CBSE Math XII Board Paper – SET 1 (Delhi)
MATHEMATICS
Time allowed : 3 hours] [Maximum marks : 100
i. All questions 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 requirements 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.
1. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)}
not to be transitive.
Solution:
It is known that a relation R in a set A is transitive if (a1, a2) R and (a2, a3) R
(a1, a3) R a1, a2, a3 A.
It can be observed that (1, 2), (2, 1) R, but (1,1) R.
Thus, the relation R in the set {1, 2, 3} is not transitive.
2. Write the value of 1π 1sin sin
3 2
Solution:
1
1
1
π 1sin sin
3 2
1Let sin
2
1sin
2
π π πsin sin sin sin 2π
6 6 6
π2π
6
π 1 π πsin sin sin 2π
3 2 3 6
9π 9πsin sin
6 6
3π πsin sin π
2 2
x
x
x
x
πsin 1
2
Thus, the value of the given expression is 1.
3. For a 2 2 matrix, A = [aij] whose elements are given by ij
ia
j , write the value of
a12.
Solution:
Any element in the ith
row and jth
column is ij
ia
j
For a12, the value of i = 1 and j = 2.
12
1
2a
Thus, the value of a12 is 1
2.
4. For what value of x, the matrix 5 1
2 4
x x
is singular?
Solution:
The matrix 5 1
2 4
x xA
is singular
0
5 10
2 4
4 5 2 1 0
20 4 2 2 0
6 18
183
6
A
x x
x x
x x
x
x
Thus, the required value of x is 3.
5. Write A–1
for 2 5
1 3A
Solution:
1
1
2 5A
1 3
3 51A
1 2det A
3 51
1 22 3 1 5
3 51
1 26 5
3 5A
1 2
6. Write the value of sec sec tanx x x dx
Solution:
secx (secx + tanx) dx
= (sec2x + secx tanx)dx
= sec2x dx + secx tanx dx
= tanx + secx + c , where c is an arbitrary constant
7. Write the value of 2 16
dx
x
Solution:
2
2 2
16
4
dx
x
dx
x
11tan
4 4
xC
, where C is the arbitrary constant
8. For what value of ‘a’ the vectors ˆˆ ˆ2 3 4i j k and ˆˆ ˆ6 8ai j k are collinear?
Solution:
Two vectors x and y are collinear if x y , where is a constant.
It is given that ˆ ˆˆ ˆ ˆ ˆ2 3 4 and 6 8i j k ai j k are collinear.
ˆ ˆˆ ˆ ˆ ˆ2 3 4 6 8i j k ai j k for some constant .
2 , 3 6 ,4 8
13 6 or 4 8 gives
2
a
Now, 2 = a
12
2
4
a
a
Thus, the value of a is –4.
9. Write the direction cosines of the vector ˆˆ ˆ2 5i j k .
Solution 9:
The given vector is ˆˆ ˆ2 5i j k .
The direction cosines of the given vector is given by
2 2 2 2 2 2 2 2 2
2 1 5, ,
2 1 5 2 1 5 2 1 5
2 1 5, ,
30 30 30
10. Write the intercept cut off by the plane 2x + y – z = 5 on x-axis.
Solution:
2 5 ... 1x y z
Dividing both sides of the equation (1) by 5, we obtain
21
5 5 5
1 ... 25 5 5
2
y zx
x y z
It is known that the equation of a plane in intercept form is 1x y z
a b c , where a, b
and c are the intercepts cut off by the plane at x, y, and z- axes respectively.
Therefore, for the given equation, 5
, 5, and 52
a b c
Thus, the intercept cut off by the given plane on the x-axis is 5
.2
SECTION – B
Question numbers 11 to 22 carry 4 marks each.
11. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min.
{a, b}. Write the operation table of the operation *.
