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KENDRIYA VIDYALAYA SANGATHAN, CHENNAI REGION
CLASS XII- FIRST COMMON PRE-BOARD EXAMINATION-2013-2014
Q No.
Value points
Marks
1 to
10
4
6 
 2



1)  .2) 3)   3  6  9  4) a=4,b=2;
3
4
 4
8 12 

dy
y
7 13 1
3 6 2
5) 
6)k=  , 7) (3iˆ  6 ˆj  2kˆ), , , 8)Show that
dx 2 x
2 2
7
7 7 7
Each

1 mark
9) 1 10)1.5 cm/sec
11
12
For proving commutative
1.5
For proving associative
1.5
For proving 0 is the identity and it is not in N
1
For cross multiplying by 2
0.5
For applying the formula and getting 2=6x2
2
For rejecting x<0
0.5
For getting x =
1
1
3
OR
For tan-1 =

and the remaining term in terms of cot-1
4
1.5
Converting into cot-1 to tan-1
1
For applying tan-1x+tan-1y
1
For getting the value π
.5
Page 1 of 6
13
For applying R1  R1-R2 and R2  R2-R3
2
a 1 1 0
For getting (a-1)2 2
1 0
3
3 1
1
1
For getting (a-1)
14
For each differentiation 1.5 mark each
For
15
3
3
dy
 tan x

 x tan x 
 sec 2 x log x   sin x cos x (cot x cos x  sin x log sin x)
dx
 x

For substituting x  tan  in each of the term and simplifying
For differentiating and getting the answer
16
For converting the limit x 
For getting a 

2
1
3
1
2
3
to h  0 .5 mark each
1
3
1
and b  4 for each 1.5 mark
2
(OR)
17
For finding first derivative
1
For cross multiplication,squaring and substituting each 0.5mark
1.5
For finding second derivative and proving the result 1.0+0.5
1.5
For dividing by cos4x in the Nr.and Dr.
1
Substituting tan2x=t and getting the answer as
1
tan 1 (tan 2 x)  C
2
(OR)
1
For splitting (6x-5) as -3(3-2x)+4
1.5+1.5
For getting correct integrated value for each

 2 8  3x  x 2

3
2
 (2 x  3) 8  3x  x 2 
Page 2 of 6
1.5+1.5
41 1  2 x  3 
sin 
 +C
2
 41 
18
Put x+b=t and expanding cos(t-b+a)
1
Splitting in to two terms and integrating
19
Ans.cos(a-b)log sin(x + b) - sin(a-b).(x+b)+c
1+1+1
For expressing in terms of cos and sin
1
For applying F(a +b-x)=f(x)
1
For I+I and getting the answer
20

1+1
12

  


For getting a  b  c  3 a  3 b  3 c
1
  
  
 1 
For getting angle between a, b , c with a  b  c as cos 1  
 3
21
Let A be the point (2,3,4) and let B(x1,y1,z1)be the point in the plane.
To find the equation of AB
To find (x1,y1,z1)=(3λ+2,6 λ+3,2 λ+4)
1+1+1
1
1
Substituting the point in the plane and getting λ=-1
(x1,y1,z1)=(-1,-3,2) and d=7
1
1
OR

For finding N  10iˆ  18 ˆj  22kˆ
2
For finding the equation 5x+9y+11z=8
22
23
For P(E)=0.6,P(F)=0.4,P(E  F)=0.2
2
3x0.5
1.5
P(F/E)=0.33
.5
For each question on VBQ
1+1
For correct diagram and SA=πr2+
2
2 v 4v

r r
2
2 3
For differentiating and equating to 0 and to get π r =v(π+2)
For deciding minimum
1
For proving h:2r=π : π+2
1
(OR)
Page 3 of 6
Let C   . Then AC= a sec  +b cosec

2
1
b 3
For finding first derivative and getting tan    
a
2
For second derivative and showing minimum

2
3
2
3

For proving the minimum length of AC=  a  b 

1
3
2
1

24
1
For correct diagram and shading
1
a
b
For getting  { a 2  x 2  (a  x)}dx
a0
For correct integration and getting the answer as
25
2
 ab
4

1
ab
2
For identifying homogeneous and putting y=vx
For getting up to
1
1 v
dx
dv   
2
x
 4v
2
 2v
1+1+1
4
For integrating and simplifying to get the answer y2+2xy=
Page 4 of 6
2
c
x3
26
1
For finding (x1,y1,z1)=(10λ+11,-4 λ-2,-11 λ-8)
For finding d.rs. of AB 10λ+9,-4 λ-1,-11 λ-13
1
Substituting in the equation of plane and to find λ=-1
1
For (x1,y1,z1)=(1,2,3) and distance= 14
1
For image 0,5,1
(OR)
1
For equation of plane x+2y+3z-4+ λ(2x+y+z+5)=0
1
Using Dr’s of normal to the planes, finding λ=-1
1
Equation plane: x-y-2z+9=0
4
1
27
Let E1: the number greater than 4, E2: the number not greater than
4,A: he reports it is greater than 4
1
3
P(E1)= ,P(E2)=
2
3
2
, P(A/ E1)= ,P(A/ E2)=
3
5
5
1
3
3
7
1
1
P(E1/A)=
Page 5 of 6
2
28
2
1
1
Let x no. of packets of food P and Y no. of food Q be prepared.
Constraints: x, y  0,4 x  y  80, x  5 y  115 and 3x  2 y  150
Objective function: Z=6x+3y
Corner points: (2,72),(15,20),(40,15)
It is when x=15,y=20 minimum .Minimum amount of Vit.A=150units
29
2 2
 4


1
For det.A=10, A 1    5 0 5 
10 

 1  2 3
Using ( A )  ( A)
1
To get x 
1
3
1
2
18
4
14
,y  ,z 
10
10
10
Page 6 of 6
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