Question

If $A=\{ϵR|\frac{\Pi }{4}\le x\le \frac{\Pi }{3}\}$, and $f\left(x\right)=\sin x-x$, then $f\left(A\right)=$

• A. $\frac{\sqrt{3}}{2}-\frac{\Pi }{3}$,$\frac{1}{\sqrt{2}}-\frac{\Pi }{4}$
• B. $-\frac{1}{\sqrt{2}}-\frac{\Pi }{4}$ , $\frac{\sqrt{3}}{2}-\frac{\Pi }{3}$
• C. $-\frac{\Pi }{3}$, $\frac{-\Pi }{4}$
• D. $\frac{\Pi }{4}$, $\frac{\Pi }{3}$

Answer 1

The correct option is A $\frac{\sqrt{3}}{2}-\frac{\Pi }{3}$,$\frac{1}{\sqrt{2}}-\frac{\Pi }{4}$

$A=\{x\in R|\frac{\prod }{4}\le x\le \frac{\prod }{3}\}$

$f\left(x\right)=\sin x-x$

$f\left(A\right)=?$

Which is decreasing function in $\frac{\prod }{4}$ and $\frac{\prod }{3}$ .

$\frac{\prod }{4}\le x\le \frac{\prod }{3}$

$f\left(\frac{\prod }{4}\right)\ge f\left(x\right)\ge f\left(\frac{\prod }{3}\right)$

$\frac{1}{\sqrt{2}}-\frac{\prod }{4}\ge f\left(x\right)\ge \frac{\sqrt{3}}{2}-\frac{\prod }{3}$

$\Rightarrow f\left(A\right)=(\frac{\sqrt{3}}{2}-\frac{\prod }{3},\frac{1}{\sqrt{2}}-\frac{\prod }{4})$

Hence, it is the correct answer.