Solution:
The binary operation on the set {1, 2, 3, 4, 5} is defined by a b = min {a, b}
1 1 = min {1, 1} = 1
1 2 = min {1, 2} = 1
2 3 = min {2, 3} = 2
4 5 = min {4, 5} = 4
The operation table for the given operation on the given set can be written as:
1 2 3 4 5
1 1 1 1 1 1
2 1 2 2 2 2
3 1 2 3 3 3
4 1 2 3 4 4
5 1 2 3 4 5
12. Prove the following:
1 1 sin 1 sin πcot , 0,
2 41 sin 1 sin
x x xx
x x
OR
Find the value of 1 1tan tanx x y
y x y
Solution:
2
2 2
2
2 2
1 sin cos sin 2sin cos cos sin2 2 2 2 2 2
1 sin cos sin 2sin cos cos sin2 2 2 2 2 2
x x x x x xx
x x x x x xx
1
2 2
1
2 2
1
1 sin 1 sincot
1 sin 1 sin
cos sin cos sin2 2 2 2
cot
cos sin cos sin2 2 2 2
cos sin cos sin2 2 2 2
cot
cos sin cos sin2 2 2
x x
x x
x x x x
x x x x
x x x x
x x x
1
1
2
2cos2cot
2sin2
cot cot2
2
x
x
x
x
x
Thus, the result is proved.
OR
1 1
1 1
1 1 1 1 1
1 1 1
1
tan tan
1
tan tan
1
1
tan tan tan tan tan1
1 1
tan tan tan 1
tan
x x y
y x y
x
x y
xy
y
x
x a bya b
xy ab
y
x x
y y
x
1 1
1
tan tan 1
tan 1
π
4
x
y y
Thus, the value of the given expression is π
.4
13. Using properties of determinants, prove that
2
2 2 2 2
2
4
a ab ac
ba b bc a b c
ca cb c
Solution:
2
2
1 2 3
2 2 2
1 2 3
Taking out factors , , from R , R , and R respectively
1 1 1
1 1 1 Taking out factors , , from C , C , and C respectively
1 1 1
a ab ac
ba b bc
ca cb c
a b c
abc a b c a b c
a b c
a b c a b c
Applying R2 R2 + R1 and R3 R3 + R1, we have:
2 2 2
2 2 2
2 2 2 2 2 2
1 1 1
0 0 2
0 2 0
0 21
2 0
0 4 4
a b c
a b c
a b c a b c
Hence, the result is proved.
14. Find the value of ‘a’ for which the function f defined as
3
πsin 1 , 0
2
tan sin, 0
a x x
f xx x
xx
is continuous at x = 0
Solution:
The given function f is defined as
3
πsin 1 , 0
2
tan sin, 0
a x x
f xx x
xx
The given function is defined for all real numbers.
By, definition, f is continuous at x = 0, if 0 0
lim lim 0x x
f x f x f
00
π πlim lim sin 1 sin
2 2xxf x a x a a
300
30
30
2
30
2
0 0 0
02
tan sinlim lim
sinsin
coslim
sin 1 coslim
cos
sin 2sin2lim
cos
sin1 sin 22lim lim lim
cos
sin1 22 1 1 lim4
2
12 1 1 1
4
1
2
xx
x
x
x
x x x
x
x x
x
xx
x
x
x
x x
xx
x x
xx
x x x
x
x
Now, π π
0 sin 0 1 sin 12 2
f a a a a
Since f is continuous at x = 0, 0 0
lim lim 0x x
f x f x f
1
2a
Thus, for 1
2a , the given function f is continuous at x = 0.
