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Question

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

A

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

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B

$\left(-\infty ,2\right)$

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C

$\right]-1,2\right]$

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D

None of these

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Solution

## The correct option is A $\left(-\infty ,2\right]-\left\{-1\right\}$Explanation for correct option:Step 1: Set the terms inside the root symbol to be greater than or equal to $0$.We have given$f\left(x\right)=\sqrt{\left[\frac{2}{\left({x}^{2}-x+1\right)}-\frac{1}{\left(x+1\right)}-\frac{2x-1}{\left({x}^{3}+1\right)}\right]}$.$\frac{2}{\left({x}^{2}-x+1\right)}-\frac{1}{\left(x+1\right)}-\frac{2x-1}{\left({x}^{3}+1\right)}\ge 0\phantom{\rule{0ex}{0ex}}⇒\frac{2}{\left({x}^{2}-x+1\right)}-\frac{1}{\left(x+1\right)}-\frac{2x-1}{\left(x+1\right)\left({x}^{2}-x+1\right)}\ge 0\phantom{\rule{0ex}{0ex}}⇒\frac{2\left(x+1\right)-\left({x}^{2}-x+1\right)-\left(2x-1\right)}{\left(x+1\right)\left({x}^{2}-x+1\right)}\ge 0\phantom{\rule{0ex}{0ex}}⇒\frac{2x+2-{x}^{2}+x-1-2x+1}{\left(x+1\right)\left({x}^{2}-x+1\right)}\ge 0\phantom{\rule{0ex}{0ex}}⇒\frac{-{x}^{2}+x+2}{\left(x+1\right)\left({x}^{2}-x+1\right)}\ge 0\phantom{\rule{0ex}{0ex}}⇒-\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left({x}^{2}-x+1\right)}\ge 0\phantom{\rule{0ex}{0ex}}$ From above$x\ne -1$….$\left(1\right)$.Step 2: Find the points where the function changes the signs.$⇒-\frac{\left(x-2\right)}{\left({x}^{2}-x+1\right)}\ge 0\left\{\left({x}^{2}-x+1\right)>0\because a>0&D<0\right\}\phantom{\rule{0ex}{0ex}}⇒-\left(x-2\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\le 0\phantom{\rule{0ex}{0ex}}⇒x\le 2.......\left(2\right)$So,$x\in \left(-\infty ,2\right]-\left\{-1\right\}$Hence, option$\mathbf{\left(}\mathbit{A}\mathbf{\right)}$ is the correct option.

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