+4점 [3.1] 직선 l 1 \displaystyle l_1 l 1 x \displaystyle x x θ 1 \displaystyle \theta_1 θ 1 l 2 \displaystyle l_2 l 2 x \displaystyle x x θ 2 \displaystyle \theta_2 θ 2 θ = θ 1 − θ 2 \displaystyle \theta=\theta_1-\theta_2 θ = θ 1 − θ 2
+5점 P 1 \displaystyle \mathrm{P}_1 P 1 P 2 \displaystyle \mathrm{P}_2 P 2 t − x 2 = t x \displaystyle t-x^2=tx t − x 2 = t x α \displaystyle \alpha α β \displaystyle \beta β α = − t 2 + t 2 + 4 t 2 , β = − t 2 − t 2 + 4 t 2 \displaystyle \alpha=-\frac t2+\frac{\sqrt{t^2+4t}}2,\quad\beta=-\frac t2-\frac{\sqrt{t^2+4t}}2 α = − 2 t + 2 t 2 + 4 t , β = − 2 t − 2 t 2 + 4 t 이다. P 1 ( α , t α ) \displaystyle \mathrm{P}_1(\alpha,t\alpha) P 1 ( α , t α ) P 2 ( β , t β ) \displaystyle \mathrm{P}_2(\beta,t\beta) P 2 ( β , tβ ) y = t − x 2 \displaystyle y=t-x^2 y = t − x 2 − 2 α \displaystyle -2\alpha − 2 α − 2 β \displaystyle -2\beta − 2 β tan θ 1 = − 2 α \displaystyle \tan\theta_1=-2\alpha tan θ 1 = − 2 α tan θ 2 = − 2 β \displaystyle \tan\theta_2=-2\beta tan θ 2 = − 2 β tan 2 θ = tan 2 ( θ 1 − θ 2 ) = ( tan θ 1 − tan θ 2 1 + tan θ 1 tan θ 2 ) 2 = ( − 2 α + 2 β 1 + 4 α β ) 2 = 4 ( α − β ) 2 ( 1 + 4 α β ) 2 = 4 { ( α + β ) 2 − 4 α β } ( 1 + 4 α β ) 2 \displaystyle \tan^2\theta=\tan^2(\theta_1-\theta_2)=\left(\frac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}\right)^2=\left(\frac{-2\alpha+2\beta}{1+4\alpha\beta}\right)^2=\frac{4(\alpha-\beta)^2}{(1+4\alpha\beta)^2}=\frac{4\{(\alpha+\beta)^2-4\alpha\beta\}}{(1+4\alpha\beta)^2} tan 2 θ = tan 2 ( θ 1 − θ 2 ) = ( 1 + tan θ 1 tan θ 2 tan θ 1 − tan θ 2 ) 2 = ( 1 + 4 α β − 2 α + 2 β ) 2 = ( 1 + 4 α β ) 2 4 ( α − β ) 2 = ( 1 + 4 α β ) 2 4 {( α + β ) 2 − 4 α β } 이므로 근과 계수의 관계에서 α + β = − t \displaystyle \alpha+\beta=-t α + β = − t α β = − t \displaystyle \alpha\beta=-t α β = − t tan 2 θ = 4 ( t 2 + 4 t ) ( 1 − 4 t ) 2 = 4 t 2 + 16 t 16 t 2 − 8 t + 1 \displaystyle \tan^2\theta=\frac{4(t^2+4t)}{(1-4t)^2}=\frac{4t^2+16t}{16t^2-8t+1} tan 2 θ = ( 1 − 4 t ) 2 4 ( t 2 + 4 t ) = 16 t 2 − 8 t + 1 4 t 2 + 16 t
+4점 · 직선 l 1 \displaystyle l_1 l 1 y = ( t − t 2 + 4 t ) x − ( t − t 2 + 4 t ) t 2 + 4 t 2 \displaystyle y=(t-\sqrt{t^2+4t})x-\frac{(t-\sqrt{t^2+4t})\sqrt{t^2+4t}}2 y = ( t − t 2 + 4 t ) x − 2 ( t − t 2 + 4 t ) t 2 + 4 t x \displaystyle x x t 2 + 4 t 2 \displaystyle \frac{\sqrt{t^2+4t}}2 2 t 2 + 4 t y \displaystyle y y ( t 2 + 4 t − t ) t 2 + 4 t 2 \displaystyle \frac{(\sqrt{t^2+4t}-t)\sqrt{t^2+4t}}2 2 ( t 2 + 4 t − t ) t 2 + 4 t A 1 ( t ) = 1 2 × t 2 + 4 t 2 × ( t 2 + 4 t − t ) t 2 + 4 