Q. The solution of the inequation $\log_{1/3}\left( x^{2} + x + 1 \right) + 1 < 0$ is
A) $( - \infty, - 2) \cup (1,\infty)$
B) $\lbrack - 1,\ 2\rbrack$
C) $( - 2,\ 1)$
D) $( - \infty,\infty)$
Explanation
(a)
$\log_{1/3}{\left( x^{2} + x + 1 \right) < - 1 = \log_{1/3}\left( \frac{1}{3} \right)^{- 1}}$
$\Rightarrow x^{2} + x + 1 > \left( \frac{1}{3} \right)^{- 1}$
$(\because$ where $0 < a < 1$, then $\log_{a}{x < \log_{a}{y \Rightarrow x > y)}}$
$\Rightarrow x^{2} + x - 2 > 0 \Rightarrow (x + 2)(x - 1) > 0$
$\Rightarrow x \in ( - \infty, - 2) \cup (1,\infty)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.