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Question

Let ddx(F(x))=esinxx, x>0. If 412esinx2xdx=F(k)F(1), then the possible value of k is

A
10
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B
14
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C
16
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D
18
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Solution

The correct option is D 16
Substitute x2=z
Then 412esinx2xdx=161esinzzdz=161ddz{F(z)}dz
=(F(z))161=F(16)F(1)
F(16)F(k)=F(16)F(1)
Gives k=16

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