+20점 · 자연수 n \displaystyle n n a n \displaystyle a_{n} a n a n − a 1 \displaystyle a_{n} - a_{1} a n − a 1 1. a 2 = a 0 + a 1 2 , a 3 = a 1 + a 2 2 , a 4 = a 2 + a 3 2 , ⋯ \displaystyle a_2=\frac{a_0+a_1}2,\ a_3=\frac{a_1+a_2}2,\ a_4=\frac{a_2+a_3}2,\cdots a 2 = 2 a 0 + a 1 , a 3 = 2 a 1 + a 2 , a 4 = 2 a 2 + a 3 , ⋯ a 2 − a 1 = a 0 − a 1 2 , a 3 − a 2 = a 1 − a 2 2 , a 4 − a 3 = a 2 − a 3 2 , ⋯ \displaystyle a_2-a_1=\frac{a_0-a_1}2,\ a_3-a_2=\frac{a_1-a_2}2,\ a_4-a_3=\frac{a_2-a_3}2,\cdots a 2 − a 1 = 2 a 0 − a 1 , a 3 − a 2 = 2 a 1 − a 2 , a 4 − a 3 = 2 a 2 − a 3 , ⋯ a 2 − a 1 = − a 1 − a 0 2 , a 3 − a 2 = − a 2 − a 1 2 = a 1 − a 0 2 2 , a 4 − a 3 = − a 1 − a 0 2 3 , ⋯ \displaystyle a_2-a_1=-\frac{a_1-a_0}2,\ a_3-a_2=-\frac{a_2-a_1}2=\frac{a_1-a_0}{2^2},\ a_4-a_3=-\frac{a_1-a_0}{2^3},\cdots a 2 − a 1 = − 2 a 1 − a 0 , a 3 − a 2 = − 2 a 2 − a 1 = 2 2 a 1 − a 0 , a 4 − a 3 = − 2 3 a 1 − a 0 , ⋯ 자연수 n ≥ 2 \displaystyle n\ge2 n ≥ 2 a n − a 1 = − a 1 − a 0 2 ( 1 − 1 2 + 1 2 2 + ⋯ + ( − 1 ) n − 2 1 2 n − 2 ) = a 1 − a 0 3 ( ( − 1 2 ) n − 1 − 1 ) \displaystyle a_n-a_1=-\frac{a_1-a_0}2\left(1-\frac12+\frac1{2^2}+\cdots+(-1)^{n-2}\frac1{2^{n-2}}\right)=\frac{a_1-a_0}3\left(\left(-\frac12\right)^{n-1}-1\right) a n − a 1 = − 2 a 1 − a 0 ( 1 − 2 1 + 2 2 1 + ⋯ + ( − 1 ) n − 2 2 n − 2 1 ) = 3 a 1 − a 0 ( ( − 2 1 ) n − 1 − 1 )
+10점 · 극한값 lim n → ∞ a n \displaystyle \lim_{n \to \infty} a_{n} n → ∞ lim a n a n = 2 a 1 + a 0 3 + ( − 1 ) n − 1 a 1 − a 0 3 ⋅ 2 n − 1 \displaystyle a_n=\frac{2a_1+a_0}3+(-1)^{n-1}\frac{a_1-a_0}{3\cdot2^{n-1}} a n = 3 2 a 1 + a 0 + ( − 1 ) n − 1 3 ⋅ 2 n − 1 a 1 − a 0 lim n → ∞ a n = 2 a 1 + a 0 3 = 40 + 5 3 = 15 \displaystyle \lim_{n\to\infty}a_n=\frac{2a_1+a_0}3=\frac{40+5}3=15 n → ∞ lim a n = 3 2 a 1 + a 0 = 3 40 + 5 = 15
+10점 · ( 1 − 2 x ) n = ∑ k = 0 n n C k ( − x ) k ( 1 − x ) n − k \displaystyle (1-2x)^n=\sum_{k=0}^n{}_n\mathrm C_k(-x)^k(1-x)^{n-k} ( 1 − 2 x ) n = k = 0 ∑ n n C k ( − x ) k ( 1 − x ) n − k 2. ( 1 − 2 x ) n = ( 1 − x − x ) n = ∑ k = 0 n n C k ( − x ) k ( 1 − x ) n − k \displaystyle (1-2x)^n=(1-x-x)^n=\sum_{k=0}^n{}_n\mathrm C_k(-x)^k(1-x)^{n-k} ( 1 − 2 x ) n = ( 1 − x − x ) n = k = 0 ∑ n n C k ( − x ) k ( 1 − x ) n − k ∑ k = 0 n n C k ( − 1 ) k ∫ − 1 2 x k ( 1 − x ) n − k d x = ∫ − 1 2 ( 1 − 2 x ) n d x \displaystyle \sum_{k=0}^n{}_n\mathrm C_k(-1)^k\int_{-1}^2x^k(1-x)^{n-k}\,dx=\int_{-1}^2(1-2x)^n\,dx k = 0 ∑ n n C k ( − 1 ) k ∫ − 1 2 x k ( 1 − x ) n − k d x = ∫ − 1 2 ( 1 − 2 x ) n d x
