MATHEMATICSClass: XII

Sample Paper -5Time Allowed : 3 Hrs Maximum Marks: 100

_______________________________________________________________1. All questions are compulsory.2. The question paper consist of 29 questions divided into three sections A, B

and C. Section A comprises of 10 questions of one mark each, section Bcomprises of 12 questions of four marks each and section C comprises of07 questions of six marks each.

3. All questions in Section A are to be answered in one word, one sentence oras per the exact requirement of the question.

4. There is no overall choice. However, internal choice has been provided in 04questions of four marks each and 02 questions of six marks each. Youhave to attempt only one of the alternatives in all such questions.

5. Use of calculators is not permitted.

SECTION – A

1. Evaluate 1 1sin sin3 2

2.This graph does not represent a function . Make the required changes inthis graph, and draw the graph , so that it does represent a function.

3. For what value of , are the vectors 2a i j k and 2 3b i j k

orthogonal?

4. Find the value of ‘’ for which (î + ĵ + k̂ ) is a unit vector.

5.Find the Obtuse angle of inclination to the Z-axis of a line that is inclined toX-axis at 45o and to Y-axis at 60o.6.Find the slope of the tangent to the curve 3 1y x x at the point where thecurve cuts y-axis.

7. Evaluate:1

1

2 xlog dx2 x

.

8.This 3 x 2 matrix gives information about the number of men and womenworkers in three factories I, II and III who lost their jobs in the last 2months. What do you infer from the entry in third row and second column ofthis matrix?

Men workers Women workers

Factory I 40 15

Factory II 35 40

Factory III 72 64

9. If A and B are two matrices of the same order ,under what conditions is(A-B)(A+B) = A2 - B2

x sin cos

10.Evaluate the determinant : sin 1

cos 1

x

x

SECTION – B

3

5 3

x 511. Differentiate , w . r .t x

7 3 8 5

x

x x

OR2 '' '11.If y = a cos (log x) + b sin (log x) , prove that x 0 ,y xy y

12. ( ) 3,g(x) = x -3; ,

Show that (i) f is not an onto function (ii) gof is an onto function

Let f x x x N

13.Find the distance between the parallel planes

. 2 1 3 4and . 6 3 9 13 0r i j k r i j k

2 2 2 2

14. plane is at a distance of p units from the origin.

It makes an intercept of a,b,c with the x , y and z axis repectively.Show that it satisfies the equation:

1 1 1 1

a

A

b c p

15. Four cards are drawn successively with replacement from a well shuffleddeck of 52 cards . What is the probability that

(i) All the four cards are spades?(ii) Only 3 cards are spades(iii) None is a spade

21 2 1 2 5

16.Solve the equation:(tan ) (cot )8

x x

17. Find the equation of tangent to the curve given by3 3sin , cosx a t y b t at

a point, where2

t

.

18. Evaluate:4

0

sinx cosxdx9 16sin2x

12-3x

0

19.Evaluate e dx as limit of sum

20. Ifcos sin

sin cosA

then prove thatcos sin

,sin cos

n n nA n N

n n

OR

2

2

2

is one of the cube roots of unity , evaluate the given determinant

1

1

1

If

21. Show that the function f defined by f(x) = 1-x + x , x R is continous .

ORShow that a logarithmic function is continuous at every point in its domain.

2

(3 sin 2) cos22.Evaluate

5 4sind

cos

OR

1132

dxEvaluate

x x

Section C

23.Show that the right circular cone of least curved surface and given volumehas an altitude equal to 2 times the radius of the base.

OR4 3.Find the points at which the function f given by f(x) = (x - 2 ) ( 1)

(i)local maxima (ii) local minima (iii) point of inflexion .

