Q60 / 90 · JEE Main 2011 · Mathematics

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

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

Solution

$f^{\prime}(x)=\sqrt{x} \sin x$ Given $x \in\left(0, \frac{5 \pi}{2}\right)$ $f^{\prime}(x)$ changes sign from +ve to -ve at $\pi$ $f^{\prime}(x)$ changes sign from -ve to +ve at $2 \pi$ $f$ has local max at $\pi$, local min at $2 \pi$

Hence, the correct answer is option: 'C'.

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