5 maths cbse_2012-13_12th_20-03-13

-(1)-

Series : SKS/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 title page ofthe 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 distributed at10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will notwrite 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 2012-2013

-(2)-

STUDYmate

SECTION-A

Question numbers 1 to 10 carry 1 mark each.

1. Write the principal value of 1 1 1tan 1 cos

2

Ans. tan–1 (1) + cos–1 1

2

=1 1tan tan cos cos

4 3

=1cos cos

4 3

=2

4 3

=3 8

12

=

11

12

2. Write the value of 1 1tan 2tan

5

Ans.1 1

tan 2tan5

= 12

12

5tan tan1

15

=1 2 5

tan tan24

=1 5

tan tan12

=5

12

3. Find the value of a if 2 1 5

2 3 0 13

a b a c

a b c d

Ans. a – b = –1

2a – b = 0

–a = –1

a = 1

4. If 1 1 4 1

3 2 1 3

x x

x x

, then write the value of x.

Ans. (x + 1) (x + 2) – (x – 3) (x – 1) = 12 + 1

(x2 + 3x + 2) – (x2 – 4x + 3) = 13

7x – 1 = 13

7x = 14

x = 2

-(3)-

STUDYmate

5. If 9 1 4 1 2 1

2 1 3 0 4 9A

, then find the martix A.

Ans. A = 9 1 4 1 2 1

2 1 3 0 4 9

A = 8 3 5

2 3 6

6. Write the degree of the differential equation

2 423

20

d y dyx x

dxdx

Ans. Degree = 2

7. If ˆ ˆ ˆ2a xi j zk

and ˆ ˆ ˆ3b i yj k

are two equal vectors, then write the value of x + y + z.

Ans. ˆ ˆ ˆ ˆ ˆ ˆ2 3a xi j zk i yj k

x = 3 y = –2 z = –1

x + y + z = 0

8. If a unit vector a makes angles

3

with ,

4i

with j and an acute angle with k , then find

the value of .

Ans.2 2 2cos cos cos 1

3 4

21 1cos

4 2 = 1

cos2 = 1

4

cos = 1

2, as is an acute angle.

= 3

9. Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is

parallel to the line 3 4 8

3 5 6

x y z .

Ans. The required equation of the line

2 4 5

3 5 6

x y z

10. The amount of pollution content added in air in a city due to x-diesel vehicles is given by P(x)

= 0.005x3 + 0.02x2 + 30x. Find the marginal increase in pollution content when 3 dieselvehicles are added and write which value is indicated in the above question.

Ans. P(x) = 0.005x3 + 0.02x2 + 30x

P'(x) = 3 × 0.005x2 + 2(0.02)x + 30

P'(3) = 3 × 0.005 × 9 + 2(0.02)(3) + 30 = 30.255

-(4)-

STUDYmate

SECTION-B

Question numbers 11 to 22 carry 4 marks each.

11. Show that the function f in A = 2

3 defined as f(x) =

4 3

6 4

x

x

is one-one and onto. Hence

find f–1.

Ans.

4 3( )

6 4

xf x

x

Let

1 2 1 22

, ;3

x x A x x

Consider, f (x1) = f (x

2)

i.e.

21

1 2

4 34 3

6 4 6 4

xx

x x

i.e. (4x1 + 3) (6x

2 – 4) = (4x

2 + 3) (6x

1 – 4)

i.e. 24x1x

2 – 16x

1 + 18x

2 – 12 = 24 x

1x

2 – 16x

2 + 18 x

1 – 12

i.e. –34x1 = –34x

2

i.e. x1 = x

2

f is one - one.

Let

4 3

6 4

xy

x

6xy – 4y = 4x + 3

i.e. (6y – 4) x = 3+ 4y

3 4

6 4

yx

y

4 2

. .6 3

y i e

2

3y

f is onto

Since f is one-one and onto

1 3 4

6 4

yx f y

y

i.e. y (6x – 4) = 4x + 3

1 3 4

6 4

xf x

x

12. Find the value of the following:

21 1

2 2

1 2 1tan sin cos

2 1 1

x y

x y

, |x| < 1, y > 0 and xy < 1.

