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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
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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FAQs on CBSE Past Year Paper Session (2015) Solutions, Math Class 12 - Mathematics (Maths) Class 12 - JEE
| 1. How can I access the CBSE past year paper solutions for the Math Class 12 exam held in 2015? |  |
Ans. You can access the CBSE past year paper solutions for the Math Class 12 exam held in 2015 by searching for them online. Many websites and educational platforms provide these solutions for free or for a nominal fee. You can also check the official CBSE website or refer to popular study material books for these solutions.
| 2. Are the CBSE past year paper solutions for Math Class 12 exam in 2015 helpful for exam preparation? | |
Ans. Yes, the CBSE past year paper solutions for the Math Class 12 exam held in 2015 are extremely helpful for exam preparation. These solutions provide a clear understanding of the paper pattern, important topics, and the level of difficulty of questions asked in the actual exam. By practicing these solutions, you can familiarize yourself with the exam format and improve your problem-solving skills.
| 3. How can solving CBSE past year paper solutions for Math Class 12 exam in 2015 benefit me? | |
Ans. Solving CBSE past year paper solutions for the Math Class 12 exam held in 2015 can benefit you in several ways. It helps you identify your strengths and weaknesses, understand the marking scheme, and improve time management skills. Additionally, it allows you to practice a wide range of questions and enhances your problem-solving abilities. By analyzing your performance in these past year papers, you can also gauge your preparedness for the actual exam.
| 4. Can I rely solely on CBSE past year paper solutions for Math Class 12 exam in 2015 for my exam preparation? | |
Ans. While CBSE past year paper solutions for the Math Class 12 exam held in 2015 are valuable resources for exam preparation, it is not advisable to rely solely on them. These solutions should be used in conjunction with other study materials, textbooks, and reference books. It is important to have a comprehensive understanding of the entire syllabus, practice different types of questions, and seek clarification from teachers or mentors when required.
| 5. Where can I find additional practice questions apart from CBSE past year paper solutions for Math Class 12 exam in 2015? | |
Ans. Apart from CBSE past year paper solutions for the Math Class 12 exam held in 2015, you can find additional practice questions in various study materials, textbooks, and reference books specifically designed for the Class 12 Math syllabus. Additionally, online platforms, educational websites, and mobile applications offer a wide range of practice questions, quizzes, and mock tests for Math Class 12. You can also join coaching institutes or attend online classes that provide additional practice material and question banks.
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