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CBSE XII | Mathematics
Sample Paper – 7 Solution
Mathematics
Class XII
Sample Paper – 7 Solution
SECTION – A
1. Element at 3
rd
column and 2
nd
row
So i =2 and j =3
Substituting in
ij
i 2j
a
2
?
? we get
23
2 2 3 8
a4
22
??
? ? ?
2. y + sin y = cos x
differentiating w.r.t. x, we get,
? ?
dd
y sin y cosx
dx dx
dy dy
cosy sin x
dx dx
dy sin x
dx 1 cosy
??
? ? ?
?
?
?
3.
2
3
dy 1
2
dy
dx
dx
rearranging
dy dy
12
dx dx
??
??
??
??
??
??
??
??
Order: 1
Degree:3
4. The vector equation of the line passing through the point (5, 2,-4) and parallel to
ˆ ˆˆ
3i 2j 8k ?? .
? ? ? ?
ˆˆ ˆ ˆ ˆ ˆ
r 5i 2j 4k 3i 2j 8k ? ? ? ? ? ? ?
Page 2
CBSE XII | Mathematics
Sample Paper – 7 Solution
Mathematics
Class XII
Sample Paper – 7 Solution
SECTION – A
1. Element at 3
rd
column and 2
nd
row
So i =2 and j =3
Substituting in
ij
i 2j
a
2
?
? we get
23
2 2 3 8
a4
22
??
? ? ?
2. y + sin y = cos x
differentiating w.r.t. x, we get,
? ?
dd
y sin y cosx
dx dx
dy dy
cosy sin x
dx dx
dy sin x
dx 1 cosy
??
? ? ?
?
?
?
3.
2
3
dy 1
2
dy
dx
dx
rearranging
dy dy
12
dx dx
??
??
??
??
??
??
??
??
Order: 1
Degree:3
4. The vector equation of the line passing through the point (5, 2,-4) and parallel to
ˆ ˆˆ
3i 2j 8k ?? .
? ? ? ?
ˆˆ ˆ ˆ ˆ ˆ
r 5i 2j 4k 3i 2j 8k ? ? ? ? ? ? ?
CBSE XII | Mathematics
Sample Paper – 7 Solution
OR
The vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6) is
? ? ? ?
ˆˆ ˆ ˆ ˆ
r i 2k 4i 4j 4k ? ? ? ? ? ? ?
SECTION – B
5.
A = {1, 2, 3, 4, 5}
R = {(a, b): ab ? is even}
For R to be an equivalence relation it must be
??
? ? ? ?
(i) Reflexive, a a 0
(a,a) R for a A
So R is reflexive.
(ii) Symmetric,
if (a,b) R a b is even
b a is also even
? ? ?
??
So R is symmetric.
(iii) Transitive
If (a, b) ?R (b, c) ? R then (a, c) ?R
(a, b) ? ? ? R a b is even
(b, c) ? R ?? bc is even
Sum of two even numbers is even
So, a b b c
a b b c a c iseven since, a b and b c are even
? ? ?
? ? ? ? ? ? ? ?
So (a, c) ? R
Hence, R is transitive.
Therefore, R is an equivalence relation.
Page 3
CBSE XII | Mathematics
Sample Paper – 7 Solution
Mathematics
Class XII
Sample Paper – 7 Solution
SECTION – A
1. Element at 3
rd
column and 2
nd
row
So i =2 and j =3
Substituting in
ij
i 2j
a
2
?
? we get
23
2 2 3 8
a4
22
??
? ? ?
2. y + sin y = cos x
differentiating w.r.t. x, we get,
? ?
dd
y sin y cosx
dx dx
dy dy
cosy sin x
dx dx
dy sin x
dx 1 cosy
??
? ? ?
?
?
?
3.
2
3
dy 1
2
dy
dx
dx
rearranging
dy dy
12
dx dx
??
??
??
??
??
??
??
??
Order: 1
Degree:3
4. The vector equation of the line passing through the point (5, 2,-4) and parallel to
ˆ ˆˆ
3i 2j 8k ?? .
? ? ? ?
ˆˆ ˆ ˆ ˆ ˆ
r 5i 2j 4k 3i 2j 8k ? ? ? ? ? ? ?
CBSE XII | Mathematics
Sample Paper – 7 Solution
OR
The vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6) is
? ? ? ?
ˆˆ ˆ ˆ ˆ
r i 2k 4i 4j 4k ? ? ? ? ? ? ?
SECTION – B
5.
A = {1, 2, 3, 4, 5}
R = {(a, b): ab ? is even}
For R to be an equivalence relation it must be
??
