Question

If $f(x)=\pi \left(\frac{\sqrt{x+7}-4}{x-9}\right)$, then the range of function $y=\sin (2f(x))$ is

• A. $[0,1]$
• B. $\left(0,\frac{1}{\sqrt{2}}\right]$
• C. $\left(0,\frac{1}{\sqrt{2}}\right)\cup \left(\frac{1}{\sqrt{2}},1\right]$
• D. $(0,1]$

Answer 1

Answer: (C)

funcisc Given, $f(x)=\pi \left(\frac{\sqrt{x+7}-4}{x-9}\right)$
Clearly, domain of $f(x)=[-7,\infty )-\{9\}$.
Now, $f(x)=\frac{\pi (x+7-16)}{(x-9)(\sqrt{x+7}+4)}$ (Rationalise $)=\frac{\pi }{\sqrt{x+7}+4}$.
So, range of $f(x)$ is $\left(0,\frac{\pi }{4}\right]-\left\{\frac{\pi }{8}\right\}$.
Hence, range of $y=\sin (2f(x))$ is $(0,1]-\left\{\frac{1}{\sqrt{2}}\right\}$. Ans.]