ข้อสอบ PAT 1 - ตุลาคม 2559

$$\begin{gathered} \left(\mathrm{ก}\right) \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\in \left(-1,1\right) \\ \mathrm{จะได้} -1<\mathrm{x}<1 ; \mathrm{ยกกำลัง} 2 \mathrm{ตลอด} \\ 0\le {\mathrm{x}}^2<1 ; \mathrm{คูณ} -1 \mathrm{ตลอด} \\ 0\ge -{\mathrm{x}}^2>-1 ; \mathrm{บวก} 1 \mathrm{ตลอด} \\ 1\ge 1-{\mathrm{x}}^2>0 ; \mathrm{ใส่} \sqrt{ } \mathrm{ตลอด} \\ 1\ge \sqrt{1-{\mathrm{x}}^2}>0 ; \mathrm{หาร} \sqrt{1-{\mathrm{x}}^2} \mathrm{ตลอด} \\ \frac{1}{\sqrt{1-{\mathrm{x}}^2}}\ge 1>0 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{1}{\sqrt{1-{\mathrm{x}}^2}}\ge 1 \\ \mathrm{g}\left(\mathrm{x}\right)\ge 1 \\ \mathrm{จากโจทย์} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)=3 ; \mathrm{x}\ge 1 \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=3 ; \mathrm{g}\left(\mathrm{x}\right)\ge 1 \\ \mathbf{โดย}\hspace{0.17em}\left(\mathrm{fog}\right)\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right) ; \\ \mathrm{ดังนั้น} \left(\mathrm{fog}\right)\left(\mathrm{x}\right)=3 \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ข}\right) \mathrm{เนื่องจาก} {\mathrm{D}}_{\mathrm{f}}\cap {\mathrm{D}}_{\mathrm{g}}=\left(-1,1\right) \\ \mathrm{จะได้} \mathrm{x}\in {\mathrm{D}}_{\mathrm{f}}\cap {\mathrm{D}}_{\mathrm{g}}=\left(-1,1\right) \\ \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{fg}\right)\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} =\left(\mathrm{x}+1\right)\left(\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} =\frac{\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}+\frac{1}{\sqrt{1-{\mathrm{x}}^2}} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ผิด} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ค}\right) \mathrm{เนื่องจาก} {\mathrm{D}}_{\mathrm{f}}\cap {\mathrm{D}}_{\mathrm{g}}=\left(-1,1\right) \\ \mathrm{จะได้} \mathrm{x}\in {\mathrm{D}}_{\mathrm{f}}\cap {\mathrm{D}}_{\mathrm{g}}=\left(-1,1\right) \\ \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \left(\frac{\mathrm{f}}{\mathrm{g}}\right)\left(\mathrm{x}\right)=\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} =\frac{\mathrm{x}+1}{{\displaystyle \frac{1}{\sqrt{1+{\mathrm{x}}^2}}}} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} =\left(\mathrm{x}+1\right)\sqrt{1-{\mathrm{x}}^2} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ข้อ} \left(\mathrm{ค}\right) \mathrm{ถูก} \end{gathered}$$