OR

Prove that: 1 1 11 1 1

tan tan tan2 5 8 4

Ans.2

1 12 2

1 2 1 1tan sin cos

2 1 2 1

x y

x y

-(5)-

STUDYmate

= 1 11 1

tan (2tan ) (2tan )2 2

x y

= tan (tan–1 x + tan–1 y)

= 1

x y

xy

OR

Ans. 1 1

1 112 5tan tan

1 8110

1 1 1[using tan tan tan , 1]1

x yx y xy

xy

= 1 17 1tan tan

9 8

=1

7 19 8tan

71

72

=1 56 9

tan72 7

=1 65

tan65

= tan–1 (1)

=4

13. Using properties of determinants, prove the following:

2

22 3

2

1

1 1

1

x x

x x x

x x

Ans. Operating C1 C

1 + C

2 + C

3, we obtain

L.H.S. =

2 2

2

2 2

1

1 1

1 1

x x x x

x x x

x x x

Take out 1 + x + x2 from C1

= (1 + x + x2)

2

2

1

1 1

1 1

x x

x

x

Operate R2 R

2 – R

1 : R

3 R

3 – R

1

= (1 + x + x2)

2

2

1

0 1 (1 )

0 ( 1) 1

x x

x x x

x x x

Expand with C1

= (1 + x + x2) 2

1 (1 )

(1 ) (1 )

x x x

x x x ,

Take out 1 – x from C1 and same from C

2

-(6)-

STUDYmate

= (1 + x + x2) (1 – x)21

1 x

x x

= (1 + x + x2) (1 – x)2 (1 + x + x2)

= {(1 + x + x2) (1 – x)}2

= (1 – x3)2 = R.H.S.

14. Differentiate the following function with respect to x:

(log x)x + xlog x

Ans. Let y = (log x)x + xlog x,

Thenlog{(log ) } x xdy d

x xdx dx

=log(log ) ( )x xd d

x xdx dx

=log(log ) {( log(log )} (log log )x xd d

x x x x x xdx dx

( ) ( log )

v vd du u v u

dx dx

= log1 1 1(log ) log(log ) 2(log )

logx xx x x x x

x x x

2(log log ) ((log ) )

d dx x x

dx dx

=log1 log

(log ) log(log ) 2log

x xxx x x

x x

15. If 2 2log[ ],y x x a show that 2

2 22

( ) 0.d y dy

x a xdxdx

Ans. 2 2log [ ]y x x a

2 2 2 2

1 21

2

dy x

dx x x a x a

2 2

12 2 2 2

____1____

x a xy

x x a x a

2 21( ) 1y x a

2 2 1

2 2 2

1.2 .( ) 0

2

x yy x a

x a

y2 (x2 + a2) + xy

1 = 0

16. Show that the function ( ) | 3|, ,f x x x is continuous but not differentiable at x = 3.

OR

If x = a sin t and y = a (cot t + log tan t/2), find 2

2.

d y

dx

Ans. f(x) = |x – 3|; x = 3

LHD = f '(3–) =

00

3 3limhh

f h f

h

=00

|3 3| |3 3|limhh

h

h

-(7)-

STUDYmate

=00

| |limhh

h

h

= 00

limhh

h

h

= –1

RH D = f '(3+) =

00

3 3limhh

f h f

h

=00

|3 3| |3 3|limhh

h

h

=00

| |limhh

h

h

= 00

limhh

h

h

= 1

As LHD RHD f is not differentiable.

Again, LHL =3

lim | 3|x

x

(at x = 3)

= 0lim|3 3|h

h

= 0lim| | 0h

h

RHL =3

lim | 3|x

x

(at x = 3)

= 0lim|3 3|h

h

= 0lim| | 0h

h

Since, LHL = RHL

f is continuous at x = 3.

