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. Construct a 2 2× matrix whose elements are given by 2ija i j= − .

2. If A is a square matrix of order 3 such that 144adj A = , write the value of A .

3. For what value of k,the matrix

432K

has no inverse.

4. If :f R R→ is defined byf(x) = 2x + 3, write the value of ( ( ))f f x .

5. Write the value of )6

5(tantan 1 π− .

6. Evaluate dxxxx

∫ ++

sincos1

7. Write the order and degree of the differential equation ( a + x) 2

2

dxyd

+y=0.

8 .Find the value of λ ,such that i2 -λ j + k and i + j + k are perpendicular

9. Write the value of a b−

, if two vectors anda b

are such that

2, 3 and . 4a b a b= = =

.

10.Let A =(1,2,3) and B = (2,-3,5).Find the direction ratios of AB.

SECTION – B

11. Show that the function f: RR → defined by f(x) =3

12 −x, x∈R, is one –one and onto

function .Also find the inverse of the function f.

OR

Prove that the relation R on the set Z of all integers defined by ( ),a b R a b∈ ⇔ − is

divisible by 5 is an equivalence relation.

12. Prove that: 1 1 11 5 2 12 tan sec 2 tan5 7 8 4

π− − − + + =

.

13. Using properties of determinants show that:

( )( )3

3 33

a a b a cb a b b c a b c ab bc cac a c b c

− + − +− + − + = + + + +− + − +

.

14. Find all the points of discontinuity of the function f defined by

f(x) =

≥<<−

≤+

20212

12

xxx

xx

15. Find dydx

, if xxy cos= .+ x2

OR

If 11

xyx

−=

+, show that ( )21 0dyx y

dx− + = .

16. Find the equation of the normal to the curve 3 2 6y x x= + + at point (0, 6).

17. Evaluate dxxx

x∫ +−

+)32)(1(

23

18 Evaluate dxxx

x∫ +

2/

0 cossinsinπ

19 Solve the following differential equation:

( )22 2

2 1 ; (0) 01 1

dy x y ydx x x

+ = =+ +

OR

Solve the following differential equation:

( )3 3 2 0x y dy x y dx+ − = .

20. Find the area of the parallelogram whose diagonals are determined by the vectors

∧∧∧

+− kji2 and∧

∧∧

−+ kji 43 ..

21. Find the shortest distance between the lines:

( ) ( ) ( ) ( )3 and 4 2 3r i j k i j r i k i kλ µ= + − + − = − + +

.

22. A factory has two machines A and B. Past record shows that machine A produced 60% of the

items of output and machine B produced 40% of the items. Further, 2% of the items produced by

machine A and 1% produced by machine B were defective. All the items are put into one

stockpile and then one item is chosen at random from this and is found to be defective. What is

the probability that it was produced by machine B?

SECTION – C

23. Using matrix method, solve the following system of linear equations:

.23,0,42 =+−=−−=++ zyxzyxzyx

24.. . A wire of length 36cm is cut into two pieces is turned in the form of a circle and the other in

the form of an equilateral triangle .find the length of each piece so that the sum of area of the two

be minimum.

OR

Show that the right circular cylinder of given volume open at the top has minimum total

surface area, provided its height is equal to the radius of its base.

25. Evaluate the following integral as limit of sums: dxxx )( 24

1 ∫ −

26. Using integration, find area of ABC∆ whose vertices have

coordinates ( ) ( ) ( )2,0 , 4,5 and 6,3A B C .

27. Find the equation of the plane which contains line of intersection of planes

( ). 2 3 4 0r i j k+ + − = , ( ). 2 5 0r i j k+ − + =

and which passes through the

Point (1,0,-2).

OR

28. Equation of the plane containing the lines

Findkjijirandkjijir ).2()2(∧∧∧∧∧∧∧∧∧∧∧→

−+−++=−+−++= λλ the distance of this plane from

the origin and also from the point (1,1,1).

28. A furniture firm manufactures chairs and tables, each requiring the use of three machines A,

B and C. Production of one chair requires 2 hours on machine A, 1 hour on machine B and 1 hour

on machine C. Each table requires 1 hour each on machine A and B and 3 hours on machine C.

