CBSE Past Year Paper Session (2015) Solutions, Math Class 12 JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : CBSE Past Year Paper Session (2015) Solutions, Math Class 12 JEE Notes | EduRev

 Page 1


  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper – 2015  Solution 
All India   
      
SECTION – A 
  
1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k 
? ? ? ? ? ?
?
??
?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
2 2 2
We need to find a b
i j k
a b 2 1 3
3 5 2
i 2 15 j 4 9 k 10 3
17i 13j 7k
Hence, a b 17 13 7
a b 507
 
 
2. ? ? ? ? Let a i j; b j k 
? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ?
? ? ?
? ? ?
?
? ? ? ?
? ? ? ?
2
22
2
22
a b i j j k 1 0 1 1 0 1 1
a 1 1 0 2
b 0 1 1 2
We know that a b a b cos
a b 1 1
Thus, cos =
2
22 ab
cos cos120
120
 
  
 
 
 
 
 
 
 
Page 2


  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper – 2015  Solution 
All India   
      
SECTION – A 
  
1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k 
? ? ? ? ? ?
?
??
?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
2 2 2
We need to find a b
i j k
a b 2 1 3
3 5 2
i 2 15 j 4 9 k 10 3
17i 13j 7k
Hence, a b 17 13 7
a b 507
 
 
2. ? ? ? ? Let a i j; b j k 
? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ?
? ? ?
? ? ?
?
? ? ? ?
? ? ? ?
2
22
2
22
a b i j j k 1 0 1 1 0 1 1
a 1 1 0 2
b 0 1 1 2
We know that a b a b cos
a b 1 1
Thus, cos =
2
22 ab
cos cos120
120
 
  
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
3. Consider the vector equation of the plane. 
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
?
? ? ?
1 1 1
r 6i 3j 2k 4
xi yj zk 6i 3j 2k 4
6x 3y 2z 4
6x 3y 2z 4 0
Thus the Cartesian equation of the plane is
6x 3y 2z 4 0
Let d be the distance between the point 2, 5, 3
to the plane.
ax by cz d
Thus, d=
a
? ?
? ?
??
? ? ? ? ? ? ?
??
? ? ?
? ? ?
??
??
?
??
??
2 2 2
2
22
bc
6 2 3 5 2 3 4
d
6 3 2
12 15 6 4
d
36 9 4
13
d
49
13
d units
7
 
  
4. Given that of a ij = e
2ix
sin(jx) 
? ? ? ?
??
? ? ?
2 1 x 2x
12
Substitute i = 1 and j = 2
Thus, a e sin 2 x e sin 2x
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Page 3


  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper – 2015  Solution 
All India   
      
SECTION – A 
  
1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k 
? ? ? ? ? ?
?
??
?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
2 2 2
We need to find a b
i j k
a b 2 1 3
3 5 2
i 2 15 j 4 9 k 10 3
17i 13j 7k
Hence, a b 17 13 7
a b 507
 
 
2. ? ? ? ? Let a i j; b j k 
? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ?
? ? ?
? ? ?
?
? ? ? ?
? ? ? ?
2
22
2
22
a b i j j k 1 0 1 1 0 1 1
a 1 1 0 2
b 0 1 1 2
We know that a b a b cos
a b 1 1
Thus, cos =
2
22 ab
cos cos120
120
 
  
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
3. Consider the vector equation of the plane. 
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
?
? ? ?
1 1 1
r 6i 3j 2k 4
xi yj zk 6i 3j 2k 4
6x 3y 2z 4
6x 3y 2z 4 0
Thus the Cartesian equation of the plane is
6x 3y 2z 4 0
Let d be the distance between the point 2, 5, 3
to the plane.
ax by cz d
Thus, d=
a
? ?
? ?
??
? ? ? ? ? ? ?
??
? ? ?
? ? ?
??
??
?
??
??
2 2 2
2
22
bc
6 2 3 5 2 3 4
d
6 3 2
12 15 6 4
d
36 9 4
13
d
49
13
d units
7
 
  
4. Given that of a ij = e
2ix
sin(jx) 
? ? ? ?
??
? ? ?
2 1 x 2x
12
Substitute i = 1 and j = 2
Thus, a e sin 2 x e sin 2x
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
5. Consider the equation, y = mx, where m is the parameter. 
Thus, the above equation represents the family of lines which pass through the origin. 
y mx....(1)
y
m....(2)
x
?
??
 
