Math XII Set 1

CBSE Pariksha Solution (Mathematics-XII) 1

CBSE Pariksha-2012 Solution (Mathematics–XII)

Section-AQ1. Show that the binary operation : R R R defined by 3a b a b , is not commutative. 1Sol. a * b = a + 3b

and b * a = b + 3aAs a + 3b b + 3a a * b b * aNot commutative.

Q2. Write the principal value of 1 3sin sin5

. 1

Sol. sin–1 3sin5

= sin–1 2sin

5

= sin–12sin5

2 ,5 2 2

Q3. Evaluate : 1

[ ] [ ]

ac b

a b a b c dd c

d

Sol.

ac b

a b a b c dd c

d

= [ac + bd] + [a2 + b2 + c2 + d2]= [ac + bd + a2 + b2 + c2 + d2]

Q4. Evaluate : 1

2

2

11

x dxx

Sol.2

2

11

x dxx =

2

2

1 21

x dxx

= 2

211

dxx = x – 2tan–1 x + c

RSPL/1

2 CBSE Pariksha Solution (Mathematics-XII)

Q5. Given matrix 11 12

21 22

a aA

a a

, such that |A| = –20, find 11 21 12 22a A a A , where AAij represents cofactor of an

element aij of a determinant. 1Sol. a11 A21 + a12 A22 = a11 (– a12) + a12 (a11) = 0Q6. Evaluate : 1

1 dxx x .

Sol.1 dx

x x=

1(1 )

dxx x

=12 dtt = 2 log | t | + c

Let 11

21 2

x t

dx dtx

dx dtx= 2 log | 1 + x | + c

Q7. Find the value of x, if2 4 2 45 1 6

xx

. 1

Sol. Consider2 4 2 45 1 6

xx

2 – 20 = 2x2 – 24

2x2 = 6 x2 = 3 x = 3Q8. Find the distance of the plane 3x – 3y + 3z = 0, from the point (1, 1, 1). 1Sol. Distance of the plane 3x – 3y + 3z = 0 from point (1, 1, 1)

=3 3 3 0 3 1 units

9 9 9 3 3 3

Q9. If a

is a unit vector and ( ).( ) 15,x a x a

find | |x

. 1

Sol. Consider ( ) ( – )

x a x a = 15

2 2–

x a = 15 2 2| | – | |

x a = 15

2| |

x – |1|2 = 15 2| |

x = 16 | |

x = 4

Q10. Find the angle between vectors ˆ ˆˆ ˆ ˆ ˆ .a i j k and b i j k

1

Sol. Angle between vectors ˆˆ ˆ–

a i j k and b ˆˆ ˆ –i j k is

cos =. 1 1 1 1

31 1 1 1 1 1| | | |

a b

a b

=1 1cos

3


Section-BQ11. A die is thrown twice and the sum of the numbers appearing is observed to be 7. What is the conditional

probability that the number 2 has appeared at least once ? 4Sol. n(S) = 36

A : Sum of the numbers appearing is 7 i.e. {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}B : Number 2 has appeared at least once. i.e. {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}A B = {(2,5), (5,2)}

We have to find P(B/A) = P(A B) 2 / 36 1P(A) 6 / 36 3

Q12. Let A be the set of all 46 students of class XII in a school. Let f : A N be a function defined by f (x) = Rollnumber of the student x. Check if, function f is bijective. 4

Sol. For function to be bijective, it should be one-one & onto.For one-one : Let x1x2 A

Thus x1 x2 students are different Their roll numbers are different f(x1) f(x2)Hence one-one.For onto : For every y N there should exist x A such that f(x) = y. This may not be true if y is greater than46, e.g. for 47 N. There is no student x A associated with 47 i.e. f(x) 47, for any x A.Hence not onto.Hence function is not bijective.

Q13. Prove that : 4

1 1 112 4 63sin cos tan13 5 16

.

OR

Write the simplest form of :

1 costan1 sin

xx

.

