-(1)-

Series : SMA/1

Roll No.

Code No. 65/1/1Candidates must write the Code onthe title page of the answer-book.

Please check that this question paper contains 8 printed pages.

Code number given on the right hand side of the question paper should be written on the titlepage of the answer-book by the candidate.

Please check that this question paper contains 29 questions.

Please write down the Serial Number of the questions before attempting it.

15 minutes time has been allotted to read this question paper. The question paper will be distributedat 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only andwill not write any answer on the answer script during this period.

MATHEMATICS

[Time allowed : 3 hours] [Maximum marks : 100]

General Instructuions:

(i) All questions are compulsory.

(ii) Questions numbered 1 to 10 are very short-answer questions and carry 1 mark each.

(iii) Questions numbered 11 to 22 are short-answer questions and carry 4 marks each.

(iv) Questions numbered 23 to 29 are also short-answer questions and carry 6 marks each.

(v) Use of calculators is not allowed.

Studymate Solutions to CBSE Board Examination 2011-2012

-(2)-

STUDYmate

SECTION-A

Question numbers 1 to 10 carry 1 mark each.1. If a line has direction ratios 2, –1, –2, then what are its direction cosines?

Ans. d.r.’s = <2, –1, –2>

d.c.’s = 2 1 2

, ,3 3 3

2. Find ‘’ when the projection of 4

a i j k on 2 6 3

b i j k is 4 units.

Ans. Projection of on a b =

.

a b

b

4 = 2 6 12

4 36 9

4 = 2 18

7

28 – 18 = 2

5 =

3. Find the sum of the vectors 2 , 2 4 5 a i j k b i j k and 6 7

c i j k .

Ans. 0 4 a b c i j k

4. Evaluate:

3

2

1 dx

x

Ans. 3

3

22

1log | | dx x

x = log 3 – log 2 = 3

log2

5. Evaluate (1 ) x xdx

Ans. 32( ) x x dx

=3 5

2 2

3 52 2

C x x

=3 5

2 22 2

C3 5

x x

6. If =

5 3 8

2 0 1

1 2 3, write the minor of the element a

23.

Ans. Minor of a23

= 5 3

1 2 = 10 – 3 = 7.

-(3)-

STUDYmate

7. If 2 3 1 3 4 6

5 7 2 4 9

x

, write the value of x.

Ans.2 3 1 3 4 6

5 7 2 4 9

x

2 6 6 12 4 6

5 14 15 28 9

x

4 6 4 6

9 13 9

x

x = 13

8. Simplify: cos sin sin cos

cos sinsin cos cos sin

Ans. = 2 2

2 2

cos cos sin sin cos sin

cos sin cos sin cos sin

= 1 0

0 1

9. Write the principal value of 1 11 1cos 2sin

2 2

Ans. cos–112

– 2 sin–112

cos–1 cos3

+ 2 sin–1 sin6

23 6

23 3 3

10. Let * be a ‘binary’ operation on N given by a * b = LCM (a, b) for all a, b N. Find 5 * 7.

Ans. a * b = LCM of (a, b), a, b N

5 * 7 = LCM of (5, 7) = 35

SECTION-B

Question numbers 11 to 22 carry 4 marks each.

11. If (cos x)y = (cos y)x, find dydx

.

OR

If sin y = x sin (a + y), prove that 2sin ( )

sin

dy a y

dx a

Ans. Given (cos x)y = (cos y)x, taking logarithms on the two sides, we have

log (cos x)y = log (cos y)x

-(4)-

STUDYmate

or y log (cos x) = x log (cos y)

Differentiating both sides w.r.t. x, we get

1 1( sin ) log(cos ) ( sin ) log(cos ).1

cos cos

dy dyy x x x y y

x dx y dx

log (cos x) dydx

+ x tan ydydx

= log (cos y) + y tan x

log(cos ) tandy

x x ydx

= log (cos y) + y tan x

log(cos ) tanlog(cos ) tan

dy y y xdx x x y

OR

Ans. sin y = x sin (a + y)

sin

sin( )y

a y = x

2

sin( )cos sin cos( )sin ( )

a y y y a y dx

a y dy

sin( a y y

2

)

sin ( )

dx

a y dy

2

sinsin ( )

a dx

a y dy

2sin ( )sin

a y dya dx

12. How many times must a man toss a fair coin, so that the probability of having at least one head is more

than 80%?

Ans. Let the man throws the coin ‘n’ times.

Probability of getting a head in a single throw, 1

2p

Probability of not getting a head in a single throw, 1

2q

Given: Probability that the man gets atleast one head is more than 80%.

i.e. P(atleast 1 Head) > 80%

i.e. 1 – P (no head) 80

100

i.e.0 0

0

1 1 81

2 2 10

nnC

i.e.1 4

12 5

n

i.e.1 1

2 5

n

i.e.1 1

2 5

n

i.e. 2n > 5

-(5)-

STUDYmate

i.e. n = 3 or more.

