Past Year Paper - Solutions, Mathematics (Set - 4, 5 and 6), 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/E, 65/2/E, 65/3/E
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/E, 65/2/E, 65/3/E
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/E
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. R
1
 ? R
1
 + R
2
 + R
3
 or C
1
 ? C
1
 + C
2
 + C
3
1
2
Ans. 0
1
2
2. b
21
 = –16, b
23
 = –2             [For any one correct value]
1
2
b
21
 + b
23
 = –16 + (–2) = –18
1
2
3. 2
6
or 64 1
4. (a, –ß, ?) 1
5.
1(a b) 3(a 3b)
4
- + +

 


 

 (i.e., using correct formula)
1
2
= 
a 2b +




1
2
6. Finding 
3
cos
2
? =
1
2
|a b| 6 × =




1
2
SECTION B
7.
( )
1 1 2 x 1 x
tan tan
2 x 2 2
- - -
=
+
?
( )
1 1 2 x x
2 tan tan
2 x 2
- - -
=
+
1
2
?
( )
( )
1 1
2
2 x
2
x 2 x
tan tan
2
2 x
1
2 x
- -
-
+
=
-
-
+
1
1
2
?
2
1 1 4 x x
tan tan
4x 2
- -
  
-
=  
  
1
?
2
4 x
4x
-
 = 
x
2
1
2
? x = 
2
3
  (? x > 0)
1
2
65/1/E (1)
65/1/E
Page 3


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/E, 65/2/E, 65/3/E
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/E
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. R
1
 ? R
1
 + R
2
 + R
3
 or C
1
 ? C
1
 + C
2
 + C
3
1
2
Ans. 0
1
2
2. b
21
 = –16, b
23
 = –2             [For any one correct value]
1
2
b
21
 + b
23
 = –16 + (–2) = –18
1
2
3. 2
6
or 64 1
4. (a, –ß, ?) 1
5.
1(a b) 3(a 3b)
4
- + +

 


 

 (i.e., using correct formula)
1
2
= 
a 2b +




1
2
6. Finding 
3
cos
2
? =
1
2
|a b| 6 × =




1
2
SECTION B
7.
( )
1 1 2 x 1 x
tan tan
2 x 2 2
- - -
=
+
?
( )
1 1 2 x x
2 tan tan
2 x 2
- - -
=
+
1
2
?
( )
( )
1 1
2
2 x
2
x 2 x
tan tan
2
2 x
1
2 x
- -
-
+
=
-
-
+
1
1
2
?
2
1 1 4 x x
tan tan
4x 2
- -
  
-
=  
  
1
?
2
4 x
4x
-
 = 
x
2
1
2
? x = 
2
3
  (? x > 0)
1
2
65/1/E (1)
65/1/E
OR
( ) ( )
1 1 3 17
2sin tan
5 31
- -
-
= 
( ) ( )
1 1 3 17
2 tan tan
4 31
- -
- 1
= 
( )
( )
1 1
2
3
2
17
4
tan tan
31
3
1
4
- -
×
-
-
1
= 
1 1 24 17
tan tan
7 31
- -
-
= 
1
24 17
7 31
tan
24 17
1
7 31
-
? ?
-
? ?
? ?
+ × ? ?
? ?
1
= tan
–1
 (1) 1
= 
4
p
8. Let the number of children be x and the amount distributed by Seema for one student be 
`
 y.
So, (x – 8)(y + 10) = xy
? 5x – 4y = 40 ...(i)
1
2
and (x + 16)(y – 10) = xy
? 5x – 8y = –80 ...(ii)
1
2
Here A = 
5 4 x 40
, X , B
5 8 y 80
-      
= =
      
- -
      
AX = B ? X = A
–1
B
A
–1
 = 
8 4
1
20
5 5
-  
-
  
-
  
1
?
x 32
y 30
    
=
    
    
? x = 32, y = 30 1
No. of students = 32
Amount given to each student = ` 30.
Value reflected: To help needy people. 1
9.
cos 2 t
dx
e ( 2sin 2t)
dt
= -
 or –2x sin 2t 1
sin 2t
dy
e 2cos 2t
dt
=
 or 2y cos 2t 1
65/1/E (2)
65/1/E
Page 4


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/E, 65/2/E, 65/3/E
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/E
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. R
1
 ? R
1
 + R
2
 + R
3
 or C
1
 ? C
1
 + C
2
 + C
3
1
2
Ans. 0
1
2
2. b
21
 = –16, b
23
 = –2             [For any one correct value]
1
2
b
21
 + b
23
 = –16 + (–2) = –18
1
2
3. 2
6
or 64 1
4. (a, –ß, ?) 1
5.
1(a b) 3(a 3b)
4
- + +

 


 

