Math Set1 SectionA Delhi Merged

7/30/2019 Math Set1 SectionA Delhi Merged

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CBSE Math XII Board PaperSET 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 asper the exact requirements of the question.

iv. There is no overall choice. However, internal choice has been provided in 4questions 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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1. State the reason for the relationR in the set {1, 2, 3} given by R = {(1, 2), (2, 1)}

not to be transitive.

Solution:

It is known that a relationR in a setA is transitive if (a1, a2) R and (a2, a3) R

(a1, a3) Ra1, a2, a3A.It can be observed that (1, 2), (2, 1) R, but (1,1) R.Thus, the relationR in the set {1, 2, 3} is not transitive.

2. Write the value of 1 1

sin sin3 2

Solution:

1

1

1

1

sin sin3 2

1Let sin

2

1sin

2

sin sin sin sin 2

6 6 6

2

6 1

sin sin sin 23 2 3 6

9 9sin sin

6 6

3 sin sin

2 2

x

x

x

x

sin 1

2

Thus, the value of the given expression is 1.

3. For a 2 2 matrix,A = [aij] whose elements are given by iji

aj

, write the value of

a12.

Solution:

Any element in the ith

row andjth

column is iji

aj

For a12, the value ofi = 1 and j = 2.


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12

1

2a

Thus, the value ofa12 is1

2.

4. For what value ofx, the matrix5 1

2 4

x x

is singular?

Solution:

The matrix5 1

2 4

x xA

is singular

0

5 1

02 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 ofx is 3.

5. WriteA1 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


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6. Write the value of sec sec tanx x x dx

Solution:

secx (secx + tanx) dx

= (sec2x + secx tanx)dx

= sec2xdx + secx tanxdx

= tanx + secx + c , where c is an arbitrary constant

7. Write the value of2 16

dx

x

Solution:

2

2 2

16

4

dx

x

dx

x

11 tan4 4

xC

, where Cis the arbitrary constant

8. For what value of a the vectors

2 3 4i j k and

6 8ai j k are collinear?

Solution:

Two vectors x and y are collinear ifx y , where is a constant.

It is given that 2 3 4 and 6 8i j k ai j k are collinear.

2 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 = a1

22

4

a

a

Thus, the value ofa is4.

9. Write the direction cosines of the vector 2 5i j k .

Solution 9:

The given vector is 2 5i j k .


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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 +yz = 5 onx-axis.

Solution:

2 5 ... 1x y z Dividing both sides of the equation (1) by 5, we obtain

2

15 5 5

1 ... 25 5 5

2

y z

x

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 atx,y, andz- axes respectively.

Therefore, for the given equation,

5

, 5, and 52a b c

Thus, the intercept cut off by the given plane on thex-axis is5

.2


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MATHEMATICS


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 asper the exact requirements of the question.

iv. There is no overall choice. However, internal choice has been provided in 4questions of four marks each and 2 questions of six marks each. You have to

attempt only one of the alternatives in all such questions.



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SECTIONB

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 ab = 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 cot , 0,

2 41 sin 1 sin

x x xx

x x

OR

Find the value of1 1

tan 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


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1

2 2

1

2 2

1

1 sin 1 sincot

1 sin 1 sin

cos sin cos sin2 2 2 2

cotcos sin cos sin

2 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 tan 11 1

tan tan tan 1

tan

x x y

y x y

x

x y

xy

y

x

x a by

a bxy ab

y

x x

y y

x

1 1

1

tan tan 1

tan 1

4

x

y y

Thus the value of the given expression is


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13. Using properties of determinants, prove that2

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 respective

1 1 1

1 1 1 Taking out factors , , from C , C , and C respectivel

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 functionfdefined as

3

sin 1 , 02

tan sin, 0

a x xf x

x xx

x

is continuous atx = 0

Solution:

The given functionfis defined as

3

sin 1 , 0

2

tan sin, 0

a x x

f x x xx

x


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The given function is defined for all real numbers.

By, definition,fis continuous atx = 0, if 0 0

lim lim 0x x

f x f x f

00

lim 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 14

1

2

xx

x

x

x

x x x

x

x x

x

xx

x

x

x

x x

xx

x x

x

x

x x x

x

x

Now,

0 sin 0 1 sin 12 2

f a a a a

Sincefis continuous atx = 0, 0 0

lim lim 0x x

f x f x f

1

2a

Thus, for1

2a , the given functionfis continuous atx = 0.

