Question

If $A=\left\{x \in R / \frac{\pi}{4} \leq x \leq \frac{\pi}{3}\right\}$ and $f(x)=\sin x-x$, then $f(A)$ is equal to

• A. $\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3}, \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right]$
• B. $\left[\frac{-1}{\sqrt{2}}-\frac{\pi}{4}, \frac{\sqrt{3}}{2}-\frac{\pi}{3}\right]$
• C. $\left[-\frac{\pi}{3},-\frac{\pi}{4}\right]$
• D. $\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$

Answer 1

The Correct Option is A

Given, $f(x)=\sin x-x$
which is decreasing function in the interva $\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$.
$\therefore \frac{\pi}{4} \leq x \leq \frac{\pi}{3}$
$\Rightarrow f\left(\frac{\pi}{4}\right) \geq f(x) \geq f\left(\frac{\pi}{3}\right)$
$\Rightarrow \sin \frac{\pi}{4}-\frac{\pi}{4} \geq f(x) \geq \sin \frac{\pi}{3}-\frac{\pi}{3}$
$\Rightarrow \frac{1}{\sqrt{2}}-\frac{\pi}{4} \geq f(x) \geq \frac{\sqrt{3}}{2}-\frac{\pi}{3}$
$\Rightarrow f(A) \in\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3}, \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right]$