OR

Ans. y = cos log tan2

ta t

; x = a sin t

dy

dt= 21 1

sin sec2 2

tan2

ta t

t

= 2

cot1 12sin2

sin cos2 2

t

a tt t

= 1

sinsin

a tt

= 21 sin

sin

ta

t

= 2cos

sin

ta

t

dx

dt= a cos t

-(8)-

STUDYmate

2cossincos

a tdy tdx a t

= cot t

2

2

d y d dy

dx dxdx

= 2cosecdt

tdx

= 2

2 1 cosec .seccosec .

cos

t tt

a t a

= 2cosec

cos

t

a t

17. Evaluate :

sin( )

sin(

x adx

x a

OR

Evaluate :

2

5 2

1 2 3

xdx

x x

Ans.sin( )

Isin( )

x a

dxx a

sin( 2 )I

sin

t a

dxt

sin cos2 cos sin2sin sin

t a t a

dt dtt t

= t cos 2a – sin 2a . cot dt

= t cos 2a – sin 2a ln | sin t | + C

= (x + a) cos 2a – sin 2a ln | sin (x + a) | + C

OR

Ans. 2

5 2I

1 2 3

xdx

x x

25 2 A (1 2 3 ) B d

x x xdx

5x – 2 = A (2 + 6x) + B

Comparing the co-efficients we get

5 = 6A

A = 56

– 2 = 2A + B

B = – 2A – 2

= 5

23

= 113

-(9)-

STUDYmate

21

2 2

5 11(2 6 )

6 3I1 2 3 1 2 3

II

x

dx dxx x x x

1 2

5 2 6I

6 1 2 3

xdx

x x

Let 1 + 2x + 3x2 = t

(2 + 6x)dx = dt

1

5 5I ln

6 6

dtt

t

= 2

1

5ln|1 2 3 | C

6 x x

2 2

11I

3 3 2 1

dx

x x

2

112 1 1 193 3 9 9

dx

x x

2

119 1 2

3 9

dx

x

12

111 1 3. tan C9 2 2

3 3

x

= I = I1 – I

2

= 2 15 11 3 1

ln|1 2 3 | tan C6 3 2 2

xx x

18. Evaluate :

2

2 2( 4) ( 9)

xdx

x x

Ans. I = 2

2 2

1 2 4 9 4 9

2 4 9

xdx

x x

= 2 2 2

2 2 2 2 2 2 2 2

1 4 1 9 1 4 1 13

2 2 2 24 9 4 9 4 9 4 9

x x x dxdx dx dx dx

x x x x x x x x

= 1 1

2 2

1 1 1 1 1 3 1 1. tan . tan .

2 3 3 2 2 2 2 5 4 9

x xdx

x x

= 1 1 1 11 1 13 13

tan tan tan tan6 3 4 2 20 2 30 3

x x x xC

= 1 11 13 1 13

tan tan3 6 30 2 4 20

x xC

= 1 13 2

tan tan5 3 5 2

x xC

-(10)-

STUDYmate

19. Evaluate : (| | | 2| | 4|)x x x dx

Ans. 4

0

| | | 2| | 4|x x x dx I

0 < x < 4 |x| = + x

0 < x < 2 |x – 2| = – (x – 2)

0 < x < 4 |x – 2| = + (x – 2)

0 < x < 4 |x – 4| = – (x – 4)

I = 4 4 4 4

0 0 0 0

2 ( 2) 4x x x x

4 2 4 42 2 2 2

0 0 2 0

2 2 42 2 2 2

x x x xx x x

= (8) + [(4 – 2) – 0] + [(8 – 8) – (2 – 4)] + [16 – 16/2]

= 8 + 2 + 2 + 16/2

= 20

20. If a and b are two vectors such that

| | | |,a b a then prove that vector

2a b is

perpendicular to vector .b

Ans. 2 2| | | |a b a

2 2 2| | 2 . | | | |a a b b a

2 . . 0a b b b

(2 ). 0a b b

2a b b

21. Find the coordinates of the point, where the line

2 1 2

3 4 2

x y z intersects the plane

x – y + z – 5 = 0. Also, find the angle between the line and the plane.

