Question

The domain of the function $f\left(x\right)=\frac{\sqrt{-{\log }_{0.3}\left(x-1\right)}}{\sqrt{-x^2+2x+8}}$is

• A. $\left(1,4\right)$
• B. $\left(-2,4\right)$
• C. $[2,4)$
• D. None of these

Answer 1

The correct option is C

$[2,4)$

Explanation for correct option:

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

Given,

$f\left(x\right)=\frac{\sqrt{-{\log }_{0.3}\left(x-1\right)}}{\sqrt{-x^2+2x+8}}$

Now, to find the domain, we need to identify the conditions

When we consider the term in the denominator, $\sqrt{-x^2+2x+8}$, we can say,

$\sqrt{-x^2+2x+8}\ne 0$, as the denominator cannot be $0$.

Also, the term with the square roots will always be $>0$.

Now, when we consider the term in the numerator, $\sqrt{-{\log }_{0.3}\left(x-1\right)}$,

$-{\log }_{0.3}\left(x-1\right)>0$. This can be obtained by plotting the graph.

Hence, from the above three conditions, it is clearly understood that

$-x^2+2x+8>0$

$$\begin{gathered} \Rightarrow x^2-2x-8<0 \\ \Rightarrow x^2-4x+2x-8<0 \\ \Rightarrow x\left(x-4\right)+2\left(x-4\right)<0 \\ \Rightarrow \left(x+2\right)\left(x-4\right)<0 \\ \Rightarrow x\in \left(-2,4\right) \end{gathered}$$

and $-{\log }_{0.3}\left(x-1\right)\ge 0$

$$\begin{gathered} \Rightarrow -{\log }_{0.3}\left(x-1\right)\ge 0 \\ \Rightarrow {\log }_{0.3}\left(x-1\right)\le 0 \\ \Rightarrow \left(x-1\right)\ge 1 \\ \Rightarrow x\ge 1+1 \\ \Rightarrow x\ge 2 \end{gathered}$$

Thus, $x\in [2,4)$

Hence, the domain of the given function is $[2,4)$

Therefore, option (C) is the correct answer.