Question

The points of extrema of $f\left(x\right)={\int }_0^x\frac{\sin t}{t}dt$ in the domain $x>0$ are

• A. $n\pi ;n=1,2,..$
• B. $(2n+1)\frac{\mathrm{\pi }}{2};n=1,2,..$
• C. $(4n+1)\frac{\mathrm{\pi }}{2};n=1,2,..$
• D. $(2n+1)\frac{\mathrm{\pi }}{2};n=1,2,<...$

Answer 1

The correct option is A

$n\pi ;n=1,2,..$

The explanation for the correct answer.

Solve for the points of extrema of $f\left(x\right)={\int }_0^x\frac{\sin t}{t}dt$

$f\text{'}\left(x\right)=\frac{\sin x}{x}$

Put the first derivative equal to zero.

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{\sin x}{x}=0 \\ x=n\pi ,n\in I,x\ne 0 \end{gathered}$$

Calculate the second derivative

$f\text{'}\text{'}\left(x\right)=\frac{(x\cos x–\sin x)}{x^2}$

$f\text{'}\text{'}\left(n\pi \right)=\frac{(n\pi \cos n\pi –\sin n\pi )}{{\left(n\mathrm{\pi }\right)}^2}$ [If $\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{k}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{\left(}\mathit{n}\mathit{\pi }\mathbf{\right)}\mathbf{}\mathbf{<}\mathbf{0}$]

$f\text{'}\text{'}\left(x\right)<0$ ( where $x\in n\pi ,n\in 2k-1,k\in I$ )

$f\text{'}\text{'}\left(x\right)>0$ ( where$x\in n\pi ,n=2k$ )

Hence, option(A) is the correct answer.