Class 12 Cbse Maths Sample Paper 2013-14

SAMPLE QUESTION PAPER
MATHEMATICS
CLASS –XII (2013-14)
BLUE PRINT
Unit
I.

VSA (1)

LA (6)

Total

1 (1)

Relations and Functions

SA (4)
4 (1)

–

5 (2)
10 (4)

II.

Matrices

2 (2)

Determinants
III.

–

–

VB 6 (1)

5 (2)
8 (3)
13 (5)

–

5 (2)

–

8 (2)

–

18 (4)

*4 (1)

6 (1)

2 (2)

*4 (1)

6 (1)

18 (5)

–

–

6 (1)

–

–

8 (2)

–

8 (2)

2 (2)

*4 (1)

–

6 (3)

1 (1)

4 (1)

*6 (1)

11 (3)

du

4 (1)

–

Continuity and Differentiability

.e

Integrals

w
w

Application of Integrals

w

Differential equations
Vectors

*4 (1)

1 (1)

Application of Derivatives

IV.

1 (1)

rit
e.
co
m

Inverse Trigonometric Functions

3–dim geometry

44 (11)

17 (6)

V.

Linear Programming

–

–

VB 6 (1)

–

6 (1)

VI.

Probability

–

VB 4 (1)

*6 (1)

–

10 (2)

SECTION–A
Question number 1 to 10 carry 1 mark each.
1.

Write the smallest equivalence relation R on Set A = {1, 2, 3}.

2.

If | ⃗|

3.

If ⃗ and ⃗⃗ are two unit vectors inclined to x–axis at angles 300 and 1200

, then find the value of | ⃗ ̂̇|

respectively, then write the value of | ⃗

–

√ –

–

(– )

5.

Evaluate

6.

If A = (

7.

For what value of k, the matrix (

8.

If |

9.

If ∫ (

10.

Evaluate: ∫

̂|

⃗⃗|

Find the sine of the angle between the line
– 5 = 0.

|⃗

–

–

–

and the plane 2x – 2y + z

rit
e.
co
m

4.

| ⃗ ̂̇|

.e

) is skew symmetric?

, where ,  are acute angles, then write the value of +.

w
w

|

du

), then what is A. (adj. A)?

)

write the value of k.

– )

w

(

SECTION–B

Questions numbers 11 to 22 carry 4 marks each.
11.

Let S be the set of all rational numbers except 1 and * be defined on S by
a * b = a + b – ab,

a, b S. Prove that:

a)

* is a binary on S.

b)

* is commutative as well as associative. Also find the identity element of *.

12.

If a + b + c  o and |

|

, then using properties of determinants, prove

that a = b = c.
13.

Evaluate: ∫(

)√ –

–

–

dx

OR
Evaluate: ∫ (

)

Find a unit vector perpendicular to the plane of triangle ABC where the vertices

rit
e.
co
m

14.

)(

are A (3, –1, 2) B (1,–1, –3) and C (4, –3, 1).
OR

Find the value of , if the points with position vectors ̂̇– ̂̇– ̂ ̂̇

15.

̂ are coplanar.
̂

(̂̇

Show that the lines ⃗

̂– ̂ )

( ̂̇– ̂) and ⃗

du

̂ and ̂̇

( ̂̇ – ̂ )

( ̂̇

̂̇– ̂ – ̂̇

̂̇

̂ ) intersect.

–

–

–

w
w

Prove that

–

OR

w

16.

.e

Also find their point of intersection.

Find the greatest and least values of (
17.

Show that the differential equation

–

–

)

(
√

–

)
is homogeneous, and

solve it.
18.

Find the particular solution of the differential equation cos x dy = sin x (cos x – 2y)
dx, given that y = 0 when

19.

.

Out of a group of 8 highly qualified doctors in a hospital, 6 are very kind and
cooperative with their patients and so are very popular, while the other two

remain reserved. For a health camp, three doctors are selected at random. Find
the probability distribution of the number of very popular doctors.
What values are expected from the doctors?
20.

Show that the function g(x) = |x–2|, xR, is continuous but not differentiable at
x = 2.
(

) with respect to x.

21.

Differentiate

22.

Show that the curves xy = a2 and x2 + y2 = 2a2 touch each other.

Separate the interval [

rit
e.
co
m

OR

] into sub–intervals in which f (x) =

increasing or decreasing.

is

SECTION–D

Find the equation of the plane through the points A (1, 1, 0), B (1, 2, 1) and C (–2, 2,

.e

23.

du

Question numbers 23 to 29 carry 6 marks each.

w
w

–1) and hence find the distance between the plane and the line

–

–
–

.

