Question

Find the domain of the function$f\left(x\right)=\sqrt{[\frac{2}{(x^2-x+1)}-\frac{1}{(x+1)}-\frac{2x-1}{(x^3+1)}]}$

• A. $(-\infty ,2]-\{-1\}$
• B. $(-\infty ,2)$
• C. $]-1,2]$
• D. None of these

Answer 1

The correct option is A

$(-\infty ,2]-\{-1\}$

Explanation for correct option:

Step 1: Set the terms inside the root symbol to be greater than or equal to $0$.

We have given$f\left(x\right)=\sqrt{[\frac{2}{(x^2-x+1)}-\frac{1}{(x+1)}-\frac{2x-1}{(x^3+1)}]}$.

$$\begin{gathered} \frac{2}{(x^2-x+1)}-\frac{1}{(x+1)}-\frac{2x-1}{(x^3+1)}\ge 0 \\ \Rightarrow \frac{2}{(x^2-x+1)}-\frac{1}{(x+1)}-\frac{2x-1}{(x+1)(x^2-x+1)}\ge 0 \\ \Rightarrow \frac{2(x+1)-(x^2-x+1)-(2x-1)}{(x+1)(x^2-x+1)}\ge 0 \\ \Rightarrow \frac{2x+2-x^2+x-1-2x+1}{(x+1)(x^2-x+1)}\ge 0 \\ \Rightarrow \frac{-x^2+x+2}{(x+1)(x^2-x+1)}\ge 0 \\ \Rightarrow -\frac{(x+1)(x-2)}{(x+1)(x^2-x+1)}\ge 0 \end{gathered}$$

From above$x\ne -1$….$\left(1\right)$.

Step 2: Find the points where the function changes the signs.

$$\begin{gathered} \Rightarrow -\frac{(x-2)}{(x^2-x+1)}\ge 0\left\{\right(x^2-x+1)>0\because a>0\&D<0\} \\ \Rightarrow -(x-2)\ge 0 \\ \Rightarrow (x-2)\le 0 \\ \Rightarrow x\le 2.......\left(2\right) \end{gathered}$$

So,

$x\in (-\infty ,2]-\{-1\}$

Hence, option$\mathbf{\left(}\mathit{A}\mathbf{\right)}$ is the correct option.