 (i.e., using correct formula)
1
2
= 
a 2b +




1
2
6. Finding 
3
cos
2
? =
1
2
|a b| 6 × =




1
2
SECTION B
7.
( )
1 1 2 x 1 x
tan tan
2 x 2 2
- - -
=
+
?
( )
1 1 2 x x
2 tan tan
2 x 2
- - -
=
+
1
2
?
( )
( )
1 1
2
2 x
2
x 2 x
tan tan
2
2 x
1
2 x
- -
-
+
=
-
-
+
1
1
2
?
2
1 1 4 x x
tan tan
4x 2
- -
  
-
=  
  
1
?
2
4 x
4x
-
 = 
x
2
1
2
? x = 
2
3
  (? x > 0)
1
2
65/1/E (1)
65/1/E
OR
( ) ( )
1 1 3 17
2sin tan
5 31
- -
-
= 
( ) ( )
1 1 3 17
2 tan tan
4 31
- -
- 1
= 
( )
( )
1 1
2
3
2
17
4
tan tan
31
3
1
4
- -
×
-
-
1
= 
1 1 24 17
tan tan
7 31
- -
-
= 
1
24 17
7 31
tan
24 17
1
7 31
-
? ?
-
? ?
? ?
+ × ? ?
? ?
1
= tan
–1
 (1) 1
= 
4
p
8. Let the number of children be x and the amount distributed by Seema for one student be 
`
 y.
So, (x – 8)(y + 10) = xy
? 5x – 4y = 40 ...(i)
1
2
and (x + 16)(y – 10) = xy
? 5x – 8y = –80 ...(ii)
1
2
Here A = 
5 4 x 40
, X , B
5 8 y 80
-      
= =
      
- -
      
AX = B ? X = A
–1
B
A
–1
 = 
8 4
1
20
5 5
-  
-
  
-
  
1
?
x 32
y 30
    
=
    
    
? x = 32, y = 30 1
No. of students = 32
Amount given to each student = ` 30.
Value reflected: To help needy people. 1
9.
cos 2 t
dx
e ( 2sin 2t)
dt
= -
 or –2x sin 2t 1
sin 2t
dy
e 2cos 2t
dt
=
 or 2y cos 2t 1
65/1/E (2)
65/1/E
sin 2t
cos 2t
dy e 2cos 2t y cos 2t
or
dx x sin 2t e 2sin 2t
-
= -
1
      = 
y log x
x log y
-
1
OR
f(x) = 2 sin x + sin 2x on [0, p]
f(x) is continuous in [0, p]
f(x) is differentiable in (0, p)
1
? Mean value theorem is applicable
f(0) = 0, f(p) = 0
f'(x) = 2 cos x + 2 cos 2x 1
f'(c) = 2 cos c + 2 cos 2c
f'(c) = 
f( ) f(0)
0
0
p -
=
p -
1
? 2 cos c + 2 cos 2c = 0
? cos c + 2 cos2 c – 1 = 0
? (2 cos c – 1)(cos c + 1) = 0
? cos c = 
1
2
? c = (0, )
3
p
? p
Hence mean value theorem is verified.
1
2
 + 
1
2
10. f(x) = 
1
x
1
x
e 1
x 0
e 1
1 x 0
 
- 
?
 
 
+
 
 - =
 
LHL: 
1
x
1
x 0
x
e 1
lim
e 1
-
?
-
+
= 
1
h
1
h 0
h
e 1 0 1
lim 1
0 1
e 1
-
? -
- -
= = -
+
+
2
RHL: 
1 1
h h
1 1
h 0 h 0
h h
e 1 1 e
lim lim 1
e 1 1 e
-
? ? -
- -
= =
+ +
2
LHL ? RHL
? f(x) is discontinuous at x = 0
65/1/E (3)
65/1/E
?
?
?
?
?
Page 5


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/E, 65/2/E, 65/3/E
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/E
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. R
1
 ? R
1
 + R
2
 + R
3
 or C
1
 ? C
1
 + C
2
 + C
3
1
2
Ans. 0
1
2
2. b
21
 = –16, b
23
 = –2             [For any one correct value]
1
2
b
21
 + b
23
 = –16 + (–2) = –18
1
2
3. 2
6
or 64 1
4. (a, –ß, ?) 1
5.
1(a b) 3(a 3b)
4
- + +

 


 

 (i.e., using correct formula)
1
2
= 
a 2b +




1
2
6. Finding 
3
cos
2
? =
1
2
|a b| 6 × =




1
2
SECTION B
7.
( )
1 1 2 x 1 x
tan tan
2 x 2 2
- - -
=
+
?
( )
1 1 2 x x
2 tan tan
2 x 2
- - -
=
+
1
2
?
( )
( )
1 1
2
2 x
2
x 2 x
tan tan
2
2 x
1
2 x
- -
-
+
=
-
-
+
1
1
2
?
2
1 1 4 x x
tan tan
4x 2
- -
  
