Question

Consider a function $f\left(x\right)=\frac{\sin x}{2}$. Let $g\left(x\right)=\int f\left(x\right)dx$, where constant of integration is zero. The number of local maxima of $g\left(x\right)$ in $(0,8\pi )$ are

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (B)

$4$

Given $f\left(x\right)=\frac{\sin x}{2}$
Also, $g\left(x\right)=\int f\left(x\right)dx$
$g\left(x\right)=\int \frac{\sin x}{2}dx=-\frac{\cos x}{2}+C$
$\Rightarrow g\left(x\right)=-\frac{\cos x}{2}$ ($\because C=0$, given)
For maxima or minma,
$g^{\prime }\left(x\right)=0$
$\Rightarrow \frac{\sin x}{2}=0$
$\Rightarrow \sin x=0$
$\Rightarrow x=\pi ,2\pi$ in $(0,2\pi )$
Now, $g^{\prime \prime }\left(x\right)=\frac{\cos x}{2}$
$g^{\prime \prime }\left(\pi \right)=-\frac{1}{2}<0$
$g^{\prime \prime }\left(2\pi \right)=\frac{1}{2}>0$
Hence, $g\left(x\right)$ has a local maxima at $x=\pi$ in $(0,2\pi ]$
So, $g\left(x\right)$ has $4$ local maxima in $(0,8\pi )$ at $x=\pi ,3\pi ,5\pi ,7\pi$