? ? ? ?
(i) Reflexive, a a 0
(a,a) R for a A
So R is reflexive.
(ii) Symmetric,
if (a,b) R a b is even
b a is also even
? ? ?
??
So R is symmetric.
(iii) Transitive
If (a, b) ?R (b, c) ? R then (a, c) ?R
(a, b) ? ? ? R a b is even
(b, c) ? R ?? bc is even
Sum of two even numbers is even
So, a b b c
a b b c a c iseven since, a b and b c are even
? ? ?
? ? ? ? ? ? ? ?
So (a, c) ? R
Hence, R is transitive.
Therefore, R is an equivalence relation.
CBSE XII | Mathematics
Sample Paper – 7 Solution
6.
Let
23
B
14
??
?
??
?
??
and
36
C
38
? ??
?
??
?
??
.Then, the given matrix equation is A + B = C
now,
A + B – B = C – B
A = C – B
3 6 2 3
A
3 8 1 4
3 2 6 3
A
3 1 8 4
19
A
24
? ? ? ? ?
??
? ? ? ?
??
? ? ? ?
? ? ? ??
?
??
? ? ?
??
? ??
?
??
?
??
7.
It is known that,
1
sin AsinB cos A B cos A B
2
1
sin xsin2xsin3x dx sinx cos 2x 3x cos 2x 3x
2
1
sin xcos x sin xcos5x dx
2
1
sin xcosx sin xcos5x dx
2
1 sin2x 1
dx sin xcos5x
2 2 2
1 cos2x 1 1
sin x 5x sin x 5x dx
4 2 2 2
cos2x 1
sin6x sin 4x dx
84
cos2x 1 cos6x cos4x
C
8 4 6 4
cos2x 1 cos6x cos4x
C
8 8 3 2
6cos2x 1 2cos6x 3cos4x
C
48 8 6
1
2cos6x 3cos4x 6cos2x C
48
Page 4
CBSE XII | Mathematics
Sample Paper – 7 Solution
Mathematics
Class XII
Sample Paper – 7 Solution
SECTION – A
1. Element at 3
rd
column and 2
nd
row
So i =2 and j =3
Substituting in
ij
i 2j
a
2
?
? we get
23
2 2 3 8
a4
22
??
? ? ?
2. y + sin y = cos x
differentiating w.r.t. x, we get,
? ?
dd
y sin y cosx
dx dx
dy dy
cosy sin x
dx dx
dy sin x
dx 1 cosy
??
? ? ?
?
?
?
3.
2
3
dy 1
2
dy
dx
dx
rearranging
dy dy
12
dx dx
??
??
??
??
??
??
??
??
Order: 1
Degree:3
4. The vector equation of the line passing through the point (5, 2,-4) and parallel to
ˆ ˆˆ
3i 2j 8k ?? .
? ? ? ?
ˆˆ ˆ ˆ ˆ ˆ
r 5i 2j 4k 3i 2j 8k ? ? ? ? ? ? ?
CBSE XII | Mathematics
Sample Paper – 7 Solution
OR
The vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6) is
? ? ? ?
ˆˆ ˆ ˆ ˆ
r i 2k 4i 4j 4k ? ? ? ? ? ? ?
SECTION – B
5.
A = {1, 2, 3, 4, 5}
R = {(a, b): ab ? is even}
For R to be an equivalence relation it must be
??
? ? ? ?
(i) Reflexive, a a 0
(a,a) R for a A
So R is reflexive.
(ii) Symmetric,
if (a,b) R a b is even
b a is also even
? ? ?
??
So R is symmetric.
(iii) Transitive
If (a, b) ?R (b, c) ? R then (a, c) ?R
(a, b) ? ? ? R a b is even
(b, c) ? R ?? bc is even
Sum of two even numbers is even
So, a b b c
a b b c a c iseven since, a b and b c are even
? ? ?
? ? ? ? ? ? ? ?
So (a, c) ? R
Hence, R is transitive.
Therefore, R is an equivalence relation.
CBSE XII | Mathematics
Sample Paper – 7 Solution
6.
Let
23
B
14
??
?
??
?
??
and
36
C
38
? ??
?
??
?
??
.Then, the given matrix equation is A + B = C
now,
A + B – B = C – B
A = C – B
3 6 2 3
A
3 8 1 4
3 2 6 3
A
3 1 8 4
19
A
24
? ? ? ? ?
??
? ? ? ?
??
? ? ? ?
? ? ? ??
?
??
? ? ?
??
? ??
?
??
?
??
7.
