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EXAMINATION PAPERS – 2010

MATHEMATICS CBSE (Delhi)CLASS – XII

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 eachand 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.

Set–I

SECTION–A

Question numbers 1 to 10 carry 1 mark each.

1. What is the range of the function f xx

x( )

| |

( )?=

-

-

1

1

2. What is the principal value of sin - -æ

èç

ö

ø÷

1 3

2 ?

3. If A =-æ

èç

ö

ø÷

cos

sin

sin

cos

a

a

a

a , then for what value of a is A an identity matrix?

4. What is the value of the determinant

0 2 0

2 3 4

4 5 6

?

5. Evaluate : log

.x

xdxò

6. What is the degree of the following differential equation?

5 62 2

2x

dy

dx

d y

dxy x

æ

èç

ö

ø÷ - - = log

Downloaded from www.studiestoday.com




7. Write a vector of magnitude 15 units in the direction of vector $ $ $.i j k- +2 2

8. Write the vector equation of the following line:x y z-

=+

=-5

3

4

7

6

2

9. If 1 2

3 4

3 1

2 5

7 11

23

æ

èç

ö

ø÷

æ

èç

ö

ø÷ =

æ

èç

ö

ø÷

k, then write the value of k.

10. What is the cosine of the angle which the vector 2$ $ $i j k+ + makes with y-axis?

SECTION–B

11. On a multiple choice examination with three possible answers (out of which only one iscorrect) for each of the five questions, what is the probability that a candidate would get fouror more correct answers just by guessing?

12. Find the position vector of a point R which divides the line joining two points P and Q whose

position vectors are ( )2 a b® ®

+ and ( )a b® ®

- 3 respectively, externally in the ratio 1 : 2. Also,

show that P is the mid-point of the line segment RQ.

13. Find the Cartesian equation of the plane passing through the points A ( , , )0 0 0 and

B( , , )3 1 2- and parallel to the line x y z-

=+

-=

+4

1

3

4

1

7.

14. Using elementary row operations, find the inverse of the following matrix :

2 5

1 3

æ

èç

ö

ø÷

15. Let Z be the set of all integers and R be the relation on Z defined as R a b a b Z= Î{( , ) ; , , and ( )a b- is divisible by 5.} Prove that R is an equivalence relation.

16. Prove the following:

tan cos , ( , )- -=-

+

æ

èç

ö

ø÷ Î1 11

2

1

10 1x

x

xx

OR

Prove the following :

cos sin sin- - -æèç

öø÷ + æ

èç

öø÷ = æ

èç

öø÷

1 1 112

13

3

5

56

65

17. Show that the function f defined as follows, is continuous at x = 2, but not differentiable:

f x

x

x x

x

x

x

x

( )

,

,

,

=

-

-

-

< £

< £

>

ì

íï

îï

3 2

2

5 4

0 1

1 2

2

2

OR

Find dy

dx, if y x x x x= - - --sin [ ].1 21 1

148 Xam idea Mathematics – XII





18. Evaluate : ex

xdxx

ò-

-

æ

èç

ö

ø÷

sin

cos.

4 4

1 4

OR

Evaluate : 1

1 2

2-

-òx

x xdx

( ).

19. Evaluate : p

p

/

/ sin cos

sin.

6

3

2ò+x x

xdx

20. Find the points on the curve y x= 3 at which the slope of the tangent is equal to the

y-coordinate of the point.

21. Find the general solution of the differential equation

x xdy

dxy

xxlog . log+ = ×

2

OR

Find the particular solution of the differential equation satisfying the given conditions:

dy

dxy x= tan , given that y = 1 when x = 0.

22. Find the particular solution of the differential equation satisfying the given conditions:

x dy xy y dx2 2 0+ + =( ) ; y = 1 when x = 1.

SECTION–C

Question numbers 23 to 29 carry 6 marks each.

23. A small firm manufactures gold rings and chains. The total number of rings and chainsmanufactured per day is atmost 24. It takes 1 hour to make a ring and 30 minutes to make achain. The maximum number of hours available per day is 16. If the profit on a ring is Rs. 300 and that on a chain is Rs 190, find the number of rings and chains that should bemanufactured per day, so as to earn the maximum profit. Make it as an L.P.P. and solve itgraphically.

24. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards aredrawn at random and are found to be both clubs. Find the probability of the lost card beingof clubs.

OR

From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random.Find the probability distribution of the number of defective bulbs.

25. The points A B( , , ), ( , , )4 5 10 2 3 4 and C ( , , )1 2 1- are three vertices of a parallelogram ABCD.Find the vector equations of the sides AB and BC and also find the coordinates of point D.

26. Using integration, find the area of the region bounded by the curve x y2 4= and the line

x y= -4 2.

Examination Papers – 2010 149





OR

Evaluate: 0

p

ò +

x x

x xdx

tan

sec tan.

27. Show that the right circular cylinder, open at the top, and of given surface area andmaximum volume is such that its height is equal to the radius of the base.

28. Find the values of x for which f x x x( ) [ ( )]= - 2 2 is an increasing function. Also, find the

points on the curve, where the tangent is parallel to x-axis.

29. Using properties of determinants, show the following:

( )

( )

( )

( )

b c

ab

ac

ab

a c

bc

ca

bc

a b

abc a b c

+

+

+

= + +

2

2

2

32

Set-II

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

3. What is the principal value of cos ?- -æ

èç

ö

ø÷

1 3

2

7. Find the minor of the element of second row and third column ( )a23 in the followingdeterminant:

2

6

1

3

0

5

5

4

7

-

-

11. Find all points of discontinuity of f, where f is defined as follows :

f x

x

x

x

x

x

x

( )

| | ,

,

,

=

+

-

+

£ -

- < <

³

ì

íï

îï

3

2

6 2

3

3 3

3

OR

Find dy

dx, if y x xx x= +(cos ) (sin ) ./1

12. Prove the following:

tan cos ,- -=-

+

æ

èç

ö

ø÷1 11

2

1

1x

x

x x Î( , )0 1

OR

Prove the following:


öø÷ +

3æèç

öø÷ = æ

èç

öø÷

1 1 112

13 5

56

65






14. Let * be a binary operation on Q defined by

a bab

* =3

5

Show that * is commutative as well as associative. Also find its identity element, if it exists.

18. Evaluate: 0 1

p

ò +

x

xdx

sin.

20. Find the equations of the normals to the curve y x x= + +3 2 6 which are parallel to the line

x y+ + =14 4 0.

23. Evaluate 1

3 23 2ò +( )x x dx as limit of sums.

