(Solution)

Class XII Mathematics

Sol.1 If a and b are reciprocal, then a = b , R + and | a || b |= 1

| a || || b |
|a| 1
| |= = 2
|b| |b|
1
| a |= b
| b |2
1
a b = | b || b | cos 0 = 1
| b |2

Sol.2 Given, R = {(a, a3)! A is prime number less than 5}

We know that, 2 and 3 are the prime No. less than 5.
a can take values 2 and 3
Then R = {(2, 23), (3, 33)}
= {(2, 8), (3, 27)}
Hence, the range of R is {8, 27}

Sol.3 Let
2 3 4
= 5 6 8
6x 9x 12 x
On taking 3x common from R3 we get
2 3 4
= 5 6 8
2 3 4
==0
[ R1 and R3 are identical].

Sol.4 we know that the principal value branch of tan x is ,
2 2

tan 1 ( 1) = tan 1 tan
4

= tan 1 tan
4

= ,
4 2 2

Hence, tan 1 ( 1) =
4

Sol.5 Area of parallelogram

1
= ( AC BD )
2
i j k
1
= 1 2 1
2
1 2 5
1
= (8i + 4 j )
2
= 4i + 2 j
Area of parallelogram
=| 4i + 2 j |= 2 5 sq.units.

Sol.6 [ a + b , b + c, c + a ]
= (a + b) (b + c) (c + a )

= (a + b) b c + b a + c a

= a b c b c a

= 2 a b c

Sol.7 Given
a + 4 3b 2a + 2 b + 2
= =
8 6 8 a 8b
We know that two matrices are equal, if its corresponding elements are equal
a + 4 = 2a + 2 (1)
3b = b + 2 ..(2)
On solving equation 1, 2 and 3, we get
a = 2 and b = 1
now a 2b = 0

1
Sol.8 Given, f(x) = 27x3 and g(x) = x 3
g of (x) = g[f(x)]
= g (27x3)
1
= (27 x 3 ) 3
 1 1
= (27 3 ) 3 ( x 3 ) 3
1 1
= (33 ) 3 ( x 3 ) 3
= 3x
g of (x) = 3x

A P( A B)
Sol.9 P =
B P(B)
P( A B)
P=
1
3
P
= P( A B) (1)
3
P ( AUB ) = P ( A) + P ( B ) P ( A B )
5 P 1 P
= +
9 1 3 3
5 3P + 1 P
=
9 3
5
1 = 2P
3
53
= 2P
3
2
= 2P
3
2 1
P= =
6 3

Sol.10 we have to prove

x 3 3x 2
cos 1 + cos 1 + =
2 2 3
L.H.S.
x 3 3x 2
= cos 1 ( x) + cos 1 +
2 2
Let cos 1 ( x) = x = cos
Then, L.H.S. = + cos-1
3
cos cos + 1 cos 2
3 2

= + cos 1 cos cos + sin sin
3 3
3
sin = 1 cos , sin =
2

3 2

= + cos 1 cos
3
[ cos(A B) = cosAcosB + sinAsinB]

= +
3 3

Sol.11 The circumference of a circle with radius is given by

C = 2r
Therefore, the rate of change of circumference with respect to time is given
by :-
dc dc dr
=
dt dr dt
d dr
= (2 r )
dr dt
dt
= 2
dt
= 2 (0.7)
= 1.4 cm/s.

Sol.12 Given that,

e (tan x + 1)sec x)dx = e f ( x) + c
x x

e (sec x + sec x tan x)dx = e f ( x) + c

x x

e x sec x + c = e x f ( x) + c

e x { f ( x) + f 1 ( x)} dx = e x f ( x) and
d
(sec x) = sec tan x
dx
On comparing both sides, we get
F(x) = secx.

Sol.13 L.H.S:-
1 + sin x + 1 sin x 1 + sin x + 1 sin x
cot
1 + sin x 1 sin x 1 + sin x + 1 sin x
( 1 + sin x + 1 sin x ) 2
= cot 1 2
( 1 + sin x ) ( 1 sin x )
2
 1 + sin x + 1 sin x + 2 1 sin 2 x
= cot 1
1 + sin x 1 + sin x
2 + 2 cos x
= cot 1
2sin x
x
2 cos 2
1 + cos x
= cot 1 = cot 1 2
sin x x x
2sin cos
2 2
2 x x x
cos x = 2 cos 2 1 and sin x = 2 sin 2 cos 2