15. Differentiate 2
cos
2
1
1
x x xx
x
w.r.t. x
OR
If x = a ( – sin), y = a (1 + cos), find 2
2
d y
dx
Solution: 2
cos
2
1
1
x x xx
x
Let y = xxcosx
and 2
2
1
1
xz
x
y = xxcosx
Taking log of both sides:
coslog log
log cos log
x xy x
y x x x
Differentiating with respect to x, we obtain:
cos
1 1cos log cos
1cos log cos sin
cos log cos sin
cos log cos sin ... 1x x
dy dx x x x x
y dx x dx
dyx x x x x
y dx
dyy x x x x x
dx
dyx x x x x x
dx
2
2
1
1
xz
x
Differentiating with respect to x, we obtain:
2
2
2 2
2 2
2
22 2
2
22 2
2
22
22
22
22
1
1
1 11 1
1 1
1 22 1
1 1
2 21
1 1
2 11
11
2 2
11
4
1
4... 2
1
dz d x
dx dx x
d dx x
x dx dx x
xx x
x x
x xx
x x
x x
xx
x
xx
x
x
dz x
dx x
On adding (1) and (2):
cos
22
2cos cos
22 2
4cos log cos sin
1
1 4cos log cos sin
1 1
x x
x x x x
dy dz xx x x x x x
dx dx x
d x xx x x x x x x
dx x x
OR
sin , 1 cosx a y a
On differentiating x and y with respect to , it is obtained:
1 cos ... 1
and 0 sin
sin .. 2
dxa
d
dya
d
dya
d
Dividing equation (2) by equation (1), we obtain:
2
2
sin
1 cos
sin
1 cos
2sin cos2 2
1 1 2sin2
2sin cos2 2
2sin2
cos2
sin2
cot2
dy
ad
dx a
d
dy d
d dx
dy
dx
dy
dx
dy
dx
dy
dx
On differentiating again with respect to x, we obtain:
22
2
2
22
2
2
2
2
4
1cosec
2 2
1 1cosec
2 2
1 1cosec
2 2 1 cos
cosec2
2 1 cos
cosec2
2 2sin2
1.cosec
4 2
d y d
dx dx
dx
d
d y
dx a
a
a
a
16. Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base. How fast is the height of the sand cone increasing when the height
is 4 cm?
OR
Find the points on the curve x2 + y
2 – 2x – 3 = 0 at which the tangents are parallel to x-
axis.
Solution:
The volume (V) of a cone with radius (r) and height (h) is given by,
21π
3V r h
It is given that,
16
6h r r h
2 31
π 6 12π3
V h h h
The rate of change of the volume with respect to time (t) is given by,
312πdV d dh
hdt dh dt
[By chain rule]
2
2
12π 3
36π
dhh
dt
dhh
dt
It is also given that 312 cm / sdV
dt .
Therefore, when h = 4 cm, we have:
212 36π 4
12 1
36π 16 48π
dh
dt
dh
dt
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate
of1
cm/s48π
.
OR
Let P , x y be any point on the given curve x2 + y
2 – 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by dy
dx.
Differentiating the equation of the curve on both the sides with respect to x, we get
2 2 2 0
2 2 1
2
dyx y
dx
dy x x
dx y y
Let 1 1P ,x y be the point on the given curve at which the tangents are parallel to the
x-axis.
1 1,
1
1
1
1
0
10
1 0
1
x y
dy
dx
x
y
x
x
When x1 = 1,
2 2
1
2
1
2
1
1
1 2 1 3 0
4 0
4
2
y
y
y
y
So, the required points are (1, 2) and (1, –2).
Thus, the points on the given curve at which the tangents are parallel to x-axis are (1,
2) and (1, –2).
17. Evaluate: 2
5 3
4 10
xdx
x x
OR
Evaluate: 2 2
2
1 3
xdx
x x
Solution:
2
2
2 2
2 2
1 2
12
5 3
4 10
5 3 4 10
5 3 2 4
5 3 2 4
2 5 and 4 3
5
2
54 3
2
10 3
3 10 7
52 4 7
5 3 2
4 10 4 10
5 2 47
2 4 10 4 10
57 ... 1
2
2 4
4
x
x x
dx A x x B
dx
x A x B
x Ax A B
A A B
A
B
B
B
xx
dxx x x x
x dxdx
x x x x
I I
xI
x
10dx
x
Put x2 + 4x + 10 = z
2
1
1
2
1 1
22
2
22
22
2
2
2
1 2
2
2 4 2
2
2 C
2 4 10 C
4 10
4 4 6
2 6
log 2 2 6 C
log 2 4 10 C
Substituting the values of and in equation (1), it is obtained:
52 4 10
2
x dx zdz
zdzI
z
z
I x x
dxI
x x
dx
x x
dx
x
x x
x x x
I I
I x x
2
1 2
2 2
1 2
C 7 log 2 4 10 C
5 4 10 7 log 2 4 10 C
5Where C C 7C
2
x x x
x x x x x
OR
2 2
2
2
1 3
Put
2
1 3
1
1 3 1 3
1 3 1
xI dx
x x
x z
xdx dz
dz
z z
A B
z z z z
A z B z
Putting z = –3, it is obtained: 1 2
1
2
B
B
Putting z = – 1, it is obtained:
2 2
2 2
1 2
1
2
111 22
1 3 1 3
1 1
1 3 2 1 2 3
1 1log 1 log 3 C
2 2
2 1 1log 1 log 3 C
2 21 3
A
A
z z z z
dz dz dz
z z z z
z z
xdxx x
x x
18. Solve the following differential equation:
ex tan y dx + (1 – e
x) sec
2y dy = 0
Solution:
The given differential equation is:
ex tan y dx + (1 – e
x) sec
2y dy = 0
(ex –1) sec
2y dy = e
x tan y dx
2sec
tan –1
x
x
y edy dx
y e
On integrating on both sides, we get
2sec
. ...(i)tan 1
x
x
y edy dx
y e
2sec
tan
ydy
y
Put tan y = u
sec2y dy = du 2sec
tan
y dudy
y u = log |u| = log tan y … (ii)
–1
x
x
edx
e
Put ex –1 = v
ex dx = dv
–1
x
x
e dvdx
ve
= log v
= log (ex –1) … (iii)
Form (i), (ii) and (iii), we obtain:
log tan y = log (ex –1) + log C
log tan y = log C (ex –1)
tan y = C (ex –1)
Thus, the solution of the given differential equation is tan y = C (ex –1).
