For x∈(0$$,$$5$$π$$/2), define $$f(x)=$$∫$$0x$$ tsin⁡tdt Then f has

f′$$(x)=xsin$$⁡x, so f′$$(x)=0$$⇒$$x=$$π,$$2$$π but f′′$$(x)=xcos$$⁡x−$$12xsin$$⁡x, so f′′$$($$π$$)=$$−π $$0$$. Hence f has local maximumat $$x=$$π and local minimum at $$x=2$$π.

For x∈(0$$,$$5$$π$$/2), define $$f(x)=$$∫$$0x$$ tsin⁡tdt Then f has

Correct option is 1local maximum at π and local minimum at 2π.

Questions

For $x \in (0, 5 π / 2)$ , define $f (x) = \int_{0}^{x} \sqrt{t} \sin t d t$ Then $f$ has

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detailed solution

Correct option is A

f′(x)=xsin⁡x, so f′(x)=0⇒x=π,2π but f′′(x)=xcos⁡x−12xsin⁡x, so f′′(π)=−π<0 and f′′(2π)=2π>0. Hence f has local maximumat x=π and local minimum at x=2π.