JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

The range of the function

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

is:

View answer and explanation

Correct option: A

[\sqrt{5}, \sqrt{10}]

Option A (

[\sqrt{5}, \sqrt{10}]

) is correct because the range of the function is determined by finding the minimum and maximum values within its domain.

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