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…

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
  1. $\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3}, \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right]$
  2. $\left[\frac{-1}{\sqrt{2}}-\frac{\pi}{4}, \frac{\sqrt{3}}{2}-\frac{\pi}{3}\right]$
  3. $\left[-\frac{\pi}{3},-\frac{\pi}{4}\right]$
  4. $\left[\frac{\pi}{4}, \frac{\pi}{3}\right]$

Solution

Given, $f(x)=\sin x-x$ Here, $\quad \frac{\pi}{4} \leq x \leq \frac{\pi}{3}$ $ f\left(\frac{\pi}{4}\right) \geq f(x) \geq f\left(\frac{\pi}{3}\right) $ $\left(\because f(x)\right.$ is decreasing function in $\left.x \in\left[\frac{\pi}{4}, \frac{\pi}{3}\right]\right)$ $ \begin{aligned} & \sin \frac{\pi}{4}-\frac{\pi}{4} \geq f(x) \geq \sin \frac{\pi}{3}-\frac{\pi}{3} \\ & \frac{1}{\sqrt{2}}-\frac{\pi}{4} \geq f(x) \geq \frac{\sqrt{3}}{2}-\frac{\pi}{3} \\ & \therefore \quad f(A) \in\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3} \cdot \frac{1}{\sqrt{2}}-\frac{\pi}{4}\right] \end{aligned} $

Asked in: AP EAMCET 2015

Practice more Functions questions on Aicharya