Question

If $\left|\frac{12x}{4x^{2}+9}\right| \ge1$ 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)

$\left|\frac{12x}{4x^{2}+9}\right| \ge1 \Rightarrow \frac{12\left|x\right|}{4x^{2}+9} \ge1$
$\because 4x^{2}+9>0$
$\Rightarrow 4x^{2}-12\left|x\right|+9 \le0$
$\Rightarrow 4\left|x\right|^{2}-12\left|x\right|+9 \le0$
$\Rightarrow \left(2\left|x\right|-3\right)^{2}=0 \Rightarrow \left|x\right|=\frac{3}{2}$