Sample Solution Paper 7 - Math, Class 12 JEE Notes | EduRev

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JEE : Sample Solution Paper 7 - Math, Class 12 JEE Notes | EduRev

 Page 1


  
 
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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