Past Year Paper - Solutions, Mathematics (Set - 7, 8 and 9 ), Outside Delhi, 2016, Class 12, Maths | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/N, 65/2/N, 65/3/N
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers
given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has
given any other answer which is different from the one given in the Marking Scheme, but conveys
the meaning, such answers should be given full weightage
2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done
according to one’s own interpretation or any other consideration — Marking Scheme should be
strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted first should
be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full marks if the
answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain
photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out strictly as
per value points for each answer as given in the Marking Scheme.
Page 2


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/N, 65/2/N, 65/3/N
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers
given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has
given any other answer which is different from the one given in the Marking Scheme, but conveys
the meaning, such answers should be given full weightage
2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done
according to one’s own interpretation or any other consideration — Marking Scheme should be
strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted first should
be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full marks if the
answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain
photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out strictly as
per value points for each answer as given in the Marking Scheme.
QUESTION PAPER CODE 65/1/N
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Finding A
T
 = 
( )
cosa –sina
sina cosa
1
2
Getting a = 
4
p
 or 45°
1
2
2. k = 27 1
3. For a unique solution
1 1 1
2 1 –1
3 2 k
 ? 0
1
2
? k ? 0
1
2
4. Getting equation as 
5
2
x y z
+ +
5 –5
 = 1
1
2
Sum of intercepts 
5
5 – 5
2
+ = 
5
2
1
2
5. Getting ? = – 9 and µ = 27
1
2
each
6.
ˆ ˆ ˆ
a b 6i 3j 2k + = - +




1
2
Unit vector parallel to a b +




 is 
1
ˆ ˆ ˆ
(6i 3j 2k)
7
- +
1
2
SECTION B
7. tan
–1
 (x – 1) + tan
–1
 (x + 1) = tan
–1
 3x – tan
–1
 x
1
2
?
–1
2
2x
tan
2 – x
? ?
? ?
? ?
? ?
 = 
–1
2
2x
tan
1 + 3x
? ?
? ?
? ?
? ?
1
1
2
2
2x
2 – x
 = 
2
2x
1 + 3x
1
2
2x(1 + 3x
2
 – 2 + x
2
) = 0
1
2
x = 0, 
1 1
, –
2 2
1
65/1/N (1)
65/1/N
Page 3


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/N, 65/2/N, 65/3/N
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers
given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has
given any other answer which is different from the one given in the Marking Scheme, but conveys
the meaning, such answers should be given full weightage
2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done
according to one’s own interpretation or any other consideration — Marking Scheme should be
strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted first should
be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full marks if the
answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain
photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out strictly as
per value points for each answer as given in the Marking Scheme.
QUESTION PAPER CODE 65/1/N
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Finding A
T
 = 
( )
cosa –sina
sina cosa
1
2
Getting a = 
4
p
 or 45°
1
2
2. k = 27 1
3. For a unique solution
1 1 1
2 1 –1
3 2 k
 ? 0
1
2
? k ? 0
1
2
4. Getting equation as 
5
2
x y z
+ +
5 –5
 = 1
1
2
Sum of intercepts 
5
5 – 5
2
+ = 
5
2
1
2
5. Getting ? = – 9 and µ = 27
1
2
each
6.
ˆ ˆ ˆ
a b 6i 3j 2k + = - +




