Question

The solution set of $\frac{1}{x-1}+\frac{1}{x+1}\le \frac{1}{x}$ is

• A. $(-\infty ,-1)\cup (0,1)$
• B. $(-\infty ,-1]\cup [0,1)$
• C. $(-\infty ,-1)\cup (0,1]$
• D. $(-\infty ,-1)\cup (0,1)$

Answer 1

The correct option is A $(-\infty ,-1)\cup (0,1)$

Given $\frac{1}{x-1}+\frac{1}{x+1}\le \frac{1}{x}$
$\Rightarrow \frac{1}{x-1}+\frac{1}{x+1}-\frac{1}{x}\le 0$
$\Rightarrow \frac{x(x+1)+x(x-1)-(x-1)(x+1)}{x(x-1)(x+1)}\le 0\Rightarrow \frac{x^2+1}{x(x-1)(x+1)}\le 0$
Since, $x^2+1>0\forall x\in R$,
$\therefore$ the inequality holds true if
$x(x-1)(x+1)<0$


From wavy curve method,
$x\in (-\infty ,-1)\cup (0,1)$