Question

Solve the following inequality:
$\left|\frac{3x}{x^2-4}\right|⩽1$

Answer 1

Sol:- By $|A|\le 1$

$-1\le A\le 1$

So, $|\frac{3x}{x^2-4}|\le 1$

$=-1\le \frac{3x}{x^2-4}\le 1$

Now, $\frac{3x}{x^2-4}\le 1$ & $\frac{3x}{x^2-4}\ge 1$

$\frac{3x}{x^2-4}-1\le 0$ & $\frac{3x}{x^2-4}+1\le 0$

$\frac{3x-x^2-4}{x^2-4}<0$ & $\frac{3x+x^2-4}{x^2-4}\ge 0$

$\frac{-\left[3\right(x^2-4)-3x]}{x^2-4}\le 0$ & $\frac{x^2+3x-4}{x^2-4}\ge 0$

$\frac{x^2-3x-4}{x^2-4}\ge 0$ & $\frac{x^2+3x-4}{x^2-4}\ge 0$

$\frac{(x-4)(x+1)}{(x-2)(x+2)}\ge 0$ & $\frac{(x+4)(x-1)}{(x-2)(x+2)}\ge 0$