Cbse Class 12 Maths Sample Paper 2013 Model 3

BLUE PRINT: CLASS XII MATHS
1 MARK
1

4 MARKS
1

6 MARKS

TOTAL
5

1
1
1
2

5
1

7
6
8

1

1

10

1

1
1

rit
e

2
3

12
6

.c

2

om

1
2

1
1

8
7

du

1

.e

1
12 ( 48 )

1
1
7 ( 42 )

6
10
100

w

w

10 ( 10)

10

w

CHAPTER’S NAME
1. RELATIONS AND
FUNCTIONS
2. INVERSE
TRIGONOMETRIC
FUNCTIONS
3. MATRICES
4. DETERMINANTS
5.
DIFFERENTIATION
6 APPLICATION OF
DERIVATIVES
7. INTEGRALS
8. APPLICATION OF
INTEGRALAS
9. DIFFERENTIALS
EQUATION
10 . VECTORS
11 THREE
DIMENSAIONAL
GEOMETRY
12. LINEAR
PROGRAMMING
13. PROBABILITY
TOTAL

Model question paper

Mathematics
Class: 12
Time: 3hrs

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 1 mark each, Section B comprises of 12
questions of 4 marks each and Section C comprises of 7 questions of 6 marks each.
3. Use of calculators is not permitted.
SECTION A
1. If A is square matrix of order 3 such that
= 64 , find

x
2
18 x

=

6 2
18 6

. when A +
.Find x

9. If

. find the values of a and b.

.e

=i+j ,
=

+

= j + k , find the unit vector in the direction of
,

w

8. If

+ bx ) dx =

= 2 and angle between

w

7. If

dx
.
xx



du

6. Evaluate

}.

. Write the value of 3 * 4.

rit
e

4. Let * be a binary operation on R given by a * b =
5. Evaluate : Sin {π/3 -

= I.

om

3. If

, find ; 0<

.c

2. If A

and

+

.

is 60. Find

w

10. What is the cosine of the angle which the vector √2 i + j + k makes with y-axis.
SECTION B
11. Prove that

sin 1

5
7
253
 sin 1
 cos 1
.
13
25
325

OR

12 Consider

f : R  [5, ) given by f  x   9 x 2  6 x  5.Showthat

.

f is invertible . Find inverse of f ?
13. Show that
1 a
1

1

1

1 b

1

1

 1 1 1
1  abc 1    
 a b c
1 c

.c

om

d2y
14. If x=a sin pt and y=b cos pt, find the value of
at t  0 .
dx 2

1 x 
dy
x
15. If y = sin  2 tan 1

 Pr ove that

1 x 
dx
1  x2


16 Find the intervals in which the function f(x) = sinx –cosx ; 0 ≤ x ≤ 2π is (i) increasing
(ii) decreasing
x2  4
17. Evaluate  4
dx.
x  x 2  16



Evaluate

2

rit
e

OR
x  sin x

 1  cos x dx.

du

0

.e

18. Form the differential equation representing the family of ellipse having foci on x-axis
and centre at the origin.
OR

w

Form the differential equation of the family of circles having radii 3.

w

w

19. Find the equation of the plane determined by the points A(3,-1,2), B(5,2,4) and C(-1,-1,6)
Also find the distance of the point P( 6,5,9) from the plane
dy
20. Solve the differential equation
 y sec 2 x  tan x.sec 2 x ; y (0) 1.
dx
21. If a and b are unit vectors and θ is the angle between them, then prove that
cos θ/2 = ½ |a + b|
22. Find the probability distribution of the number of heads in a single throw of three
coins .
OR
Three balls are drawn one by one without replacement from a bag containing 5 white and 4
green balls. Find the probability distribution of number of green balls drawn.

SECTION C
23. Find the matrix A satisfying the matrix equation

 2 1   3 2   1 0 
 3 2  A  5 3  0 1 

 
 


24. Evaluate
25. Show that the semi vertical angle of right circular cone of given surface area and maximum
1
volume is sin 1 .
3
26. Using method of integration find the area of the region bounded by the lines
2x + y =4 , 3x - 2y = 6 and x - 3y + 5 = 0.

om

OR

du

rit
e

.c

.
27. Find the distance of the point (3,4,5) from the plane x  y  z  2 measured parallel to
the line 2x= y =z.
OR
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the
point (1,3,4) from the plane 2x – y + z + 3 = 0.
And also find the image of the point in the plane.

w

w

w

.e

28. A dealer in rural area wishes to purchase a number of sewing machines. He has only rupees
5760.00 to invest and has space for at most 20 items. Electronic sewing machines cost him
rupees 360 and manually operated sewing machine rs.240 . He can sell an electronic sewing
machine at a profit of rupees 22 and a manually operated sewing machine at a profit of Rs.18.
Assuming that he can sell all the items he can buy, how should he invest his money in order
to maximize his profit. Make it as a Linear Programming Problem and solve it graphically.
Justify the values promoted for the selection of the manually operated machines.
29. A student is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six.
Find the probability that it is actually a six.
To get the probability as 1 , Which value to be promoted among students.

