The range of the function (f(x)=\sqrt{3-x}+\sqrt{2+x}) is: | Mathematics | SolveeIt

The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is:

$[2 \sqrt{2}, \sqrt{11}]$

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

$[\sqrt{5}, \sqrt{13}]$

$[\sqrt{2}, \sqrt{7}]$

Answer

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

Explanation

y2=3−x+2+x+2(3−x)(2+x)
=5+26+x−x2
y2=5+2425−(x−21)2
ymax=5+5=10
ymin=5
So , the correct option is (B) : $[\sqrt{5}, \sqrt{10}]$

Mathematics Question on Functions