JEE Exam > JEE Notes > Additional Study Material for JEE > Past Year Paper - Solutions, Mathematics (Set - 3), Delhi, 2014, Class 12, Maths
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Page 1
26
QUESTION PAPER CODE 65/1/3
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x
3/2
+ 2 x + c
5.
5
1
x =
6. x = 25 7. 5
8.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
9. 1
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
10.
k
ˆ
13
12
– j
ˆ
13
3
i
ˆ
13
4
+
1×10 = 10 m
SECTION - B
11.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
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
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
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
Marks
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Page 2
26
QUESTION PAPER CODE 65/1/3
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x
3/2
+ 2 x + c
5.
5
1
x =
6. x = 25 7. 5
8.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
9. 1
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
10.
k
ˆ
13
12
– j
ˆ
13
3
i
ˆ
13
4
+
1×10 = 10 m
SECTION - B
11.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
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
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
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
Marks
PDF created with pdfFactory trial version www.pdffactory.com
27
? ? ? ? ? ? ?
= + ? = + + 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
12. 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
13.
( ) ] [
? ?
·
+ +
=
·
+
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
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Page 3
26
QUESTION PAPER CODE 65/1/3
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x
3/2
+ 2 x + c
5.
5
1
x =
6. x = 25 7. 5
8.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
9. 1
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
10.
k
ˆ
13
12
– j
ˆ
13
3
i
ˆ
13
4
+
1×10 = 10 m
SECTION - B
11.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
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
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
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
Marks
PDF created with pdfFactory trial version www.pdffactory.com
27
? ? ? ? ? ? ?
= + ? = + + 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
12. 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
13.
( ) ] [
? ?
·
+ +
=
·
+
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
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28
( )
?
+ = 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
14. (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
– 1 0 2
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Page 4
26
QUESTION PAPER CODE 65/1/3
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x
3/2
+ 2 x + c
5.
5
1
x =
6. x = 25 7. 5
8.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
9. 1
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
10.
k
ˆ
13
12
– j
ˆ
13
3
i
ˆ
13
4
+
1×10 = 10 m
SECTION - B
11.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
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
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
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
Marks
PDF created with pdfFactory trial version www.pdffactory.com
27
? ? ? ? ? ? ?
= + ? = + + 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
12. 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
13.
( ) ] [
? ?
·
+ +
=
·
+
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
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28
( )
?
+ = 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
14. (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
– 1 0 2
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29
? 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
15. 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
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Page 5
26
QUESTION PAPER CODE 65/1/3
EXPECTED ANSWERS/V ALUE POINTS
SECTION - A
1-10. 1. 10 2. x = 2 3. x = + 6 3. 2x
3/2
+ 2 x + c
5.
5
1
x =
6. x = 25 7. 5
8.
( ) { } ( ) 0 k
ˆ
j
ˆ
i
ˆ
k
ˆ
c j
ˆ
b i
ˆ
a – r = + + · + +
9. 1
or
( ) c b a k
ˆ
j
ˆ
i
ˆ
r + + = + + ·
10.
k
ˆ
13
12
– j
ˆ
13
3
i
ˆ
13
4
+
1×10 = 10 m
SECTION - B
11.
?
?
?
?
?
?
?
?
?
?
?
?
+ ×
?
?
?
?
?
?
+ ·
?
?
?
?
?
?
+ =
?
?
?
?
?
?
+ + +
? ? ? ? ? ? ? ? ? ? ? ?
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
= ?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× · +
?
?
?
?
?
?
× ·
? ? ? ? ? ? ? ? ? ? ? ?
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
Marks
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27
? ? ? ? ? ? ?
= + ? = + + 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
12. 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
13.
( ) ] [
? ?
·
+ +
=
·
+
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
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28
( )
?
+ = 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
14. (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
– 1 0 2
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29
? 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
15. 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
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30
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
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. 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
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FAQs on Past Year Paper - Solutions, Mathematics (Set - 3), Delhi, 2014, Class 12, Maths - Additional Study Material for JEE
| 1. What is the importance of solving past year papers for the Mathematics JEE exam? |  |
Ans. Solving past year papers is important for the Mathematics JEE exam as it helps students become familiar with the exam pattern, understand the types of questions asked, and improve their time management skills. It also allows students to identify their strengths and weaknesses, enabling them to focus on the topics that need more attention.
| 2. Can solving past year papers guarantee success in the Mathematics JEE exam? | |
Ans. While solving past year papers is an effective preparation strategy, it cannot guarantee success in the Mathematics JEE exam. Success depends on various factors such as understanding the concepts, regular practice, and thorough revision. Solving past year papers is just one component of a comprehensive preparation plan.
| 3. How can I effectively utilize the solutions provided in the past year papers for Mathematics JEE? | |
Ans. To effectively utilize the solutions provided in the past year papers for Mathematics JEE, it is important to solve the questions on your own first before referring to the solutions. Once you have attempted a question, compare your solution with the provided one to identify any errors or alternative approaches. Analyze the solutions to understand the reasoning and problem-solving techniques used.
| 4. Are the questions in the Mathematics JEE exam similar to those in the past year papers? | |
Ans. The questions in the Mathematics JEE exam may not be exactly similar to those in the past year papers, but they are often based on similar concepts and difficulty levels. Solving past year papers helps students develop a strong foundation in these concepts and familiarize themselves with the type of questions that can be expected in the exam.
| 5. Can I solely rely on solving past year papers for Mathematics JEE preparation? | |
Ans. Solving past year papers is an important part of the preparation process, but it should not be the sole focus. It is essential to thoroughly understand the concepts, practice a wide range of questions from various sources, and revise regularly. Combining these strategies with solving past year papers will provide a more comprehensive preparation for the Mathematics JEE exam.
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