OR

Find the vector equation of the plane which contains the line of intersection of the planes

ˆ ˆ ˆ. ( 2 3 ) 4 0r i j k and

ˆ ˆ ˆ. (2 ) 5 0r i j k and which is perpendicular to the plane

ˆ ˆ ˆ. (5 3 6 ) 8 0.r i j k

Ans.2 1 2

3 4 2

x y z [x = 2 + 2, y = 4 – 1; z = 2 + 2]

This point lies on the plane x – y + z – 5 = 0

(3 + 2) – (4 – 1) + (2 + 2) – 5 = 0

3 + 2 – 4 + 1 + 2 + 2 – 5 = 0

= 0

-(11)-

STUDYmate

Point of Intersection = [2, –1, 2]

sin = 3 1 4 1 2 1

9 16 4 1 1 1

sin = 3 4 2

29 3

sin = 1

87

= 1 1

sin87

OR

Ans. Equation of plane through 1 &

2

(x + 2y + 3z – 4) + k(2x + y – z + 5) = 0

x (1 + 2k) + y(2 + k) + z (3 – k) – 4 + 5k = 0

This plane is perpendicular to 5x + 3y – 6z + 8 = 0

5 (1 + 2k) + 3 (2 + k) – 6 (3 – k) = 0

5 + 10 k + 6 + 3k – 18 + 6k = 0

19 k – 7 = 0

719

k

Required plane is will be

14 7 71 2 3

19 19 19

x y z

854 0

19

33x + 45y + 50z + 9 = 0

22. A speak truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are

they likely to contradict each other in stating the same fact? In the cases of contradiction doyou think, the statement of B will carry more weight as he speaks truth in more number of

cases than A?

Ans. P (AT) =

60

100P (A

C) =

40

100

P (BT) =

90

100P (B

C) =

10

100

P (contradiction) 60 10 400 90

100 100 100 100

600 3600 4200

10000 10000

% P (contradiction) = 42 %

In general parlace we understand who speaks truth, will have a better say and accepted

more.

-(12)-

STUDYmate

SECTION-C

Question numbers 23 to 29 carry 6 marks each.

23. A school wants to award its students for the values of Honesty, Regularity and Hard work witha total cash award of ` 6,000. Three times the award money for Hardwork added to that given

for honesty amounts to ` 11,000. The award money given for Honesty and Hard work together

is double the one given for Regularity. Represent the above situation algebraically and findthe award money for each value, using matrix method. Apart from these values, namely,

Honesty, Regularity and Hard work, suggest one more value which the school must include

for award.

Ans. Let Honesty award = ` x

Regularity award = ` y

Hard work award = ` z

According to question;

x + y + z = 6000

3z + x = 11000

x + z = 2y

x + y + z = 6000

x + 0y + 3z = 11000

x – 2y + z = 0

We can represent these equations using Matrices as:

1 1 1 6000

1 0 3 11000

1 2 1 0

x

y

z

AX = B

X = A–1B

Now |A| =

1 1 1

1 0 3

1 2 1

= 6 + 2 – 2 = 6 0

A–1 exists

adj. A =

6 3 3

2 0 2

2 3 1

A–1 = 1adj. A

|A|

=

6 3 31

2 0 26

2 3 1

X = A–1B =

6 3 3 60001

2 0 2 110006

2 3 1 0

=

36000 33000 01

12000 0 06

12000 33000 0

-(13)-

STUDYmate

=

30001

120006

21000

=

500

2000

3500

Award for honesty = ` 500

Award for regularity = ` 2000

Award for hardwork = 3500

One more value can be punctuality.

24. Show that the height of the cylinder of maximum value, that can be inscribed in a sphere of

radius R is 3R

.3

Also find the maximum volume.

OR

Find the equation of the normal at a point on the curve x2 = 4y which passes through thepoint(1, 2). Also find the equation of the corresponding tangent.

Ans. Let x be the radius of the base and h be the height of the cylinder ABCD which is inscried ina sphere of radius R. It is obvious that for maximum volumke the axis of the cylinder must be

along OA2 = OL2 + AL2

AL = 2 2R x

Let V be the volume of the cylinder. Then, D M C

O

A L B

a

V = (AL)2 × LM

V = (AL)2 × 2 (OL)

V = (R2 – x2) × 2x

V = 2 (R2x – x3)

2

2 22

V V2 ( 3 ) 12

d dR x and x

dx dx

For maximum or minimum values of V, we must have

V0

d

dx 2(R2 – 3x2) = 0

x = 3

R

Clearly,

2

2

3

V12 0

3

Rx

d R

dx

Hence, V is maximum when 3

R

x and hence LM = 2x = 2

3

R.