MARKING SCHEME
CLASS XII
1. AdjA  A
2. /3

n 1

 A  8

        (1)

3. x = +6 or -6
4. 5/3
5. 1
6. 2 log 1  x +c.
7. a = 4 and b = 3
8. i+ 2j + k / √6
9. √3
10. ½

5
7
,   sin 1
,
13
25
12
24
th e n c o s  
,cos  
13
25
253
 c o s     
325
5
7
253
 sin 1
 s in 1
 c o s 1
13
25
325
1

.c

(1)
(1 )

rit
e

  s in

om

11. Let ,

(1 )

du

1 

w

w

w

.e

OR

12. let x1 ,x2 R+ s.t f (x1) = f (x 2)
f is 1-1

prove that x1 = x2

(1)

9 x2  6x  5  y
 (3 x  1) 2  6  y
3x  1 

y6

(y  6) 1
 f 1 ( y )
3
Therefore for every y
) ,
(
f(
-1) / 3 ) = y. f is onto.
x

-1) / 3

R+ s.t
(1 )

since f is 1-1 and onto , f is invertible

Therfore

(x): from

) to

(

R+

defined as

1
1
b
1
b

1
1
c

du

C1  C2  C3 

1
c
1
c

1

.e

.c

1
b

-1)/3

1 

rit
e

1
1
a
1
  abc
a
1
a

(x) = (

om

13.

)

1
b

1
c
1
c

(1)

1 1 1 1 0
  )1
a b c 0 1

w

   abc(1 

w

w

1
 1 1 1
  abc 1     1
1
b
 a b c
1
1
1
1
b
c

 a b c (1 

14.

1
1
1

 )
a
b
c

(1)

(1 )

(1)

dx
dy
 ap cos pt
 bp sin pt
dt
dt
dy
b
  tan pt
dx
a
2
d y
b
  2 sec3 pt
2
dx
a
2
d y
b
 2
2
dx at t  0
a

1
(1)

1
1

om

15.
.Let x = cos t
=sin

(1 )

.c

Y=sin

rit
e

Y=sin t=

du

=

17.

(1 )

(1)
(1)

(2)

w

w

w

.e

16. f ‘(x)=cos(x) + sin(x)
f ‘ (x)=0 → tan(x) = -1
x=3π/4 and 7π/4
Intervals are [0,3π/4),(3π/4,7π/4) , (7π/4,2π]
[0,3π/4)- f ‘ is positive, so f(x) is increasing
(3π/4,7π/4)- f ’ is negative, so f(x) is decreasing
(7π/4,2π]-f ‘ is positive, so f(x) is increasing

(1)

Consider

4
1 2
x2  4
 x 4  x 2  16dx   2 x 16 dx
x 1 2
x
dt
4
=  2 2 , wheret  x 
t 3
x
1
t
= tan 1  c
3
3
1 1 x 2  4
= tan
c
3
3x

OR

           (1)

           (1)
           (1)

           (1)

consider


2
0



2
x  sin x
dx  
1  cos x
0

x
x
x  2sin cos
2
2 dx
x
2cos 2
2

        (1)


2
1 
x
x
=  2 x sec 2 dx   tan dx
2 0
2
2
0
By solving further we get given =


 
tan 
2
4 2

               (2)

w

w

.e

rit
e

du

OR

.c

om

The equation of the ellipse is + = 1
2x/a2 + 2y/b2dy/dx = 0
(y/x)dy/dx=-b2/a2
Differentiating again and getting the differential equation as
(xy)d 2y/dx2 + x(dy/dx) 2 – ydy/dx = 0

w

18.

