Question

Let $f\left(x\right)={\int }_0^x\frac{\cos t}{t}dx(x>0);$ then for $x=(2n+1)\frac{\mathrm{\pi }}{2},f\left(x\right)$ has?

• A. Maxima When $n=–2,–4,–6,\dots$
• B. Maxima when, $n=–1,–3,–5,\dots$
• C. Maxima when, $n=2,4,6,\dots$
• D. Maxima when, $n=1,3,5,\dots$
• E. $2\&3$

Answer 1

The correct option is C

Maxima when, $n=2,4,6,\dots$

Explanation for the correct option:

Finding the maximum values of $f\left(x\right)$:

We have,

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

On differentiating w.r.t $x$

$f^{\text{'}}\left(x\right)$ will be zero because for $x=(2n+1)\frac{\mathrm{\pi }}{2}$. $\frac{\cos x}{x}$ is always zero.

$$\begin{gathered} \therefore f^{\text{'}}\left(x\right)=0 \\ \Rightarrow \cos x=0 \\ \Rightarrow x=(2n+1)\frac{\mathrm{\pi }}{2},n\in I. \end{gathered}$$

Also,

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& \frac{-x\sin x-\cos x}{x^2}\\ & \Rightarrow & {\left[f\text{'}\text{'}\left(x\right)\right]}_{x=(2n+1)\frac{\mathrm{\pi }}{2}}=\frac{-\left(2n+1\right){\displaystyle \frac{\mathrm{\pi }}{2}}\sin (2n+1){\displaystyle \frac{\mathrm{\pi }}{2}}-0}{{\left[(2n+1){\displaystyle \frac{\mathrm{\pi }}{2}}\right]}^2}\\ & =& \frac{-2{(-1)}^n}{(2n+1)\mathrm{\pi }}<0\end{array}$

For $n=0,2,4,6...$

$f\left(x\right)hasmaximawhenn=0,2,4,6...$

Also, $forn=-1,-3,-5...$

$f\left(x\right)$ attains minima at $n=-1,-3,-5...$

Hence, the correct answer is option (B).