Past Year Paper - Solutions, Maths(Set - 10, 11 and 12 ), Outside Delhi, 2016, Class 12, Maths Commerce Notes | EduRev

Mathematics (Maths) Class 12

Commerce : Past Year Paper - Solutions, Maths(Set - 10, 11 and 12 ), Outside Delhi, 2016, Class 12, Maths Commerce Notes | EduRev

 Page 1


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S
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/S, 65/2/S, 65/3/S
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/S
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Getting sin ? = 
1
1
3
1 4
2
.
2 3
=
1
2
Hence 
| . | a b


 = 
1 4 3
. . 1
2 2 3
=
1
2
2.
2
| 2 | 1 a b - =


 ? 
1
. .
2
a b =

 1
2
? Angle between a

 and b

 = 
4
p 1
2
3. Writing or using, that given planes are parallel
1
2
d = 
| 4 10 |
2 units
4 9 36
+
=
+ +
1
2
4.
T T 2
| AA | |A| | A | |A| = =
1
2
 
= 25
1
2
5. Getting AB = 
7 8
0 10
- ? ?
? ?
-
? ?
1
2
|AB| = –70
1
2
6. k(2) = –8  ?  k = –4
1
2
–4(3) = 4a  ?  a = –3
1
2
SECTION B
7.
1
(sin 2 ) sin ( 3 )
x
y x x u v
-
= + = +
?
dy du dv
dx dx dx
= +
1
2
(sin 2 ) log logsin 2
x
u x u x x = ? =
1
2
1
2 cot 2 log sin 2 = · +
du
x x x
u dx
1
65/1/S (1)
65/1/S
Page 3


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S
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/S
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Getting sin ? = 
1
1
3
1 4
2
.
2 3
=
1
2
Hence 
| . | a b


 = 
1 4 3
. . 1
2 2 3
=
1
2
2.
2
| 2 | 1 a b - =


 ? 
1
. .
2
a b =

 1
2
? Angle between a

 and b

 = 
4
p 1
2
3. Writing or using, that given planes are parallel
1
2
d = 
| 4 10 |
2 units
4 9 36
+
=
+ +
1
2
4.
T T 2
| AA | |A| | A | |A| = =
1
2
 
= 25
1
2
5. Getting AB = 
7 8
0 10
- ? ?
? ?
-
? ?
1
2
|AB| = –70
1
2
6. k(2) = –8  ?  k = –4
1
2
–4(3) = 4a  ?  a = –3
1
2
SECTION B
7.
1
(sin 2 ) sin ( 3 )
x
y x x u v
-
= + = +
?
dy du dv
dx dx dx
= +
1
2
(sin 2 ) log logsin 2
x
u x u x x = ? =
1
2
1
2 cot 2 log sin 2 = · +
du
x x x
u dx
1
65/1/S (1)
65/1/S
? (sin 2 ) [2 cot 2 log sin 2 ]
x
du
x x x x
dx
= +
1
2
1 3
1 3 2
dv
dx x x
=
-
1
?
3
(sin 2 ) [2 cot 2 logsin 2 ]
2 1 3
x
dy
x x x x
dx x x
= + +
-
1
2
OR
Let y = 
2 2
1 1 2
2 2
1 1
tan and cos
1 1
x x
z x
x x
- -
? ?
+ - -
? ?
=
? ?
+ + -
? ?
z = cos
–1
 x
2
  ?  x
2
 = cos z  ? y = 
1
1 cos 1 cos
tan
1 cos 1 cos
z z
z z
-
? ?
+ - -
? ?
? ?
+ + -
? ?
1
? y = 
1 1
cos sin 1 tan
2 2 2
tan tan
cos sin 1 tan
2 2 2
z z z
z z z
- -
? ? ? ?
- -
? ? ? ?
=
? ? ? ?
? ? ? ?
+ +
? ? ? ?
1
2
 + 
1
2
? y = 
1
tan tan
4 2 4 2
z z
-
? p ? p ? ?
- = -
? ? ? ?
? ? ? ?
1
2
 + 
1
2
?
dy
dz
 = 
1
2
- 1
8. LHL = 
0
lim .sin ( 1)
2
-
?
p
+ =
x
k x k 1
RHL = 
3
0
tan (1 cos )
lim
x
x x
x
+
?
-
1
= 
2
0
tan sin /2 1
lim . 2
2 /2 2
x
x x
x x
+
?
? ?
=
? ?
? ?
1
? k = 
1
2
1
9. When x = am
2
, we get y = ± am
3
1
ay
2
 = x
3
 ? 
2
2
3
2 3
2
dy dy x
ay x
dx dx ay
= ? = 1
slope of normal = 
3
2 4
2 2
3 3
a am
m
a m
= ± ± 1
? Equation of normal is y ± am
3
 = 
2
2
( )
3
x am
m
- ±
1
[Full marks may be given, if only one value for point, slope and equation is derived]
65/1/S (2)
65/1/S
Page 4


