Sample Solution Paper 5 - Math, Class 12 JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : Sample Solution Paper 5 - Math, Class 12 JEE Notes | EduRev

 Page 1


  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper Solution 
Sample Paper – 5 
      
SECTION – A 
 
1. Matrix A is a matrix of order 2. 
Identity matrix of second order is 
10
01
??
??
??
 
For A to be an identity matrix,  
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
2. We have, 
 
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
 
 
 
 
 
 
 
 
 
 
 
 
Page 2


  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper Solution 
Sample Paper – 5 
      
SECTION – A 
 
1. Matrix A is a matrix of order 2. 
Identity matrix of second order is 
10
01
??
??
??
 
For A to be an identity matrix,  
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
2. We have, 
 
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
 
OR 
 
? Since f(x) is continuous at x 3. 
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
 
 
 
3. We have, 
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
 
 
4. Equation of line through (-2, 1, 3) and parallel to 
x 3 y 4 z 8
3 5 6
? ? ?
?? is  
      
x 2 y 1 z 3
3 5 6
? ? ?
?? 
  
Page 3


  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper Solution 
Sample Paper – 5 
      
SECTION – A 
 
1. Matrix A is a matrix of order 2. 
Identity matrix of second order is 
10
01
??
??
??
 
For A to be an identity matrix,  
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
2. We have, 
 
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
 
OR 
 
? Since f(x) is continuous at x 3. 
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
 
 
 
3. We have, 
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
 
 
4. Equation of line through (-2, 1, 3) and parallel to 
x 3 y 4 z 8
3 5 6
? ? ?
?? is  
      
x 2 y 1 z 3
3 5 6
? ? ?
?? 
  
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
Section-B 
 
5. We shall prove by principle of mathematical induction 
Here, let 
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
 
? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??
  
? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?
  
Assuming result to be true for n = k i.e. P(k) to be true 
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??
  
We have to prove P (k +1) is true, 
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?
   
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??
 
? P (k + 1) is true.
  
Thus by principle of mathematical induction  
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
 for all nN ?   
     
 
 
 
 
 
 
 
 
 
 
Page 4


  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper Solution 
Sample Paper – 5 
      
SECTION – A 
 
1. Matrix A is a matrix of order 2. 
Identity matrix of second order is 
10
01
??
??
??
 
For A to be an identity matrix,  
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
2. We have, 
 
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
 
OR 
 
? Since f(x) is continuous at x 3. 
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
 
 
 
3. We have, 
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
 
 
4. Equation of line through (-2, 1, 3) and parallel to 
x 3 y 4 z 8
3 5 6
? ? ?
?? is  
      
x 2 y 1 z 3
3 5 6
? ? ?
?? 
  
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
Section-B 
 
5. We shall prove by principle of mathematical induction 
Here, let 
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
 
? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??
  
? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?
  
Assuming result to be true for n = k i.e. P(k) to be true 
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??
  
We have to prove P (k +1) is true, 
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?
   
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??
 
? P (k + 1) is true.
  
Thus by principle of mathematical induction  
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
 for all nN ?   
     
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
OR 
 
? Since A is skew symmetric matrix. 
??
? ? ?
? ? ?
? ? ?
? ? ? ? ?
??
??
T
T
T3
T
T
Therefore,A A
AA
A ( 1) A
AA
A A ......(Since A A )
2 A 0
A0
 
 
 
6. Let P(x, y) be any point on the given curve x
2
 + y
2
 – 2x – 3 = 0. 
       Tangent to the curve at the point (x, y) is given by 
dy
dx
. 
       Differentiating the equation of the curve w .r. t. x we get  
     
dy
2x 2y 2 0
dx
dy 2 2x 1 x
dx 2y y
? ? ?
??
??
  
       Let P(x1, y1) be the point on the given curve at which the tangents are parallel to the x 
axis 
    
? ?
dy
dx
x ,y
11
?
?
?
?
 = 0 
      ?
1x
1
y
1
?
 = 0 
      ? 1 – x1 = 0 
      ? x1 = 1 
     To get the value of y1 just substitute x1 = 1 in the equation x
2
 + y
2
 – 2x – 3 = 0, we get 
? ?
2
2
1 y 2 1 3 0
1
2
y 4 0
1
2
y4
1
y2
1
? ? ? ? ?
? ? ?
??
? ? ?
  
         So the points on the given curve at which the tangents are parallel to the x-axis are 
         (1,  2) and (1,  -2).  
 
Page 5


  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper Solution 
Sample Paper – 5 
      
SECTION – A 
 
1. Matrix A is a matrix of order 2. 
Identity matrix of second order is 
10
01
??
??
??
 
For A to be an identity matrix,  
1 0 cos sin
0 1 sin cos
cos 1 andsin 0
o
cos cos0 and sin sin0
0
Thus, for 0 , A is an identity matrix
? ? ? ? ? ? ?
?
? ? ? ?
??
? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ?
 