x x
= cot 1 cot = = R.H .S
2 2

1 x x2
Sol.14 L.H .S . = x 2 1 x
2
x x 1
On applying C1 C1 + C2 + C3, we get
1 + x + x2 x x2
L.H .S . = 1 + x + x 2 1 x
1 + x + x2 x2 1
On taking common (1 + x + x2), from
1 x x2
= (1 + x + x ) 1
2
1 x
1 x2 1
On applying R2 R2 R1, R3 + R3 R1
1 x x2
= (1 + x + x 2 ) 0 1 x x(1 x)
0 x( x 1) 1 x 2
On expanding along c1, we get
1 x x(1 x)
= (1 + x + x 2 )
x(1 x) (1 x)(1 + x)
Taking common
= (1 + x + x 2 )(1 x) 2 (1 + x + x 2 )
2
= (1 + x + x 2 )(1 x)
= (1 x3 ) 2 = R.H.S
Sol.15 Let

( x 3) x 2 + 3 x 18dx
d 2
Here, we can write x 3 = A ( x + 3 x 18) + B
dx
x 3 = A(2 x + 3) + B
On equation the coefficients of x and constant term from both sides, we get
2A = 1 and 3A + B = -3
1 1
A = and 3 + B = 3
2 2
1 9
A = and B =
2 2
Thus, the given integral reduces in the following form
1 9
I = (2 x + 3 x 2 + 3 x 18dx
2 2
1 9
I = (2 x + 3) x 2 + 3 x 18dx x 2 + 3 x 18dx
2 2
1 9
= I1 I 2
2 2
Where = I1 = (2 x + 3) x 2 + 3 x 18dx
Put x 2 + 3 x 18 = t
(2 x + 3) dx = dt
1 3
2
I1 = t dt = t 2 + C1
2
3
3
2 2
= ( x 3 x 18) 2 + C1
3
and I 2 = x 2 + 3 x 18dx
2
3 9
= x + 18 dx
2 4
2
3 81
= x + dx
2 4
2 2
3 9
= x + dx
2 2
3
x+
= 2 x 2 + 3x 18 81 log x + 3 + x 2 + 3 x 18 + c
2
2 8 2
2x + 3 2 81 2x + 3
= x + 3 x 18 log + x 2 + 3 x 18
4 8 2
 On putting the volume of I1, and I2 in I then
1 2 9 2x + 3 2
3
I = ( x 2 + 3x 18) 2 + c1 x + 3 x 18
2 3 2 4
81 2x + 3
log + x 2 + 3 x 18 + c2
2 2
3 3
1 9
I = ( x3 + 3 x 18) 2 (2 x + 3) x3 + 3 x 18) 2
3 8
729 2x + 3
+ log + x 2 + 3 x 18 + c
16 2
c1 9c2
Where c =
2 2

Sol.16 The required line passes through the origin. Therefore, its position vector is
given by,
a=o ..(1)
The dir. Ratios of the line through origin and (5, -2, 3) are (5 0) = 5, (-2, -0) =
-2
(3 0) = 3
The line is parallel to the vector given by the equation of the b = 5i 2 j + 3k .
The equation of the line in vector form through a point with position vector a
and parallel to b is r = a + b; R
r = 0 + (5i 2 j + 3k )
r = (5i 2 j + 3k)
The equation of the line passes through the point (x1 y1 z1) and dir. Ratios the
point a, b, c is given by
x x1 y y1 z z1
= =
a b c
Therefore, the equation of the required line in the Cartesian form is
x0 y0 z0
= =
5 2 3
x y z
= =
5 2 3

Sol.17 Given
2 0 1

A = 2 1 3
1 1 0
We have to find the value of A2 3A + 2I.
 Now A2 = A : A
2 0 1 2 0 1

= 2 1 3 2 1 3

1 1 0 1 1 0
5 1 2
A = 9
2
2 5
0 1 2
2 0 1 6 0 3
3 A = 3 2 1 3 6
3 9
1 1 0 3 3 0
1 0 0
And 2 I = 2 0 1 0
0 0 1
2 0 0
= 0 2 0
0 0 2
A2 3A + 2I
5 1 2 6 0 3 2 0 0

