Question

What is the domain and range of : $y=\sqrt{2-x}+\sqrt{1+x}$

• A. $D:-1\le x\le 2,R:[\sqrt{3},\sqrt{6}]$
• B. $D:-2\le x\le 1,R:[\sqrt{3},\sqrt{6}]$
• C. $D:-1\le x\le 2,R:[\sqrt{3},\sqrt{12}]$
• D. none of these

Answer 1

The correct option is A $D:-1\le x\le 2,R:[\sqrt{3},\sqrt{6}]$

For domain $2-x\ge 0$
Or $2\ge x$ and
$1+x\ge 0$
Or $x\ge -1$.
Hence $xϵ[-1,2]$

Now $y^{\prime }=0$ implies
$\frac{-\sqrt{1+x}+\sqrt{2-x}}{\sqrt{2-x}\sqrt{1+x}}=0$
Or $(1+x)=2-x$
Or $2x=1$
Or $x=\frac{1}{2}$
$f\left(\frac{1}{2}\right)=2\sqrt{\frac{3}{2}}$
$=\sqrt{6}$.

Hence $maximum=\sqrt{6}$.
And $f(-1)=f\left(2\right)=\sqrt{3}=minimum$.
Hence range is $Rϵ[\sqrt{3},\sqrt{6}]$.