sample question paper 2 with solution

Sample Question Paper

MATHEMATICS

Class XII

Time: 3 Hours Max. Marks: 100

General Instructions

1. All questions are compulsory.

2. The question paper consists of 29 questions divided into three sections A, B and C. Section

A comprises of 10 questions of one mark each, section B comprises 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. You have to attempt only one of the

alternatives in all such questions.

5. Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION-A

1. If f(x) is an invertible function, find the inverse of f(x)=3x-2.

2. find the principal value of 1 1sin2

− −

3. If3 4 1

7 4 0 4x y y

x+ −

= − find the value of x and y.

4. A matrix A of order 3X3 has determinant 5. what is the value of 3A .

5 .Find 2

cos 6 ( )3 sin 6x x d xx x++∫

6. Evaluate xxe dx∫ .

Pratima Nayak,KV Teacher

7. If p

is a unit vector and ( ) ( ). 80x p x p+ − =

.

8.write the direction cosine of a line equally inclined the three co-ordinate axes.

9.write the value of the following determinant.

a b b c c ab c c a c ac a a b b c

− − −− − −− − −

10. find the value of p, if ( ) ( )ˆ ˆˆ ˆ ˆ ˆ2 6 27 3 0i j k X i j pk+ + + + =

SECTION-B

11.Let the value of 1 1 11 5 2 12 tan sec 2 tan5 7 8

− − − + +

OR

Solve for x, 1 11 1tan tan , 12 2 2

x x xx x

π− −− + + = ≤ − +

13.Uasing properties of determinants prove that

( )32 2

2 22 2

a b c a ab b c a b a b cc c c a b

− −− − = + +

− −

14.Show that the function f(x)= 2x + is continuous at every x∈R but fails to be

differentiable at x=-2.

OR

Verify lagrange`smean value theorem for the following function f(x)=

[ ]2 2 3 4,6x x for+ +

15.Differential following function w.r.to x

21 1 1tan x

x− + −

OR

Differentiate ( ) ( )sin coscos sinx xx x+ with respect to x


16.Evalute

20

sin1 (cos )

x x dxx

π

+∫

17.Solve the following differential equation

2 2( ) 2 0, (1) 1x y dx xydx y− + = =

18.The volume of a spherical balloon is increasing at the rate of 25 3 / sec.cm Find the rate

of change of its surface area at the instant when its radius is 5cm.

19.Form the differential equation representing the family of parabola having vertex at

origin and axis along positive of X-axis.

20.Find the projection of b c+

on a

when ˆ ˆˆ ˆ ˆ ˆ2 2 , 2 2a i j k b i j k= − + = + −

and

ˆˆ ˆ2 4 .c i j k= − +

21.Find the equation of the plan passing through the points (0,-1,0),(1,1,1) and (3,3,0).

22.The probability of A solving a problem is 3/7 and that B solving it is 1/3 what is the

probability

(i) at least one of them solve the problem?

(ii)only one of them will solve the problem?

SECTION-C

(23) Using matrices solve the following system of linear equations-

2x-y+z=3, -x+2y-z=-4, x-y+2z=1

OR

Using elementary transformations, find the inverse of the following matrix-

−−−

052503231

(24)Determine the points on the curve y =4

2x which are nearest to point (0, 5)

OR


Show that the surface area of a closed cuboid with square base and given volume is

maximum when it is a cube.

(25) Find the area of the region included between parabola y 2 =x and the line x +y =2.

(26) Prove that- ∫−

+−a

a xaxa dx = π a

(27) Mona wants to invest at most Rs.12000 in Saving certificate (SC) and National

saving bonds (NSB).She has to invest at least Rs.2000 in SC and at least Rs.4000 in NSB.

If the rate of interest on SC is 8 pa. and the rate of interest on NSB 10pa, how much

money should she invest to earn maximum yearly income? Also find the maximum

income.

(28) Find the foot of perpendicular drawn from the point A (1, 0, 3) to the join of the

points B (4, 7, 1) and C (3, 5, 3) and also find the perpendicular distance.

(29) Urn A contains 1 white, 2 black, 3 red balls; Urn B Contains 2 white ,1 black ,1red

ball; Urn C contains 4 white ,5 black and 3 red balls. One urn is chosen at random and

two balls are drawn, these happens to be one white and one red. What is the probability

that they come from urn C.

