Question

Let $f:X\to Y$ be a function such that $f\left(x\right)=\sqrt{x-2}+\sqrt{4-x}$. If $f\left(x\right)$ is both injective as well as surjective, then set $X$ and $Y$ is

• A. $[2,4]$ and $[\sqrt{2},2]$
• B. $[3,4]$ and $[\sqrt{2},2]$
• C. $[2,4]$ and $[1,2]$
• D. $[2,3]$ and $[1,2]$

Answer 1

The correct option is B $[3,4]$ and $[\sqrt{2},2]$

$f\left(x\right)=\sqrt{x-2}+\sqrt{4-x}$
For the square root to exist,
$x-2\ge 0\Rightarrow x\ge 24-x\ge 0\Rightarrow x\le 4$
So, $D_f\in [2,4]$
Now,
$f^{\prime }\left(x\right)=0\Rightarrow \frac{1}{2\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}=0\Rightarrow (x-2)=4-x\Rightarrow x=3$
$f\left(3\right)=2,f\left(2\right)=\sqrt{2},f\left(4\right)=\sqrt{2}$,
The graph of the function,


$\therefore$ Range $[\sqrt{2},2]$

Now, for the function to be both injective and surjective the domain has to be restricted in either $[2,3]$ or $[3,4]$.

Hence, from the given options,
$X=[3,4],Y=[\sqrt{2},2]$