Question

If ${\log }_{(x+1)}(x^2-2x)>0,$ then $x\in$

• A. $(1-\sqrt{2},0)\cup (1+\sqrt{2},\infty )$
• B. $(1-\sqrt{2},1+\sqrt{2})-\left\{0\right\}$
• C. $R-\left\{0\right\}$
• D. no solution

Answer 1

The correct option is A $(1-\sqrt{2},0)\cup (1+\sqrt{2},\infty )$

$x^2-2x>0,x+1>0,\&x+1\ne 0,1\Rightarrow x\in (-\infty ,0)\cup (2,\infty ),x>-1,x\ne -1,0\Rightarrow x\in (-1,0)\cup (2,\infty )$

${\log }_{b}a>0\Rightarrow a,b\in (0,1)\text{or}a,b\in (1,\infty )$

$\text{Case -1:}$
Let $0<x^2-2x<1$ and $x+1\in (0,1)$
$\Rightarrow 0<x(x-2)<1$ and $x\in (-1,0)$
$\Rightarrow x(x-2)>0$ and $x(x-2)<1$
$\Rightarrow x\in (-\infty ,0)\cup (2,\infty )$ and
$x(x-2)<1\Rightarrow x^2-2x-1<0\Rightarrow (x-1{)}^2<2\Rightarrow x\in (1-\sqrt{2},1+\sqrt{2})$
$x\in (1-\sqrt{2},0)\cdots \left(1\right)$

$\text{Case -2:}$
$x^2-2x>1$ and $x+1>1$
$\Rightarrow x^2-2x-1>0$ and $x>0$
$\Rightarrow x\in (1+\sqrt{2},\infty )\cdots \left(2\right)$

From $\left(1\right)\&\left(2\right)$,
$x\in (1-\sqrt{2},0)\cup (1+\sqrt{2},\infty )$