Class... · CBSE Examination Paper, Delhi-2014 Time allowed: 3 hours Maximum marks: 100 ¡ General Instructions: 1. All questions are compulsory. 2. The question paper …

CBSE ExaminationPaper, Delhi-2014

Time allowed: 3 hours Maximum marks: 100

¡ General Instructions:

1. All questions are compulsory.

2. The question paper consists of 29 questions divided into three Sections A, B and C. Section Acomprises 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.

3. All questions in Section A are to be answered in one word, one sentence or as per the exactrequirement of the question.

4. There is no overall choice. However, internal choice has been provided in 4 questions of four markseach and 2 questions of six marks each. You have to attempt only one of the alternatives in all suchquestions.

5. Use of calculator is not permitted. You may ask for logarithmic tables, if required.

SET–I

SECTION–A

Question numbers 1 to 10 carry 1 mark each.

1. Let * be a binary operation, on the set of all non-zero real numbers, given by a bab

* =5

for all a,

b RÎ - { }.0 Find the value of x, given that 2 5 10* ( * ) .x =

2. If sin sin cos- -+æèç ö

ø÷ =1 11

51x , then find the value of x.

3. If 23 4

5

1

0 1

7 0

10 5x

yé

ëê

ù

ûú +

é

ëê

ù

ûú =

é

ëê

ù

ûú, find (x – y).

4. Solve the following matrix equation for x x O: [ ] .11 0

2 0-

é

ëê

ù

ûú =

5. If 2 5

8

6 2

7 3

x

x=

- , write the value of x.

6. Write the antiderivative of 31

xx

+æèç

öø÷.

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7. Evaluate: dx

x9 20

3

+ò .

8. Find the projection of the vector $ $ $i j k+ +3 7 on the vector 2 3 6$ $ $i j k- + .

9. If a®

and b®

are two unit vectors such that a b® ®

+ is also a unit vector, then find the angle

between a®

and b®

.

10. Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the

plane r®

. ($ $ $)i j k+ + = 2.

SECTION–BQuestion numbers 11 to 22 carry 4 marks each.

11. Let A = {1, 2, 3, …, 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for(a, b), (c, d) in A × A. Prove that R is an equivalence relation. Also obtain the equivalence class[(2, 5)].

12. Prove that cotsin sin

sin sin; ,- + + -

+ - -

æ

èç

ö

ø÷ = Îæ

èç1 1 1

1 1 20

4

x x

x x

xx

p öø÷.

OR

Prove that 21

5

5 2

72

1

8 4

1 1 1tan sec tan- - -æèç ö

ø÷ +

æ

èç

ö

ø÷ + æ

èç ö

ø÷ =

p.

13. Using properties of determinants, prove that

2 2

2 2

2 2

3

y y z x y

z z z x y

x y z x x

x y z

- -

- -

- -

= + +( ) .

14. Differentiate tan - -æ

è

çç

ö

ø

÷÷

121 x

x with respect to cos ( ),- -1 22 1x x when x ¹ 0.

15. If y xx= , prove that d y

dx y

dy

dx

y

x

2

2

21

0-æèç

öø÷ - = .

16. Find the intervals in which the function f x x x x( ) = - - +3 4 12 54 3 2 is

(a) strictly increasing

(b) strictly decreasing

OR

Find the equations of the tangent and normal to the curve x a= sin 3 q and y a= cos 3 q at

qp

=4

.

17. Evaluate:sin cos

sin . cos

6 6

2 2

x x

x xdx

+ò

OR

Evaluate: ( )x x x dx- + -ò 3 3 182

502 Xam idea Mathematics–XII



18. Find the particular solution of the differential equation e y dxy

xdyx 1 02- + = , given that

y = 1 when x = 0.

19. Solve the following differential equation:

( )xdy

dxxy

x

2

21 2

2

1- + =

-

20. Prove that, for any three vectors a b c® ® ®

, ,

[ , , ] [ , , ]a b b c c a a b c® ® ® ® ® ® ® ® ®

+ + + = 2

OR

Vectors a b® ®

, and c®

are such that a b c® ® ® ®

+ + = 0 and | | ,| |a b® ®

= =3 5 and | |c®

= 7. Find the angle

between a®

and b®

.

21. Show that the lines x y z+

=+

=+1

3

3

5

5

7 and

x y z-=

-=

-2

1

4

3

6

5 intersect. Also find their

point of intersection.

22. Assume that each born child is equally likely to be a boy or a girl. If a family has two

children, what is the conditional probability that both are girls given that

(i) the youngest is a girl?

(ii) atleast one is a girl?

SECTION–C

Question numbers 23 to 29 carry 6 marks each.

23. Two schools P and Q want to award their selected students on the values of Discipline,

Politeness and Punctuality. The school P wants to award x each, y each and z each for

the three respective values to its 3, 2 and 1 students with a total award money of 1,000.

School Q wants to spend 1,500 to award its 4, 1, and 3 students on the respective values (by

giving the same award money for the three values as before). If the total amount of awards

for one prize on each value is 600, using matrices find the award money for each value.

Apart from the above three values, suggest one more value for awards.

24. Show that the semi-vertical angle of the cone of the maximum volume and of given slant

height is cos .-1 1

3

25. Evaluate: dx

x16

3

+ò cot/

/

p

p

26. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the

circle x y2 2 32+ = .

27. Find the distance between the point (7, 2, 4) and the plane determine by the points A(2, 5, –3),

B(–2, –3, 5) and C ( , , ).5 3 3-

OR

Find the distance of the point ( , , )- - -1 5 10 from the point of intersection of the line

r®

= - + + + +2 2 3 4 2$ $ $ ( $ $ $)i j k i j kl and the plane r i j k®

- + =.($ $ $) .5

Examination Papers – 2014 503



28. A dealer in rural area wishes to purchase a number of sewing machines. He has only 5,760

to invest and has space for at most 20 items for storage. An electronic sewing machine cost

him 360 and a manually operated sewing machine 240. He can sell an electronic sewing

machine at a profit of 22 and a manually operated sewing machine at a profit of 18.

Assuming that he can sell all the items that he can buy, how should he invest his money in

order to maximise his profit? Make it as a LPP and solve it graphically.

29. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack threecards are drawn at random (without replacement) and are found to be all spades. Find theprobability of the lost card being a spade.

OR

From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by onewith replacement. Find the probability distribution of number of defective bulbs. Hence findthe mean of the distribution.

SET–II

Only those questions, not included in Set I, are given.

9. Evaluate: cos (sin ) .-ò

1 x dx

10. If vectors a®

and b®

are such that, | | ,| |a b® ®

= =32

3 and a b

® ®´ is a unit vector, then write the

angle between a®

and b®

.

19. Prove the following using properties of determinants:

a b c a b

c b c a b

c a c a b

a b c

+ +

+ +

+ +

= + +

2

2

2

2 3( )

20. Differentiate tan -

-

1

21

x

x with respect to sin ( )- -1 22 1x x .

21. Solve the following differential equation:

cosec x ydy

dxx ylog + =2 2 0.

22. Show that the lines 5

4

7

4

3

5

-

-=

-=

+

-

x y z and

x y z-=

-=

-8

7

2 8

2

5

3 are coplanar.

28. Evaluate: x x

x xdx

tan

sec ..

cosec0

p

ò

29. Prove that the semi-vertical angle of the right circular cone of given volume and least curved

surface area is cot .-1 2

SET–III

Only those questions, not included in Set I and Set II, are given.