The man must throw the coin atleast 3 times.

13. Find the Vector and Cartesian equations of the line passing through the point (1, 2, –4) and perpendicular

to the two lines 8 19 10

3 16 7

x y z

and 15 29 5

3 8 5

x y z

.

Ans. Any line through (1, 2, –4) can be written as

1 2 4

x y za b c

...(i)

(i) is at right angles to given lines with d.n. < 3, –16, 7 > and < 3, 8, –5 >

if 3a – 16b + 7c = 0 ...(ii)

and 3a + 8b – 5c = 0 ...(iii)

By cross-multiplication, we have

80 56 21 15 24 48

a b c

or24 36 72

a b c

or2 3 6

a b c...(iv)

From (i) and (iv), we find that required line is

1 2 4

2 3 6

x y z

In vector form, this line can be written as

2 4 2 3 6

r i j k i j k ( d.n. of the line are < 2, 3, 6 > and line passes through < 1, 2, –4 >)

14. If , , a b c are three vectors such that 5, 12, 13

a b c , and 0

a b c , find the value of

. . . a b b c c a .

Ans. 0 a b c

( ) · ( ) 0·0 a b c a b c

i.e.2

0 a b c

i.e. 2 2 2| | | | | | 2( · · · ) 0 a b c a b b c c a

i.e. 52 + 122 + 132 + 2( · · · ) 0 a b b c c a

i.e. 338 2( · · · ) 0 a b b c c a

i.e. · · · 169 a b b c c a

15. Solve the following differential equation:

2 22 2 0 dy

x xy ydx

Ans. 2 22 2 dy

x xy ydx

-(6)-

STUDYmate

2

2

22

dy xy ydx x

2

2 2

22 2

dy xy ydx x x

2

2

12

dy y ydx x x

...(i)

Put y = vx and dy dv

v xdx dx

in (i) we get,

v dv

x vdx

212

v

2 22

dv dxv x

2 1

2 log | | C2 1

vx

2

log | | C

xv

2

log | | C

x

xy

2log | | C

xx

y; (y 0)

16. Find the particular solution of the following differential equation;

2 2 2 21 , dy

x y x ydx

given that y = 1 when x = 0

Ans.2 2 2 21

dyx y x y

dx

2 2 2 2(1 ) (1 ) (1 )(1 ) dy

x y x x ydx

22

(1 )1

dy

x dxy

31tan C

3

xy x

Now, when x = 0, y = 1

3

1 0tan 1 0 C

3 C

4

Solution is 3

1tan3 4

xy x

17. Evaluate : sin sin 2 sin 3 x x x dx

OR

Evaluate : 2

2

(1 )(1 ) dxx x

Ans. Now, sin x sin 2x sin 3x

-(7)-

STUDYmate

=12

{2sin 3x sin x} sin 2x

=1

2{cos (2x) – cos (4x)} sin 2x

=1

4{2 sin 2x cos 2x – 2 cos 4x sin 2x}

=14

{sin 4 x – (sin 6x – sin 2x)}

sin sin 2 sin3 x x xdx

=1

(sin 4 sin 6 sin 2 )4

x x x dx

=1 cos4 cos6 cos2

C4 4 6 2

x x x

=cos4 cos6 cos2

C16 24 8

x x x

OR

Ans. Let 2 2

2 A B C(1 )(1 ) 1 1

x

x x x x ...(i)

Multiplying both sides by (1 – x) (1 + x2), we have

2 = A (1 + x2) + (Bx + C) (1 – x)

2 = x2 (A – B) + x (B – C) + (A + C) ...(ii)

Equating coefficients on the two sides of (ii), we get

A – B = 0, B – C = 0, A + C = 2

A = B = C = 1

2 2

2 1 1(1 )(1 ) 1 1

x

x x x x

2

2(1 )(1 ) dx

x x = 2

1 1

1 1

x

dx dxx x

= 2 2

log |1 | 1 2 11 2 1 1

x xdx dx

x x

= –log |x – 1| + 1

2log (1 + x2) + tan–1 x + C.

18. Find the point on the curve y = x3 – 11x + 5 at which the equation of tangent is y = x – 11.

OR

Using differentials, find the approximate value of 49.5

Ans. Given curve is y = x3 – 11x + 5 ...(i)

Given line is y = x – 11 ...(ii)

Slope of line (ii) = 1 ( (ii) is of the form y = mx + b)

From (i),dy

dx = 3x2 – 111

Since, slope of tangent = 1

-(8)-

STUDYmate

dydx

= 1, i.e., when 3x2 – 11 = 1

x = 2 ( 3x2 – 11 = 1 3x2 = 12 x2 = 4)

When x = 2, then

from (i), y = 23 – 11 × 2 + 5 = –9

When x = –2, then

from (i), y = (–2)3 – 11 (–2) + 5 = 19

Thus, we find that at the points (2, –9) and (–2, 19), the slope of tangent is 1.