-
=  
  
1
?
2
4 x
4x
-
 = 
x
2
1
2
? x = 
2
3
  (? x > 0)
1
2
65/1/E (1)
65/1/E
OR
( ) ( )
1 1 3 17
2sin tan
5 31
- -
-
= 
( ) ( )
1 1 3 17
2 tan tan
4 31
- -
- 1
= 
( )
( )
1 1
2
3
2
17
4
tan tan
31
3
1
4
- -
×
-
-
1
= 
1 1 24 17
tan tan
7 31
- -
-
= 
1
24 17
7 31
tan
24 17
1
7 31
-
? ?
-
? ?
? ?
+ × ? ?
? ?
1
= tan
–1
 (1) 1
= 
4
p
8. Let the number of children be x and the amount distributed by Seema for one student be 
`
 y.
So, (x – 8)(y + 10) = xy
? 5x – 4y = 40 ...(i)
1
2
and (x + 16)(y – 10) = xy
? 5x – 8y = –80 ...(ii)
1
2
Here A = 
5 4 x 40
, X , B
5 8 y 80
-      
= =
      
- -
      
AX = B ? X = A
–1
B
A
–1
 = 
8 4
1
20
5 5
-  
-
  
-
  
1
?
x 32
y 30
    
=
    
    
? x = 32, y = 30 1
No. of students = 32
Amount given to each student = ` 30.
Value reflected: To help needy people. 1
9.
cos 2 t
dx
e ( 2sin 2t)
dt
= -
 or –2x sin 2t 1
sin 2t
dy
e 2cos 2t
dt
=
 or 2y cos 2t 1
65/1/E (2)
65/1/E
sin 2t
cos 2t
dy e 2cos 2t y cos 2t
or
dx x sin 2t e 2sin 2t
-
= -
1
      = 
y log x
x log y
-
1
OR
f(x) = 2 sin x + sin 2x on [0, p]
f(x) is continuous in [0, p]
f(x) is differentiable in (0, p)
1
? Mean value theorem is applicable
f(0) = 0, f(p) = 0
f'(x) = 2 cos x + 2 cos 2x 1
f'(c) = 2 cos c + 2 cos 2c
f'(c) = 
f( ) f(0)
0
0
p -
=
p -
1
? 2 cos c + 2 cos 2c = 0
? cos c + 2 cos2 c – 1 = 0
? (2 cos c – 1)(cos c + 1) = 0
? cos c = 
1
2
? c = (0, )
3
p
? p
Hence mean value theorem is verified.
1
2
 + 
1
2
10. f(x) = 
1
x
1
x
e 1
x 0
e 1
1 x 0
 
- 
?
 
 
+
 
 - =
 
LHL: 
1
x
1
x 0
x
e 1
lim
e 1
-
?
-
+
= 
1
h
1
h 0
h
e 1 0 1
lim 1
0 1
e 1
-
? -
- -
= = -
+
+
2
RHL: 
1 1
h h
1 1
h 0 h 0
h h
e 1 1 e
lim lim 1
e 1 1 e
-
? ? -
- -
= =
+ +
2
LHL ? RHL
? f(x) is discontinuous at x = 0
65/1/E (3)
65/1/E
?
?
?
?
?
11. y 5x 3 5 = - - ...(i)
dy
5
dx
2 5x 3
=
-
1
Slope of line 4x – 2y + 5 = 0 is 
4
2.
2
-
=
-
1
2
?
5 73
2 x
80
2 5x 3
=   
-
1
Putting x = 
73
80
 in eqn. (i), we get y = 
15
4
-
1
2
Equation of tangent
( )
15 73
y 2 x
4 80
+ = -
1
or 80x – 40y – 223 = 0
12. { }
5
1
| x 1| | x 2 | | x 3| dx - + - + -
?
= 
5 2 5 3 5
1 1 2 1 3
(x 1) dx (2 x) dx (x 2) dx (3 x) dx (x 3) dx - + - + - + - + -
? ? ? ? ?
2
1
2
= 
5 2 5 3 5
2 2 2 2 2
1 1 2 1 3
x x x x x
x 2x 2x 3x 3x
2 2 2 2 3
? ? ? ? ? ? ? ? ? ?
- + - + - + - + -
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
1
= 17
1
2
OR
Let I = 
2
0
x sin x
dx
1 3cos x
p
+
 
                                                    ...(i)
I = 
2
0
( x)sin( x)
dx
1 3cos ( x)
p
p - p -
+ p -
 
1
= 
2 2
0 0
sin x x sin x
dx dx
1 3cos x 1 3cos x
p p
p
-
+ +
  
...(ii)
Adding (i) & (ii), we have
2I = 
2
0
sin x
dx
1 3cos x
p
p
+
?
1
Put cos x = t
–sin x dx = dt, when x = 0 ? t = 1, for x = p ? t = –1
2I = 
1
2
1
dt
1 3t
-
-p
+
?
1
65/1/E (4)
65/1/E