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CBSE XII | Mathematics
Board Paper 2015 – All India Set – 1 Solution
CBSE Board
Class XII Mathematics
Board Paper – 2015 Solution
All India Set – 1
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 Set – 1
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 Set – 1
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 Set – 1
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
4
? ? ? ?
? ?
?
??
?
? ? ?
?
?
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 Set – 1
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
4
? ? ? ?
? ?
?
??
?
? ? ?
?
?
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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FAQs on Past Year Paper - Solutions, Mathematics (Set - 1), 2015, Class 12, Maths - Additional Study Material for JEE
| 1. What is the format of the Mathematics JEE exam for Class 12? |  |
Ans. The Mathematics JEE exam for Class 12 typically follows a multiple-choice question format. Students are required to choose the correct option from the given choices.
| 2. How many questions are there in the Mathematics JEE exam for Class 12? | |
Ans. The number of questions in the Mathematics JEE exam for Class 12 can vary from year to year. However, on average, there are around 25-30 questions in the exam.
| 3. What topics are covered in the Mathematics JEE exam for Class 12? | |
Ans. The Mathematics JEE exam for Class 12 covers a wide range of topics including algebra, calculus, trigonometry, coordinate geometry, and statistics. It is important for students to have a thorough understanding of these topics to perform well in the exam.
| 4. Are calculators allowed in the Mathematics JEE exam for Class 12? | |
Ans. No, calculators are not allowed in the Mathematics JEE exam for Class 12. Students are expected to solve the problems using their mathematical skills and knowledge.
| 5. How can I prepare effectively for the Mathematics JEE exam for Class 12? | |
Ans. To prepare effectively for the Mathematics JEE exam for Class 12, it is important to practice solving a variety of problems from different topics. Additionally, referring to past year papers and solving sample papers can help in understanding the exam pattern and identifying areas that need improvement. Seeking guidance from teachers or joining coaching classes can also be beneficial in preparing for the exam.
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