cbse 2015 delhi

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CBSE 2015 ANNUAL EXAMINATION DELHI

SET 1/2/3 With Complete Explanations

Note that all the sets have same questions. Only their sequence of appearance is different.

Max. Marks : 100 Time Allowed : 3 Hours SECTION A

Q01. If a 7i j 4k and b 2i 6 j 3k

, then find the projection of a on b

.

Sol. Given a 7i j 4k and b 2i 6j 3k

.

Projection of a on b a .b 2i 6 j 3k 14 6 12 8 (7i j 4k).

7 74 36 9

.

Q02. Find , if the vectors a i 3j k, b 2i j k and c j 3k

are coplanar.

Sol. Since the given vectors are coplanar so, [a b c] 0

That is,

1 3 1

2 1 1 0

0 3

1( 3 ) 3(6 0) 1(2 0) 0

3 21 7 .

Q03. If a line makes angles o o90 , 60 and with x, y and z-axis respectively, where is acute angle,

then find .

Sol. Here o o90 , 60 and Since 2 2 2cos cos cos 1

So, 2 o 2 o 2cos 90 cos 60 cos 1 21

0 cos 14

23

cos4

3

cos2

3

cos2

[Since is an acute angle

cos cos6

6

.

Q04. Write the element 23a of a 3 3 matrix ijA (a ) whose elements ija are given by ij| i j |

a2

.

Sol. We have ij| i j |

a2

23

| 2 3 | 1a

2 2

.

Q05. Find the differential equation representing the family of curves A

v Br

, where A and B are

arbitrary constants.

Sol. Given A

v Br

2

dv A0

dr r 2

dvr A

dr

2

2

2

d v dvr (2r) 0

dr dr

2

2

d v dvr 2 0

dr dr is the required differential equation.

Q06. Find the integrating factor of the differential equation 2 xe y dx

1dyx x

.

Sol. We have 2 xe y dx

1dyx x

2 xdy y e

dx x x

dy

On comparing to P(x)y Q(x)dx

, we observe 2 x1 e

P(x) and Q(x)x x

1

dx2 xxIntegrating Factor e e

.

CBSE 2015 Annual Exam Paper (Delhi) Compiled By OP Gupta (+91-9650350480)


SECTION B

Q07. If

2 0 1

A 2 1 3

1 1 0

find 2A 5A 4I and hence find a vector X such that 2A 5A 4I X O .

OR If

1 2 3

A 0 1 4

2 2 1

, find 1(A ) .

Sol. We have

2 0 1

A 2 1 3

1 1 0

22 0 1 2 0 1 5 1 2

A A.A 2 1 3 2 1 3 9 2 5

1 1 0 1 1 0 0 1 2

25 1 2 2 0 1 1 0 0

So, A 5A 4 I 9 2 5 5 2 1 3 4 0 1 0

0 1 2 1 1 0 0 0 1

25 1 2 10 0 5 4 0 0

A 5A 4I 9 2 5 10 5 15 0 4 0

0 1 2 5 5 0 0 0 4

21 1 3

A 5A 4I 1 3 10

5 4 2

.

Also 21 1 3 0 0 0

A 5A 4I X O 1 3 10 X 0 0 0

5 4 2 0 0 0

1 1 3

X 1 3 10

5 4 2

.

OR

1 2 3

A 0 1 4

2 2 1

1 0 2

A 2 1 2

3 4 1

A 1( 9) 0 2( 5) 1

Consider ijA as the cofactor of corresponding element ija of matrix A .

11 12 13A 9, A 8, A 5 ,

21 22 23A 8, A 7, A 4 ,

31 32 33A 2, A 2, A 1

Note that we can alternatively use the property, 1 1(A ) (A ) .

Q08. If 2

a 1 0

f (x) ax a 1

ax ax a

, using properties of determinants find the value of f (2x) f (x) .

Sol. We have 2

a 1 0

f (x) ax a 1

ax ax a

By 1 1 2C C xC

a x 1 0

f (x) 0 a 1

0 ax a

Expanding along 1C

2f (x) (a x)(a ax) 0 0 2f (x) a(a x)

Now 2 2f (2x) f (x) a(a 2x) a(a x) 2 2 2 2f (2x) f (x) a[a 4x 4ax a x 2ax]

2f (2x) f (x) a[3x 2ax] f (2x) f (x) ax(3x 2a) .

