Question

Solve ${\left(\frac{1}{2}\right)}^{x^{2-2x}}<\frac{1}{4}$

Then $xϵ(-\infty ,1-\sqrt{k})\cup (1+\sqrt{k},\infty )$, what is k?

Answer 1

We have ${\left(\frac{1}{2}\right)}^{x^{2-2x}}<{\left(\frac{1}{2}\right)}^2$

It means $x^2-2x>2$

$\Rightarrow x^2-2x-2>0$

Roots of this equation are $1+\sqrt{3}$ and $1-\sqrt{3}$

$\Rightarrow$ $(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))>0$

$\Rightarrow$ $x>1+\sqrt{3}$ or $x<1-\sqrt{3}$

$\Rightarrow$ $xϵ(-\infty ,1-\sqrt{3})\cup (1+\sqrt{3},\infty )$

Comparing with $xϵ(-\infty ,1-\sqrt{k})\cup (1+\sqrt{k},\infty )$

We get, K=3