Question

If $f\left(x\right)=\frac{x}{\sin x}$ and $g\left(x\right)=\frac{x}{\tan x}$ where $0<x\le 1$ then in the interval

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

Answer 1

Answer: (C)

We know that if $f^{\prime }\left(x\right)>0$ then $f$ is increasing on that interval and if $f^{\prime }\left(x\right)<0$ then $f$ is decreasing on the interval.

Given $f\left(x\right)=\frac{x}{sinx}$

$⟹f^{\prime }\left(x\right)=\frac{sinx-xcosx}{si{n}^{2}x}=0$

$⟹sinx-xcosx=0$

$⟹x=\frac{sinx}{cosx}=tanx$

In the interval $(0,1)$ there is no solution for $x=tanx$

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

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

Given $g\left(x\right)=\frac{x}{tanx}$

$⟹g^{\prime }\left(x\right)=\frac{tanx-xse{c}^{2}x}{ta{n}^{2}x}=0$

$⟹tanx-xse{c}^{2}x=0$

$⟹x=\frac{se{c}^{2}x}{tanx}=\frac{1}{sinxcosx}$

In the interval $(0,1)$, $g\left(x\right)$ is not an increasing function.