9 8 2

adj(A ) 8 7 2

5 4 1

19 8 2

adj(A )So, (A ) 8 7 2

| A |5 4 1



Q09. Find : dx

sin x sin 2x. OR Integrate

2

2

x 3x 1

1 x

w.r.t. x.

Sol. Let dx

Isin x sin 2x

dx

Isin x 2sin x cos x

dx

Isin x(1 2cos x)

2

sin x dxI

sin x(1 2cos x)

sin x dxI

(1 cos x)(1 cos x)(1 2cos x)

Put cos x t sin xdx dt dt

I(1 t)(1 t)(1 2t)

Consider 1 A B C

(1 t)(1 t)(1 2t) 1 t 1 t 1 2t

21 A(1 t)(1 2t) B(1 t)(1 2t) C(1 t )

On comparing the coefficient of like terms on both the sides, we get : 1 1 4

A , B , C6 2 3

1 1 1 1 4 1

I dt6 1 t 2 1 t 3 1 2t

1 1 1 4 1

I log 1 t log 1 t log 1 2t C6 1 2 3 2

1 1 2

I log 1 cos x log 1 cos x log 1 2cos x C6 2 3

.

OR Let 2

2

x 3x 1I dx

1 x

2

2

1 x 3x 2I dx

1 x

2

2 2

1 x 3x 2I dx dx

1 x 1 x

2

2 2

3x 2I 1 x dx dx dx

1 x 1 x

2 1 12

x 1 3 2xI 1 x sin x dx 2sin x

2 2 2 1 x

2 1 2x 3 3

I 1 x sin x 2 1 x C2 2 2

i.e., 2 1 2x 3

I 1 x sin x 3 1 x C2 2

.

Q10. Evaluate : 2(cos ax sin bx) dx

.

Sol. Let 2I (cos ax sin bx) dx

2 2I (cos ax sin bx 2cosax sin bx)dx

2 2I cos ax dx sin bx dx 2cos ax sin bxdx

Consider f (x) 2cosax sin bx f ( x) [2cosax sin bx] f (x)

And, 2 2g(x) cos ax g( x) cos ax g(x) also 2 2h(x) sin bx h( x) sin bx h(x)

Clearly, f is an odd function whereas g and h are even functions.

Therefore by using

a

a

0

a

2 f (x)dx, if f is even functionf (x)dx

0, if f is odd function

We have 0 0

1 cos 2ax 1 cos 2bxI 2 dx 2 dx 0

2 2

0 0

1 1I x sin 2ax x sin 2bx

2a 2b

1 1 1 1

I sin 2a 0 sin 0 sin 2b 0 sin 02a 2a 2b 2b

1 1

I 2 sin 2a sin 2b2a 2b

.



Q11. A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black. OR An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained. Sol. Let E be the event that a red ball and another black ball is drawn. Let E1 be the event that 1 or 2 appears on the die. Let E2 be the event that 3, 4, 5 or 6 appears on the die.

Then 12

P(E )6

= , 24

P(E )6

= , 4 6

1 11 10

2

C CP(E|E )

C

= and

7 31 1

2 102

C CP(E|E )

C

= .

Required probability, 1 1 2 22 4 6 2 4 7 3 2 22

P(E) P(E )P(E|E ) P(E )P(E|E )6 10 9 6 10 9 45

= + = .

OR Let X denote the number of heads so, it can take values 0, 1, 2, 3, 4.

Let E : head appears on the coin. 1 1

P(E) , P(E)2 2

.

X P(X) 0 4

40

1 1C

2 16

1 34

1

1 1 4C

2 2 16

2 2 24

2

1 1 6C

2 2 16

3 2 24

3

1 1 4C

2 2 16

4 44

4

1 1C

2 16

Q12. If r xi yj zk

, find (r i).(r j) xy.

Sol. We have r xi yj zk

i j k r i x y z 0i zj yk

1 0 0

And,

i j k r j x y z zi 0 j xk

0 1 0

(r i).(r j) xy (zj yk).( zi xk) xy

(r i).(r j) xy 0 xy 0 xy 0

.

Q13. Find the distance between the point (1, 5, 10) and the point of intersection of the line

x 2 y 1 z 2

3 4 12

and the plane x y z 5 .

Sol. Let the line L : x 2 y 1 z 2

3 4 12

and the plane : x y z 5 . Let P(1, 5, 10).

coordinates of random point on L : M(3 2, 4 1, 12 2) .

If the line & plane intersect then M must lie on plane i.e., 3 2 4 1 12 2 5 0

M(2, 1, 2) . Therefore, 2 2 2PM (2 1) ( 1 5) (2 10) 13 units .

