Question

The set of all solution of the inequality ${\left(1/2\right)}^{x^2-2x}<1/4$ contains the set:

• A. $(-\infty ,0)$
• B. $(-\infty ,1)$
• C. $(1,\infty )$
• D. $(3,\infty )$

Answer 1

The correct option is B $(-\infty ,1)$

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

$(\frac{1}{2}{)}^{x^2-2x}<(\frac{1}{2}{)}^2$

on equating the power

$x^2-2x<2$

$x^2-2x-2<0$

$x=\frac{2\pm \sqrt{4-4(-2)}}{2}$

$x=\frac{2\pm \sqrt{4+8}}{2}=1\pm \sqrt{3}$

$\therefore (x-1+\sqrt{3})(x-1-\sqrt{3})<0$

$\therefore (-\infty ,1)$ option (B)