       (1)

(1)
(1)
(2)

19. Equation of the plane
x  3 y 1 z  2
2
3
2 0
4
0
4

i.e., 3 x  4 y  3 z  19  0
(2)
Now the perp. Distance from (6,5,9)to this plane is


3.6  4.5  3.9  19
9  16  9

------------------- (2)

om

6

units
34

rit
e

.c

20. Which is in linear differential equation
sec2 x dx
For finding I.F.= e
 e tan x                  (1)
Solution y.I.F. =
(1)
tan x
tan x
=tan x. e  e  c
             (1)
When x =0, y =1  c  2 and writing the completed solution          (1)
2

 
21. Consider a  b  1  1  2 cos 

du

           (1)

=2(1+cos  )

=4 cos 2
2
 1  
 cos  a  b
2 2

.e

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

w

             (2)

w

22. Let X be the number of heads ,X=0,1,2,3

w

P(having head)=p=1/2

Now P(X=0)= 3C0 p 0 q 3  q 3 

q=1/2 ,

n=3 .- --------------------------(1)

1
8

1

Probability distribution
X
P(X)

0
1/8

1
3/8

2
3/8

3
1/8

-------------------------------------(2)
OR

SECTION C

om

23.

2
3


rit
e

2 
         (3)
3 


du

 2 1
 3
adjB  
 adjC  5
 3 2 

 2 1
3
1
B 1  
 and C  5
 3 2 


.c

2 1
 3 2 
Let B  
 and C   5 3
 3 2


B  1  0 and c  1  0

.e

 2 1 3 2  1 1 
 A  B 1C 1  


 .          (3)
 3 2  5 3  1 0 

w

24.

(1)

w

I=

w

I=

2I=π

(1)

I=π
I=π
For getting the answer as

(1)

(3)

25.

Let r, h, l ,S and V be the radius, height, slant
height. surface area and the volume of the cone.



h

l

r

S=
 rl   r 2

S  r2
----------------------------------------- (1)
r
1
1
and V   r 2 h  V 2   2 r 4 h 2
3
9
----------------------------------------- (1)
2
dv
0
dr
1
(2rS 2  8S r 3 )  0
9
r 1
For max or min  
---------------------------(2)
l 3
d 2V 2
now
( at S  4 r 2 )  o
dr 2
V 2 is max imum.
V is max imum.
----------------------------------- (1)
r 1
sin   
l 3
Now
1
  sin 1 ( ).
3
----------------------------------------- (1)

w

w

w

.e

du

rit
e

.c

om

l

26. Let AB→ 2x + y =4
BC→ 3x-2y=6
and AC→ x-3y+5=0
Solving 1 and 2
B(2,0), C(4,3) and A(1,2)

(1 )

For the correct figure
Area of triangle =

(1)
dx -

dx

(1 )

For integrating and getting area=7/2 sq.units

(2)

w

w

w

.e

du

rit
e

.c

om

OR

27. Line is x/1=y/2=z/2

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

Line PQ through P(3,4,5)and II to the given line is
(x-3)/1=(y-4)/2 =(z-5)/2= 
General point on the line is Q(  3, 2  4, 2  5)
If this point lies on the plane x+y+z=2,

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

   3  2  4  2  5  2
   2
 Q(1, 0,1)

-------------------- (2)

2

2

 PQ  (3  1)  (4  0)  (5  1)
 6.

-------------------- (2)
2

.e

du

rit
e

.c

om

OR

w

w

x+y < 20 - - - (i)

w

28. Suppose number of electronic operated machine = x and number of manually
operated sewing machines = y 2

and, 360 x + 240y < 5760 or 3x+2y < 48
x>0, y>0
To maximise Z=22x + 18y
Corners of feasible region are A (0,20), P(8,12),
B(16,0)
ZA = 18x20 = 360, ZP = 22x8+18x12 = 392, ZB = 352 ½

- - - (ii)
2

Z is maximum at x=8 and y=12
The dealer should invest in 8 electric and 12 manually operated machines (½)
Keeping the ‘save environment’ factor in mind the manually operated machine
should be promoted so that energy could be saved.
(1)

29.
Let E be the event that the student reports that 6 occurs in the throwing of die and let S1 be
the event that 6 occurs and S2 be the event that 6 does not occur.
P(S1)=1/6, P(S2)=5/6,
(1)
P(E/S1)=3/4, P(E/S2)=1/4
(2)
P(S1/E)=

=3/8

(2)

w

w

w

.e

du

rit
e

.c

om

To get the probability as one, the student should always speak truth.
The value to be promoted among students is truth value.

(1)

Cbse Class 12 Maths Sample Paper 2013 Model 3

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