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Past Year Paper - Solutions, Maths(Set - 10, 11 and 12 ), Outside Delhi, 2016, Class 12, Maths | Mathematics (Maths) Class 12 - JEE PDF Download

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 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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FAQs on Past Year Paper - Solutions, Maths(Set - 10, 11 and 12 ), Outside Delhi, 2016, Class 12, Maths - Mathematics (Maths) Class 12 - JEE

1. What are the important topics covered in the Maths exam for Class 12, Set 10, 11, and 12, Outside Delhi, 2016?
Ans. The important topics covered in the Maths exam for Class 12, Set 10, 11, and 12, Outside Delhi, 2016 include algebra, calculus, coordinate geometry, and probability. These topics form the core of the syllabus and are expected to be thoroughly understood by the students.
2. How can I prepare effectively for the Maths exam for Class 12, Set 10, 11, and 12, Outside Delhi, 2016?
Ans. To prepare effectively for the Maths exam, students can follow these tips: a) Understand the concepts thoroughly by referring to textbooks and class notes. b) Practice a lot of numerical problems and equations to strengthen problem-solving skills. c) Solve past year papers and sample papers to get an idea of the exam pattern and question types. d) Seek help from teachers or classmates in case of any doubts or difficulties. e) Create a study schedule and allocate specific time for each topic to ensure comprehensive preparation.
3. Are there any specific tips for scoring well in the Maths exam for Class 12, Set 10, 11, and 12, Outside Delhi, 2016?
Ans. Yes, here are some specific tips for scoring well in the Maths exam: a) Understand the marking scheme and allocate time accordingly to each question. b) Start with the easier questions and then move on to the difficult ones. c) Show all the steps and write answers neatly to ensure clarity and avoid confusion. d) Revise all the formulas, theorems, and concepts thoroughly before the exam. e) Manage your time well during the exam and avoid spending too much time on a single question.
4. How can I improve my problem-solving skills for the Maths exam for Class 12, Set 10, 11, and 12, Outside Delhi, 2016?
Ans. Improving problem-solving skills for the Maths exam can be achieved through consistent practice. Here are some suggestions: a) Solve a variety of mathematical problems from different sources such as textbooks, reference books, and online resources. b) Break down complex problems into smaller, manageable steps. c) Understand the underlying concepts and formulas behind each problem. d) Seek guidance from teachers or classmates when facing difficulties in solving problems. e) Analyze your mistakes and learn from them to avoid repetition in the future.
5. Where can I find the solutions for the past year Maths exam papers for Class 12, Set 10, 11, and 12, Outside Delhi, 2016?
Ans. The solutions for the past year Maths exam papers for Class 12, Set 10, 11, and 12, Outside Delhi, 2016 can be found in various sources. Some of the options include: a) Official websites of educational boards or institutions that conduct the exams. b) Online platforms or forums where students share study materials and solutions. c) Reference books or guide books specifically designed for the Maths exam of Class 12. d) Coaching institutes or tuition centers that provide study materials and solutions. e) Offline or online bookstores that sell comprehensive solution books for past year papers.
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