Solution

SECTION-A

1. f(x)=3x-2

Let y =f(x)

y =3x-2

3x+2 =y

1( )f x− =3x+2

2. y = 1 1sin ( 1/ 2) sin (1/ 2) / 6π− −− = − = −

3. 3 4 1

3 4, 1,7 07 4 0 4x y y

x y y xx

+ − = ⇒ + = = − − = −

7, 1x y⇒ = = −

4. 33 3 27 5 135A A= = × =


5. ( )22

cos 6 1 log 3 sin 63 sin 6 6x xI dx x xx x+

= = ++∫ +c

6. 1.x x x x xI xe dx x e dx e dx xe e c= = − = − +∫ ∫ ∫

7. ( ) ( ) 2 2 2. 80 80 81x p x p x p x+ − = ⇒ − = ⇒ = 9x⇒ =

8. 13

l m n= = = ±

9. Taking operation 1 1 2 3C C C C→ + +

a b b c c ab c c a c ac a a b b c

− − −− − −− − −

= 0

10 ( ) ( )ˆ ˆˆ ˆ ˆ ˆ2 6 27 3 0i j k X i j pk+ + + + =2 6 27 271 3 2

PP

⇒ = = ⇒ =

SECTION-B

11. 1 1 11 5 2 12 tan sec 2 tan5 7 8

− − − + + =

1 1 1 1 1

1 11 1 5 2 5 82 tan tan sec 2 tan tan1 15 8 7 1 .

5 8

− − − − −

+ + + = + −

25 2 1

7

−

= 1 1 1 1

21 1 132 tan tan tan tan13 7 71

9

− − − −

+ = +

−

= 1 1

3 14 7tan tan (1)3 1 41

4 7

π− −

+ = =

− ×

.

Value=4π

OR

1 11 1tan tan2 2 4

x xx x

π− −− + + = − +


1

1 12 2tan1 1 412 2

x xx x

x xx x

π−

− ++ − + ⇒ =

− + − − +

( )( ) ( )( )( )( ) ( )( )

1 1 2 2 1tan

2 2 1 1 4x x x xx x x x

π− − + + − +⇒ = − + − − +

2 2 2

2 2

2 2 2 41 14 1 3

x x x x xx x+ − + − − −

⇒ = ⇒ =− − + −

12

x⇒ = ±

12. R={ }1 2 1 2( , ) :T T T T≅

(i) R is reflexive:-

1 1 1 1( , )T T T T R≅ ⇒ ∈

∴ R is reflexive

(ii) R is symmetric:-

Let 1 2 1 2 2 1 2 1( , ) ( , )T T R T T T T T T R∈ ⇒ ≅ ⇒ ≅ ⇒ ∈

⇒ R is symmetric.

(iii) R is Transitive:-

Let 1 2 2 3 1 2 2 3 1 3 1 3( , ) & ( , ) , ( , )T T R T T R T T T T T T T T R∈ ∈ ⇒ ≅ ≅ ⇒ ≅ ⇒ ∈

∴R is Transitive.

∴R is equivalence relation.

13. 2 2

2 22 2

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

− −∆ = − −

− −

1 1 2 3R R R R→ + +

2 22 2

a b c a b c a b cb b c a bc c c a b

+ + + + + +∆ = − −

− −

Taking (a + b + c) common

baccc

bacbbcba−−

−−++=∆22

22111

)(


322211 , CCCCCC −→−→

baccba

bcbacbacba−−++

++−++++=∆0

2)(100

)(

abcbcba−−

−++=∆10

211100

)( 3 =

−+−++

1011

100)( 3cba

3)( cba ++=∆

14. 2)( += xxf

)(xf =x+2 if x>-2

f(x) =0 if x=-2

f(x)=-(x+2) if x<-2

For continuity

Case(i)

When c>-2

)(2)2(limlim cfcxcxcx =+=+= →→

isxf )(∴ continuous for all c > -2

Case(ii)

When c <- 2 then

)()2()2(lim)(lim cfcxxf cxcx =+−=+−= →→

∴ f(x) is continuous for all c <- 2

Case(iii)

When c=-2

0)22(lim)2(lim)(lim 002 =++−=+−= →→−→ + hhfxf hhx

0)22(lim)2(lim)(lim 002 =+−−−=−−= →→−→ − hhfxf hhx

)2()(lim 2 −=∴ −→ fxfx

isxf )(∴ continuous at x=-2

isxf )(∴ continuous for all Rx∈


For differentiability at x=-2

00` lim)2()2(lim)2( →→ =

−−+−=− hh h

fhfRf 10)22(=

−−+−h

h

Lf 10)22(lim)2()2(lim)2( 00 −=−

−−+=

−−−−−

=− →→ hh

hfhf

hh

)2(` −Lf ⇒−≠ )2(`Rf f(x) is not diff. at x=-2

OR

f(x)= [ ]2 2 3 4,6x x for+ +

22)(` +=⇒ xxf

F(x) is a polynomial function

∴f(x) is continuous for [ ]6,4∈x and differentiabl for )6,4(∈x

Now )6,4(∈=∃ cx s.t.

ab

afbfcf−−

=)()()(` )6,4(51222 ∈=⇒=+⇒ cc

∴Lagranges theorem satisfied.