19. Solve the following differential equation:
2 πcos tan 0
2
dyx y x x
dx
Solution:
The given differential equation is:
2
2 2
cos tan
sec sec tan
dyx y x
dx
dyx y x x
dx
This equation is in the form of:
2
2 2
sec tan
(where sec and sec tan )
Now, I.F .pdx xdx x
dypy Q p x Q x x
dx
e e e
The general solution of the given differential equation is given by the relation,
tan tan 2
2
2
I.F. Q I.F. C
sec tan C ... 1
Let tan .
tan
sec
sec
x x
y dx
y e e x x dx
x t
d dtx
dx dx
dtx
dx
x dx dt
Therefore, equation (1) becomes:
tan
tan
tan
tan
tan
tan
tan tan
tan
1
tan 1
tan 1 ,where is an arbitrary constant.
x t
x t
x t t
x t t
x t t
x t
x x
x
y e e t dt C
y e t e dt C
dy e t e dt t e dt dt C
dt
y e t e e dt C
y e t e e C
ye t e C
ye x e C
y x Ce C
20. Find a unit vector perpendicular to each of the vector a b and a b , where ˆˆ ˆ3 2 2a i j k and ˆˆ ˆ2 2b i j k .
Solution:
We have, ˆˆ ˆ3 2 2a i j k and ˆˆ ˆ2 2b i j k
2 22
2 2 2 2 2
2 2
ˆˆ ˆ ˆ4 4 , 2 4
ˆˆ ˆ
ˆ ˆˆ ˆ ˆ ˆ4 4 0 16 16 8 16 16 8
2 0 4
16 16 8
2 8 2 8 8
8 2 2 1 8 9 8 3 24
a b i j a b i k
i j k
a b a b i j k i j k
a b a b
Hence, the unit vector perpendicular to each of the vectors a b and a b is given by
the expression,
ˆ ˆˆ ˆ ˆ ˆ16 16 8 2 2 2 2 1 ˆˆ ˆ24 3 3 3 3
a b a b i j k i j ki j k
a b a b
1
1
2
1 1
22
2
22
22
2
2
2
1 2
2
1
2 4 2
2
2 C
2 4 10 C
4 10
4 4 6
2 6
log 2 2 6 C
log 2 4 10 C
Substituting the values of and in equation (1), we obtain:
52 4 10 C 7
2
x dx zdz
zdzI
z
z
I x x
dxI
x x
dx
x x
dx
x
x x
x x x
I I
I x x
2
2
2 2
1 2
log 2 4 10 C
5 4 10 7 log 2 4 10 C
5Where C C 7C
2
x x x
x x x x x
OR
2 2
2
2
1 3
Put
2
1 3
1
1 3 1 3
1 3 1
xI dx
x x
x z
xdx dz
dz
z z
A B
z z z z
A z B z
Putting z = –3, we obtain: 1 2
1
2
B
B
Putting z = – 1, we obtain:
2 2
2 2
1 2
1
2
111 22
1 3 1 3
1 1
1 3 2 1 2 3
1 1log 1 log 3 C
2 2
2 1 1log 1 log 3 C
2 21 3
A
A
z z z z
dz dz dz
z z z z
z z
xdxx x
x x
21. Find the angle between the following pair of lines:
2 1 3
2 7 3
x y z
and
2 2 8 5
1 4 4
x y z
and check whether the lines are parallel or perpendicular.
