JEE Exam > JEE Notes > Additional Study Material for JEE > Past Year Paper - Solutions, Mathematics (Set - 1), Delhi, 2014, Class 12, Maths
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Page 1
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. x = 25 2.
5
1
x =
3. 10 4. x = 2 5. x = + 6
6. 2x
3/2
+ 2 x + c 7.
12
p
8. 5 9.
3
2p
10.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
1×10 =10 m
SECTION - B
11. A A b) (a, × ? ?
a + b = b + a ? (a, b) R (a, b) ? R is reflexive 1 m
For (a, b), (c, d) A A × ?
If (a, b) R (c, d) i.e. a + d = b + c ? c + b = d + a
then (c, d) R (a, b) ? R is symmetric 1 m
For (a, b), (c, d), (e, f) A A × ?
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e ? a + f = b + e
then (a, b) R (e, f) ? R is transitive 1 m
? R is reflexive, symmetric and transitive
hence R is an equivalance relation ½ m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} ½ m
Q. No. Marks
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Page 2
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. x = 25 2.
5
1
x =
3. 10 4. x = 2 5. x = + 6
6. 2x
3/2
+ 2 x + c 7.
12
p
8. 5 9.
3
2p
10.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
1×10 =10 m
SECTION - B
11. A A b) (a, × ? ?
a + b = b + a ? (a, b) R (a, b) ? R is reflexive 1 m
For (a, b), (c, d) A A × ?
If (a, b) R (c, d) i.e. a + d = b + c ? c + b = d + a
then (c, d) R (a, b) ? R is symmetric 1 m
For (a, b), (c, d), (e, f) A A × ?
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e ? a + f = b + e
then (a, b) R (e, f) ? R is transitive 1 m
? R is reflexive, symmetric and transitive
hence R is an equivalance relation ½ m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} ½ m
Q. No. Marks
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3
12. cot
–1
?
?
?
?
?
?
?
?
?
?
+
+ +
x sin – 1 – x sin 1
x sin – 1 x sin 1
= cot
–1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
?
?
?
?
?
?
+
?
?
?
?
?
?
- +
?
?
?
?
?
?
+
2 2
2 2
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2½ m
2
x
2
x
cot cot
2
x
sin 2
2
x
cos 2
cot
1 1
= ?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
=
- -
1½ m
OR
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
+ =
- - -
7
2 5
sec
8
1
tan
5
1
tan 2 LHS
1 1 1
7
1
tan
40
1
– 1
8
1
5
1
tan 2
1 1 - -
+
?
?
?
?
?
?
?
?
?
?
?
?
+
=
1½+½ m
7
1
tan
3
1
– 1
3
1
2
tan
7
1
tan
3
1
tan 2
1
2
1 1 1 - - - -
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
·
= + =
1 m
RHS
4
p
) 1 ( tan
25
25
tan
7
1
tan
4
3
tan
1 1 1 1
= = = = + =
- - - -
1 m
13. LHS =
2x 2x z – y – x
y – x – z 2z 2z
2y – x z – y 2y
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Page 3
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. x = 25 2.
5
1
x =
3. 10 4. x = 2 5. x = + 6
6. 2x
3/2
+ 2 x + c 7.
12
p
8. 5 9.
3
2p
10.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
1×10 =10 m
SECTION - B
11. A A b) (a, × ? ?
a + b = b + a ? (a, b) R (a, b) ? R is reflexive 1 m
For (a, b), (c, d) A A × ?
If (a, b) R (c, d) i.e. a + d = b + c ? c + b = d + a
then (c, d) R (a, b) ? R is symmetric 1 m
For (a, b), (c, d), (e, f) A A × ?
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e ? a + f = b + e
then (a, b) R (e, f) ? R is transitive 1 m
? R is reflexive, symmetric and transitive
hence R is an equivalance relation ½ m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} ½ m
Q. No. Marks
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3
12. cot
–1
?
?
?
?
?
?
?
?
?
?
+
+ +
x sin – 1 – x sin 1
x sin – 1 x sin 1
= cot
–1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
?
?
?
?
?
?
+
?
?
?
?
?
?
- +
?
?
?
?
?
?
