[1-1]
+3점 · f ′ ( x ) \displaystyle f^{\prime}(x) f ′ ( x ) f ′ ( x ) = 1 x 2 − 4 \displaystyle f^{\prime}(x)=\frac{1}{\sqrt{x^2-4}} f ′ ( x ) = x 2 − 4 1 f ( x ) = ln ( x + x 2 − 4 ) ( x > 2 ) \displaystyle f(x)=\ln\left(x+\sqrt{x^2-4}\right)\ (x>2) f ( x ) = ln ( x + x 2 − 4 ) ( x > 2 ) f ′ ( x ) = 1 + 2 x 2 x 2 − 4 x + x 2 − 4 = x 2 − 4 + x ( x + x 2 − 4 ) x 2 − 4 = 1 x 2 − 4 \displaystyle f^{\prime}(x)=\frac{1+\frac{2x}{2\sqrt{x^2-4}}}{x+\sqrt{x^2-4}}=\frac{\sqrt{x^2-4}+x}{\left(x+\sqrt{x^2-4}\right)\sqrt{x^2-4}}=\frac{1}{\sqrt{x^2-4}} f ′ ( x ) = x + x 2 − 4 1 + 2 x 2 − 4 2 x = ( x + x 2 − 4 ) x 2 − 4 x 2 − 4 + x = x 2 − 4 1 이다.
+5점 · 부분적분을 할 수 있다. ∫ ( 1 × x 2 − 4 ) d x = x x 2 − 4 − ∫ x 2 x 2 x 2 − 4 d x \displaystyle \int\left(1\times\sqrt{x^2-4}\right)dx=x\sqrt{x^2-4}-\int x\frac{2x}{2\sqrt{x^2-4}}\,dx ∫ ( 1 × x 2 − 4 ) d x = x x 2 − 4 − ∫ x 2 x 2 − 4 2 x d x 한편 I = ∫ x 2 − 4 d x = ∫ ( 1 × x 2 − 4 ) d x = x x 2 − 4 − ∫ x 2 x 2 x 2 − 4 d x = x x 2 − 4 − ∫ x 2 − 4 + 4 x 2 − 4 d x = x x 2 − 4 − I − ∫ 4 x 2 − 4 d x \displaystyle \begin{aligned}I&=\int\sqrt{x^2-4}\,dx\\&=\int\left(1\times\sqrt{x^2-4}\right)dx=x\sqrt{x^2-4}-\int x\frac{2x}{2\sqrt{x^2-4}}\,dx\\&=x\sqrt{x^2-4}-\int\frac{x^2-4+4}{\sqrt{x^2-4}}\,dx=x\sqrt{x^2-4}-I-\int\frac{4}{\sqrt{x^2-4}}\,dx\end{aligned} I = ∫ x 2 − 4 d x = ∫ ( 1 × x 2 − 4 ) d x = x x 2 − 4 − ∫ x 2 x 2 − 4 2 x d x = x x 2 − 4 − ∫ x 2 − 4 x 2 − 4 + 4 d x = x x 2 − 4 − I − ∫ x 2 − 4 4 d x 이다.
+7점 · 식의 변형을 통해 부정적분을 구할 수 있다. I = 1 2 ( x x 2 − 4 − 4 ln ∣ x + x 2 − 4 ∣ ) + C \displaystyle I=\frac{1}{2}\left(x\sqrt{x^2-4}-4\ln\left|x+\sqrt{x^2-4}\right|\right)+C I = 2 1 ( x x 2 − 4 − 4 ln x + x 2 − 4 ) + C ∫ 4 x 2 − 4 d x \displaystyle \int\frac{4}{\sqrt{x^2-4}}\,dx ∫ x 2 − 4 4 d x x = 2 sec θ \displaystyle x=2\sec\theta x = 2 sec θ d x = 2 sec θ tan θ d θ \displaystyle dx=2\sec\theta\tan\theta\,d\theta d x = 2 sec θ tan θ d θ ∫ 4 x 2 − 4 d x = ∫ 4 2 sec 2 θ − 1 × 2 sec θ tan θ d θ = 4 ∫ sec θ d θ = 4 ∫ sec θ ( sec θ + tan θ ) sec θ + tan θ d θ = 4 ln ∣ sec θ + tan θ ∣ + C 1 \displaystyle \begin{aligned}\int\frac{4}{\sqrt{x^2-4}}\,dx&=\int\frac{4}{2\sqrt{\sec^2\theta-1}}\times2\sec\theta\tan\theta\,d\theta\\&=4\int\sec\theta\,d\theta\\&=4\int\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}\,d\theta\\&=4\ln|\sec\theta+\tan\theta|+C_1\end{aligned} ∫ x 2 − 4 4 d x = ∫ 2 sec 2 θ − 1 4 × 2 sec θ tan θ d θ = 4 ∫ sec θ d θ = 4 ∫ sec θ + tan θ sec θ ( sec θ + tan θ ) d θ = 4 ln ∣ sec θ + tan θ ∣ + C 1 C 1 \displaystyle C_1 C 1 ∫ 4 x 2 − 4 d x = 4 ln ∣ x 2 + x 2 − 4 2 ∣ + C 1 \displaystyle \phantom{\int\frac{4}{\sqrt{x^2-4}}\,dx}=4\ln\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right|+C_1 ∫ x 2 − 4 4 d x = 4 ln 2 x + 2 x 2 − 4 + C 1 이다. 그러므로 I = x x 2 − 4 − I − 4 ln ∣ x + x 2 − 4 ∣ + C ′ \displaystyle I=x\sqrt{x^2-4}-I-4\ln\left|x+\sqrt{x^2-4}\right|+C^{\prime} I = x x 2 − 4 − I − 4 ln x + x 2 − 4 + C ′ C ′ \displaystyle C^{\prime} C ′ ∴ I = 1 2 ( x x 2 − 4 − 4 ln ∣ x + x 2 − 4 ∣ ) + C \displaystyle \therefore I=\frac{1}{2}\left(x\sqrt{x^2-4}-4\ln\left|x+\sqrt{x^2-4}\right|\right)+C ∴ I = 2 1 ( x x 2 − 4 − 4 ln x + x 2 − 4 ) + C C \displaystyle C C 이다.
