Question

What is the domain of definition of $f\left(x\right)=\frac{\left[{\log }_2\left(x+3\right)\right]}{\left(x^2+3x+2\right)}$?

• A. $R-\left\{-1,-2\right\}$
• B. $\left(-2,\infty \right)$
• C. $R-\left\{-1,-2,-3\right\}$
• D. $\left(-3,\infty \right)-\left\{-1,-2\right\}$

Answer 1

The correct option is D

$\left(-3,\infty \right)-\left\{-1,-2\right\}$

Explanation of correct answer :

Finding the domain of the function :

Given function, $f\left(x\right)=\frac{\left[{\log }_2\left(x+3\right)\right]}{\left(x^2+3x+2\right)}$

For $f\left(x\right)$ to be defined, $\left(x^2+3x+2\right)\ne 0$

$$\begin{gathered} \Rightarrow \left(x^2+3x+2\right)\ne 0 \\ \Rightarrow x^2+2x+x+2\ne 0 \\ \Rightarrow \left(x+2\right)\left(x+1\right)\ne 0 \\ \Rightarrow x\ne -1,-2 \end{gathered}$$

As we know, for ${\log }_{a}b$, $b>0$ always as logarithmic of any negative number is not defined.

Then,

$$\begin{gathered} \left(x+3\right)>0 \\ \Rightarrow x>-3 \\ \Rightarrow x\in \left(-3,\infty \right) \end{gathered}$$

From above we got the value of $x\in \left(-3,\infty \right)$ but $-1,-2$are not possible.

Hence, the domain of the function is $\left(-3,\infty \right)-\left\{-1,-2\right\}$.

Thus, the correct answer is Option(D).