1
2
Unit vector parallel to a b +




 is 
1
ˆ ˆ ˆ
(6i 3j 2k)
7
- +
1
2
SECTION B
7. tan
–1
 (x – 1) + tan
–1
 (x + 1) = tan
–1
 3x – tan
–1
 x
1
2
?
–1
2
2x
tan
2 – x
? ?
? ?
? ?
? ?
 = 
–1
2
2x
tan
1 + 3x
? ?
? ?
? ?
? ?
1
1
2
2
2x
2 – x
 = 
2
2x
1 + 3x
1
2
2x(1 + 3x
2
 – 2 + x
2
) = 0
1
2
x = 0, 
1 1
, –
2 2
1
65/1/N (1)
65/1/N
OR
Let 2x = tan ? 1
L. H. S = tan
–1
3
–1
2 2
3tan – tan 2 tan
– tan
1– 3tan 1– tan
? ?
? ? ? ? ?
? ?
? ?
? ?
? ? ? ?
? ?
1
= tan
–1
(tan 3?) – tan
–1
 (tan 2?) 1
= 3? – 2?
= ? or tan
–1
 2x
? L. H. S = R. H. S 1
8. Getting matrix equation as 
( ) ( )
10 3 E
3 10 H
 = 
( )
145
180
1
1
2
?
( )
E
H
 = 
( ) ( )
–1
10 3 145
3 10 180
?
( )
E
H
 = 
( )
10
15
? E = 10, H = 15 1
1
2
The poor boy was charged ` 65 less
V alue: Helping the poor 1
9. L.H.L = a + 3 1
1
2
R.H.L = b/2 1
1
2
f(x) is continuous at x = 0. So, a + 3 = 2 = b/2
1
2
? a = –1 and b = 4
1
2
10.
dx
dy
 = 
2
sin a
cos (a + y)
1
1
2
? 
dy
dx
 = 
2
cos (a y)
sin a
+ 1
2
2
2
d y
dx
 = 
–2cos(a + y) sin(a + y) dy
sin a dx
= 
sin 2(a y) dy
sin a dx
- +
1
1
2
? 
2
2
d y dy
sin a + sin 2(a + y)
dx
dx
 = 0
1
2
65/1/N (2)
65/1/N
Page 4


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/N, 65/2/N, 65/3/N
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers
given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has
given any other answer which is different from the one given in the Marking Scheme, but conveys
the meaning, such answers should be given full weightage
2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done
according to one’s own interpretation or any other consideration — Marking Scheme should be
strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted first should
be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full marks if the
answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain
photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out strictly as
per value points for each answer as given in the Marking Scheme.
QUESTION PAPER CODE 65/1/N
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Finding A
T
 = 
( )
cosa –sina
sina cosa
1
2
Getting a = 
4
p
 or 45°
1
2
2. k = 27 1
3. For a unique solution
