KENDRIYA VIDYALAYA SANGATHAN, CHENNAI REGION
CLASS-XII-REVISION EXAMINATION-2012-2013-MATHEMATICS
SCORING KEY
1
3 AB  27 A B  54
2
 i 0  i 0   i 2

 
B 2  
 0 i  0 i   0
1
0    1 0 
   I
 
i 2   0  1
1
B4  I 2  I
B 4n  I , n  N
3
 e 7 dx   7e dx 
x
x
x
7ex
log( 7e)
4
a
5 
cos 1    sec 1   
5
4 2
a
3 
 cos 1    sin 1   
5
5 2
a3
5
I 
4 x
3
1
4 x 
3
I 
1
x
x
x  4 x
1
C
1
1
dx      1
dx      2 
 4 x 

(1)  (2)  2 I    4  x 
3
1
x
3
 dx
3
x   2 I   dx  2 I  x 1  2  I  1
1
6
1
Condition for coplanarity
a

 
b c 0
1
 1
p
1
1
1
2 0
p  2 1
 11  2( p  2  1 1  2 p   1 p  2  p   0
 5  2 p 1 2 p  2 p  2  0
 2 p  4  0  p  2
1
7
x  3y 
 is a skew symmetric matrix then x= 0 ( diagonal elements of a skew
x 
If  0
12

1
symmetric matrix is 0 ) and aij  a ji  a12  a 21  x  3 y  12  y  4
(or) AT   A for a skew symmetric matrix
8
A(1,2,3), B(1,2,1)
1

AB  (2,4,4)

AB  (2) 2  (4) 2  (4) 2  36  6

  2  4 4   1  2 2 
direction cosines of AB  
,
,  ,
, 
 6 6 6  3 3 3
9
 
a  b ,   60 
  
a.b  a b cos   8
1
2 1
 a  8
2
2
 a  16


 a 4 b
10
11
R= (1,1), (2,2), (3,3), (1,2), (2,1), (1,3)(3,1) is the only relation which is reflexive ,
symmetric but not transitive .

y  log x  x 2  a 2



dy
1
2x

1 

2
2
2
2
dx x  x  a  2 x  a 


dy
1
x


1


dx x  x 2  a 2 
x2  a2 
 x2  a2  x 
dy
1




dx x  x 2  a 2 
x 2  a 2 
dy
1


dx
x2  a2

1
1
2
 dy 
 x 2  a 2    1
 dx 
2
dy d 2 y
 dy 
 x 2  a 2 2 . 2  2 x   0
dx dx
 dx 

 x2  a2
d
½
1
 dy 
 x   0
dx
 dx 
2
1
y
½
2
2
11
(OR)
substitute
½
x  cos 2
2
u  tan
1
1 x2
1 x2
1  cos 2
2 sin 2 
 tan 1
 tan 1 tan    
2
1  cos 2
2 cos 
1
1
 cos 1 x 2  v where v  cos 1 x 2
2
2
1
 u v
2
du 1

dv 2
 tan 1
12
1½
1
1
From 0 to 9 the prime numbers are 2 , 3 , 5 , 7
X = 0 , 1,2
6 5 30 10
 

10 9 90 30
 4 6  48 16
P(x=1)= 2   

 10 9  90 30
4 3 12
4
 

p(x=3)=
10 9 90 30
p( x = 0 ) =
X
p(x)
0
1
2
10
30
16
30
4
30
16
4 24 4
 2


30
30 30 5
16
4 32 16
E ( x 2 )   x 2 P( x)  0  1 
 4


30
30 30 15
Mean =  xp(x ) = 0  1 
1
1
1
2
32  4 
16 16 32
var( x)  E ( x )  E ( x) 
  


