지수함수 그래프와 사각형

주어진 그림에서 두 점 $\mathrm{A}$, $\mathrm{C}$의 $x$좌표를 각각 $k$ $\left( k > 0 \right)$, $m$ $\left( m < 0 \right)$이라 하면 $\mathrm{A} \mathit{\left( k , 1 - 2 ^{-k} \right)}$, $\mathrm{B} \mathit{\left( k , 2 ^{k} \right)}$, $\mathrm{C} \mathit{\left( m , 2 ^{m} \right)}$, $\mathrm{D} \mathit{\left( m , 1 - 2 ^{-m} \right)}$ 두 점 $\mathrm{A}$, $\mathrm{C}$의 $y$좌표가 같으므로 $1 - 2 ^{-k} = 2 ^{m}$ $\cdots$$\cdots$ ㉠ $\displaystyle \overline{\mathrm{AB}} = {2 ^{k} + 2 ^{-k} - 1}$, $\displaystyle \overline{\mathrm{CD}} = 2 ^{m} + 2 ^{-m} - 1$이고 $\displaystyle \overline{\mathrm{AB}} = 2 \overline{\mathrm{CD}}$에서 $2 ^{k} + 2 ^{-k} - 1 = 2 \left( 2 ^{m} + 2 ^{-m} - 1 \right)$ $\cdots$$\cdots$ ㉡ ㉠을 ㉡에 대입하면 $\displaystyle 2 ^{k} + 2 ^{-k} - 1 = 2 \left( 1 - 2 ^{-k} + \frac{1}{1 - 2 ^{-k}} - 1 \right)$ $2 ^{k} = a$ $\left( a > 0 \right)$이라 하면 $\displaystyle a + \frac{1}{a} - 1 = 2 \left( \frac{1}{1 - \frac{1}{a}} - \frac{1}{a} \right)$, $\displaystyle \frac{a ^{2} - a + 1}{a} = 2 \times \frac{a ^{2} - a + 1}{a} \times \frac{1}{a - 1}$ 이때 $a \neq 0$, $a ^{2} - a + 1 \neq 0$이므로 $\displaystyle 1 = \frac{2}{a - 1}$, $a - 1 = 2$ $\therefore$ $a = 3$ 즉 $2 ^{k} = 3$에서 $k = \log _{2} 3$이고 $\displaystyle 2 ^{-k} = \frac{1}{3}$이므로 ㉠에서 $\displaystyle 2 ^{m} = 1 - \frac{1}{3} = \frac{2}{3}$, $\displaystyle m = \log _{2} \frac{2}{3}$ 따라서 $\displaystyle \mathrm{A} \mathit{\left( \log _{2} 3 , \frac{2}{3} \right)}$, $\mathrm{B} \mathit{\left( \log _{2} 3 , 3 \right)}$, $\displaystyle \mathrm{C} \mathit{\left( \log _{2} \frac{2}{3} , \frac{2}{3} \right)}$, $\displaystyle \mathrm{D} \mathit{\left( \log _{2} \frac{2}{3} , - \frac{1}{2} \right)}$이므로 $\displaystyle \overline{\mathrm{AC}} = \log _{2} 3 - \log _{2} \frac{2}{3} = \log _{2} \frac{9}{2}$, $\displaystyle \overline{\mathrm{AB}} = 3 - \frac{2}{3} = \frac{7}{3}$, $\displaystyle \overline{\mathrm{CD}} = \frac{2}{3} - \left( - \frac{1}{2} \right) = \frac{7}{6}$ 삼각형 $\mathrm{ABC}$, $\mathrm{ACD}$의 넓이를 각각 $S _{1}$, $S _{2}$라 하면 $S _{1}$$\displaystyle = \frac{1}{2} \times \log _{2} \frac{9}{2} \times \frac{7}{3}$$\displaystyle = \frac{7}{3} \log _{2} 3 - \frac{7}{6}$, $S _{2}$$\displaystyle = \frac{1}{2} \times \log _{2} \frac{9}{2} \times \frac{7}{6}$$\displaystyle = \frac{7}{6} \log _{2} 3 - \frac{7}{12}$ 따라서 구하려는 사각형 $\mathrm{ABCD}$의 넓이는 $S _{1} + S _{2}$$\displaystyle = \left( \frac{7}{3} \log _{2} 3 - \frac{7}{6} \right) + \left( \frac{7}{6} \log _{2} 3 - \frac{7}{12} \right)$ $\displaystyle = \frac{7}{2} \log _{2} 3 - \frac{7}{4}$