t 2 = ( t 2 + 4 t ) ( t 2 + 4 t − t ) 8 \displaystyle A_1(t)=\frac12\times\frac{\sqrt{t^2+4t}}2\times\frac{(\sqrt{t^2+4t}-t)\sqrt{t^2+4t}}2=\frac{(t^2+4t)(\sqrt{t^2+4t}-t)}8 A 1 ( t ) = 2 1 × 2 t 2 + 4 t × 2 ( t 2 + 4 t − t ) t 2 + 4 t = 8 ( t 2 + 4 t ) ( t 2 + 4 t − t ) [3.2] 직선 l 1 \displaystyle l_1 l 1 y = ( t − t 2 + 4 t ) x − ( t − t 2 + 4 t ) t 2 + 4 t 2 \displaystyle y=(t-\sqrt{t^2+4t})x-\frac{(t-\sqrt{t^2+4t})\sqrt{t^2+4t}}2 y = ( t − t 2 + 4 t ) x − 2 ( t − t 2 + 4 t ) t 2 + 4 t x \displaystyle x x t 2 + 4 t 2 \displaystyle \frac{\sqrt{t^2+4t}}2 2 t 2 + 4 t y \displaystyle y y ( t 2 + 4 t − t ) t 2 + 4 t 2 \displaystyle \frac{(\sqrt{t^2+4t}-t)\sqrt{t^2+4t}}2 2 ( t 2 + 4 t − t ) t 2 + 4 t A 1 ( t ) = 1 2 × t 2 + 4 t 2 × ( t 2 + 4 t − t ) t 2 + 4 t 2 = ( t 2 + 4 t ) ( t 2 + 4 t − t ) 8 \displaystyle A_1(t)=\frac12\times\frac{\sqrt{t^2+4t}}2\times\frac{(\sqrt{t^2+4t}-t)\sqrt{t^2+4t}}2=\frac{(t^2+4t)(\sqrt{t^2+4t}-t)}8 A 1 ( t ) = 2 1 × 2 t 2 + 4 t × 2 ( t 2 + 4 t − t ) t 2 + 4 t = 8 ( t 2 + 4 t ) ( t 2 + 4 t − t ) 이다.
+8점 · lim t → ∞ A 1 ( t ) t k = lim t → ∞ { ( t 2 + 4 t ) ( t 2 + 4 t − t ) 8 t k × t 2 + 4 t + t t 2 + 4 t + t } = lim t → ∞ ( t 2 + 4 t ) 8 t k 4 t t 2 + 4 t + t \displaystyle \lim_{t\to\infty}\frac{A_1(t)}{t^k}=\lim_{t\to\infty}\left\{\frac{(t^2+4t)(\sqrt{t^2+4t}-t)}{8t^k}\times\frac{\sqrt{t^2+4t}+t}{\sqrt{t^2+4t}+t}\right\}=\lim_{t\to\infty}\frac{(t^2+4t)}{8t^k}\frac{4t}{\sqrt{t^2+4t}+t} t → ∞ lim t k A 1 ( t ) = t → ∞ lim { 8 t k ( t 2 + 4 t ) ( t 2 + 4 t − t ) × t 2 + 4 t + t t 2 + 4 t + t } = t → ∞ lim 8 t k ( t 2 + 4 t ) t 2 + 4 t + t 4 t lim t → ∞ 4 t t 2 + 4 t + t = lim t → ∞ 4 1 + 4 t + 1 = 2 \displaystyle \lim_{t\to\infty}\frac{4t}{\sqrt{t^2+4t}+t}=\lim_{t\to\infty}\frac4{\sqrt{1+\frac4t}+1}=2 t → ∞ lim t 2 + 4 t + t 4 t = t → ∞ lim 1 + t 4 + 1 4 = 2 lim t → ∞ t 2 + 4 t 8 t k = { ∞ ( k = 1 ) 1 8 ( k = 2 ) 0 ( k ≥ 3 ) \displaystyle \lim_{t\to\infty}\frac{t^2+4t}{8t^k}=\begin{cases}\infty&(k=1)\\\frac18&(k=2)\\0&(k\geq3)\end{cases} t → ∞ lim 8 t k t 2 + 4 t = ⎩ ⎨ ⎧ ∞ 8 1 0 ( k = 1 ) ( k = 2 ) ( k ≥ 3 ) lim t → ∞ A 1 ( t ) t k = { ∞ ( k = 1 ) 1 4 ( k = 2 ) 0 ( k ≥ 3 ) \displaystyle \lim_{t\to\infty}\frac{A_1(t)}{t^k}=\begin{cases}\infty&(k=1)\\\frac14&(k=2)\\0&(k\geq3)\end{cases} t → ∞ lim t k A 1 ( t ) = ⎩ ⎨ ⎧ ∞ 4 1 0 ( k = 1 ) ( k = 2 ) ( k ≥ 3 ) lim t → ∞ A 1 ( t ) t k = lim t → ∞ { ( t 2 + 4 t ) ( t 2 + 4 t − t ) 8 t k × t 2 + 4 t + t t 2 + 4 t + t } = lim t → ∞ ( t 2 + 4 t ) 8 t k 4 t t 2 + 4 t + t = 1 4 lim t → ∞ ( t 