Question

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

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

Answer 1

Answer: (A)

$y=2\ln (1+\cos x)$
$y_1=\frac{-2\sin x}{1+\cos x}$
$y_2=-2\left[\frac{(1+\cos x)\cos x-\sin x(-\sin x)}{(1+\cos x{)}^2}\right]=-2\left[\frac{\cos x+1}{(1+\cos x{)}^2}\right]=\frac{-2}{(1+\cos x)}$
$\therefore 2e^{-y/2}=2\cdot e^{-\frac{\ln (1+\cos x{)}^2}{2}}=\frac{2}{(1+\cos x)}$
$\therefore y_2+\frac{2}{e^{y/2}}=0$