Also find the (iv)local maximum value and the (v) local m

x has

inimum value

24. Obtain the inverse of the following matrix using elementary operations.

0 1 2

A = 1 2 3

3 1 1

2 2

2 2

2 2

2 2

x y25. Calculate the area (i) between the curves + 1,and the x-axis between x = 0 to x =aa b

x y(ii) Triangle AOB is in the first quadrant of the ellipse + 1,where OA = a and OB = b.a b

Fi

nd the area enclosed between the chord AB and the arc AB of the ellipse (iii) Find the ratio of the two areas found .

OR2 2 2 2Find the smaller of the two areas in which the circle x y 2a is divided by the parabola y ax,a 0

26.Find the equation of a plane that is parallel to the X-axis and passes through the line

common to two intersectiing planes r. i+j+k 1 0and r. 2i+3j-k 4

27.In a bank, principal increases continuously at the rate of 5% per year. Inhow many years will Rs 1000 double itself?

28. Two trainee carpenters A and B earn Rs 150 and Rs 200 per day respectively. A can make 6frames and 4 stools per day while B can make 10 frames and 4 stools per day . How many daysshall each work , if it is desired to produce atleast 60 frames and 32 stools at a minimum

labour cost ? Solve the problem graphically.

29.A random variable X has the following probability distribution :X 0 1 2 3 4 5 6 7P(X) 0 k 2k 2k 3k k2 2k2 7k2 +k

Determine(i) k (ii) P(X<3) (iii) P(X>6) (iv) P(1 X<3)

1 11. sin sin sin3 2 3 6

sin sin 1 (1 Mark)3 6 2

2. If a vertical line intersects the graph of a relation in two or more points,then the relation is not a function.

Graph should have no vertical lines

3.Two vectors anda b

are orthogonal/perpendicular, if a.b 0

(2i j k).(i 2j 3k) 02 2 3 0

52

(1 Mark)4. Magnitude of : (î + ĵ + k )=1 , for it to be a unit vector.

2 2 2 2.( i+ j+k ) 3 3 1

13

(1 mark)o o

2 2 2 2

o

1 15.Let cos45 , m cos6022

1 1 1 1Also, m n 1 n 1 n .2 4 4 2

1Let n cos2

1For to be obtuse , we take the negative value of n2

120 [1mark]

3

2

2

x 0x 0

6.y x x 1dy 3x 1dxAt a point where the curve cuts y-axis, x = 0dy 3x 1 1 (1 Mark)dx

1

1

1

1

2 x7.I log dx ...(1)2 x

2 xLet f(x) log and2 x

2 x 2 xf( x) log log f(x).2 x 2 x

f(x) is an odd function.2 xThus, log dx 0 (2 x

1 Mark)

8. 64 Women lost their jobs in the Factory III in the last two months. (1Mark)

9. (A-B)(A+B) = A2 +AB –BA - B2== A2 - B2 iff AB = BA (1 Mark)

2

3 2 2

3 2 2 3 3

x sin cos10. sin x 1 =x(-x -1 ) - sin ( x sin cos ) cos ( sin x cos )

cos 1 x

x x x sin sin cos sin cos x cosx x x(sin cos ) x x x x (1 Mark)

Section B

3

5 3

x 5 x11.Let y =7 3x 8 5x

Taking log on both sides , we get1 1log y = 3 log x + log 5 x 5log 7 3x log 8 5x (1 Mark)2 3

1 dy 1 1 1 1 1 1. 3. . .1 5. 3 . .5y dx x 2 5 x 7 3x 3 8 5x1 dy 3 1 15 5. (1 My dx x 2 5 x 7 3x 3 8 5x

3

5 3

ark)

dy 3 1 15 5ydx x 2 5 x 7 3x 3 8 5x

dy x 5 x 3 1 15 5 (2 Marks)dx x 2 5 x 7 3x 3 8 5x7 3x 8 5x

OR

'

'' '

y = a cos (log x) + b sin (log x)dy 1 1a sin logx . b. cos logx . (1 Mark)dx x x

dyx xy a sin logx b. cos logx (1 Mark)dx

Differentiating,

xy y a cos

2 ''

2 ''

'

'

