Question

The domain of the real-valued function $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{5-4\mathrm{x}-{\mathrm{x}}^2}+{\mathrm{x}}^2\log \left(\mathrm{x}+4\right)$ is

• A. $-5\le \mathrm{x}\le 1$
• B. $-5\le \mathrm{x}\mathrm{and}\mathrm{x}\ge 1$
• C. $-4<\mathrm{x}\le 1$
• D. $\mathrm{\varphi }$
• E. $0\le \mathrm{x}\le 1$

Answer 1

The correct option is C

$-4<\mathrm{x}\le 1$

Explanation for correct answer:

Given function is

$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{5-4\mathrm{x}-{\mathrm{x}}^2}+{\mathrm{x}}^2\log \left(\mathrm{x}+4\right)$

In the logarithmic part of the function

$$\begin{gathered} \log \left(\mathrm{x}+4\right)\Rightarrow \mathrm{x}+4>0 \\ \Rightarrow \mathrm{x}>-4 \end{gathered}$$

Also, for $\sqrt{\mathrm{a}}$,

$\Rightarrow \mathrm{a}\ge 0$

$$\begin{gathered} \Rightarrow 5-4\mathrm{x}-{\mathrm{x}}^2\ge 0 \\ \Rightarrow \left(\mathrm{x}+5\right)\left(1-\mathrm{x}\right)\ge 0 \\ \mathrm{x}\in \left[-5,1\right] \end{gathered}$$

After combining both solution of $\mathrm{x}$

$\mathrm{x}\in (-4,1]$

$-4<\mathrm{x}\le 1$

Hence, option $\left(\mathrm{C}\right)$ is the correct answer.