1 1 1
2 1 –1
3 2 k
 ? 0
1
2
? k ? 0
1
2
4. Getting equation as 
5
2
x y z
+ +
5 –5
 = 1
1
2
Sum of intercepts 
5
5 – 5
2
+ = 
5
2
1
2
5. Getting ? = – 9 and µ = 27
1
2
each
6.
ˆ ˆ ˆ
a b 6i 3j 2k + = - +




1
2
Unit vector parallel to a b +




 is 
1
ˆ ˆ ˆ
(6i 3j 2k)
7
- +
1
2
SECTION B
7. tan
–1
 (x – 1) + tan
–1
 (x + 1) = tan
–1
 3x – tan
–1
 x
1
2
?
–1
2
2x
tan
2 – x
? ?
? ?
? ?
? ?
 = 
–1
2
2x
tan
1 + 3x
? ?
? ?
? ?
? ?
1
1
2
2
2x
2 – x
 = 
2
2x
1 + 3x
1
2
2x(1 + 3x
2
 – 2 + x
2
) = 0
1
2
x = 0, 
1 1
, –
2 2
1
65/1/N (1)
65/1/N
OR
Let 2x = tan ? 1
L. H. S = tan
–1
3
–1
2 2
3tan – tan 2 tan
– tan
1– 3tan 1– tan
? ?
? ? ? ? ?
? ?
? ?
? ?
? ? ? ?
? ?
1
= tan
–1
(tan 3?) – tan
–1
 (tan 2?) 1
= 3? – 2?
= ? or tan
–1
 2x
? L. H. S = R. H. S 1
8. Getting matrix equation as 
( ) ( )
10 3 E
3 10 H
 = 
( )
145
180
1
1
2
?
( )
E
H
 = 
( ) ( )
–1
10 3 145
3 10 180
?
( )
E
H
 = 
( )
10
15
? E = 10, H = 15 1
1
2
The poor boy was charged ` 65 less
V alue: Helping the poor 1
9. L.H.L = a + 3 1
1
2
R.H.L = b/2 1
1
2
f(x) is continuous at x = 0. So, a + 3 = 2 = b/2
1
2
? a = –1 and b = 4
1
2
10.
dx
dy
 = 
2
sin a
cos (a + y)
1
1
2
? 
dy
dx
 = 
2
cos (a y)
sin a
+ 1
2
2
2
d y
dx
 = 
–2cos(a + y) sin(a + y) dy
sin a dx
= 
sin 2(a y) dy
sin a dx
- +
1
1
2
? 
2
2
d y dy
sin a + sin 2(a + y)
dx
dx
 = 0
1
2
65/1/N (2)
65/1/N
   OR
Let 2x = sin ? 1
? y = 
2
–1
6x – 1 – 4x
sin
5
? ?
? ?
? ?
? ?
= 
–1
3 4
sin sin – cos
5 5
? ?
? ?
? ?
? ?
= sin
–1
 (cos a sin ? – sin a cos ?)               [cos a = 
3
5
; sin a = 
4
5
] 1
= sin
–1
(sin(? – a))
= ? – a 1
= sin
–1
 (2x) – a
?
dy
dx
 = 
2
2
1– 4x
1
11. Slope of the tangent = 3x
2
 + 2 = 14 1
Points of contact (2, 8) and (–2, –16) 1
Equations of tangent
14x – y – 20 = 0 1
and 14x – y + 12 = 0 1
12. Let 2x = t
I = 
t
3
1 (t – 5)
e dt
2
(t – 3)
?
1
= 
t
2 3
1 1 2
e dt
2
(t 3) (t 3)
? ?
-
? ?
- -
? ?
?
2
= 
t 2x
2 2
1 1 1 1
e C e C
2 2
(t 3) (2x 3)
+ = +
- -
1
OR
Writing 
2
2
x + x +1
(x +1) (x + 2)
 = 2
Ax + B C
+
x + 2
x +1
? A = 
2 1 3
, B = , C =
5 5 5
1
? I = 
2 2
1 2x 1 dx 3 dx
dx + +
5 5 5 x + 2
x + 1 x + 1
? ? ?
2
? I = 
2 –1
1 1 3
log x + 1 + tan x + log x + 2 + C
5 5 5
1
65/1/N (3)
65/1/N
Page 5


