Q. $|2x - 3| < |x + 5|,$ then $x$ belongs to

A) $( - 3,5)$
B) (5, 9)
C) $\left( - \frac{2}{3},8 \right)$
D) $\left( - 8,\frac{2}{3} \right)$
Explanation

(c)
We have, $|2x - 3| < |x + 5|$
$\Rightarrow |2x - 3| - |x + 5| < 0$
$\Rightarrow \left\{ \begin{matrix}
3 - 2x + x + 5 < 0,x \leq - 5 \\
3 - 2x - x - 5 < 0, - 5 < x \leq \frac{3}{2} \\
2x - 3 - x - 5 < 0,x > \frac{3}{2}
\end{matrix} \right.\$
$\Rightarrow \left\{ \begin{matrix}
x > 8,x \leq - 5 \\
x > - \frac{2}{3}, - 5 < x \leq \frac{3}{2} \\
x < 8,x > \frac{3}{2}
\end{matrix} \right.\$
$\Rightarrow x \in \left( - \left. \ \frac{2}{3},\frac{3}{2} \right\rbrack \cup \left( \frac{3}{2},8 \right.\ \right)$
$\Rightarrow x \in \left( - \frac{2}{3},8 \right)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.