= 9 2 5 6 3 9 + 0 2 0
0 1 2 3 3 0 0 0 2
1 1 1
= 3 3 4
3 2 0

Sol.18 Given y logx = x y .(1)

diff. wrt x
1 dy dy
y + log x = 1
x dx dx
y dy dy dy
+ log x + + =1
x dx dx dx
dy y x y
(1 + log x) = 1 =
dx x x
dy x y
(1 + log x ) = ..(2)
dx x
From 1st y log x = x y
Use in 2nd
 dy y log x
(1 + log x) =
dx x
Solve equation Ist
y log x + y = x
y (log x + 1) = x
x
(1 + log x) = (3)
y
x
Use value of in equation
y
dy log x
(1 + log x) =
dx (1 + log x)
dy log x
=
dx (1 + log x) 2
= R.H.S

Sol.19 x = a (cos + sin )

Diff. wrt
dx
, a ( sin + cos + sin )
d
= a cos
dy
= b(cos + sin cos )
d
= b sin
dy
dy d
=
dx dx
d
b sin
=
a cos
dy b
= tan
dx a
Diff. wrt x
d2 y b d
2
= sec 2
dx a dx
b 1
= sec2
a a cos
d2 y b
== 2 sec3
dx 2
a

Sol.20 The given diff. equation is:

 sec 2 x tan ydx + sec 2 y tan xdy = 0
sec 2 x tan dx + sec 2 y tan xdy
=0
tan x tan y
sec 2 x sec 2 y
dx + dy = 0
tan x tan y
sec 2 x sec 2 y
dx = dy
tan x tan y
Integrating both sides of this equation , we get :-
sec 2 sec 2 y
tan x dx = tan y dy ..(1)
Let tanx = t
d dt
(tan x) =
dx dx
dt
sec2 x = , sec 2 xdx = dt
dx
sec2 x 1
Now, tan x dx = t dt
= log t
= log(tan x )
sec 2 y
Log; tan y dy = log(tan y )
Substituting value in 1st
log(tan x ) = log(tan y ) + log c
c
log(tan x) = log
tan y
c
tan x =
tan y
tan x tan y = c
This is the required solution of the Given equation

1
Sol.21 probability of solving the problem by A, P ( A) =
2
1
Probability of solving the problem by B, P ( B ) =
3
Since the problem is solved independently by A and B,
1 1 1
P ( AB ) = P ( A) P ( B ) = =
2 3 6
1 1
P ( A1 ) = 1 P ( A) = 1 =
2 2
 1 2
P ( B1 ) = 1 P ( B ) = 1 =
3 3
i) Probability that the problem is solved = P ( A B )
= P ( A) + P ( B ) P ( AB )
1 1 1 4
= + =
2 3 6 6
2
=
3
ii) Probability that exactly one of them solved the problem is given by,
1 2 1 1
= +
2 3 2 3
1 1
= +
3 6
1
=
2

1
Sol.22 Let I = dx
1 tan x
1
= dx
sin x
1
cox
cos x
= dx
cos x sin x
1 2 cos x
= dx
2 cos x sin x
1 (cos x sin x) + (cos x + sin x)
= dx
2 cos x sin x
1 1 cos x + sin xdx
= 1dx +
2 2 cos x sin x
x 1 cos x + sin x
= + dx
2 2 cos x sin x
Put cosx sinx = t
( sin x cos x ) dx = dt
x 1 dt
2 2 t
I= +

x 1
= log | t | + c
2 2
x 1
= log cos x sin x + c
2 2

Sol.23 Let r be the radius of circle and x be the side of a square then, given that
 Perimeter of square + circumference of circle = k
i.e. 4 x + 2 r = k
k 2 r
x= (1)
4
Let A denotes the sum of their areas
A = x2 + r 2 (2)
On putting the value of x we get
k 2 r
2

A= +r
2

4
On diff. wrt r
dA k 2 r 2
= 2 + 2 r
dr 4 4

= ( k 2 r ) + 2 r
4
dA
For max. and min. =0
dr

( k 2 ) + 2 r = 0
4
2r
k+ + 2 r = 0
4 2
k
r= (3)
2 + 8
d2A 2 2
Now . = 2 +
dr 2 4
2
= 2 + >0
2
d2A
> 0 A is min.
dr 2
from equation 3rd, we get
k
r=
2 + 8
2 r + 8r = k
2 r + 8r = 4 x + 2 r
8r = 4 x
Or x = 2r
Side of sq. = double the radius of circle

Sol.24 The area add by the curves, (x 1)2 + y2 = 1 and x2 + y2 1, is represented by

On solving the equation
(x 1)2 + y2 = 1 and x2 + y2 = 1
We obtain the point of intersection
1 3 1 3
A = , and B = ,
2 2 2 2
Area OBCAO = 2 area OCAO
12 1