+20점 · 적분값 ∫ − 1 2 ( 1 − 2 x ) n d x \displaystyle \int_{-1}^2(1-2x)^n\,dx ∫ − 1 2 ( 1 − 2 x ) n d x 위의 적분에서 1 − 2 x = u \displaystyle 1-2x=u 1 − 2 x = u ∫ − 1 2 ( 1 − 2 x ) n d x = 1 2 ∫ − 3 3 u n d u = { 0 , n = 홀수 3 n + 1 n + 1 , n = 짝수 \displaystyle \int_{-1}^2(1-2x)^n\,dx=\frac12\int_{-3}^3u^n\,du=\begin{cases}0,&n=\text{홀수}\\\frac{3^{n+1}}{n+1},&n=\text{짝수}\end{cases} ∫ − 1 2 ( 1 − 2 x ) n d x = 2 1 ∫ − 3 3 u n d u = { 0 , n + 1 3 n + 1 , n = 홀수 n = 짝수
+15점 · r ( t ) \displaystyle r ( t ) r ( t ) 3. 주어진 두 조건을 만족하려면 타원위의 점 ( x , y ) \displaystyle (x,y) ( x , y ) ( t , 0 ) \displaystyle (t,0) ( t , 0 ) r \displaystyle r r 즉, { x 2 + 2 y 2 = 1 ( x − t ) 2 + y 2 ≥ r 2 \displaystyle \begin{cases}x^2+2y^2=1\\(x-t)^2+y^2\ge r^2\end{cases} { x 2 + 2 y 2 = 1 ( x − t ) 2 + y 2 ≥ r 2 ( x , y ) \displaystyle (x,y) ( x , y ) { x 2 + 2 y 2 = 1 ( x − t ) 2 + y 2 = r 2 \displaystyle \begin{cases}x^2+2y^2=1\\(x-t)^2+y^2=r^2\end{cases} { x 2 + 2 y 2 = 1 ( x − t ) 2 + y 2 = r 2 연립하여 x \displaystyle x x 1 2 x 2 − 2 t x + t 2 − r 2 + 1 2 = 0 \displaystyle \frac12x^2-2tx+t^2-r^2+\frac12=0 2 1 x 2 − 2 t x + t 2 − r 2 + 2 1 = 0 문제의 조건을 만족하려면, 연립방정식의 근이 중근으로 나타나야 한다. 따라서 식 ①에서 x \displaystyle x x D / 4 = 1 2 t 2 + 1 2 r 2 − 1 4 = 0 ⇒ r ( t ) = 1 2 − t 2 \displaystyle D/4=\frac12t^2+\frac12r^2-\frac14=0\quad\Rightarrow\quad r(t)=\sqrt{\frac12-t^2} D /4 = 2 1 t 2 + 2 1 r 2 − 4 1 = 0 ⇒ r ( t ) = 2 1 − t 2
+25점 · 적분값 ∫ 0 2 4 r ( t ) d t \displaystyle \int_{0}^{\frac{\sqrt{2}}{4}} r ( t ) d t ∫ 0 4 2 r ( t ) d t 구하고자 하는 적분값을 계산하면 다음과 같다. ∫ 0 2 4 r ( t ) d t = ∫ 0 2 4 1 2 − t 2 d t = ∫ 0 π 6 1 2 − 1 2 sin 2 θ 1 2 cos θ d θ = 1 2 ∫ 0 π 6 cos 2 θ d θ = 1 2 ∫ 0 π 6 1 2 ( cos 2 θ + 1 ) d θ = 1 4 [ 1 2 sin 2 θ + θ ] 0 π 6 = 1 4 ( 3 4 + π 6 ) = 3 16 + π 24 \displaystyle \begin{aligned}\int_0^{\frac{\sqrt2}4}r(t)\,dt&=\int_0^{\frac{\sqrt2}4}\sqrt{\frac12-t^2}\,dt=\int_0^{\frac\pi6}\sqrt{\frac12-\frac12\sin^2\theta}\frac1{\sqrt2}\cos\theta\,d\theta=\frac12\int_0^{\frac\pi6}\cos^2\theta\,d\theta\\&=\frac12\int_0^{\frac\pi6}\frac12(\cos2\theta+1)\,d\theta=\frac14\Biggl[\frac12\sin2\theta+\theta\Biggr]_0^{\frac\pi6}=\frac14\left(\frac{\sqrt3}4+\frac\pi6\right)=\frac{\sqrt3}{16}+\frac\pi{24}\end{aligned} ∫ 0 4 2 r ( t ) d t = ∫ 0 4 2 2 1 − t 2 d t = ∫ 0 6 π 2 1 − 2 1 sin 2 θ 2 1 cos θ d θ = 2 1 ∫ 0 6 π cos 2 θ d θ = 2 1 ∫ 0 6 π 2 1 ( cos 2 θ + 1 ) d θ = 4 1 [ 2 1 sin 2 θ + θ ] 0 6 π = 4 1 ( 4 3 + 6 π ) = 16 3 + 24 π
수리논술, 이게 다다.