 
2. We have, 
 
? ?
? ? ? ?
2
x 25
f(x) x 5
x5
k x 5
2
x 25
lim f 5
x5 x5
x 5 x 5
lim k
x5 x5
lim x 5 k
x5
5k
?
??
?
??
?
?
? ?
??
?
? ?
??
?
?
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
 
OR 
 
? Since f(x) is continuous at x 3. 
?
?
?
?
?
??
??
??
?
? ? ? ?
??
?
??
??
?
? ? ? ? ? ?
? ? ?
??
x3
2
x3
x3
x3
x3
lim f(x) f(3)
(x 3) 36
lim k
x3
(x 3 6)(x 3 6)
lim k
x3
(x 9)(x 3)
lim k
x3
lim (x 9) k ( x 3, x 3 0)
3 9 k
k 12
 
 
 
3. We have, 
1
I= dx
2
9 25x
11
dx
52
3
2
x
5
1 5 x
1
sin c
3
53
5
1 5x
1
sin c
33
?
?
?
?
??
?
??
??
??
??
?
? ? ?
??
??
??
??
?
??
??
??
 
 
4. Equation of line through (-2, 1, 3) and parallel to 
x 3 y 4 z 8
3 5 6
? ? ?
?? is  
      
x 2 y 1 z 3
3 5 6
? ? ?
?? 
  
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
Section-B 
 
5. We shall prove by principle of mathematical induction 
Here, let 
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
 
? ?
cosn sinn
n
P n : A ,
sinn cosn
?? ??
?
??
? ? ?
??
  
? ?
cos sin
1
So, A
sin cos
P 1 is true.
?? ??
?
??
? ? ?
??
?
  
Assuming result to be true for n = k i.e. P(k) to be true 
? ?
cosk sink
k
P k : A
sink cosk
?? ??
?
??
? ? ?
??
  
We have to prove P (k +1) is true, 
? ?
k 1 1 k
P k 1 : A A A
cos sin cosk sink
k1
A
sin cos sink cosk
?
??
? ? ? ? ? ? ? ?
?
??
? ? ? ?
? ? ? ? ? ?
? ? ? ?
   
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
cos cosk sin sink cos sink sin cosk
sin cosk cos sink sin sink cos cosk
cos k sin k
sin k cos k
cos k 1 sin k 1
sin k 1 cos k 1
? ? ? ? ? ? ? ? ? ? ??
?
??
? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
?
??
? ? ? ? ?
??
 
? P (k + 1) is true.
  
Thus by principle of mathematical induction  
cosn sinn
n
A
sinn cosn
?? ??
?
??
? ? ?
??
 for all nN ?   
     
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
OR 
 
? Since A is skew symmetric matrix. 
??
? ? ?
? ? ?
? ? ?
? ? ? ? ?
??
??
T
T
T3
T
T
Therefore,A A
AA
A ( 1) A
AA
A A ......(Since A A )
2 A 0
A0
 
 
 
6. Let P(x, y) be any point on the given curve x
2
 + y
2
 – 2x – 3 = 0. 
       Tangent to the curve at the point (x, y) is given by 
dy
dx
. 
       Differentiating the equation of the curve w .r. t. x we get  
     
dy
2x 2y 2 0
dx
dy 2 2x 1 x
dx 2y y
? ? ?
??
??
  
       Let P(x1, y1) be the point on the given curve at which the tangents are parallel to the x 
axis 
    
? ?
dy
dx
x ,y
11
?
?
?
?
 = 0 
      ?
1x
1
y
1
?
 = 0 
      ? 1 – x1 = 0 
      ? x1 = 1 
     To get the value of y1 just substitute x1 = 1 in the equation x
2
 + y
2
 – 2x – 3 = 0, we get 
? ?
2
2
1 y 2 1 3 0
1
2
y 4 0
1
2
y4
1
y2
1
? ? ? ? ?
? ? ?
??
? ? ?
  
         So the points on the given curve at which the tangents are parallel to the x-axis are 
         (1,  2) and (1,  -2).  
 
  
 
CBSE XII | Mathematics 
Sample Paper  5 Solution 
 
  
7. Here, 
       
33
x asin t , y bcos t ??    
 Differentiating (1) wrt t 
      
dx
2
3asin t cost
dt
?? and 
       
dy
2
3bcos t sint
dt
? ? ? 
      
dy
2
dy 3bcos t sint b
dt
cot t
2
dx
dx a
3asin t cost
dt
??
? ? ? ?
?
 
 
 
? Slope of the tangent at t
2
?
? 
 
dy b
cot 0
dx a 2
2
? ?
? ? ?
?
?
?
  
         Hence, equation of tangent is given by 
  
3
y bcos 0  or  y 0
2
?
? ? ?   
 
8. The volume of a sphere(V) with radius (r) is given by, 
     
4
3
Vr
3
?? 
       ?Rate of change of volume (V) w.r.t. (t) is given by, 
      
?
??
??
??
??
??
?
? ? ?
??
??
?
?
dV dV dr
.
dt dr dt
d 4 dr
3
r.
dr 3 dt
dr
2
4 r .
dt
dV
3
It is given that 900 cm / s
dt
dr 900 225
22
dt
4 r r
Therefore, when radius = 15 cm
dr 225 1
2
dt
(15)
Hence,the rate at which the radius of the balloon 
increases whe
?
1
n the radius is 15 cm is cm / s.
 
 
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