Question

The extreme values of the function $f\left(x\right)=\frac{1}{\sin x+4}-\frac{1}{\cos x-4}$, where $xϵR$ is

• A. $\frac{4}{8-\sqrt{2}}$
• B. $\frac{2\sqrt{2}}{8-\sqrt{2}}$
• C. $\frac{2\sqrt{2}}{4\sqrt{2}+1}$
• D. $\frac{4\sqrt{2}}{8+\sqrt{2}}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{4}{8-\sqrt{2}}$
C $\frac{2\sqrt{2}}{4\sqrt{2}+1}$

Given function is $f\left(x\right)=\frac{1}{\sin x+4}-\frac{1}{\cos x-4}$

To get the extreme values, $f^{\prime }\left(x\right)=0$

$f^{\prime }\left(x\right)=\frac{-\cos x}{(\sin x+4{)}^2}-\frac{\sin x}{(\cos x-4{)}^2}=0\Rightarrow {\sin }^{3}x+{\cos }^{3}x+16(\sin x+\cos x)+8({\sin }^{2}x-{\cos }^{2}x)=0$

$(\sin x+\cos x)(1-\sin x\cos x+16+8\sin x-8\cos x)=0$

$\sin x=-\cos x\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}$

$f\left(\frac{3\pi }{4}\right)=\frac{2\sqrt{2}}{4\sqrt{2}+1}$

$f\left(\frac{7\pi }{4}\right)=\frac{4}{8-\sqrt{2}}$