OR

Ans. Let ( ) f x x so that 1

( ) .2

f xx

Now ( ) ( ) ( ) f x x f x xf x

2

xx x x

x

Taking x = 49 and x = 0.5, we obtain

49.50.5 0.5

49 7 7 0.036142 49

49.5 7.036

19. If y = (tan–1x)2, show that 2 2 22 1( 1) 2 ( 1) 2 x y x x y .

Ans. Given y = (tan–1 x)2 we get

12

12(tan )

1

dy

xdx x

or (1 + x2) y1 = 2 tan–1 x

Again differentiating w.r.t. x we get

22 1 2

2(1 ) (0 2 )

1

x y y x

x

2 2 22 1( 1) 2 ( 1) 2 x y x x y

20. Using properties of determinants, prove that

2

b c q r y z a p x

c a r p z x b q y

a b p q x y c r z

Ans. LHS =

b c q r y z

c a r p z x

a b p q x y

Interchanging rows and columns, we get,

b c c a a b

q r r p p q

y z z x x y

Operating C3 C

3 – C

1 – C

2, we have

-(9)-

STUDYmate

2

2

2

b c c a a b b c c a c

q r r p p q q r r p r

y z z x x y y z z x z(Take out –2 from C

3 i.e., operate C

3

12

C3)

= 2

b c c a c

q r r p r

y z z x zNow, Operate C

1 C

1 – C

3 and C

2 C

2 – C

3

= 2b a c

q p r

y x z

, operate C1 C

2

= 2

a b c

p q r

x y z

= 2

a p x

b q y

c r z

(by interchanging rows and columns again)

21. Prove that 1 costan , ,

1 sin 4 2 2 2

x xx

x

OR

Prove that 1 1 18 3 36sin sin cos

17 5 85

Ans.2

sin 2sin coscos 2 4 2 4 2

1 sin1 cos 2cos

2 4 2

x xx

xxx

x

sin4 2

tan4 2

cos4 2

xx

x

1 cos

tan .1 sin 4 2

x x

x Hence proved

OR

Ans. Let x = 1 8

sin17

and y =

1 3sin

5

sin x = 8

17 and sin y =

35

cos x = 21 sin x and cos y = 21 sin y

cos x = 15

17 and cos y =

4

5Now, cos (x + y) = cos x cos y – sin x sin y

=15 4 8 317 5 17 5

-(10)-

STUDYmate

=3685

x + y = cos–1 36

85

1 1 18 3 36sin sin cos

17 5 85 Hence Proved

22. Let A {3} and B {1} . Consider the function f : A A B defined by 2

( ) .3

xf x

x Show that

f is one-one and onto and hence find f –1.

Ans.2

( )3

x

f xx

Let 1 2, x x A

Let 1 2( ) ( )f x f x

i.e.1 2

1 2

2 2

3 3

x x

x x

i.e. (x1 – 2) (x

2 – 3) = (x

1 – 3) (x

2 – 2)

i.e. x1x

2 – 3x

1 – 2x

2 + 6 = x

1x

2 – 2x

1 – 3x

2 + 6

i.e. –x1 = – x

2

i.e. x1 = x

2

f is one - one. ... (i)

Also2

3

x

yx

i.e. y (x – 3) = (x – 2)

xy – 3y = x – 2

x (y – 1) = 3y – 2

3 2

1

y

xy

f is onto, y 1, y B ... (ii)

from (i) and (ii), f is bijective

it is invertible.

For inverse : Interchanging x and y in y = 2

3

x

x, we get

y = 3 2

1

x

x or f –1(x) =

3 2

1

x

x

SECTION-C

Question numbers 23 to 29 carry 6 marks each.23. Find the equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6) and

hence find the distance between the plane and the point P (6, 5, 9).

Ans. Equation of plane in three point form is

-(11)-

STUDYmate

3 1 2

5 3 2 1 4 2 0

1 3 1 1 6 2

x y z

i.e.,

3 1 2

2 3 2 0

4 0 4

x y z

i.e. (x – 3) (12) – (y + 1) (8 + 8) + (z – 2) (12) = 0

i.e., 3(x – 3) – 4 (y + 1) + 3 (z – 2) = 0

i.e., 3x – 4y + 3z – 19 = 0 is the equation of the required plane.

and d = 18 20 27 19

9 16 9

d = 6 34

34 34

d = 3 34

17.

24. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (no residing

in hostel). Previous year results report that 30% of all students who reside in hostel attain ‘A’ grade and

20% of day scholars attain ‘A’ grade in their annual examination. At the end of the year, one student is

chosen at random from the college and he has an ‘A’ grade, what is the probability that the student is a

hostlier?

