Question

If $f:A \rightarrow B$ defined as $f\left(x\right)=2sin x-2cos ⁡ x+3\sqrt{2}$ is an invertible function, then the correct statement can be

• A. $A=\left[\frac{\pi }{4} , \frac{5 \pi }{4}\right],B=\left[2 \sqrt{2} , 5 \sqrt{5}\right]$
• B. $A=\left[\frac{- \pi }{4} , \frac{5 \pi }{4}\right],B=\left[\sqrt{2} , 5 \sqrt{2}\right]$
• C. $A=\left[\frac{- \pi }{4} , \frac{3 \pi }{4}\right],B=\left[\sqrt{2} , 4 \sqrt{2}\right]$
• D. $A=\left[\frac{- \pi }{4} , \frac{3 \pi }{4}\right],B=\left[\sqrt{2} , 5 \sqrt{2}\right]$

Answer 1

Answer: (D)

$f\left(x\right)=2\sqrt{2}\left(\frac{1}{\sqrt{2}} sin x - \frac{1}{\sqrt{2}} cos ⁡ x\right)+3\sqrt{2}=2\sqrt{2}sin⁡\left(x - \frac{\pi }{4}\right)+3\sqrt{2}$
Clearly, for options $A$ and $B$ $f\left(x\right)$ is not injective and for option $C$ $f\left(x\right)$ is not surjective
$f\left(x\right)$ is bijective only for option $D$