Question

Let $y=\ln (1+\cos x{)}^2.$ Then the value of $\frac{d^{2}y}{d{x}^2}+\frac{2}{e^{y/2}}$ is

• A. $0$
• B. $\frac{2}{1+\cos x}$
• C. $\frac{4}{1+\cos x}$
• D. $\frac{4}{(1+\cos x{)}^2}$

Answer 1

The correct option is A $0$

$y=2\ln (1+\cos x)$
Differentiating with respect to $x$
$y_1=\frac{-2\sin x}{1+\cos x}$
Again differentiating with respect to $x$
$y_2=-2\left[\frac{(1+\cos x)\cos x-\sin x(-\sin x)}{(1+\cos x{)}^2}\right]=-2\left[\frac{\cos x+1}{(1+\cos x{)}^2}\right]=\frac{-2}{1+\cos x}$

$2e^{-y/2}=2\cdot e^{-\frac{\ln (1+\cos x{)}^2}{2}}=\frac{2}{1+\cos x}$
$\therefore y_2+\frac{2}{e^{y/2}}=0$