CBSE 2014

1. Given that A2 = A.

We need to find the value of 7A – (I + A)3, where I is the identity matrix.

Thus,

7A - (I + A)3 = 7A - (I3 + 3I2A + 3I2A + A3)

⟹ 7A - (I + A)3 = 7A – (I3 + 3A + 3A2 + A2 x A) [I3 = I, I2A = A, IA2 = A2]

⟹ 7A – (I + A)3 = 7A – (I + 3A + 3A + A) [∵A2 = A]

⟹ 7A – (I + A)3 = 7A – I – 3A – 3A – A

⟹ 7A – (I + A)3 = 7A – I – 7A

⟹ 7A – (I + A)3 = - I.

2. Given that

50

41

2 wyx

zyx

We need to find the value of x + y.

50

41

2 wyx

zyx

Two matrices A and B are equal to each other, if they have the same dimensions and the same

elements aij = bij, for I = 1, 2, …, n and j = 1, 2, …,m.

x – y = -1…(1)

2x – y = 0…(2)

Equation (2) – (1) is x = 1

Substituting the value of x = 1 in equation (1), we have

1 – y = -1

⟹ y = 2

Therefore, x + y = 1 + 2 = 3

3. Given that tan-1x + tan-1y = 4

and xy < 1.

We need to find the value of x + y + xy.

tan-1x + tan-1y = 4

CBSE SAMPLE PAPER SOLUTIONSCLASS-XII

MATHS SET-1

CBSE

11

11

4tan

1tantan

141

tan

1

1

xyyxxyyx

xy

yx

xy

yx

xyxy

yx

4. Given that46

78

42

73

x

We need to find the value of x

46

78

42

73

x

⟹ 12x – (– 14) = 32 – 42

⟹ 12x + 14 = – 10

⟹ 12x = – 10 – 14

⟹12x = – 24

⟹ x = – 2

5. Since differentiation operation is the inverse operation of integration, we have xxxf sin)('

Let x

tdttxf0

sin)(

Let us do this by integration by parts.

Therefore assume u = t; du = dt

dvtdtsin

−cost = v

Therefore,

Cxxxxf

dttttxf

xx

sincos

coscos0

0

Differentiating the above function with respect to x,

xxxxxxxf sincoscossin'

6. Since the vectors are parallel, we have

2014

CBSE

kjpikji

kjpikji

ba

ˆ3ˆ2ˆˆ9ˆ2ˆ3

)ˆ3ˆ2ˆ(ˆ9ˆ2ˆ3

Comparing the respective coefficients, we have

⟹ λ = 3;

−2 λp = 2

⟹ −2 x 3 x p = 2

⟹ p = 3

1

7. The set of natural numbers, N = {1,2,3,4,5,6…}

The relation is given as

R = {(x, y) : x + 2y = 8}

Thus, R = {(6,1),(4,2),(2,3)}

Domain = {6,4,2}

Range = {1,2,3}.

8. Given that the cartesian equation of the line as

4

62

7

4

5

3

zyx

That is,

2

3

7

)4(

5

3

4

)3(2

7

)4(

5

)3(

zyx

zyx

Any point on the line is of the form :

32,47,35

Thus, the vector equation is of the form:

bar

, where a

is the position vector of any point on the line and b

is the vector parallel to

the line.

Therefore, the vector equation is

kjikjir

kjikjir

kjir

ˆ2ˆ7ˆ5ˆ3ˆ4ˆ3

ˆ3ˆ4ˆ3ˆ2ˆ7ˆ5

ˆ32ˆ47ˆ35

9. Given that

a

x

dx

0

2 84

We need to find the value of a.

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CBSE

Let I =

a

x

dx

0

2 84

Thus , I = 82

tan2

1

0

1

ax

212

42tan

82

2tan

82tan

2

1

1

1

1

aa

a

a

a

10. Given that ba

and are two perpendicular vectors.

Thus, 0. ba

Also given that, .5 and 13 aba

We need to find the value of b

.