Solution:
Let 1b and 2b be the vectors parallel to the pair of lines,
2 1 3 2 2 8 5 and
2 7 3 1 4 4
x y z x y z
, respectively.
2 1 3Now,
2 7 3
2 1 3
2 7 3
x y z
x y z
2 2 8 5
1 4 4
2 4 5
1 2 4
x y z
x y z
1ˆˆ ˆ2 7 3b i j k and 2
ˆˆ ˆ2 4b i j k
2 2 2
1
2 2 2
2
1 2
2 7 3 62
1 2 4 21
ˆ ˆˆ ˆ ˆ ˆ2 7 3 2 4
2 1 7 2 3 4
2 14 12
0
b
b
b b i j k i j k
The angle, Q, between the given pair of lines is given by the relation,
1 2
1 2
1
cos
0cos 0
62 21
πcos 0
2
b bQ
b b
Q
Q
Thus, the given lines are perpendicular to each other.
22. Probabilities of solving a specific problem independently by A and B are 1
2 and
1
3 respectively. If both try to solve the problem independently, find the probability
that (i) the problem is solved (ii) exactly one of them solves the problem.
Solution:
The probability of solving the problem independently by A and B are given as
1 1 and
2 3respectively.
i.e. = 1 1
P , P2 3
A B .
P P P Since the events corresponding to A and B are independent
1 1 1
2 3 6
A B A B
(i) Probability that the problem is solved
P
P P P
1 1 1
2 3 6
3 2 1
6
4
6
2
3
A B
A B A B
Thus, the probability that the problem is solved is 2
3.
(ii) Probability that exactly one of them solves the problem
P P
P P P P
1 1 1 1
2 6 3 6
3 1 2 1
6
3
6
1
2
A B B A
A A B B A B
Section - C
Question numbers 23 to 29 carry 6 marks each.
23. Using matrix method, solve the following system of equations:
2 3 10 4 6 5 6 9 20
4, 1, 2; , , 0x y zx y z x y z x y z
OR
Using elementary transformations, find the inverse of the matrix
1 3 2
3 0 1
2 1 0
Solution:
The given system of equation is
2 3 10 4 6 5 6 9 204, 1, 2; , , 0x y z
x y z x y z x y z
The given system of equation can be written as
1
2 3 10 41
4 6 5 1
6 9 20 21
or , where
1
2 3 10 41
4 6 5 , and 1
6 9 20 21
x
y
z
AX B
x
A X By
z
2 3 10
Now, 4 6 5
6 9 20
A
2 120 45 3 80 30 10 36 36
2 75 3 110 10 72
150 330 720
1200 0
So, the given system of equation has unique solution and is given by 1X A B
Let Cij be the cofactors of elements aij in A = [aij], then
2 3 4
11 12 13
3 4 5
21 22 23
4 5 6
31 32 33
C 1 120 45 75, C 1 80 30 110, C 1 36 36 72
C 1 60 90 150, C 1 40 60 100, C 1 18 18 0
C 1 15 60 75, C 1 10 40 30, C 1 12 12 24
75 110 72 75 150 75
150 100 0 110
75 30 24
T
AdjA
1
1
100 30
72 0 24
75 150 751
110 100 301200
72 0 24
75 150 75 41
110 100 30 11200
72 0 24 2
300 150 1501
440 100 601200
288 0 48
6001
4001200
240
60
AdjAA
A
X A B
0 11 1
1200 2 2
400 1 1 1
1200 3 3
240 1 11
1200 5 5
1 1 1 1 1 1, and
2 3 5
2, 3 and 5
x
y
z
x y z
x y z
Thus, the solution of given system of equation is given by
x =2, y = 3 and z = 5
OR
Let the given matrix be denoted as
1 3 2
3 0 1
2 1 0
A
.
We have A = IA.
