ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2564 (หลักสูตรใหม่)

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{อนุกรมอนันต์} \left(\mathrm{เรขาคณิต}\right) \\ S_{\infty } = \frac{a_1}{1-r} \to \overline{)1} \mathrm{โดย} \left|r\right| < 1 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} 1+\frac{5^x}{1+5^x}+\frac{5^{2x}}{{\left(1+5^x\right)}^2}+\frac{5^{3x}}{{\left(1+5^x\right)}^3}+\dots = \frac{26}{25} \\ 1+\frac{5^x}{1+5^x}+\frac{{\left(5^x\right)}^2}{{\left(1+5^x\right)}^2}+\frac{{\left(5^x\right)}^3}{{\left(1+5^x\right)}^3}+\dots = \frac{26}{25} \\ 1+\frac{5^x}{1+5^x}+{\left(\frac{5^x}{1+5^x}\right)}^2+{\left(\frac{5^x}{1+5^x}\right)}^3+\dots = \frac{26}{25} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} A = \frac{5^x}{1+5^x} \\ \mathrm{จะได้} 1+A+A^2+A^3+\dots = \frac{26}{25} ; r=A \\ \mathrm{จาก} \overline{)1} \frac{1}{1-A} = \frac{26}{25} \\ 25 = 26\left(1-A\right) \\ 25 = 26-26A \\ A = \frac{1}{26} ; \mathrm{แทน} A = \frac{5^x}{1+5^x} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{5^x}{1+5^x} = \frac{1}{26} \\ \mathrm{กำหนดให้} 5^x = a \\ \mathrm{จะได้} \frac{a}{1+a} = \frac{1}{26} \\ 26a = 1+a \\ a = \frac{1}{25} = \frac{1}{5^2} \\ 5^x = 5^{-2} \\ \mathrm{ดังนั้น} x = -2 \end{gathered}$$