Functions Question 15
Question 15 - 30 January - Shift 2
The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is
(1) $[\sqrt{5}, \sqrt{10}]$
(2) $[2 \sqrt{2}, \sqrt{11}]$
(3) $[\sqrt{5}, \sqrt{13}]$
(4) $[\sqrt{2}, \sqrt{7}]$
Show Answer
Answer: (1)
Solution:
Formula: Operations on functions, Range of function
$y^{2}=3-x+2+x+2 \sqrt{(3-x)(2+x)}$
$=5+2 \sqrt{6+x-x^{2}}$
$y^{2}=5+2 \sqrt{\frac{25}{4}-(x-\frac{1}{2})^{2}}$
$y _{\max }=\sqrt{5+5}=\sqrt{10}$
$y _{\min }=\sqrt{5}$
