JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relations and Functions-I

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

A) \[(-\infty ,2]-\{-1\}\]

B) \[(-\infty ,2)\]

C) \[]-1,2]\]

D) None of these

Correct Answer: A

Solution :

[a] \[f(x)=\sqrt{\left( \frac{2}{{{x}^{2}}-x+1}-\frac{1}{x+1}-\frac{2x-1}{{{x}^{3}}+1} \right)}\]

We must have \[\frac{2}{{{x}^{2}}-x+1}-\frac{1}{x+1}-\frac{2x-1}{{{x}^{3}}+1}\ge 0\]

Or \[\frac{2(x+1)-({{x}^{2}}-x+1)-(2x-1)}{(x+1)({{x}^{2}}-x+1)}\ge 0\]

Or \[\frac{-({{x}^{2}}-x-2)}{(x+1)({{x}^{2}}-x+1)}\ge 0\]

Or \[\frac{-(x-2)(x+1)}{(x+1)({{x}^{2}}-x+1)}\ge 0\]

Or \[\frac{2-x}{{{x}^{2}}-x+1}\ge 0,\] where \[x\ne -1\]

Or \[2-x\ge 0,x\ne -1\] (as\[{{x}^{2}}-x+1>0\forall x\in R\])

Or \[x\le 2,x\ne -1\]

Hence. Domain of the function is \[(-\infty ,-1)\cup \left( -1,2 \right].\]or \[\left( -\infty ,2 \right]-\{-1\}\]