A) \[\left( \frac{8}{3},\infty \right)\]
B) \[\left( -\infty ,\frac{8}{3} \right)\]
C) \[\left( -\infty ,\infty \right)\]
D) \[\left( 0,\infty \right)\]
Correct Answer: A
Solution :
[a] Since \[{{x}^{2}}+4>0\]for each real x. \[\therefore \]We must have \[3x-8>0,i.e.,x>\frac{8}{3}\] and \[lo{{g}_{0.5}}\left( \frac{3x-8}{{{x}^{2}}+4} \right)\ge 0\] \[\Rightarrow 0<\frac{3x-8}{{{x}^{2}}+4}\le 1\Rightarrow 0<3x\le {{x}^{2}}+4\] \[\Rightarrow {{x}^{2}}-3x+12\ge 0\] Now, \[{{x}^{2}}-3x+12={{\left( x-\frac{3}{2} \right)}^{2}}+12-\frac{9}{4}={{\left( x-\frac{3}{2} \right)}^{2}}+\frac{39}{4}>0,\forall x\in R\]\[\therefore \]Domain is \[\left( \frac{8}{3},\infty \right)\] Hence, option [a] is correct.
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