9. Evaluate: e x x dxx (sin cos ) ./

-ò0

p 2




10. Write a unit vector in the direction of the sum of the vectors a®

= + -2 2 5$ $ $i j k and

b i j k®

= + -2 7$ $ $.

19. Using properties of determinants, prove the following:

x xy xz

xy y yz

xz yz z

x y z

2

2

2

2 2 2

1

1

1

1

+

+

+

= + + + .

20. Differentiate tan - + -æ

è

çç

ö

ø

÷÷

121 1x

x with respect to sin ,-

+

æ

èç

ö

ø÷

1

2

2

1

x

x when x ¹ 0.

21. Find the particular solution of the differential equation dy

dx

x x

y y y=

+

+

( log )

sin cos

2 1 given that

y =p

2 when x = 1.

22. Show that lines r i j k i j®

= + - + -($ $ $) ( $ $)l 3 and r i k i k®

= - + +( $ $) ( $ $)4 2 3m intersect. Also find

their point of intersection.

28. Evaluate: x x x

x xdx

sin cos

sin cos

/

4 40 +ò

p 2

.

29. Of all the closed right circular cylindrical cans of volume 128 p cm3, find the dimensions ofthe can which has minimum surface area.




SET–I

SECTION–A

1. Given 2 5 10* ( * )x =

Þ 25

5*

x ´ = 10 Þ 2 * x = 10

Þ2

510

´=

xÞ x =

´10 5

2Þ x = 25.

2. Given sin(sin cos )- -+ =1 11

51x

Þ sin cos sin- - -+ =1 1 11

5x 1 Þ sin cos- -+ =1 11

5 2x

p

Þ sin -1 1

5 = - -p

2

1cos x Þ sin sin- -=1 11

5x Þ x =

1

5.

3. Given 2 3

5

4 1

0 1x

yé

ëê

ù

ûú +

é

ëê

ù

ûú =

7

10

0

5

é

ëê

ù

ûú

Þ6

10

8

2

1

0 1

7

10

0

5x

yé

ëê

ù

ûú +

é

ëê

ù

ûú =

é

ëê

ù

ûú Þ

7

10

8

2 1

7

10

0

5

+

+

é

ëê

ù

ûú =

é

ëê

ù

ûú

y

x

Equating we get 8 0+ =y and 2 1 5x + =

Þ y = -8 and x = 2 Þ x y- = + =2 8 10

4. Given [ ]x 11

2

0

0-

é

ëê

ù

ûú = 0

Þ [ ] [ ]x - =2 0 0 0

Þ x - =2 0 Þ x = 2

5. Given 2 5

8

6 2

7 3

x

x=

-

Þ 2 40 18 142x - = - -( ) Þ 2 40 322x - =

Þ 2 722x = Þ x2 36= Þ x = ± 6

6. Antiderivative of 31

31

xx

xx

dx+æèç

öø÷ = +

æèç

öø÷ò

= + òò31

xdxx

dx = +ò ò-3 1 2 1 2x dx x dx/ /

=+

+- +

++ - +

31

21

1

21

1 2 1 1 2 1

./ /x x

c

= ´ + +32

323 2 1 2x x c/ /

= + +2 23 2x x c/


SolutionsSolutionsSolutions



7. Let Idx

x=

+ò

9 20

3

=+

òdx

x32 20

3

= é

ëêù

ûú-1

3 3

1

0

3

tanx

= -- -1

31 01 1[tan ( ) tan ( )] = -é

ëêù

ûú1

3 40

p =

p

12

8. Let a i j k®

= + +$ $ $3 7

b i j k®

= - +2 3 6$ $ $

Now projection of a®

on b®

= a b

b

® ®

®

.

| |

=+ + - +

- +

($ $ $).( $ $ $)

| $ $ $|

i j k i j k

i j k

3 7 2 3 6

2 3 6

=- +

+ - +

2 9 42

2 3 62 2 2( ) = =

35

49

35

7 = 5.

9. | | ( ).( )a b a b a b® ® ® ® ® ®

+ = + +2

Þ 1 22 2 2= + +® ® ® ®

| | . | |a a b b [Q | |a b® ®

+ = 1]

Þ 1 1 2 1= + +® ®a b. [Q a

® and b

® are unit vector, hence | |a

® = 1 and | |b

® = 1]

Þ 1 2 2= +® ®a b. Þ a b

® ®= -.

1

2

Þ | |.| |cosa b® ®

= -q1

2, where q is angle between a

® and b

®

Þ 1.1 cos q = -1

2[Q | | | |a b

® ®= = 1]

Þ cosq = -1

2

Þ cos cosqp

= -3

Þ cos cosq pp

= -æèç ö

ø÷

3

Þ cos cosqp

=2

3Þ q

p=

2

3.

10. Since, the required plane is parallel to plane r i j k®

+ + =.($ $ $) 2

\ Normal of required plane is normal of given plane.

Þ Normal of required plane = $ $ $.i j k+ +

\ Required vector equation of plane

{ r ai bj ck i j k®

- + + + + =( $ $ $)}.($ $ $) 0




SECTION–B

11. Given, R is a relation in A × A defined by

( , ) ( , )a b R c d a d b cÛ + = +

(i) Reflexivity: " Îa b A,

Q a b b a+ = + Þ ( , ) ( , )a b R a b

So, R in reflexive.

(ii) Symmetry: Let (a, b) R (c, d)

Q ( , ) ( , )a b R c d Þ a d b c+ = +

Þ b c d a+ = + [Q a, b, c, d ÎN and N is commutative under addition]

Þ c b d a+ = +

Þ ( , ) ( , )c d R a b

So, R is symmetric.

(iii) Transitivity: Let ( , ) ( , )a b R c d and ( , ) ( , )c d R e f

Now, ( , ) ( , )a b R c d and ( , ) ( , )c d R e f Þ a d b c+ = + and c f d e+ = +

Þ a d c f b c d e+ + + = + + +

Þ a f b e+ = +

Þ ( , ) ( , ).a b R e f

R is transitive.

Hence, R is an equivalence relation.

2nd Part: Equivalence class: [( , )] {( , ) : ( , ) ( , )}2 5 2 5= Î ´a b A A a b R

= Î ´ + = +{( , ) : }a b A A a b5 2

= Î ´ - ={( , ) : }a b A A b a 3

= {( , ), ( , ), ( , ), ( , ), ( , ), ( , )}1 4 2 5 3 6 4 7 5 8 6 9

12. LHS =+ + -

+ - -

æ

èçç

ö

ø÷÷

-cotsin sin

sin sin

1 1 1

1 1

x x

x x, x Îæ

èç ö

ø÷0

4,

p

=+ + -

+

-cot(cos / sin / ) (cos / sin / )

(cos / sin

12 22 2 2 2

2

x x x x

x x x x/ ) (cos / sin / )2 2 22 2- -

æ

è

çç

ö

ø

÷÷

=+ + -

+ -

-cotcos / sin / cos / sin /

cos / sin / cos

1 2 2 2 2

2 2

x x x x

x x x x/ sin /2 2+

æ

èç

ö

ø÷

=æ

èç

ö

ø÷

-cotcos /

sin /

1 2

2

x

x = =-cot (cot / )1 2

2x

x = R.H.S.