Consider : 2

ba

12

144

25169

25169

02513

. 2

2

2

2

222

222

b

b

b

b

b

bbaaba

SECTION – B 11. Given differential equation is:

2

tan

2

tan2

11

1

1

1

x

e

x

y

dx

dy

eydx

dyx

x

x

This a linear equation of the form

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CBSE

x

e

xP

QPydx

dy

x

1Q and

1

1 Where

1tan

2

Therefore,

I.F = xPdx

ee1tan

Thus the solution is,

y(I.F) = dxFIQ .

; Substitute

11

1̀

1

1

tan

tan

2

tan

tan

te

dxex

eey

x

x

x

x

Thus,

C

eey

Ct

ey

tdtey

xx

x

x

2

22

tantan

2tan

tan

1

1

1

1

12. Given position vectors of four A, B, C and D are:

kjiOD

kjiOC

kjOB

kjiOA

ˆˆˆ4

ˆ4ˆ9ˆ3

ˆˆ

ˆˆ5ˆ4

These points are coplanar, if the vectors, ADACAB and , are coplanar.

kjikjikji

OAODAD

kjikjikji

OAOCAC

kjikjikj

OAOBAB

ˆ3ˆˆ8ˆˆ5ˆ4ˆˆˆ4

ˆ3ˆ4ˆˆˆ5ˆ4ˆ4ˆ9ˆ3

ˆ2ˆ6ˆ4ˆˆ5ˆ4ˆˆ

These vectors are coplanar if and only if, they can be expressed as a linear combination of other

two.

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CBSE

kyxjyxiyxkji

ADACxAB

ˆ33ˆ4ˆ8ˆ2ˆ6ˆ4

Comparing the coefficeints, we have,

−x – 8y = – 4;4x – y =– 6; 3x + 3y = –2

Thus, solving the first two equations, we get

3

2 and

3

4

yx

These values of x and y satisfy the equation 3x + 3y = –2.

Hence the vectors are coplanar.

OR

Given that

kjic

kjib

ˆ3ˆ2ˆ

ˆ5ˆ4ˆ2

Now consider the sum of the vectors :cb

kjicb

kjikjicb

ˆ2ˆ6ˆ2

ˆ3ˆ2ˆˆ5ˆ4ˆ2

Let n̂ be the unit vector along the sum of vectors cb

:

222262

ˆ2ˆ6ˆ2ˆ

kjin

The scalar product of na

and is 1. Thus,

1

88

3612444

3612444

6402

6262

262262

262

2.16.1211

262

ˆ2ˆ6ˆ2.ˆˆˆˆ.

22

22

222

222

222

222

kjikjina

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kjin

kjin

kjin

kjin

kjin

n

ˆ7

2ˆ7

6ˆ7

3ˆ

7

ˆ2ˆ6ˆ3ˆ

49

ˆ2ˆ6ˆ3ˆ

263

ˆ2ˆ6ˆ3ˆ

2612

ˆ2ˆ6ˆ12ˆ

:is ˆ Thus,

222

222

13. We need to evaluate the integral

0

2)1...(

cos1

sin4dx

x

xxI

Using the property dxxfdxxaf , we have

0

2

0

2

0

2

)2...(cos1

sin4

cos1

sin4

cos1

sin4

dxx

xdx

x

xI

dxxx

xxI

Adding equations (1) and (2), we have,

0

2

0

2

0

2

0

2

0

2

cos1

sin42

cos1

sin42

cos1

sin4

cos1

sin4

cos1

sin42

dxx

xI

dxx

xI

dxx

xdx

x

xdx

x

xxI

Subsititute t = cosx; dt = - sinxdx

When x = , t = 1

When x = π, t = -1

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CBSE

2

1

1

0

1

1

0

2

1

1

2

1

1

2

44

1tan4

tan22

122

12

1

142

I

I

tI

t

dtI

t

dtI

t

dtI

OR

We need to evaluate the integral

dxxx

x

dxxx

x

65

2ILet

65

2

2

2

Consider the integrand as follows:

BAxAxBxAx

xx

Bxxdx

dA

xx

x

522522

65

65

65

2

2

2

2

Comparing the coefficients, we have

2A = 1; 5A + B = 2

Solving the above equations, we have

A = 2

1and B =

2

1

Thus,

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CBSE

dxxx

I

I

xx

t

t

dtI

dtdxxtxx

dxxx

xI

I

dxxx

I

dxxx

xI

III

dxxx

dxxx

x

dxxx

x

dxxx

xI

65

1

:consider Now

652

2

52;65

Subsitute

65

52

:consider Now

65

1 and

65

52 Where

,2

1,

2

1

65

1

2

1

65

52

2

1

65

2

1

2

52

65

2

22

2

2

1

2

21

1

22

21

21

22

2

2

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CxxxxxI

III

CxxxI

dx

x

x

dx

x

dx

xx

652

5log

2

165

2

1

2

1 Thus,

652

5log

2

1

2

5

2

4

256

2

5

1

2

56

2

55

1

22

21

2

2

22

2

22

2

14. Given function is

f(x) = [x(x – 2) ]2

⟹f’(x) = x2 x 2(x – 2) + (x – 2)2 x 2x

⟹f’ (x) = 2x (x – 2)[x+(x – 2)]

⟹f’(x) = 2x(x – 2)[2x – 2]

⟹f’(x) = 2x (x – 2)[2(x – 1)]

⟹f’(x) = 4x(x – 1) (x – 2)

Since f’(x) is an increasing function, f’(x) > 0.

⟹ f’(x) = 4x (x – 1) (x – 2) > 0

⟹x(x – 1) (x – 2) > 0

⟹0 < x < 1 or x > 2

⟹ x ∈(0, 1) ∪ (2, ∞)

OR

Let 12

2

2

2

b

y

a

xbe the equation of the curve.

Rewriting the above equation as,

22

2

22

2

2

2

2

1

bxa

by

a

x

b

y

Differentiating the above function w.r.t.x, we get,

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CBSE

a

b

b

a

a

b

dx

dy

y

x

a

b

dx

dy

xa

b

dx

dyy

ba

22

22

2

2

,2

2

2

2

2

Slope of the tangent is m = a

b2

Equation of the tangent is

(y – y1) = m (x – x1)

02

022

22

22

abaybx

ababaybx

axbbya

axa

bby

Slope of the normal is

a

b2

1

Equation of the normal is

022

0222

22

22

22

22

11

babyax

abbyax

axabyb

axb

aby

xxmyy

15. Given that f(x) = x2 + 2 and g(x) =1x

x

Let us find f °g:

f °g = f(g(x))

⟹ f °g = (g(x))2 + 2

⟹ f °g = 21

2

x

x

⟹ f °g =2

22

)1(

)1(2

x

xx

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CBSE

⟹ f °g = 12

)12(22

22

xx

xxx

⟹ f °g = 12

2432

2

xx

xx

Therefore, (f°g)(2) = 1222

224232

2

⟹ (f °g)(2) = 6144

2812

Now let us find g °f:

g °f = g (f(x))

⟹ g °f = 1)(

)(

xf

xf

⟹ g °f = 12

22

2

x

x

⟹ g °f = 1

22

2

x

x

Therefore, (g°f)(-3) = 10

11

19

29

1)3(

2)3(2

2

16. We need to prove that

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CBSE

R.H.Scos2

1

4

4

4tantan

tan4

tan1

tan4

tan

tan

tan1

tan1tan

sin2cos2

sin2cos2tan

2cos12cos1

2cos12cos1tan L.H.S

;2cosConsider

12

1,cos

2

1

411

11tan

1

1

1

1

1

1

11

x

t

t

t

t

t

t

tt

tt

tt

tt

tx

xxxx

xx

OR

Given that 44

2tan

4

2tan 11

x

x

x

x

We need to find the value of x.

2

2

42

12162

112

162

1416

8282

1)2)(2()4)(4(

)4)(2()4)(2(

4tan

4

2

4

21

4

2

4

2

4

4

2

4

21

4

2

4

2

tan

44

2tan

4

2tan

2

2

2

2

22

22

1

11

x

x

x

x

x

xx

xxxx

xxxx

xxxx

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

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CBSE

17. An experiment succeeds thrice as often as it fails. Therefore, there are 3 successes and 1 failure.

Thus the probability of success =4

3

And the probability of failure =4

1

We need to find the probability of atleast 3 successes in the next five trials.

