Question

A function f: $R\to R$ satisfies $\sin x\cos y\left[f\right(2x+2y)-f(2x-2y\left)\right]=\cos x\sin y\left[f\right(2x+2y)+f(2x-2y\left)\right]$.

If $f^{\prime }\left(0\right)=\frac{1}{2}$, then

• A. $f^{\prime \prime }\left(x\right)=f\left(x\right)=0$
• B. $4f^{\prime \prime }\left(x\right)+f\left(x\right)=0$
• C. $f^{\prime \prime }\left(x\right)+f\left(x\right)=0$
• D. $4f^{\prime \prime }\left(x\right)=f\left(x\right)=0$

Answer 1

The correct option is B $4f^{\prime \prime }\left(x\right)+f\left(x\right)=0$

$f(2x+2y)[\sin x\cos y-\cos x\sin y]=f(2x-2y)[\cos x\sin y+\sin x\cos y]$

We have, $\frac{f(2x+2y)}{f(2x-2y)}=\frac{sin(x+y)}{sin(x-y)}$
$\Rightarrow \frac{f\left(\alpha \right)}{\sin \frac{\alpha }{2}}=\frac{f\left(\beta \right)}{\sin \frac{\beta }{2}}=K$
$\Rightarrow f\left(x\right)=K\sin \frac{x}{2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{K}{2}\cos \frac{x}{2}$
and $f^{\prime \prime }\left(x\right)=\frac{-K}{4}\sin \frac{x}{2}$
$\Rightarrow 4f^{\prime \prime }\left(x\right)+f\left(x\right)=0$
Hence, $B$ is the correct answer.