30  5 
15 25 75

13 let the adjacent sides of the parallelogram be a  2iˆ  4 ˆj  5kˆ , b  iˆ  2 ˆj  3kˆ
 
area of parallelogram = a  b
2
2
iˆ ˆj kˆ


 
a  b = 2 4  5  22iˆ  11 ˆj  a  b  11 5
1 2 3




vectors parallel to diagonals are a  b and a  b
3
1
2
 

Let a  b = c


 c  3iˆ  6 ˆj  2kˆ , c  49  7

 c 1 ˆ
unit vector parallel to c    3i  6 ˆj  2kˆ
c 7
  
Let a  b  d


d  iˆ  2 ˆj  8kˆ , d  69

 d
1 ˆ
unit vector parallel to d   
i  2 ˆj  8kˆ
69
d


1

1

14
x


2 ye dx   y  2 xe y dy  0




x
y
x
y
y  2 xe
dx

x
dy
2 ye y
1
x
dx
 x 
dy
y
2e y
½
1
½
substitute x= vy
dx
dv
v y
dy
dy
v y
2
dv
1
dv
1
dy
  v y
    2vdv   
dy
2v
dy
2v
y
v2
  log y  c  v 2   log y  c 
2
x2
y2
x= 0 , y = 1  0 =  log 1 + c  c  0
  log y  c
1
½
½
x2
hence 2   log y
y
(OR)
 dy 
  3x  4 y
 dx 
dy

 e3 x4 y
dx
dy

 e 3 x .e 4 y
dx
log 
1
½
4

dy
  e 3 x dx
4y
e
½
  e  4 y dy   e 3 x dx
e 4 y e 3x

c
4
3
x=0,y=0


1
1 1
1 1
7
 cc     
4 3
4 3
12
½
hence
e 4 y e 3 x 7


4
3 12
15
½
(a, b) * (c, d )  (a  c, b  d )  c  a, d  b  (c, d ) * (a, b)
1
hence * is commutative
(a, b) * (c, d )* (e, f )  (a  c, b  d ) * (e, f )  (a  c  e, b  d  f )
 a  (c  e), b  (d  f )
 (a, b) * (c  e, d  f )  (a, b) * (c, d ) * (e, f )
hence * is associative
let ( c, d) be the identity element of ( a, b ) then ( a, b ) * ( c, d ) = ( a, b )
 (a  c, b  d )  (a, b)  a  c  a., b  d  b  a  0, b  0  N
2
1
identity element for A on * does not exist
(OR)
1
For any a  A , a  a  0 is even  (a, a)  R ,hence R is reflexive
( a, b)  R  a  b is even  b - a is even  (b.a)  R
1
hence R is symmetric
( a, b)  R  a  b is even  a – b is even
( b, c)  R  b  c  b-c is even
a- b + b-c is even  a-c is even  (a, c)  R
hence R is transitive
since R is reflexive , symmetric and transitive it is an equivalence relation
16
1½
½
5𝑥−2
∫ 3𝑥²+2𝑥+1 dx
d
5x-2 =A dx (1+2x+3x2 ) +B = A( 2+ 6x) + B
5
11
,B= 
6
3
5
2  6x
11
1
I  2
dx   2
dx
6 3x  2 x  1
3 3x  2 x  1
1
A=
5
1
I=
5
11
I1  I2
6
3
I 1 = log |3x² + 2x + 1|
1
dx
1
dx
dx
 
I 2 =∫ 3x²+2x+1 == 
2
1 3 
3 2 2
1
2
x  x
x


 
3
3
3
9

dx
=1

2
3 
1 
 x    
3 


1
 x 
3x+1
1 1
)
1 
3   1 tan−1 (


tan
√2
2
 2 
3
2
2
2



3
 3 
3 
I = 5/6 log |3x² + 2x + 1|-
11
3x+1
tan−1
+c
3√2
√2
(OR)
dx
x4

 x( x 5  1)  x 5 ( x 5  1) dx
substitute x 5  t
17
½
4
½
1
dt
I 
5 t (t  1)
1
A
B
 
t (t  1) t t  1
1  A(t  1)  Bt
A  1, B  1
I
1
½
1
5 x dx  dt  x dx  dt
5
4
½
½
1
x5 
1  dt
dt  1
1  log t  1 