OR

w

A plane meets the x, y and z axes at A, B and C respectively, such that the centroid
of the triangle ABC is (1, –2, 3). Find the Vector and Cartesian equation of the
plane.
24.

A company manufactures two types of sweaters, type A and type B. It costs Rs. 360
to make one unit of type A and Rs. 120 to make a unit of type B. The company can
make atmost 300 sweaters and can spend atmost Rs. 72000 a day. The number of
sweaters of type A cannot exceed the number of type B by more than 100. The
company makes a profit of Rs. 200 on each unit of type A but considering the

difficulties of a common man the company charges a nominal profit of Rs. 20 on a
unit of type B. Using LPP, solve the problem for maximum profit.
(

–

)

25.

Evaluate: ∫

26.

Using integration find the area of the region {(

27.

A shopkeeper sells three types of flower seeds A1, A2 and A3. They are sold as a

)

}

mixture where the proportions are 4:4:2 respectively. The germination rates of

rit
e.
co
m

three types of seeds are 45%, 60% and 35%. Calculate the probability.
a)

of a randomly chosen seed to germinate.

b)

that it is of the type A2, given that a randomly chosen seed does not
germinate.

OR

du

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.
One ball is transferred from Bag I to Bag II and then two balls are drawn at

.e

random (without replacement) from Bag II. The balls so drawn are found to be

28.

w
w

both red in colour. Find the probability that the transferred ball is red.
Two schools A and B want to award their selected teachers on the values of

w

honesty, hard work and regularity. The school A wants to award Rs. x each, Rs. y
each and Rs. z each for the three respective values to 3, 2 and 1 teachers with a
total award money of Rs. 1.28 lakhs. School B wants to spend Rs. 1.54 lakhs to
award its 4, 1 and 3 teachers on the respective values (by giving the same award
money for the three values as before). If the total amount of award for one prize on
each value is Rs. 57000, using matrices, find the award money for each value.
29.

A given rectangular area is to be fenced off in a field whose length lies along a
straight river. If no fencing is needed along the river, show that the least length of
fencing will be required when length of the field is twice its breadth.

MARKING SCHEME
SECTION–A
1.

R = {(

2.

2a2

1

3.

√

1

4.

)(

)(

)}

1

1

√

–

6.

–
(

7.

k=–

1

rit
e.
co
m

5.

)

–

8.

10.

–cot x ex + C

du

k = –2

1
1
1
1

.e

9.

1

a)

let

–

w

11.

w
w

SECTION–B

( – )(

Since

–

–

– )
–
1

b)

is commutative
also, (

)

(
=

–
–

)
–

–

1

(

)

(

–
–

=
)

(

)

)

(

=

–

–

is associative

1

Let e be the identity,
Then a * e = a  a + e – ae = a  e (1–a) = 0

1

Since 1 – a  0  e = 0.
|

|

(

)|

(

–

|

)|

du

–

rit
e.
co
m

12.

= (a + b + c) (–b2–c2 + 2bc –a2 + ac + ab – bc) = 0

–
–

1

– |
–

1

w
w

.e

 = – (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0
= – (a + b + c) [(a–b)2 + (b–c)2 + (c–a)2] = 0

1

w

Since a + b + c = 0  a–b = 0, b–c = 0, c–a = 0
or a = b = c.
13.

∫(

∫(

–

– ) (√

∫( – )√ –

1
) (√ –

–

–

) dx

)

where sin x = t

1

½

∫ √( – )

= ∫( – )√ –
(

=

) ⁄

–

= [

[

–

√ –

]

–

[√

½

|( – )

⁄

(

–

√ –

√

–

|

]

1

– )

– )|

|(

]

1

∫(
)(

)

–

)

– ∫
|

(

–

)

1
1

w
w

A vector perpendicular to the plane of ABC
= ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗

w

14.

|

2

.e

–

∫

du

(

)(

rit
e.
co
m

OR

̂

̂

̂

= |–

½

– |

–

̂– ̂

̂

1+1

√

1

–
or

̂̇

|⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗|

̂– ̂
√

 Unit vector  to plane of ABC =

√

(
OR

̂

̂ – ̂)

½

Let the points be A (3, –2, –1), B (2, 3, –4), C (–1, 1, 2) and D (4, 5, )
⃗⃗⃗⃗⃗⃗

̂– ̂

– ̂̇

⃗⃗⃗⃗⃗⃗
̂

̂
̂

)̂

1½

A, B, C, D are coplanar if [⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗] = 0

½

̂

̂

[⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗]

(

–
|–

–
|

½

rit
e.
co
m

⃗⃗⃗⃗⃗⃗

 1(15+9) – 7 (–3 – 12) + (+1) (–3+20) = 0
=–
(̂

̂– ̂ )

du

Let, any point on the line ⃗

( ̂– ̂ )

( ̂

̂ ) be Q (

.e

and any point on line ⃗

( ̂– ̂) be P (



–

– )

½

)

½
½

w
w

If the lines intersect, P and Q must coincide for some  and .

