Question

Which of the following inequalities are valid:

• A. $|{\tan }^{-1}x-{\tan }^{-1}y|\le |x-y|\forall x,y\in R$
• B. $|{\tan }^{-1}x-{\tan }^{-1}y|\ge |x-y|$
• C. $|\sin x-\sin y|\le |x-y|$
• D. $|\sin x-\sin y|\ge |x-y|$

Answer 1

The correct option is A $|{\tan }^{-1}x-{\tan }^{-1}y|\le |x-y|\forall x,y\in R$

Applying LMVT

$c$ in $(x,y)$

$f^{{}^{\prime }}\left(c\right)=\frac{f\left(y\right)-f\left(x\right)}{y-x}$

Let $f\left(x\right)={\tan }^{-1}x,f\left(y\right)={\tan }^{-1}y$

$\therefore f^{{}^{\prime }}\left(c\right)=\frac{1}{1+c^2}=\frac{f\left(y\right)-f\left(x\right)}{y-x}$

As, $c^2\ge 0$

$\therefore 1+c^2\ge 1$

$\therefore \frac{1}{1+c^2}\le 1$

$\therefore \frac{f\left(y\right)-f\left(x\right)}{y-x}\le 1$

$\Rightarrow f\left(y\right)-f\left(x\right)\le y-x$

$\Rightarrow \mid f\left(y\right)-f\left(x\right)\mid \le \mid y-x\mid$

$\Rightarrow \mid f\left(x\right)-f\left(y\right)\mid \le \mid x-y\mid$

As, $f\left(x\right)={\tan }^{-1}x$ and $f\left(y\right)={\tan }^{-1}y$

$\therefore \mid {\tan }^{-1}x-{\tan }^{-1}y\mid \le \mid x-y\mid \forall x,y\in R$

Similarly,

Let $f\left(x\right)=\sin x,f\left(y\right)=\sin y$

$\therefore f^{{}^{\prime }}\left(x\right)=\cos x$

$\therefore f^{{}^{\prime }}\left(c\right)=\frac{f\left(y\right)-f\left(x\right)}{y-x}$

$\cos c=\frac{\sin y-\sin x}{y-x}$

As $\mid \cos c\mid \le 1$

$\mid \frac{\sin y-\sin x}{y-x}\mid \le 1$

$\mid \sin y-\sin x\mid \le \mid y-x\mid$

$\therefore \mid \sin x-\sin y\mid \le \mid x-y\mid$