Question

Which of the following is/are not the solution of ${\log }_3\left(x^2-2\right)<{\log }_3\left(\frac{3}{2}\left|x\right|-1\right)$

• A. $\left(\sqrt{2},2\right)$
• B. $\left(-2,-\sqrt{2}\right)$
• C. $\left(-\sqrt{2},2\right)$
• D. $\left(-2,2\right)$

Answer 1

Answer: (C), (D)

The correct options are
C $\left(-\sqrt{2},2\right)$
D $\left(-2,2\right)$

We have,${\log }_3\left(x^2-2\right)<{\log }_3\left(\frac{3}{2}\left|x\right|-1\right)$For ${\log }_3\left(x^2-2\right)$ and $\frac{3}{2}\left|x\right|-1$ to be defined, we must have $\left(x^2-2\right)>0and\frac{3}{2}\left|x\right|-1>0\Rightarrow x<-\sqrt{2}orx>\sqrt{2}and\left|x\right|>\frac{2}{3}......\left[1\right]Also,{\log }_3\left(x^2-2\right)<{\log }_3\left(\frac{3}{2}\left|x\right|-1\right)\Rightarrow \left(x^2-2\right)<\left(\frac{3}{2}\left|x\right|-1\right)\Rightarrow 2x^2-3\left|x\right|-2<0\Rightarrow 2|x{|}^2-3|x|-2<0\Rightarrow (2\left|x\right|+1\left)\right(\left|x\right|-2)<0\Rightarrow |x|<2\Rightarrow -2<x<2....[2\left]From\right[1\left]and\right[2]wegetx\in (-2,-\sqrt{2})\cup (\sqrt{2},2)$