ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \mathrm{Expo} 0<\mathrm{a}<1 \mathrm{ถ้า} {\mathrm{a}}^{\mathrm{x}}<{\mathrm{a}}^{\mathrm{y}} \mathrm{จะได้} \mathrm{x}>\mathrm{y} \\ \mathrm{a}>1 \mathrm{ถ้า} {\mathrm{a}}^{\mathrm{x}}<{\mathrm{a}}^{\mathrm{y}} \mathrm{จะได้} \mathrm{x}<\mathrm{y} \\ \mathrm{จากโจทย์}\hspace{0.17em} {\left(x-2\right)}^{x^2+2}<{\left(x-2\right)}^{2x+10} \\ \mathrm{แบ่งได้} 2 \mathrm{กรณี} \\ \underline{\mathrm{กรณี} 1} 0<\mathrm{a}<1 \mathrm{จะได้} \mathrm{x}>\mathrm{y} \\ 0<\mathrm{x}-2<1 {\mathrm{x}}^2+2>2\mathrm{x}+10 \\ 2<\mathrm{x}<3 {\mathrm{x}}^2-2\mathrm{x}-8>0 \\ \left(2,3\right) \left(\mathrm{x}+2\right)\left(\mathrm{x}-4\right)>0 \end{gathered}$$
                                      
$$\begin{gathered} \left(-\infty ,-2\right)\cup \left(4,\infty \right) \\ \mathrm{ดังนั้น} \left(2,3\right)\cap \left[\left(-\infty ,-2\right)\cup \left(4,\infty \right)\right]=\varnothing \\ \underline{\mathrm{กรณี} 2} \mathrm{a}>1 \mathrm{จะได้} \mathrm{x}<\mathrm{y} \\ \mathrm{x}-2>1 {\mathrm{x}}^2+2<2\mathrm{x}+10 \\ \mathrm{x}>3 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{x}^2-2x-8<0 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(3,\infty )\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(x+2)(x-4)<0 \end{gathered}$$
                                     
$$\begin{gathered} \left(-2,4\right) \\ \mathrm{ดังนั้น} \left(3,\infty \right)\cap \left(-2,4\right)=\left(3,4\right)\subset \left(2.5,4\right) \end{gathered}$$