Required Probability=P(x = 3) + P(x = 4) + P( x = 5)

= 5C3p3q2 + 5C4p

4q1 + 5C5p5q0

051323

05

5

5

14

4

5

23

3

5

4

1

4

3

4

1

4

35

4

1

4

310

4

1

4

3

4

1

4

3

4

1

4

3

CCC

512

459

1024

918

18. Given that

y = Peax + Qebx

Differentiating the above function w.r.t.x,

bxax QbePaedx

dy

Differentiating once again, we have,

bxax eQbePadx

yd 22

2

2

Let us now find (a + b) dx

dy:

(a + b)dx

dy= (a + b) bxax QbePae

bxbxax

bxbxbxax

eQbabeQPePadx

dyba

eQbPabeQabeePadx

dyba

22

22

)()(

)(

Also we have,

aby = ab(Peax + Qebx)

Thus, abydx

dyba

dx

yd )(

2

2

= Pa2eax + Qb2ebx – Pa2eax – (P + Q)abebx – Qb2ebx + abPeax + abQebx

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= 0

19. Consider the determinant

c

b

a

111

111

111

Taking abc common outside, we have

1111

11

11

111

1

cba

cba

cba

abc

Apply the transformation , C1 → C1 + C2 + C3,

111

1

11

11

111

1111

111111

1

11

11111

111111

cb

cb

cb

cbaabc

cbcba

cbcba

cbcba

abc

Apply the transformations, R2 → R2 – R1 and R3 → R3 – R1

100

010

111

1111

cb

cbaabc

Expanding along C1, we have

cabcababccba

abc

cbaabc

1111

10

011

1111

20. x = cost(3 – 2cos2t)

and

y = sint (3 – 2sin2t)

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CBSE

We need to find dx

dy:

dt

dxdt

dy

dx

dy

Let us find dt

dx:

X = cost (3 – 2cos2t)

dt

dx=cost(4costsint) + (3 – 2cos2t) (- sint)

tttttdt

dycossin2cossin4cos3 22

14

cot

cot

sin211cos2sin

cos

)cos21(sin3

)sin21cos(3

sincos6sin3

cossin6cos3

sincos2sincos4sin3

cossin2cossin4cos3Thus,

4

22

2

2

2

2

22

22

tdx

dy

tdx

dy

ttt

t

dx

dy

tt

t

dx

dy

ttt

tt

dx

dy

ttttt

ttttt

dt

dy

21. Consider the differential equation,

yxdx

dy43log

Taking exponent on both sides, we have

dxee

dy

eedx

dy

edx

dy

ee

x

y

yx

yx

yxdx

dy

3

4

43

43

43log

.

Integration in both the sides, we have

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CBSE

Cee

dxee

dy

xy

x

y

34

34

3

4

We need to find the particular solution.

We have, y = 0, when x = 0

12

7

12

43

3

1

4

1

3

1

4

1

C

C

C

Thus, the solution is 12

7

43

43

yx ee

22. The equation of line L1:

)1....(2

3

7

2

3

1

2

3147

3

1

z

p

yx

z

p

yx

The equation of line L2:

)2...(5

6

1

5

7

3

1

5

6

1

5

3

77

zy

p

x

zy

p

x

Since line L1 and L2 are perpendicular to each other, we have

7

7010

1077

9

052177

33

p

p

pp

pp

Thus equations of lines L1 and L2 are:

5

6

1

5

3

1

2

3

1

2

3

1

zyx

zyx

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CBSE

Thus the equation of the line passing through the point (3, 2, -4) and parallel to the line L1 is:

2

4

1

2

3

3

zyx

SECTION – C

23. Equation of the plane passing through the intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5

is:

(x + y + z – 1) + λ (2x + 3y + 4z – 5)

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

This plane has to be perpendicular to the plane x – y + z = 0

Thus,

( 1 +2 λ) x + (1 + 3 λ)(-1) + ( 1 + 4 λ)1 = 0

1 +2 λ – 1 – 3 λ + 1 + 4 λ = 0

1 + 3 λ = 0

λ = 3

1

Thus, the equation of the plane is:

22

2

101

)2(

222

OR

Any point in the line is

2 + λ, -4 + 4 λ, 2 + 2 λ

The vector equation of the plane is given as

0ˆˆ2ˆ. kjir

Thus the cartesian equation of the plane is x – 2y + z = 0

Since the point lies in the plane

(2 + 3 λ)1 + (-4 + 4 λ)(-2)+(2 + 2 λ) = 0

⟹ 2 + 8 +2 + 3 λ – 8 λ + 2 λ = 0

⟹12 - 3 λ

⟹ λ = 4

Thus, the point of intersection of the line and the plane is: 2 + 3 x 4, -4 + 4 x 4, 2 + 2 x 4

⟹ 14, 12, 10

Distance between (2, 12,5) and (14, 12, 10)is:

units 13

169

25144

5101212214222

d

d

d

d

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24. Consider the vertices, A(-1, 2), B(1, 5) and C(3, 4)

Let us find the equation of the sides of ∆ABC.