1 1logx . b. sin logx . (1 Mark)x x

x y xy a cos logx b. sin logx y

x y xy y 0 (1 Mark)

12.f(x) x 3,x N,Domainof f {1,2,3,...}Range {4,5,6,...} Codomain of f ={1,2,3,...}

f is not an onto function (2 Marks)f(x)=x+3g(x) = x -3

g[f(x)] x 3 -3

=x

Domainof gof {1,2,3,...}Range {1,2,3,4,5,6,...} Codomain of gof ={1,2,3,...}

gof is an onto function (2 Marks)

13. Distance between the parallel planesd-k

isgiven by (1 Mark)n

r. 6i 3j 9k 13 0

13r. 2i j 3k (1 Mark)3

the distance between the given parallel planes

r. 2i 1j 3k

22 2

134 and r. 2i j 3k3

134-3

is (1 Mark)2 1 3

13 254+3 253= (1 Mark)

4 1 9 14 3 14

2 2

14. The equation of the plane in the intercept form isx y z 1a b c

1 1 1x y z 1 0a b c

Plane is at a distance of p units from the origin

1 1 10 0 0 1a b c

1 1 1a b

2

2 2 2

2 2 2 2

=p (2 Marks)

c

1 1 1 1p a b c1 1 1 1 (2 Marks)a b c p

15.This is a case of bernoulli trials.13 1p = P(Success) = P(getting a spade in a single draw ) =52 4

1 3q P(Failure) 1 p 1 (1 Mark)4 4

(i)All the

44 4 0

4

4 3 13

44 0 4

0

1 1 four cards are spades =P(X =4) = C p q (1 Mark)4 256

12 3(ii)Only 3 cards are spades=P(X=3)= C p q (1 Mark)256 64

3 81(iii)None is a spade=P(X=0) = C p q4 256

(1 Mark)

16.Here,2

1 2 1 2 5(tan ) (cot )

8x x

21 1 2 1 1

2 21 1

2 21 1

2 21 1

5(tan x cot x) 2 tan x.cot x8

52 tan x.cot x (1mark)2 8

52 tan x.cot x4 8(2 5) 32 tan x.cot x (1mark)

8 8

21 1

21 1 2

1 2 1 2

32tan x. tan x 02 8

3tan x 2(tan x) 08

16(tan x) 8 tan x 3 0

(1 mark)

2 21

21

1

1

8 64 4 16 3tan x2 16

8 256 8 16tan x32 32

3tan x and4 4

x so tan x2 2 4

x tan4

x 1

(1 mark)

17.Here, 3 3sin , cosx a t y b t …(1)Differentiating (1) w.r.t. t

2dx 3asin t.cos tdt

and

2dy 3bcos t.sintdt

(1 Mark)

2

2

dydy 3bcos t.sint bdt cot t

dxdx a3asin t.cos tdt

(1Mark)

Slope of the tangent at2

t

2

dy b cot 0dx a 2

( 1 Mark)

Hence, equation of tangent is given by3y bcos 0 or y 02

(1 mark)

0

20

sinx cosx18.Let I dx9 16sin2x

sinx cosx dx(1 mark)9 16 1 sinx cosx

Put sinx cosx tcosx sinx dx dt

For x , t 0 andFor x 0, t 1

(1 mark)

0

21

0

21

0

221

2 2

0

1

1

9 16 1

1

25 16

1

5 4

1log

2

1 5 4log

40 5 4

1 1log

1409

1log 9

40

I dtt

dtt

dtt

dx a xc

a x a a x

t

t

(2 Marks)

1

2-3. 0 2-3. 0+h 2-3. 0+2h 2-3. 0+(n-1)h2-3x

h 00

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

h 0

h

119. Here a =0, b=1 , b-a =1-0 = 1 h (1/2Mark)n

e dx limh e e e ... e

limh e e e ... e (1Mark)

lim

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

0

n-3h

2-3hh 0

-32

-3hh 0

-32

-3hh 0

-32

-3hh 0

he 1 e e ... e

1 elimhe ,but nh = 1 (1Mark)

