mathematics · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

34
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.

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Page 1: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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.

Page 2: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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.

Page 3: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 4: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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 .

Page 5: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 6: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 7: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 8: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 9: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 10: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 11: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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:

Page 12: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 13: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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).

Page 14: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 15: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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:

Page 16: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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)

Page 17: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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:

Page 18: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 19: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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:

Page 20: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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,

Page 21: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 22: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 23: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 24: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 25: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 26: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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:

Page 27: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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:

Page 28: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 29: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 30: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 31: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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

Page 32: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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.

Page 33: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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.

Page 34: MATHEMATICS · 2 2 2 2 2 2 2 2 2 2 1 5, , 2 1 5 2 1 5 2 1 5 2 1 5

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