OR

Differentiating x2 = 4y with respect to x, we get

2

dy x

dx

Let (h, k) be the coordinates of the point of contact of the normal to the curve x2 = 4y. Now, slope

of the tangent at (h, k) is given by

-(14)-

STUDYmate

( , ) 2 h k

dy h

dx

Hence, slope of the normal at (h, k) = 2

h

Therefore, the equation of normal at (h, k) is

2( )

y k x h

h… (i)

Since it passes through the point (1, 2), we have

22 (1 )

k h

h

or2

2 (1 ) k hh

… (ii)

Since (h, k) lies on the curve x2 = 4y, we have

h2 = 4k … (iii)

From (ii) and (iii), we have h = 2 and k = 1. Substituting the values of h and k in (i), we get the

required equation of normal as

21 ( 2)

2

y x

or x + y = 3

25. Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.

OR

Using integration, find the area of the region enclosed between the two circles

x2 + y2 = 4 and (x – 2)2 + y2 = 4.

Ans. The equations of the given curves are

x2 = 4y … (i)

and x = 4y – 2 … (ii)

Eq. (i) represents a parabola with vertex at the originand axis along positive direction of y-axis. Eq. (ii)

represents a straight line which meets the

coordinate axes at (– 2, 0) and (2, 0) respectively.

To find the points of intersection of the given parabola

and the line, we solve (i) and (ii) simultaneously.Solving the two equations simultaneously.

We obtain that the points of intersection of the givenparabola and the line are (2, 1) and (–1, 1/4).

The region whose area is to be found out is shaded in figure.

Let us slice the shaded region into vertical strips. We find that each vertical strip runs from

parabola (i) to the line (ii). So, the approximating rectangle shown in figure has

Width = x, Length = (y2 – y

1), and the Area (y

2 – y

1) x.

Since the approximating rectangle can move from x = – 1 to x = 2.

Required area A is given by

2

2 1

1

A ( )

y y dx

2 2

1

2A

4 4

x xdx

2 1

2

2 1

P( , ) and Q( , ) lie on (ii) and (i) respec.

2and

4 4

x y x y

x xy y

-(15)-

STUDYmate

22 31A

8 2 12

x xx

4 2 8 1 1 1 9

A8 2 12 8 2 12 8

sq. units

OR

Ans. Equations of the given circles are

x2 + y2 = 4 ...(1)

and (x – 2)2 + y2 = 4 ...(2)

Equation (1) is a circle with centre O at the origin

and radius 2. Equation (2) is a circle with centre C(2, 0) and radius 2. Solving equations (1) and (2), we

have

(x –2)2 + y2 = x2 + y2

or x2 – 4x + 4 + y2 = x2 + y2

or x = 1 which gives y = 3

Thus, the points of intersection of the given circles are A 1, 3 and A' 1, 3 as shown in

the figure.

Required area of the enclosed region OACA'O between circles

= 2 [area of the region ODCAO] (Why?)

= 2 [area of the region ODAO + area of the region DCAD]

=

1 2

0 1

2 ydx ydx

= 1 2

2 2

0 1

2 4 2 4x dx x dx

(Why?)

= 1 2

2 1 2 1

10

1 1 2 1 12 2 4 2 4sin 2 4 4sin

2 2 2 2 2 2

x xx x x x

= 1 2

2 1 2 1

10

22 4 2 4sin 4 4sin

2 2

x xx x x x

= 1 1 1 11 13 4sin 4sin 1 4sin 1 3 4sin

2 2

= 3 4 4 4 3 46 2 2 6

= 2 2

3 2 2 33 3

= 8

2 33

26. Show that the differential equation 2yex/y dx + (y – 2x ex/y) dy = 0 is homogeneous. Find the

particular solution of this differential equation, given that x = 0 when y = 1.