The profit obtained by selling one chair is Rs. 30 while by selling one table the profit is Rs. 60.

The total time available per week on machine A is 70 hours, on machine B is 40 hours and on

machine C is 90 hours. How many chairs and tables should be made per week so as to maximize

profit? Formulate the problem as L.P.P. and solve it graphically.

29. If two cards are drawn at random from a deck of 52 cards and X is the number of aces

obtained then find the value of mean and variance.

SOLUTION PAPER -A

SECTION A

1. Required matrix is

2221

1211

aaaa

=

−−−−

22*212*221*211*2

=

2301

2. adjA =144 => 13−A =144 => A =+12 or -12

3. 432k

=o =>4k-6=0 => k=23 .

4. f(x)= 2x+3 => f(f(x))= f(2x+3) = 2(2x +3) + 3 =4x +6+3 = 4x+9

5. )6

5(tantan 1 Π− =6

)6

tan(tan)6

tan(tan 11 Π−=

Π−=

Π−Π −−

6. Let x+sinx = t => (1+cosx)dx = dt

I = cxxcttdt

++=+=∫ sinloglog

7. Order -2 deg- 1

8. 2- 301 =⇒++ λλ .

9. .2222 →→→→→→

⋅−+=− bababa

= 4 + 9 - 2 4×

= 5

10. Direction ratios are (2-1, -3-2, 5-3)

That is (1,-5, and 2).

SECTION B

11. Let Rxx ∈21, such that

)()( 21 xfxf =

21

21

312

312

xx

xx

=⇒

−=

−⇒

⇒ f is1-1

Let ‘y’ be any arbitrary element of co domain R

Then 3

12 −x =y

⇒x=2

13 +y

Rx∈ Such that f(x) =

+

213yf

=3

12

132 −

+y

= y

∴ f is onto.

f-1 (x)= 2

13 +x

(OR) Reflexive : (a,a) 50 ebyisdivisiblaaR =−∈

Symmetric: (a,b) R∈ => (a-b) is divisible by 5

=> (b-a) is divisible by 5

=> (b,a) R∈

Transitive: let (a,b),(b,c) R∈

=> a-b and b-c are divisible by5

=> (a-b) + (b-c) = a-c is divisible by 5

As R is reflexive ,symmetric and transitive then then R is equivalence.

12. (71tan)

40/118/15/1(tan2

725sec)

81tan

51(tan2 11111 −−−−− +

−+

=++

=

41tan

2528

2825tan

71tan

43tan

71tan

9/113/12tan

71tan

31tan2 11111111 Π

==×=+=+−×

=+ −−−−−−−−

13

))((3)](3)[(

)]2)(()2)(2)[((2

2)(

)133()122(20

201

)()(31

311

)3211(3

3

cabcabcbacabcabcba

cabacabacbacaca

babacba

RRRandRRRcaca

babacaba

cbacbacbc

cbbcaba

cccccbccba

cbbcbacabacba

++++=++++

+−−++++=+−−+

++

−→−→+−−++−+−

++=+++−

+−+−+−

++→+−++

+−+++−+−++

14. For x 21 ≥≤ andx f(x) is a polynomial function so f(x) is continuous

At x=1 L.H.L is 3)21(lim)2(lim01

=+−=+→→ −

hxhx

R.H.L is 1)21(lim)2(lim01

−=−+=−→→

++hx

hx

As L.H.L ≠ R.H.L

X=1 is point of discountinity

At x = 2 L.h.L is 0)22(lim)2(lim22

=−−=−+− →→

hxxx

R.H.L is 00lim2

=+→x

As L.H.L = R.H.L

X=2 is f(x) is continuous.

1=∴ x is the only point of discontinuity.

