Question

Solution set of $\frac{2}{x^2-x+1}-\frac{1}{x+1}\ge \frac{2x-1}{x^3+1}$ is

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

Answer 1

The correct option is B $(-\infty ,-1)\cup (-1,2]$

$\frac{2}{x^2-x+1}-\frac{1}{x+1}\ge \frac{2x-1}{x^3+1}$
$\Rightarrow \frac{2}{x^2-x+1}-\frac{1}{x+1}-\frac{2x-1}{x^3+1}\ge 0$
$\Rightarrow \frac{x^2-x-2}{(x+1)(x^2-x+1)}\le 0$
$\Rightarrow \frac{(x+1)(x-2)}{(x+1)(x2-x+1)}\le 0$
$\Rightarrow \frac{x-2}{x^2-x+1}\le 0,$ where $x\ne -1$

$x^2-x+1>0$ as $D=1-4=-3<0$
Hence, $x-2\le 0\Rightarrow x\le 2$
$\therefore x\in (-\infty ,-1)\cup (-1,2]$