Question

$A=\left\{\begin{array}{c}xϵR/\frac{\pi }{4}\le x\le \frac{\pi }{3}\end{array}\right\}$ and $f\left(x\right)=sinx-x$ then $f\left(A\right)=$

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

Answer 1

The correct option is A $\left[\begin{array}{c}\frac{\sqrt{3}}{2}-\frac{\pi }{3},\frac{1}{\sqrt{2}}-\frac{\pi }{4}\end{array}\right]$

1 $f\left(x\right)=sinx-x$
Which is decreasing function in $\left[\begin{array}{c}\frac{\pi }{4},\frac{\pi }{3}\end{array}\right]$
$\therefore \frac{\pi }{4}\le x\le \frac{\pi }{3}$
$\Rightarrow f\left(\begin{array}{c}\frac{\pi }{4}\end{array}\right)\ge f\left(x\right)\ge f\left(\begin{array}{c}\frac{\pi }{3}\end{array}\right)$
$\Rightarrow \frac{1}{\sqrt{2}}-\frac{\pi }{4}\ge f\left(x\right)\ge \frac{\sqrt{3}}{2}-\frac{\pi }{3}$
$f\left(A\right)\in \left[\begin{array}{c}\frac{\sqrt{3}}{2}-\frac{\pi }{3},\frac{1}{\sqrt{2}}-\frac{\pi }{4}\end{array}\right]$