Q14. If 1 1sin[cot (x 1)] cos(tan x) , then find x.

Mean1 4 6 4 1

XP(X) 0 1 2 3 416 16 16 16 16

4 12 12 4

i.e., 0 216 16 16 16

2 2 2 2 2

2 2 2

1 4 6and, Variance X P(X) 0 1 2

16 16 16

4 13 4 2

16 16

2Variance ( ) 1



OR If 2

1 2 1 2 5(tan x) (cot x)8

, then find x.

Sol. Weve 1 1sin[cot (x 1)] cos(tan x) 1 12 2

1 1sin sin coscos

1 (x 1) 1 x

2 2

1 1

1 (x 1) 1 x

2 22 x 2x 1 x

1x

2 .

OR We have 2

1 2 1 2 5(tan x) (cot x)8

2

21 1 1 1 5tan x cot x 2 tan x cot x8

2 2

1 1 52 tan x tan x2 2 8

2 2 2

1 1 5 32 tan x tan x2 8 4 8

2

1 1 2 3tan x (tan x)2 16

2

1 2 1 3(tan x) tan x 02 16

1 2 1 216(tan x) 8 tan x 3 0 1 2 1 1 216(tan x) 12 tan x 4 tan x 3 0

1 1 14 tan x 4 tan x 3 4 tan x 3 0 1 14 tan x 3 4 tan x 0 1 14 tan x 3 0 or 4 tan x 0

3

x tan , tan i.e., x 1, 14 4

So, value of x is 1 .

Q15. If 2 2

1 2

2 2

1 x 1 xy tan , x 1

1 x 1 x

, then find dy

dx.

Sol. Given 2 2

1 2

2 2

1 x 1 xy tan , x 1

1 x 1 x

Put 2 1 21

x cos 2 cos x ...(i)2

11 cos2 1 cos 2

y tan1 cos 2 1 cos2

1

2 cos 2 siny tan

2 cos 2 sin

11 tan

y tan1 tan

1tan( /4) tan

y tan1 tan( /4) tan

1y tan tan4 4

1 21

y cos x4 2

[By (i)

4 4

dy 1 1 x0 2x

dx 2 1 x 1 x

.

Q16. If x a cos bsin and, y a sin bcos , show that 2

2

2

d y dyy x y 0

dx dx .

Sol. Given x a cos bsin and, y a sin bcos

dx

a sin bcosd

and, dy

a cos bsin d

dy dy d a cos bsin

dx d dx b cos a sin

x

y (i)

dy

y xdx

2

2

d y dy dyy 1

dx dx dx By (i), we get :

2

2

d y x dyy 1 0

dx y dx

2

2

2

d y dyy x y 0

dx dx .

Q17. The side of an equilateral triangle is increasing at the rate of 2cm/s. At what rate is its area increasing when the side of the triangle is 20cm?



Sol. Let a denote the side length of the triangle so, 1da

2cmsdt

.

Since area of the equilateral triangle is 23 a

A4

dA 3 a da

dt 2 dt

2 1

at a 20cm

dA 3 202 20 3 cm s

dt 2

.

Q18. Find : 2(x 3) 3 4x x dx .

Sol. Let 2I (x 3) 3 4x x dx Consider 2dx 3 A [3 4x x ] B

dx

x 3 A[ 4 2x] B On equating the coefficients of like terms, we get : 1

A , B 12

21

I [ 4 2x] 1 3 4x x dx2

2 21

I [ 4 2x] 3 4x x dx 3 4x x dx2

2Put 3 4x x t ( 4 2x)dx dt in 1st integral.

2 3/2

2 21 [3 4x x ]I ( 7) (x 2) dx2 3/2

2 3/2

2 1[3 4x x ] x 2 7 x 2Therefore, I 3 4x x sin C3 2 2 7

.

Q19. Three schools A, B and C organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand made fans, mats and plates from recycled material at a cost of `25, `100 and `50 each. The number of articles sold are given below :

School Article

A B C

Hand fans 40 25 35 Mats 50 40 50 Plates 20 30 40

Find the funds collected by each school separately by selling the above articles. Also find the total funds collected for the purpose. Write one value generated by the above situation. Sol. Let the funds collected by schools A, B and C be x, y and z (in `) respectively.

