Solve this following


For $x \in\left(0, \frac{5 \pi}{2}\right)$, define $f(x)=\int_{0}^{x} \sqrt{t} \sin t d t$. Then $f$ has :-


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

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

  3. local maximum at $\pi$ and $2 \pi$

  4. local minimum at $\pi$ and $2 \pi$

Correct Option: , 2


$f^{\prime}(x)=\sqrt{x} \sin x$

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