OR

Using integration, find the area of the following region:

( , ) ;x yx y x y2 2

9 41

3 2+ £ £ +

ìíï

îï

üýï

þï

29. Write the vector equations of the following lines and hence determine the distance betweenthem:

x y z-=

-=

+1

2

2

3

4

6;

x y z-=

-=

+3

4

3

6

5

12

Set-III

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

1. Find the principal value of sin cos .- --æèç

öø÷ + -æ

èç

öø÷

1 11

2

1

2

9. If A is a square matrix of order 3 and | | | |,3A K A= then write the value of K.

11. There are two Bags, Bag I and Bag II. Bag I contains 4 white and 3 red balls while another Bag II contains 3 white and 7 red balls. One ball is drawn at random from one of the bags and it isfound to be white. Find the probability that it was drawn from Bag I.

14. Prove that : tan ( ) tan ( ) tan ( ) .- - -+ + =1 1 11 2 3 p

OR

If tan tan ,- --

-

æ

èç

ö

ø÷ +

+

+

æ

èç

ö

ø÷ =1 11

2

1

2 4

x

x

x

x

p find the value of x.

17. Show that the relation S in the set R of real numbers, defined as S a b a b R= Î{( , ) : , and a b£ 3}

is neither reflexive, nor symmetric nor transitive.

19. Find the equation of tangent to the curve yx

x x=

-

- -

7

2 3( ) ( ) , at the point, where it cuts the

x-axis.

23. Find the intervals in which the function f given by

f x x x x( ) sin cos ,= - £ £0 2p

is strictly increasing or strictly decreasing.






24. Evaluate 1

4 2ò -( )x x dx as limit of sums.

OR

Using integration find the area of the following region :

{( , ) :| | }x y x y x- £ £ -1 5 2

SOLUTIONS

Set–I

SECTION–A

1. We have given

f xx

x( )

| |

( )=

-

-

1

1

| |( ),

( ),x

x if x or x

x if x or x- =

- - > >

- - - < <

ìíî

11 1 0 1

1 1 0 1

(i) For x > 1, f xx

x( )

( )

( )=

-

-=

1

11

(ii) For x < 1, f xx

x( )

( )

( )=

- -

-= -

1

11

\ Range of f xx

x( )

| |

( )=

-

-

1

1 is { , }.-1 1

2. Let x = -æ

èç

ö

ø÷

-sin 1 3

2

Þ sin x = -3

2 Þ sin sinx = -æ

èç

öø÷

p

3Q

3

2 3=

é

ëê

ù

ûúsin

p

Þ x = -p

3

The principal value of sin - -æ

èç

ö

ø÷

1 3

2 is - ×

p

3

3. We have given

A =-é

ëê

ù

ûú

cos

sin

sin

cos

a

a

a

a

For the identity matrix, the value of A11 and A12 should be 1 and value of A12 and A21should be 0.

i.e., cos a = 1 and sin a = 0

As we know cos 0 1° = and sin 0 0° =

Þ a = °0






4.

0 2 0

2 3 4

4 5 6

= 03

5

4

62

2

4

4

60

2

4

3

5- + (expanding the given determinant by R1)

= - 22

4

4

6

= - - =2 12 16 8( )

The value of determinant is 8.

5. We have givenlog x

xdxò

Let log x t= Þ 1

xdx dt=

Given integral = t dtò

= +t

c2

2=

(log )xc

2

2+

6. 5 62 2

2x

dy

dx

d y

dxy x

æ

èç

ö

ø÷ - - = log

Degree of differential equation is the highest power of the highest derivative. In above d y

dx

2

2 is

the highest order of derivative.

\ Its degree = 1.

7. Let A i j k®

= - +$ $ $2 2

Unit vector in the direction of A®

is $$ $ $

( ) ( ) ( )A

i j k=

- +

+ - +

2 2

1 2 22 2 2 = - +

1

32 2($ $ $)i j k

\ Vector of magnitude 15 units in the direction of A®

= 15 152 2

3$ ($ $ $)

Ai j k

=- +

= - +15

3

30

3

30

3$ $ $i j k

= - +5 10 10$ $ $i j k

8. We have given line asx y z-

=+

=-

-

5

3

4

7

6

2

By comparing with equationx x

a

y y

b

z z

c

-=

-=

-1 1 1 ,






We get given line passes through the point ( , , )x x x1 2 3 i.e., ( , , )5 4 6- and direction ratios are ( , , )a b c i.e., (3, 7, –2).

Now, we can write vector equation of line as

A i j k i j k®

= - + + + -( $ $ $) ( $ $ $)5 4 6 3 7 2l

9.1 2

3 4

3 1

2 5

7 11

23

é

ëê

ù

ûú

é

ëê

ù

ûú =

é

ëê

ù

ûúk

LHS = 1 2

3 4

3 1

2 5

é

ëê

ù

ûú

é

ëê

ù

ûú

= ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) (

1 3 2 2

3 3 4 2

1 1 2 5

3 1 4

+

+

+

+ 5)

é

ëê

ù

ûú =

7 11

17 23

é

ëê

ù

ûú

Now comparing LHS to RHS, we get

\ k = 17

10. We will consider

a i j k®

= + +2$ $ $

Unit vector in the direction of a®

is $$ $ $

( ) ( ) ( )a

i j k=

+ +

+ +

2

2 1 12 2 2

=+ +

=+ +2

4

2

2

$ $ $ $ $ $i j k i j k

= + +2

2

1

2

1

2$ $ $i j k = + +

1

2

1

2

1

2$ $ $i j k

The cosine of the angle which the vector 2$ $ $i j k+ + makes with y-axis is 1

2

æèç

öø÷ .

SECTION–B

11. No. of questions = =n 5

Option given in each question = 3

p = probability of answering correct by guessing =1

3

q = probability of answering wrong by guessing = -1 p = 11

3

2

3- =

This problem can be solved by binomial distribution.

P r Cnr

n r r

( ) = æèç

öø÷

1æèç

öø÷

-2

3 3

where r = four or more correct answers = 4 or 5

(i) P C( )42

3

1

3

54

4

= æèç

öø÷

æèç

öø÷ (ii) P C( )5

1

3

55

5

= æèç

öø÷






\ P P P= +( ) ( )4 5

= æèç

öø÷

æèç

öø÷ + æ

èç

öø÷

54

45

5

52

3

1

3

1

3C C

= æèç

öø÷ +

1é

ëêù

ûú=

´ ´ ´

é

ëêù

ûú1

3

10

3 3

1

3 3 3 3

11

3

4

= = ×11

2430 045

12. The position vector of the point R dividing the join of P and Q externally in the ratio 1 : 2 is

ORa b a b®® ® ® ®

=- - +

-

1 3 2 2

1 2

( ) ( )

=- - -

-

® ® ® ®a b a b3 4 2

1 =

- -

-= +

® ®® ®3 5

13 5

a ba b

Mid-point of the line segment RQ is

( ) ( )3 5 3

22

a b a ba b

® ® ® ®® ®+ + -

= +

As it is same as position vector of point P, so P is the mid-point of the line segment RQ.

