JEE Exam > JEE Notes > Additional Study Material for JEE > Past Year Paper - Solutions, Mathematics (Set - 2), Delhi, 2014, Class 12, Maths
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Page 1
14
QUESTION PAPER CODE 65/1/2
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 5 2.
( ) { } ( ) 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 + + = + + ·
3. 2x
3/2
+ 2 x + c 4. 10 5. x = 2
6. x = + 6 7.
5
1
x =
8. x = 25
9. c
2
x
–
2
x p
2
+ 10.
6
p
1×10 = 10 m
SECTION - B
11. (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
Marks
– 1 0 2
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Page 2
14
QUESTION PAPER CODE 65/1/2
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 5 2.
( ) { } ( ) 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 + + = + + ·
3. 2x
3/2
+ 2 x + c 4. 10 5. x = 2
6. x = + 6 7.
5
1
x =
8. x = 25
9. c
2
x
–
2
x p
2
+ 10.
6
p
1×10 = 10 m
SECTION - B
11. (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
Marks
– 1 0 2
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15
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
12.
( ) ] [
? ?
·
+ +
=
·
+
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
?
?
?
?
?
?
?
·
+
= 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
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Page 3
14
QUESTION PAPER CODE 65/1/2
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 5 2.
( ) { } ( ) 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 + + = + + ·
3. 2x
3/2
+ 2 x + c 4. 10 5. x = 2
6. x = + 6 7.
5
1
x =
8. x = 25
9. c
2
x
–
2
x p
2
+ 10.
6
p
1×10 = 10 m
SECTION - B
11. (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
Marks
– 1 0 2
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15
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
12.
( ) ] [
? ?
·
+ +
=
·
+
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
?
?
?
?
?
?
?
·
+
= 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
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16
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
13. Given differential equation can be written as
( )
2
2
2
1 – x
2
y
1 – x
2x
dx
dy
= +
1 m
Integrating factor = 1 – x e e
2 1) – (x log
dx
1 –
x 2
2 2
= =
?
x
1 m
c dx 1) – (x
1) – (x
2
1) – (x y is Solution
2
2 2
2
+ · = · ?
? 1 m
c dx
1 – x
1
2 1) – (x y
2
2
+ = ?
?
c
1 x
1 – x
log 1) – (x y
2
+
+
= ?
1 m
14. 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
0
x
y
–
dx
dy
y
1
–
dx
y d
2
2
2
=
?
?
?
?
?
?
?
½ m
15.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
a c c b b a a c , c b , b a ½ m
=
?
?
?
?
?
?
× + × + × + ×
?
?
?
?
?
?
+
? ? ? ? ? ? ? ? ? ?
a c c c a b c b b a
1 m
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Page 4
14
QUESTION PAPER CODE 65/1/2
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 5 2.
( ) { } ( ) 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 + + = + + ·
3. 2x
3/2
+ 2 x + c 4. 10 5. x = 2
6. x = + 6 7.
5
1
x =
8. x = 25
9. c
2
x
–
2
x p
2
+ 10.
6
p
1×10 = 10 m
SECTION - B
11. (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
Marks
– 1 0 2
PDF created with pdfFactory trial version www.pdffactory.com
15
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
12.
( ) ] [
? ?
·
+ +
=
·
+
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
?
?
?
?
?
?
?
·
+
= 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
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16
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
13. Given differential equation can be written as
( )
2
2
2
1 – x
2
y
1 – x
2x
dx
dy
= +
1 m
Integrating factor = 1 – x e e
2 1) – (x log
dx
1 –
x 2
2 2
= =
?
x
1 m
c dx 1) – (x
1) – (x
2
1) – (x y is Solution
2
2 2
2
+ · = · ?
? 1 m
c dx
1 – x
1
2 1) – (x y
2
2
+ = ?
?
c
1 x
1 – x
log 1) – (x y
2
+
+
= ?
1 m
14. 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
0
x
y
–
dx
dy
y
1
–
dx
y d
2
2
2
=
?
?
?
?
?
?
?
½ m
15.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
a c c b b a a c , c b , b a ½ m
=
?
?
?
?
?
?
× + × + × + ×
?
?
?
?
?
?
+
? ? ? ? ? ? ? ? ? ?
a c c c a b c b b a
1 m
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17
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
c b b a c a a b a c b a
1½ m
+
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ?
a c b a b b
?
?
?
?
?
?
=
?
?
?
?
?
?
× · =
?
?
?
?
?
?
× · =
?
?
?
?
?
?
× ·
?
?
?
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
0 a b b c b b a c a , a b a
=
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
× ·
? ? ? ? ? ?
c , b , a 2 c b a 2 1 m
OR
? ? ? ? ? ? ?
= + ? = + + c – b a 0 c b a
½ m
2 2 2
c c – b a
?
?
?
?
?
?
=
?
?
?
?
?
?
=
?
?
?
?
?
?
+ ?
? ? ? ?
½ m
2 2 2
c b a 2 b a
? ? ? ? ?
= · + + ?
1 m
? ? ? ?
