Question

For $x\in (0,\frac{\pi }{2})$, define $f\left(x\right)={\int }_0^x\sqrt{t}\sin tdt$. Then $f$ has:

• A. local minimum at $\pi$ and local maximum at $2\pi$
• B. local maximum at $\pi$ and local minimum at $2\pi$
• C. local maximum at $\pi$ and $2\pi$
• D. local minimum at $\pi$ and $2\pi$

Answer 1

The correct option is A local minimum at $\pi$ and local maximum at $2\pi$

Mistake : given interval is wrong $(0,2\pi ]$ must be there

$f\left(x\right)={\int }_0^x\sqrt{t}\sin tdt$

$f^1\left(x\right)=\sqrt{x}\sin x=0⟹x=\pi ,2\pi$

$f^{11}\left(x\right)=\sqrt{x}\cos x+\frac{\sin x}{2\sqrt{x}}$

$f^{11}\left(\pi \right)=-\sqrt{\pi }<0⟹$ local minimum at $\pi$

$f^{11}\left(2\pi \right)=\sqrt{2\pi }>0⟹$ local maximum at $2\pi$