Ans. Let E1 : ‘A student is residing in hostel’

and E2 : ‘A student is a day scholar’

then E1 and E

2 are mutually exclusive and exhaustive. Moreover,

P(E1) =

60 3100 5

and P(E2) =

40 2100 5

Let E : Student attains ‘A’ grade;

then P(E/E1) =

30 3

100 10 and P(E/E

2) =

20 2

100 10

Required probability = P (E1/E)

=1 1

1 1 2 2

P(E/E )P(E )P(E/E )P(E )+P(E/E )P(E ) [By Baye’s Theorem]

=

3 39 910 5

3 3 2 2 9 4 1310 5 10 5

25. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine

B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a

package of bolts. He earns a profit of ` 17.50 per package on nuts and ` 7 per package of bolts. How

many packages of each should be produced each day so as to maximize his profits if he operates his

machines for at the most 12 hours a day? Form the above as a linear programming problem and solve it

graphically.

-(12)-

STUDYmate

Ans. Let the manufacturer produces x Packages of nuts and y Packages of bolts each day. We construct the

following table:

Item Number of packages Total Time on machine A Total Time on machine B Profits

Nuts 1 hrs 3 hrs 17.50

Bolts 3 hrs 1 hr 7

Total 3 hrs 3 hrs 17.50 7

x x x x

y y y y

x y x y x y

We have to maximize profit, i.e.,

P = 17.50x + 7y = 35

72

x y ...(i)

subject to the constraints

x + 3y 12 ...(ii)

Y

OX

D(0,12)

B(0,4) E(3,3)

C(4,0)

A(12,0)

3x + y 12 ...(iii)

x, y 0 ...(iv)

First of all, we locate the region represented by (2),

(3) and (4). For this, we consider the line x + 3y =

12, which passes through A (12, 0) and B (0, 4) and

the line 3x + y = 12, which passes through the points

C (4, 0) and D (0, 12). The two lines meet at E (3, 3).

The feasible region is shown shaded in the figure.

Note that O (0, 0) lies in this region.

The corner points, which are to be examined for optimum solution are C (4, 0), E (3, 3) and B (0, 4).

At C (4, 0) P = 35

4 7 0 702

At E (3, 3) P = 35

3 7 3 73.52

At B (0, 4) P = 35

0 7 4 282

Hence the profit is maximum equal to ̀ 73.50 when 3 packages of each of nuts and bolts are manufactured.

26. Prove that

/4

0

( tan cot ) 2.2

x x dx

OR

Evaluate : 3

2

1

(2 5 ) x x dx as a limit of a sum.

Ans. Let I = sin cos

cos sin

x xdx

x x

=sin cos

sin cos

x xdx

x x

Now let (sin x – cos x) = t ...(i)

(cos x + sin x)dx = dt

Squaring (i), sin2 x + cos2 x – 2 sin x cos x = t2

21

sin cos2

tx x

-(13)-

STUDYmate

When x = 0, t = 0 – 1 = –1

When x = 4

, t = 1 1

02 2

I =

0

21

21 dt

t

= 01

12 sin

t

= 1 12 sin (0) sin 1

= 2 02

= 22

OR

Ans. I =

32

1

(2 5 ) x x dx

a = 1, b = 3 h = 3 1

b a

n n nh = 2

Now, f(x) = 2x2 + 5x

f(a) = f(1) = 2(1)2 + 5(1) = 7

f(a + h) = f(1 + h) = 2(1 + h)2 + 5(1 + h)

= 2(1 + h2 + 2h) + 5 + 5h

= 2h2 + 9h + 7

f(a + 2h) = f(1 + 2h) = 2(1 + 2h)2 + 5(1 + 2h)

= 2(1 + 4h2 + 4h) + 5 + 10h

= 8h2 + 18h + 7

... ... ... ...

... ... ... ...

1 1 1 f a n h f n h = 22 1 1 5 1 1 n h n h

= 2 22 1 1 2 ( 1) 5 5 1 n h h n h n

= 2(n – 1)2 h2 + 9h (n – 1) + 7

Since ( )b

a

f x dx = 0

( ) ( ) ( 2 ) ... 1

hLt h f a f a h f a h f a n h

I = 2 2 2 2

07 {7 9 2 } {7 18 8 } ... {7 9 ( 1) 2( 1) }

h

Lt h h h h h h n n h

= 22 2 2

07 9 (1 2 ... 1) 2 1 2 ... 1

hLt h n h n h n

= 2

0

( 1) ( )( 1)(2 1)7 9 2

2 6

h

n n n n nLt h n h h

= 0

9 ( ) ( )(2 1)7

2 3

h

hn hn h hn hn h hnLt h hn

-(14)-

STUDYmate

= 9 2(2 0) 2(2 0)(4 0)

7 22 3

= 16

14 183

= 42 54 16

3

= 112

3

27. Using the method of integration, find the area of the region bounded by the lines 3x – 2y + 1 = 0, 2x + 3y

– 21 = 0 and x – 5y + 9 = 0.