It is known that,
1
sin AsinB cos A B cos A B
2
1
sin xsin2xsin3x dx sinx cos 2x 3x cos 2x 3x
2
1
sin xcos x sin xcos5x dx
2
1
sin xcosx sin xcos5x dx
2
1 sin2x 1
dx sin xcos5x
2 2 2
1 cos2x 1 1
sin x 5x sin x 5x dx
4 2 2 2
cos2x 1
sin6x sin 4x dx
84
cos2x 1 cos6x cos4x
C
8 4 6 4
cos2x 1 cos6x cos4x
C
8 8 3 2
6cos2x 1 2cos6x 3cos4x
C
48 8 6
1
2cos6x 3cos4x 6cos2x C
48
CBSE XII | Mathematics
Sample Paper – 7 Solution
8.
22
2
22
2 A Bx C
Let
1x
1-x 1 x 1 x
2 A 1 x Bx C 1 x
2 A Ax Bx Bx C Cx
Equating the coefficient of x
2
, x, and constant term, we obtain
A - B = 0
B - C = 0
A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
2
2
22
2
2 1 x 1
1x
1x
1 x 1 x
2 1 x 1
dx dx dx dx
1x
1 x 1 x
1 x 1 x
22
21
1 1 2x 1
dx dx dx
x 1 2
1 x 1 x
1
log x 1 log 1 x tan x C
2
OR
sin x a
I dx
sin x a
Let (x + a) = t dx = dt
sin t 2a
I dt
sin t
sin t cos2a cost sin2a
dt
sin t
cos2a cot t sin2a dt
cos2a t sin2a log sin t C
cos2a x a sin2a log sin x a C
Page 5
CBSE XII | Mathematics
Sample Paper – 7 Solution
Mathematics
Class XII
Sample Paper – 7 Solution
SECTION – A
1. Element at 3
rd
column and 2
nd
row
So i =2 and j =3
Substituting in
ij
i 2j
a
2
?
? we get
23
2 2 3 8
a4
22
??
? ? ?
2. y + sin y = cos x
differentiating w.r.t. x, we get,
? ?
dd
y sin y cosx
dx dx
dy dy
cosy sin x
dx dx
dy sin x
dx 1 cosy
??
? ? ?
?
?
?
3.
2
3
dy 1
2
dy
dx
dx
rearranging
dy dy
12
dx dx
??
??
??
??
??
??
??
??
Order: 1
Degree:3
4. The vector equation of the line passing through the point (5, 2,-4) and parallel to
ˆ ˆˆ
3i 2j 8k ?? .
? ? ? ?
ˆˆ ˆ ˆ ˆ ˆ
r 5i 2j 4k 3i 2j 8k ? ? ? ? ? ? ?
CBSE XII | Mathematics
Sample Paper – 7 Solution
OR
The vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6) is
? ? ? ?
ˆˆ ˆ ˆ ˆ
r i 2k 4i 4j 4k ? ? ? ? ? ? ?
SECTION – B
5.
A = {1, 2, 3, 4, 5}
R = {(a, b): ab ? is even}
For R to be an equivalence relation it must be
??
? ? ? ?
(i) Reflexive, a a 0
(a,a) R for a A
So R is reflexive.
(ii) Symmetric,
if (a,b) R a b is even
b a is also even
? ? ?
??
So R is symmetric.
(iii) Transitive
If (a, b) ?R (b, c) ? R then (a, c) ?R
(a, b) ? ? ? R a b is even
(b, c) ? R ?? bc is even
Sum of two even numbers is even
So, a b b c
a b b c a c iseven since, a b and b c are even
? ? ?
? ? ? ? ? ? ? ?
So (a, c) ? R
Hence, R is transitive.
Therefore, R is an equivalence relation.
CBSE XII | Mathematics
Sample Paper – 7 Solution
6.
Let
23
B
14
??
?
??
?
??
and
36
C
38
? ??
?
??
?
??
.Then, the given matrix equation is A + B = C
now,
A + B – B = C – B
A = C – B
3 6 2 3
A
3 8 1 4
3 2 6 3
A
3 1 8 4
19
A
24
? ? ? ? ?
??
? ? ? ?
??
? ? ? ?
? ? ? ??
?
??
? ? ?
??
? ??
?
??
?
??
7.
