question_answer

Determine the range of values of x for which \[\frac{{{x}^{2}}-2x+5}{3{{x}^{2}}-2x-5}>\frac{1}{2}\].

A)  \[\left( -5,-1 \right)\cup \left( 3,\infty  \right)\]         

B)  \[\left( -5,-1 \right)\cup \left( \frac{5}{3},3 \right)\]

C)  \[\left( -1,\frac{4}{3} \right)\cup \left( \frac{5}{3},3 \right)\]           

D)  \[(-5,-1)\cup \left( \frac{1}{2},\frac{5}{3} \right)\]

Correct Answer: B

Solution :

(b): \[\frac{{{x}^{2}}-2x+5}{3{{x}^{2}}-2x-5}>\frac{1}{2}\]. Or, \[\frac{{{x}^{2}}-2x+5}{3{{x}^{2}}-2x-5}>\frac{1}{2}>0\] Or, \[\frac{{{x}^{2}}+2x-15}{3{{x}^{2}}-2x-5}<0\] Or, \[\frac{\left( x+5 \right)\left( x-3 \right)}{\left( x+1 \right)\left( 3x-5 \right)}<0\] Or, \[\frac{\left( x+5 \right)\left( x-3 \right){{\left( x+1 \right)}^{2}}{{\left( 3x-5 \right)}^{2}}}{\left( x+1 \right)\left( 3x-5 \right)}<0\] Or, \[\left( x+5 \right)\left( x-3 \right)\left( x+1 \right)\left( 3x-5 \right)\] \[\therefore -5<x<-1\] or \[\frac{5}{3}<x<3\] i.e., \[\left( -5,-1 \right)\cup \left( \frac{5}{3},3 \right)\]