Sol. Consider 1 1 112 4 63sin cos tan13 5 16

Let =1 12 12 5sin sin cos

13 13 13

65

16

63

1 12sin13

= 1 5cos13

Also let = 1 63tan16

63tan16

cos =116 16cos

65 65

Consider

1 1 112 4 63sin cos tan

13 5 16


1 1 112 63 4sin tan cos

13 16 5

1 1 15 16 4cos cos cos

13 65 5

1 15 16 25 256 4cos 1 1 cos

13 65 169 4225 5

1 180 12 63 4cos cos

13 65 13 65 5

1 180 – 756 4cos cos

13 65 5

1 1–676 4cos cos

13 65 5

=1 1–4 4cos cos

5 5

= 1 14 4cos cos5 5

=

OR

Consider, 1 costan1 sin

xx

=1

sin2tan

1 cos2

x

x

= 1

2

2sin cos4 2 4 2tan

2cos4 2

x x

x

= 1sin

4 2tancos

4 2

x

x

= 1tan tan4 2

x =

4 2x

Q14. If1 01 7

A

, find k such that AA2 = 8A + kI. 4

Sol. Consider, A =1 01 7

A2 =1 0 1 0 1 01 7 1 7 8 49

...(i)


8A + kI =1 0 1 0

81 7 0 1

k

=8 0 0 8 08 56 0 8 56

k kk k

...(ii)

From (i) & (ii), we get

1 08 49

=8 0

8 56k

k

1 = 8+k; 49 = 56 + k k = – 7.

Q15. If , ,a b c

be the vectors such that 0, | | 3, | | 4 and | | 5,a b c a b c

find . . . .a b b c c a

4

OR

Let ˆ ˆˆ ˆ ˆ ˆ, 3 7 .a i j b j k and c i k

Find a vector d

which is perpendicular to both a

and b

and

. 1.c d

Sol. Given a b c

= 0

2( )a b c

=2

0

2 2 2

2 . 2 . 2 .a b c a b b c c a

= 0

2 2 2| | | | | | 2( . . . )a b c a b b c c a

= 0

9 + 16 + 25 + 2( . . . )a b b c c a

= 0

2( . . . )

a b b c c a = –50

. . .a b b c c a

= – 25

OR

Given a

= ˆ ˆi j , b

= ˆˆ3 j k and c

= ˆˆ7 i k , d

is perpendicular to both a

and b

d

= ( )k a b

a b

=

ˆˆ ˆ

1 1 00 3 1

i j k

= ˆ ˆ(1) ( 1) (3)i j k

= ˆˆ ˆ 3i j k

d

= ˆˆ ˆ( 3 ) k i j k ...(i)


Also c d

= 1

ˆ ˆˆ ˆ ˆ(7 ) ( 3 )i k k i j k = 1

k(7 – 3) = 1 k =14

Substituting in (i), we get

d

=1 1 1 3ˆ ˆˆ ˆ ˆ ˆ( 3 )4 4 4 4 i j k i j k .

Q16. Find the image of the point (1, 2, –1) in the plane 2x + y – z = 2. 4

OR

Cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find the fixed point through which it passes, itsdirection ratios and also its vector equation.

Sol. Let P(, , ) be image of point A(1, 2, –1) in plane

R

A(1, 2, –1)

P( , , )

2 + – = 2x y z

2x + y – z = 2.Let AP meets plane at R.Then (i) R is mid point of AP.(ii) R is foot of perpendicular from A to plane as R is mid point of AP

Coordinates of R are1 2 1, ,

2 2 2 ...(i)

Also AR plane 2x + y – z = 2 Equations of AR are

12

x =

2 11 1

y z =

General point on line is R(2 + 1, + 2, – – 1) ...(ii)If this lies on plane 2x + y – z = 2

2(2 + 1) + ( + 2) – (–– 1) = 2

6 = – 3 =1–2

Substituting in (ii), foot of perpendicular is

3 1R 0, ,2 2

...(iii)

From (i) and (iii), we get

12

= 0,2 3 1 1,

2 2 2 2

= –1, = 1, = 0 Image is P(–1, 1, 0).

ORCartesian equations of line are

3x + 1 = 6y – 2 = 1 – z


133

x =16 ( 1)3

y z

Dividing by 6, we get

13

2

x =

13

1

y =

16

z

Fixed point through which line passes is1 1, ,13 3

and direction ratios of line are 2, 1, –6.

Vector equation of line is

r =

1 1 ˆ ˆˆ ˆ ˆ ˆ(2 6 )3 3

i j k i j k

Q17. Solve the differential equation : 4

2log log .dyx x y xdx x

OR

Find the particular solution of the differential equation :

0,x y dy x y dx given that y = 1 when x = 1.

