Question

If $x$ is real and $\frac{x-1}{4x+5}<\frac{x-3}{4x-3}$, then $x$ lies in the interval

• A. $(-\frac{3}{4},\frac{5}{4})$
• B. $(-\frac{5}{4},-\frac{3}{4})$
• C. $(-\frac{5}{4},\frac{3}{4})$
• D. $(-\frac{3}{4},\frac{7}{5})$

Answer 1

Answer: (C)

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

$x\in R$
The inequality is $\frac{x-1}{4x+5}<\frac{x-3}{4x-3}$

$\Rightarrow \frac{(x-1)(4x-3)-(x-3)(4x+5)}{(4x+5)(4x-3)}<0$
$\Rightarrow \frac{18}{(4x+5)(4x-3)}<0$
$\Rightarrow (4x+5)(4x-3)<0$
$\therefore x=(-\frac{5}{4},\frac{3}{4})$
Hence, option C is correct.