Thus, the equation of AB is:

0723

11

1

25

2

yx

xy

Similarly, the equation of BC is:

0112

13

1

54

5

yx

xy

Also, the equation of CA is:

052

31

3

42

4

yx

xy

Now the area of ABC=Area of ∆ADB + Area of ∆BDC

1

12

5

2

73ADB of Area dx

xx

Similarly, Area of ∆ADB =

3

12

5

2

11dx

xx

Thus, Area of ∆ADB + Areaof ∆BDC

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CBSE

units square 4

2

5

2

92

2

13

2

992

23

2

31

2

26

2

22

2

5

2

11

2

5

2

73

3

1

21

1

2

3

1

1

1

1

1

3

1

1

1

3

1

xxx

x

dxxdxx

dxx

dxx

dxxx

dxxx

25. Let x be the number of pieces manufactured of type A and y be the number of pieces manufactured

of type B. Let us summarise the data given in the problem as follows:

Product Time for

Fabricating

(in hours)

Time for

Finishing (in

hours)

Maximum labour

hours available

Type A 9 1 180

Type B 12 3 30

Maximum

Profit (in

Rupees)

80 120

Thus,the mathematical form of above L.P.P. is

Maximise Z = 80x + 120y subject to

9x + 12y ≤ 180

x + 3y ≤ 30

Also, we have x≥ 0, y ≥ 0

Let us now find the feasible region, which is the set of all points whose coordinates satisfy all

constraints. Consider the following figure.

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CBSE

Thus, the feasible region consists of the points A, B and C. The values of the objective function at the

corner points are given below in the following table:

Points Value of Z

A(12, 6) Z = 80 x 12+120 x 6 = Rs. 1680

B(0, 10) Z = 80 x 0+12 x 10 = Rs. 1200

C(20, 0) Z = 80 x 20+120 x 0 = Rs. 1600

Clearly,Z is maximum at x = 12 and y = 6 and the maximum profit is Rs. 1680.

26. Let E1 ,E2 ,E3 and A be the events defined as follows:

E1 = Choosing 2 headed coin

E2 = Choosing coin with 75% chance of getting heads

E3 = Choosing coin with 40% chance of getting heads

A= Getting heads

Then P(E1) = P(E2) = P(E3) = 3

1

Also, P(A/E1) = 1, P(A / E2) = 4

3

100

75 , P(A / E3) =

5

2

100

40

Required probability

= P (E1 / A)

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CBSE

43

20

60

433

1

15

2

4

1

3

13

1

5

2

3

1

4

3

3

11

3

1

13

1

///

/)(

332211

11

EAPEPEAPEPEAPEP

EAPEP

OR

If 1 is the smallest number,

the other numbers are:2,3,4,5,6

If 2 is the smallest number,

the other numbers are:3,4,5,6

If 3 is the smallest number,

the other numbers are:4,5,6

If 4 is the smallest number,

the other numbers are:5,6

If 5 is the smallest number,

the other number is:6

Thus, the sample space is S =

56

46,45

36,35,34

26,25,24,23

16,15,14,13,12

Thus, there are 15 set of numbers in the sample space.