15. y = x xx 2cos +

Let u = x xcos xxu logcoslog =⇒

)logsincos1(cos1logsin1 cos xxxx

xdxdux

xxx

dxdu

ux ×−=⇒+−=⇒

2log2)logcos1()2( cos xxx xSinxxx

xdxd

dxdu

dxdy

+×−=+=

OR

yxx

−+

=112

=> (1+x) y )1(2 x−=

=> (1+x) . 2y 12 −=+ ydxdy

=>2y(1+x) 111

−=+−

+xx

dxdy

=> 2y(1+x) 1

2111

+−=

+−

−−=xx

xdxdy

yy

xxdx

dyx −=−

×+

−=−⇒1

11)1( 2

=> (1-x )2 0=+ ydxdy

16. y = x 623 ++ x 223)6,0(

2 =⇒+=dxdyx

dxdy

Slope of the normal is -21

Equation of the normal is y-6 = 0122)0(21

=+−⇒−− xyx

17. 3 2( 1)(2 3) 1 2 3

x A Bx x x x

+= +

− + − +

3x+2=A(2x+3)+B(x-1) ⇒A=1 B=1

I = 1 2 3

dx dxx x

+− +∫ ∫

cxx +++−⇒ 32log211log

∴

18 dxxx

xdxxx

xI ∫∫ +

=−+−

−=

2

0

2

0 cossincos

)2

sin()2

cos(

)2

sin( ππ

ππ

π

2I = dxxxxxdx

xxxdx

xxx

∫∫∫ ++

=+

++

2

0

2

0

2

0 cossincossin

cossincos

cossinsin

πππ

2I = 2

1 20

2

0

πππ

==∫ xdx => I = 4π

19. 222 )1(1

12

+=

++

xy

xx

dxdy

2)1log(12

1.22 xeeFI x

dxxx

+==∫=⇒ ++

dxxx

xy )1()1(

1)1.( 222

2 +×+

=+⇒ ∫

cxdxx

+=+

⇒ −∫ 12 tan

11

When x=0 ,y=0

0 = tan c+− 01 => c=0

xxy 12 tan)1( −=+∴

(OR)

33

2

yxyx

dxdy

+=

Let y=vx dxdvxv

dxdy

+=⇒

vv

vdxdvx

vv

dxdvxv −

+=⇒

+=+⇒ 33 11

cxxy

yxcxv

vcxvv

cxdvdvv

dxx

dxv

vv

vdxdvx

+−=+−

⇒+−=+−⇒+−=+−

⇒

+−=+⇒−=+

⇒+

−=⇒

−

∫∫∫∫2

2

2

33

3

3

3

log21log

2

log1111

20.∧Λ→

+−= kjid 21 and ∧∧∧→

−+= kjid 432

Area of parallelogram = ∧∧∧

∧∧∧

→→

++−=−

−=× kjikji

dd 1153143

11221

21

21.

→→→→

−+= kjia1 →→→

−= jib 31

→→→

−= kia 42 →→→

+= kib 322

S.D =

)

))

21

2112

(

((→→

→→→→

×

ו−

bbbbaa

→→

−aa 12 = 3

→→

− ji

→→→

→→→

→→

+−−=−=× kjikji

bb 29330201321

9421 =×→→

bb .

S.D = (3→→

− ji ) • (→→→

+−− kji 293 ) / 94

= 0

22. E1 - Items from Machine A and E2

- Items from Machine B

E- Choosing a defective item

53)( 1 =EP

52)( 2 =EP

P(E/ )1E = 501 P(E/ )2E =

1001

)/( 2 EP E = ) )P( P(E/ ) )P( P(E/

) )P( P(E/

EEEEEE

2211

22

+

=

501

53

1001

52

52

1001

×+×

×

= 41

SECTION C

24. 23. Let A =

−−−131111

121

A = -10

AdjA =

−−−−−−−

352202114

−−−

−−−

=−

352202114

1011A

X = A 1− B

=

−−−

−−−

352202114

101

204

X = 10

1−

−−−

144

18

=

575259

24.. 2πr+3x=36 3236 rx π−

=⇒

s = π r2+ 2

43 x

=π r2+2

3236

43

− rπ

2=drds π r+ 23 144 8 0

36rπ π − + =

rdr

sd∨+= 08

3632 2

2

2

ππ

for minimum 1803 3

ds rdr π

= ⇒ =+

x =33

3108+π

(OR)V = 2rπ

h = 2rVπ

S = 22 rrh ππ +

S = 22 rrV π+

2

22rVr

drds

−= π

30 rVdrds π=⇒=

h = rrV

=2π

32

2 42rV

drSd

+= π > 0

∴ h= r is a minima.

