Question

Let f and g be real functions defined by $f\left(x\right)=\sqrt{x+2}\mathrm{and}g\left(x\right)=\sqrt{4-x^2}.\mathrm{Then},\mathrm{the}\mathrm{domain}\mathrm{of}\mathrm{the}\mathrm{function}\left(\frac{f}{g}\right)\left(x\right)\mathrm{is}$

• A. $\left(-2,2\right]$
• B. $\left(-2,2\right)$
• C. $[-2,2)$
• D. $(-2,2]$

Answer 1

The correct option is B $\left(-2,2\right)$

$\mathrm{Let}h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\mathrm{then}h\left(x\right)=\frac{\sqrt{x+2}}{\sqrt{4-x^2}}\mathrm{Since}\mathrm{square}\mathrm{root}\mathrm{functions}\mathrm{are}\mathrm{defined}\mathrm{for}\mathrm{which}\mathrm{the}\mathrm{argument}\mathrm{is}\ge 0\mathrm{and}\mathrm{rational}\mathrm{functions}\mathrm{are}\mathrm{defined}\mathrm{where}\mathrm{denominator}\ne 0,h\mathrm{is}\mathrm{defined}\mathrm{for}\mathrm{all}x\mathrm{for}\mathrm{which}x+2\ge 0\mathrm{and}\sqrt{4-x^2}>0\Rightarrow x\ge -2\mathrm{and}4-x^2>0\Rightarrow x\ge -2\mathrm{and}x\in \left(-2,2\right)\Rightarrow x\in \left(-2,2\right)$