It is known that,
1
sin AsinB cos A B cos A B
2
1
sin xsin2xsin3x dx sinx cos 2x 3x cos 2x 3x
2
1
sin xcos x sin xcos5x dx
2
1
sin xcosx sin xcos5x dx
2
1 sin2x 1
dx sin xcos5x
2 2 2
1 cos2x 1 1
sin x 5x sin x 5x dx
4 2 2 2
cos2x 1
sin6x sin 4x dx
84
cos2x 1 cos6x cos4x
C
8 4 6 4
cos2x 1 cos6x cos4x
C
8 8 3 2
6cos2x 1 2cos6x 3cos4x
C
48 8 6
1
2cos6x 3cos4x 6cos2x C
48
CBSE XII | Mathematics
Sample Paper – 7 Solution
8.
22
2
22
2 A Bx C
Let
1x
1-x 1 x 1 x
2 A 1 x Bx C 1 x
2 A Ax Bx Bx C Cx
Equating the coefficient of x
2
, x, and constant term, we obtain
A - B = 0
B - C = 0
A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
2
2
22
2
2 1 x 1
1x
1x
1 x 1 x
2 1 x 1
dx dx dx dx
1x
1 x 1 x
1 x 1 x
22
21
1 1 2x 1
dx dx dx
x 1 2
1 x 1 x
1
log x 1 log 1 x tan x C
2
OR
sin x a
I dx
sin x a
Let (x + a) = t dx = dt
sin t 2a
I dt
sin t
sin t cos2a cost sin2a
dt
sin t
cos2a cot t sin2a dt
cos2a t sin2a log sin t C
cos2a x a sin2a log sin x a C
CBSE XII | Mathematics
Sample Paper – 7 Solution
9. y
2
= a(b – x)(b + x)
y
2
= a(b
2
– x
2
)
There are two arbitrary constants so we have to differentiate it two times
? ?
? ?
2
2
2
2
2
2
Differentiating w.r.t. x
dy
2y 2ax
dx
y dy
a....... i
x dx
dy
ya
dx
Differentiating again
d y dy
ya
dx dx
putting value of -a
d y dy y dy
y from i
dx dx x dx
??
??
??
??
? ? ?
??
??
??
??
??
??
10.
Let the angle between ? a and b be .
We know that a b a b cos
Given a b 60
a b cos 60
13 5 cos 60
60 12
cos
13 5 13
? ? ?
??
? ? ?
? ? ? ? ?
? ? ? ?
?
144 5
sin 1
169 13
Also we know that, a b a b sin
5
a b 5 13 25
13
? ? ? ? ?
? ? ?
? ? ? ? ?
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FAQs on Sample Solution Paper 7 - Math, Class 12 - Mathematics (Maths) Class 12 - JEE
| 1. What are the major topics covered in the Class 12 Math exam? |  |
Ans. The major topics covered in the Class 12 Math exam include calculus, algebra, coordinate geometry, probability, and statistics. These topics are essential for understanding advanced mathematical concepts and solving complex problems.
| 2. How can I prepare effectively for the Class 12 Math exam? | |
Ans. To prepare effectively for the Class 12 Math exam, it is important to understand and practice the fundamental concepts. Start by reviewing the textbooks and class notes, and then solve plenty of practice problems. Additionally, make use of online resources, such as video tutorials and practice tests, to further enhance your understanding and test your knowledge.
| 3. What are some common mistakes students make in the Class 12 Math exam? | |
Ans. Some common mistakes students make in the Class 12 Math exam include not showing all the steps of their solutions, misinterpreting the question, and rushing through the exam without carefully reading each question. It is important to take your time, read the instructions thoroughly, and double-check your answers before submitting the exam.
| 4. Are there any helpful tips for solving calculus problems in the Class 12 Math exam? | |
Ans. Yes, here are some helpful tips for solving calculus problems in the Class 12 Math exam:
- Understand the basic concepts of differentiation and integration.
- Practice different types of problems, including finding derivatives and integrals, solving optimization problems, and applying the chain rule.
- Pay attention to the given conditions and constraints in the problem, as they may impact the solution.
- Draw diagrams or graphs whenever possible to visualize the problem and make it easier to solve.
- Practice using calculators or software tools for numerical calculations and graphing, if allowed in the exam.
| 5. How can I improve my problem-solving skills for the Class 12 Math exam? | |
Ans. Improving problem-solving skills for the Class 12 Math exam requires regular practice and a systematic approach. Start by solving a variety of problems from different topics, including both basic and advanced ones. Analyze the solution techniques used in each problem and try to understand the underlying concepts. Additionally, seek guidance from teachers or tutors, participate in study groups, and make use of online resources that provide step-by-step solutions to problems. With consistent practice and exposure to different types of problems, your problem-solving skills will gradually improve.
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Nov 22, 2024 Last updated |
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