Download - Jee Advanced Maths Paper- 2 Solutions
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JEE ADVANCED - 2015
Paper - II
MATHEMATICS
CODE - 0
Solutions
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PART -III(MATHEMATICS)
41.
Sol : 7
Required coefficient = coefficient of 9x in ( )2 91 ..... .....x x x+ + + + +( ) ( ) ( ) ( ) ( )3 5 6 7 91 1 1 1 1x x x x x+ + + + +
= 5+1+1 = 7
42.
Sol : 4
ay mxm
= + , a = 2
11
2y m xm
= + this is passing through (-4, 0)
21 2m =
22
4y m xm
= + Passing Through (-2, 0)
22 2m = i.e 2 2
1 2
1m m+
43.
Sol : 2
( )( )cos 10
sin .Lt
2
n n n
m n
e n e
m
=
Comparing 2, 1m n= =
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44.
Sol : 9
19 3tanx x t+ =
2
212 9
1x dx dt
x
+=
+
( )1 1 9 39 3 tan 40
1x xe e
pi+ +
= =
( ) 3log 14pi
+
3 39 94 4pi pi
= + =
45.
Sol : 7
( )( )1
114x
F xLt
G x=
( )
( )( )1
1
1
114
x
xx
f t dtLt
t f f t dt
=
( )( )( )1
114.x
f xLt
x f f x =
( )( )( )
1 1 1 1114 141. 1 2.2
ff f f
= = 1 72
f =
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46.
Sol : 9
4 3 5b p q r= + +
( ) ( ) ( )b x p q r y p q r z p q r= + + + + + +4x y z + = 2 7z =
3x y z =72
z =
5x y z+ + = 2 9y =
9 7 52 2
x + =92
y =
4x =
9 78 8 1 92 2
+ = + =
47.
Sol : 4
cos sin7 7k
k kipi pi = + 12
1 2 1 3 2 13 121
......k kk
+=
= + + + 1/22 2
2 12 2
cos cos sin sin7 7 7 7
i pi pi pi pi
= + / 2
2 2 c o s .7
1pi
=
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1/212
11
12 2 2cos7k kK
+=
pi =
3
4 1 4 2 3 2 7 6 11 101
k kK
=
= + + 1/2
3 2 2cos7pi
= 12
11
3
4 1 4 21
12 43
k kk
k kk
+=
=
= =
48.
Sol : 9
( )( )
7 2 6 6211 112 102
a d
a d
+=
+
( )7 3 65
a da d
+ =
+7 21 6 30a d a d + = +
( )9 ......... 1a d =as 7th term lies in between 130 and 140
130 6 140a d< + < 130 15 140d