topper 12 sample paper

7/29/2019 topper 12 sample paper

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MATHEMATI CS

Class: XI ISa m p l e Pa p er - 5

Time 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 of

07 questions of six marks each.3. All questions in Section A are to be answered in one word, one sentence or

as per the exact requirement of the question.4. There is no overall choice. However, internal choice has been provided in 04

questions 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.

SECTI ON 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 ( + + k ) 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 the

curve cuts y-axis.


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7. Evaluate:1

1

2 xlog dx

2 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

SECTI ON 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 shuffled

deck of 52 cards . What is the probability that


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

21 2 1 2 516.Solve the equation:(tan ) (cot )

8x 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 cos xdx

9 16 sin2x

12-3x

019.Evaluate e dx as limit of sum

20. Ifcos sin

sin cosA

then prove thatcos sin

,sin cos

nn n

A n Nn 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

11

32

dx

Evaluatex x

Sect ion C

23.Show that the right circular cone of least curved surface and given volume

has an altitude equal to 2 times the radius of the base.OR

4 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


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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 =a

a 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

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. In

how 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(X6) (iv) P(1 X


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Graph should have no vertical lines

3.Two vectors anda b

are orthogonal/perpendicular, ifa.b 0

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

2 2 3 0

5

2

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

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

1

3

(1 mark)

o o

2 2 2 2

o

1 15.Let cos45 , m cos60

22

1 1 1 1Also, m n 1 n 1 n .

2 4 4 2

1Let n cos

2

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

2

120 [1mark]


6/26

3

2

2

x 0x 0

6.y x x 1

dy3x 1

dxAt a point where the curve cuts y-axis, x = 0

dy3x 1 1 (1 Mark)

dx

1

1

1

1

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

2 x

2 xLet f(x) log and

2 x

2 x 2 x

f( 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 cos

10. 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 cos

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

Sect ion B


7/26

3

5 3

x 5 x11.Let y =

7 3x 8 5x

Taking log on both sides , we get1 1

log 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 . .5

y dx x 2 5 x 7 3x 3 8 5x

1 dy 3 1 15 5. (1 M

y dx x 2 5 x 7 3x 3 8 5x

3

5 3

ark)

dy 3 1 15 5y

dx 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)


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13. Distance between the parallel planes

d-kisgiven 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 3k

3

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 is

x y z1

a 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 1

a b c

1 1 1

a b

2

2 2 2

2 2 2 2

=p (2 Marks)

c

1 1 1 1

p a b c

1 1 1 1(2 Marks)

a b c p


9/26

15.This is a case of bernoulli trials.

13 1p = P(Success) = P(getting a spade in a single draw ) =

52 41 3q P(Failure) 1 p 1 (1 Mark)

4 4

(i)All the

4

4 4 04

4 3 13

4

4 0 40

1 1four 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 q

4 256

(1 Mark)

16.Here,

21 2 1 2 5

(tan ) (cot ) 8x x

2

1 1 2 1 1

2 21 1

2 21 1

2 21 1

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

8

52 tan x.cot x (1mark)

2 8

52tan x.cot x

4 8

(2 5) 32 tan x.cot x (1mark)

8 8

21 1

21 1 2

1 2 1 2

32tan x. tan x 0

2 8

3tan x 2(tan x) 0

8

16(tan x) 8 tan x 3 0

(1 mark)


10/26

2 21

21

1

1

8 64 4 16 3tan x

2 16

8 256 8 16tan x32 32

3tan x and

4 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 cott

dxdx a3asin t.costdt

(1Mark)

Slope of the tangent at2

t

2

dy bcot 0

dx a 2

( 1 Mark)

Hence, equation of tangent is given by

3y b cos 0 or y 02

(1 mark)


11/26

0

20

sinx cos x18.Let I dx

9 16 sin2x

sinx cos xdx(1 mark)

9 16 1 sinx cosx

Put sinx cos x t

cos x sinx dx dt

For x , t 0 and

For x 0, t 1

(1 mark)

0

21

0

2

1

0

221

2 2

0

1

1

9 16 1

1

25 16

1

5 4

1log

2

1 5 4log40 5 4

1 1log

140

9

1log 9

40

I dtt

dtt

dtt

dx a xc

a x a a x

tt

(2 Marks)


12/26

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

-3

2

-3hh 0

-3

2

-3hh 0

-32

-3hh 0

he 1 e e ... e

1 elimhe ,but nh = 1 (1Mark)

1 e

1 elimhe

1 e1 e

e lim1 e

h

1 eelim

-3 1 e

3h

2-3

-3h3h 0

2 2 1-3

(1/2Mark)

e 11 e lim

-3 e 13h

e e e1 e (1Mark)

-3 3

2 0 : We shall prove by principle of mathematical induction

Here, letcos sin

sin cos

nn n

An n

cos sin

: ,sin cos

nn n

P n An n

( mark )

1cos sin

So, Asin cos

1 is true.P

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

cos sin

:sin cos

kk k

P k Ak k


13/26

( 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

(mark )

cos cosk sin sink cos sink sin cosk

sin cosk cos sink sin sink cos cosk

cos k sin ksin k cos k

cos k 1 sin k 1

sin k 1 cos k 1

1P k is true.