13. Equation of plane is given by

a x x b y y c z z( ) ( ) ( )- + - + - =1 1 1 0

Given plane passes through (0, 0, 0)

\ a x b y c z( ) ( ) ( )- + - + - =0 0 0 0 …(i)

Plane (i) passes through (3, –1, 2)

\ 3 2 0a b c- + = …(ii)

Also plane (i) is parallel to the linex y z-

=+

-=

+4

1

3

4

1

7

a b c- + =4 7 0 …(iii)

Eliminating a b c, , from equations (i), (ii) and (iii), we get

x y z

3

1

1

4

2

7

0-

-

=

Þ x y z-

-- +

-

-=

1

4

2

7

3

1

2

7

3

1

1

40

Þ x y z( ) ( ) ( )- + - - + - + =7 8 21 2 12 1 0

Þ x y z- - =19 11 0 , which is the required equation

14. Given, A =é

ëê

ù

ûú

2 5

1 3

We can write, A IA=






i.e., 2

1

5

3

1

0

0

1

é

ëê

ù

ûú =

é

ëê

ù

ûú A

1

1

2

3

1

0

1

1

é

ëê

ù

ûú =

-é

ëê

ù

ûú A [ ]R R R1 1 2® -

1

0

2

1

1

1

1

2

é

ëê

ù

ûú =

-

-é

ëê

ù

ûú A [ ]R R R2 2 1® -

1

0

0

1

3

1

5

2

é

ëê

ù

ûú =

-

-é

ëê

ù

ûú A [ ]R R R1 1 22® -

A- =-

-é

ëê

ù

ûú

1 3

1

5

2

15. We have provided

R a b a b Z a b= Î -{( , ) : , , and ( ) is divisible by 5}

(i) As ( )a a- = 0 is divisible by 5.

\ ( , )a a R a RÎ " Î

Hence, R is reflexive.

(ii) Let ( , )a b RÎ

Þ ( )a b- is divisible by 5.

Þ - -( )b a is divisible by 5. Þ ( )b a- is divisible by 5.

\ ( , )b a RÎ

Hence, R is symmetric.

(iii) Let ( , )a b RÎ and ( , )b c ZÎ

Then, ( )a b- is divisible by 5 and ( )b c- is divisible by 5.

( ) ( )a b b c- + - is divisible by 5.

( )a c- is divisible by 5.

\ ( , )a c RÎ

Þ R is transitive.

Hence, R is an equivalence relation.

16. We have to prove

tan cos ,- -=-

+

æ

èç

ö

ø÷1 11

2

1

1x

x

x x Î( , )0 1

L.H.S. = tan [ tan ]- -=1 11

22x x

=-

+

é

ë

êê

ù

û

úú

-1

2

1

1

12 2

2 2cos

( ) ( )

( ) ( )

x

x

=-

+

æ

èç

ö

ø÷ =-1

2

1

1

1cosx

x R.H.S. Hence Proved.






OR


öø÷ + æ

èç

öø÷ = æ

èç

öø÷

1 1 112

13

3

5

56

65

LHS =1æ

èç

öø÷ + æ

èç

öø÷

- -cos sin1 12

13

3

5

= æèç

öø÷ + æ

èç

öø÷

- -sin sin1 15

13

3

5Q cos sin- -æ

èç

öø÷ = æ

èç

öø÷

é

ëêù

ûú1 112

13

5

13

= ´ - æèç

öø÷ + ´ - æ

èç

öø÷

é

ë

êê

ù

û

úú

-sin 12 25

131

3

5

3

51

5

13

= ´ + ´é

ëêù

ûú=- -sin sin1 15

13

4

5

3

5

12

13

56

65 = RHS

LHS = RHS Hence Proved

17. We have given, f x

x

x x

x

x

x

x

( )

,

,

,

=

-

-

-

< £

< £

>

ì

íï

îï

3 2

2

5 4

0 1

1 2

2

2

At x = 2,(i) RHL LHL

= lim ( )x

f x® +2

=® -lim ( )

xf x

2

= lim ( )h

f h®

+0

2 =® -lim ( )

xf x

2

= lim { ( ) }h

h®

+ -0

5 2 4 = - - -®

lim { ( ) ( )}h

h h0

22 2 2

= 10 – 4 = 6 = lim {( ) ( )}h

h h®

- - -0

2 4 2 1 = 2 × 3 = 6

Also, f ( )2 = 2 2 2 8 2 62( ) - = - =

Q LHL = RHL = f ( )2

\ f x( ) is continuous at x = 2

(ii) LHD RHD

= lim( ) ( )

h

f h f

h®

- -

-0

2 2= lim

( ) ( )

h

f h f

h®

+ -

0

2 2

= lim[ ( ) ( )] ( )

h

h h

h®

- - - - -

-0

22 2 2 8 2= lim

[ ( ) ] ( )

h

h

h®

+ - - -

0

5 2 4 8 2

= lim[ )

h

h h h

h®

+ - - + -

-0

28 2 8 2 6= lim

h

h

h® 0

5

= limh

h h

h®

-

-0

22 7=

®lim ( )

h 05

= lim ( )h

h®

- +0

2 7 = 7 = 5






Q LHD ¹ RHD

\ f x( ) is not differentiable at x = 2

OR

We have given

y x x x x= - - --sin [ ].1 21 1

= - - --sin [ ( ) ]1 2 21 1x x x x

Þ y x x= -- -sin sin1 1

[using sin sin sin [ ]- - -- = - - -1 1 1 2 21 1x y x y y x

Differentiating w.r.t. x, we get

dy

dx x x

d

dxx=

--

-

1

1

1

12 2( )( )

=-

--

=-

--

1

1

1

1

1

2

1

1

1

2 12 2x x x x x x.

( )

18. ex

xdxx

ò-

-

æ

èç

ö

ø÷

sin

cos

4 4

1 4

=-æ

è

çç

ö

ø

÷÷ò e

x x

xdxx 2 2 2 4

2 22

sin cos

sin [sin sin cos4 2 2 2x x x= and 1 4 2 22- =cos sin ]x x

= -ò e x x dxx (cot )2 2 2cosec2

= -ò òcot .2 2 2x e dx e x dxx x cosec2

= - -ò[cot . ( ) . ]2 2 2x e x e dxx xcosec2 - ò2 2e x dxx cosec2

= + - òòcot . . .2 2 2 2 22x e x e dx x e dxx x xcosec cosec2 = +e x cx cot 2

OR

We have given

1

1 2

2-

-òx

x xdx

( ) =

-

-ò

1

2

2

2

x

x xdx

=-

-ò

x

x xdx

2

2

1

2 =

1

2

2 2

2

2

2ò-

-

æ

è

çç

ö

ø

÷÷

x

x xdx

=- + -

-ò

1

2

2 2

2

2

2

( ) ( )x x x

x xdx

= +-

-

æ

è

çç

ö

ø

÷÷ò

1

21

2

2 2

x

x xdx …(i)






