Question

For $0<x_1<x_2<\frac{\pi }{2}$ .

• A. $\frac{\tan x_2}{\tan x_1}<\frac{x_2}{x_1}$
• B. $\frac{\tan x_2}{\tan x_1}>\frac{x_2}{x_1}$
• C. $\frac{\tan x_2}{\tan x_1}=\frac{x_2}{x_1}$
• D. None of these

Answer 1

The correct option is B $\frac{\tan x_2}{\tan x_1}>\frac{x_2}{x_1}$

Consider a function $f\left(x\right)=\frac{\tan x}{x}$

$f^{\prime }\left(x\right)=\frac{x{\sec }^{2}x-\tan x}{x^2}$

$={\sec }^{2}x\frac{x-\sin x\cos x}{x^2}$

$=\frac{x-\sin x\cos x}{x^2{\cos }^{2}x}$

$=\frac{x-\frac{2\sin x\cos x}{2}}{x^2{\cos }^{2}x}$

$=\frac{x-\frac{\sin 2x}{2}}{x^2{\cos }^{2}x}$

$=\frac{2x-\sin 2x}{2x^2{\cos }^{2}x}$

Given:$0<x<\frac{\pi }{2}$

or $0<2x<\pi$

$\Rightarrow 2x>\sin 2x$

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

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

$x\in (0,\frac{\pi }{2})$

If $x_1<x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right)$

$\Rightarrow \frac{\tan x_1}{x_1}<\frac{\tan x_2}{x_2}$

or $\frac{x_2}{x_1}<\frac{\tan x_2}{\tan x_1}$

or $\frac{\tan x_2}{\tan x_1}>\frac{x_2}{x_1}$