25. = 1, b = 4 and h = n3

f(x) = x2 – x.

f(a) = f(1) = 12- 1 = 0

f(a+h)= f(1+h) = (1+h)2 – (1+h) = 12h2 + h

f(a+2h) = f(1+2h) = (1+2h)2-(1+2h) =22h2 +2h

…

….

f(a+(n-1)h) = (1+(n-1)h)2-(1+(n-1)h) = (n-1)2 h2 – (n-1)h

1mark

r

∫∞→

=b

a nLimdxxf )( h{f(a)+ f(a+h)+ f(a+2h)+…..+ f(a+(n-1)h)}

= lim∞→n n

3 (0 + 12h2 + h+22h2 +2h+…. +(n-1)2 h2 – (n-1)h)

= lim∞→n n

3 {h2 (12+22+32+… (n-1)2) + h (1+2+3+….. (n-1)}

=lim

∞→n n3

{ ( )( )( )

2)1(

61212 nnhnnnh −+

−−

}

= lim∞→n n

3 23

nn.n.n

6

)12)(11(nn

−− +

n3

n3 n.n.

2

)11(n

−

1mark

=29

6227+

× =2

27

26. Equation of AB is y = )2(

25

−x , equation of BC is y = -x + 9, equation

of AC is

y = )2(

43

−x

Required area = dxxdxxdxx )2(

93)9()2(

25 6

4

4

2

−−+−+− ∫ ∫∫ =7 square unit

f(x)=2.5(x-2)

f(x)=(-x+9)

f(x)=(1/3)(x-2)

-8 -6 -4 -2 2 4 6 8

-5

5

x

y

B(4,5)

A(2,0)

C(6,3)

27. Equations of plane in Cartesian form are x+2y+3z-4=0 and 2x+y-z+5=0

Required equation is (x+2y+3z-4)+k(2x+y-z+5)=0----------(1)

It passes through (1,0,-2)

=> (1+0-6-4) + k ( 2+0+2+5)=0

1=⇒ k

equation is (x+2y+3z-4) + 1(2x+y-z+5)=0

3x+3y+2z+1=0

(OR) Given lines are

)2()()2()(

∧→∧∧∧∧∧∧∧∧∧→

−+−++=−+++= kjijirandkjijir µλ

Normal perpendicular to the plane containing two lines is

∧∧∧

∧∧∧

→

++−=−−−= kjikji

n 333211121

Plane is )333)}.(2(){(.

∧∧∧∧∧

∧∧∧→→

++−−+++= kjikjijinr λ

0)333.( =++−

∧∧∧→

kjir---------- (1)

Distance from origin is 0

Distance from (1, 1, 1) to the plane (1) is {(∧∧∧

++− kji) (.

∧∧∧

++ kji)-

0}/111 ++

= 31

. 28. x chairs &y tables are made per week

LPP is max z=30x+60y

x≥0 , y≥0

2x+y ≤ 70, x+y≤ 40 , x+3y ≤ 90

FIGURE &SHADING

A(35,0) Z=1050

B(30,10) Z=1500

C(15,25) Z=1950

D(0,30) Z=1800

Profit is max at C(15,25) ie x=15, y=25

Max profit is 1950

29. Range of the random variable X is {0, 1, and 2}

P(x = 0)221188

)2,52()2,48(=

CC

P (x = 1) = 22132

)2,52()1,48()1,4(=

×C

CC

P(x = 2) = 2211

)2,52()2,4(=

CC

X 0 1 2

P(X) 221188

22132

2211

Mean = E(X) = ∑ ×=2211880.px + 1

22134

221121

22132

=×+×

E(x 2 ) = 0 2213221

221188

+× + 221122 × =

22136

Variance = E(X )()2 XE− = 36/221 – (34/221) 2 =0.372