By partial fractionx

x x

x

x x

A

x

B

x

-

-=

-

-= +

-

2

2

2

2 1 2 12 ( )

x A x Bx- = - +2 2 1( ) …(ii)

Equating co-efficient of x and constant term, we get

2 1A B+ = and - = -A 2

Þ A B= = -2 3,

\x

x x x x

-

-= +

-

2

2

2 3

1 22

From equation (i)

1

1 2

1

21

1

2

2 3

1 2

2-

-= + +

-

æ

èç

ö

ø÷ò òò

x

x xdx dx

x xdx

( )

= + - - +1

2

3

41 2x x x clog| | log| |

19. Given integral can be written as

I = p

p / 3

/

sin cos

( sin )6 1 1 2ò+

- -

x x

xdx =

+

- -òp

p / 3

/

sin cos

( sin cos )6 21

x x

x xdx

Put sin cosx x t- =

so that, (cos sin )x xdt

dx+ =

when x =p

6, t = - = -sin cos

p p

6 6

1

2

3

2

when x =p

3, t = - = -sin cos

p p

3 3

3

2

1

2

Þ I = [ ]1

2

3

2

3

2

1

22

1

1

2

3

2

3

2

1

2

1-

- -

-

-

ò-

=dt

ttsin

= sin sin- --é

ëê

ù

ûú - -

é

ëê

ù

ûú

1 13

2

1

2

1

2

3

2

= sin sin- --é

ëê

ù

ûú + -

é

ëê

ù

ûú

1 13

2

1

2

3

2

1

2 = --2

1

23 11sin ( )

20. Let P x y( , )1 1 be the required point. The given curve is

y x= 3 …(i)

dy

dxx= 3 2

dy

dxx

x y

æ

èç

ö

ø÷ =

1 1

123

,






Q the slope of the tangent at ( ),x y y1 1 1=

3 12

1x y= …(ii)

Also, ( , )x y1 1 lies on (i) so y x1 13= …(iii)

From (ii) and (iii), we have

3 12

13x x= Þ x x1

213 0( )- =

Þ x1 0= or x1 3=

When x y1 130 0 0= = =, ( )

When x y1 133 3 27= = =, ( )

\ the required points are (0, 0) and (3, 27).

21. x xdy

dxy

xxlog log+ =

2

Þdy

dx x xy

x+ =

1 22log

...(i)

This is a linear differential equation of the form

dy

dxPy Q+ =

where Px x

=1

log and Q

x=

22

\ I.F. = òePdx

=ò

e x xdx

1

log

[Let log x tx

dx dt= \ =1

]

=ò

= =e e ttdt

t1

log = log x

\ y xx

x dx Clog log= +ò22

[ \ solution is y Q dx C( . .) ( . .) ]I F I F= +ò

Þ y x x x dx clog log .= +ò-2 2

I II

Þ y x xx

x

xdxlog log=

-

é

ëêê

ù

ûúú

--

é

ëêê

ù

ûúú

é

ëêê

ù

ûú

- -

ò21

1

1

1 1

ú+ C

Þ y xx

xx dx Clog

log= - +

é

ëê

ù

ûú +-

ò2 2

Þ y xx

x xClog

log= - -

é

ëê

ù

ûú +2

1

Þ y xx

x Clog ( log )= - + +2

1 , which is the required solution






OR

dy

dxy x= tan Þ

dy

yx dx= tan

By integrating both sides, we get

dy

yx dx= òò tan .

log log|sec |y x C= + ...(i)

By putting x = 0 and y = 1 (as given), we get

log log (sec )1 0= + C

C = 0

\ (i) Þ log log|sec |y x=

Þ y x= sec

22. x dy y x y dx2 0+ + =( )

x dy y x y dx2 = - +( )

dy

dxy

x y

x= -

+( )

2

dy

dx

xy y

x= -

+æ

è

çç

ö

ø

÷÷

2

2…(i)

Putting y vx= and dy

dxv x

dv

dx= + in equation (i)

v xdv

dx

vx v x

x+ = -

+æ

è

çç

ö

ø

÷÷

2 2 2

2Þ v x

dv

dxv v+ = - +( )2

Þ x dv

dxv v= - -2 2

Þdv

v v

dx

x2 2+= - (by separating variable)

Þ1

2

12v v

dvx

dx+

= -ò ò (Integrating both sides)

Þ1

2 1 1

12v v

dvx

dx+ + -

= - òò

Þ1

1 1

12 2( )v

dvx

dx+ -

= - òò

Þ1

2

1 1

1 1log log log

v

vx C

+ -

+ += - +

Þ1

2 2log log log

v

vx C

+= - +






Þ log log logv

vx C

++ =

22 2

Þ log log log ,v

vx k

++ =

2

2 where k C= 2

Þ log logvx

vk

2

2+= Þ

vx

vk

2

2+=

Þ

y

xx

y

x

k×

+

=

2

2

Qy

xv=é

ëêù

ûú

Þ x y k y x2 2= +( ) …(ii)

It is given that y = 1 and x = 1, putting in (ii), we get

1 3= k Þ k =1

3

Putting k =1

3 in (ii), we get

x y y x2 1

32= æ

èç

öø÷ +( )

Þ 3 22x y y x= +( )

SECTION–C

23. Total no. of rings & chain manufactured per day = 24.

Time taken in manufacturing ring = 1 hour

Time taken in manufacturing chain = 30 minutes

One time available per day = 16

Maximum profit on ring = Rs 300

Maximum profit on chain = Rs 190

Let gold rings manufactured per day = x

Chains manufactured per day = y

L.P.P. is

maximize Z x y= +300 190

Subject to x y³ ³0 0,

x y+ £ 24

x y+ £1

216

Possible points for maximum Z are

(16, 0), (8, 16) and (0, 24).

At (16, 0), Z = + =4800 0 4800


32

28

24

20

16

12

8

4

0 4 8 12 16 20 24

x+y=24

x + y=16

(8,16)

(0,24)

Y

X

(16,0)

12





At (8, 16), Z = + =2400 3040 5440 ¬ Maximum

At (0, 24), Z = + =0 4560 4560

Z is maximum at (8, 16).

\ 8 gold rings & 16 chains must be manufactured per day.

24. Let A E1 1, and E2 be the events defined as follows:

A : cards drawn are both club

E1: lost card is club

E2 : lost card is not a club

Then, P E( )113

52

1

4= = , P E( )2

39

52

3

4= =

P A E( / )1 = Probability of drawing both club cards when lost card is club = 12

51

11

50´

P A E( / )2 = Probability of drawing both club cards when lost card is not club = 13

51

12

50´

To find : P E A( / )1

By Baye’s Theorem,

P E A( / )1 = P E P A E

P E P A E P E P A E

( ) ( / )

( ) ( / ) ( ) ( / )1 1

1 1 2 2+

=´ ´

´ ´ + ´ ´

1

4

12

51

11

501

4

12

51

11

50

3

4

13

51

12

50

= 12 11

12 11 3 13 12

´

´ + ´ ´ =

11

11 39

11

50+=

OR

There are 3 defective bulbs & 7 non-defective bulbs.