Sol. Consider, equation log dyx x ydx =

2 log xx

1

logdy ydx x x =

2 loglog

xx x x

1

logdy ydx x x = 2

2x

Here P(x) =1

logx x , 2

2( )Q xx

Integrating factor (I.F) =1

log(log )log log

dxxx xe e x

Solution is

(I.F.)y = {(I.F.).Q( )}x dx

log x · y = 2

2log . x dxx

= 22 log x x

=1 112 log1 1

x xx dx

x


= 212 log x x dxx

log x·y =1 12 log x cx x

is the required solution

ORConsider the equation

(x + y)dy + (x – y)dx = 0 (x + y)dy = –(x – y)dx

dydx

=( )x yx y ...(i)

Let y = vx

dydx

=dvv xdx

Substituting in (i), we get

dvv xdx

=1

1x vx v

x vx v

dvxdx

=2 21 1 (1 )

1 1 1v v v v vv

v v v

2

11

v dvv

=1 dxx

Integrating both sides, we get

2

11

v dvv = –

1 dxx

2 2

11 1

vdv dvv v

=1 dxx

1 21tan log |1 |2

v v = – log |x| + c

2

12

1tan log 12

y yx x

= – log |x| + c

1 2 21tan log | |2

y x yx

= c ...(ii)

Given y =1 when x = 1

tan–1(1) +1 log 22

= c

c =1 log 2

4 2


Substituting in (ii), we get

1 2 21tan log | |2

y x yx

=1 log 2

4 2 as the particular solution.

Q18. Form the differential equation representing the family of parabolas having vertex at origin and axis alongpositive direction of x-axis. 4

Sol. Family of parabolas having vertex at origin and axis along positive direction of x-axis, isy2 = 4ax ...(i), a is constant.

Differentiating w.r.t x, we get2yy = 4a

Substituting in (i), we gety2 = 2yy·x or 2xyy– y2 = 0

or 2xy – y = 0 as required equation.Q19. Evaluate : 4

3cosec .x dx

Sol. Let I = 3cosec x dx

= 2cosec cosec x x dx

= cosec x.(– cot x) – ( cosec cot ( cot ) x x x dx

= 2cosec cot cosec cot x x x x dx

I = 2cosec cot cosec (cosec 1) x x x x dx

= 3cosec cot cosec cosec x x x dx x dx

I = cosec cot I log | cosec cot | x x x x

2I = – cosec x cotx + log |cosec x – cot x|

I = 1 cosec cot log | cosec cot |2 x x x x c

Q20. Evaluate : 4

4 2

2

.2 1

x x dxx

Sol. Consider,2

2 1x x dx

x

= 2 1/ 2( ) (2 1) x x x dx


=1/ 21/ 2

2 (2 1) (2 1)( ) (2 1)1 12 22 2

x xx x x dx

= 2 3/ 2( ) 2 1 (2 1)x x x x dx

=5 / 2

2 (2 1)( ) 2 1 5 22

xx x x

4 2

2 2 1

x x dxx = 2 5 / 2 4

21( ) 2 1 (2 1) ]5

x x x x

= 5 / 2 5 / 21 1{(16 4) 8 1 (9) } {(4 2) 5 (5) }5 5

= 1 5760 243 6 5 5 5 55 5

Q21. If 1 1 0,x y y x for 1 1,x < <

Prove that 2

1 .1

dydx x

4

Sol. Consider, 1 1x y y x = 0

1x y = 1y x

Squaring, we getx2(1 + y) = y2(1 + x)

x2 + x2y – y2 – xy2 = 0

(x2 – y2) + (x2y – xy2) = 0

(x – y) (x + y) + xy(x – y) = 0

(x – y) (x + y + xy) = 0

x – y = 0 or x + y + xy = 0

x = y (rejected) or x + y + xy = 0

y(1 + x) = – x

y =1

xx

dydx

= 2

(1 )( 1) 1(1 )x x

x

= 2

1(1 )x


Q22. Differentiate, 1 sin costancos sin

a x b xa x b x

with respect to x. 4

Sol. Consider y =1 sin costan

cos sina x b xa x b x

Dividing numerator & denominator by a cos x, we get

y =1

tantan

1 tan

bxa

b xa

y = tan–1(tan x) + tan–1(b/a) y = x + tan–1 (b/a)

dydx

= 1 + 0 = 1

Section-CQ23. Using properties of determinants, prove that 6

2 .a b b c c a a b cb c c a a b b c ac a a b b c c a b

OR

If1 1 12 1 31 1 1

A

, find AA–1 and use it to solve the system of equations,

2 4x y z

0x y z

3 2x y z

Sol. Consider

a b b c c ab c c a a bc a a b b c

=

2( ) 2( ) 2( )a b c a b c a b cb c c a a bc a a b b c

[By performing R1 R1+ R2 + R3]