(1mark )

Thus by principle of mathematical induction

cos sinsin cos

n n nAn n

for all n N

(1 mark )OR


14/26

2

2

2

1 1 2 3

2 2

2 2

2

2

1

1

1C C C C

1

1 1 (1Mark)

1 1

But, is one of the cube roots of unity

1 0

2

2

(1Mark)

0

0 1 (1mark)

0 1

Since I column of the determinant is zero therefore, value of the given determinant is

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 continuous

x ,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


15/26

x c x c

h 0

h 0

h 0 h 0

h 0

Domainof logx is(0, )

f(x) logx,x (0, )

Let c (0, )limf(x) limlogx

limlog c h ,c 0 (1Mark)

hlimlog c 1 ,c 0

c

hlimlog c limlog 1

c

hlog c limlog 1

c

log c lim

h 0

h 0 h 0

x 0

h 0

h 0

hlog 1

hc. (1Mark)

h cc

hlog 1

hclog c lim .lim

h cc

log(1 x)But lim 1

x

hlog 1c

lim 1h

c

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

c

f(c)

logarithmic function

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


16/26

2

2

2 2

2

(3 sin 2)cos22. Let I = d

5 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

BA 3,B 4 (1Mark)

(t 2)

3 4I dt

t 2 (t 2)

3 4I dt dt (1Mark)

t 2 (t 2)

43log sin 2 C (1Mark)

sin 2

OR

11

32

1 1

3 6

16 56

5 3

2

3

3

2

dxLet I

x 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 1

3 6 6

t)

dt6 t t 1 dt 6 (1Mark)

(1 t)

t t6 t 6log 1 t

3 2

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


17/26

Sect ion C

2 3 .

Here, Volume V of the cone is1 3V2 2

V r h r

3 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

2

1 S S

, then by (1)2

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

Differentiating S1 with respect to h, we get

2213 9

3

dS

V V

dh h

(1 mark)


18/26

dS1 0 for maxima/minima

dh

223 V 9V 03

h

223 V 9V

3h

6V3h

(1 mark)2 2d S 54V1

42hdh

2d S 6V310 at h

2 2dh

Therefore curved surface area is minimum at3

hV

6

3h 1 2 2 2

Thus, r h h 2r6 3

h 2r

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


19/26

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

(i) x = -1 :

When x

'

'

'

0

When 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) x

72

When x< ; f (x)>072

When x> ; f (x)


20/26

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 -3R

1 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 - 2R

1 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 31

R R2

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


21/26

1 1 1

1 0 0 2 2 2

0 1 2 1 0 0

0 0 1 5 3 1

2 2 2

.(1/2 mark)

Applying 2 2 3R R - 2R

1 1 1

1 0 0 2 2 2

0 1 0 4 3 1 A

0 0 1 5 3 1

2 2 2

.(1/2 mark)

Hence1

1 1 1

2 2 2

A 4 3 1

5 3 1

2 2 2

. (1 mark)25.


22/26

2 2x y

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

a b

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

20 0a a 2 2a0

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 4a0

1ab

4Ratio (2Marks)2 2

ab4


23/26

OR

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

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

4

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

2 2x ax 2a 0

x 2a x a 0

x a, 2a

x a (1Mark)2 2

y 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

a2a

3/2 21/2x x 2a x2 2 12 a 2 2a x sin

3 2 2 2a0

2 0

4 3a a

3

2a1/2 x/2 2 2 2 1

x 2a x 2a sin2a 0

24a 2

a sq units (3Marks)3


24/26

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 get

1 1 1 1x(1 2 ) y(1 3 ) z(1 ) ( 1 4 ) 02 2 2 2

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

2 2

y 3z 6 0 y 3z 6

0 (2Marks)

2 7 .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

20

1log20

...(2)t k t t

k

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)


25/26

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 constraints

6x + 10y 60 3x +5y30

4x+4y 32 x+ y 8

And

x0,y0..(2 Mark)

Feasible region is shown shaded ..

(2 Marks)


26/26

This region is unbounded.corner points objective funcA( 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 0

10k(k 1) (k 1) 0

(k 1)(10k 1) 0

1k 1,k

10

(1Mark)

k,also being a probability cannot be negative

1

k (1Mark)103

(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 10 100

3(iii)P(1 X 3) P(1) P(2) k 2k 3k (1Mark)

10

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