Functions
Range of a Function
JEE Main 2023
Grade 12
Question:
The range of $f(x) = \sqrt{3 - x} + \sqrt{2 + x}$ is:
(1) $[\sqrt{5}, \sqrt{10}]$
(2) $[1, \sqrt{10}]$
(3) $[\sqrt{5}, 3]$
(4) $(0, \sqrt{10}]$
Step-by-Step Solution
Key Concept: Square $y = f(x)$ to get $y^2 = 5+2\sqrt{(3 - x)(2 + x)}$. Maximise $(3-x)(2+x)$ using AM-GM or completing the square, then find the range of $y$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)