log
t

log(
t

1
)


log

C


5   t  t  1 5
5  log( t  1)  5 
( x 5  1) 
x  sin t 
dx
dy
dy
sin t
 cos t , y  cos t 
  sin t 

  tan t
dt
dt
dx
cos t
1½
1
dy

 1 at t 
dx
4

1

1
, y1  cos 
x1  sin 
4
4
2
2
1
equation of tangent
1
y
1 

 1 x 
  2x  2 y  2
2
2

1
equation of normal
1
1 

y
 1 x 
  2x  2 y  0
2
2

1
6
18
x
y
z
x2 1  x3
y2 1 y3  0
z2 1 z3
x
 y
z
x2 1 x
y2 1  y
z2 1 z
x
 y
z
x2 1
1 x
2
y 1  xyz 1 y
z2 1
1 z
x2
y2
z2
½
x3
y3  0
z3
x2
y2  0
z2
1
x2
y2  0
z2
1 x
 (1  xyz 1 y
1 z
R1  R1  R2 , R2  R2  R3
0 x y
 (1  xyz) 0 y  z
1
z
x2  y2
0 1 xy
2
2
y  z  (1  xyz)(x - y)(y - z) 0 1 y  z
z2
1 z
z2
1
 (x - y)(y - z)(z - x)(1  xyz)  0
since x  y  z  1  xyz  0  xyz  1
19
½

 cos x  
tan 1 
,  x 
2
2
 1  sin x 

 cos 2 x  sin 2 x

2
2
tan 1 
2
  cos x  sin x 

2
2





  cos x  sin x  cos x  sin x  

2
2 
2
2
1  
  tan 

2
x
x




 cos  sin 



2
2






x
  cos  sin
2
 tan 1  

x
  cos  sin
2

20
1
lt
x0
lt
x0

 lt
f ( x)  lt
x0
f ( x)  lt
x0 16 
x0
x
x


1  tan 

2
2  tan 1  tan    tan x     x
 tan 1 


x 
x
2  4 2
4


1

tan


2 
2
f ( x)  f (0)
x

16  x  4
x
x  16
 lt
x
x0

16  x  4
 16  x  4  lt
x0
x
x
7
 lt

x0
1+1
4(½)
16  x  4
16  x  4
16  x  4  8
1½
1  cos 4 x
2 sin 2 2 x
2 sin 2 2 x
 sin 2 x 
lt f ( x)  lt
 lt
 lt
 4  8 lt 
 8
2
2
2
x 0 
x 0 
x

0

x

0

2
x

0

x
x
4x
 2x 
 f (0)  a  8
2
Thoughts affecting the continuity while writing exam- ant two factors
Lack of preparation, movement of students in the corridor, friend asking for answer,
announcement of teacher regarding time, noise of the bell , any other distractions. The
thoughts affecting continuity while writing exam are unavoidable at many occasions. We
should learn and practice to avoid external disturbances and concentrate on our work to
reach our goal.
21
Any point on the line
x 1 y z 1
 

3
2
7
1
½
1
½
Q(3  1,2 ,7  1)
1
P lies in the plane x + y – z = 8
1½
3  1  2  7  1  8    3 , hence Q(-8,-6,-22)
equation of line through P( 1 , 3 , 2), Q(-8,-6,-22)
x  x1 y  y1 z  z1
x 1 y  3 z  2 x 1 y  3 z  2








x  x2 y  y 2 z  z 2
9
9
 24
3
3
8
22
tan

1

1

y  x dy  1  y 2 dx
dx tan 1 y  x

dy
1 y2
dx tan 1 y
x


2
dy 1  y
1 y2
½
dx
x
tan 1 y


dy 1  y 2 1  y 2
1
tan 1 y
,
Q

1 y2
1 y2
1
1
 Pdy   1  y 2 dy  tan y
P
e
Pdy
 e tan
1
½
y
1
solution:
xe
  Qe 
Pdy
xetan
1
y

