Q. The points of extrema of $f(x) = \int_{0}^{x}\frac{\sin t}{t}dt$ in the domain $x > 0$ are
A) $(2n + 1)\frac{\pi}{2},\ n = 1,\ 2,\ldots$
B) $(4n + 1)\frac{\pi}{2},\ n = 1,\ 2,\ldots$
C) $(2n + 1)\frac{\pi}{4},\ n = 1,\ 2,\ldots$
D) $n\pi,\ n = 1,\ 2,\ \ldots$
Explanation
(d)
Given curve is $f(x) = \int_{0}^{x}\frac{\sin t}{t}dt$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{\sin x}{x}$
For point of extreme, put $f^{'}(x) = 0$
$\Rightarrow \ \frac{\sin x}{x} = 0\ \ \Rightarrow \ \sin x = 0$
$\Rightarrow \ \ x = n\pi,\ n = 1,\ 2,\ 3,\ldots$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.