Question

If $S$ is the set of all real $x$ such that $\frac{2}{x^2-x+1}-\frac{1}{x+1}-\frac{2x-1}{x^3+1}\ge 0$, then $S$ contains

• A. $(-\infty ,2)$
• B. $(-\infty ,2]-\{-1\}$
• C. $(-\infty ,2)-\{-1\}$
• D. $(-\infty ,-1)\cup (2,\infty )$

Answer 1

The correct option is B $(-\infty ,2]-\{-1\}$

$\frac{2}{x^2-x+1}-\frac{1}{x+1}-\frac{2x-1}{x^3+1}\ge 0$

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

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

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

$\Rightarrow \frac{2-x}{x^2-x+1}\ge 0$, where $x\ne 1$

$\Rightarrow 2-x\ge 0,x\ne -1$
[As $x^2-x+1>0\forall x\in R$]

$\Rightarrow x\le 2,x\ne -1$