15.y=

−+−

xx 11tan

21

Let x= θtan

Y=

x11112

1 tan21

22tantan

sincos1tan

tan1sectan

tan1tan1tan −−−−− ==

=

−

=

−

=

−+ θθθθ

θθ

θθ

)1(21tan

21

21

xdxdyxy

+=⇒= −

OR

Let y= ( ) ( )sin coscos sinx xx x+

y = u + v


)sin)(log(sincossincoslog)log(sincoslog)(sin cos xxx

xx

dxudxxuxu x −+=⇒=⇒=

xxdxduxxxx

dxdu

ucos)(sinsinlog.sincot.cos1

=⇒−= [ ( ) ( )sin coscos sinx xx x+ ]

( ) )log(coscossincos

1sin1)log(cossinlog)(cos sin xxxx

xdxdv

vxxvxv x +−×=⇒=⇒=

[ ]xxxxxdxdv x coslogcostansin)(cos sin +−=

dxdv

dxdu

dxdy

+=

=⇒dxdy xx cos)(sin [ ( ) ( )sin coscos sinx xx x+ ] + [ ]xxxxx x coslogcostansin)(cos sin +−

16. I= ∫ ∫ ∫ +−

=⇒−+−−

=⇒+

π π π ππππ

0 0 0222 cos1

sin)()(cos1)sin()(

cos1sin dx

xxxIdx

xxxIdx

xxx

I= ∫ ∫ +−

+

π ππ

0 022 )(cos1

)sin(cos1sin dx

xxxdx

xx

2I= ∫ ∫ +

π π

02 )(cos1

)sin( dxx

x∫ +

=⇒π π

02cos1

sin21 dx

xxI

Let cosx=t dtxdx =−⇒ sin

I= [ ] ( ) ( )[ ]1tan1tantan1

1 1111

11

12 −−=⇒

+− −−

−−

−

∫ πππ tdtt

I=4

2π

17. xy

yxdxdy

2)( 22 −−

=

Let y=vxdxdvxv

dxdy

+=⇒

x

dxdvvv

vxvx

dxdvxv −=

+⇒

−−=+⇒ 22

22

12

2)1(


cxyxcxvx

dxdvvv

=−+⇒+−=+⇒−=+

⇒ ∫ ∫ log)log(log)1log(1

2 2222 …….(1)

Now put x=1,y=1

C=log2

From (1)

xyxx

yx 2)(2loglog 2222

=+⇒=

+

18.Let any time t,radius=r,volume=V,surface area=S

g.t. ?sec,/25 3 ==dtdscm

dtdV

πππ

414

34

5

23 =

⇒=⇒=

=rdtdr

dtdrr

dtdVrV

24 rS π=

dtdrr

dtds π8=⇒

sec/10 2

5

cmdtds

r

=

⇒

=

19. Equ. Of parabola

axy 42 = ………(i)

d.w.r.to x

dxdyya

ya

dxdy

224

=⇒=

From equ. (i)

dxdyxy 2=

20. ˆ ˆˆ ˆ ˆ ˆ2 2 , 2 2a i j k b i j k= − + = + −

, ˆˆ ˆ2 4 .c i j k= − +

kjicb ˆ2ˆˆ3 ++=+⇒

Projection of cb + on a = ( )

aacb

.+ =

142 Ans.

21. points (o,-1,0) ,(1,1,1) ,(3,3,0).


Equ. Of plan passing through point (0,-1,0)

A(x-0) + B(y+1) + C(z-0) = 0 ……….(i)

Point (1, 1, 1)and (3,3,0) satisfy equ.(i) we get

A + 2B + C = 0……….(ii)

3A + 4B + 0.C = 0…….(iii)

Solve eque. (i)@ (ii)

ACBA⇒=

−==

−λ

234

A = - 4λ , B = 3λ , C = - 2λ

Putting in (1)

-4λ x + 3λ (y + 1) - 2λ z = 0

-4x + 3 (y + 1) - 2z = 0

4x + 3y + 2z-3 = 0 Ans.

