the sale of x units of a
product is given by R(x) = 3x2 + 36x + 5, find the marginal revenue, when x = 5, and write which value does
the equations indicate.
Sol. R(x) = 3x2 + 36x + 5
dR
MR   6x 2  36
dx
when x  5
MR  30  36  66

11. Consider f : R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f–1 of f given
by
f –1(y) = y  4, where R+ is the set of all non-negative real numbers.
Sol. f : R   [4.)
f (x)  x 2  4
f (x)  x 2  4  (one  one)
As f (x)  x 2  4  4
 Rage  [4.)  co  domain
 onto
Further : y  x 2  4
so f is invertible.
 y  4  x2  x   y  4
As x  0 so x  y  4
 y  x  4  f 1 (x)
Or f 1 (y)  y  4

12. Show that :

1 3 4 7
tan  sin 1  
2 4 3
OR
Solve the folowing equation :
 3
cos (tan 1 x)  sin  cot 1 
 4
Solution:
1 3 3
Let sin 1   then  sin 2
2 4 4
1 3
Now tan  sin 1   tan 
2 4

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 CBSE-XII-2013 EXAMINATION

3 2 tan  3
if sin 2  then 
4 1  tan  4
2

8 tan   3  3 tan 2 
3 tan 2   8 tan   3  0
8  64  4  3  3 |
tan  
6
8  28 4  7
tan   
6 3
4 7 4 7
tan   or
3 3
1 3 4 7
tan  sin 1   Hence proved.
2 4 3
OR
cos(tan 1 x)
LHS. lrt tan 1 x    x  tan 
1 1
cos   
1  tan 
2
1 x2
1
Hence cos ( tan 1 x) 
1 x2
3 3
R.H.S Let cot 1     cot 
4 4
1 1 4
then sin    
1  cot 2  1
9 5
16
Now LHS  RHS
1 4

1 x2 5
25  16  16x 2
9 3
x2  x
16 4

13. Using properties of determinants, prove the following :

x x  y x  2y
x  2y x x  y  9y2 (x  y)
x  y x  2y x
Solution:

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 CBSE-XII-2013 EXAMINATION

x x  y x  2y
x  2y x x  y  9y 2 (x  y)
x  y x  2y x
x x  y x  2y
LHS x  2y x xy
x  y x  2y x
Now, apply C1  C1  C2  C3
3x  3y x  y x  2y
3x  3y x xy
3x  3y x  2y x
3x  3y x  y x  2y
3(x  y) 3x  3y x xy
3x  3y x  2y x
 y 2y
3(x  y)
2y y
1 2
3y2 (x  y)
2 1
3y2 (x  y) (1  4)  9y2 (x  y). Hence proved.

dy (1  log y)2
14. If yx = ey – x, prove that 
dx log y
Sol. yx  e yx
 x log e y  y  x ....(1)
Differentiating w.r.t.x
1 dy dy
 log e y  x.  1
y dx dx
dy  x  x 1
 log e y  1  1  {form(1)  }
dx  y  y 1  log e y
dy  1 
 log e y  1  1  
dx  1  log e y 
dy  log e y 
 (log e y  1)  1  
dx  1  log e y 
dy (1  log e y) 2
 
dx log e y

15. Differentiate the following with respect to x :

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 CBSE-XII-2013 EXAMINATION

 2x 1.3x 
sin 1  x 
 1  (36) 
Solution:
1  2 .3 
x 1 x
y  sin  x 
1  (36) 
 2x.2.3x 
y  sin 1  x 
1  (36) 
 2.(6) x 
y  sin 1  2x 
1  (6) 
y  2 tan 1 (6) x
dy 2
 .6x log 6
dx 1  (6) 2x
dy 2.6x log 6

dx 1  (36) x

 1  kx  1  kx if  1  x  0

16. Find the value of k, for which f (x)   x , is continuous at x  0.
 2x  1
, if 0  x  1
 x 1
OR
d2 y 
If x = a cos3θ and y = a sin3θ, then find the value of 2
at   .
dx 6
Solution:
 1  kx  1  kx if  1  x  0

f (x)   x ,
 2x 1
, if 0  x  1
 x 1
function f(x) is continuous at x = 0

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 CBSE-XII-2013 EXAMINATION

 f (0)  lim f (x)

x 0

0 1  1  kx  1  kx 
  lim  
0  1 x 0  x 
 1  kx  1  kx  1  kx  1  kx 
  1  lim  
 1  kx  1  kx 
x 0 x
  