OR

L.H.S. = æèç ö

ø÷ +

æ

èç

ö

ø÷ + æ

èç ö

ø÷- - -2

1

5

5 2

72

1

8

1 1 1tan sec tan

= æèç ö

ø÷ + æ

èç ö

ø÷

ìíî

üýþ

+æ

èç

ö

ø÷

- - -21

5

1

8

5 2

7

1 1 1tan tan sec


Given

04

02 8

20

80

< <

Þ < <

Þ Îæèç ö

ø÷ Ì

é

ë

êêêêêê

ù

û

úúúúúú

x

x

x

p

p

pp, ( , )



=+

-

ì

íï

îï

ü

ýï

þï

+æ

èç

ö

ø÷ -- -2

1

5

1

8

11

5

1

8

5 2

711 1

2

tan.

tan [ sec tan ]Q - -= -1 1 2 1x x

= + -- -2

13

4039

40

50

4911 1tan tan = ´ +- -2

13

40

40

39

1

49

1 1tan tan

= æèç ö

ø÷ + æ

èç ö

ø÷- -2

1

3

1

7

1 1tan tan =´

- æèç ö

ø÷

æ

è

çççç

ö

ø

÷÷÷÷

+ æèç ö

ø÷- -tan tan1

2

12

1

3

11

3

1

7 Q2

2

1

1 1

2tan tan- -=

-

é

ëê

ù

ûúx

x

x

=

æ

è

ççç

ö

ø

÷÷÷

+ æèç ö

ø÷- -tan tan1 1

2

38

9

1

7 = ´æ

èç ö

ø÷ + æ

èç ö

ø÷- -tan tan1 12

3

9

8

1

7 = æ

èç ö

ø÷ + æ

èç ö

ø÷- -tan tan1 13

4

1

7

=+

- ´

æ

è

ççç

ö

ø

÷÷÷

-tan 1

3

4

1

7

13

4

1

7

= ´æèç ö

ø÷-tan 1 25

28

28

25 = -tan ( )1 1 =

p

4 = R.H.S.

13. L.H.S. D =

- -

- -

- -

2 2

2 2

2 2

y y z x y

z z z x y

x y z x x

Applying R R2 3« then R R1 2« , we have

D =

- -

- -

- -

x y z x x

y y z x y

z z z x y

2 2

2 2

2 2

Applying R R R R1 1 2 3® + + , we have

D =

+ + + + + +

- -

- -

x y z y z x z x y

y y z x y

z z z x y

2 2

2 2

Taking out ( )x y z+ + from first row, we have

D = + + - -

- -

( )x y z y y z x y

z z z x y

1 1 1

2 2

2 2

Applying C C C1 1 2® - and C C C2 2 3® - , we have

D = + + + + - + +

+ + - -

( ) ( )x y z y z x y z x y

z x y z x y

0 0 1

2

0

Expanding along first row, we have

D = + + + +( ) ( )x y z x y z 2 = + +( )a b c 3 = R.H.S.




14. Let ux

x=

-æ

è

çç

ö

ø

÷÷

-tan 121

and v x x= --cos ( )1 22 1

We have to determine du

dv

Let x x= Þ = -sin sinq q 1

Now, u =-æ

è

çç

ö

ø

÷÷

-tansin

sin

121 q

q

Þ u = æèç ö

ø÷-tan

cos

sin

1 q

q Þ u = -tan (cot )1 q

Þ u = -æèç ö

ø÷

é

ëêù

ûú-tan tan1

2

pq Þ u =

pq

2–

Þ u x= -p

2

1– sin Þdu

dx x=

-0

1

1 2–

Þdu

dx x=

-–

1

1 2

Again, v x x= --cos ( )1 22 1

Q x = sin q

\ v = --cos ( sin sin )1 22 1q q

Þ v = -cos ( sin . cos )1 2 q q

Þ v = -cos (sin )1 2q

Þ v = -æèç ö

ø÷

æèç

öø÷

-cos cos1

22

pq

Þ v = -p

q2

2

Þ v x= - -p

22 1sin

Þdv

dx x= -

-0

2

1 2Þ

dv

dx x= -

-

2

1 2

\du

dv

du

dxdv

dx

x

x

= = -

-

–

–

1

12

1

2

2

=1

2.

[Note: Here the range of x is taken as - < <1

2

1

2x ]

15. Given y xx=

Taking logarithm of both sides, we get

log . logy x x=


Q - < <1

2

1

2x Þ sin sin sin-æ

èç ö

ø÷ < < æ

èç ö

ø÷

pq

p

4 4

Þ - < <p

qp

4 4

Þ - < <p

qp

22

2

Þ p

qp

22

2> >– –

Þ pp

q> æèç ö

ø÷ >

22 0–

Þ p

q p p2

2 0 0– ( , ) [ , ]æèç ö

ø÷ Î Ì



Differentiating both sides, we get

Þ 1 1

y

dy

dxx

xx. . log= + Þ

dy

dxy x= +( log )1 …(i)

Again differentiating both sides, we get

d y

dxy

xx

dy

dx

2

2

11= + +. ( log ).

Þd y

dx

y

x y

dy

dx

dy

dx

2

2

1= + . . [From (i)]

Þd y

dx

y

x y

dy

dx

2

2

21

= +æèç

öø÷ Þ

d y

dx y

dy

dx

y

x

2

2

21

0-æèç

öø÷ - =

16. Given f x x x x( ) = - - +3 4 12 54 3 2

Þ ¢ = - -f x x x x( ) 12 12 243 2 Þ ¢ = - -f x x x x( ) ( )12 22

Þ ¢ = - + -f x x x x x( ) { }12 2 22 Þ ¢ = - + -f x x x x x( ) { ( ) ( )}12 2 1 2

Þ ¢ = - +f x x x x( ) ( )( )12 2 1 … (i)

For critical points

¢ =f x( ) 0 Þ 12 2 1 0x x x( )( )- + =

Þ x = -0 1 2, , (critical points)

These critical points divide the real number line into 4 disjoint intervals ( , ), ( , ), ( , )- ¥ - -1 1 0 0 2

and ( , )2 ¥ .

For ( , )- ¥ - 1

¢ =f x( ) +ve × –ve × –ve × –ve = –ve [From (i)]

Þ f(x) is decreasing in ( , )- ¥ - 1

For (– 1, 0)

¢ =f x( ) +ve × –ve × –ve × +ve = +ve

Þ f(x) is increasing in (–1, 0)

For (0, 2)

¢ =f x( ) +ve × +ve × –ve × +ve = –ve

Þ f(x) is decreasing in (0, 2).

For (2, ¥)

¢f x( ) = +ve × +ve × +ve × +ve = +ve

Þ f(x) is increasing in (2, ¥).

Hence, f(x) is strictly increasing in (–1, 0) U (2, ¥)

and f(x) is strictly decreasing in ( , ) ( , ).- ¥ - 1 0 2U

OR

Q x a= sin 3 q and y a= cos 3 q

Þdx

da

qq q= 3 2sin . cos and

dy

da

qq q= -3 2cos sin




Þdy

dx

dy

ddx

d

a

a= =

-q

q

q q

q q

3

3

2

2

cos . sin

sin . cos = - cotq

Þdy

dx= - cotq

Þ Slope of tangent to the given curve at qp

qp

= =é

ëêù

ûú =44

dy

dx = - cot

p

4 = -1.