Pdy
½
dy  C
tan 1 y tan 1 y
e
dy
1 y2
8
RHS
tan 1 y tan 1 y
 1  y 2 e dy
substitute
tan 1 y  t
1
dy  dt
1 y2
tan 1 y tan 1 y
t
 1  y 2 e dy   te dt      int egration by parts
u  t  du  dt , dv  e t dt   dv   e t dt  v  e t
 udv  uv   vdu  te   e dt  te
 xetan
1
y
 e tan
1
y
tan
t
t
1
y  1  C
t
 e t  e t t  1  e tan
1
y
tan
1
y  1
1
½
23
½
s  r 2
Surface area = rl  r  l 
r
1
V  r 2 h
3
2
½
2

1 2 4 2 1 2 4 2
1 2 4  s  r 2 
2
  r 2 
 T   r h   r l  r   r 
9
9
9
 r 

2
2
2 4
2 4
 1
1 2 4  s 2  2sr 2   2 r 4 
2
2 4 s  2 sr   r   r


  r 
  r   9 r
9
 2r 2
 2r 2



 s 2  2sr 2 
1
1 2 2
1 2 2
2
4
 T   2r 4 
  T  r s  2 sr  T  s r  2sr
2 2
9

r
9
9










dT 1 2
 2 s r  8sr 3  0  2s 2 r  8sr 3  S  4r 2
dr 9
9



2
1
s  r 2 4r 2  r 2 3r 2


 3r
r
r
r
r
r
1
sin   
   sin 1
l 3r
3
2
d T 1 2
1 2
1 2
4s 2
2
2
2



2
s

24
s

r

2
s

6
s
4

r

2
s

6
s


 0 max
9
9
9
dr 2 9
l





The volume of the cone is maximum when   sin 1

1
1
1
3
(OR)
let length = breadth = x . height = h
volume = 1024cm 3  lbh  1024  x 2 h  1024  h 
1024
x2
10240
 1024 
C  52lb   2.502hl  b   C  10 x 2  5 2 2 x  C  10 x 2 
x
 x 
dC
10240
 1 
 20 x  10240  2   0 
 20 x  x 3  512  x  8
2
dx
x
 x 
d 2C
20480
 2 
 20  10240  3   20 
 0 min imum
2
dx
x3
 x 
 10240 
 =640 + 1280 = 1920
Least cost =10(64) + 
 8 
24
1
1
2
1
1

x sin x
dx
1  sin x
0
I 


(  x) sin(   x)
(  x) sin x
dx  
dx
1  sin(   x)
1  sin x
0
0
I 

2I  
0
 sin x
1  sin x
1
dx
sin x 1  sin x
sin x1  sin x 


dx
2
1

sin
x
1

sin
x
1

sin
x
0
0
½
sin x1  sin x 
dx
2
cos
x
0
½

2I   


2I   


 sin x sin 2 x 
sin x  sin 2 x

dx


0  cos 2 x  cos 2 x d x
cos 2 x
0
10
1




2
    tan x sec x  tan x) d x      tan x sec x  (sec 2 x  1) d x 
0

0





1
2 I   sec x  tan x  x0

  sec   sec 0  tan   tan 0   

2 I    1  1        2  I    2
2
2
(OR)

I 
2
x  sin x
 1  cos x dx
0

I 
2

x
x
cos
2
2 dx
2 x
2 cos
2
x  2 sin
0

I 

2

0
x
x
2
2 cos 2
dx 
2


2
x
x
cos
2
2 dx
2 x
2 cos
2
2 sin
0

1
I 
2
1
½

2
x
0 x sec 2 dx 
2
2
x
 tan 2 dx
1
0
x
 tan 2 dx     Integration by parts
0
x
1
x
u  tan  du  sec 2 dx ,
2
2
2
1
 dv   dx  v  x
 udv  uv   vdu