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S
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/S
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Getting sin ? = 
1
1
3
1 4
2
.
2 3
=
1
2
Hence 
| . | a b


 = 
1 4 3
. . 1
2 2 3
=
1
2
2.
2
| 2 | 1 a b - =


 ? 
1
. .
2
a b =

 1
2
? Angle between a

 and b

 = 
4
p 1
2
3. Writing or using, that given planes are parallel
1
2
d = 
| 4 10 |
2 units
4 9 36
+
=
+ +
1
2
4.
T T 2
| AA | |A| | A | |A| = =
1
2
 
= 25
1
2
5. Getting AB = 
7 8
0 10
- ? ?
? ?
-
? ?
1
2
|AB| = –70
1
2
6. k(2) = –8  ?  k = –4
1
2
–4(3) = 4a  ?  a = –3
1
2
SECTION B
7.
1
(sin 2 ) sin ( 3 )
x
y x x u v
-
= + = +
?
dy du dv
dx dx dx
= +
1
2
(sin 2 ) log logsin 2
x
u x u x x = ? =
1
2
1
2 cot 2 log sin 2 = · +
du
x x x
u dx
1
65/1/S (1)
65/1/S
? (sin 2 ) [2 cot 2 log sin 2 ]
x
du
x x x x
dx
= +
1
2
1 3
1 3 2
dv
dx x x
=
-
1
?
3
(sin 2 ) [2 cot 2 logsin 2 ]
2 1 3
x
dy
x x x x
dx x x
= + +
-
1
2
OR
Let y = 
2 2
1 1 2
2 2
1 1
tan and cos
1 1
x x
z x
x x
- -
? ?
+ - -
? ?
=
? ?
+ + -
? ?
z = cos
–1
 x
2
  ?  x
2
 = cos z  ? y = 
1
1 cos 1 cos
tan
1 cos 1 cos
z z
z z
-
? ?
+ - -
? ?
? ?
+ + -
? ?
1
? y = 
1 1
cos sin 1 tan
2 2 2
tan tan
cos sin 1 tan
2 2 2
z z z
z z z
- -
? ? ? ?
- -
? ? ? ?
=
? ? ? ?
? ? ? ?
+ +
? ? ? ?
1
2
 + 
1
2
? y = 
1
tan tan
4 2 4 2
z z
-
? p ? p ? ?
- = -
? ? ? ?
? ? ? ?
1
2
 + 
1
2
?
dy
dz
 = 
1
2
- 1
8. LHL = 
0
lim .sin ( 1)
2
-
?
p
+ =
x
k x k 1
RHL = 
3
0
tan (1 cos )
lim
x
x x
x
+
?
-
1
= 
2
0
tan sin /2 1
lim . 2
2 /2 2
x
x x
x x
+
?
? ?
=
? ?
? ?
1
? k = 
1
2
1
9. When x = am
2
, we get y = ± am
3
1
ay
2
 = x
3
 ? 
2
2
3
2 3
2
dy dy x
ay x
dx dx ay
= ? = 1
slope of normal = 
3
2 4
2 2
3 3
a am
m
a m
= ± ± 1
? Equation of normal is y ± am
3
 = 
2
2
( )
3
x am
m
- ±
1
[Full marks may be given, if only one value for point, slope and equation is derived]
65/1/S (2)
65/1/S
10. Writing
1 sin (1 sin ) 2sin
sin (1 sin ) sin (1 sin )
x x x
dx dx
x x x x
- + -
=
+ +
? ?
1
= 
1 1
2
sin 1 sin
dx dx
x x
-
+
? ?
1
= 
2
(1 sin )
cosec 2
cos
x
x dx dx
x
-
-
? ?
1
= 
2
log | cosec cot | 2 (sec sec tan ) x x x x x dx - - -
?
1
2
= log | cosec cot | 2(tan sec ) C - - - + x x x x
1
2
11. I = 
2
1
log(log )
(log )
x dx
x
? ?
+
? ?
? ?
?
= 
2
1
log(log ).1
(log )
x dx dx
x
+
? ?
1
= 
1 1
log(log )
log
x x
x x
· - · x ·
2
1
(log )
dx dx
x
+
? ?
2
= 
2 2
1 1 1 1
log(log )
log
(log ) (log )
? ?
-
· - · - · · +
? ?
? ?
? ?
x x x x dx dx
x x
x x
1
2
= 
log (log ) C
log
x
x x
x
- +
1
2
12. I = 
2
/2
0
sin
sin cos
x
dx
x x
p
+
?
...(i)
I = 
2 2
/2 /2
0 0
sin ( /2 ) cos
sin ( /2 ) cos ( /2 ) cos sin
x x
dx dx
x x x x
p p p -
=
p - + p - +
? ?
...(ii) 1
2I = 
/2
0
1
sin cos
dx
x x
p
+
?
1
? I = 
/2 /2
0 0
1 1 1
sec
1 1
4 2 2 2 2
cos sin
2 2
p p p ? ?
= -
? ?
? ?
+
? ?
dx x dx
x x
1
= 
/2
0
1
log sec tan
4 4 2 2
x x
p
? ? p p ? ? ? ?
- + -
? ? ? ? ? ?
? ? ? ?
? ?
1
2
= 
1 2 1 1
log or log | 2 1|
2 2 2 1 2
+
+
-
1
2
65/1/S (3)
65/1/S
Page 5


