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

(C) local maximum at $\pi$ and local minimum at $2 \pi$

(A) local minimum at $\pi$ and $2 \pi$

(B) local minimum at $\pi$ and local maximum at $2 \pi$

(D) local maximum at $\pi$ and $2 \pi$

JEE Main · 2011 · 10 Apr 2011 · Offline · Application of Derivatives

For

x \in (0, \frac{5 π}{2})

, define

f (x) = \int_{0}^{x} t sin t d t

. Then

f

has

Answer: (C)

By the fundamental theorem of calculus,

f^{'} (x) = x sin x

, and on

(0, \frac{5 π}{2})

the factor

x

is positive, so the sign of

f^{'}

follows

sin x

. Thus

f^{'}

is positive on

(0, π)

, negative on

(π, 2 π)

and positive on

(2 π, \frac{5 π}{2})

. The derivative changes from positive to negative at

π

, a local maximum, and from negative to positive at

2 π

, a local minimum.

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option