+
2 2
2 2
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2½ m
2
x
2
x
cot cot
2
x
sin 2
2
x
cos 2
cot
1 1
= ?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
=
- -
1½ m
OR
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
+ =
- - -
7
2 5
sec
8
1
tan
5
1
tan 2 LHS
1 1 1
7
1
tan
40
1
– 1
8
1
5
1
tan 2
1 1 - -
+
?
?
?
?
?
?
?
?
?
?
?
?
+
=
1½+½ m
7
1
tan
3
1
– 1
3
1
2
tan
7
1
tan
3
1
tan 2
1
2
1 1 1 - - - -
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
·
= + =
1 m
RHS
4
p
) 1 ( tan
25
25
tan
7
1
tan
4
3
tan
1 1 1 1
= = = = + =
- - - -
1 m
13. LHS =
2x 2x z – y – x
y – x – z 2z 2z
2y – x z – y 2y
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4
= 3 2 1 1
R R R R
2x 2x z – y – x
y – x – z 2z 2z
z y x z y x z y x
+ + ?
+ + + + + +
1 m
=
( ) ;
z y x z y x z – y – x
z y x – 0 2z
0 0 z y x
+ + + +
+ +
+ +
1 3 3
1 2 2
C – C C
C – C C
?
?
2 m
= (x + y + z)
.
{0
.
(x + y + z) + (x + y + z)
2
} = (x + y + z)
3
1 m
14. let ( ) x cos ? ? cos x , x – 1 2x cos v ,
x
x – 1
tan u
1 – 2 1 –
2
1 –
= ? = =
?
?
?
?
?
?
?
?
=
( ) x cos ? ? tan tan
? cos
? cos – 1
tan u
1 – 1 –
2
1 –
= = =
?
?
?
?
?
?
?
?
= ?
1 m
( ) ( )
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= = = ? 2 –
2
p
cos cos ? 2 sin cos ? cos – 1 ? cos 2 cos v and
1 – 1 – 2 1 –
= x cos 2 –
2
p
? 2 –
2
p
1 –
= 1 m
2 2
x – 1
2
dx
dv
,
x – 1
1 –
dx
du
= =
1 m
2
1 –
2
x – 1
x – 1
1 –
dv
du
2
2
= × = ?
1 m
( In case, If x = sin ? then answer is
2
1
)
15. y = x
x
?
log y = x log x, Taking log of both sides ½ m
dx
dy
y
1
?
= log x + 1, Diff. w r t “x” 1½ m
,
x
1
dx
dy
y
1
–
dx
y d
y
1
2
2 2
2
=
?
?
?
?
?
?
? Diff. w r t “x” 1½ m
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Page 4
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. x = 25 2.
5
1
x =
3. 10 4. x = 2 5. x = + 6
6. 2x
3/2
+ 2 x + c 7.
12
p
8. 5 9.
3
2p
10.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
1×10 =10 m
SECTION - B
11. A A b) (a, × ? ?
a + b = b + a ? (a, b) R (a, b) ? R is reflexive 1 m
For (a, b), (c, d) A A × ?
If (a, b) R (c, d) i.e. a + d = b + c ? c + b = d + a
then (c, d) R (a, b) ? R is symmetric 1 m
For (a, b), (c, d), (e, f) A A × ?
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e ? a + f = b + e
then (a, b) R (e, f) ? R is transitive 1 m
? R is reflexive, symmetric and transitive
hence R is an equivalance relation ½ m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} ½ m
Q. No. Marks
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3
12. cot
–1
?
?
?
?
?
?
?
?
?
?
+
+ +
x sin – 1 – x sin 1
x sin – 1 x sin 1
= cot
–1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
?
?
?
?
?
?
+
?
?
?
?
?
?
- +
?
?
?
?
?
?
+
2 2
2 2
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2½ m
2
x
2
x
cot cot
2
x
sin 2
2
x
cos 2
cot
1 1
= ?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
=
- -
1½ m
OR
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
+ =
- - -
7
2 5
sec
8
1
tan
5
1
tan 2 LHS
1 1 1
7
1
tan
40
1
– 1
8
1
5
1
tan 2
1 1 - -
+
?
?
?
?
?
?
?
?
?
?
?
?
