Q58 / 160 · WBJEE 2015 · Mathematics
Given that, $x$ is a real number satisfying $\frac{5 x^{2}-26 x+5}{3 x^{2}-10 x+3} < 0,$ then
- A$x < \frac{1}{5}$
- B$\frac{1}{5} < x < 3$
- C$x>5$
- Correct answerD$\frac{1}{5} < x < \frac{1}{3}$ or $3 < x < 5$
Solution
We have, $\frac{5 x^{2}-26 x+5}{3 x^{2}-10 x+3} < 0$
$\Rightarrow \quad \frac{5 x^{2}-25 x-x+5}{3 x^{2}-9 x-x+3} < 0$
$\Rightarrow \quad \frac{5 x(x-5)-1(x-5)}{3 x(x-3)-1(x-3)} < 0 \Rightarrow \frac{(x-5)(5 x-1)}{(x-3)(3 x-1)} < 0$
$\therefore \quad x \in\left(\frac{1}{5}, \frac{1}{3}\right) \cup(3,5)$
Hence, the correct answer is option: 'D'.