자연계열 A 2번

(a) 원점을 지나고 $\displaystyle x$축과 이루는 각이 $\displaystyle \frac{\pi}{2n+2}$라디안인 직선 $\displaystyle y=f_{n}(x)$의 방정식은 $\displaystyle y=f_{n}(x)=x\cdot\tan\left(\frac{\pi}{2n+2}\right)$이므로$\displaystyle y=f_{1}(x)=x\cdot\tan\frac{\pi}{4}=x$$\displaystyle y=f_{2}(x)=x\cdot\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}x$이다.[답] $\displaystyle y=f_{1}(x)=x$, $\displaystyle y=f_{2}(x)=\frac{\sqrt{3}}{3}x$

(b)

(ⅰ) 타원 $\displaystyle x^{2}+3y^{2}=1$과 직선 $\displaystyle y=f_{1}(x)=x$의 교점을 $\displaystyle \mathrm{P}_{1}$이라 하면$\displaystyle x^{2}+3x^{2}=1$, $\displaystyle 4x^{2}=1$, $\displaystyle x^{2}=\frac{1}{4}$, $\displaystyle x=\frac{1}{2}$ ($\displaystyle \because x>0$)이므로$\displaystyle \therefore\ \mathrm{P}_{1}\left(\frac{1}{2},\frac{1}{2}\right)$(ⅱ) 타원 $\displaystyle x^{2}+3y^{2}=1$과 직선 $\displaystyle y=f_{2}(x)=\frac{\sqrt{3}}{3}x$의 교점을 $\displaystyle \mathrm{P}_{2}$라 하면$\displaystyle x^{2}+x^{2}=1$, $\displaystyle 2x^{2}=1$, $\displaystyle x^{2}=\frac{1}{2}$, $\displaystyle x=\frac{\sqrt{2}}{2}$ ($\displaystyle \because x>0$)이므로$\displaystyle \therefore\ \mathrm{P}_{2}\left(\frac{\sqrt{2}}{2},\frac{\sqrt{6}}{6}\right)$(ⅲ) 두 점 $\displaystyle \mathrm{P}_{1}$, $\displaystyle \mathrm{P}_{2}$에서 $\displaystyle x$축에 내린 수선의 발을 각각 $\displaystyle \mathrm{Q}_{1}$, $\displaystyle \mathrm{Q}_{2}$라 하고 도형 $\displaystyle \mathrm{P}_{1}\mathrm{Q}_{1}\mathrm{Q}_{2}\mathrm{P}_{2}$의 넓이를 $\displaystyle A_{1}$이라 하면$\displaystyle A_{1}=\int_{\frac{1}{2}}^{\frac{\sqrt{2}}{2}}\frac{1}{\sqrt{3}}\sqrt{1-x^{2}}\,dx$ ($\displaystyle \because$ $\displaystyle x^{2}+3y^{2}=1$에서 $\displaystyle y=\frac{1}{\sqrt{3}}\sqrt{1-x^{2}}$)$\displaystyle x=\sin\theta$라 하면 $\displaystyle dx=\cos\theta\,d\theta$$\displaystyle x=\frac{1}{2}$일 때 $\displaystyle \theta=\frac{\pi}{6}$이고, $\displaystyle x=\frac{\sqrt{2}}{2}$일 때 $\displaystyle \theta=\frac{\pi}{4}$이므로$\displaystyle A_{1}=\frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\cos^{2}\theta\,d\theta=\frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{1+\cos2\theta}{2}\,d\theta$ ($\displaystyle \because$ $\displaystyle \cos^{2}\theta=\frac{1+\cos2\theta}{2}$)$\displaystyle =\frac{1}{2\sqrt{3}}\Biggl[\theta+\frac{1}{2}\sin2\theta\Biggr]_{\frac{\pi}{6}}^{\frac{\pi}{4}}=\frac{1}{2\sqrt{3}}\left\{\left(\frac{\pi}{4}+\frac{1}{2}\sin\frac{\pi}{2}\right)-\left(\frac{\pi}{6}+\frac{1}{2}\sin\frac{\pi}{3}\right)\right\}=\frac{\sqrt{3}}{6}\left(\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}\right)$삼각형 $\displaystyle \mathrm{P}_{1}\mathrm{O}\mathrm{Q}_{1}$의 넓이를 $\displaystyle A_{2}$라 하면 $\displaystyle A_{2}=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$삼각형 $\displaystyle \mathrm{P}_{2}\mathrm{O}\mathrm{Q}_{2}$의 넓이를 $\displaystyle A_{3}$라 하면 $\displaystyle A_{3}=\frac{1}{2}\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{6}}{6}=\frac{\sqrt{3}}{12}$따라서, 구하는 넓이 $\displaystyle S_{1}$은$\displaystyle S_{1}=A_{1}+A_{2}-A_{3}=\frac{\sqrt{3}}{6}\left(\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}\right)+\frac{1}{8}-\frac{\sqrt{3}}{12}=\frac{\sqrt{3}\pi}{72}$[답] $\displaystyle \frac{\sqrt{3}\pi}{72}$

(d)

점 $\displaystyle \mathrm{Q}$의 좌표를 $\displaystyle (1,0)$이라 하면 $\displaystyle \sum_{n=1}^{\infty}S_{n}$은 도형 $\displaystyle \mathrm{P}_{1}\mathrm{O}\mathrm{Q}$의 넓이와 같으므로 도형 $\displaystyle \mathrm{P}_{1}\mathrm{Q}_{1}\mathrm{Q}$의 넓이를 $\displaystyle A_{4}$라 하면$\displaystyle A_{4}=\int_{\frac{1}{2}}^{1}\frac{1}{\sqrt{3}}\sqrt{1-x^{2}}\,dx$ ($\displaystyle \because$ $\displaystyle x^{2}+3y^{2}=1$에서 $\displaystyle y=\frac{1}{\sqrt{3}}\sqrt{1-x^{2}}$)$\displaystyle x=\sin\theta$라 하면 $\displaystyle dx=\cos\theta\,d\theta$$\displaystyle x=\frac{1}{2}$일 때 $\displaystyle \theta=\frac{\pi}{6}$이고, $\displaystyle x=1$일 때 $\displaystyle \theta=\frac{\pi}{2}$이므로$\displaystyle A_{4}=\frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cos^{2}\theta\,d\theta=\frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{1+\cos2\theta}{2}\,d\theta$ ($\displaystyle \because$ $\displaystyle \cos^{2}\theta=\frac{1+\cos2\theta}{2}$)$\displaystyle =\frac{1}{2\sqrt{3}}\Biggl[\theta+\frac{1}{2}\sin2\theta\Biggr]_{\frac{\pi}{6}}^{\frac{\pi}{2}}=\frac{1}{2\sqrt{3}}\left\{\left(\frac{\pi}{2}+\frac{1}{2}\sin\pi\right)-\left(\frac{\pi}{6}+\frac{1}{2}\sin\frac{\pi}{3}\right)\right\}=\frac{\sqrt{3}}{6}\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$따라서$\displaystyle \sum_{n=1}^{\infty}S_{n}=A_{2}+A_{4}=\frac{1}{8}+\frac{\sqrt{3}}{6}\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)=\frac{\sqrt{3}\pi}{18}$($\displaystyle \because$ $\displaystyle A_{2}$는 삼각형 $\displaystyle \mathrm{P}_{1}\mathrm{O}\mathrm{Q}_{1}$의 넓이이고, (b)에서 $\displaystyle A_{2}=\frac{1}{8}$)[답] $\displaystyle \frac{\sqrt{3}\pi}{18}$