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If ` int _(0)^(x^(2)) f(t) dt = x cos pix` , then the value of f(4) isA. 1B. `1/4`C. `-1`D. `(-1)/4` |
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Answer» Correct Answer - B `int_(0)^(x^(2))f(t)dt=xcospix` On differentiating both sides , we get `2xf(x^(2))=(-xsinpix)/(pi)+cospix` `rArrf(x^(2))=-(xsinpix)/(2pix)+(cospix)/(2x)` `thereforef(4)=f(2^(2))=(-2sin2pi)/(4pi)+(cos^(2)pi)/(4)=(1)/(4)` |
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