Ans. The given differential equation can be written as

2

2

x

y

x

y

dx xe y

dyye

...(i)

-(16)-

STUDYmate

Let F(x, y) = 2

2

x

y

x

y

xe y

ye

Then F(x, y) = 02F ,

2

x

y

x

y

xe yx y

ye

Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differentialequation is a homogeneous differential equation.

To solve it, we make the substitution

x = vy ...(ii)

Differentiating equation (ii) with respect to y, we get

dx dvv y

dy dy

Substituting the value of x and dx

dy in equation (i), we get

2 1

2

v

v

dv vev y

dy e

or2 1

2

v

v

dv vey v

dy e

or1

2 v

dvy

dy e

or 2evdv = dy

y

or 2 .v dye dv

y

or 2ev = –log |y| + C

and replacing v by x

y , we get

2 log| | C

x

ye y ...(iii)

Substituting x = 0 and y = 1 in equation (iii), we get

2e0 + log |1| = C = C = 2

Substituting the value of C in equation (iii), we get

2 log| | 2

x

ye y

which is particular solution of the given differential equation.

27. Find the vector equation of the plane passing thorugh the three points with position vectors

ˆ ˆ ˆ ˆ2 , 2i j k i j k and ˆ ˆ ˆ2 .i j k Also find the coordinates of the point of intersection of this

plane and the line ˆ ˆ ˆ ˆ ˆ ˆ3 (2 2 ).r i j k i j k

Ans.

( ).( ) 0r a b c

ˆ ˆ ˆ

ˆ ˆ ˆ2 1 1 3 5

1 2 1

i j k

n b c i j k

-(17)-

STUDYmate

ˆ ˆ ˆ ˆ ˆ ˆ( 2 ) . ( 3 5 ) 0r i j k i j k

Equation of plane is

ˆ ˆ ˆ.(3 5 ) 14r i j k

or ˆ ˆ ˆ.(3 5 ) 14r i j k

Now ˆ ˆ ˆ ˆ ˆ ˆ3 (2 2 )r i j k i j k

This line intersects the plane

[(3 + 2l) ˆ ˆ ˆ ˆ ˆ ˆ[(3 2 ) ( 1 2 ) ( 1 ) ]. (3 5 ) 14)i j k i j k

– = 1

= – 1

Point of intersection is

ˆ ˆ ˆ2 .i j k

28. A cooperative society of farmers has 50 hectares of land to grow two crops A and B. The profits

from crops A and B per hectare are estimated as ` 10,500 and ` 9,000 respectively. To control

weeds, a liquid herbicide has to be used for crops A and B at the rate of 20 litres and 10 litresper hectare, respectively. Further not more than 800 litres of herbicide should be used in

order to protect fish and wildlife using a pond which collects drainage from this land. Keeping

in mind that the protection of fish and other wildlife is more important than earning profit,how much land should be allocated to each crop so as to maximize the total profit? Form an

LPP from the above and solve it graphically. Do you agree with the message that the protection

of wildlife is utmost necessary to preserve the balance in environment?

Ans. Let x hectares for crop A and y hectares for crop B be allocated.

A B

x

1050020 L/H

y

900010 L/H

50 hectaresProfit800 litres atmost

Y

C50

O A40

X

B(30, 20)

50x y + = 50

2x y + = 80

-(18)-

STUDYmate

A.T.Q.

We need to maximise profit given by

Z = 10500x + 9000y subject to

x + y 50 (Area constraint)

20x + 10y 800 (Herbicide constraint)

x, y 0 (Non-negative constraint)

After plotting graph, the corner points are

Corner Point Value of Z

O(0, 0)A(40, 0)B(30, 20)C(0, 50)

Z = 0Z = 420000Z = 315000 + 180000 = 495000Z = 450000

Maximum

30 hectares for crop A and 20 hectares for crop B.

29. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditationand yoga course reduces the risk of heart attack by 30% and prescription of certain drug

reduces its chance by 25%. At a time a patient can choose any one of the two options with

equal probabilities. It is given that after going through one of the two options, the patientselected at random suffers a heart attack. Find the probability that the patient followed a

course of meditation and yoga. Interpret the result and state which of the above stated methods

if more beneficial for the patient.