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S
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/S
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. Getting sin ? = 
1
1
3
1 4
2
.
2 3
=
1
2
Hence 
| . | a b


 = 
1 4 3
. . 1
2 2 3
=
1
2
2.
2
| 2 | 1 a b - =


 ? 
1
. .
2
a b =

 1
2
? Angle between a

 and b

 = 
4
p 1
2
3. Writing or using, that given planes are parallel
1
2
d = 
| 4 10 |
2 units
4 9 36
+
=
+ +
1
2
4.
T T 2
| AA | |A| | A | |A| = =
1
2
 
= 25
1
2
5. Getting AB = 
7 8
0 10
- ? ?
? ?
-
? ?
1
2
|AB| = –70
1
2
6. k(2) = –8  ?  k = –4
1
2
–4(3) = 4a  ?  a = –3
1
2
SECTION B
7.
1
(sin 2 ) sin ( 3 )
x
y x x u v
-
= + = +
?
dy du dv
dx dx dx
= +
1
2
(sin 2 ) log logsin 2
x
u x u x x = ? =
1
2
1
2 cot 2 log sin 2 = · +
du
x x x
u dx
1
65/1/S (1)
65/1/S
? (sin 2 ) [2 cot 2 log sin 2 ]
x
du
x x x x
dx
= +
1
2
1 3
1 3 2
dv
dx x x
=
-
1
?
3
(sin 2 ) [2 cot 2 logsin 2 ]
2 1 3
x
dy
x x x x
dx x x
= + +
-
1
2
OR
Let y = 
2 2
1 1 2
2 2
1 1
tan and cos
1 1
x x
z x
x x
- -
? ?
+ - -
? ?
=
? ?
+ + -
? ?
z = cos
–1
 x
2
  ?  x
2
 = cos z  ? y = 
1
1 cos 1 cos
tan
1 cos 1 cos
z z
z z
-
? ?
+ - -
? ?
? ?
+ + -
? ?
1
? y = 
1 1
cos sin 1 tan
2 2 2
tan tan
cos sin 1 tan
2 2 2
z z z
z z z
- -
? ? ? ?
- -
? ? ? ?
=
? ? ? ?
? ? ? ?
+ +
? ? ? ?
1
2
 + 
1
2
? y = 
1
tan tan
4 2 4 2
z z
-
? p ? p ? ?
- = -
? ? ? ?
? ? ? ?
1
2
 + 
1
2
?
dy
dz
 = 
1
2
- 1
8. LHL = 
0
lim .sin ( 1)
2
-
?
p
+ =
x
k x k 1
RHL = 
3
0
tan (1 cos )
lim
x
x x
x
+
?
-
1
= 
2
0
tan sin /2 1
lim . 2
2 /2 2
x
x x
x x
+
?
? ?
=
? ?
? ?
1
? k = 
1
2
1
9. When x = am
2
, we get y = ± am
3
1
ay
2
 = x
3
 ? 
2
2
3
2 3
2
dy dy x
ay x
dx dx ay
= ? = 1
slope of normal = 
3
2 4
2 2
3 3
a am
m
a m
= ± ± 1
? Equation of normal is y ± am
3
 = 
2
2
( )
3
x am
m
- ±
1
[Full marks may be given, if only one value for point, slope and equation is derived]
65/1/S (2)
65/1/S
10. Writing
1 sin (1 sin ) 2sin
sin (1 sin ) sin (1 sin )
x x x
dx dx
x x x x
- + -
=
+ +
? ?
1
= 
1 1
2
sin 1 sin
dx dx
x x
-
+
? ?
1
= 
2
(1 sin )
