Question

If $\left|\frac{12x}{4x^2+9}\right|\ge 1$ for all real values of $x$, the inequality being satisfied only if $|x|$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$\left|\frac{12x}{4x^2+9}\right|\ge 1\Rightarrow \frac{12|x|}{4x^2+9}\ge 1$
$\because 4x^2+9>0$
$\Rightarrow 4x^2-12\left|x\right|+9\le 0$
$\Rightarrow 4{\left|x\right|}^2-12\left|x\right|+9\le 0$
$\Rightarrow {\left(2\left|x\right|-3\right)}^2=0\Rightarrow \left|x\right|=\frac{3}{2}$