(1  2k)  (1  kx)
  1  lim
x 0 x[ 1  kx  1  kx ]

2k
  1  lim
x 0 1  kx  1  kx

2k
 1   k  1
2
OR
x  a cos  3
and y  a sin 
3

dx dy
 3a cos2 sin  and  3a sin 2  cos 
d d
x  a cos 
3
and y  a sin 3 
dx dy cos 
 3a cos 2  sin  and  3a sin 2 
d d
dy
dy d
 
dx dx
d
dy
   tan 
dx
d2 y 1
 2   sec2 
dx (3a cos 2 .sin )
d2 y 1
 2
 sec4  cosec 
dx 3a
 d2 y 
4
1 2  32
 2    .2 
 dx   3a  3  27a
6

17. Evaluate :
cos 2x  cos 2
 cos x  cos 
OR
Evaluate :
x2
 x 2  2x  3 dx
Solution:

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 CBSE-XII-2013 EXAMINATION

cos 2x  cos 2
 cos x  cos 
dx

(2cos 2 x  1)  (2cos 2   1)
 dx
cos x  cos 
2(cos 2 x  1)(2 cos 2   1)
 dx
cos x  cos 
 2 (cos x  cos )dx
 2(sin x  x cos )  c
OR
x2
I dx
x  2x  3
2

 (x  1)  1 
I   dx
 x 2
 2x  3 
 (x  1)   1 
I   dx    dx
 x 2
 2x  3   x 2
 2x  3 
I  I1  I 2
In I1let x 2  3  t 2
 (2x  2)dx  2t 2t
 (x  1)dx  tdt
t.dt
 I1   t
t
I1  x 2  2x  3
1 dx
Now in I1   dx  
x 2  2x  3 x 2  2x  3
I2  log[(x  1)2  (x  1) 2  2]
Now I  I1  I 2
 I  x 2  2x  3  log(x  1  x 2  2x  | 3)  c

18. Evaluate :
dx
 x(x5  3)
Solution:

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 CBSE-XII-2013 EXAMINATION

dx
I
x(x 5  3)
x 4dx
I
x(x 5  3)
Let x 5  t  5x 4dx  dt
1 dt
I 
5 t(t  3)
1 1 1 1 
I  .   dt
5 3  t t 3
1
I  {log t  log(t  3)}  c
15
1  t 
I  log   c
15  t 3
1  x5 
I log   5 c
15  x 3

19. Evaluate
2 1

0 1  esin x
dx

Solution:
2 1
I dx ....(1)
0 1  esin x

2 1
I  x)
dx
0 1 e sin(2

2 1
I dx
0 1  e  sin x

sin x
2 e
I   sin x dx
0 e 1
Adding (1) & (2) we get
2  1  e 
sin x
 21     dx
 1 e
sin x
0

 21  [x]02 
 21  2
I

20. If a  i  j  7k and b  5i  j  k , then find the value of λ, so that a  b and a  b are perpendicular
vectors.
Solution:

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 CBSE-XII-2013 EXAMINATION

a  i  j  7k
b  5i  j   k
a  b  6i  2 j  (7  )k
a  b  4i  0 j  (7  )k
given (a  b) and (a  b) are perpendicular
 (a  b).(a  b)  0
{6i  2 j  (7  )k}  {4i  0 j  (7  )k}  0
6(4)  0(2)  (7  )(7  )  0
 24  49   2  0
 2  25    5

21. Show that the lines

r  3i  2j  4k  (i  2j  2k);
r  3i  2j  (3i  2j  6k)
are intersecting. Hence find their point of intersection.
OR
Find the vector equation of the plane through the points (2, 1, –1) and (–1, 3, 4) and perpendicular to the
plane x – 2y + 4z = 10.
Sol. If the given lines are intersecting then the shortest distance between the lines is zero and also they have
same
common point r  3i  2j  4k  (i  2j  2k)
x 3 y2 z  4
   ( )(Let)
1 2 2
Let P is (  3, 2  2, 2  4)
Also, r  5i  2 j  (3i  2 j  6k)
x 5 y 2 z 0
   ( )(Let)
3 2 6
Let Q is (  5, 2  2, 6)
If lines are intersecting then P and Q will be same.
  3  3  5 ....(1)
2  2  2  2 ....(2)
2  4  6 ....(3)