15. Differentiate2

cos

2

1

1

x x xxx

w.r.t.x

OR

Ifx = a (sin),y = a (1 + cos), find

2

2d ydx


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Solution:2

cos

2

1

1

x x xxx

Lety =x

xcosx

and

2

2

1

1

x

z x

y =xxcosx

Taking log of both sides:

coslog loglog cos log

x xy x

y x x x

Differentiating with respect tox, we obtain:

cos

1 1cos log cos

1 cos log cos sin

cos log cos sin

cos log cos sin ... 1x x

dy dx x x x x

y dx x dx

dy x x x x xy 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

222

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 ... 21

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 xdx x


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On adding (1) and (2):

cos

22

2cos cos

222

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 differentiatingx andy with respect to, it is obtained:

1 cos ... 1

and 0 sin

sin .. 2

dxa

d

dy ad

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 tox, we obtain:


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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 2sin

21.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 curvex2

+y22x3 = 0 at which the tangents are parallel tox-

axis.

Solution:

The volume (V) of a cone with radius (r) and height (h) is given by,

21

3V r h

It is given that,16

6h r r h

2 31

6 123

V h h h

The rate of change of the volume with respect to time (t) is given by,

312dV d dh

hdt dh dt

[By chain rule]

2

2

12 3

36

dhh

dt

dhhdt


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It is also given that 312 cm /sdV

dt .

Therefore, when h = 4 cm, we have:

212 36 4

12 1

36 16 48

dh

dt

dh

dt

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate

of1

cm/s48

.

OR

Let P ,x y be any point on the given curvex2 +y22x3 = 0.

Tangent to the curve at the point (x,y) is given bydy

dx.

Differentiating the equation of the curve on both the sides with respect tox, 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 thex-axis.

1 1,

1

1

1

1

0

10

1 0

1

x y

dy

dx

x

y

x

x

Whenx1 = 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 tox-axis are (1,

2) and (1,2).


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17. Evaluate:2

5 3

4 10

xdx

x x

OR

Evaluate: 2 2

21 3

x dxx x

Solution:

2

2

2 2

2 2

1 2

12

5 3

4 10

5 3 4 10

5 3 2 45 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 Bx 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

10

dx

x

Putx2

+ 4x + 10 =z2


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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 xdx

Ix 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

Puttingz =3, it is obtained:1 2

1

2

B

B

Putting z =1, it is obtained:


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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

tany dx + (1ex) sec

2y dy = 0

Solution:

The given differential equation is:

ex

tanydx + (1ex) sec

2ydy = 0

(ex1) sec

2y dy = e

xtany dx

2sec

tan 1

x

x

y edy dx

ye

On integrating on both sides, we get

2sec. ...(i)

tan 1

x

x

y edy dx

y e

2sec

tan

ydy

y

Put tany = u

sec2y dy = du2sec

tan

y dudy

y u = log |u| = log tany (ii)

1

x

x

edx

e

Put ex1 = v

exdx = dv

1

x

x

e dvdx

ve

= log v

l ( x 1) (iii)


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Form (i), (ii) and (iii), we obtain:

log tany = log (ex1) + log C

log tany = log C(ex1)

tany = C (ex1)

Thus, the solution of the given differential equation is tany = C(e

x

1).

19. Solve the following differential equation:

2 cos tan 02

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 x dx 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

dtxdx

x dx dt

Therefore, equation (1) becomes:


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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 3 2 2a i j k and 2 2b i j k .

Solution:

We have, 3 2 2a i j k and 2 2b i j k

2 22

2 2 2 2 2

2 2

4 4 , 2 4

4 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,

16 16 8 2 2 2 2 1 24 3 3 3 3

a b a b i j k i j k i j k

a b a b


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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 xdx

Ix 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

Puttingz =3, we obtain:1 2

1

2

B

B

Putting z =1, we obtain:


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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 5and

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

1

2 7 3b i j k

and 2

2 4b i j k

2 2 2

1

2 2 2

2

1 2

2 7 3 62

1 2 4 21

2 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,


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1 2

1 2

1

cos

0cos 0

62 21

cos 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 are1

2and

1

3respectively. 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 byA andB are given as

1 1and

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 is2

3.

(ii) Probability that exactly one of them solves the problem


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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


24/36


MATHEMATICS


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 theexact requirements of the question.

iv. There is no overall choice. However, internal choice has been provided in 4 questions offour marks each and 2 questions of six marks each. You have to attempt only one of thealternatives in all such questions.



