Question

Consider the function $y=f\left(x\right)=\ln (1+\sin x)$ with $-2\pi \le x\le 2\pi .$

Find local maxima and minima of f(x)

• A. maxima at $x=\frac{\pi }{2}$ and no minima,
• B. maxima at $-\frac{3\pi }{2}$ and no minima,
• C. maxima at $-\frac{3\pi }{2}$ and minima at $x=\frac{\pi }{2}$
• D. maxima at $x=\frac{\pi }{2}$ and minima at $-\frac{3\pi }{2}$

Answer 1

Answer: (A), (B)

The correct options are
A maxima at $x=\frac{\pi }{2}$ and no minima,
B maxima at $-\frac{3\pi }{2}$ and no minima,

$y=\ln (1+\sin x)$

$\frac{dy}{dx}=\frac{\cos x}{1+\sin x}=0$ for extremum.

$\Rightarrow x=\pm \frac{\pi }{2},\pm \frac{3\pi }{2}$

But at $x=-\frac{\pi }{2},\frac{3\pi }{2}$ $\ln (1+\sin x)$ is not defined

Thus $x=\frac{\pi }{2},-\frac{3\pi }{2}$

$\frac{d^{2}y}{d{x}^2}=\frac{-(1+\sin x)\sin x-\cos x.\cos x}{(1+\sin x{)}^2}=\frac{-1}{1+\sin x}<0\forall x\in$ the domain of $f$.

Hence both $x=\frac{\pi }{2},$ and $x=-\frac{3\pi }{2}$ are point of local maxima.