22. Given that probability of A solving the question P(A) = 37

Probability of B solving the question P (B) = 13

∴ P( A−

) =1 - 37

= 47

and P( B−

) = 1 - 13

= 23

(1) Probability that at least one of them solve the problem = P(A B−

) + P( A−

B) +

P(AB)

= 37

. 23

+ 47

. 13

+ 37

. 13

= 6 4 321+ + = 13

21


(2) Probability that only one of them will solve the problem = P(A B−

) + P( A−

B)

= 37

. 23

+ 47

. 13

= 6 421+

= 1021

Ans.

SECTION-C

(23) Given equations can be written as

−−−

−

211121

112

zyx

=

−14

3

⇒AX=B→ (1)

A =4≠ 0

⇒ A 1− exits.

A 11 =(-1) 11+ (4-1)=33

Similarly other cofactors can be obtained.

Adj. A=A

adjA = 41

−

−

311131113

From (1)

X= A 1− B


⇒

ZYX

= 41

−

−

311131113

−14

3

=

−−

12

1⇒ x=1, y=-2, z=-1

OR

A =AI

R1 →R 1 +3R1

−−

0521190231

=

100013001

A Applying R 3 →R 3 -2R1

R1 →R 1 +3R 3 , R 2 → 91 R 2 R 3 →R 3 +R 2 R 3 → 25

9 R 3 R 2 →R 2 +911R 3

R1 →R 1 -10 R 3

⇒

100010001

=

−

−

−−

259

251

53

2511

254

52

53

521

⇒ I=BA⇒B is inverse of A

(24)let (x,y) be the foot of perpendicular on the given curve which is nearest to (0,5)

D= [ ]22 )5()0( −+− yx 21


If D is minimum then D 2 = x 2 +(y-5) 2 =E is also minimum

dYdE = 2y-6

For minimum distance dYdE = 0⇒y=3

At y=3 2

2

dyEd =2 ⟩ 0

⇒ for y=3 ,distance is minimum

⇒x= ± 2 3

Reqd. point is ( 3,32± )

OR

Let x be the side of the square base & y be the height of cuboid

V=x 2 y 2xVy =⇒

Surface area (S)=2x 22 /4 xxv+ =ds/dx=4x-4v/x 2

For minimum ds/dx=0 => x 3 =v ,x=v 31

S’’ =4+8v/x 3 when x =v 31

=4+8 >0 therefore S is ,minimum

When x 3 =v=>x3

=x 2 y =>x =y

There when S is mini mum cuboid is cube.

Q25 given curves are y 2 =x and x+y =2 solving points of intersection aare (1,1),(4,-2)

The area of shaded region ∫−

−1

2

12 )( dyxx

= ∫−

−−1

2

2 )2( dyyy

=9/2 sq units.


(26)

Given integral = ∫∫−− −

−−

a

a

a

a xa

xdxdxxa

a

)()( 2222

Putting x=asinθ ,dx=acos θθd ,and changing limits in second integral

Given integral= a

Π

−−Π

2(

2+a [ ]Π

Π−2

2cosθ =aΠ

(27)

Let she invests Rs. x in saving certificates and Rs. y in National saving bonds

Then LPPis

To maximize Z=0.08x+0.1y

Subject to constraints

x 12000,4000,2000 ≤+≥≥ yxy

corner points of feasible region ABC are A(2000,4000), B(8000,4000),c(2000,10000)

at A, Z=160+400=560

at B, Z=640+400=1040

at C,Z=160+1000=1160

thus Rs. 2000 should be invested in saving certificates and Rs.10000 in National saving

bonds. Maximum yearly income is Rs. 1160.


(28)Let A(1,0,3),B(4,7,1),C(3,5,3) be the given points . Let P be the foot of perpendicular

from A on BC. If P divides BC in k:1 then coordinates of P are (113,

175,

143

++

++

++

kk

kk

kk )

d.r.’s of BC are 1,2,-2

d.r.’s of AP are 1

2,175,

132

+−

++

++

kkk

kk

since AP BC⊥

therefore k= 47−

thus coordinates of P are (3

17,37,

35 )

reqd. ==⊥ APcedis tan

−+−+− 222 )3

317()0

37()1

35( =

3117

(29)

Let E1,E2 ,E3 be the events that the balls are drawn from urn A, urn B, urn C

respectively and let E be the event that balls drawn are one white and one black

Then P(E1)=P(E2)=P(E3)=31

P(E/E1)= )2,6(

)1,3(*)1,1(C

CC = 31

Similarly P(E/E2)=31 , P(E/E3)=

112

Using Bayes’ Theorem reqd. probability= P(E3/E)

= (

332

91

151

332

++) =

5915


sample question paper 2 with solution

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