Or,
1 3 2 1 0 0
3 0 1 0 1 0
2 1 0 0 0 1
A
Applying R2 R2 + 3R1 and R3 R3 – 2R1
2 2
1 3 2 1 0 0
0 9 7 3 1 0
0 5 4 2 0 1
1Applying R R
9
1 3 2 1 0 0
7 1 10 1 0
9 3 9
0 5 4 2 0 1
A
A
Applying R1 R1 – 3R2 and R3 R3 + 5R2
1 11 0 0 0
3 3
7 1 10 1 0
9 3 9
1 1 50 0 1
9 3 9
A
Applying R3 9R3
1 11 0 0 0
3 3
7 1 10 1 0
9 3 9
0 0 1 3 5 9
A
Applying R1 R1 – 1
3R3 and R2 R2 +
7
9R3
1
1 0 0 1 2 3
0 1 0 2 4 7
0 0 1 3 5 9
1 2 3
2 4 7
3 5 9
1 2 3
2 4 7
3 5 9
A
I A
A
Therefore, the inverse of the matrix
1 3 2
3 0 1
2 1 0
is
1 2 3
2 4 7
3 5 9
.
24. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum
area.
Solution:
Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.
Then, the diagonal passes through the centre and is of length 2a cm.
Now, by applying the Pythagoras Theorem, we have:
2 2 2
2 2 2
2 2
2
4
4
a l b
b a l
b a l
Area of the rectangle, 2 24A l a l
22 2 2 2
2 2 2 2
2 2
2 2
2 2 2 2
2 2 2
2 2 2
2 2 2 2
32 2 2
2 22 3
3 32 2 2 22 2
2 2
2 2
14 2 4
2 4 4
4 2
4
24 4 4 2
2 4
4
4 4 4 2
4
2 612 2
4 4
Now, 0 gives 4 2 2
4 2 2
dA la l l l a l
dl a l a l
a l
a l
la l l a l
d A a l
dl a l
a l l l a l
a l
l a la l l
a l a l
dAa l l a
dl
b a a
2 2a a
2 22 3
2 3 3
Now, when 2 ,
2 2 6 2A 8 24 0
2 2 2 2
l a
a a ad a
dl a a
By the second derivative test, when 2l a , then the area of the rectangle is the maximum.
Since 2l b a , the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square
has the maximum area.
25. Using integration find the area of the triangular region whose sides have equations
y = 2x + 1, y = 3x + 1 and x = 4.
Solution:
The triangular region bounded by the lines y = 2x + 1, y = 3x + 1 and x = 4 is represented
graphically as:
Equations of the lines are y = 2x + 1, y = 3x + 1 and x = 4
Let y1 = 2x + 1, y2 = 3x + 1
Now, area of the triangle bounded by the given lines
4
2 1
0
4
0
4
0
42
0
2 2
3 1 2 1
1
2
14 0
2
116
2
8 square units
y y dx
x x dx
xdx
x
Thus, the area of the required triangular region is 8 sq. units.
26. Evaluate:
π
21
0
2sin cos tan sinx x x dx
OR
Evaluate:
π
2
4 4
0
sin cos
sin cos
x x xdx
x x
Solution:
π
12
02sin cos tan sinx x x dx
Let t = sin x
dt = cos x dx
When x =π
2,
πsin 1
2t
When x = 0, t = sin 0 = 0
1
1
1 1
21 2
2
22 1
2
2 1
2
2 1 1
Now, 2sin cos tan sin
2 tan
tan 2 tan 2
1tan 2
2 1
tan1
1tan 1
1
tan tan
x x x dx
t t dt
dt t dt t tdt dt
dt
tt t dt
t
tt t dt
t
t t dtt
t t t t
π
12
02sin cos tan sinx x x dx
12 1 1
0tan tant t t t
2 1 1 2 1 11 tan 1 1 tan 1 0 tan 0 0 tan 0
π π1 1 0
4 4
π1
2
OR
π
2
4 4
0
π
2
4 40 0 0
π
2
4 4
0
sin cos1
sin cos
π π πsin cos
2 2 2
π πsin cos
2 2
πcos sin
22
cos sin
a a
x x xI
x x
x x x
I dx f x dx f a x dx
x x
x x x
I dxx x
Adding (1) and (2), we get π
2
4 4
0
π
2
4 4
0
π
2 4
4
04
π
22
4
0
πsin cos
22sin cos
π sin cos
4 sin cos
sin cos
π cos
sin41
cos
π tan sec
4 tan 1
x x
I dxx x
x xI dx
x x
x x
x dxx
x
x xdx
x
Put tan2 x = z
2 tan x sec2 xdx = dz
2tan sec2
dzx xdx
When 0, 0
πwhen ,
2
x z
x z
2
0
2
0
1
0
1 1
2
π
22
4 4
0
π 2
4 1
π
8 1
πtan
8
πtan tan 0
8
π π0
8 2
π
16
sin cos πThus,
sin cos 16
dz
Iz
dzI
z
z
x x xdx
x x
27. Find the equation of the plane which contains the line of intersection of the planes
ˆˆ ˆ2 3 4 0r i j k , ˆˆ ˆ2 5 0r i j k and which is perpendicular to the plane
ˆˆ ˆ5 3 6 8 0r i j k .
