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Question

For x ϵ (0,5π2), define f(x)=x0tsin t dt. Then f has

A
local minimum at π and 2 π
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B
local minimum at π and local maximum at 2π
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C
local maximum at π and local minimum at 2π
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D
local maximum at π and 2π
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Solution

The correct option is C local maximum at π and local minimum at 2π
f(x)=xsin x
Given x ϵ (0,5π2)
f ' (x) changes sign from +ve to – ve at π
f ' (x) changes sign from - ve to + ve at 2 π
f' has local max at π, local min at 2 π

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