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/N, 65/2/N, 65/3/N
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers
given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has
given any other answer which is different from the one given in the Marking Scheme, but conveys
the meaning, such answers should be given full weightage
2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done
according to one’s own interpretation or any other consideration — Marking Scheme should be
strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted first should
be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full marks if the
answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain
photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out strictly as
per value points for each answer as given in the Marking Scheme.
QUESTION PAPER CODE 65/1/N
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Finding A
T
 = 
( )
cosa –sina
sina cosa
1
2
Getting a = 
4
p
 or 45°
1
2
2. k = 27 1
3. For a unique solution
1 1 1
2 1 –1
3 2 k
 ? 0
1
2
? k ? 0
1
2
4. Getting equation as 
5
2
x y z
+ +
5 –5
 = 1
1
2
Sum of intercepts 
5
5 – 5
2
+ = 
5
2
1
2
5. Getting ? = – 9 and µ = 27
1
2
each
6.
ˆ ˆ ˆ
a b 6i 3j 2k + = - +




1
2
Unit vector parallel to a b +




 is 
1
ˆ ˆ ˆ
(6i 3j 2k)
7
- +
1
2
SECTION B
7. tan
–1
 (x – 1) + tan
–1
 (x + 1) = tan
–1
 3x – tan
–1
 x
1
2
?
–1
2
2x
tan
2 – x
? ?
? ?
? ?
? ?
 = 
–1
2
2x
tan
1 + 3x
? ?
? ?
? ?
? ?
1
1
2
2
2x
2 – x
 = 
2
2x
1 + 3x
1
2
2x(1 + 3x
2
 – 2 + x
2
) = 0
1
2
x = 0, 
1 1
, –
2 2
1
65/1/N (1)
65/1/N
OR
Let 2x = tan ? 1
L. H. S = tan
–1
3
–1
2 2
3tan – tan 2 tan
– tan
1– 3tan 1– tan
? ?
? ? ? ? ?
? ?
? ?
? ?
? ? ? ?
? ?
1
= tan
–1
(tan 3?) – tan
–1
 (tan 2?) 1
= 3? – 2?
= ? or tan
–1
 2x
? L. H. S = R. H. S 1
8. Getting matrix equation as 
( ) ( )
10 3 E
3 10 H
 = 
( )
145
180
1
1
2
?
( )
E
H
 = 
( ) ( )
–1
10 3 145
3 10 180
?
( )
E
H
 = 
( )
10
15
? E = 10, H = 15 1
1
2
The poor boy was charged ` 65 less
V alue: Helping the poor 1
9. L.H.L = a + 3 1
1
2
R.H.L = b/2 1
1
2
f(x) is continuous at x = 0. So, a + 3 = 2 = b/2
1
2
? a = –1 and b = 4
1
2
10.
dx
dy
 = 
2
sin a
cos (a + y)
1
1
2
? 
dy
dx
 = 
2
cos (a y)
sin a
+ 1
2
2
2
d y
dx
 = 
–2cos(a + y) sin(a + y) dy
sin a dx
= 
sin 2(a y) dy
sin a dx
- +
1
1
2
? 
2
2
d y dy
sin a + sin 2(a + y)
dx
dx
 = 0
1
2
65/1/N (2)
65/1/N
   OR
Let 2x = sin ? 1
? y = 
2
–1
6x – 1 – 4x
sin
5
? ?
? ?
? ?
? ?
= 
–1
3 4
sin sin – cos
5 5
? ?
? ?
? ?
? ?
= sin
–1
 (cos a sin ? – sin a cos ?)               [cos a = 
3
5
; sin a = 
4
5
] 1
= sin
–1
(sin(? – a))
= ? – a 1
= sin
–1
 (2x) – a
?
dy
dx
 = 
2
2
1– 4x
1
11. Slope of the tangent = 3x
2
 + 2 = 14 1
Points of contact (2, 8) and (–2, –16) 1
Equations of tangent
14x – y – 20 = 0 1
and 14x – y + 12 = 0 1
12. Let 2x = t
I = 
t
3
1 (t – 5)
e dt
2
(t – 3)
?
1
= 
t
2 3
1 1 2
e dt
2
(t 3) (t 3)
? ?
-
? ?
- -
? ?
?
2
= 
t 2x
2 2
1 1 1 1
e C e C
2 2
(t 3) (2x 3)
+ = +
- -
1
OR
Writing 
2
2
x + x +1
(x +1) (x + 2)
 = 2
Ax + B C
+
x + 2
x +1
? A = 
2 1 3
, B = , C =
5 5 5
1
? I = 
2 2
1 2x 1 dx 3 dx
dx + +
5 5 5 x + 2
x + 1 x + 1
? ? ?
2
? I = 
2 –1
1 1 3
log x + 1 + tan x + log x + 2 + C
5 5 5
1
65/1/N (3)
65/1/N
13. Using property: 
b b
a a
f (x) dx f (a b x) dx = + -
? ?
1
 I = 
2
2
x
–2
x
dx
1 5
? ?
? ?
? ?
+
? ?
?
 = 
2
2
–x
–2
x
dx
1 5
? ?
? ?
? ?
+
? ?
?
1
2I = 
2
2
–2
x dx
?
1
2I = 
16 8
or I
3 3
=
1
14. Writing x + 3 = A(–4 – 2x) + B
? A = 
1
– ,
2
 B = 1 1
? I = 
( )
2
2 2
1
– (– 4 – 2x) 3 – 4x – x dx + 7 – (x + 2) dx
2
? ?
1
I = 
2 3/2 2 –1
1 x + 2 7 x + 2
– (3 – 4 – x ) + 3 – 4x – x + sin + C
3 2 2 7
2
15. Writing linear equation 
dy cos x x
+ y = –
dx 1 + sin x 1 + sin x
1
I.F = 
cos x
dx
1 + sin x
e
?
 = 1 + sin x 1
General solution is: y(1 + sin x) = 
2
x
– C
2
+ 1
Particular solution is: y (1 + sin x) =  
2
x
1 –
2
1
16.
v
v
dx 2xe y
dy
2ye
-
=
x dx dv
v, then v y
y dy dy
= = +
1
v
v
dv 2vye y
v y
dy
2ye
-
+ =
v
dy
2 e dv
y
= -
? ?
1
General solution is: 
v x/y
2e log |y| C or 2e log |y| C = - + = - + 1
Particular solution is: 
x/y
2e log | y| 2 + = 1
17.
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
AB = – 4i – 6j – 2k, AC = – i + 4j + 3k, AD= – 8i – j + 3k
 
  
 

1
1
2
For 4 points to be coplanar, 
[AB AC AD] = 0
 
  
 

65/1/N (4)
65/1/N