JEE Advanced 2015 Paper 2

Section: Mathematics

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Question : 49 of 60

Marks: +1, -0

Let f′(x)=

192x3

2+sin4πx

for all x∊ℝ with f(

1

2

)=0.If m≤

1

∫

1

2

f(x)dx≤M,then the possible values of m and M are

[JEE Adv 2015 P2]

m=13,M=24
m=
1
4
,M=
1
2
m=−11,M=0
m=1,M=12

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