Ans. L1: 3x – 2y + 1 = 0 y =

3 1

2

x1

30

0.5 0

x

y

L2: 2x + 3y – 21 = 0 y =

21 23 x 0 10.5

7 0

x

y

L3: x – 5y + 9 = 0 y =

9

5

x 0 9

1.8 0

x

y

L1

L2

L3

(–9, 0) (10.5, 0)

(3, 5)

AB

C

(1,2)

(6,3)

Point of Intersection:

L1 and L

2 :

3 1 21 22 3

x x

9x + 3 = 42 – 4x

13x = 39

x = 3

y = 5

L2 and L

3 :

3 1 9

2 5

x x

15x + 5 = 2x + 18

13x = 19

x = 1

y = 2

L2 and L

3 :

21 2 9

3 5

x x

-(15)-

STUDYmate

105 – 10x = 3x + 27

78 = 13x

x = 6

y = 3

A = (1, 2) ; B = (6, 3) ; C = (3, 5)

Thus, required area: A =

3 6 6

( 1) ( 2) ( 3)1 3 1

Line Line Lineydx ydx ydx

=

3 6 6

1 3 1

1 1 1(3 1) (21 2 ) ( 9)

2 3 5 x dx x dx x dx

=

3 62 2623

1 1

1 3 1 121 9

2 2 3 5 2

x xx x x x

= 1 27 3 1 1 13 1 126 36 63 9 18 54 9

2 2 2 3 5 2

= 1 1 114 36

2 3 5

12525

2

=25 25 38 25 13

7 12 192 2 2 2

sq. units.

28. Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal

to the diameter of its base.

Ans. Let r be the radius of the circular base, h the height and S, the total surface area of a right circular

cylinder, then S = 2r2 + 2rh is given to be a constant.

Let V be the volume of the cylinder with the above dimensions, then

V = r2h = 2

2 2S 2(S 2 )

2 2

r rr r

r[ S = 2r2 + 2rh, h =

2S 2

2

r

r]

V = 3S S,0

2 2

r

r r ...(i) [ h = 2S 2

2

r

r > 0, r <

S2

]

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

2V S3

2

dr

dr

and2

2

V6

dr

dr

Now Vd

dr exists at all points in

S0,

2

and

V0

d

dr 2S

3 02 r r2 =

S

6

r = S6

[ 0 < r < S2

]

Also, 2

2S/(6 )

V S6 0

6

r

d

dr

V has a local maximum value at r = S6

-(16)-

STUDYmate

Since V is continuous in S0,

2

and has only one extremum at S S

0,6 2

, therefore, V is

absolutely maximum for r = S6

.

When r = S6

, then h = 2

SS 2

S 2 2S 662 6 SS

26

rr

= i.e., h = S2

6 = 2 radius = diameter..

So, volume is maximum when the height is equal to the diameter.

29. Using matrices, solve the following system of linear equations:

x – y + 2z = 7

3x + 4y – 5z = – 5

2x – y + 3z = 12

OR

Using elementary operations, find the inverse of the following matrix.

1 1 2

1 2 3

3 1 1

Ans. The given system can be written as AX = B,

where

1 1 2

3 4 5

2 1 3

A

x

X y

z

and

7

5

12

B

Here

1 1 2

| | 3 4 5

2 1 3

A

= 1 (12 – 5) – (– 1) (9 + 10) + 2(–3 – 8)

= 7 + 19 – 22 = 4 0

A–1 exists

Therefore, the given system is consistent and has a unique solution given by

X = 1 1A B = ( . A) B

| |

adjA

-(17)-

STUDYmate

=

7 19 11 71

1 1 1 54

3 11 7 12

t

7 1 3 71

19 1 11 54

11 1 7 12

49 5 361

133 5 1324

77 5 84

8 21

4 14

12 3

x

y

z

x = 2 , y = 1, z = 3

OR

Ans. A = IA

1 1 2 1 0 0

1 2 3 0 1 0 A

3 1 1 0 0 1

R

1 R2

1 2 3 0 1 0

1 1 2 1 0 0 A

3 1 1 0 0 1

R

2 R

2 + R

1 ; R

3 R

3 – 3R

1

1 2 3 0 1 0

0 3 5 1 1 0 A

0 5 8 0 3 1

R

3 R

3 + 2R

2

1 2 3 0 1 0

0 3 5 1 1 0 A

0 1 2 2 1 1

R

2 R3

1 2 3 0 1 0

0 1 2 2 1 1 A

0 3 5 1 1 0

R

1 R

1 – 2R

2

R3 R

3 – 3R

2

1 0 1 4 3 2

0 1 2 2 1 1 A

0 1 1 5 4 3

R

3 – R

3

-(18)-

STUDYmate

1 0 1 4 3 2

0 1 2 2 1 1 A

0 0 1 5 4 3

R

1 R

1 + R

3

R2 R

2 – 2R

3

1 0 0 1 1 1

0 1 0 8 7 5 A

0 0 1 5 4 3

1

1 1 1

A 8 7 5

5 4 3

× · × · × · × · ×

-(19)-

STUDYmate

SECTION-A

Question numbers 1 to 10 carry 1 mark each.9. Find the sum of the following vectors:

ˆˆ ˆ ˆ ˆ ˆ2 , 2 3 , 2 3 a i j b i j c i k

Ans. a b c ˆˆ ˆ ˆ ˆ ˆ( 2 ) (2 3 ) (2 3 ) i j i j i k

ˆˆ ˆ5 5 3 i j k

10. If

5 3 8

2 0 1 ,

1 2 3

write the cofactor of the element a32

.

Ans. 3 232

5 8M ( 1) (5 16) 11

2 1

SECTION-B

Question numbers 11 to 22 carry 4 marks each.19. Using properties of determinants, prove the following

3 3 3

1 1 1

( )( )( )( ) a b c a b b c c a a b c

a b c

Ans.3 3 3

1 1 1

a b c

a b c

Applying C1 C

1 – C

3(Given)

C2 C

2 – C

3

3 3 3 3 3

0 0 1

a c b c c

a c b c c

= 2 2 2 2 3

0 0 1

( )( ) ( ) ( )

a c b c c

a c a ac c b c b bc c c

Code No. 65/1/2

Studymate Solutions to CBSE Board Examination 2011-2012

UNCOMMON QUESTIONS ONLY

Series : SMA/1

-(20)-

STUDYmate

=

2 2 2 2 3

0 0 1

( ) ( ) 1 1

( ) ( )

a c b c c

a ac c b bc c c

= 2 2 2 2( ) ( ) ( ) (a c b c b bc c a ac c Expanding along R1

= (a – c) (b – c) {(b2 – a2) + (bc – ac)}

= (a – c) (b – c) {(b2 – a2) + c(b – a)}

= (a – c) (b – c) (b – a) {(b + a) + c}

= (a – b) (b – c) (c – a) (a + b + c)

= RHS. Hence proved

20. If y = 3 cos (log x) + 4 sin (log x), show that

22

20

d y dyx x y

dxdx

Ans. Given y = 3 cos (log x) + 4 sin (log x) … (i)

Differentiating w.r.t. x, we get

dy

dx = – 3 sin (log x)

1

x + 4 cos (log x)

1

xMultiplying by x, we get

xy1 = – 3 sin (log x) + 4 cos (log x) … (ii)

Again differentiating w.r.t. x, we obtain

xy2 + y

1 . 1 = – 3 cos (log x)

1

x – 4 sin (log x)

1

xMultiplying throughout by x, we have

x2y2 + xy

1 = – (3 cos (log x) + 4 sin (log x))

or x2y2 + xy

1 = – y (using (i))

or x2y2 + xy

1 + y = 0

21. Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the lines

1 2 3

x y z and

2 1 1

3 2 5

x y z

Ans. Let the equation of line passing through (–1, 3, –2) be

( 1) 3 ( 2)

x y z

a b c... (i) where (a, b, c) are dr’s of the line

Since, the required line is to the lines:

1 2 3

x y z and

2 1 1

3 2 5

x y z

a + 2b + 3c = 0 and – 3a + 2b + 5c = 0 (using a1a

2 + b

1b

2 + c

1c

2 = 0)

Solving, 2 3 3 1 1 2

2 5 5 3 3 2

a b c

4 14 8

a b c

-(21)-

STUDYmate

or2 7 4

a b c

a = 2, b = –7, c = 4

Putting the values of a, b and c in equation (i) we get

1 3 2

2 7 4

x y z

or1 3 2

2 7 4

x y z

[Note: Using is not compulsory]

22. Find the particular solution of the following differential equation

( 1) 2 1; 0 ydyx e y

dx when x = 0

Ans. Given differential equation is

( 1) 2 1 ydyx e

dx...(i)

12 1

y

dy dx

xe

,12

y

y

e dy dx

xe

Integrating, we obtain

12

y

y

e dy dxC

xe

–12

y

y

e dy dxC

xe

– log |2 – ey| = log | x + 1| + C

log (x + 1) (2 – ey) = –C

|(x + 1) (2 – ey) | = e–c

(x + 1) (2 – ey) = ± e–c = A (say)

(x + 1) (2 – ey) = A ... (ii)

Also, when x = 0, y = 0,

(0 + 1) (2 – e0) = A

1(2 – 1) = A

A = 1

Substituting this value of A in (ii), we obtain, the required particular solution as

(x + 1) (2 – ey) = 1.