2
x
x 1

2
 tan 2 dx  x tan 2  2  x sec
0
0

2
1½
2
x
dx
2

2

1
x
x 1
x
x 2   

I   x sec 2 dx  x tan   x sec 2 dx   x tan   tan 
20
2
2 20
2
20 2 4 2

11
1
25
Point of intersection of x = 4y- 2 and x 2  4 y
x 2  x  2  x 2  x  2  0  ( x  2)( x  1)  0  x  1,2
1
1
Area of the shaded region
1
x2
x2
dx

 4
 4 dx
1
1
2
2
1
2
1  x2
x3 
   2x  
4 2
3  1
26

1 
1
8 1 
2    22  1     
4 
2
 3 3

1 3
 1 3  1 9 9
 6  3    3     square units

4 2
 4 2  4 2 8
2
 1 2  3  x    4 

   
2  y    2 
2 3
 3  3  4  z   11 

   
AX  B
A  1(6)  2(14)  3(15)  6  28  45  67
1
 6

adjA   14
  15

2
13 

 8
9
 1


6
17
13 


adjA
1 
1
A


5
 8
 14
A
67 
9
 1
  15

17
13   4 
 6


1 
X  A 1 B 
5
 8  2 
 14
67 


9
 1
  15
 11 
 201 
 3 



1 


134



  2
67 

 1 
 67 


x  3, y  2, z  1
17
5
12
½
1
1
½
27
Let A denote the missing card as red, B missing card as black, C denote the first 13 cards
½
drawn are red
P( A)  P( B) 
P B C  
 
 
25
26
C
C
26 1 C
 ,P
 51 13 , P C  51 13
A
B
52 2
C13
C13
 B
PB P C
 A PB PC B 
P ( A) P C
1
1 26C13
26!
 51
26
2
C13
C13
13 !13!
2

 25


25
26
26
25!
26!
3
C
C
C13  C13
1
1

 51 13   51 13
13! 12! 13 !13!
2
C13 2
C13
If the probability rate is high the player decides to go for it in the game of gambling.But
the success rate will not be constant. By understanding the imbalance between success
and failure rate one should not go for gambling game.
28
1½
2
1
Let the no of kilograms of fertilizer of type I used be x kg and type II be y kg
Minimize : z= 2x + 3y
1
subject to
10
5
x
y  14  10 x  5 y  1400  2 x  y  280
100
100
6
10
x
y  14  6 x  10 y  1400  3 x  5 y  700
100
100
1
x, y  0
1½
13
corner points
Z=2x+3y in rupees
(100,80)
440------ minimum
(0,280)
840
(700/3, 0)
1400/3
Hence 100kg of type I and 80 kg of type Ii have to be used to minimize the cost.
Natural fertilizers- any two
bird seed bags, seaweed, fish emulsion,seaweed,animals and human
waste,earthworms,ashes,vegetable waste
29
1
½
1
Equation of plane passing through the points A ( 2 , 1 , 1) , B ( 1 , 2, 1 ) and C ( 1 , 1 , 2).
x  x1
x 2  x1
x3  x1
x2
1
1
y  y1
y 2  y1
y 3  y1
z  z1
z 2  z1  0
z 3  z1
1
y 1 z 1
1
0 0
0
1
2
 x y z40
Equation of line from P (1,1,1)
x 1 y 1 z 1



1
1
1
Q (  1,   1,   1)
1
Q lies in the plane x + y + z – 4 = 0
  1  1  1 4  0
1
 3  1  0   
3
4 4 4
foot of perpendicu lar is Q , , 
3 3 3
½
½
distance from P(1,1,1,) on the plane x + y + z – 4 =0

111 4
3

1
1
3
14