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$\frac{x-1}{3 x+4} < \frac{x-3}{3 x-2}$ holds, for all $\mathrm{x}$ in the interval
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Verified Answer
The correct answer is:
$\left(\frac{-4}{3}, \frac{2}{3}\right)$
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 $\left(-\frac{4}{3}, \frac{2}{3}\right)$.
$\begin{aligned}
& \frac{(x-1)(3 x-2)-(x-3)(3 x+4)}{(3 x+4)(3 x-2)} < 0 \\
& \frac{\left(3 x^2-2 x-3 x+2\right)-\left(3 x^2+4 x-9 x-12\right)}{(3 x+4)(3 x-2)} < 0 \\
& \frac{-5 x+2+5 x+12}{(3 x+4)(3 x-2)} < 0 \\
& \frac{14}{(3 x+4)(3 x-2)} < 0
\end{aligned}$
Thus, $(3 x+4)(3 x-2) < 0$
$\therefore \quad x$ lies in the interval $\left(-\frac{4}{3}, \frac{2}{3}\right)$.
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