SECTION-C

Question numbers 23 to 29 carry 6 marks each.28. A girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she

gets 1, 2, 3 or 4, she tosses a coin two times and notes the number of heads obtained. If she obtained

exactly two heads, what is the probability that she threw 1, 2, 3 or 4 with the die?

Ans. Let E1 : ‘1, 2, 3, or 4 is shown on die’, and

-(22)-

STUDYmate

E2 : ‘5 or 6 is shown on die’,

then E1 and E

2 are mutually exclusive and exhaustive. Moreover,

1

4 2P(E )

6 3

and 2

2 1P(E )

6 3

Let E : ‘exactly one head shows up’,

then1

EP P

E

(exactly one head shows up when coin is tossed twice) = P({HT, TH}) =

2 1

4 2

and2

EP P

E

(exactly one head shows up when coin is tossed thrice)

= P ({HTT, THT, TTH})

= 38

Required probability = 1EP

E

11

1 21 2

EP P(E )

E

E EP P(E ) P P(E )

E E

(using Bayes Theorem)

1 2 18 82 3 3

1 2 3 1 1 1 8 3 112 3 8 3 3 8

29. Using the method of integration, find the area of the region bounded by the following lines:

3x – y – 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0

Ans. 3x – y – 3 = 0 ... (i)

2x + y – 12 = 0 ... (ii)

x – 2y – 1 = 0 ... (iii)

Solving (i) and (ii) 5x – 15 = 0 i.e. x = 3 i.e. y = 9 – 3 = 6

Point of intersection (3, 6)

Solving (ii) and (iii)

2 12 0

2 4 2 0

5 10 0

2

x y

x y

y

y

x = 5

Point of intersection (5, 2)

-(23)-

STUDYmate

Solving (iii) and (i)

3 3 0

3 6 3 0

0 and 15 0

x y

x y

y xy

Point of intersection is (1, 0)

0

y

xA(1, 0) E(3,0)

B(3, 6)

C(5, 2)

( )i( )ii

( )iii

D(5,0)

Required area

= Area of region ABCA

= area of region ABCDA – Area of ACDA

3 5 5

( ) ( ) ( )1 3 1

i ii iiiy dx y dx y dx

3 5 5

1 3 1

( 1)(3 3) ( 2 12)

2

xx dx x dx dx

3 5

2 252

31 1

33 12

2 4 2

x x xx x x

27 3 25 5 1 19 3 ( 25 60) ( 9 36)

2 2 4 2 4 2

24 24 46 [35 27]

2 4 2

= 10 sq. units

× · × · × · × · ×

-(24)-

STUDYmate

SECTION-A

Question numbers 1 to 10 carry 1 mark each.9. Find the sum of the following vectors:

ˆ ˆ ˆˆ ˆ ˆ ˆ3 , 2 , 2 3 2 a i k b j k c i j k

Ans. ˆˆ ˆ3 2 a b c i j k

10. If

1 2 3

2 0 1 ,

5 3 8

write the minor of the element a22

.

Ans. 22

1 31 8 3 5 8 15 7

5 8 M

SECTION-B

Question numbers 11 to 22 carry 4 marks each.19. Using properties of determinants, prove the following

1 1 1

1 1 1

1 1 1

a

b ab bc ca abc

c

Ans. Taking out factors a, b, c common from R1, R

2 and R

3, we get

L.H.S. =

1 1 11

1 1 11

1 1 11

a a a

abcb b b

c c c

Applying R1 R

1 + R

2 + R

3, we have

1 1 1 1 1 1 1 1 11 1 1

1 1 11

1 1 11

a b c a b c a b c

abcb b b

c c c

Code No. 65/1/3

Studymate Solutions to CBSE Board Examination 2011-2012

UNCOMMON QUESTIONS ONLY

Series : SMA/1

-(25)-

STUDYmate

1 1 1

1 1 1 1 1 11 1

1 1 11

abca b c b b b

c c c

Now applying C2 C

2 – C

1, C

3 C

3 – C

1, we get

1 0 0

1 1 1 11 1 0

10 1

abca b c b

c

1 1 11 [1(1 0)]

abca b c

1 1 11 R.H.S.

abc abc bc ca aba b c

20. If y = sin–1 x, show that 2

22

(1 ) 0. d y dy

x xdxdx

Ans. We have y = sin–1x. Then

2

1

(1 )

dy

dx x

or 21 1 dy

xdx

So 21 ). 0

d dyx

dx dx

2

2 22

1 ). . 1 ) 0d y dy d

x xdx dxdx

22

2 2

21 ). . 0

2 1

d y dy xx

dxdx x

or

22

2 21 ). – . 0

1

d y dy xx

dxdx x

Hence2

22

(1 ) 0d y dy

x xdxdx

21. Find the particular solution of the following differential equation

( 2)( 2); 1 dy

xy x y ydx

when x = 1

Ans. Given differential equation is

( 2) ( 2)dy

xy x ydx

...(i)

