Question

The domain of definition of $f\left(x\right)=\frac{\sqrt{(x+3)}}{x^2+3x+2}$ is

• A. $(-3,-2)$
• B. $(-2,-1]$
• C. $[-3,\infty )-\{-1,-2\}$
• D. $[0,\infty )$

Answer 1

The correct option is C $[-3,\infty )-\{-1,-2\}$

$f\left(x\right)=\frac{\sqrt{(x+3)}}{x^2+3x+2}$

we need $x+3⩾0$ and $x^2+3x+2\ne 0$

for $f\left(x\right)$ to be defined

$\begin{array}{cc}x+3⩾0\Rightarrow x⩾-3& \\ x\in [-3\infty )& \end{array}$

$x^2+3x+2\ne 0\Rightarrow x^2+2x+x+2\ne 0$

$\Rightarrow (x+2)(x+1)\ne 0$

$\therefore x\ne -2$ and $x\ne -1$

so finally $x\in [-3\infty )-\{-1,-2\}$

$\therefore$ Answer: option (c)