Q. If $\sqrt{9x^{2} + 6x + 1} < (2 - x),$ then
A) $x \in \left( - \frac{3}{2},\ \frac{1}{4} \right)$
B) $x \in \left( - \frac{3}{2},\ \frac{1}{4} \right)$
C) $x \in \left\lbrack - \frac{3}{2},\ \frac{1}{4} \right\rbrack$
D) $x < \frac{1}{4}$
Explanation
(d)
Given, $\sqrt{(3x + 1)^{2}} < (2 - x)$
$\Rightarrow \ \ (3x + 1) < 2 - x$
$\Rightarrow \ \ 3x + 1 < 2 - x\ \Rightarrow x < \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.