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 CBSE-XII-2013 EXAMINATION

Solve(2) & (3)

 1   1
2  2  3
  
3  2  1
4  2
  2
put   2 ....(3)
2  4  6(2)
2  12  4
2  8
  4
put  & in (1)
  3  3  5
 4  3  3 (2)  5
 1  1
 from   4 then P is (1, 6, 12)
from   2 then Q is (1, 6, 12)
as P and Q are same
∴ lines are intersecting lines and their point of intersection is (–1, –6, –12).
OR

AB  OB  OA
AB  3i  2j  5k
given plane x  2y  4z  10
 n1  i  2j  4k
The required plane is perpendicular to given plane.
Therefore n r of required plane will be perpendicular to n1 and AB.

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 CBSE-XII-2013 EXAMINATION

 n || (n1  AB)
n1  i  2j  4k
AB  3i  2j  5k
 n1  AB  18i  17 j  4k
 required plane is
r. n  a. n
r. (18i  17 j  4k)  (2i  j  k).(18i  17 j  4k)
r. (18i  17 j  4k)  36  17  4
r. (18i  17 j  4k)  49
18x  17 y  4z  49

3 5
22. The probabilities of two students A and B coming to the school in time are and respectively.
7 7
Assuming that the events, ‘A coming in time’ and ‘B coming in time’ are independent, find the probability
of only one of them coming to the school in time. Write at least one advantage of coming to school in time.
Sol. If P(A come in school time) = 3/7
P (B come in school time) = 5/7
P (A not come in school time) = 4/7
P (B not come in school time) = 2/7
P (only one of them coming school in time)
 P(A)  P(B)  P(A).P(B)
3 2 5 4 26
    
7 7 7 7 49

x 2 y2
23. Find the area of the greatest rectangle that can be inscribed in an ellipse  1
a 2 b2
OR
4 
Find the equations of tangents to the curve 3x2 – y2 = 8, which pass through the point  ,0 
3 
x 2 y2
Sol. Given ellipse is  1
a 2 b2

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 CBSE-XII-2013 EXAMINATION

Area of reactangle
A  2a  cos . 2b sin 
A  2ab. sin 2
 Amax  2ab
OR
Let a po int (x1 , y1 )
3x 2  y 2  8  6x  2y.y' = 0

 y' 
3x
y
3x1
 Tangent y  y1  (x  x1 )
y1
4 
It pas sin g throgh  , 0 
3 
3x  4 
 y1  1   x1 
y1  3 
  y12  4x1  3x12  y12  4x1  3x12
 3x12  8  3x12  4x1
 x1  2
So 12  y 2  8
 y 2  4  y1  2

24. Find the area of the region bounded by the parabola y = x2 and y = | x |.
Sol.

Required area = 2[area of ΔOAB – Area of curve OCBA]

1 
A  2  (1)(1)   x 2dx 
1

2 0

1 1 1 1
A  2    A  2  
 2 3 6 3

25. Find the particular solution of the differential equation (tan–1y – x)dy = (1 + y2)dx, given that when x =
0, y=0

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 CBSE-XII-2013 EXAMINATION

Sol. (tan–1 y – x)dy = (1 + y2)dx

dx tan 1 y x
  
dy 1  y 2
1  y2
dx x tan 1 y
  
dy 1  y 2 1  y 2
1
 1 y2 dy
IF  e
1
IF  e tan y

x. IF   Q.IF dy  c
tan 1 y tan 1 y

1
 x. e tan y
.e dy  c
(1  y 2 )
Put tan 1 y  t
1
 x. e tan y
 (t.e t )  (e t )  c
1 1 1
 x. e tan y
 tan 1 y. e tan y
 e tan y
c

26. Find the equation of the plane passing through the line of intersection of the planes r.(i  3j)  6  0 and
r .(3j  j  4k)  0, whose perpendicular distance from origin is unity.

