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

1. $\left(\frac{-4}{3}, \frac{2}{3}\right)$
2. $\left(-\infty, \frac{-5}{4}\right)$
3. $(-\infty)$
4. $\left(-\infty, \frac{-5}{4}\right) \cup\left(\frac{3}{4},-\infty\right)$

Solution

Given that, $\frac{x-1}{3 x+4}-\frac{x-3}{3 x-2} < 0$ $\begin{aligned} & \frac{(x-1)(3 x-2)-(x-3)(3 x+4)}{(3 x+4)(3 x-2)} < 0 \\ & \frac{\left(3 x^2-2 x-3 x+2\right)-\left(3 x^2+4 x-9 x-12\right)}{(3 x+4)(3 x-2)} < 0 \\ & \frac{-5 x+2+5 x+12}{(3 x+4)(3 x-2)} < 0 \\ & \frac{14}{(3 x+4)(3 x-2)} < 0 \end{aligned}$ Thus, $(3 x+4)(3 x-2) < 0$ $\therefore \quad x$ lies in the interval $\left(-\frac{4}{3}, \frac{2}{3}\right)$.

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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