Since for qp

=4

, x a= sin 3

4

p and y a= cos 3

4

p

Þ x a=æèç

öø÷

1

2

3

and y a=æèç

öø÷

1

2

3

Þ xa

=2 2

and ya

=2 2

i.e., co-ordinates of the point of contact = a a

2 2 2 2,

æèç

öø÷

\ Equation of tangent is

ya

xa

-æèç

öø÷ = - -

æèç

öø÷

2 21

2 2( ). Þ y

ax

a- = - +

2 2 2 2

Þ x ya

+ =2

Also slope of normal (at qp

=4

) = -1

slope of tangent = -

-=

1

11

\ Equation of normal is

ya

xa

-æèç

öø÷ = -

æèç

öø÷

2 21

2 2( ).

Þ ya

xa

- = -2 2 2 2

Þ y x- = 0

17. Let Ix x

x xdx=

+ò

sin cos

sin . cos

6 6

2 2

Þ Ix x

x xdx=

+ò

(sin ) (cos )

sin . cos

2 3 2 3

2 2

Þ Ix x x x x x

x=

+ - +(sin cos )(sin sin . cos cos )

sin . cos

2 2 4 2 2 4

2 2 xdxò

Þ Ix x x x

x xdx=

- +ò

sin sin . cos cos

sin . cos

4 2 2 4

2 2 = - +ò ò òtan cot2 2x dx dx xdx

Þ I x dx x x dx= - - + -ò ò(sec ) ( )2 21 1cosec

Þ I x dx x dx x x x c= + - - - +ò òsec2 2cosec = - - +tan cotx x x c3




OR

Let I x x x dx= - + -ò ( )3 3 182 … (i)

Let x Ad

dxx x B- = + - +3 3 182( ) Þ x A x B- = + +3 2 3( ) … (ii)

Þ x Ax A B- = + +3 2 3( )

Equating the co-efficient, we get

2 1A = and 3 3A B+ = - Þ A =1

2 and 3

1

23´ + = -B

Þ A =1

2 and B = - - = -3

3

2

9

2

\ I x x x= + -æèç ö

ø÷ + -ò ò

1

22 3

9

23 182( ) dx [From (i) and (ii)]

I x x x dx x x dx= + + - - + -òò1

22 3 3 18

9

23 182 2( )

Þ I I I= -1

2

9

21 2 … (iii) where I x x x dx122 3 3 18= + + -ò ( )

and I x x dx22 3 18= + -ò

Now I x x x dx122 3 3 18= + + -ò ( )

Let x x z2 3 18+ - =

Þ ( )2 3x dz+ =

\ I z dzz

c z c1

1

21

1

3

211

21

2

3= =

++ = +ò

+

( )

Þ I x x c12

3

21

2

33 18= + - +( ) … (iv)

Again I x x dx22 3 18= + -ò

= + + æèç ö

ø÷ - -ò x x dx2

2

23

2

3

2

9

418. . = +æ

èç ö

ø÷ - æ

èç ö

ø÷ò x

3

2

9

2

2 2

I x x x x x x22 21

2

3

23 18

81

4 2

3

23 1= +æ

èç ö

ø÷ + - -

´+æ

èç ö

ø÷ + + -log 8

Þ I x x x x x x22 21

2

3

23 18

81

8

3

23 18= +æ

èç ö

ø÷ + - - +æ

èç ö

ø÷ + + - +log c2 … (v)

Putting the value of I 1 and I 2 in (iii), we get

I x x x x x x= + - - +æèç ö

ø÷ + - + +

1

33 18

9

4

3

23 18

729

16

3

2

23

2 2( ) log æèç ö

ø÷ + + -x x2 3 18 + c

where cc

c= -é

ëê

ù

ûú

122

9

2




18. We have,

e y dxy

xdyx 1 02- + =

Þ e y dxy

xdyx 1 2- = - Þ xe dxx = -

-

y

ydy

1 2

Þ x e dxy

ydy

I

x

IIò ò= -

-1 2

Þ xe e dxdt

t

x x- =ò ò1

2, where t y= -1 2 (Using I LATE on LHS)

Þ xe et

Cx x- =æ

èç

ö

ø÷ +

1

2 1 2

1 2/

/Þ xe e t Cx x- = +

Þ xe e y Cx x- = - +1 2 , where x RÎ is the required solution.

Putting y = 1 and x = 0

0e e C0 0 21 1- = - + Þ C = -1

Therefore required particular solution is xe e yx x- = - -1 12 .

19. The given differential equation is

( )xdy

dxxy

x

2

21 2

2

1- + =

-

Þdy

dx

x

xy

x+

-=

-

2

1

2

12 2 2( )… (i)

This is a linear differential equation of the form dy

dxPy Q+ = ,

where Px

x=

-

2

12 and Q

x=

-

2

12 2( )

\ I F e e dx e xPdx x x x. . ( )

/( ) log( )= ò = ò = = -- -2 2 1 2 1 2 1

y xx

x dx C( )( )

( )2

2

212

12 1- =

-´ - +ò [Using: y I F Q I F dx C( . .) .( . .)= +ò ]

\ y xx

dx C( )2

21

2

1- =

-+ò

Þ y x( )2 1 21

2- = ´ log

x

xC

-

++

1

1Þ y x

x

xC( ) log2 1

1

1- =

-

++

This is the required solution.

20. L.H.S. = + + +® ® ® ® ® ®

[ , , ]a b b c c a = + + ´ +® ® ® ® ® ®

( ).{( ) ( )}a b b c c a

= + ´ + ´ + ´ + ´® ® ® ® ® ® ® ® ® ®

( ).{ }a b b c b a c c c a

= + ´ + ´ + ´® ® ® ® ® ® ® ®

( ).{ }a b b c b a c a [Q c c® ® ®

´ = 0]




= ´ + ´ + ´ + ´ +® ® ® ® ® ® ® ® ® ® ® ® ®a b c a b a a c a b b c b b.( ) .( ) .( ) .( ) .(

® ® ® ® ®´ + ´a b c a) .( )

= + + + +® ® ® ® ® ® ® ® ® ® ® ® ®

[ , , ] [ , , ] [ , , ] [ , , ] [ ,a b c a b a a c a b b c b b® ® ® ® ®

+, ] [ , , ]a b c a

= + + + + +® ® ® ® ® ®

[ , , ] [ , , ]a b c b c a0 0 0 0 [By property of scalar triple product]

= +® ® ® ® ® ®

[ , , ] [ , , ]a b c b c a

= +® ® ® ® ® ®

[ , , ] [ , , ]a b c a b c [By property of circularly rotation]

=® ® ®

2[ , , ]a b c

OR

a b c® ® ®

+ + = 0 Þ ( ) ( )a b c® ® ®

+ = -2 2

Þ ( ) . ( ) .a b a b c c® ® ® ® ® ®

+ + =

Þ | | | | . | |a b a b c® ® ® ® ®

+ + =2 2 22 Þ 9 25 2 49+ + =® ®a b.

Þ 2 49 25 9a b® ®

= - -.

Þ 2 15| || |cosa b® ®

=q Þ 30 15cos q =

Þ cos cosq = = °1

260 Þ q = 60°

21. Given lines arex y z+

=+

=+1

3

3

5

5

7…(i)

x y z-=

-=

-2

1

4

3

6

5…(ii)

Let two lines (i) and (ii) intersect at a point P( , , )a b g .