= 1 ( x 1) dx + 1 x dx
2 2

0 1
2

1
x 1
1
1 2 x 1
= 1 ( x 1) 2 + sin 1 ( x 1) + 1 x 2 + sin 1 x
2 2 0 2 2 1
2

1 1
2
1 1
2
1 1 1 1
= 1 + sin 1 1 sin 1 ( 1) + sin 1 (1) 1
4 2 2 2 2 2 4 2

1 1
+ sin 1
2 2

3 1 1 1 3 1
= + +
8 2 6 2 2 2 2 8 2 6
3
= + +
4 12 4 4 12
3
= +
4 6 2
2 3
=
6 4
Therefore, required area
2 3
OBCAO = 2
6 4
 2 3
= unit
3 2

Sol.25 Let H denotes the students who read Hindi newspaper and E denote the
students who read English newspaper.
It is Given that,
6 3
P ( H ) = 60% = =
10 5
40 2
P ( E ) = 40% = =
100 5
1
P ( H E ) = 20% =
5
i) Probability that a student reads Hindi or English newspaper is
P ( H E )1 = 1 P ( H E )
= 1 { P( H ) + P( E ) P( H E )}
4 1
= 1 =
5 5
E P( E H )
ii) P =
H P( H )
1
1
=5 =
3 3
5
H P( H E )
iii) P =
E P( E )
1
1
= 5 =
2 2
5

Sol.26 Let be the semi vertical angle of the cone:

It is clear that = 0,
2
Let r, h and l be the radius height and slant height of the cone respectively.
The slant height of cone is constant.
 R = l sin, h = l cos
1
v = r2h
3
1
= l 2 sin 2 )(l cos
3
1 3 2
= l sin cos
3
dx l 3
= sin 3 + 2 sin cos 2

d 3
d 2 v l 3
= 2 cos3 7 sin 2 cos
d 2
3
du
Now =0
d
sin 3 = 2 sin cos 2
tan 2 = 2
tan = 2
= tan 1 2
When = tan 1 2, then tan 2 = 2 or sin 2 = 2 cos 2
d 2 v l 2
= 2 cos3 14 cos3
d 2 3

= 4 l 3 cos 3 < 0 for 0,
2
By 2nd derivative test,
The volume is maximum
= tan 1 2 .
Hence, for a given height the semi-vertical angle of the cone of the maximum
volume is tan 1 2 .

Sol.27 Given that

2x + 3y + 10z = 4
4x 6y + 5z = 1
Ax = B where
2 3 10 x 4

A = 4 6 5 , x = y B = 1

6 9 20 z 2
| A |= 150 + 330 + 720 = 1200
A11 = 75, A12 = 110,
A13 = 72, A21 = 150, A22 = -100
A23 = 0, A31 = 75
 A32 = 30, A33 = -24.
1
A1 = adj A
| A|
75 100 75
1
= 110 100 30
1200
72 0 24
Now
X = A-1 B
x 75 150 75 4
y = 1 110 100 30 1
1200
z 72 0 24 2
600
1
= 400
1200
240
x 1 / 2
y = 1 / 3

z 1 / 5

Sol.28 The equation of the plane through the point (-1, 3, 2) is

a(x + 1) + b(y 3) + C(z 2) = 0 (1)
where a, b, c are the direction ratios of normal to the plane
It is known that 2 Planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0
Are , if a1a2 + b1b2+c1c2 = 0 Plane (1) if perpendicular to plane
3x + 3y + 3 = 0
a3+b3+c1=0
3a + 3b +c = 0 .(2)
Perpendicular to plane x + 2y + 3z = 5
a1 + b2 + c3 = 0
a + 2b + 3c = 0 (3)
From 2 & 3 equation
a b c
= = =k
7 8 3
a = -7k, b = 8k, c = -3k
Use a, b, c in equation 1st 7k(x + 1) + 8k(y 3) -3k(z 2) = 0
-7x + 8y 3z 25 = 0
7x 8y + 3z + 25 = 0
This is required equation of plane.

Sol.29 Let the mixture contain xkg of food and ykg of food minimize
Z = 16x + 20y ..(1)
X + 2y 10 ..(2)
X+y6 ..(3)
3x + y 8 ..(4)
X, y 0 ..(5)

The corner point of the feasible region A(10, 0) B (2, 4) C(1, 5), D=(0, 8)
at Z = 16x + 20y
A(10, 0) 160
B(2, 4) 112 Minimum
C(1, 5) 116
D(0, 8) 160
Minimum value of z = 112
At (2, 4)