A) \[\frac{3}{2}\]
B) \[\frac{2}{3}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
[a] \[\left| \frac{12x}{4{{x}^{2}}+9} \right|\ge 1\Rightarrow \frac{12\left| x \right|}{4{{x}^{2}}+9}\ge 1\] \[\because 4{{x}^{2}}+9>0\] \[\Rightarrow 4{{x}^{2}}-12\left| x \right|+9\le 0\] \[\Rightarrow 4{{\left| x \right|}^{2}}-12\left| x \right|+9\le 0\] \[\Rightarrow {{(2\left| x \right|-3)}^{2}}=0\Rightarrow \left| x \right|=\frac{3}{2}\]
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