question_answer

Find the range of real values of x for which \[\frac{x-1}{(4x+5)}<\frac{x-3}{4x-3}\]

A)  \[\left( \frac{3}{4},3 \right)\]                  

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

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

D)  \[\left( -\frac{5}{4},\frac{3}{4} \right)\]

Correct Answer: D

Solution :

(d): \[\frac{x-1}{\left( 4x+5 \right)}<\frac{x-3}{4x-3}\] or \[\frac{x-1}{4x+5}-\frac{x-3}{4x-3}<0\] Or, \[\frac{\left( x-1 \right)\left( 4x-3 \right)-\left( x-3 \right)\left( 4x+5 \right)}{\left( 4x-3 \right)\left( 4x+5 \right)}<0\] Or, \[\frac{\left( 4{{x}^{2}}-7x+3 \right)-\left( 4{{x}^{2}}-7x-15 \right)}{\left( 4x+5 \right)\left( 4x-3 \right)}<0\] Or, \[\frac{18}{\left( 4x+5 \right)\left( 4x-3 \right)}<0\] Since 18 is positive, \[\therefore \left( 4x+5 \right)\left( 4x-3 \right)<0\] Or,        \[16\left( x+\frac{5}{4} \right)\left( x-\frac{3}{4} \right)<0\] \[\therefore \frac{-5}{4}<x<\frac{3}{4}\]