Question

The set of all solution of the inequality (1/2)^{xsquare - 2x}< (1/4) contains which set .

Answer 1

$$\begin{gathered} {\left(\frac{1}{2}\right)}^{{\mathrm{x}}^2-2\mathrm{x}}<\frac{1}{4} \\ {\left(\frac{1}{2}\right)}^{{\mathrm{x}}^2-2\mathrm{x}}<{\left(\frac{1}{2}\right)}^2 \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}<2 \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}-2<0 \\ \mathrm{now}\mathrm{x}=\frac{2\pm \sqrt{4-4\left(-2\right)}}{2}=\frac{2\pm \sqrt{12}}{2}=1\pm \sqrt{3} \\ \mathrm{now}\mathrm{consider}3\mathrm{intervals} \\ \left(-\infty ,1-\sqrt{3}\right),\left(1-\sqrt{3},1+\sqrt{3}\right),\left(1+\sqrt{3},\infty \right) \\ \mathrm{For}\mathrm{x}\in \left(-\infty ,1-\sqrt{3}\right),\mathrm{value}\mathrm{of}{\mathrm{x}}^2-2\mathrm{x}-2 \\ \mathrm{consider}-1 \\ {\left(-1\right)}^2-2\left(-1\right)-2=1+2-2=1>0 \\ \mathrm{so},\mathrm{x}\in \left(-\infty ,1-\sqrt{3}\right),{\mathrm{x}}^2-2\mathrm{x}-2>0 \\ \mathrm{For}\mathrm{x}\in \left(1-\sqrt{3},1+\sqrt{3}\right),\mathrm{value}\mathrm{of}{\mathrm{x}}^2-2\mathrm{x}-2 \\ \mathrm{consider}0 \\ {\left(0\right)}^2-2\left(0\right)-2=10-2=-2<0 \\ \mathrm{so},\mathrm{x}\in \left(-\infty ,1-\sqrt{3}\right),{\mathrm{x}}^2-2\mathrm{x}-2<0 \\ \mathrm{For}\mathrm{x}\in \left(1+\sqrt{3},\infty \right),\mathrm{value}\mathrm{of}{\mathrm{x}}^2-2\mathrm{x}-2 \\ \mathrm{consider}3 \\ {\left(3\right)}^2-2\left(3\right)-2=9-6-2=1>0 \\ \mathrm{so},\mathrm{x}\in \left(-\infty ,1-\sqrt{3}\right),{\mathrm{x}}^2-2\mathrm{x}-2>0 \\ \mathrm{therefore}{\mathrm{x}}^2-2\mathrm{x}-2<0\mathrm{for}\mathrm{x}\in \left(-\infty ,1-\sqrt{3}\right)\mathrm{ANS}... \end{gathered}$$