2 + 4 t ) t k \displaystyle \begin{aligned}\lim_{t\to\infty}\frac{A_1(t)}{t^k}&=\lim_{t\to\infty}\left\{\frac{(t^2+4t)(\sqrt{t^2+4t}-t)}{8t^k}\times\frac{\sqrt{t^2+4t}+t}{\sqrt{t^2+4t}+t}\right\}\\&=\lim_{t\to\infty}\frac{(t^2+4t)}{8t^k}\frac{4t}{\sqrt{t^2+4t}+t}\\&=\frac14\lim_{t\to\infty}\frac{(t^2+4t)}{t^k}\end{aligned} t → ∞ lim t k A 1 ( t ) = t → ∞ lim { 8 t k ( t 2 + 4 t ) ( t 2 + 4 t − t ) × t 2 + 4 t + t t 2 + 4 t + t } = t → ∞ lim 8 t k ( t 2 + 4 t ) t 2 + 4 t + t 4 t = 4 1 t → ∞ lim t k ( t 2 + 4 t ) 위 극한은 k < 2 \displaystyle k<2 k < 2 k > 2 \displaystyle k>2 k > 2 0 \displaystyle 0 0 k = 2 \displaystyle k=2 k = 2 1 4 \displaystyle \frac14 4 1 0 , 1 4 \displaystyle 0,\frac14 0 , 4 1
+4점 [3.3] 직선 l 2 \displaystyle l_2 l 2 y = ( t + t 2 + 4 t ) x + ( t + t 2 + 4 t ) t 2 + 4 t 2 \displaystyle y=(t+\sqrt{t^2+4t})x+\frac{(t+\sqrt{t^2+4t})\sqrt{t^2+4t}}2 y = ( t + t 2 + 4 t ) x + 2 ( t + t 2 + 4 t ) t 2 + 4 t x \displaystyle x x − t 2 + 4 t 2 \displaystyle -\frac{\sqrt{t^2+4t}}2 − 2 t 2 + 4 t y \displaystyle y y ( t + t 2 + 4 t ) t 2 + 4 t 2 \displaystyle \frac{(t+\sqrt{t^2+4t})\sqrt{t^2+4t}}2 2 ( t + t 2 + 4 t ) t 2 + 4 t A 2 ( t ) \displaystyle A_2(t) A 2 ( t ) A 2 ( t ) = ( t 2 + 4 t ) ( t + t 2 + 4 t ) 8 \displaystyle A_2(t)=\frac{(t^2+4t)(t+\sqrt{t^2+4t})}8 A 2 ( t ) = 8 ( t 2 + 4 t ) ( t + t 2 + 4 t )
+4점 · 곡선과 직선 사이의 넓이 S ( t ) \displaystyle S(t) S ( t ) S ( t ) = ∫ β α ( t − x 2 − t x ) d x = [ t x − 1 3 x 3 − 1 2 t x 2 ] β α = ( t 2 + 4 t ) t 2 + 4 t 6 \displaystyle S(t)=\int_\beta^\alpha(t-x^2-tx)\,dx=\Biggl[tx-\frac13x^3-\frac12tx^2\Biggr]_\beta^\alpha=\frac{(t^2+4t)\sqrt{t^2+4t}}6 S ( t ) = ∫ β α ( t − x 2 − t x ) d x = [ t x − 3 1 x 3 − 2 1 t x 2 ] β α = 6 ( t 2 + 4 t ) t 2 + 4 t 곡선과 직선 사이의 넓이 S ( t ) \displaystyle S(t) S ( t ) S ( t ) = ∫ β α ( t − x 2 − t x ) d x = [ t x − 1 3 x 3 − 1 2 t x 2 ] β α = ( t 2 + 4 t ) t 2 + 4 t 6 \displaystyle S(t)=\int_\beta^\alpha(t-x^2-tx)\,dx=\Biggl[tx-\frac13x^3-\frac12tx^2\Biggr]_\beta^\alpha=\frac{(t^2+4t)\sqrt{t^2+4t}}6 S ( t ) = ∫ β α ( t − x 2 − t x ) d x = [ t x − 3 1 x 3 − 2 1 t x 2 ] β α = 6 ( t 2 + 4 t ) t 2 + 4 t 이다.
+4점 따라서 lim t → ∞ S ( t ) A 2 ( t ) = lim t → ∞ ( t 2 + 4 t ) t 2 + 4 t 6 8 ( t 2 + 4 t ) ( t + t 2 + 4 t ) = 2 3 \displaystyle \lim_{t\to\infty}\frac{S(t)}{A_2(t)}=\lim_{t\to\infty}\frac{(t^2+4t)\sqrt{t^2+4t}}6\frac8{(t^2+4t)(t+\sqrt{t^2+4t})}=\frac23 t → ∞ lim A 2 ( t ) S ( t ) = t → ∞ lim 6 ( t 2 + 4 t ) t 2 + 4 t ( t 2 + 4 t ) ( t + t 2 + 4 t ) 8 = 3 2 이다.
수리논술, 이게 다다.