Question

$\frac{x-1}{3 x+4}<\frac{x-3}{3 x-2}$ holds, for all $x$ in the interval

• A. $\left(\frac{-4}{3}, \frac{2}{3}\right)$
• B. $\left(-\infty, \frac{5}{4}\right)$
• C. $\left(\frac{3}{3}, \infty\right)$
• D. $\left(-\infty, \frac{-5}{4}\right) \cup\left(\frac{3}{4},-\infty\right)$

Answer 1

Answer: (A)

We have, $\frac{x-1}{3 x+4}-\frac{x-3}{3 x-2}<0$
$\Rightarrow \frac{(x-1)(3 x-2)-(x-3)(3 x+4)}{(3 x+4)(3 x-2)}<0$
$\Rightarrow \frac{\left(3 x^{2}-2 x-3 x+2\right)-\left(3 x^{2}+4 x-9 x-12\right)}{(3 x+4)(3 x-2)}<0$
$\Rightarrow \frac{-5 x+2+5 x+12}{(3 x+4)(3 x-2)}<0$
$\Rightarrow \frac{14}{(3 x+4)(3 x-2)}<0$
$\Rightarrow (3 x + 4) (3 x -2)<0$

$\Rightarrow x \in\left(-\frac{4}{3}, \frac{2}{3}\right)$