Question

All real values of $x$ for which $\frac{x-2}{3x+1}<\frac{x-3}{3x-2}$ are

• A. $[\frac{-1}{4},\frac{2}{3}]$
• B. $[\frac{1}{3},\frac{2}{3}]$
• C. $[\frac{-1}{3},\frac{2}{3}]$
• D. $R$

Answer 1

The correct option is C $[\frac{-1}{3},\frac{2}{3}]$

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

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

$\Rightarrow \frac{(x-2)(3x-2)-(x-3)(3x+1)}{(3x+1)(3x-2)}<0$

$\Rightarrow \frac{({3x}^2-8x+4)-({3x}^2-8x-3)}{(3x+1)(3x-2)}<0$

$\Rightarrow \frac{7}{(3x+1)(3x-2)}<0$

$\Rightarrow \frac{7(3x+1)(3x-2)}{{(3x+1)}^2{(3x-2)}^2}<0⟶\left(1\right)$

(Denominator is a positive quantity)

$\therefore$ the values of ${}^{\prime }{x}^{\prime }$ for which $\left(1\right)$ holds is :

$\therefore$ $xϵ(\frac{-1}{3},\frac{2}{3})$

$xϵ[\frac{-1}{3},\frac{2}{3}]$

$\therefore$ Option C.