Q. If $x$ is real, then the value of $\frac{x + 2}{2x^{2} + 3x + 6}$ is equal to
A) $\left( \frac{1}{13},\ \frac{1}{3} \right)$
B) $\left( - \frac{1}{13},\ \frac{1}{3} \right)$
C) $\left( - \frac{1}{3},\ \frac{1}{13} \right)$
D) None of these
Explanation
(b)
Let $y = \frac{x + 2}{2x^{2} + 3x + 6}$
$\Rightarrow \ \ 2yx^{2} + (3y - 1)x + x + 6y - 2 = 0$
But $x$ is real, then
$(3y - 1)^{2} - 4(2y)(6y - 2) \geq 0\ \ \ \ \ \lbrack\because D \geq 0\rbrack$
$\Rightarrow \ \ (13y + 1)(3y - 1) \leq 0$
$\Rightarrow \ - \frac{1}{13} \leq y \leq \frac{1}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.