Mathematics · 2023
JEE Main · 30 January 2023, Shift 2 · Q61
The range of the function f(x)=√(3-x)+√(2+x) is:
The range of the function $\displaystyle f(x)=\sqrt{3-x}+\sqrt{2+x}$ is:
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle [\sqrt{5}, \sqrt{10}]$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.