Ans. Let E1: the patient follows meditation and yoga

E2: tha patient uses drug

then E1 and E

2 are mutually exclusive and P(E

1) = P(E

2) =

1

2

Let E: the selected patient suffers a heart attack

then P(E/E1) =

40 30 281

100 100 100

and P(E/E2) =

40 25 301

100 100 100

Hence, P (patient who suffers heart attack follows meditation and yoga).

= P(E1/E) =

1 1

1 1 2 2

P E/E P E

P E/E P E P E/E P E (Using Bayes Theorem)

=

28 128 14100 2

28 1 30 1 58 29100 2 100 2

× · × · × · × · ×

-(19)-

STUDYmate

SECTION-A

9. Write the degree of the differential equation

4 2

23 0.

dy d yx

dx dx

Ans.

4 2

23 0

dy d yx

dx dx

Degree = 1

16. P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are

they likely to agree in stating the same fact?

Do you think, when they agree, means both are speaking truth?

Ans. P(PT) =

70

100P(P

L) =

30

100

P(QT) =

80

100P(Q

L) =

20

100

P(agreement) = (PL & Q

T) or (P

L & Q

L)

= 5600 600

10000 10000

= 6200

62%10000

No, agreement does not mean they are speaking truth.

18. If ˆ ˆ ˆa i j k and

ˆ ˆ ˆ,b j k k find a vector

,c such that

a c b and

. 3a c

Ans. Let ˆ ˆ ˆc xi yj zk

ˆ ˆ ˆ

1 1 1

i j k

a c

x y z

ˆ ˆ ˆ( ) ( ) ( )i z y j z x k y x

Now,

a c b (given)

ˆ ˆ( )b j k

z – y = 0, x – z = 1, y – x = – 1

y = z and x – y = 1 ... (i)

Again, . 3a c x + y + z = 3

x + y + y = 3

x + 2y = 3 ... (ii)

Code No. 65/1/2

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTIONS ONLY

Series : SKS/1

-(20)-

STUDYmate

Solving (i) and (ii)

2 3

1

3 2/3

x y

x y

y

2

3y z

x = 1 + y = 5

3

So, 5 2 2ˆ ˆ ˆ.

3 3 3c i j k

19. Evaluate : 3

1

| 1| | 2| | 3| .x x x dx

Ans. I = 3

1

| 1| | 2| | 3|x x x dy

2 3

1 2

| 1 2 3 ( 1 2 3 )x x x dx x x x dx

2 3

1 2

(4 )x dx x dx

2 32 2

1 2

42 2

x xx

1 96 4 2

2 2

7 5

62 2

5 5

52 2

20. Evaluate :

2

2 2

1

( 4)( 25)

xdx

x x

Ans. I

2

2 2

1

( 4)( 25)

xdx

x x2

2 2 2 2

1

( 4) ( 25) ( 4) 25

x A B

x x x x

x = 0

1

25 4 14 25 100

A BA B ... (i)

x = 1

2

5 26 5 26

A B

1

5 26 65

A B

13A + 5

2 B = 1

-(21)-

STUDYmate

26A + 5B = 2 ... (ii)

1 8

,7 7

A B

I = 2 2

1 8

7 74 25

dx dx

x x

1 11 1 8 1. tan . tan

7 2 2 7 5 5

x xc

28. Show that the differential equation

sin sin 0dy y y

x x ydx x x is homogeneous. Find the

particular solution of this differential equation, given that x = 1 when

.2

y

Ans. sin sindy y y

x y xdx x x

1

sin

dy yydx xx

Let F (x, y) = 1

sin

yyxx

F (x, y) =

1

sin

y

yxx

1

sin

yyxx

= ° F (x, y)

Given d.e. is homogeneous

Let y = vx

dy dv

v xdx dx

Now

sin sin 0dv

x v x v x vx vdx

2sin sin sin 0

dvvx v x v x vx v

dx

sindx

v dvx

– cos v = – ln |x| + c

– cos

y

x = – ln |x| + c

1,

2x y

0 = c

cos ln| |y

xx

cos( / )y xx e

cos logy

xx

-(22)-

STUDYmate

29. Find the vector equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) andC (–1, –1, 6). Also find the distance of Point P (6, 5, 9) from this plane.