OR
Find the vector equation of the line passing through the point (1, 2, 3) and parallel to the planes
r.(3i  j  2k)  5 and r.(3i  j  k)  6
Sol. P1 is r.(3i  j)  6  0
1 x  3y  6  0
Pis
P2is r .(3j  j  4k)  0
P2is 3x  y  4z  0
Equation of plane passing through intersection of P 1 and P2 is P1 + λP2 = 0
(x  3y  6)  (3x  y  4z)  0
(1  3)x  (3  )y  (4)z  (6)  0
Its distance from (0, 0, 0) is 1.
0006
(1  3)2  (3  )2  (4)2
36 = (1 + 3λ)2 + (3 – λ)2 + (–4λ)2
36 = 1 + 9λ2 + 6λ + 9 + λ2 – 6λ + 16λ2
36  262 10  262  26  2  1    1
Hence required plane is
For λ = 1, (x + 3y – 6) + 1 (3x – y – 4z) = 0
4x + 2y – 4z – 6 = 0

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 CBSE-XII-2013 EXAMINATION

For λ = –1, (x + 3y – 6) – 1 (3x – y – 4z) = 0

– 2x + 4y + 4z – 6 = 0
OR
Pis
1 r .(i  j  2k)  5
 n1  i  j  2k
P2is r .(3i  j  2k)  6
n 2  3i  j  k
The line parallel to plane P1 & P2 will be perpendicular to n1 & n 2
 b || (n1  n 2 )
n1  i  j  2k
n 2  3i  j  k
n1  n 2  3i  5j  4k
 b  3i  5j  4k
Point is (1, 2, 3)
 a  i  j  3k
 required line is r  a |  b
r  (i  2j  3k)  (3i  5j  4| k)

27. In a hockey match, both teams A and B scored same number of goals up to the end of the game, so to
decide the winner, the referee asked both the captains to throw a die alternately and decided that the team,
whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, find
the irrespective probabilities of winning the match and state whether the decision of the referee was fair or
not.
Sol. P(6 get) = 1/6
P(6 not get) = P (6get) = 5/6
P.(A win)  P(Aget6)  P(6get).P(6get)P(6get)  P(6get).P(6get) .P(6get). .P(6get)P(6get)  …..+ 
1 5 5 1 5 5 5 5 1
P.(A win)           ...  
6 6 6 6 6 6 6 6 6
2 4
1 5 1 5 1
          ...  
6 6 6 6 6
a
S 
1 r
1
 
   
6 36 6

 25  11 6 11
1  
 36 
Similarly winning for B

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 CBSE-XII-2013 EXAMINATION

P(B win) = 1 – P(A win)

6 5
 1 
11 11

28. A manufacturer considers that men and women workers are equally efficient and so he pays them at the
same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to
produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required
while 3workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at j− 100 and
j− 120per unit respectively, how should he use his resources to maximise the total revenue ? Form the above
as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women
workers are equally efficient and so should be paid at the same rate ?
Sol. if zmax. = 100x + 120y
typeA typeB
worker 2 3 30
capitl 3 1 17
Subject to,
2x+3y≤30
3x+y≤17
x≥0
Let object of type A = x
Object of type B = y

pts coordinate Zmax=100x+ 120y

O (0,0) Z=0
A  17  Z
1700
 ,0 
3  3
E (3.8) Z=300+960=1260
C (0.10) Z=1200
maximum revenue = 1260.

29. The management committee of a residential colony decided to award some of its members (say x) for
honesty, some (say y) for helping others and some other (say z) for supervising the workers to keep the
colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation

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 CBSE-XII-2013 EXAMINATION

and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of
awardees for honesty and supervision is twice the number of awardees for helping others, using matrix
method, find the number of awardees of each category. Apart from these value, namely, honesty,
cooperation and supervision, suggest one more value which the management of the colony must include for
awards.
Sol. Given
x + y + z = 12 …..(1)
3(y + z) + 2x = 33 …..(2)
(x + z) = 2y …..(3)
x + y + z = 12
2x + 3y + 3z = 33
x – 2y + z =
1 1 1  x  12 
2 3 3  y   33
    
1 2 1  z   0 
AX = B
A–1(AX) = A–1(B)
I ⋅ X = A–1⋅B
X = A–1.B
(Adj.A) . B
X
|A|
 1 1 1
| A |  2 3 3
1 2 1
| A | = 1(3 + 6) – 1(2 – 3) + 1 (–4 – 3)
|A|=9+1–7=3
|A|≠0
 9 3 0 
(Adj. A)   1 0 1
7 3 1 
 9 3 0  12 
(Adj. A) .B   1 0 1 33
7 3 1 33  0 33
9
(Adj. A) .B  12
15
(Adj.A) . B
 X
|A|

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 CBSE-XII-2013 EXAMINATION

9  3  x   3
X  12  X  4   y   4
1 
3
15 5  z  5
x = 3, y = 4, z = 5.

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