Þ ( , , )a b g satisfy line (i)

Þa b g

l+

=+

=+

=1

3

3

5

5

7(say)

Þ a l= -3 1, b l= -5 3, g l= -7 5 …(iii)

Again ( , , )a b g also lie on (ii)

\ a b g-

=-

=-2

1

4

3

6

5

Þ3 1 2

1

5 3 4

3

7 5 6

5

l l l- -=

- -=

- -

Þ3 3

1

5 7

3

7 11

5

l l l-=

-=

-

I II III




From I and II From II and III

3 3

1

5 7

3

l l-=

- 5 7

3

7 11

5

l l-=

-

Þ 9 9 5 7l l- = - Þ 25 35 21 33l l- = -

Þ 4 2l = Þ 4 2l =

Þ l =1

2Þ l =

1

2

Since, the value of l in both the cases is same

Þ Both lines intersect each other at a point.

\ Intersecting point = = - - -æèç ö

ø÷( , , ) , ,a b g

3

21

5

23

7

25 [From (iii)]

= --æ

èç ö

ø÷

1

2

1

2

3

2, ,

22. A family has 2 children,

then Sample space = S = { , , , },BB BG GB GG where B stands for Boy and G for Girl.

(i) Let A and B be two event such that

A = Both are girls = {GG}

B = the youngest is a girl = {BG, GG}

PA

B

P A B

P B

æèç ö

ø÷ =

( )

( )

I[ { }]Q IA B GG=

PA

B

æèç ö

ø÷ =

1

42

4

= 1

2

(ii) Let C be event such that

C = at least one is a girl = { , , }BG GB GG

Now P(A/C) = P A C

P C

( )

( )

I[Q A C GGI = { }]

=

1

43

4

=1

3.

SECTION–C

23. According to question

3 2 1000x y z+ + =

4 3 1500x y z+ + =

x y z+ + = 600

The given system of linear equations may be written in matrix form as AX B= where

A =

é

ë

êêê

ù

û

úúú

3 2 1

4 1 3

1 1 1

, X

x

y

z

=

é

ë

êêê

ù

û

úúú

and B =

é

ë

êêê

ù

û

úúú

1000

1500

600

Q AX B= Þ X A B= -1 … (i)




Now for A-1

| | ( ) ( ) ( )A = = - - - + -

3 2 1

4 1 3

1 1 1

3 1 3 2 4 3 1 4 1

= - - + = - + = - ¹6 2 3 8 3 5 0

Hence, A-1 exists.

Also, A11

1 3

1 11 3 2= = - = - A12

4 3

1 14 3 1= - = - - = -( )

A13

4 1

1 14 1 3= = - = A21

2 1

1 12 1 1= - = - - = -( )

A22

3 1

1 13 1 2= = - = A23

3 2

1 1= - = - - = -( )3 2 1

A31

2 1

1 36 1 5= = - = A32

3 1

4 39 4 5= - = - - = -( )

A33

3 2

4 13 8 5= = - = -

\ Adj A =

- -

- -

- -

é

ë

êêê

ù

û

úúú

2 1 3

1 2 1

5 5 5

T

=

- -

- -

- -

é

ë

êêê

ù

û

úúú

2 1 5

1 2 5

3 1 5

\ Aadj A

A

- = =-

- -

- -

- -

é

ë

êêê

ù

û

úúú

1 1

5

2 1 5

1 2 5

3 1 5

( )

| |

Putting the value of X A B, ,-1 in (i), we get

x

y

z

é

ë

êêê

ù

û

úúú

= -

- -

- -

- -

é

ë

êêê

ù

û

úúú

1

5

2 1 5

1 2 5

3 1 5

1000

150. 0

600

é

ë

êêê

ù

û

úúú

= -

- - +

- + -

- -

é

ë

1

5

2000 1500 3000

1000 3000 3000

3000 1500 3000

êêê

ù

û

úúú

= -

-

-

-

é

ë

êêê

ù

û

úúú

1

5

500

1000

1500

Þ

x

y

z

é

ë

êêê

ù

û

úúú

=

é

ë

êêê

ù

û

úúú

100

200

300

Þ x y z= = =` ` `100 200 300, ,

i.e., 100 for discipline

200 for politeness and

300 for punctuality

One more value like sincerity or truthfulness can be awarded.

24. Let ABC be cone having slant height l and semi-vertical angle q.

If V be the volume of cone then.

V DC AD= ´ ´1

3

2.p = ´ ´p

q q3

2 2l lsin cos

Þ Vl

=p

q q3

2

3sin cos




ÞdV

d

l

q

p=

3

3 [ sin sin . cos ]- +3 22q q q

For maximum value of V.

dV

dq= 0

Þpl 3

3[ sin sin . cos ]- + =3 22 0q q q

Þ - + =sin sin . cos3 22 0q q q

Þ - - =sin (sin cos )q q q2 22 0

Þ sin q = 0 or 1 2 02 2- - =cos cosq q

Þ q = 0 or 1 3 02- =cos q

Þ q = 0 or cosq =1

3

Now d V

d

l2

2

3

3q

p= { sin . cos sin . cos cos }- - +3 4 22 2 3q q q q q

Þd V

d

l2

2

3

3q

p= { sin cos cos }- +7 22 3q q q

Þd V

d

2

20q q

ù

ûúú

=

=

+ve

andd V

d

2

2 1

3

q q

ù

ûúú

=

=cos

–ve [Putting cosq =1

3 and sin q = 1

1

3

2

3

2

-æèç

öø÷ = ]

Hence for cosq =1

3 or q =

æèç

öø÷

-cos ,1 1

3 V is maximum.

25. Let Idx

x=

+ò 1

6

3

cotp

p

=

+ò

dx

x

x1

6

3

cos

sinp

p

Ix

x xdx=

+òsin

sin cosp

p

6

3

… (i)

=

-æèç ö

ø÷

-æèç ö

ø÷ + -æ

èç ö

ø÷

òsin

sin cos

p

p

p

2

2 26

3 x

xx

xx

dx Q f x dx f a b x dx

a

b

a

b

( ) ( )ò ò= + -é

ë

êê

ù

û

úú

Ix

x xdx=

+òcos

sin cosp

p

6

3

… (ii)


BD

A

C

l

q



Adding (i) and (ii), we get

2

6

3

Ix x

x xdx=

+

+òsin cos

sin cosp

p

= ò dxp

p

6

3

2

6

3I x= [ ]p

p

= -p p

3 6 =

-2

6

p p =

p

6

Þ I =p

12.

26. The given equations are

y x= ...(i)

and x y2 2 32+ = ...(ii)

Solving (i) and (ii), we find that the line and the circle meetat B ( , )4 4 in the first quadrant. Draw perpendicular BM tothe x-axis.

Therefore, the required area = area of the region OBMO +area of the region BMAB.

Now, the area of the region OBMO

= =ò ò0

4

0

4y dx x dx = =

1

282

0

4[ ]x ...(iii)

Again, the area of the region BMAB

= = -ò òy dx x dx4

4 2

4

4 2 232 = - + ´ ´é

ëêù

ûú-1

232

1

232

4 2

2 1

4

4 2

x xx

sin

= ´ + ´ ´æèç ö

ø÷ - - + ´ ´

æè

- -1

24 2 0

1

232 1

4

232 16

1

232

1

2

1 1sin sinçöø÷

= - + = -8 8 4 4 8p p p( ) . ...(iv)

Adding (iii) and (iv), we get the required area = 4p sq units.