Differentiating the above equation (1) with respect to x,  
? ?
?
??
??
??
? ? ?
??
y mx
dy
m1
dx
dy
m
dx
dy y
from equation (2)
dx x
dy y
0
dx x
Thus we have eliminated the constant, m.
The required differential equation is
dy y
0
dx x
 
  
6. Consider the given differential equation: 
dy
xlog x y 2log x
dx
Dividing the above equation by xlogx, we have,
xlog x dy y 2log x
xlog x dx xlog x xlog x
dy y 2
....(1)
dx xlog x x
Consider the general linear differential equation,
dy
Py Q,where P and Q are funct
dx
??
??
? ? ?
?? ions of x
 
? ? ? ?
Pdx
dx
Pdx
x log x
Comparing equation (1) and the general equation, we have,
12
P x and Q x
xlog x x
The integrating factor is given by the formula e
Thus,I.F. e e
??
?
?
?
??
 
Page 4


  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper – 2015  Solution 
All India   
      
SECTION – A 
  
1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k 
? ? ? ? ? ?
?
??
?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
2 2 2
We need to find a b
i j k
a b 2 1 3
3 5 2
i 2 15 j 4 9 k 10 3
17i 13j 7k
Hence, a b 17 13 7
a b 507
 
 
2. ? ? ? ? Let a i j; b j k 
? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ?
? ? ?
? ? ?
?
? ? ? ?
? ? ? ?
2
22
2
22
a b i j j k 1 0 1 1 0 1 1
a 1 1 0 2
b 0 1 1 2
We know that a b a b cos
a b 1 1
Thus, cos =
2
22 ab
cos cos120
120
 
  
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
3. Consider the vector equation of the plane. 
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
?
? ? ?
1 1 1
r 6i 3j 2k 4
xi yj zk 6i 3j 2k 4
6x 3y 2z 4
6x 3y 2z 4 0
Thus the Cartesian equation of the plane is
6x 3y 2z 4 0
Let d be the distance between the point 2, 5, 3
to the plane.
ax by cz d
Thus, d=
a
? ?
? ?
??
? ? ? ? ? ? ?
??
? ? ?
? ? ?
??
??
?
??
??
2 2 2
2
22
bc
6 2 3 5 2 3 4
d
6 3 2
12 15 6 4
d
36 9 4
13
d
49
13
d units
7
 
  
4. Given that of a ij = e
2ix
sin(jx) 
? ? ? ?
??
? ? ?
2 1 x 2x
12
Substitute i = 1 and j = 2
Thus, a e sin 2 x e sin 2x
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
5. Consider the equation, y = mx, where m is the parameter. 
Thus, the above equation represents the family of lines which pass through the origin. 
y mx....(1)
y
m....(2)
x
?
??
 
Differentiating the above equation (1) with respect to x,  
? ?
?
??
??
??
? ? ?
??
y mx
dy
m1
dx
dy
m
dx
dy y
from equation (2)
dx x
dy y
0
dx x
Thus we have eliminated the constant, m.
The required differential equation is
dy y
0
dx x
 
  
6. Consider the given differential equation: 
dy
xlog x y 2log x
dx
Dividing the above equation by xlogx, we have,
xlog x dy y 2log x
xlog x dx xlog x xlog x
dy y 2
....(1)
dx xlog x x
Consider the general linear differential equation,
dy
Py Q,where P and Q are funct
dx
??
??
? ? ?
?? ions of x
 
? ? ? ?
Pdx
dx
Pdx
x log x
Comparing equation (1) and the general equation, we have,
12
P x and Q x
xlog x x
The integrating factor is given by the formula e
Thus,I.F. e e
??
?
?
?
??
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
? ? ? ?
? ?
?
??
?
? ? ?
?
?
dx
log log x
x log x
dx
Consider I=
xlog x
dx
Substituting logx=t; dt
x
dt
Thus I= log t log log x
t
Hence,I.F. e e log x
 
 
SECTION – B 
7.    
2
1 2 2
A 2 1 2
2 2 1
1 2 2 1 2 2
A 2 1 2 2 1 2
2 2 1 2 2 1
 
      
1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1
2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1
2 1 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1
1 4 4 2 2 4 2 4 2
2 2 4 4 1 4 4 2 2
2 4 2 4 2 2 4 4 1
9 8 8
8 9 8
8 8 9
 
     
2
Consider A 4A 5I
9 8 8 1 2 2 1 0 0
8 9 8 4 2 1 2 5 0 1 0
8 8 9 2 2 1 0 0 1
9 8 8 4 8 8 5 0 0
8 9 8 8 4 8 0 5 0
8 8 9 8 8 4 0 0 5
 
 
9 9 8 8 8 8
8 8 9 9 8 8
8 8 8 8 9 9
 
Page 5


  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
CBSE Board 
Class XII Mathematics 
Board Paper – 2015  Solution 
All India   
      
