Question

If f(x) = $\frac{x}{sinx}$ and g(x) = $\frac{x}{tanx}$ where 0<x $\le$ 1, then in this interval $f\left(x\right)$ is

• 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(x) is an increasing function

We have f(x) $=\frac{x}{sinx}$ and g(x) $=\frac{x}{tanx}$
$\therefore f^{\prime }\left(x\right)=\frac{sinx-xcosx}{si{n}^{2}x}$ and
$g^{\prime }\left(x\right)=\frac{tanx-xse{c}^{2}x}{ta{n}^{2}x}$
Let $\Phi \left(x0\right)=sinx-xcosx$ and $\Psi \left(x\right)=tanx-xse{c}^{2}x$
Then, f'(x) = $\Phi \left(x\right)/si{n}^{2}x$ and $g^{\prime }\left(x\right)=\Psi \left(x\right)/ta{n}^2$
Now, $\Phi \left(x\right)=cosx-cosx+xsinx=xsinx$
and $\Psi \left(x\right)=se{c}^{2}x-se{c}^{2}x-2xse{c}^{2}xtanx=-2xse{c}^{2}xtanx$
For $0<x\le 1$, we have x > 0 , sin x >0, tan x >0, sec x >0
$\therefore \Phi \left(x\right)=xsinx>x$ and $\Psi \left(x\right(<0$ for $0<x\le 1$
$\Rightarrow \Phi \left(x\right)$ is increasing on (0,1) and $\Psi \left(x\right)$ is decreasing on (0,1)
$\Rightarrow \Phi \left(x\right)>Phi\left(0\right)$ and $\Psi \left(x\right)<\Psi \left(0\right)\Rightarrow \Phi \left(x\right)>0$ and $\Psi \left(x\right)<0$
$\therefore f\left(x\right)=\Phi \left(x\right)/si{n}^{2}x>0$ & g(x) = $\Phi \left(x\right)/ta{n}^{2}x<0$
$\Rightarrow$ f(x) is increasing on (0,1) and g(x) is decreasing on (0,1)