Let X be

X : 2 3 4 5 6

P(X) = 15

5

15

4

15

3

15

2

15

1

We know that,

E(X) = XiP(Xi)

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66.4

15

70

15

30201262

15

56

15

45

15

34

15

23

15

12

27. From the given data, we write the following equations:

900

1

1

1

2300

3

1

4

1600

1

2

3

zyx

zyx

zyx

From above system, we get:

3x+ 2y + z = 1600

4x + y + 3z = 2300

x + y + z = 900

Thus we get:

900

2300

1600

111

314

123

z

y

x

AX = B, where

900

2300

1600

and ;

111

314

123

B

z

y

x

XA

05326141342313

111

314

123

A

We need to find A-1 :

C11 = -2; C12 = -1; C13 = 3

C21 = -1; C22 = 2; C23 = -1

C31 = 5; C32 = -5; C33 = -5

2014

CBSE

Therefore, adj A =

513

521

512

555

121

312T

Thus, A-1 =

513

521

512

5

1

A

adjA

Therefore, X = A-1B

400

300

200

2000

1500

1000

5

1

90052300116003

90052300216001

90052300116002

5

1

900

2300

1600

513

521

512

5

1

z

y

x

z

y

x

z

y

x

z

y

x

Awards can be given for discipline.

28. Let ∆ABC be the right angled triangle with base b and hypotenuse h.

Given that b + h = k

Let A be the area of the right triangle.

4

22

4

24

4

4

1

2

1

32222

22

2222

2

222

2

2222

22

kbkbAkbk

bA

bkbbkb

A

bkhbbkb

A

bhbA

bhbA

Differentiating the above function w.r.t.x, we have

2014

CBSE

A

kbbk

db

dA

kbbk

db

dAA

2

3

)1....(4

622

22

22

For the area to be maximum, we have

3

3

03

0

2

22

kb

bbk

kbbk

db

dA

Again differentiating the function in equation (1), with respect to b, we have

)2....(4

12222

2

2

22kbk

db

AdA

bd

dA

Now substituting db

dA=0 and b =

3

kin equation (2), we have

04

22

12

1262

4

3122

2

2

2

2

2

2

2

22

2

2

2

2

2

A

k

db

Ad

k

db

AdA

kk

db

AdA

kkk

db

AdA

Thus area is maximum at 3

kb .

Now, h = k 3

2

3

kk

Let θ be the angle between the base of the triangle and the hypotenuse of the right triangle.

Thus, cosθ = 2

1

3

23 k

k

h

b

32

1cos 1

2014

CBSE

29. We need to evaluate xxxx

dx4224 coscossinsin

Let I = xxxx

dx4224 coscossinsin

Multiply the numerator and the denominator by sec 4x, we have

1tantan

secsec

1tantan

sec

24

24

24

4

xx

xdxxI

xx

xdxI

We know that sec2x = 1 + tan2x

Thus,

1tantan

sectan124

22

xx

xdxxI

Now substitute t = tanx; dt = sec2xdx

Therefore,

42

2

1

1

tt

dttI

Let us rewrite the integrand as

11

1

1

122

2

42

2

tttt

t

tt

t

Using partial fractions, we have

11

1

1

11

1

1

11

11

1

1

111

1

22

23

42

2

22

223223

42

2

22

22

42

2

2242

2

tttt

DBDCBAtDCBAtCAt

tt

t

tttt

DDtDtCtCtCtBtBtAtAtAt

tt

t

tttt

ttDCtttBAt

tt

t

tt

DCt

tt

BAt

tt

t

So we have,

A + C = 0; A + B – C + D = 1; A + B+ C – D = 0; B + D = 1

Solving the above equations, we have

2014

CBSE

A = C = 0 and B = D = 2

1

3

1tan2tan

3

1

3

12tan

3

1

4

3

2

1

tan

4

3

1

2

1

4

3

2

12

1

4

11

4

12

1

12

1

:Consider

12

1 and

12

1 Where,

1212

1/

1

12

1

1

1

1

1

1

2

2

21

1

2221

22

22

22

42

2

x

t

t

t

dt

tt

dt

tt

dtI

I

tt

dtI

tt

dtI

II

tt

dt

tt

dt

tttt

tt

dttI

Similarly,

Consider I2:

2014

CBSE

4

3

2

12

1

4

11

4

12

1

12

1

2

2

22

t

dt

tt

dt

tt

dtI

Cxx

I

xxI

III

x

t

t

3

1tan2tan

3

1tan2tan

3

1

3

1tan2tan

3

1

3

1tan2tan

3

1

Thus,

3

1tan2tan

3

1

3

12tan

3

1

4

3

2

1

tan

4

3

1

2

1

11

11

21

1

1

1

2014