+
=
1½+½ m
7
1
tan
3
1
– 1
3
1
2
tan
7
1
tan
3
1
tan 2
1
2
1 1 1 - - - -
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
·
= + =
1 m
RHS
4
p
) 1 ( tan
25
25
tan
7
1
tan
4
3
tan
1 1 1 1
= = = = + =
- - - -
1 m
13. LHS =
2x 2x z – y – x
y – x – z 2z 2z
2y – x z – y 2y
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4
= 3 2 1 1
R R R R
2x 2x z – y – x
y – x – z 2z 2z
z y x z y x z y x
+ + ?
+ + + + + +
1 m
=
( ) ;
z y x z y x z – y – x
z y x – 0 2z
0 0 z y x
+ + + +
+ +
+ +
1 3 3
1 2 2
C – C C
C – C C
?
?
2 m
= (x + y + z)
.
{0
.
(x + y + z) + (x + y + z)
2
} = (x + y + z)
3
1 m
14. let ( ) x cos ? ? cos x , x – 1 2x cos v ,
x
x – 1
tan u
1 – 2 1 –
2
1 –
= ? = =
?
?
?
?
?
?
?
?
=
( ) x cos ? ? tan tan
? cos
? cos – 1
tan u
1 – 1 –
2
1 –
= = =
?
?
?
?
?
?
?
?
= ?
1 m
( ) ( )
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= = = ? 2 –
2
p
cos cos ? 2 sin cos ? cos – 1 ? cos 2 cos v and
1 – 1 – 2 1 –
= x cos 2 –
2
p
? 2 –
2
p
1 –
= 1 m
2 2
x – 1
2
dx
dv
,
x – 1
1 –
dx
du
= =
1 m
2
1 –
2
x – 1
x – 1
1 –
dv
du
2
2
= × = ?
1 m
( In case, If x = sin ? then answer is
2
1
)
15. y = x
x
?
log y = x log x, Taking log of both sides ½ m
dx
dy
y
1
?
= log x + 1, Diff. w r t “x” 1½ m
,
x
1
dx
dy
y
1
–
dx
y d
y
1
2
2 2
2
=
?
?
?
?
?
?
? Diff. w r t “x” 1½ m
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5
0
x
y
–
dx
dy
y
1
–
dx
y d
2
2
2
=
?
?
?
?
?
?
?
½ m
16. (x) f ' = 12 x
3
– 12 x
2
– 24 x = 12 x (x + 1) (x – 2) 1+½ m
(x) f ' > 0, ) , 2 ( ) 0 , 1 (– 8 ? ? U x
- - - - - ? ?
+ + – –
1 m
(x) f ' < 0, ) 2 , 0 ( ) 1 – , (– U 8 ? ? x 1 m
f(x) ? is strictly increasing in ) , 2 ( ) 0 , 1 (– 8 U
½ m
and strictly decreasing in ) 2 , 0 ( ) 1 – , (– U 8
OR
Point at ?
?
?
?
?
?
=
2 2
a
,
2 2
a
is
4
p
? ½ m
? cos ? sin 3a
d?
dx
?; sin ? cos 3a –
d?
dy
2 2
= =
1 m
?
slope of tangent at
4
p
?
2
2
4
p
?
? cos ? sin 3a
? sin ? cos 3a –
dx
dy
is
4
p
?
=
=
?
?
?
=
?
?
?
=
=
1 –
4
p
cot – =
1 m
Equation of tangent at the point :
0
2
a
– y x
2 2
a
– x 1 –
2 2
a
– y = + ? ?
?
?
?
?
?
=
1 m
Equation of normal at the point :
0 y – x
2 2
a
– x 1
2 2
a
– y = ? ?
?
?
?
?
?
=
½ m
17.
( ) ] [
? ?
·
+ +
=
·
+
dx
x cos x sin
x cos x 3sin – x) cos x sin ( x cos x sin
dx
x cos x sin
x cos x sin
2 2
2 2 2 2 2 2 2
2 2
6 6
1½ m
? ?
?
?
?
?
?
·
= dx 3 –
x cos x sin
1
2 2
– 1 0 2
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Page 5
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. x = 25 2.
5
1
x =
3. 10 4. x = 2 5. x = + 6
6. 2x
3/2
+ 2 x + c 7.
12
p
8. 5 9.
3
2p
10.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
1×10 =10 m
SECTION - B
11. A A b) (a, × ? ?
a + b = b + a ? (a, b) R (a, b) ? R is reflexive 1 m
For (a, b), (c, d) A A × ?
