Question

Given that, $x$ is a real number satisfying $\frac{5x^2-26x+5}{3x^2-10x+3}<0$, then

• A. $x<\frac{1}{5}$
• B. $\frac{1}{5}<x<3$
• C. $x>5$
• D. $\frac{1}{5}<x<\frac{1}{3}$ or $3<x<5$

Answer 1

The correct option is D $\frac{1}{5}<x<\frac{1}{3}$ or $3<x<5$

We have,

$\frac{5x^2-26x+5}{3x^2-10x+3}<0$

$\Rightarrow \frac{5x^2-25x-x+5}{3x^2-9x-x+3}<0$

$\Rightarrow \frac{5x(x-5)-1(x-5)}{\left(3x\right(x-3)-1(x-3)}<0$

$\Rightarrow \frac{(5x-1)(x-5)}{\left(\right(3x-1\left)\right(x-3)}<0$

$\therefore x\in (\frac{1}{5},\frac{1}{3})\cup (3,5)$