QJEEM23Ma0d09: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0d09 Solution: The range of the function is :

$f(x) = \sqrt {3 - x} + \sqrt {x + 2}$

$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$

$\Rightarrow \sqrt x + 2 = \sqrt 3 - x$

$\Rightarrow x = {1 \over 2}$

$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10}$

${y_{\min }}$ at $x = - 2$ or $x = 3$ , i.e., $\sqrt 5$

$\therefore$ $f(x) \in [\sqrt 5 ,\sqrt {10} ]$