So,

Fans Mats Plates Cost

Funds by school A x 40 50 20 25

Funds by school B y 25 40 30 100

Funds by school C z 35 50 40 50

x 8 10 4 1 8 40 8

y 5 25 5 8 6 4 125 5 32 12

z 7 10 8 2 7 40 16

x 56 7000

y 125 49 6125

z 63 7875

By equality of matrices, we get : x 7000, y 6125, z 7875 .

So the funds collected by school A is `7000, by school B is `6125 and by school C is `7875. And, total funds collected = `(7000 + 6125 + 7875) = `21000.

SECTION C Q20. Let N denote the set of all natural numbers and R be the relation on NN defined by (a,b)R(c,d)



if ad(b c) bc(a d) . Show that R is an equivalence relation.

Sol. We have (a,b)R(c,d) given as ad(b c) bc(a d)

Reflexivity : Let (a, b) N N ab(b a) ba(a b) (a,b)R(a,b)

So, R is reflexive. Symmetry : Let (a, b), (c,d) N N Let (a, b)R(c,d) ad(b c) bc(a d)

cb(d a) da(c b) (c,d)R(a,b) . So, R is symmetric.

Transitivity : Let (a, b), (c,d), (e, f ) N N

Since 1 1 1 1

(a, b)R(c,d) ad(b c) bc(a d)b c a d

[Dividing by abcd both sides

Now 1 1 1 1

(a, b)R(c,d)a d b c

and 1 1 1 1

(c,d)R(e, f )c f d e

Adding these two equations, 1 1 1 1 1 1 1 1

a d c f b c d e

1 1 1 1

a f b e (a, b)R(e, f )

Hence R is transitive. Since R is reflexive, symmetric and transitive so, it is an equivalence relation. Q21. Using integrals, find the area of the triangle formed by positive x-axis, and the tangent and the

normal to the circle 2 2x y 4 at (1, 3 ) .

OR Evaluate 3

2 3x 2

1

(e x 1)dx as a limit of a sum.

Sol. Given circle 2 2x y 4 (i) Let P (1, 3 )

dy

2x 2y 0dx

dy x

dx y T

at P

dy 1m

dx 3

Equation of tangent : 1

y 3 (x 1)3

4 x

y3

(i)

And, equation of normal : y 3 3(x 1) y 3x (ii)

Also, x axis means y 0 (iii)

Solving these equations simultaneously we get : A( 3,0), O(0,0), P(1, 3)

Now ar(AOP) 1 4

(ii ) (i)

0 1

y dx y dx

1 4

0 1

13 x dx (4 x)dx

3

1 42 2

0 1

3 1x (4 x)

2 2 3

42 2 2 2

1

3 11 0 0 3

2 2 3

3 3 3

2 3 sq.units2 2

OR 3

2 3x 2

1

Let I (e x 1)dx

We know b

na

f x dx lim h f a f a h f a 2h ... f a (n 1)h

,

As n , h 0 nh b a 3 1 2 b n 1

nr 0a

f x dx lim h f a rh

(i)

Here 2 3x 2f (x) e x 1, a 1, b 3 .

2 3a 3rh 2f (a rh) e (a rh) 1



1 3rh 2f (1 rh) e (1 rh) 1 1 3rh 2 2f (1 rh) e e 2 r h 2rh .

3 n 1

2 3x 2 1 3rh 2 2

nr 01

By (i), (e x 1)dx lim h e e 2 r h 2rh

n 1 n 1 n 1 n 1

1 3rh 2 2

nr 0 r 0 r 0 r 0

I lim h e e 2 h r 2h r

1 3h 6h 3(n 1)h 2n

n(n 1)(2n 1) n(n 1)I lim h e 1 e e ....e 2n h 2h

6 2

1 3nh3hn

h nh(nh h)(2nh h)I lim e [e 1] 2nh nh(nh h)

e 1 6

3nh

1

3hn

e 1 3h nh(nh h)(2nh h)I lim e 2nh nh(nh h)

3 e 1 6

1 7 1 7e e 8 e e 32

I 83 3 3

.

Q22. Solve the differential equation : 1 2(tan y x)dy (1 y )dx .

OR Find the particular solution of the differential equation 2 2

dy xy

dx x y

given that

y 1 when x 0 .