= + + ? b & a between angle being ? 49, ? cos b a 2 25 9
1 m
3
p
?
2
1
5 3 2
15
? cos = ? =
· ·
= ?
1 m
16. 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
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Page 5
14
QUESTION PAPER CODE 65/1/2
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 5 2.
( ) { } ( ) 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 + + = + + ·
3. 2x
3/2
+ 2 x + c 4. 10 5. x = 2
6. x = + 6 7.
5
1
x =
8. x = 25
9. c
2
x
–
2
x p
2
+ 10.
6
p
1×10 = 10 m
SECTION - B
11. (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
Marks
– 1 0 2
PDF created with pdfFactory trial version www.pdffactory.com
15
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
12.
( ) ] [
? ?
·
+ +
=
·
+
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
?
?
?
?
?
?
?
·
+
= 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
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16
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
13. Given differential equation can be written as
( )
2
2
2
1 – x
2
y
1 – x
2x
dx
dy
= +
1 m
Integrating factor = 1 – x e e
2 1) – (x log
dx
1 –
x 2
2 2
= =
?
x
1 m
c dx 1) – (x
1) – (x
2
1) – (x y is Solution
2
2 2
2
+ · = · ?
? 1 m
c dx
1 – x
1
2 1) – (x y
2
2
+ = ?
?
c
1 x
1 – x
log 1) – (x y
2
+
+
= ?
1 m
14. 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
0
x
y
–
dx
dy
y
1
–
dx
y d
2
2
2
=
?
?
?
?
?
?
?
½ m
15.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
a c c b b a a c , c b , b a ½ m
=
?
?
?
?
?
?
× + × + × + ×
?
?
?
?
?
?
+
? ? ? ? ? ? ? ? ? ?
a c c c a b c b b a
1 m
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17
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
c b b a c a a b a c b a
1½ m
+
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ?
a c b a b b
?
?
?
?
?
?
=
?
?
?
?
?
?
× · =
?
?
?
?
?
?
× · =
?
?
?
?
?
?
× ·
?
?
?
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
0 a b b c b b a c a , a b a
=
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
× ·
? ? ? ? ? ?
c , b , a 2 c b a 2 1 m
OR
? ? ? ? ? ? ?
= + ? = + + c – b a 0 c b a
½ m
2 2 2
c c – b a
?
?
?
?
?
?
=
?
?
?
?
?
?
=
?
?
?
?
?
?
+ ?
? ? ? ?
½ m
2 2 2
c b a 2 b a
? ? ? ? ?
= · + + ?
1 m
? ? ? ?
= + + ? b & a between angle being ? 49, ? cos b a 2 25 9
1 m
3
p
?
2
1
5 3 2
15
? cos = ? =
· ·
= ?
1 m
16. 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
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18
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
17. 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
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FAQs on Past Year Paper - Solutions, Mathematics (Set - 2), Delhi, 2014, Class 12, Maths - Additional Study Material for JEE
| 1. What are the important topics to study for the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE? |  |
Ans. Some important topics to study for the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE include:
- Matrices and Determinants
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Differential Equations
| 2. How can I effectively prepare for the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE? | |
Ans. To effectively prepare for the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE, you can follow these steps:
- Create a study schedule and allocate specific time slots for each topic.
- Understand the concepts thoroughly by referring to your textbook and additional study materials.
- Practice solving a variety of problems from each topic to enhance your problem-solving skills.
- Solve the past year papers and mock tests to get familiar with the exam pattern and time management.
- Seek help from your teachers or join coaching classes if you're facing difficulties in understanding any topic.
| 3. Are there any specific tips for solving the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE efficiently? | |
Ans. Yes, here are some specific tips for solving the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE efficiently:
- Read the question carefully and understand what is being asked before attempting to solve it.
- Manage your time effectively by setting a time limit for each question and moving on if you're stuck.
- Practice using shortcuts and tricks to solve problems faster, especially in topics like calculus and algebra.
- Double-check your calculations and answers to avoid any silly mistakes.
- Focus on accuracy rather than attempting all the questions. It's better to solve fewer questions correctly than attempting more with errors.
| 4. Can you provide some additional resources for studying Mathematics for the Class 12 JEE exam? | |
Ans. Yes, here are some additional resources for studying Mathematics for the Class 12 JEE exam:
- NCERT textbooks for Class 12 Mathematics
- Reference books like "RD Sharma Mathematics for Class 12" and "Arihant's Handbook of Mathematics for Class 12"
- Online platforms and websites that offer video lectures and practice questions, such as Khan Academy, BYJU'S, and Vedantu.
- Joining coaching institutes that specialize in JEE preparation can provide comprehensive study materials and guidance from experienced teachers.
| 5. What is the weightage of the Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE? | |
Ans. The Mathematics (Set - 2) exam in Delhi, 2014 for Class 12 JEE carries a weightage of 100 marks. It is an important subject and contributes significantly to the overall JEE score. It is essential to perform well in this exam to secure a good rank in the JEE Main and JEE Advanced exams.
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