Let X denote the random variable of “the no. of defective bulb.”

Then X can take values 0, 1, 2 since bulbs are replaced

p P D= =( )3

10 and q P D= = - =( ) 1

3

10

7

10

We have

P XC C

C( )= =

´=

´

´=0

7 6

10 9

7

15

72

30

102

P XC C

C( )= =

´=

´ ´

´=1

7 3 2

10 9

7

15

71

31

102

P XC C

C( )= =

´=

´ ´

´=

12

1 3 2

10 9

1

15

70

32

02

\ Required probability distribution is

X 0 1 2

P x( ) 7/15 7/15 1/15






25. The points A B( , , ), ( , , )4 5 10 2 3 4 and C ( , , )1 2 1- are three vertices of parallelogram ABCD.

Let coordinates of D be ( , , )x y z

Direction vector along AB is

a i j k®

= - + - + -( ) $ ( ) $ ( ) $2 4 3 5 4 10 = - - -2 2 6$ $ $i j k

\ Equation of line AB, is given by

b i j k i j k®

= + + + + +( $ $ $) ( $ $ $)4 5 10 2 2 6l

Direction vector along BC is

c i j k®

= - + - + - -( ) $ ( ) $ ( ) $1 2 2 3 1 4 = - - -$ $ $i j k5

\ Equation of a line BC, is given by .

d i j k i j k®

= + + + + +( $ $ $) ($ $ $)2 3 4 5m

Since ABCD is a parallelogram AC and BD bisect each other

\ 4 1

2

5 2

2

10 1

2

2

2

3

2

4

2

+ + -é

ëê

ù

ûú =

+ + +é

ëê

ù

ûú, , , ,

x y z

Þ 2 5+ =x , 3 7+ =y , 4 9+ =z

Þ x y z= = =3 4 5, ,

Co-ordinates of D are (3, 4, 5).

26. Given curve

x y2 4= …(i)

Line equation

x y= -4 2 …(ii)

Equation (i) represents a parabola withvertex at the origin and axis along (+)vedirection of y-axis.

Equation (ii) represents a straight linewhich meets the coordinates axes at

(–2, 0) and 01

2,æ

èç

öø÷ respectively.

By solving two equations, we obtain

x x= -2 2

Þ - - =x x2 2 0 (by eliminating y)

Þ ( ) ( )x x- + =2 1 0

Þ x = - 1 2,

The point of intersection of given

parabola & line are (2, 1) and -æèç

öø÷1

1

4, .


2 x = 4y

(-1, )

(-2,0) (-1,0) (2,6)

x = 4

y – 2

X

(2,1)

Q(x, y )1

P(x, y

)2

X'

Y'

Y

14





\ required area = --ò 1

2

2 1( ) .y y dx ...(iii)

QP x y( , )2 and Q x y( , )1 lies on (ii) and (i) respectively

\ yx

yx

2 1

22

4 4=

+=and

\ (iii) Þ Area = x x

dx+

-æ

èçç

ö

ø÷÷

-ò

2

4 4

2

1

2

= x

dx dx x dx4

1

2

1

41

2

1

22

1

2

+ -

-- -òò ò = + -

é

ëêê

ù

ûúú -

xx

x2 3

1

2

8

1

2 12

= + -é

ëêù

ûú- - +é

ëêù

ûú=

4

8

2

2

8

12

1

8

1

2

1

12

9

8 sq. units.

OR

Ix x

x xdx=

+ò0p tan

sec tan

I

xx

x

x

x

x

dx=

+ò0 1

p

sin

cos

cos

sin

cos

=+ò0 1

p x x

xdx

sin

sin...(i)

Ix x

xdx=

- -

+ -ò0 1

p p p

p

( ) sin ( )

sin ( )Q

0 0

a af x dx f a x dxò ò= -é

ëêùûú

( ) ( )

Ix x

xdx=

-

+ò0 1

p p( ) sin

sin...(ii)

\ 210

Ix

xdx=

+òp p sin

sin[Using (i) and (ii)]

Þ 21

1 10I

x x

x xdx=

-

+ -òpp sin ( sin )

( sin ) ( sin )

=-

-òp

p

0

2

21

sin sin

sin

x x

xdx = -

é

ë

êê

ù

û

úú

òpp

0 2

2

2

sin

cos

sin

cos

x

x

x

xdx

= -ò òp pp p

0 0

2tan sec tanx x dx x dx

= - -ò òp pp p

0 0

2 1tan sec (sec )x x dx x dx

= - +ò ò òp p pp p p

0 0

2

0sec tan secx x dx x dx dx

[ ] [ ] [ ]= - + +p p pp p psec tanx x x C0 0 0 = p p p[ ] [ ]- - - + -1 1 0 0 = -p p( )2

Þ I = -p

p2

2( )






27. Let r be the radius and h be the height of the cylinder of given surface s. Then,

s r hr= +p p2 2

hs r

r=

- p

p

2

2…(i)

Then v r h= p 2 =-é

ë

êê

ù

û

úú

pp

pr

s r

r

22

2[From eqn. (i)]

vsr r

=- p 3

2

dv

dr

s r=

- 3

2

2p…(ii)

For maximum or minimum value, we have

dv

dr= 0

Þs r-

=3

20

2pÞ s r= 3 2p

Þ p pr rh2 2+ = 3 2pr

Þ r h=

Differentiating equation (ii) w.r.t. r, we get

d v

drr

2

23 0= - <p

Hence, when r h= , i.e., when the height of the cylinder is equal to the radius of its base v ismaximum.

28. We have given

y x x= -[ ( )]2 2 …(i)

= - +x x x2 2 4 4( ) = - +x x x4 3 24 4

dy

dxx x x= - +4 12 83 2

For the increasing function,

dy

dx> 0

Þ 4 12 8 03 2x x x- + > Þ 4 3 2 02x x x( )- + >

Þ 4 1 2 0x x x( ) ( )- - >

For 0 1< <x , dy

dx= + - - = +( ) ( ) ( ) ( ) ve

For x > 2,dy

dx= + + + = +( ) ( ) ( ) ( ) ve






The function is increasing for 0 1< <x and x > 2

If tangent is parallel to x-axis, then dy

dx= 0

Þ 4 1 2 0x x x( ) ( )- - =

Þ x = 0 1 2, ,

For x = 0, f ( )0 0=

For x = 1, f ( ) [ ( )]1 1 1 2 12= - =

For x = 2, f ( ) [ ]2 2 0 02= ´ =

\ Required points are (0, 0), (1, 1), (2, 0).