cosec 2
cos
x
x dx dx
x
-
-
? ?
1
= 
2
log | cosec cot | 2 (sec sec tan ) x x x x x dx - - -
?
1
2
= log | cosec cot | 2(tan sec ) C - - - + x x x x
1
2
11. I = 
2
1
log(log )
(log )
x dx
x
? ?
+
? ?
? ?
?
= 
2
1
log(log ).1
(log )
x dx dx
x
+
? ?
1
= 
1 1
log(log )
log
x x
x x
· - · x ·
2
1
(log )
dx dx
x
+
? ?
2
= 
2 2
1 1 1 1
log(log )
log
(log ) (log )
? ?
-
· - · - · · +
? ?
? ?
? ?
x x x x dx dx
x x
x x
1
2
= 
log (log ) C
log
x
x x
x
- +
1
2
12. I = 
2
/2
0
sin
sin cos
x
dx
x x
p
+
?
...(i)
I = 
2 2
/2 /2
0 0
sin ( /2 ) cos
sin ( /2 ) cos ( /2 ) cos sin
x x
dx dx
x x x x
p p p -
=
p - + p - +
? ?
...(ii) 1
2I = 
/2
0
1
sin cos
dx
x x
p
+
?
1
? I = 
/2 /2
0 0
1 1 1
sec
1 1
4 2 2 2 2
cos sin
2 2
p p p ? ?
= -
? ?
? ?
+
? ?
dx x dx
x x
1
= 
/2
0
1
log sec tan
4 4 2 2
x x
p
? ? p p ? ? ? ?
- + -
? ? ? ? ? ?
? ? ? ?
? ?
1
2
= 
1 2 1 1
log or log | 2 1|
2 2 2 1 2
+
+
-
1
2
65/1/S (3)
65/1/S
OR
    I = 
1 1
1 2 1
2
0 0
1
cot (1 ) tan
1
x x dx dx
x x
- -
? ?
- + =
? ?
- + ? ?
? ?
1
2
= 
1 1 1
1 1 1
0 0 0
(1 )
tan tan tan (1 )
1 (1 )
x x
dx x dx x dx
x x
- - -
? ? + -
= + -
? ?
- -
? ?
? ? ? 1
= 
1
1
0
2 tan x dx
-
?
1
2
= 
( )
1
1
1
2
0
0
2 tan .
1
-
? ?
-
? ?
+ ? ?
?
x
x x dx
x
1
2
= 
1
1 2
0
1
2 tan log |1 |
2
x x x
-
? ?
- +
? ?
? ?
1
= 
1
2 log 2 or log 2
4 2 2
p p ? ?
- -
? ?
? ?
1
2
13. The given differential equation can be written as
1
1
dy
y
dx x
-
+
 = 
2 3
( 1) .
x
x e +
1
2
Here, integrating factor = 
1
1
1
1
-
+
?
=
+
dx
x
e
x
1
? Solution is  
3
1
( 1)
1
x
y x e dx
x
= +
+
?
1
?
1
y
x +
 = 
3 3
( 1) C
3 9
x x
e e
x + - + 1
1
2
or y = 
2 3
1 1
( 1) C( 1)
3 9
x
x
x e x
+ ? ?
+ - + +
? ?
? ?
14. From the given differential equation, we can write
dx
dy
 = 
/ /
/ /
2 2 / 1
2 2
x y x y
x y x y
xe y x y e
ye e
- -
=
1
Putting
x
y
 = v ? 
dx
dy
 = 
dv
v y
dy
+
1
2
?
dv
v y
dy
+
 = 
2 1
2
v
v
ve
e
-
 ? 
dv
y
dy
 = 
1
2
v
e
-
1
? 2
v
e dv
?
 = 
dy
y
-
?
1
2
?
2 log | |
v
e y +
 = C ? 
/
2 log | | C
x y
e y + =
1
65/1/S (4)
65/1/S
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