1 e1 e

limhe1 e

1 ee lim

1 eh

1 ee lim-3 1 e

3h

2-3

-3h3h 0

2 2 1-3

(1/2Mark)

e 11 e lim-3 e 1

3he e e1 e (1Mark)-3 3

20: We shall prove by principle of mathematical induction

Here, letcos sin

sin cosn n n

An n

cos sin

: ,sin cos

n n nP n A

n n

(½ mark )

1 cos sinSo, A

sin cos

1 is true.P

(½ mark )Assuming result to be true for n = k i.e P(k) to be true

cos sin

:sin cos

k k kP k A

k k

(½ mark)

We have to prove 1P k is true,

1 1

1

1 :

cos sin cos sin

sin cos sin cos

k k

k

P k A A A

k kA

k k

(½mark )

cos cosk sin sink cos sink sin cosksin cosk cos sink sin sink cos cosk

cos k sin ksin k cos k

cos k 1 sin k 1sin k 1 cos k 1

1P k is true.(1

mark )

Thus by principle of mathematical inductioncos sin

sin cosn n n

An n

for all n N

(1 mark )OR

2

2

2

1 1 2 3

2 2

2 2

2

2

11

1

C C C C

11 1 (1Mark)1 1

But, is one of the cube roots of unity1 0

2

2

(1Mark)

00 1 (1mark)0 1

Since I column of the determinant is zero therefore, value of the given determinant is zero (1Mark)

21. f(x) = 1-x + x , x R

Let g(x) =1-x+ x ,x R

h(x) x ,x R

h[g(x)]=h[1-x+ x]

1-x + x f(x) (1Mark)

1 x,being a polynomial function is continuousx ,being a modulus function is

continuous

g(x) =1-x+ x ,x R is a continuous function (1Mark)

Again , h(x) = x ,is a continuous function (1Mark)

f(x)iscontinuous (1Mark)

OR

x c x c

h 0

h 0

h 0 h 0

h 0

Domainof logxis(0, )f(x) logx,x (0, )Let c (0, )limf(x) limlogx

limlog c h ,c 0 (1Mark)

hlimlog c 1 ,c 0c

hlimlog c limlog 1c

hlog c limlog 1c

log c lim

h 0

h 0 h 0

x 0

h 0

h 0

hlog 1hc . (1Mark)

h cc

hlog 1hclog c lim .lim

h cc

log(1 x)But lim 1x

hlog 1clim 1

hc

hlog c 1.lim log c 1.0 logc (1Mark)c

f(c)logarithmic function

is continuous at every point in its domain (1Mark)

2

2

2 2

2

(3 sin 2)cos22. Let I = d5 cos 4sin

Let t sin dt cos d(3 t 2)I dt

5 (1 t ) 4t(3 t 2) (3 t 2)I dt dt (1Mark)

t 4t 4 (t 2)Using the method of Partial fractions(3 t 2) A

t 2(t 2)

2

2

2

B A 3,B 4 (1Mark)(t 2)

3 4I dtt 2 (t 2)

3 4I dt dt (1Mark)t 2 (t 2)

43log sin 2 C (1Mark)sin 2

OR

1132

1 13 6

16 56

5 3

2

3

3

2

dxLet Ix x

dx

x 1 x

Let x t x t dx 6t dt (1Mark)6t dt 6t dtI

(1 t)t (1 t)

t 1 1 dt6

(1 t)

t 1 dt dt6 6 (1Mark)(1 t) (1 t)

t 1 t t 1 dt dt6 6(1 t) (1

2

3 2

1 1 13 6 6

t)dt6 t t 1 dt 6 (1Mark)

(1 t)

t t6 t 6log 1 t3 2

2 x 3x 6x 6log 1 x C (1Mark)

Section C

23.