29. To prove:

( )

( )

( )

( )

b c

ab

ac

ab

a c

bc

ca

bc

a b

abc a b c

+

+

+

= + +

2

2

2

32

Let D =

+

+

+

( )

( )

( )

b c

ab

ac

ab

a c

bc

ca

bc

a b

2

2

2

[Multiplying R R1 2, and R 3 by a b c, , respectively]

D =

+

+

+

1

2

2

2

2

2

2

2

2

2abc

a b c

ab

ac

b

ba

a c

bc

a c

b c

a b c

( )

( )

( )

=

+

+

+

1

2

2

2

2

2

2

2

2

2abcabc

b c

b

c

a

c a

c

a

b

a b

( )

( )

( )

[C C C1 1 3® - and C C C2 2 3® - ]

=

+ -

- +

+ -

- + +

( )

( )

( )

( ) ( )

b c a

c a b

c a b

c a b

a

b

a b

2 2

2 2

2 2

2 2

2

2

2

0

0

=

+ + + -

+ + - -

+ + + -

+ +

( ) ( )

( )( )

( )( )

(

b c a b c a

c a b c a b

c a b b c a

c a

0

0

b c a b

a

b

a b)( ) ( )- - +

2

2

2

= + +

+ -

- -

+ -

- - +

( )

( )

a b c

b c a

c a b

c a b

c a b

a

b

a b

2

2

2

2

0

0






= + +

+ -

-

+ -

-

( )a b c

b c a

b

c a b

a

a

b

ab

2

2

20

2

0

2 2

( ( ))R R R R3 3 1 2® - +

=+ +

+ -

-

+ -

-

( )a b c

ab

ab ac a

ab

bc ba b

ab

a

b

ab

22

2

2

20

2

0

2 2

=+ +

+

+( )a b c

ab

ab ac

b

a

bc ba

a

b

ab

22

2 2

2

0 0 2

[ , ]C C C C C C1 1 3 2 2 3® + ® +

=+ +

+

+( )

. .a b c

abab ab

b c

b

a

c a

a

b2

2

0 0 1

= + ++

+2 2ab a b c

b c

b

a

c a( )

= + + + + -2 2ab a b c b c c a ab( ) {( ) ( ) }

= + +2 3abc a b c( ) = RHS

Set–II

3. Let x = -æ

èç

ö

ø÷

-cos 1 3

2

Þ cos x = -3

2

Þ cos cosx = -é

ëêù

ûúp

p

6 = cos

5

6

p[as cos /p / 6 = 3 2]

Þ x =5

6

p

The principal value of cos- -æ

èç

ö

ø÷

1 3

2 is

5

6

p.

7. We have given

2

6

1

3

0

5

5

4

7

-

-

Minor of an element

a M23 23= = 2

1

3

510 3 13

-= + =






11. We have given

f x

x

x

x

x

x

x

( )

| | ,

,

,

=

+

-

+

£ -

- < <

³

ì

íï

îï

3

2

6 2

3

3 3

3

(i) For x = - 3

LHL = lim ( ) lim ( )x h

f x f h® - - ®

= -3 0

3 = lim ( )h

h®

- - = -0

2 3 6

RHL = lim ( ) lim ( )x h

f x f h® + ®

= +3 0

3 = lim ( )h

h®

+ + =0

6 3 2 20

LHL ¹ RHL

At x = 3, function is not continuous.

OR

Given, y x xx x= +(cos ) (sin ) /1

= +e ex x x xlog (cos ) / log (sin )1

By differentiating w.r.t. x

dy

dxe cos x

x

xxx x= + -

ìíî

üýþ

log (cos ) log ( )cos

(sin )

+ - +é

ëê

ù

ûúe x

x

x

x xx

x1

2

1log (sin )log (sin )

cos

sin

= - +(cos ) {log (cos ) tan } (sin ) /x x x x xx x1 - +ìíî

üýþ

12x

xx

xlog sin

cot

14. For commutativity, condition that should be fulfilled is

a b b a* *=

Consider a bab

* =3

5 =

3

5

ba = b a*

\ a b b a* *=

Hence, * is commutative.

For associativity, condition is ( * ) * * ( * )a b c a b c=

Consider ( * ) * *a b cab

cabc

=æèç

öø÷ =

3

5

9

25 =

3

5

3

5

3

5a bc a b c a b cæ

èç

öø÷ = =( * ) * ( * )

Hence, ( * ) * * ( * )a b c a b c=

\ * is associative.

Let e QÎ be the identity element,

Then a e e a a* *= =

Þ 3

5

3

5

ae eaa= = Þ e = ×

5

3






18. Ix

xdx=

+ò0 1

p

sin…(i)

Ix

xdx=

-

+ -ò0 1

p p

psin ( )Q f x dx f a x dx

aa

( ) ( )= -é

ë

êê

ù

û

úú

òò00

=-

+ò0 1

p p x

xdx

sin…(ii)

Adding equations (i) and (ii), we get

210

Ix

dx=+ò

p p

sin

=-

+ -òpp

0

1

1 1

sin

( sin ) ( sin )

x

x xdx =

-òp

p

0 2

1 sin

cos

x

xdx

= -òpp

0

2(sec sec tan )x x x dx

[ ]= -pp

tan secx x0

= p[( ) ( )]0 1 0 1+ - - = 2p

Þ 2 2I = p or I = p

20. Given equation of curve

y x x= + +3 2 6 ...(i)

Equation of line

x y+ + =14 4 0 ...(ii)

Differentiating (i) w.r.t. x, we get

dy

dxx= +3 22 Þ

dx

dy x=

+

1

3 22

\ Slope of normal =-

+

1

3 22x.

and it is parallel to equation of line.

\-

+=

-1

3 2

1

142x

Þ 3 2 142x + = Þ 3 122x =

x2 4= Þ x = ± 2

From equation of curve,

if x y= =2 18, ; if x y= - = -2 6,

\ Equation of normal at (2, 18) is

y x- = - -181

142( ) or x y+ - =14 254 0

and for (–2, –6) it is

y x+ = - +61

142( ) or x y+ + =14 86 0






23.1

3 23 2ò +( )x x dx

We have to solve this by the help of limit of sum.

So, a b= =1 3,

f x x x( ) ,= +3 22 hn

nh=-

Þ =3 1

2

Q1

3 2

03 2 1 1 1 2 1ò + = + + + + + +

®( ) lim [ ( ) ( ) ( ) ... ( (x x dx h f f h f h f

hn h- 1) )] …(i)

f ( ) ( ) ( )1 3 1 2 12= +

f h h h h h( ) ( ) ( )1 3 1 2 1 3 8 52 2+ = + + + = + +

f h h h h h( ) ( ) ( )1 2 3 1 2 2 1 2 12 16 52 2+ = + + + = + +

&: &: &: &: &: &:

&: &: &: &: &: &:

&: &: &: &: &: &:

f n h n h n h( ( ) ) ( ( ) ) ( ( ) )1 1 3 1 1 2 1 12- - = + - + + -

= - + - +3 1 8 1 52 2( ) ( )n h n h

By putting all values in equation (i), we get

1

3 2

0

2 23 2 5 3 8 5 12 16 5ò + = + + + + + + +®

( ) lim [( ) ( ) ( )x x dx h h h h hh

...