If (a, b) R (c, d) i.e. a + d = b + c ? c + b = d + a
then (c, d) R (a, b) ? R is symmetric 1 m
For (a, b), (c, d), (e, f) A A × ?
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e ? a + f = b + e
then (a, b) R (e, f) ? R is transitive 1 m
? R is reflexive, symmetric and transitive
hence R is an equivalance relation ½ m
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} ½ m
Q. No. Marks
PDF created with pdfFactory trial version www.pdffactory.com
3
12. cot
–1
?
?
?
?
?
?
?
?
?
?
+
+ +
x sin – 1 – x sin 1
x sin – 1 x sin 1
= cot
–1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
?
?
?
?
?
?
+
?
?
?
?
?
?
- +
?
?
?
?
?
?
+
2 2
2 2
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2
x
sin
2
x
cos
2½ m
2
x
2
x
cot cot
2
x
sin 2
2
x
cos 2
cot
1 1
= ?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
=
- -
1½ m
OR
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
+ =
- - -
7
2 5
sec
8
1
tan
5
1
tan 2 LHS
1 1 1
7
1
tan
40
1
– 1
8
1
5
1
tan 2
1 1 - -
+
?
?
?
?
?
?
?
?
?
?
?
?
+
=
1½+½ m
7
1
tan
3
1
– 1
3
1
2
tan
7
1
tan
3
1
tan 2
1
2
1 1 1 - - - -
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
·
= + =
1 m
RHS
4
p
) 1 ( tan
25
25
tan
7
1
tan
4
3
tan
1 1 1 1
= = = = + =
- - - -
1 m
13. LHS =
2x 2x z – y – x
y – x – z 2z 2z
2y – x z – y 2y
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4
= 3 2 1 1
R R R R
2x 2x z – y – x
y – x – z 2z 2z
z y x z y x z y x
+ + ?
+ + + + + +
1 m
=
( ) ;
z y x z y x z – y – x
z y x – 0 2z
0 0 z y x
+ + + +
+ +
+ +
1 3 3
1 2 2
C – C C
C – C C
?
?
2 m
= (x + y + z)
.
{0
.
(x + y + z) + (x + y + z)
2
} = (x + y + z)
3
1 m
14. let ( ) x cos ? ? cos x , x – 1 2x cos v ,
x
x – 1
tan u
1 – 2 1 –
2
1 –
= ? = =
?
?
?
?
?
?
?
?
=
( ) x cos ? ? tan tan
? cos
? cos – 1
tan u
1 – 1 –
2
1 –
= = =
?
?
?
?
?
?
?
?
= ?
1 m
( ) ( )
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= = = ? 2 –
2
p
cos cos ? 2 sin cos ? cos – 1 ? cos 2 cos v and
1 – 1 – 2 1 –
= x cos 2 –
2
p
? 2 –
2
p
1 –
= 1 m
2 2
x – 1
2
dx
dv
,
x – 1
1 –
dx
du
= =
1 m
2
1 –
2
x – 1
x – 1
1 –
dv
du
2
2
= × = ?
1 m
( In case, If x = sin ? then answer is
2
1
)
15. y = x
x
?
log y = x log x, Taking log of both sides ½ m
dx
dy
y
1
?
= log x + 1, Diff. w r t “x” 1½ m
,
x
1
dx
dy
y
1
–
dx
y d
y
1
2
2 2
2
=
?
?
?
?
?
?
? Diff. w r t “x” 1½ m
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5
0
x
y
–
dx
dy
y
1
–
dx
y d
2
2
2
=
?
?
?
?
?
?
?
½ m
16. (x) f ' = 12 x
3
– 12 x
2
– 24 x = 12 x (x + 1) (x – 2) 1+½ m
(x) f ' > 0, ) , 2 ( ) 0 , 1 (– 8 ? ? U x
- - - - - ? ?
+ + – –
1 m
(x) f ' < 0, ) 2 , 0 ( ) 1 – , (– U 8 ? ? x 1 m
f(x) ? is strictly increasing in ) , 2 ( ) 0 , 1 (– 8 U
½ m
and strictly decreasing in ) 2 , 0 ( ) 1 – , (– U 8
OR
Point at ?
?
?
?
?
?
=
2 2
a
,
2 2
a
is
4
p
? ½ m
? cos ? sin 3a
d?
dx
?; sin ? cos 3a –
d?
dy
2 2
= =
1 m
?
slope of tangent at
4
p
?