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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 9 204, 1, 2; , , 0x y z

x 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 B

y

z

2 3 10

Now, 4 6 5

6 9 20

A


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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 aijinA = [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 2475 150 75

1110 100 30

120072 0 24

75 150 75 41

110 100 30 11200

72 0 24 2

300 150 1501440 100 60

1200288 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


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Thus, the solution of given system of equation is given byx =2,y = 3 andz = 5

OR

Let the given matrix be denoted as

1 3 2

3 0 1

2 1 0

A

.

We haveA =IA.

Or,

1 3 2 1 0 0

3 0 1 0 1 0

2 1 0 0 0 1

A

Applying R2R2 + 3R1 and R3 R32R1

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 R1R13R2 and R3 R3 + 5R2

1 11 0 0 0

3 3

7 1 10 1 0

9 3 9

1 1 50 0 19 3 9

A

Applying R3 9R3


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1 11 0 0 0

3 3

7 1 10 1 0

9 3 9

0 0 1 3 5 9

A

Applying R1R11

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:


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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

3

2 2 2

2 22 3

3 3

2 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 32 3 3

Now, when 2 ,2 2 6 2A 8 2

4 02 2 2 2

l aa 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 squarehas the maximum area.

25. Using integration find the area of the triangular region whose sides have equationsy = 2x + 1,y = 3x + 1 andx = 4.

Solution:The triangular region bounded by the linesy = 2x + 1,y = 3x + 1 andx = 4 is represented

graphically as:


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Equations of the lines arey = 2x + 1,y = 3x + 1 andx = 4

Lety1 = 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:

21

0

2sin cos tan sinx x x dx

OR


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Evaluate:

2

4 4

0

sin cos

sin cos

x x xdx

x x

Solution:

12


Let t= sinx

dt= cosx dx

Whenx =

2,

sin 12

t

Whenx = 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 dt t

t t t t

12


1

2 1 1

0tan tant t t t

2 1 1 2 1 11 tan 1 1 tan 1 0 tan 0 0 tan 0

1 1 04 4

12

OR


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2

4 4

0

2

4 40 0 0

2

4 4

0

sin cos1

sin cos

sin cos2 2 2

sin cos2 2

cos sin2

2cos sin

a a

x x xI

x x

x x x

I dx f x dx f a x dxx x

x x x

I dxx x

Adding (1) and (2), we get

2

4 4

0

2

4 4

0

2 4

4

04

22

4

0

sin cos

22sin cos

sin cos

4 sin cos

sin cos cos

sin41

cos

tan sec

4 tan 1

x x

I dxx x

x xI dx

x x

x x

x dxx

x

x xdx

x

Put tan2x = z

2 tanx sec2xdx = dz

2tan sec2

dzx xdx

When 0, 0

when ,2

x z

x z


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2

0

2

0

1

0

1 1

2

22

4 4

0

2

4 1

8 1

tan8

tan tan 0

8

0

8 2

16

sin cos Thus,

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

2 3 4 0r i j k , 2 5 0r i j k and which is perpendicular to the plane

5 3 6 8 0r i j k .

Solution:The equations of the given planes are

2 3 4 0 ... 1

2 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

2 3 4 2 5 0

2 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, 5 3 6 8 0r i j k


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5 2 1 3 2 6 3 0

19 7 0

7

19

Substituting 719

in equation (3), we obtain

33 45 50 41 0

19 19 19 19

33 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 r xi yj zk in equation

(4).

33 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 machinetime and 3 hours of craftsmans time in its making while a cricket bat takes 3 hours of machinetime and 1 hour of craftsmans time. In a day, the factory has the availability of not more than 42

hours of machine time and 24 hours of craftsmans 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 bex andy respectively.

The given information can be compiled in a table as follows:

Tennis Racket Cricket Bat Availability

Machine Time (h) 1.5 3 42

Craftsmans Time (h) 3 1 24

The machine time is not available for more than 42 hours.1.5 3 42x y

The craftsmans 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.


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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 ofZat these corner points are as follows.

Corner point Z = 20x + 10y

A(8, 0) 160

B(4, 12) 200 MaximumC(0, 14) 140

O(0, 0) 0

The maximum value ofZis 200, which occurs atx = 4 andy = 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.


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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 numberof males and females.

Solution:

LetM, Fand 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

100

5 0.25100 100

5

5.25

20

21

M G

M G M

M G M F G F

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