+ - + - +[ ( ) ( ) ]]3 1 8 1 52 2n h n h

= + + + - + + + + - +®

lim [ { ( ) } { ( )} ]h

h h n h n n0

2 23 1 4 1 8 1 2 1 5K K

= ×- -

+-

+é

ëê

ù

ûú

®lim

( ) ( ) ( )

hh h

n n n h n nn

0

231 2 1

6

8 1

25

[Q { .... ( )( ) ( )

1 4 11 2 1

6

2+ + + - =- -

nn n n

and { ( )( )

]1 2 11

2+ + + - =

-K n

n n

=- -

+ - +é

ëê

ù

ûú

®lim

( ) ( ) ( )( ) ( )

h

nh h nh nh hnh h nh nh

0

2

24 5

=- -

+ - +é

ëê

ù

ûú

®lim

( ) ( ) ( )( ) ( )

h

h hh

0

2 2 4

24 2 2 10

=´ ´

+ ´ ´ +é

ëê

ù

ûú

2 2 4

24 2 2 10 [by applying limit] = 34

OR

We have given

( , ) ;x yx y x y2 2

9 41

3 2+ £ £ +

ìíï

îï

üýï

þï






There are two equations

(i) y1 = equation of ellipse

i.e., x y2 2

9 41+ =

Þ y x122

39= -

and y2 = equation of straight line

i.e.,x y

3 21+ =

Þ y x22

33= -( )

\ We have required area

= -ò03

1 2( )y y dx

= - - -ìíî

üýþò0

3 22

39

2

33x x dx( )

= - - -ìíî

üýþò

2

39 3

0

3 2x x dx( )

= - + - +é

ëêê

ù

ûúú

-2

3 29

9

2 33

2

2 12

0

3x

xx

xx

sin

= + - +æèç

öø÷ - + - +

é

ëêù

ûú-2

3

3

20

9

21 9

9

20 0 0 01sin ( ) ( )

= × -é

ëêù

ûú= -

2

3

9

2 2

9

2

3

22

pp( ) sq. units.

29. Let

Line 1 : x y z-

=-

=+

=1

2

2

3

4

6m …(i)

From above, a point ( , , )x y z on line 1 will be ( , , )2 1 3 2 6 4m m m+ + -

Line 2 : x y z-

=-

=+

=3

4

3

6

5

12l …(ii)

From above, a point ( , , )x y z on line 2 will be ( , , )4 3 6 3 12 5l l l+ + -

Position vector from equation (i), we get

r i j k®

= + + + + -( ) $ ( ) $ ( ) $2 1 3 2 6 4m m m

= + - + + +($ $ $) ( $ $ $)i j k i j k2 4 2 3 6m

a i j k1 2 4®

= + -$ $ $, b i j k1 2 3 6®

= + +$ $ $

Position vector from equation (ii), we get

r i j k®

= + + + + -( ) $ ( ) $ ( ) $4 3 6 3 12 5l l l = + - + + +( $ $ $) ( $ $ $)3 3 5 4 6 12i j k i j kl


X' X

Y'

(–3,0) (3,0)

(0,–2)

(0,2)

x y += 1

3 2

2

2

x y +

= 1

9 4

Y





a i j k2 3 3 5®

= + -$ $ $, b i j k2 4 6 12®

= + +$ $ $

From b1 and b2 we get b b2 12® ®

=

Shortest distance =- ´

® ® ®

®

|( ) |

| |

a a b

b

2 1

( ) ( $ $ $) ($ $ $)a a i j k i j k2 1 3 3 5 2 4® ®

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

( )a a b2 1

® ® ®- ´ =

$ $ $

$ $ $i j k

i j k2

2

1

3

1

6

9 14 4- = - +

|( ) | ( ) ( ) ( )a a b2 12 2 29 14 4

® ® ®- ´ = + - + = + + =81 196 16 293

| | ( ) ( ) ( )b®

= + + = + + =2 3 6 4 9 36 72 2 2

Shortest distance =293

7 units

Set–III

1. We have given

sin cos- --æèç

öø÷ + -æ

èç

öø÷

1 11

2

1

2

But, as we know sin cos- -+ = ×1 1

2x x

p

\ principal value is p

2×

9. Given | | | |3A K A= , where A is a square matrix of order 3. …(i)

We know that | | ( ) | |3 3 3A A= = 27| |A …(ii)

By comparing equations (i) and (ii), we get

K = 27

11. Let A, E1, E2 be the events defined as follow:

A : Ball drawn is white

E1 : Bag I is chosen, E2 : Bag II is chosen

Then we have to find P E A( / )1

Using Baye’s Theorem

P E AP E P A E

P E P A E P E P A E( / )

( ) ( / )

( ) ( / ) ( ) ( / )11 1

1 1 2 2

=+

=´

´ + ´=

+=

1

2

4

51

2

4

7

1

2

3

10

4

740 21

70

40

61






14. tan ( ) tan ( ) tan ( )- - -+ + =1 1 11 2 3 p

Consider L.H.S.

tan ( ) tan ( ) tan ( )- - -+ +1 1 11 2 3 …(i)

Let Z = -tan ( )1 1

tan Z = 1

Z =p

4 …(ii)

And we know tan tan tan- - -+ = ++

-

1 1 1

1x y

x y

xyp …(iii)

Putting value of (ii) and (iii) in equation (i), we get

LHS = p

p4 1

1+ ++

-

-tanx y

xy =

pp

4

2 3

1 2 3

1+ ++

- ´

-tan

= p

p4

11+ + --tan ( ) = p

pp

p4 4

+ - = = RHS

OR

tan tan- --

-

æ

èç

ö

ø÷ +

+

+

æ

èç

ö

ø÷ =1 11

2

1

2 4

x

x

x

x

p

Consider above equation

We know tan tan tan- - -+ =+

-

1 1 1

1x y

x y

xy

Þ tan -

-

-+

+

+

--

-

æ

èç

ö

ø÷

+

+

æ

èç

ö

ø÷

ì

í

ïï

î

ï

1

1

2

1

2

11

2

1

2

x

x

x

x

x

x

x

xï

ü

ý

ïï

þ

ïï

=p

4

Þ x x x x

x x

2 2

2 2

2 2

4 1 4

+ - + - -

- - += tan

p

Þ2 4

31

2x -

-= Þ 2 4 32x - = -

Þ 2 12x = or x = ±1

2

i.e., x = -1

2

1

2,

17. We have given

S a b a b R= Î{( , ) : , and a b£ 3}

(i) Consider a =1

2






Then ( , ) ,a a R= æèç

öø÷ Î

1

2

1

2

But1

2

1

2

3

£ æèç

öø÷ is not true

\ ( , ) ,a a RÏ for all a RÎ

Hence, R is not reflexive.