SECTION – A 
  
1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k 
? ? ? ? ? ?
?
??
?
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
2 2 2
We need to find a b
i j k
a b 2 1 3
3 5 2
i 2 15 j 4 9 k 10 3
17i 13j 7k
Hence, a b 17 13 7
a b 507
 
 
2. ? ? ? ? Let a i j; b j k 
? ? ? ?
? ? ? ?
? ?
? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ?
? ? ?
? ? ?
? ? ?
?
? ? ? ?
? ? ? ?
2
22
2
22
a b i j j k 1 0 1 1 0 1 1
a 1 1 0 2
b 0 1 1 2
We know that a b a b cos
a b 1 1
Thus, cos =
2
22 ab
cos cos120
120
 
  
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
3. Consider the vector equation of the plane. 
? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ?
? ? ? ?
?
? ? ?
1 1 1
r 6i 3j 2k 4
xi yj zk 6i 3j 2k 4
6x 3y 2z 4
6x 3y 2z 4 0
Thus the Cartesian equation of the plane is
6x 3y 2z 4 0
Let d be the distance between the point 2, 5, 3
to the plane.
ax by cz d
Thus, d=
a
? ?
? ?
??
? ? ? ? ? ? ?
??
? ? ?
? ? ?
??
??
?
??
??
2 2 2
2
22
bc
6 2 3 5 2 3 4
d
6 3 2
12 15 6 4
d
36 9 4
13
d
49
13
d units
7
 
  
4. Given that of a ij = e
2ix
sin(jx) 
? ? ? ?
??
? ? ?
2 1 x 2x
12
Substitute i = 1 and j = 2
Thus, a e sin 2 x e sin 2x
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
5. Consider the equation, y = mx, where m is the parameter. 
Thus, the above equation represents the family of lines which pass through the origin. 
y mx....(1)
y
m....(2)
x
?
??
 
Differentiating the above equation (1) with respect to x,  
? ?
?
??
??
??
? ? ?
??
y mx
dy
m1
dx
dy
m
dx
dy y
from equation (2)
dx x
dy y
0
dx x
Thus we have eliminated the constant, m.
The required differential equation is
dy y
0
dx x
 
  
6. Consider the given differential equation: 
dy
xlog x y 2log x
dx
Dividing the above equation by xlogx, we have,
xlog x dy y 2log x
xlog x dx xlog x xlog x
dy y 2
....(1)
dx xlog x x
Consider the general linear differential equation,
dy
Py Q,where P and Q are funct
dx
??
??
? ? ?
?? ions of x
 
? ? ? ?
Pdx
dx
Pdx
x log x
Comparing equation (1) and the general equation, we have,
12
P x and Q x
xlog x x
The integrating factor is given by the formula e
Thus,I.F. e e
??
?
?
?
??
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
? ? ? ?
? ?
?
??
?
? ? ?
?
?
dx
log log x
x log x
dx
Consider I=
xlog x
dx
Substituting logx=t; dt
x
dt
Thus I= log t log log x
t
Hence,I.F. e e log x
 
 
SECTION – B 
7.    
2
1 2 2
A 2 1 2
2 2 1
1 2 2 1 2 2
A 2 1 2 2 1 2
2 2 1 2 2 1
 
      
1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1
2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1
2 1 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1
1 4 4 2 2 4 2 4 2
2 2 4 4 1 4 4 2 2
2 4 2 4 2 2 4 4 1
9 8 8
8 9 8
8 8 9
 
     
2
Consider A 4A 5I
9 8 8 1 2 2 1 0 0
8 9 8 4 2 1 2 5 0 1 0
8 8 9 2 2 1 0 0 1
9 8 8 4 8 8 5 0 0
8 9 8 8 4 8 0 5 0
8 8 9 8 8 4 0 0 5
 
 
9 9 8 8 8 8
8 8 9 9 8 8
8 8 8 8 9 9
 
  
 
CBSE XII | Mathematics 
Board Paper 2015 – All India Set – 1 Solution 
 
  
 
000
000
000
 
2
2
2 1 1 1 1
1
1
Now
A 4A 5I 0
A 4A 5I
A A 4AA 5IA Postmultiply by A
A 4I 5A
1 2 2 4 0 0
2 1 2 0 4 0 5A
2 2 1 0 0 4
 
1
1
3 2 2
2 3 2 5A
2 2 3
3 2 2
555
2 3 2
A
555
2 2 3
555
 
OR 
 
1
1
1
2 0 1
A 5 1 0
0 1 3
2 3 0 0 15 0 1 5 0
6 0 5
1
0
Hence A exists.
A A I
2 0 1 1 0 0
A 5 1 0 0 1 0
0 1 3 0 0 1
 
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