Sol. We have 1 2(tan y x)dy (1 y )dx 1

2 2

dx x tan y

dy 1 y 1 y

Its linear differential equation of the formdx

P(y)x Q(y)dy

where 1

2 2

1 tan yP(y) ,Q(y)

1 y 1 y

So, integrating factor12

dy

tan y1 ye e

the solution is : 1 1

1tan y tan y

2

tan yxe e dy C

1 y

Put 12

dytan y t dt

1 y

1tan y txe te dt C

1tan y t tdxe t e dt [t] e dt dt C

dt

1tan y t txe te e C

1 1tan y tan y 1xe e (tan y 1) C or,

11 tan yx tan y Ce 1

OR Weve 2 2

dy xy

dx x y

Put

dy dvy vx v x

dx dx

2

2 2 2

dv x vv x

dx x v x

2

dv vx v

dx 1 v

3

2

dv vx

dx 1 v

2

3

1 v dxdv 0

v x

3

1 1dv log | x | C

v v

2

1log | v | log | x | C

2v

2

2

y xlog log | x | C

x 2y

2

2

xlog y log x log x C

2y

2

2

xlog y C

2y

Given that y 1 when x 0 , 2

2

0log 1 C C 0

2 1

Hence the required particular solution is : 2 2x 2y log y .

Q23. If lines x 1 y 1 z 1

2 3 4

and

x 3 y k z

1 2 1

intersect, find the value of k and hence find the

equation of the plane containing these lines.



Sol. Let 1x 1 y 1 z 1

L :2 3 4

and, 2

x 3 y k zL :

1 2 1

The coordinates of random points on these lines are : P(2 1,3 1, 4 1) and Q( 3, 2 k, )

Since 1 2L and L intersect each other so, the points P and Q must coincide for some real values of

and . That is 2 1 3...(i),3 1 2 k...(ii), 4 1 ...(iii)

Solving (i) and (iii), we get : 3

, 52

Replacing these values in (ii), we get : 3

3 1 2( 5) k2

9

k2

The d.r.s of line 1 1 L : 2, 3, 4 b 2i 3j 4k

and 2 1

L :1, 2, 1 b i 2j k

.

Therefore the d.r.s of normal to the required plane is 1 2

i j k b b 2 3 4 5i 2j k m

1 2 1

Also a point on plane is A(1, 1, 1) from line 1L so, OA a i j k

So, required equation of plane r.m a .m

r.( 5i 2j k) (i j k).( 5i 2j k)

r.( 5i 2j k) 5 2 1 6

r.(5i 2j k) 6 or, 5x 2y z 6

.

Note that we can take 9 a 3i j2

also from line 2L to get the same plane.

Q24. If A and B are two independent events such that 2

P(A B)15

and 1

P(A B)6

, then find

P(A) and P(B) .

Sol. Since A and B are two independent events so A and B, A and B are independent events.

2

P(A B) P(A)P(B)15

and, 1

P(A B) P(A)P(B)6

.

2

[1 P(A)]P(B)15

and, 1

P(A)[1 P(B)]6

Let P(A) x, P(B) y

2

[1 x]y15

and, 1

x[1 y]6

2 1

y xy ...(i), x xy ...(ii)15 6

Subtracting (ii) from (i), 2 1 1

y x15 6 30

1 1

y6 6y 30

1

x6 6y

1 1

y6 6y 30

26y 6 y 1

6y 6y 130 5

230y 29y 4 0

29 841 480

y60

29 19y

60

1 4y ,

6 5

1 4

y ,6 5

So, 1 1 5

x ,6 6y 5 6

Hence, 1 5 1 4

P(A) , and P(B) ,5 6 6 5

.

Q25. Find the local maxima and local minima, of the function f (x) sin x cos x, 0 x 2 . Also find

the local maximum and local minimum values. Sol. We have f (x) sin x cos x, 0 x 2 f (x) cos x sin x , f (x) sin x cos x

For local points of maxima and minima f (x) cos x sin x 0

tan x 1 3 7

x , 0, 24 4



3 3 3

f cos sin 2 04 4 4

and, 7 7 7

f cos sin 2 04 4 4

3

f (x) is maximum at x4

and,

7minimum at x

4

So, 3 3 3 1 1

Local maximum value f sin cos 24 4 4 2 2

And, 7 7 7 1 1

Local minimum value f sin cos 24 4 4 2 2

Q26. Find graphically, the maximum value of z = 2x + 5y, subject to constraints given below : 2x 4y 8, 3x y 6, x y 4, x 0, y 0 .

Sol. To maximize : z = 2x + 5y Subject to constraints : 2x 4y 8, 3x y 6, x y 4, x 0, y 0

Critical Points Value of z

O(0, 0) 0

A(2, 0) 4

B(0, 2) 10Max.Value

C(8/5, 6/5) 46/5

So, maximum value of z is 10.

cbse 2015 delhi

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