Solution:
The equations of the given planes are
ˆˆ ˆ2 3 4 0 ... 1
ˆˆ ˆ2 5 0 ... 2
r i j k
r i j k
The equation of the plane passing through the line intersection of the planes given in equation (1)
and equation (2) is
ˆ ˆˆ ˆ ˆ ˆ2 3 4 2 5 0
ˆˆ ˆ2 1 2 3 5 4 0 ... 3
r i j k r i j k
r i j k
The plane in equation (3) is perpendicular to the plane, ˆˆ ˆ5 3 6 8 0r i j k
5 2 1 3 2 6 3 0
19 7 0
7
19
Substituting 7
19 in equation (3), we obtain
33 45 50 41ˆˆ ˆ 019 19 19 19
ˆˆ ˆ33 45 50 41 0 ... 4
r i j k
r i j k
This is the vector equation of the required plane.
The Cartesian equation of this plane can be obtained by substituting ˆˆ ˆr xi yj zk in equation
(4).
ˆ ˆˆ ˆ ˆ ˆ33 45 50 41 0
33 45 50 41 0
xi yj zk i j k
x y z
28. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine
time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine
time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42
hours of machine time and 24 hours of craftsman’s time.
If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the number of
tennis rackets and crickets bats that the factory must manufacture to earn the maximum profit.
Make it as an L.P.P. and solve graphically.
Solution:
Let the number of rackets and the number of bats to be made be x and y respectively.
The given information can be compiled in a table as follows:
Tennis Racket Cricket Bat Availability
Machine Time (h) 1.5 3 42
Craftsman’s Time (h) 3 1 24
The machine time is not available for more than 42 hours.
1.5 3 42x y
The craftsman’s time is not available for more than 24 hours.
3 24x y
Let the total profit be Rs Z.
The profit on a racket is Rs 20 and on a bat is Rs 10.
20 10Z x y
Thus, the given problem can be mathematically formulated as:
Maximize 20 10Z x y … (1)
Subject to the constraints,
1.5x + 3y 42 … (2)
3x + y 24 … (3)
x, y 0 … (4)
The feasible region determined by the system of constraints is as follows:
The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0).
The values of Z at these corner points are as follows.
Corner point Z = 20x + 10y
A(8, 0) 160
B(4, 12) 200 → Maximum
C(0, 14) 140
O(0, 0) 0
The maximum value of Z is 200, which occurs at x = 4 and y = 12.
Thus, the factory must produce 4 tennis rackets and 12 cricket bats to earn the maximum profit.
The maximum obtained profit earned by the factory by producing these items is Rs 200.
29. Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at
random. What is the probability of this person being male? Assume that there are equal number
of males and females.
Solution:
Let M, F and G be the following events.
M: A male is selected.
F: A female is selected.
G: A person has grey hair
It is given that the number of males = the number of females
1
P P2
M F
It is also given that,
P (G|M)
= Probability of grey haired person given that they are male
= 5%
5
100
Similarly, P (G|F) 0.25
0.25%100
When a grey haired person is selected at random, probability that this person is a male
P |
P P |Baye's Theorem
P P | P P |
1 5
2 1001 5 1 0.25
2 100 2 100
5
1005 0.25
100 100
5
5.25
20
21
M G
M G M
M G M F G F