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For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$ Then $$f$$ has

JEE · Math · previous-year question

  1. A.local minimum at $$\pi $$ and $$2\pi $$
  2. B.local minimum at $$\pi $$ and local maximum at $$2\pi $$
  3. C.local maximum at $$\pi $$ and local minimum at $$2\pi $$correct
  4. D.local maximum at $$\pi $$ and $$2\pi $$

Answer

C. local maximum at $$\pi $$ and local minimum at $$2\pi $$

Explanation

$$f'\left( x \right) = \sqrt x \sin x$$ At local maxima or minima, $$f'\left( x \right) = 0$$ $$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$ $$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$ $$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$ $$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$ At $$x = \pi ,$$ $$f''\left( x \right) < 0$$ Hence, local maxima at $$x = \pi $$ At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$ Hence local minima at $$x = 2\pi $$

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