= 2a b c a b c a b c

b c c a a bc a a b b c

[By taking 2 common from R1]


= 2 – – –a b c a b c a b c

a b cb c a

[By performing R2 R2 – R1, R3 R3 – R1]

= 2 – – –c a ba b cb c a

[By performing R1 R1 + (R2 + R3)]

= 2(–1)2c a ba b cb c a

[By taking (–1) common from R2 and R3]

= 2b a cc b aa c b

[By performing C1 C3]

= 2a b cb c ac a b

[By performing C1 C2]

OR

Consider A =1 1 12 1 31 1 1

A–1 =1

| A | Adj AA ...(i)

|A| =1 1 12 1 31 1 1

= 1(1 + 3) +1(2 + 3) +1( 2 – 1) = 4 + 5 + 1 = 10 0

Matrix formed by cofactors of each element in | A |A11 = 4, A12 = –5, A13 = 1A21 = 2, A22 = 0, A23 = – 2A31 = 2, A32 = 5, A33 = 3

Adj A =

4 5 1 4 2 22 0 2 –5 0 52 5 3 1 2 3


From (i), we get

A–1 =4 2 2

1 –5 0 510

1 2 3

...(ii)

Consider equationsx + 2y + z = 4–x + y + z = 0x – 3y + z = 2

Corresponding matrix equation is

1 2 1–1 1 11 3 1

xyz

=402

AX = B is matrix equationIts solution is X = (A)–1B = (A–1)B [As (A)–1 = (A–1)]

X =4 2 2 4

1 –5 0 5 010

1 2 3 2

=4 –5 1 4

1 2 0 2 010

2 5 3 2

=16 0 2

1 8 0 – 410

8 0 6

xyz

=9 / 52 / 57 / 5

9 2 7, ,5 5 5

x y z is the solution.

Q24. In a competitive examination , an examinee either guesses or copies or knows the answer of multiple choicequestion with four choices. The probability that he makes a guess is 0.35 and probability that he copies theanswer is 0.20. The probability that his answer is correct, given that he copies it is 0.15. Find the probabilitythat he guesses the answer to the question, given that he correctly answered it. 6

Sol. A : Guesses B : copies C : knowsP(A) = 0.35 P(B) = 0.20 P(C) = 0.45

E : Correctly answered.P(E/A) = 0.25; P(E/B) = 0.15, P(E/C) = 1Using Baye’s Theorem

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

P(A)P(E/A)+P(B)P(E/B)+P(C)P(E/C)

=0.35 0.25

0.35 0.25 0.20 0.15 0.45 1

=875

875 300 4500 =

8755675

= 0.15


Q25. If a young man rides his motorcycle at 25 kmph, he has to spend ` 2 per km on petrol, if he rides at a fasterspeed of 40 kmph, the petrol cost increases to ` 5 per km. He has ` 100 to spend on petrol and wishes to findwhat is maximum distance, he can travel within one hour. Express this as an LPP and solve it graphically.

6Sol. Let he travels x km at 25 kmph and y km at 40 km/h

10

20

30

40

50

10 20 30 40 50

y

xA(25,0)

B 503

403

,

x25

y40+ = 1

2 + 5 = 100x y

0

C(0, 20)

LPP is to maximize, Z = x + ySubject to constraints

x 0, y 0,25 40x y 1

2x + 5y 100Plotting the inequations on graph we notice shaded portionis an optimum solution.

Possible points for maximum are A(25, 0), B50 40,3 3

, C(0,20)

Point Z = x + y value

A (25, 0) 25 + 0 25

50 40B ,3 3

50 403 3 30 Maximum

C(0, 20) 0 + 20 20

Z is maximum at 50 40B ,3 3

i.e. x =503

, y =403

Young man should travel50 km3

at 25km/h and40 km3

at 40km/h to cover maximum distance within

one hour.Q26. Find the shortest distance between the lines 6

ˆ ˆˆ ˆ ˆ ˆ2 3 2 3 4r i j k i j k

and ˆ ˆˆ ˆ ˆ ˆ2 4 5 3 4 5r i j k i j k

.

