Question

For the function $f\left(x\right)={\int }_0^x\frac{\sin t}{t}dt$, where $x>0$

• A. maximum occurs at $x=n\pi$, $n$ is even
• B. minimum occurs at $x=n\pi$, $n$ is odd
• C. maximum occurs at $x=n\pi$, $n$ is odd
• D. None of these

Answer 1

The correct option is C maximum occurs at $x=n\pi$, $n$ is odd

$f\left(x\right)={\int }_0^x\frac{\sin t}{t}dt$
$f^{\prime }\left(x\right)=\frac{\sin x}{x}x>0$
When $x=n\pi$
$f^{\prime }\left(x\right)=\frac{\sin x}{x}$
$f^{\prime }\left(x\right)=\frac{x\times \cos x-\sin x}{x^2}=\frac{x\cos x-\sin x}{x^2}$
$\therefore$ $x=n\pi$ all of odd coefficient of $\pi$, $\cos x$ gives negative and becomes zero.
So it will be maximum occurs at $x=n\pi$, $n$ is odd.