27. The equation of plane determined by the points A B( , , ), ( , , )2 5 3 2 3 5- - - and C( , , )5 3 3- is

x y z–

– – –

– –

2 5 3

2 2 3 5 5 3

5 2 3 5 3 3

0

- +

- +

- +

= Þ

x y z– –

–

–

2 5 3

4 8 8

3 2 0

0

+

- =

Þ ( – ) { } – ( – ) { – } ( ) { }x y z2 0 16 5 0 24 3 8 24 0+ + + + + =

Þ 16 32 24 120 32 96 0x y z- + - + + =

Þ 16 24 32 56 0x y z+ + - =

Þ 2 3 4 7 0x y z+ + - = … (i)

Now the distance of point (7, 2, 4) to plane (i) is

2 7 3 2 4 4 7

2 3 42 2 2

´ + ´ + ´ -

+ + =

+ + -14 6 16 7

29 =

29

29 = 29 unit.


B (4

, 4)

MOX¢ X

Y

Y¢

A (4 2

, 0)

y =

x



OR

Given line and plane are

r i j k i j k®

= - + + + +( $ $ $) ( $ $ $)2 2 3 4 2l …(i)

r i j k®

- + =. ($ $ $) 5 …(ii)

For intersection point, we solve equations (i) and (ii) by putting the value of r®

from (i) in (ii).

[( $ $ $) ( $ $ $)] .($ $ $)2 2 3 4 2 5i j k i j k i j k- + + + + - + =l

Þ ( ) ( )2 1 2 3 4 2 5+ + + - + =l Þ 5 5+ =l Þ l = 0

Hence, position vector of intersecting point is 2 2$ $ $i j k- + .

i.e., coordinates of intersection of line and plane is (2, –1, 2).

Hence, Required distance = + + - + + +( ) ( ) ( )2 1 1 5 2 102 2 2 = 9 16 144+ + = 169 = 13 units

28. Suppose dealer purchase x electronic sewing machines and y manually operated sewingmachines. If Z denotes the total profit. Then according to question

(Objective function) Z = 22x + 18 y

Also, x y+ £ 20

360 240 5760x y+ £ Þ 9 6 144x y+ £

x y³ ³0 0, .




We have to maximise Z subject to above constraint.

To solve graphically, at first we draw the graph of line corresponding to given inequationsand shade the feasible region OABC.

The corner points of the feasible region OABC are O(0, 0), A(16, 0), B(8, 12) and C(0, 20).

Now the value of objective function Z at corner points are obtained in table as

Corner points Z x y= +22 18

O(0, 0) Z = 0

A(16, 0) Z = ´ + ´ =22 16 18 0 352

B(8, 12) Z = ´ + ´ =22 8 18 12 392 Maximum

C(0, 20) Z = ´ + ´ =22 0 18 20 360

From table, it is obvious that Z is maximum when x = 8 and y = 12.

Hence, dealer should purchase 8 electronic sewing machines and 12 manually operatedsewing machines to obtain the maximum profit 392 under given condition.

29. Let E1, E2 , E E3 4, and A be event defined as

E1 = the lost card is a spade card.

E2 = the lost card is a heart card.

E3 = the lost card is a club card.

E4 = the lost card is diamond card.

and A = Drawing three spade cards from the remaining cards.

P E P E P E P E( ) ( ) ( ) ( )1 2 3 413

52

1

4= = = = =

PA

E

C

C1

123

513

220

20825

æ

èç

ö

ø÷ = = , P

A

E

C

C2

133

513

286

20825

æ

èç

ö

ø÷ = =

PA

E

C

C3

133

513

286

20825

æ

èç

ö

ø÷ = = , P

A

E

C

C4

133

513

286

20825

æ

èç

ö

ø÷ = =

Now, required probability = PE

A1æ

èç

ö

ø÷

PE

A1æ

èç

ö

ø÷ =

æ

èç

ö

ø÷

æ

èç

ö

ø÷ +

æ

èç

ö

ø÷

P E PA

E

P E PA

EP E P

A

E

( ).

( ). ( ).

11

11

22

+æ

èç

ö

ø÷ +

æ

èç

ö

ø÷P E P

A

EP E P

A

E( ). ( ).3

34

4

=´

´ + ´ + ´ +

1

4

220

208251

4

220

20825

1

4

286

20825

1

4

286

20825

1

4´

286

20825

=+ + +

220

220 286 286 286

=220

1078 =

10

49




OR

Let the number of defective bulbs be represented by a random variable X. X may have value0, 1, 2, 3, 4.

If p is the probability of getting defective bulb in a single draw then

p = =5

15

1

3

\ q = Probability of getting non defective bulb = 11

3

2

3- = .

Since each trial in this problem is Bernaulli trials, therefore we can apply binomialdistribution as

P X r C p qnr

r n r( ) . .= = - when n = 4

P X C( ) .= = æèç ö

ø÷ æ

èç ö

ø÷ =0

1

3

2

3

16

81

40

0 4

Now P X C( ) .= = æèç ö

ø÷ æ

èç ö

ø÷ = ´ ´ =1

1

3

2

34

1

3

8

27

32

81

41

1 3

P X C( )= = æèç ö

ø÷ æ

èç ö

ø÷ = ´ ´ =2

1

3

2

36

1

9

4

9

24

81

42

2 2

P X C( ) .= = æèç ö

ø÷ æ

èç ö

ø÷ = ´ ´ =3

1

3

2

34

1

27

2

3

8

81

43

3 1

P X C( )= = æèç ö

ø÷ æ

èç ö

ø÷ =4

1

3

2

3

1

81

44

4 0

Now probability distribution table is

X 0 1 2 3 4

P(X)16

81

32

81

24

81

8

81

1

81

Now mean E(X) = Sp xi i

= ´ + ´ + ´ + ´ + ´016

811

32

812

24

813

8

814

1

81

Mean = + + +32

81

48

81

24

81

4

81 = =

108

81

4

3.

SET–II

9. Let I x dx= -ò cos (sin )1

= -æèç ö

ø÷

æèç

öø÷

-ò cos cos1

2

px dx = -æ

èç ö

ø÷ò

p

2x dx

\ I dx xdx= -ò òp

2

= - +p

2 2

2

xx

c




10. a b a b n® ® ® ®

´ =| || |sin $q

Þ | | || |.| |sin $|a b a b n® ® ® ®

´ = q

Þ 1 = 32

3´ sin $q n Þ 1 2= sin |$|q n

Þ 1 2= sin q [ |$| ]Q n = 1

Þ sin q =1

2Þ q = °30 .

19. L.H.S. D =

a b c a b

c b c a b

c a c a b

+ +

+ +

+ +

2

2

2

Applying R R R R R R1 1 2 2 2 3® - ® -; , we get

D =

+ + - + +

+ + - + +

+ +

a b c a b c

a b c a b c

c a c a b

( )

( )

0

0

2

Taking ( )a b c+ + common along R1 and R2 , we get

D = + +

-

-

+ +

( )a b c

c a c a b

2

1 1 0

0 1 1

2

= + + -

+ + +

( )a b c

c a c c a b

2

1 0 0

0 1 1

2

[Applying C C C2 2 1® + ]

Again applying C C C3 3 2® + , we get

D = + +

+ + +

( )

( )

a b c

c a c c a b

2

1 0 0

0 1 0

2

= + + × + +( ) ( )a b c a b c2 2 (Q determinant of triangular matrix is product of

its diagonal elements)

= + +2 3( )a b c

D = R.H.S.

