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

Answer: (D)

We have, $\frac{5 x^{2}-26 x+5}{3 x^{2}-10 x+3}<0$
$\Rightarrow \frac{5 x^{2}-25 x-x+5}{3 x^{2}-9 x-x+3}<0$
$\Rightarrow \frac{5 x(x-5)-1(x-5)}{3 x(x-3)-1(x-3)}<0$
$\Rightarrow \frac{(x-5)(5 x-1)}{(x-3)(3 x-1)}< 0$
$\therefore x \in\left(\frac{1}{5}, \frac{1}{3}\right) \cup(3,5)$