Question

If f(x) = $\frac{x}{\sin x}andg\left(x\right)=\frac{x}{\tan x}$, where $0<x\le 1$, then in this interval

• A. Both $f\left(x\right)$ and $g\left(x\right)$ are increasing functions
• B. Both $f\left(x\right)$ and $g\left(x\right)$ are decreasing functions
• C. $f\left(x\right)$ is an increasing function
• D. $g\left(x\right)$ is an increasing function

Answer 1

Answer: (C)

$f\left(x\right)$ is an increasing function

Both are continuous function in the given interval.

Hence considering $f\left(x\right).$

$f^{\prime }\left(x\right)>0$ implies

$\sin \left(x\right)\left(1\right)-\cos \left(x\right)\left(x\right)>0$

$\sin x>\cos x.x$

$\tan x>x$

Or

$\frac{\tan x}{x}>1$

Or

$\frac{1}{g\left(x\right)}>1$

Or

$g\left(x\right)<1$ for $0<x\le 1$

g(x) is less than 1, for all $xϵ(0,1]$ and $g\left(x\right)$ is only equal to $1$, at $x=0.$

Since $g\left(x\right)<1$ in the given interval, hence $f\left(x\right)$ is increasing in the above given interval.