question_answer

Solve the in equality \[\frac{x-2}{x+2}>\frac{2x-3}{4x+1}.\]

A)  \[(2,3)\]            

B)  \[\left( -\infty ,-2 \right)\cup \left( \frac{1}{4}-1 \right)\cup (4,\infty )\]

C)  \[\left( \frac{-1}{4},\frac{3}{2} \right)\]

D)  \[(-\infty ,-2)\cup (4,\infty )\]

Correct Answer: B

Solution :

(b): The range of the variable x in this inequality consists of all values of x except \[x=2\] and \[x=\frac{1}{4}\]. Hence we cannot cross multiply. So we adopt another method. \[\frac{x-2}{x+2}-\frac{2x-3}{4x-1}>0\]?????.(i) Or, \[\frac{\left( x-2 \right)\left( 4x-1 \right)-\left( x+2 \right)\left( 2x-3 \right)}{\left( x+2 \right)\left( 4x-1 \right)}>0\] Or, \[\frac{2\left( {{x}^{2}}-5x+4 \right)}{\left( x+2 \right)\left( 4x-1 \right)}>0\] Or, \[\frac{\left( x-1 \right)\left( x-4 \right)}{\left( x+2 \right)\left( x-\frac{1}{4} \right)}>0\]?????.(ii) Now, multiplying both sides of (ii) by the expression \[{{\left( x+2 \right)}^{2}}{{\left( x-\frac{1}{4} \right)}^{2}}\], which is positive for all x as square of any number is always positive except for \[x=-\frac{1}{2}\] and \[x=-\frac{1}{4}\]. \[\therefore \frac{\left( x-1 \right)\left( x-4 \right){{\left( x+2 \right)}^{2}}{{\left( x-\frac{1}{4} \right)}^{2}}}{\left( x+2 \right)\left( x-\frac{1}{4} \right)}>0\] Or, \[\left( x-1 \right)\left( x-4 \right)\left( x+2 \right)\left( x-1/4 \right)>0\] Thus, the range is \[x<-2\] or \[\frac{1}{4}<x<1\] or \[x>4\]