Question

Solution of $(2x + 1)(x - 3) (x + 7) < 0$ is

• A. $ \left(-\infty,\,-7\right) \cup \left(-\frac{1}{2},\,3\right)$
• B. $ \left(-\infty,\,-7\right) \cup \left(\frac{1}{2},\,3\right)$
• C. $ \left(-\infty,\,7\right) \cup \left(-\frac{1}{2},\,3\right)$
• D. $ \left(-\infty,\,-7\right) \cup \left(3, \, \infty \right)$

Answer 1

Answer: (A)

$\left(2 x+ 1\right)\left(x -3 \right)\left(x + 7\right) = 0$
$ \Rightarrow x = - \frac{1}{2}$, $3$, $-7$
Figure shows how the sign changes as one passes through the points $- 7$, $-\frac{1}{2}$ , $3$.

Clearly $(2x + 1) (x - 3) (x + 7) < 0$
when $x < -7$ or $-\frac{1}{2} < x < 3$
$ \therefore x \in \left(-\infty,\,-7\right) \cup \left(-\frac{1}{2},\,3\right).$