Question

Set of all real values of $x$ satisfying the inequation $\frac{{\log }_2(x^2-5x+4)}{{\log }_2(x^2+1)}>1$ is

• A. $(-\infty ,\frac{3}{5})-\left\{0\right\}$
• B. $(-\infty ,1)-\left\{0\right\}$
• C. $(\frac{3}{5},\infty )$
• D. $(-\infty ,\frac{3}{5})$

Answer 1

The correct option is A $(-\infty ,\frac{3}{5})-\left\{0\right\}$

Clearly, $x^2-5x+4>0$
$\Rightarrow (x-4)(x-1)>0$
$\Rightarrow x\in (-\infty ,1)\cup (4,\infty )\cdots \left(1\right)$
$x^2+1>0$ which is true $\forall x\in R$
Also, $x^2+1\ne 1\Rightarrow x\ne 0\cdots \left(2\right)$

$\frac{{\log }_2(x^2-5x+4)}{{\log }_2(x^2+1)}>1$
$\Rightarrow {\log }_2(x^2-5x+4)>{\log }_2(x^2+1)$
$\Rightarrow x^2-5x+4>x^2+1$
$\Rightarrow x<\frac{3}{5}\cdots \left(3\right)$

From $\left(1\right)\cap \left(2\right)\cap \left(3\right)$, we get
$x\in (-\infty ,\frac{3}{5})-\left\{0\right\}$