20. Let ux

x=

-

æ

èçç

ö

ø÷÷

-tan 1

21 and v x x= --sin ( )1 22 1

We have to determine du

dv

Let x = sin q Þ q = -sin 1 x

Now, u =-

æ

èçç

ö

ø÷÷

-tansin

sin

1

21

q

q




Þ u = æèç ö

ø÷-tan

sin

cos

1 q

qÞ u = q

Þ u x= -sin 1

Þdu

dx x=

-

1

1 2

Again, v x x= --sin (1 22 1

Þ v = --sin ( sin sin1 22 1q q

Þ v = -sin ( sin cos )1 2 q q

Þ v = -sin (sin )1 2q Þ v = 2q

Þ v x= -2 1sin Þdv

dx x=

-

2

1 2

\du

dv

du

dxdv

dx

x

x

= = -

-

=

1

12

1

1

2

2

2

[Note: Here the range of x is taken as - < <1

2

1

2x ]

21. cosec x ydy

dxx ylog + =2 2 0

cosec x ydy

dxx y. log = - 2 2

Þlog .y dy

y

x dx

x2

2

= -cosec

Þ y y dy x x dx-ò ò= -2 2. log sin

Þ log . . [ ( cos ) ( cosyy

y

ydy x x x

- + - +

- +-

- += - - - -ò

2 1 2 12

2 1

1

2 12 x dx) ]ò

Þ - + = --ò ò

122 2

yy y dy x x x x dxlog cos cos

Þ = - +- +

= - -- +

ò1

2 12

2 12

yy

yx x x x x dxlog cos [ sin sin ]

Þ = - - = - + - +1 1

2 22

yy

yx x x x x clog cos sin ( cos )

Þ = - + = - - +1

1 2 22

yy x x x x x c(log ) cos sin cos .


Q - < <1

2

1

2x

Þ sin sin sin-æèç ö

ø÷ < < æ

èç ö

ø÷

pq

p

4 4

Þ - < <p

qp

4 4

Þ - < <p

qp

22

2

Þ 22 2

qp p

Îæèç ö

ø÷– ,



22. Given lines are5

4

7

4

3

5

-

-=

-=

+

-

x y z Þ

x y z-=

-=

+

-

5

4

7

4

3

5… (i)

andx y z-

=-

--8

7

2 8

2

5

3 Þ

x y z-=

-=

-8

7

4

1

5

3… (ii)

We know that, x x

a

y y

b

z z

c

-=

-=

-1

1

1

1

1

1

and x x

a

y y

b

z z

c

-=

-=

-2

2

2

2

2

2

are coplanar iff

x x y y z z

a b c

a b c

2 1 2 1 2 1

1 1 1

2 2 2

- - -

= 0

Now

8 5 4 7 5 3

4 4 5

7 1 3

3 3 8

4 4 5

7 1 3

- - - -

- =

-

-

( )

= + + + + -3 12 5 3 12 35 8 4 28( ) ( ) ( )

= + -51 141 192 = -192 192 = 0

Hence lines (i) and (ii) are coplanar.

28. Let Ix x

x xdx= ò0

p tan

sec . cosec = ò0 1 1

px

x

x

x x

dx

.sin

cos

cos.

sin

I x x dx= ò02p

sin

= - -ò02p

p p( ) sin ( )x x dx [ ( ) ( ) ]Q0 0

a af x dx f a x dxò ò= -

I x dx x x dx= -ò ò0

2

0

2p pp sin sin Þ 2

22

0

2I x dx= òp p

sin

= -òp p

21 2

0( cos )x dx = -ò ò

p pp p

2 22

0 0dx x dxcos

[ ]= -é

ëêù

ûúp pp

p

2 2

2

200

xxsin

Þ 22

04

2 0I = - - -p

pp

p( ) (sin sin )

Þ 22

02

I = -p

Þ I =p 2

4.

29. Let r, h, q be radius, height and semi-vertical angle of cone having volume V.

If S be the surface area of cone then

S r h r= +p 2 2 Þ S r h r2 2 2 2 2= +p ( )

Þ S rV

rr2 2 2

2

2 4

29= +

æ

èç

ö

ø÷p

p

QV r h

hV

r

=

=

é

ë

êêêê

ù

û

úúúú

1

33

2

2

p

p




Þ SV

rr2

2

2

2 49= + p

Þd S

dr

V

rr

( )2 2

3

2 3184= - + p

For extremum value of S or S2 .

d S

dr

( )2

0=

Þ - + =18

4 02

3

2 3V

rrp Þ

184

2

3

2 3V

rr= p

Þ rV V6

2

2

2

2

18

4

9

2= =

p pÞ r

V3 3

2=

p

Againd S

dr

V

rr

2 2

2

2

4

2 25412

( )= + p

Þd S

drr

V

2 2

23 3

2

0( )é

ëêê

ù

ûúú

>

=p

i.e., For rV

S3 23

2=

p, or S is minimum.

Hence for minimum curve surface area

r r h3 23

2

1

3= æ

èç ö

ø÷

pp

Þ rr h3

2

2= Þ

r

h=

1

2

Þ tan q =1

2Þ cot q = 2

Þ q = -cot ( )1 2 .

SET–III

9. Let I e x x dxx= -ò (sin cos )

0

2

p

I e x x dxx= - -ò0

2

p

(cos sin ) = - + -ò e x x dxx (cos ( sin ))

0

2

p

= -[ cos ]e xx02

p

[ ]Q e f x f x dx e f x cx x( ( ) ( )) . ( )+ ¢ = +ò


h

r

A

q



= - -[ . cos . cos ]e e

pp2 0

20

= - -[ ]0 1 = 1.

10. a b i j k i j k® ®

+ = + - + + -( $ $ $) ( $ $ $)2 2 5 2 7 = + -4 3 12$ $ $i j k

\ Required vector in the direction of 4 3 12$ $ $i j k+ -

=+ -

+ + -

4 3 12

4 3 122 2 2

$ $ $

( )

i j k =

+ -4 3 12

169

$ $ $i j k =

+ -4 3 12

13

$ $ $i j k

= + -4

13

3

13

12

13$ $ $i j k.

19. L.H.S. D =

+

+

+

x xy xz

xy y yz

zx zy z

2

2

2

1

1

1

Applying C C C C1 1 2 3® + + , we have

D =

+ + +

+ + + +

+ + + +

1

1 1

1 1

2

2

x x y z xy xz

y x y z y yz

z x y z zy z

( )

( )

( )

D = +

+

+ + + +

+

1

1 1

1 1

1

1

2

2

2

2

xy xz

y yz

zy z

x y z

x xy xz

y y yz

z zy z

( )

Changing row into column, we have

D = +

+

+ + + +

+

1 1 1

1

1

1

1

2

2

2

2

xy y zy

xz yz z

x y z

x y z

xy y zy

xz yz z

( )

For I determinant we apply, C C C1 1 2® - , C C C2 2 3® -

For II determinant we take out a from 1st column, we have

D = - - + -

- - - +

+ + +

0 0 1

1 1

1 1

12 2

2 2

xy y y zy zy

xz yz yz z z

x x y z

y z

y( ) y zy

z yz z

2

2

1

1

+

+

Expanding along first row, we have

D = - - - - - - + -1 1 1 12 2 2[( ) ( ) ( ) ( )]xy y yz z xz yz y zy

+ + + + + - - + - + -x x y z y z y z y yz y z y z y z( ) [{ ( ) ( ) } ( ) (1 1 12 2 2 2 2 2 2 y z z2 - )]

On solving, we have

D = + + +1 2 2 2x y z = R.H.S.




