Q. $\left\{ x \in R:\frac{2x - 1}{x^{3} + 4x^{2} + 3x} \in R \right\}$equals
A) $R - \{ 0\}$
B) $R - \{ 0,\ 1,\ 3\}$
C) $R - \{ 0,\ - 1,\ - 3\}$
D) $R - \left\{ 0,\ - 1,\ - 3,\ + \frac{1}{2} \right\}$
Explanation
(c)
Let $A = \left\{ x \in R:\frac{2x - 1}{x^{3} + 4x^{2} + 3x} \right\}$
Now, $x^{3} + 4x^{2} + 3x = x(x^{2} + 4x + 3)$
$= x(x + 3)(x + 1)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A = R - \left\{ 0,\ - 1,\ - 3 \right\}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.