(ii) Let a b= =1

21,

Then, 1

21 3£ ( ) i.e.,

1

2£ 1

Þ ( , )a b RÎ

But 11

2

3

£ æèç

öø÷/ \ ( , )b a RÏ

Hence, (a, b) Î R but (b, a) Ï R

(iii) Let a = 3, b c= =3

2

4

3,

Then 33

2

3

£ æèç

öø÷ i.e., 3 £ 27

\ ( , )a b RÎ

Also,3

2

4

3

3

£ æèç

öø÷ i.e.,

3

2

64

27£

\ ( , )b c RÎ

But3

1

4

3

3

£ æèç

öø÷/ i.e., 3

64

27£/

\ ( , )a c RÏ

Hence, ( , )a b RÎ , ( , )b c RÎ but ( , )a c RÏ

Þ R is not transitive.

19. We have given

yx

x x=

-

- -

7

2 3( ) ( )…(i)

Let (i) cuts the x-axis at (x, 0)

thenx

x x

-

- -=

7

2 30

( ) ( ) Þ x = 7

\ the required point is (7, 0).

Differentiating equation (i) w.r.t. x, we get

dy

dx

x x x x x

x x=

- - - - - + -

- -

( ) ( ) ( ) [( ) ( )]

[( ) ( )]

2 3 1 7 2 3

2 3 2






=- + - + -

- +

x x x x

x x

2 2

2 2

5 6 2 19 35

5 6( ) =

- + -

+ -

x x

x x

2

2 2

14 29

6 5( )

dy

dx

ù

ûú =

- + -

- +=

2=

( , ) ( )7 02

49 98 29

49 35 6

0

400

1

20

\ Equation of tangent is

y y x x- = -1 21

20( )

Þ y x- = -01

207( ) or x y- - =20 7 0

23. f x x x( ) sin cos ,= - 0 2£ £x p

Differentiating w.r.t. x, we get

¢ = +f x x x( ) cos sin = +æèç

öø÷2

4sin

px

For critical points, dy

dx= 0

Þ cos sinx x+ = 0

Þ sin cosx x= - Þ tan x = - 1

Þ tan x = -æèç

öø÷tan

p

4

Þ x n= - = -pp p

4 4,

3

4

7

4

11

4

p p p, , +K

(i) For 03

4< <x

p,

p pp

4 4< + <x i.e., It lies in quadrant I, II.

Þ 22

0sinp

+æèç

öø÷ >x , Hence, function is increasing.

(ii) For 3

4

7

4

p p< <x

pp

p< + <x4

2 i.e., It lies in quadrant III, IV.

Þ 24

0sinp

+æèç

öø÷ <x , Hence, function is decreasing.

(iii) For7

42

pp< <x

24

9

4p

p p< + <x i.e., It lies in quadrant I.

Þ 24

0sinp

+æèç

öø÷ >x , Hence, function is increasing.






Internal where function is strictly increasing is

03

4

7

42, ,

p ppæ

èç

öø÷ È æ

èç

öø÷

Interval where function is strictly decreasing is

3

4

7

4

p p,æ

èç

öø÷

24.1

4 2ò -( )x x dx

We have to solve it by using limit of sums.

Here, a b= =1 4, , hb a

n n=

-=

-4 1 i.e., nh = 3

Limit of sum for 1

4 2ò -( )x x dx is

= + + + + + + + -®

lim [ ( ) ( ) ( ) .... { ( ) }]h

h f f h f h f n h0

1 1 1 2 1 1

Now, f ( )1 1 1 0= - =

f h h h h h( ) ( ) ( )1 1 12 2+ = + - + = +

f h h h h h( ) ( ) ( )1 2 1 2 1 2 4 22 2+ = + - + = +

……………………………………………

……………………………………………

……………………………………………

f n h n h n h[ ( ) ] { ( ) } { ( ) }1 1 1 1 1 12+ - = + - - + -

= - + -( ) ( )n h n h1 12 2

\1

2 2

0

2 2 2 20 4 2 1ò - = + + + + + - + -®

( ) lim [ ...: ( ) (x x dx h h h h h n h nh

1) ]h

= lim [ { .. ( ) } { ( )}]h

h h n h n®

+ + + - + + + + -0

2 21 4 1 1 2 1K

= ×- -

+-é

ëê

ù

ûú®

lim( ) ( ) ( ) ( )

hh h

n n nh

n n

0

2 1 2 1

6

1

2

[Q 1 4 11 2 1

6

2+ + + - =- -

K ( )( ) ( )

nn n n

1 2 11

2+ + + - =

- ù

ûúK ( )

( )n

n n

=- -

+-é

ëê

ù

ûú

®lim

( ) ( ) ( )

h

nh n h h nh h nh nh h

0

2

6 2

=- -

+-é

ëê

ù

ûú

®lim

( ) ( ) ( ) ( ) ( )

h

h h h

0

3 3 6

6

3 3

2

=´ ´æ

èç

ö

ø÷ +

´æ

èç

ö

ø÷ = + =

3 3 6

6

3 3

29

9

2

27

2






OR

We have provided

( , ) :| |x y x y x- £ £ -1 5 2

Equation of curve is y x= -5 2 or y x2 2 5+ = , which is a circle with centre at (0, 0) and

radius 5

2.

Equation of line is y x= -| |1

Consider, y x= - 1 and y x= -5 2

Eliminating y, we get

x x- = -1 5 2

Þ x x x2 21 2 5+ - = -

Þ 2 2 4 02x x- - =

Þ x x2 2 0- - =

Þ ( ) ( )x x- + =2 1 0

Þ x = -2 1,

The required area is

= - - - + - -- -ò ò ò1

2 2

1

1

1

25 1 1x dx x dx x dx( ) ( )

= - +é

ëê

ù

ûú - - +

é

ëêê

ù

ûúú

- --

- -

xx

x xx

xx

25

5

2 5 2 2

2 1

1

2 2

1

1 2

siné

ëêê

ù

ûúú 1

2

= +æ

èç

ö

ø÷ + - -

æ

èç

ö

ø÷ -

-+ + +æ

è- -1

5

2

2

51

5

2

1

5

1

21

1

211 1sin sin ç

öø÷ - - - +æ

èç

öø÷2 2

1

21

= +æ

èç

ö

ø÷ + - -- -5

2

2

5

1

52 2

1

2

1 1sin sin

= - + -é

ëê

ù

ûú --5

2

2

51

1

5

1

51

4

5

1

2

1sin

= +æèç

öø÷

é

ëêù

ûú--5

2

4

5

1

5

1

2

1sin

= --5

21

1

2

1sin ( )

= -æèç

öø÷

5

4

1

2

p sq. units


3

2

1

–3 –2 –1 0 1 2 3XX'

Y'

Y

y = – x + 1

2

y= 5

-x

y = x –1

(–1,0)





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