2
2
4
p
?
? cos ? sin 3a
? sin ? cos 3a –
dx
dy
is
4
p
?
=
=
?
?
?
=
?
?
?
=
=
1 –
4
p
cot – =
1 m
Equation of tangent at the point :
0
2
a
– y x
2 2
a
– x 1 –
2 2
a
– y = + ? ?
?
?
?
?
?
=
1 m
Equation of normal at the point :
0 y – x
2 2
a
– x 1
2 2
a
– y = ? ?
?
?
?
?
?
=
½ m
17.
( ) ] [
? ?
·
+ +
=
·
+
dx
x cos x sin
x cos x 3sin – x) cos x sin ( x cos x sin
dx
x cos x sin
x cos x sin
2 2
2 2 2 2 2 2 2
2 2
6 6
1½ m
? ?
?
?
?
?
?
·
= dx 3 –
x cos x sin
1
2 2
– 1 0 2
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6
?
?
?
?
?
?
?
·
+
= dx 3 –
x cos x sin
x cos x sin
2 2
2 2
½ m
( )
?
+ = dx 3 – x cosec x sec
2 2
½ m
= tan x – cot x – 3x + c 1½ m
(Accept – 2 cot 2x – 3x + c also)
OR
( )
?
+ dx 18 – 3x x 3 – x
2
( )
? ?
+ + + = dx 18 – 3x x
2
9
– dx 18 – 3x x 3 2x
2
1
2 2
1 m
( ) ( )
?
?
?
?
?
?
?
+ + · = dx
2
9
–
2
3
x
2
9
– 18 – 3x x
3
2
2
1
2
2
2
3
2
1½ m
( )
2
9
– 18 – 3x x
3
1
2
3
2
+ =
c 18 – 3x x
2
3
x log
8
81
– 18 – 3x x
2
2
3
x
2 2
+ + + + +
?
?
?
?
?
?
?
?
?
?
?
?
?
+
1½ m
or
( )
8
9
– 18 – 3x x
3
1
2
3
2
+ =
{ c 18 – 3x x
2
3
x log
2
81
– 18 – 3x x ) 3 2 (
2 2
+ + + + + + x
18.
dy
y – 1
y –
dx xe dy
x
y –
dx y – 1 e
2
x 2 x
= ? =
1 m
Integrating both sides
? ?
= dy
y – 1
2y –
2
1
dx xe
2
x
c y – 1 e – xe
2 x x
+ = ?
1+1 m
For x = 0, y = 1, c = – 1 1 – y – 1 1) – (x e : is solution
2 x
= ? ½+½ m
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FAQs on Past Year Paper - Solutions, Mathematics (Set - 1), Delhi, 2014, Class 12, Maths - Additional Study Material for JEE
| 1. What is the format of the Mathematics (Set - 1) exam in the Delhi Board 2014 Class 12 Maths JEE? |  |
Ans. The Mathematics (Set - 1) exam in the Delhi Board 2014 Class 12 Maths JEE consists of multiple-choice questions and numerical-based questions. Students are required to solve the given problems and choose the correct answer from the given options.
| 2. How can I access the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam? | |
Ans. You can access the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam by searching for them online. There are various websites and educational platforms that provide these solutions for free. You can also check with your school or coaching institute for the solutions.
| 3. Are the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam helpful for preparation? | |
Ans. Yes, the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam are extremely helpful for preparation. By going through these solutions, you can understand the question patterns, marking scheme, and the approach to solving different types of problems. It also helps in identifying your weak areas and improving your problem-solving skills.
| 4. How can I effectively use the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam? | |
Ans. To effectively use the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam, you should first attempt the questions on your own without referring to the solutions. After solving the questions, compare your answers with the solutions provided. Analyze the differences and understand the correct approach to solving the problems. Practice similar questions to reinforce your understanding and improve your speed and accuracy.
| 5. Can the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam help in scoring better marks? | |
Ans. Yes, the past year paper solutions for the Delhi Board 2014 Class 12 Maths JEE exam can definitely help in scoring better marks. By practicing these solutions, you become familiar with the exam pattern, gain confidence in solving different types of questions, and improve your time management skills. It also helps in identifying common mistakes and misconceptions, allowing you to rectify them before the actual exam.
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