Question

The domain of the function $f\left(x\right)=\frac{1}{\sqrt{\left({\left[x\right]}^2-2\left[x\right]-8\right)}}$is

• A. $(-2,5]$
• B. $\left(-2,5\right)$
• C. $\left(-\infty ,-2\right)\cup [5,\infty )$
• D. $\left(-2,4\right)$

Answer 1

The correct option is C

$\left(-\infty ,-2\right)\cup [5,\infty )$

Explanation for the correct option(s):

Finding the domain of the function $f\left(x\right)$.

Given function, $f\left(x\right)=\frac{1}{\sqrt{\left({\left[x\right]}^2-2\left[x\right]-8\right)}}$needs to be meet two conditions to be true.

1. A rational function is true only when the denominator is positive.
2. A square root function must be always $\ge 0$.

Combining the above 2 conditions, this function must be $>0$ to be true.

Solving for the above condition,

$$\begin{gathered} \Rightarrow {\left[x\right]}^2-2\left[x\right]-8>0 \\ \Rightarrow {\left[x\right]}^2-4\left[x\right]+2\left[x\right]-8>0\left(\mathrm{Rewrite}-2\left[\mathrm{x}\right]\mathrm{as}-4\left[\mathrm{x}\right]+2\left[\mathrm{x}\right]\right) \\ \Rightarrow \left[x\right]\left(\left[x\right]-4\right)+2\left(\left[x\right]-4\right)>0 \\ \Rightarrow \left(\left[x\right]+2\right)\left(\left[x\right]-4\right)>0 \\ \Rightarrow \left[x\right]<2or\left[x\right]>4 \\ \therefore x<-2\text{and}x\ge 5 \end{gathered}$$

Hence, the domain of function $f\left(x\right)$ is $\left(-\infty ,-2\right)\cup [5,\infty ]$

Therefore, the correct answer is Option (C).