Ans.

3 1 2

5 3 2 1 4 2 0

1 3 1 1 6 2

x y z

3 1 2

2 3 2 0

4 0 4

x y z

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

12x – 36 – 16y – 16 + 12 z – 24 = 0

12x – 16y + 12z – 52 – 24 = 0

12x – 16y + 12z – 76 = 0

3x – 4y + 3z – 19 = 0

ˆ ˆ ˆ.(3 4 3 ) 19r i j k

Now, distance of plane from Points P (6, 5, 9)

Distance of plane from (6, 5, 9)

(3 6) ( 4)(5) (3)(9) 19

34

18 20 27 19 6

34 34

× · × · × · × · ×

-(23)-

STUDYmate

SECTION-A

2. Write a unit vector in the direction of the sum of vectors ˆ ˆ ˆ2 2a i j k and

ˆ ˆ 3 .b i j k

Ans. ˆ ˆ ˆ0 5c a b i j k

2 2| | 1 5 26c a b

^ ˆ 5 ˆ

2 2

ia b k

b b

11. A speaks truth 75% of the cases, while B in 90 % of the cases. In what percent of cases arethey likely to contradict each other in stating the same fact? Do you think that statement of

B is true?

Ans. P(AT) =

75100

P(AL) =

25100

P(BT) =

90100

P(BL) =

10100

P (contradiction) = 75 10 25 90

100 100 100 100

= 750 2250

10000

= 3000

10000

= 3

10

% of P (Contradiction) = 3

10 × 100 = 30%

14. Evaluate : 5

2

| 2| | 3| || 5| .x x x dx

Ans. 2 < x < 5 |x – 2| + (x – 2)

2 < x < 3 |x – 3| – (x – 3)

3 < x < 5 |x – 3| x – 3

2 < x < 5 |x – 5| 5 – x

5 3 5 5

2 2 3 2

( 2) (3 ) ( 3) (5 )x x x x

5 3 5 52 2 2 2

2 2 3 2

2 3 3 52 2

x x x xx x x x

x x

25 9 25 9 2510 2 4 9 (6 2) 15 9 25 (10 2)

2 2 2 2 2

Code No. 65/1/3

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTIONS ONLY

Series : SKS/1

-(24)-

STUDYmate

5 9 5 9 252 4 8

2 2 2 2 2

9 7 4 9 39

2 2 2 2 2

15. Evaluate :

2

2 2

2 1

( 4)

xdx

x x

Ans. I =

2

2 2

2 1

( 4)

xdx

x x

Put x2 = t for PFD

2 1

( 4) 4

t A B

t t t t

2t + 1= A (t + 4) + Bt

Comp : t = 2 = A + B where 1 7

4 4A B

Const 1 = 4 A

I = 2 2

1 7

4 4 4

dx dx

x x

11 7 1tan

4 4 2 2

xC

x

11 7tan

4 8 2

xC

x

25. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses theplane, passing through the point (2, 2, 1) (3, 0, 1) and (4, –1, 0).

Ans. Equation of line

3 4 5

1 1 6

x y z

General Point [x = – + 3, y = + 4, z = 6 – 5]

Equation of plane

2 2 1

3 2 0 2 1 1 0

4 2 1 2 0 1

x y z

2 2 1

1 2 0 0

2 3 1

x y z

(x – 2) {2 – 0} – (y – 2) {– 1 – 0} + {z – 1} {– 3 + 4} = 0

2x – 4 + y – 2 + z – 1 = 0

2x + y + z – 7 = 0 _______________ ,

General point of line is a point on ,

2 ( – + 3) + ( – 4) + (6 – 5) – 7 = 0

–2 + 6 + – 4 + 6 – 5 – 7 = 0

5 – 10 = 0

= 2

Point of intersection of line and plane (1, –2, 7)

× · × · × · × · ×