Sol. Given lines are

r

= ˆ ˆˆ ˆ ˆ ˆ2 3 (2 3 4 )i j k i j k

Here 1a

= ˆˆ ˆ2 3i j k ,1b

= ˆˆ ˆ2 3 4i j k

and r

= ˆ ˆˆ ˆ ˆ ˆ2 4 5 (3 4 5 )i j k i j k

Here2a

= ˆˆ ˆ2 4 5i j k ,2b

= ˆˆ ˆ3 4 5i j k

Shortest distance = 2 1 1 2

1 2

( – ).( )

| |

a a b b

b b

...(i)

2 1–a a

= ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 4 5 – 2 3 2 2i j k i j k i j k


1 2b b

=

ˆˆ ˆ

2 3 43 4 5

i j k

= ˆ ˆˆ ˆ ˆ ˆ( 1) ( 2) ( 1) 2 i j k i j k

1 2| |b b

= 1 4 1 6

From (i), we get shortest distance,

=ˆ ˆˆ ˆ ˆ ˆ( 2 2 ).( 2 – )

6i j k i j k

=1 4 2

6

=16

Q27. Draw a rough sketch of the region : 6

2 2 2, : 2 , , 0, 0x y x y ax y ax x y .

Using the method of integration, find the area enclosed by the region.

OR

Using the method of integration, find the area of the triangle whose sides have the equations :

y = 2x + 1, y = 3x + 1, x = 4.Sol. Given region is {(x, y); x2 + y2 2ax, y2 ax, x 0, y 0}

Consider corresponding equationsx2 + y2 = 2ax and y2 = ax

(x – a)2 + y2 = a2 and y2 = axCircle with the centre (a, 0) and radius a and parabola symmetric to x-axis plotting the corresponding inequations we notice shaded portion is required areaEliminating ‘y’ from the equations

0 A(a, 0)

y

x

y ax2 =

( – ) + =x a y a2 2 2x2 + ax = 2ax x2 – ax = 0 x(x – a) = 0 x = 0, a.

Area = 1 20

( )a

y y dx y1 : (x – a)2 + y2 = a2

y2 : y2 = ax

=2 2

0

{ ( ) }a

a x a ax dx

=2

2 2 1 3 / 2

0

2( ) sin2 2 3

ax a a x a aa x a x

a

=2

2 2 1 3 / 20 20 sin (0) ( )2 2 3

a aa a

–2

2 2 1sin ( 1) 02 2a aa a

=2 2

22 2.3 2 2 4 3a a a sq.units


ORConsider equations,

2

4

6

8

10

1 2 3 4 5

y

x0

12

x = 4

yx

= 3+ 1

yx

= 2+ 1y = 2x + 1, y = 3x + 1 and x = 4

Plotting these equations on graph we notice we have to findshaded area Enclosed area

=4

0

{3 1 2 1}x x dx =

44 2

0 02xxdx

=16 8sq. units2

Q28. Prove that the surface area of a solid cuboid, of square base and given volume, is minimum when it is a cube.6

Sol. Let sides of cuboid be x, x, y.V = x·x·y

x

y

x

V = x2y ...(i)S = 2[x·x + x·y + x·y] = 2[x2 + 2xy]

= 22

V2 2x xx

from (i)

S = 2 2V2 xx

dSdx

= 2

2V2 2 –

xx

For minimum S,dSdx

= 0 2x = 2

2Vx

V = x3 ...(ii)2

2

d Sdx

= 3

4V2 2x

3

2

2V

x

d Sdx

> 0

Surface area is minimum for V = x3

x2y = x3 [from (i)] y = x

Surface area is minimum, when cuboid is a cube.Q29. Prove that the curves y2 = 4ax and xy = c2 cut at right angles, if c4 = 32a4. 6Sol. Given curves are

y2 = 4ax ...(i)and xy = c2 ...(ii)

Differentiating w.r.t. x

From (i), 2yy = 4a y =2ay ...(iii)

From (ii), xy + y1 = 0 y =y

x

...(iv)

If curves cut at right angles.From (iii) and (iv), we have

2 a yy x

= – 1 x = 2a ...(v)

Substituting in (i), we get y2 = 8a2 ...(vi)From (ii), squaring, we get

x2y2 = c4

(2a)2 × 8a2 = c4 32a4 = c4 [Using (v), (vi)]

Math XII Set 1

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Transcript of Math XII Set 1