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Question

If f(x)=cosxx0(xt)f(t)dt, then f′′(x)+f(x) is equal to

A
cosx
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B
sinx
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C
x0(xt)f(t)dt
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D
0
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Solution

The correct option is A cosx
f(x)=cosxx0(xt)f(t)dtf(x)=cosxxx0f(t)dt+x0tf(t)dt
f(x)=sinxxf(x)x0f(t)dt+xf(x)
f(x)=sinxx0f(t)dtf′′(x)=cosxf(x)f′′(x)+f(x)=cosx

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