The values of x for which x−13x+4

− ∞ , − 5 4 ∪ ( 3 / 4 , ∞ )

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The values of x for which $\frac{x - 1}{3 x + 4} < \frac{x - 3}{3 x - 2}$ holds, lie in

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$(- \frac{4}{3}, \frac{2}{3})$

$(\frac{3}{4}, \infty)$

$(- \infty, \frac{5}{4})$

$(- \infty, - \frac{5}{4}) \cup (3 / 4, \infty)$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{(x - 1) (3 x - 2) - (x - 3) (3 x + 4)}{(3 x + 4) (3 x - 2)} < 0 \\ \Rightarrow \frac{14}{(3 x + 4) (3 x - 1)} < 0 \Rightarrow x \in (\frac{- 4}{3}, \frac{2}{3}) \end{array}$

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