Question

If $f\left(x\right)=\frac{x}{sinx}$ and $g\left(x\right)=\frac{x}{tanx},0<x\le 1$, 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)

$f^{\prime }\left(x\right)=\frac{sinx-xcosx}{si{n}^{2}x}$,
$g^{\prime }\left(x\right)=\frac{tanx-xse{c}^{2}x}{ta{n}^{2}x}$
Now, $\frac{d}{dx}\left(sinx-xcosx\right)=xsinx>0$ for $0<x\le 1$
$\therefore sinx-xcosx$ is an increasing function.
Also, at $x=0$, $sinx-xcosx=0$
$\Rightarrow 0<x\le 1$, $sinx-xcosx>0$
$\therefore f\left(x\right)>0$ for $0<x\le 1$
$\therefore f\left(x\right)$ is increasing on the interval $(0,1]$.
Again, $\frac{d}{dx}\left(tanx-xse{c}^{2}x\right)=-x\cdot 2se{c}^{2}x\cdot tanx<0$
for $0<x\le 1$
$\Rightarrow tanx-xse{c}^{2}x$ is an decreasing function.
Also at $x=0$, $tanx-xse{c}^{2}x=0$
$\Rightarrow 0<x\le 1,tanx-xse{c}^{2}x<0$
$g\left(x\right)<0$ for $0<x\le 1$.
$\therefore g\left(x\right)$ is decreasing in $(0,1]$.