–

½

1+3 = 4+2 ..... (i)
1– = 0 ..... (ii)

w

15.

1

–1 = –1 + 3 ..... (iii)
Solving (ii) and (iii) we get  = 1 and  = 0

1

Putting in (i) we get 4 = 4, hence lines intersect.

½

 P or Q (4, 0, –1) is the point of intersection.

1

16.

–

LHS =

–

=

–

=

–

–

–

½

–

–

=

–

–

1

–
–

½
–

–

1½

–

= cot –1 3=RHS

½

(

–

)

–

(

)

–

=(

–

–

=( ) –
–

–
–

–

and greatest value = 2 [(
17.

x dy – y dx = √
=
x

,y

+

√

–

–

)

½
½

)

]
]

1

[ ]

w

 least value =

–

(

– )

w
w

= [(

–

( –

) –

.e

= [(

–

) –

du

=

rit
e.
co
m

OR

1

)

]

1

dx
+√

=
=

+√

( ) =f( )

½

( )
=

[
(

)

√

( ) ]

1

differential equation is homogeneous.
let

= v or y = vx

v+x

=v+x
or ∫

= v+√

log|v + √
y+√

=∫

√

1

v+√

| = log cx

= cx

1

= cx2

1

Given differential equation can be written as
cot x

+ 2y = cos x or

+ 2 tan xy = sin x
∫

Integrating factor =

=

=∫

1

½

x dx

w
w

When x = , y = 0

= sec2 x.

.e

y. sec2x = sec x + c 1

½

du

the solution is y. sec2x = ∫

y = cosx + c cos2x.

rit
e.
co
m

18.

½

0= + C

C = -2

½

19.

½

w

Hence the solution is y = cosx–2cos2 x.
let x be the random variable representing the number of very popular doctors.
x:

1

P(x)

2

6C1 . 2C2

6C2. 2C1

8C3

=

3
6C3

8C3

=

½

8C3

=

1½

It is expected that a doctor must be
 Qualified
 Very kind and cooperative with the patients
20.

2

g(x) = |x-2| = {

½

LHL = lim (2-x) = 0
x2-

lim (x-2) = 0 and g (2) = 0
x2+

1

rit
e.
co
m

RHL =

g(x) is continuous at x = 2……………………………..

RHD = lim
h0
LHD

(

(

)

)

( )

= lim
h0

(

)

)

= -1

1

=1

½

g(x) is not differentiable at x = 2

½

.e

Let y = log (xsinx + cot2 x)

(xsinx + cot2x)……………………..

1

w
w

=
Let u =

and v = cot2x.

log u = sinx. log x ,
=
or

(

= lim
h0

w

21.

RHD

( )

du

LHD = lim
h0

½

= 2 cot x (–cosec2x)

½

+ logx. cosx

= xsinx [
=

] ………………………………..
[

(

)

1
]

1½

Solving xy = a2 and x2+y2 = 2a2 to get x =

a

for x = a, y = a and x = -a, y = -a
i.e the two curves intersect at P (a, a) and Q (-a, -a)
xy = a2

+y=0

x

x2+y2 = 2a2

2x+2y

=

–

1

= -1 at P and Q

1

=

1

=0

= -1 at P and Q

rit
e.
co
m

Two curves touch each other at P
as well as at Q.

1

OR

f(x) = sin4x + cos4x

du

f ‘(x) = 4 sin3x cosx -4 cos3x sinx

.e

= -4 sinx cosx (cos2x – sin2x)

f'(x) = 0

w
w

= –2 sin2x cos2x = - sin 4x
sin4x = 0

x = 0,

,

4x = 0,

1

2 , 3 , ………………
1

w

22.