Here, Volume ‘V’ of the cone is1 3V2 2V r h r3 h

...(1)

(1 mark)

Surface area ‘S’ 2 2S rl r h r ...(2)Where h= height of the cone

r= radius of the cone

l= Slant height of the cone(1 mark)

2 2 2 2 2S r h r by (2)

Let 21 S S , then by (1)

23 V 3V 9V2S h 3 Vh1 2h h h

Differentiating S1 with respect to h, we get221 3 9 3

dS

V Vdh h

(1 mark)

dS1 0 for maxima/minimadh

223 V 9V 03h

223 V 9V 3h

6V3h

(1 mark)2 2d S 54V1

42 hdh2d S 6V31 0 at h2 2dh

Therefore curved surface area is minimum at3h

V6

3h 1 2 2 2Thus, r h h 2r6 3

h 2r

Hence for least curved surface the altitude is 2 times radius.(2 marks)

4 3

' 4 2 3 3

3 2

3 2

3 2

'

23. f(x) = (x - 2 ) (x 1)f (x) =3(x - 2 ) (x 1) 4(x 1) (x - 2 )

(x - 2 ) (x 1) 3(x - 2 ) 4(x 1)

(x - 2 ) (x 1) 3x-6 4x 4

(x - 2 ) (x 1) 7x-2

f (x) =0 (x

3 2

'

'

2 - 2 ) (x 1) 7x-2 x 1, ,2 (1Mark)7

Let us examine the behaviour of f (x) , slightly to the left and right of each of these three values of x(i) x = -1 :When x<-1; f (x)>

'

'

'

0When x>-1; f (x)>0

x =-1 is neither a point of local maxima nor minima (1Mark)It may be a point of inflexion

2(ii) x7

2When x< ; f (x)>072When x> ; f (x)<07

2x= is a point of local maxima7

4 3 8 104 3

7

8 10

7

'

'

(1Mark)

2 2 2 12 9 2 .3f ( - 2 ) ( 1)7 7 7 7 7 7

2 .3The local maximum value is (1Mark)7

(iii)x 2When x<2; f (x)<0When x>2; f (x)>0

4 3

x 2is a point of local minima (1Mark)f(2) =(2 - 2 ) (2 1) =0

The local minimum value is 0 (1Mark)

24.Since A = IA

0 1 2 1 0 0

1 2 3 0 1 0 A

3 1 1 0 0 1

.........(1/2Mark)

Applying 1 2R R

1 2 3 0 1 0

0 1 2 1 0 0 A

3 1 1 0 0 1

…………………….(1 mark)

Applying 3 3 1R R -3R1 2 3 0 1 0

0 1 2 1 0 0 A

0 5 8 0 3 1

…………….. (1/2 mark)

Applying 1 1 2R R - 2R1 0 1 2 1 0

0 1 2 1 0 0 A

0 5 8 0 3 1

…………. (1/2mark)

Applying 3 3 2R 5R R

1 0 1 2 1 0

0 1 2 1 0 0 A

0 0 2 5 3 1

………………… (1/2 mark)

Applying 3 3

1R R

2

1 0 1 2 1 0

0 1 2 1 0 0 A

0 0 1 5 3 1

2 2 2

……………………(1 mark)

Applying 1 1 3R R + R

1 1 11 0 0 2 2 20 1 2 1 0 0

0 0 1 5 3 1

2 2 2

……………………….(1/2 mark)

Applying 2 2 3R R - 2R1 1 1

1 0 0 2 2 20 1 0 4 3 1 A

0 0 1 5 3 1

2 2 2

………………….(1/2 mark)

Hence 1

1 1 1

2 2 2A 4 3 1

5 3 1

2 2 2

………………. (1 mark)25.

2 2x y(i) between the curves + 1, and the x-axis between x = 0 to x =a2 2a b

a2 2 2 2a ax b b a -x a2 2 1b 1- a -x sin ....(1Mark)20 0a a 2 2a

0

b 2 1 2 10 a sin (1) 0 a sin (0)2a

b 2a .2a 2

1

4

x xdx dx

a

ab .....(1Mark)

2 2x y(ii) Area of Triangle AOB is in the first quadrant of the ellipse + 1,where OA = a and OB = b.2 2a b =the area enclosed between the chord AB and the arc AB of the ellipse .