-(26)-

STUDYmate

or2

,2

y xdy dx

y x

integrating, we obtain

2

2

y x

dy dx Cy x

or2 2

1 12

dy dx Cy x

or y – 2 log | y + 2| = x + 2 log | x| + C ...(ii)

But, (1, –1) lies on this curve, therefore,

–1 – 2 log|–1 + 2| = 1 + 2 log 1 + C

–1 – 2 log 1 = 1 + 2 log 1 + C

C = –2 ( log 1 = 0)

Hence, from (ii), we find the required curve as

y – 2 log |y + 2| = x + 2 log |x| – 2

or y = x + 2 log | x (y + 2)| – 2.

22. Find the equation of a line passing through the point P (2, –1, 3) and perpendicular to the lines

ˆ ˆˆ ˆ ˆ ˆ( ) (2 2 ) r i j k i j k and ˆ ˆˆ ˆ ˆ ˆ(2 3 ) ( 2 2 ).

r i j k i j k

Ans. Let, the equation be

1 1 1

x x y y z z

a b cIt passes through (2, –1, 3)

2 1 3

x y z

a b c... (i)

Also (i) is to ˆ ˆˆ ˆ ˆ ˆ( ) (2 2 ) r i j k i j k

2a – 2b + c = 0 ... (ii)

Also (i) is to ˆ ˆˆ ˆ ˆ ˆ(2 3 ) µ( 2 2 ) r i j k i j k

a + 2b + 2c = 0 ... (iii)

Solving (ii) and (iii)

4 2 4 1 4 2

a b c

6 3 6

a b c

2 1 2

a b c

Taking a = 2, b = 1, c = –2

Required equation lie is

2 1 3

2 1 2

x y z

or ˆ ˆˆ ˆ ˆ ˆ(2 3 ) (2 2 ) r i j k i j k

-(27)-

STUDYmate

SECTION-C

Question numbers 23 to 29 carry 6 marks each.28. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. Two balls are transferred

at random from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red

in colour. Find the probability that the transferred balls were both black.

Ans. Bag I contains 3 red, 4 black balls

Bag II contains 4 red, 5 black balls

2 balls transferred from Bag I to Bag II

Case 1 : Event A = Both black balls transferred

Case 2 : Event B = One black balls and One red ball transferred

Case 3 : Event C = Both red balls transferred.

4

27

2

2( ) ,

7

CP A

C

3 41 17

2

4( ) ,

7

C CP B

C

32

72

1( )

7

CP C

C

Let event E = 1 ball drawn from bag II is red.

4

( / )11

P E A (Since bag II contains 4R, 7B balls now)

5( / )

11P E B (Bag II contains 5R, 6B balls now)

and6

( / )11

P E C (Bag II contains 6R, 5B balls now)

Required Probability = P (A/E)

( ) ( / )

( ) ( / ) ( ) ( / ) ( ) ( / )

P A P E A

P A P E A P B P E B P C P E C

2 4

7 112 4 4 5 1 67 11 7 77 7 11

8 8 4

8 20 6 34 17

29. Using the method of integration, find the area of the region bounded by the following lines

5x – 2y – 10 = 0, x + y – 9 = 0, 2x – 5y – 4 = 0.

Ans. 5x – 2y – 10 = 0 ... (i)

x + y – 9 = 0 ... (ii)

2x – 5y – 4 = 0. ... (iii)

Solving (i) and (ii)

5 2 10 0

9 0

7 28 0

x y

x y

x

x = 4 ; y = 5; Point of intersection is (4, 5) 0 E(4,0)

y

xA(2, 0)

B(4, 5)

C(7, 2)

( )i

( )ii

( )iii

D(7,0)

-(28)-

STUDYmate

Solving (i) and (iii)

10 4 20 0

10 25 20 0

21 0

x y

x y

y

i.e., y = 0; x = 2, therefore, Point of intersection is (2, 0)

Solving (ii) and (iii)

2 2 18 0

2 5 4 0

7 14 0

x y

x y

y

i.e., y = 2; x = 7, therefore, Point of intersection is (7, 2)

Required Area = Area of Region ABCA

= Area of Region ABCDA – Area of Region ACDA

4 7 7

( ) ( ) ( )2 4 2

i ii iiiy dx y dx y dx

4 7 7

2 4 2

5 10 2 4(9 )

2 5

x xdx x dx dx

4 7 72 2 2

2 4 2

5 45 9

4 2 5 5

x x x xx x

80 20 49 16 49 28 4 820 10 63 36

4 4 2 2 5 5 5 5

21 4[5] [38.5 28]

5 5

= 43.5 – 28 – 5

= 10.5

21

2 sq. units

× · × · × · × · ×