Question

If $\frac{(2x+3)(4-3x{)}^3(x-4)}{x^5(x+2{)}^2}\le 0$, then $x\in$

• A. $R-\{0,2\}$
• B. $[\frac{-3}{2},0)\cup [\frac{4}{3},4]$
• C. $(-\infty ,\frac{-3}{2}]\cup (0,\frac{4}{3}]\cup [4,\infty )$
• D. $(-\infty ,-2)\cup (-2,\frac{-3}{2}]\cup (0,\frac{4}{3}]\cup [4,\infty )$

Answer 1

The correct option is D $(-\infty ,-2)\cup (-2,\frac{-3}{2}]\cup (0,\frac{4}{3}]\cup [4,\infty )$

$\frac{(2x+3)(4-3x{)}^3(x-4)}{x^5(x+2{)}^2}\le 0$
Clearly, $x\ne -2,x\ne 0\cdots \left(1\right)$
$\frac{(x+\frac{3}{2}){(x-\frac{4}{3})}^3(x-4)}{x^5}\ge 0$

By wavy curve method,
$x\in (-\infty ,\frac{-3}{2}]\cup (0,\frac{4}{3}]\cup [4,\infty )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$x\in (-\infty ,-2)\cup (-2,\frac{-3}{2}]\cup (0,\frac{4}{3}]\cup [4,\infty )$