The range of the function $f(x)=2 \sqrt{x-2}+\sqrt{4-x}$ is

JEE MATHEMATICS MCQ: FUNCTIONS

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The range of the function

f(x)=2 \sqrt{x-2}+\sqrt{4-x}

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Correct option: D

[\sqrt{2}, \sqrt{10}]

Option D ([√2, √10]) is correct because it represents the range of the function, considering the domain and critical points.

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}