Question

If $\frac{{\log }_2(4x^2-x-1)}{{\log }_2(x^2+1)}>1$ , then $x$ lies in the interval

• A. $(-\infty ,-\frac{2}{3})$
• B. $(1,\infty )$
• C. $(-\frac{2}{3},0)$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $(1,\infty )$
B $(-\infty ,-\frac{2}{3})$

$\frac{{\log }_2(4x^2-x-1)}{{\log }_2(x^2+1)}>1$
Above equation is valid when $4x^2-x-1>0$ and $x\ne 0$
$\Rightarrow x\in (-\infty ,\frac{1-\sqrt{17}}{8})\cup (\frac{1+\sqrt{17}}{8},\infty )$
$\frac{{\log }_2(4x^2-x-1)}{{\log }_2(x^2+1)}>1$
$\Rightarrow {\log }_2(4x^2-x-1)>{\log }_2(x^2+1)$
$\Rightarrow 3x^2-x-2>0$
$\Rightarrow x\in (-\infty ,\frac{-2}{3})\cup (1,\infty )$
Ans: A,B