20. Let ux

x=

+ -æ

è

çç

ö

ø

÷÷

-tan 121 1

and vx

x=

+

æ

èç

ö

ø÷

-sin 1

2

2

1

Now ux

x=

+ -æ

è

çç

ö

ø

÷÷

-tan 121 1

Let x = tan q Þ q = -tan 1 x

\ x =+ -æ

è

çç

ö

ø

÷÷

-tantan

tan

121 1q

q

Þ u =-æ

èç ö

ø÷-tan

sec

tan

1 1q

q =

-æ

è

ççç

ö

ø

÷÷÷

-tan cossin

cos

1

11

qq

q

=-æ

èç ö

ø÷-tan

cos

sin

1 1 q

q

Þ u =

æ

è

ççç

ö

ø

÷÷÷

-tansin

sin . cos

1

222

22 2

q

q q =

æ

è

ççç

ö

ø

÷÷÷

-tansin

cos

1 2

2

q

q = tan tan- æ

èç ö

ø÷1

2

qÞ u =

q

2

Þ u x= -1

2

1tan

\du

dx x=

+

1

2 1 2( )… (i)

Again vx

x=

+

æ

èç

ö

ø÷

-sin 1

2

2

1

Let x = tan q Þ q = -tan 1 x

\ v =+

æ

èç

ö

ø÷

-sintan

tan

1

2

2

1

q

q

Þ v = -sin (sin )1 2q Qsintan

tan2

2

1 2q

q

q=

+

é

ëê

ù

ûú

Þ v = 2q Þ v x= -2 1tan

\dv

dx x=

+

2

1 2… (ii)

Nowdu

dv

du

dxdv

dx

= [From (i) and (ii)]

=+

+

1

2 1

2

1

2

2

( )x

x

=+

´+1

2 1

1

22

2

( )x

x =

1

4.




21. Given differential equation is

dy

dx

x x

y y y=

+

+

( log )

sin cos

2 1

Þ (sin cos ) ( log )y y y dy x x dx+ = +2 1

Þ sin cos logy dy y y dy x x dx x dxò ò ò ò+ = +2

Þ sin [ sin sin ] log .y dy y y y dy xx

x

xdxò ò ò+ - = -

é

ëêê

ù

ûúú

22

1

2

2 2

+ ò x dx

Þ sin sin sin logy dy y y y dy x x x dx x dx c+ - = - + +ò ò ò ò2

Þ y y x x csin log= +2 … (i)

It is general solution.

For particular solution we put y =p

2 when x = 1

(i) becomes p p

2 21 1sin . log= + c

p

2= c [ log ]Q 1 0=

Putting the value of c in (i), we get the required particular solution.

y y x xsin log= +2

2

p.

22. Given lines are

r i j k i j®

= + - + -($ $ $) ( $ $)l 3

r i k i k®

= - + +( $ $) ( $ $)4 2 3m

Given lines also may be written in cartesian form as

x y z-=

-

-=

+1

3

1

1

1

0…(i)

andx y z-

=-

=+4

2

0

0

1

3…(ii)

Let given lines (i) and (ii) intersect at point ( , , ).a b g

Þ Point ( , , )a b g satisfy equation (i)

Þa b g

l-

=-

-=

+=

1

3

1

1

1

0 (say)

Þ a l b l g= + = - + = -3 1 1 1, ,

Also, point ( , , )a b g satisfy equation (ii)

\a b g-

=-

=+4

2

0

0

1

3

Þ3 1 4

2

1

0

1 1

3

l l+ -=

- +=

- +




Þ3 3

2

1

00

l l-=

- +=

I II III

From I and III From II and III

3 1

20

( )l -= - + =l 1 0

3 3 0l - = l = 1

l = =3

31

The value of l in both cases are same. Hence both lines (i) and (ii) intersect at a point.

The co-ordinate of intersecting point is (4, 0, –1).

28. Let Ix x x

x xdx=

+ò

sin cos

sin cos

/

4 40

2p

Þ I

x x x

=

-æèç ö

ø÷ -æ

èç ö

ø÷ -æ

èç ö

ø÷

ò0

2

4

2 2 2p

p p p

p

/ . sin . cos

sin2 2

4-æèç ö

ø÷ + -æ

èç ö

ø÷x x

dx

cosp

By Property

0

a

ò ò= -

é

ë

êê f x dx f a x dx

a( ) ( )

0

Þ I

x x x

x xdx=

-æèç ö

ø÷

+ò0

2

4 4

2p

p/ cos . sin

cos sin

Þ Ix x

x xdx

x x x=

+-ò ò

pp p

20

2

4 40

2/ /cos . sin

sin cos

sin . cos

sin 4 4x xdx

+ cos

Þ Ix x dx

x xI=

+-ò

pp

20

2

4 4

/ sin . cos

sin cos

Þ 22

0

2

4 4I

x x dx

x x=

+ò

pp/ sin . cos

sin cos =

+ò

pp

2 10

2 4

4

/

sin . cos

cos

tan

x x

xdx

x

[Dividing numerator and denominator by cos4 x]

=´ +

òp

p

2 2

2

10

2 2

2 2

/ tan . sec

(tan )

x x dx

x

Let

If

tan ; tan . sec

, ; ,

2 22

0 02

x z x x dx dz

x z x z

= =

= = = = ¥p

\ 2I =+

¥

òp

4 102

dz

z = - ¥p

4

10[tan ]z = ¥ -- -p

401 1(tan tan )




\ 2I = p p

4 20-æ

èç ö

ø÷ Þ I =

p 2

16

29. Let r h, be radius and height of closed right circular cylinder having volume 128p cm3.

If S be the surface area then

S rh r= +2 2 2p p

S rh r= +2 2p( )

S rr

r= +æ

èç

ö

ø÷2

1282

2p .

Q V r h

r h

hr

=

Þ =

\ =

é

ë

êêêêê

ù

û

úúúúú

p

p p

2

2

2

128

128

Sr

r= +æèç ö

ø÷2

128 2p

ÞdS

dr rr= - +

æ

èç

ö

ø÷2

1282

2p

For extreme value of S

dS

dr= 0

Þ 2128

2 02

p - +æ

èç

ö

ø÷ =

rr

Þ - + =128

2 02r

r

Þ 2128

2r

r= Þ r 3 128

2=

Þ r 3 64= Þ r = 4

Again d S

dr r

2

2 32

128 22=

´+

æ

èç

ö

ø÷p

Þd S

dr r

2

24

ù

ûúú

=

=

+ve

Hence, for r = 4 cm, S(surface area) is minimum.

Therefore, dimensions for minimum surface area of cylindrical can are

radius r = 4 cm and h = 128 128

168

2r= = cm.

z z z


h

r



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Class... · CBSE Examination Paper, Delhi-2014 Time allowed: 3 hours Maximum marks: 100 ¡ General Instructions: 1. All questions are compulsory. 2. The question paper …