Question

$f\left(x\right)=\frac{3\cos x+2\sin x}{\sin +\cos x}$ .Then $f\left(x\right)$ is

• A. increasing in its domain
• B. decreasing in its domain
• C. $f\left(\frac{\pi }{2}\right)<f\left(\pi \right)$
• D. $f\left(\frac{\pi }{2}\right)>f\left(\pi \right)$

Answer 1

Answer: (B), (C)

The correct options are
B decreasing in its domain
C $f\left(\frac{\pi }{2}\right)<f\left(\pi \right)$

$f\left(x\right)=\frac{3\cos x+2\sin x}{\sin x+\cos x}=2+\frac{\cos x}{\sin x+\cos x}$
$(\sin x+\cos x\ne 0(i.e.\left)\sqrt{2}\sin \right(x+\frac{\pi }{4})\ne 0\Rightarrow x\ne n\pi -\frac{\pi }{4})$

$\therefore f^{{}^{\prime }}\left(x\right)=\frac{(\sin x+\cos x)(-\sin x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x{)}^2}$
$=\frac{-1}{(\sin x+\cos x{)}^2}<0\forall xϵR-\{n\pi -\frac{\pi }{4}\}$
$\therefore f\left(x\right)$ is decrea\sing if $n\pi -\frac{\pi }{4}<x<n\pi +\frac{3\pi }{4}$
However, $f\left(\frac{\pi }{2}\right)=2,f\left(\pi \right)=3\Rightarrow f\left(\frac{\pi }{2}\right)<f\left(\pi \right)$
(Note : \since f(x) is decrea\sing in its domain $\frac{\pi }{2}<\pi$ should imply $f\left(\frac{\pi }{2}\right)>f\left(\pi \right).$
But here $f\left(\frac{\pi }{2}\right)<f\left(\pi \right).$ because between $\frac{\pi }{2}and\pi ,f\left(x\right)$ is not defined at $x=\frac{3\pi }{4}$