= Area of Ellip

2a 2 abx 1 1 1se ( In quadrant I)- Area of AOB= b 1- dx ab ab ab (2Marks)2 2 4 2 4a01 ab4Ratio (2Marks)

2 2ab4

OR

2 2 2The circle is x y 2a C(O, 2a)

12 2The parabola is y ax, a 0 y 4 ax, a 04

2 2Their point of intersection is given by :x ax 2a2 2x ax 2a 0

x 2a x a 0

x a, 2a

x a (1Mark)2 2y a y a (1Mark)

sh

ade region is the smaller of the two areas in which the circle is divided by the parabola

1/2a 2a 2 2A=2 axdx 2a x dx (1Mark)0 0

a2a3/2 21/2x x 2a x2 2 12 a 2 2a x sin

3 2 2 2a 02 0

4 3a a3

2a1/2 x/2 2 2 2 1x 2a x 2a sin

2a 024a 2a sq units (3Marks)

3

26.r. i+j+k 1 0and r. 2i+3j-k 4

x y z 1 0 and 2x 3y z 4 0The required plane passes through the line common to two intersectiing planes

x y z 1 k(2x 3y z 4) 0

(1Mark)x(1 2k) y(1 3k) z(1 k) ( 1 4k) 0......(1)

The required plane is parallel to the X-axis whose d.cs are 1,0,0(1Mark)1.(1 2k) 0.(1 3k) 0.(1 k) 0 (1Mark)

(1 2k

1) 0 k (1Mark)2

Substituting in (1),we get1 1 1 1x(1 2 ) y(1 3 ) z(1 ) ( 1 4 ) 02 2 2 2

1 3x(0) y( ) z( ) 3 02 2

y 3z 6 0 y 3z 6

0 (2Marks)

27.Let P be the principal. According to the given problem,5

100 20

dP PP

dt

…(i)

Separating the variables in equation (i), we get

20

dP dt

P

(1 mark)

1 1 1

20 20 20

Integrating both sides, we get

201

log20

...(2)t k t tk

dP dt

P

P t k

P e e e Ce

(2marks)

1

20

Now, 1000, when 0

Substituting the values of P and t in (2), we get C = 1000

1000t

P t

P e

(1mark)

1

20

1

20

Let years be the time required to double the principal. Then

2000 1000

2

20log 2

t

t

e

t

e

e

t

………………………….(2 marks)

28.Let the two carpenters work for x days and y days respectively,then our problem is to minimise the objective functionC= 150 x + 200 ySubject to the constraints6x + 10y ≥60 3x +5y≥30

4x+4y≥ 32 x+ y ≥8And

x≥0,y≥0…………………………..(2 Mark)

Feasible region is shown shaded ..

(2 Marks)

This region is unbounded.corner points objective function valuesA( 10,0) 1500E(5,3) 1350.D(0,8) 1600

The labour cost is the least,when carpenter A works for 5 days and carpenter B works for 3 days,

(2 Marks)

7

ii 0

2 2 2

2 2 2

29.(i) P(X ) 1

0 k 2k 2k 3k k 2k 7k k 1 (1Mark)

10k 9k 1 10k 9k 1 0 10k 10k k 1 010k(k 1) (k 1) 0(k 1)(10k 1) 0

1k 1,k10

(1Mark)

k,also being a probability cannot be negative1k (1Mark)10

3(ii)P(X 3) P(0) P(1) P(2) 0 k 2k 3k (1Mark)10

(ii)P(X 6) P(

2 2

2 2 1 1 1 197) P(8) 2k 7k k 2 7 (1Mark)10 10 10 1003(iii)P(1 X 3) P(1) P(2) k 2k 3k (1Mark)

10