) (

Sub Intervals are (
f’(x) < 0 in (0,
And f’(x) > 0 in (

)

)

f(x) is decreasing in (0,
,

)

f(x) in increasing in (

)

1
,

)

1

SECTION – D
to the plane is parallel to ⃗⃗⃗⃗⃗⃗

A vector

⃗⃗ = |

̂

̂

̂

| = – ̂̇ –

̂
̂

⃗⃗⃗⃗⃗⃗

1

̂̇– ̂

̂̇

1½

Equation of plane is ⃗. (2 ̂ + 3 ̂ – 3 ̂ ) = 5
( ⃗ ⃗⃗

1

⃗ ⃗⃗)

rit
e.
co
m

Since, (2 ̂ + 3 ̂ – 3 ̂ ). (3 ̂ – ̂ + ̂ ) = 0, so the given line is parallel to the plane.

½

Distance between the point (on the line) (6, 3, -2) and the plane ⃗.
(2 ̂ + 3 ̂ – 3 ̂ ) – 5 = 0 is
d=

|

|
√

=

√

=√

1½

du

Let the coordinates of points A, B and C be (a, 0, 0), (0, b, 0) and (0, 0, c)

= 1,
a = 3,

+ = 1 and

= -2

b = -6

Equation of plane is
or

+

w
w

Equation of plane is

½

.e

respectively.

w

23.

and
and

+

1

+ =1

6x – 3y + 2z – 18 = 0

=3

1½

c=9
1
1

which in vector form is
⃗ (6 ̂ – 3 ̂ + 2 ̂ ) = 18

1

24.

let the company manufactures sweaters of type A = x, and that of type B = y. daily
LPP is Maximise P = 200x + 20y
s.t. x+y

300

360 x + 120y

x

72000

100

0y

1½

0

Correct Graph

1½

rit
e.
co
m

x–y

1

Getting vertices of the feasible region as

A (100, 0), B (175, 75), C (150, 150) and D (0, 300)
Maximum profit at B

= 200 (175) + 20 (75)

du

So Maximum Profit

let I = ∫ (
=
=

[(

) . x dx

w

25.

w
w

= Rs. 36500

1½

.e

= 35000 + 1500

)

]

1

∫

 ∫

½
x = tan

=

-∫

d

½

-∫

=

dx = sec2

∫

1

=

–[

=

–

⁄

[

1

[ ]

∫

]

+

–

=

1
log 2

1
1

.e

du

rit
e.
co
m

Correct figure:

1

w
w

Correct Figure

Solving x2 + y2 = 1 and x + = 1 to get x = and x = 1 as points of intersection

w

26.

]

Required area = ∫

=

[ √

(

[

]

(

=
=

∫ (

√

)

[

) sq. u.

)

]
]

1
1

1
1

let E1 : randomly selected seed is A1 type P(E1) =
E2 : randomly selected seed is A2 type P(E2) =

1

E3 : randomly selected seed is A3 type P (E3) =
(i)

let A : selected seed germinates
P (A/E1) =

P (A/E3) =

1

rit
e.
co
m

P (A/E2) =

½

P(A/E3) =

½

P(A) = P(E1) P(A/E1) + P(E2) P(A/E2) + P (E3). P(A/E3)
=
=

du

let A : selected seed does not germinate
P(A/E1) =

.e

P(E2/A) =

, P(A/E2) =

w
w

(ii)

or 0.49

(

) ( |

)

(

(

) ( | )
) ( | )

(

) ( |

)

=

½

1

½

1

w

27.

OR
Let

E1 : transferred ball is red.
E2 : transferred ball is black.

½

A : Getting both red from 2nd bag (after transfer)
P(E1) =

P(E2) =

1

5

P(A/E1)

=

P(A/E2)

=

P(E1/A)

=

C2
=
C2

4

C2
=
C2

1

or

10

1

) ( | )
) ( | ) ( ) ( |
(

(

=

½

)

=

1+1

The three equations are 3x+2y+z = 1.28

rit
e.
co
m

28.

or

10

4x + y + 3z = 1.54
x + y + z = 0.57
(

) ( )= (

i.e AX = B

du

)

)

1

)(

)=(

.e

w
w

w

( )= (

x = 25000, y = 21000, z = 11000
29.

½

1

|A| = -5 and X = A-1B
A-1 = (

1½

)

½
1½

let length be x m and breadth be y m.
length of fence L = x+2y
Let given area = a
L=x+

xy = a or y =
1

=1
=o
=

1
x2 = 2a

x=√

1

>o

½

for minimum length L = √
x=√

and breadth y =

√

+
=

√

√

= √
= x

1
½

w

w
w

.e

du

rit
e.
co
m

x = 2y

1

Class 12 Cbse Maths Sample Paper 2013-14

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