Question

If ${\log }_{1/3}\frac{3x-1}{x+2}$ is less than unity then $x$ must lie in the interval

• A. $(-\infty ,-2)\cup (\frac{5}{8},\infty )$
• B. $(-2,\frac{5}{8})$
• C. $(-\infty ,-2)\cup (\frac{1}{3},\frac{5}{8})$
• D. $(-2,\frac{1}{3})$

Answer 1

The correct option is A $(-\infty ,-2)\cup (\frac{5}{8},\infty )$

${\log }_{\frac{1}{3}}\left(\frac{3x-1}{x+2}\right)={\log }_3\left(\frac{x+2}{3x-1}\right)<1$

$\Rightarrow \frac{x+2}{3x-1}>0$ and $\frac{x+2}{3x-1}<3$

$\frac{x+2}{3x-1}>0$

$\Rightarrow x\in (-\infty ,-2)\cup (1/3,\infty )$ -----(1)

$\frac{x+2}{3x-1}<3$

$\Rightarrow \frac{-8x+5}{3x-1}<0$

$\Rightarrow \frac{8x-5}{3x-1}>0$

$\therefore x\in (-\infty ,1/3)\cup (5/8,\infty )$ ------(2)

From (1) and (2), we get

$x\in (-\infty ,-2)\cup (5/8,\infty )$

Hence, option A.