Let f : X rightarrow Y be a function such tha | vedclass

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(B) The domain of $f(x)$ is determined by $x-2 \ge 0$ and $4-x \ge 0$,which gives $x \in [2,4]$.To find the range,let $y = \sqrt{x-2} + \sqrt{4-x}$. S

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$[3,4]$ and $[\sqrt{2},2]$

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(B) The domain of $f(x)$ is determined by $x-2 \ge 0$ and $4-x \ge 0$,which gives $x \in [2,4]$.To find the range,let $y = \sqrt{x-2} + \sqrt{4-x}$. Squaring both sides,$y^2 = (x-2) + (4-x) + 2\sqrt{(x-2)(4-x)} = 2 + 2\sqrt{-x^2+6x-8}$.The expression $-x^2+6x-8$ has a maximum value of $1$ at $x=3$. Thus,$y^2 \in [2, 4]$,so $y \in [\sqrt{2}, 2]$.For $f(x)$ to be injective,the function must be monotonic. The function $f(x)$ increases on $[2,3]$ and decreases on $[3,4]$.Therefore,$f(x)$ is injective on either $[2,3]$ or $[3,4]$.On both intervals $[2,3]$ and $[3,4]$,the range of the function is $[\sqrt{2}, 2]$.Thus,for $f(x)$ to be both injective and surjective,$X$ can be $[2,3]$ or $[3,4]$,and $Y$ must be $[\sqrt{2}, 2]$.Comparing with the options,option $B$ is the correct choice.

Let $f : X \rightarrow Y$ be a function such that $f(x) = \sqrt{x - 2} + \sqrt{4 - x} .$ Then the set of $X$ and $Y$ for which $f(x)$ is both injective as well as surjective,is-

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