[1-2]
+3점 · 접선의 방정식 y = x \displaystyle y=x y = x 곡선 C : 3 x 2 − 4 y 2 − 6 x + 16 y = 25 \displaystyle C:3x^2-4y^2-6x+16y=25 C : 3 x 2 − 4 y 2 − 6 x + 16 y = 25 ( 5 , 5 ) \displaystyle (5,5) ( 5 , 5 ) 3 x 2 − 4 y 2 − 6 x + 16 y = 25 \displaystyle 3x^2-4y^2-6x+16y=25 3 x 2 − 4 y 2 − 6 x + 16 y = 25 y \displaystyle y y x \displaystyle x x x \displaystyle x x 6 x − 8 y ⋅ y ′ − 6 + 16 y ′ = 0 ( y ≠ 2 ) \displaystyle 6x-8y\cdot y^{\prime}-6+16y^{\prime}=0\ (y\ne2) 6 x − 8 y ⋅ y ′ − 6 + 16 y ′ = 0 ( y = 2 ) 이므로 점 ( 5 , 5 ) \displaystyle (5,5) ( 5 , 5 ) x = 5 \displaystyle x=5 x = 5 y = 5 \displaystyle y=5 y = 5 30 − 40 y ′ − 6 + 16 y ′ = 0 \displaystyle 30-40y^{\prime}-6+16y^{\prime}=0 30 − 40 y ′ − 6 + 16 y ′ = 0 이므로 y ′ = 1 \displaystyle y^{\prime}=1 y ′ = 1 l : y = x \displaystyle l:y=x l : y = x 이다.
+4점 · 접선과 y = 2 \displaystyle y=2 y = 2 S \displaystyle S S S = 9 2 − ∫ 3 5 { 3 4 ( x − 1 ) 2 − 3 } d x \displaystyle S=\frac{9}{2}-\int_3^5\left\{\sqrt{\frac{3}{4}(x-1)^2-3}\right\}dx S = 2 9 − ∫ 3 5 { 4 3 ( x − 1 ) 2 − 3 } d x 그러므로 접선과 직선 y = 2 \displaystyle y=2 y = 2 ( 2 , 2 ) \displaystyle (2,2) ( 2 , 2 ) 3 x 2 − 4 y 2 − 6 x + 16 y = 25 \displaystyle 3x^2-4y^2-6x+16y=25 3 x 2 − 4 y 2 − 6 x + 16 y = 25 ( x − 1 ) 2 4 − ( y − 2 ) 2 3 = 1 \displaystyle \frac{(x-1)^2}{4}-\frac{(y-2)^2}{3}=1 4 ( x − 1 ) 2 − 3 ( y − 2 ) 2 = 1 이므로 y = ± 3 4 ( x − 1 ) 2 − 3 + 2 \displaystyle y=\pm\sqrt{\frac{3}{4}(x-1)^2-3}+2 y = ± 4 3 ( x − 1 ) 2 − 3 + 2 따라서
S = 1 2 ⋅ 3 ⋅ 3 − ∫ 3 5 [ { 3 4 ( x − 1 ) 2 − 3 + 2 } − 2 ] d x = 9 2 − ∫ 3 5 { 3 4 ( x − 1 ) 2 − 3 } d x \displaystyle \begin{aligned}S&=\frac{1}{2}\cdot3\cdot3-\int_3^5\left[\left\{\sqrt{\frac{3}{4}(x-1)^2-3}+2\right\}-2\right]dx\\&=\frac{9}{2}-\int_3^5\left\{\sqrt{\frac{3}{4}(x-1)^2-3}\right\}dx\end{aligned} S = 2 1 ⋅ 3 ⋅ 3 − ∫ 3 5 [ { 4 3 ( x − 1 ) 2 − 3 + 2 } − 2 ] d x = 2 9 − ∫ 3 5 { 4 3 ( x − 1 ) 2 − 3 } d x 이다.
+3점 · 정적분의 치환적분법을 이용할 수 있다. ∫ 3 5 { 3 4 ( x − 1 ) 2 − 3 } d x = ∫ 2 4 3 4 t 2 − 3 d t = 3 2 ∫ 2 4 t 2 − 4 d t \displaystyle \int_3^5\left\{\sqrt{\frac{3}{4}(x-1)^2-3}\right\}dx=\int_2^4\sqrt{\frac{3}{4}t^2-3}\,dt=\frac{\sqrt{3}}{2}\int_2^4\sqrt{t^2-4}\,dt ∫ 3 5 { 4 3 ( x − 1 ) 2 − 3 } d x = ∫ 2 4 4 3 t 2 − 3 d t = 2 3 ∫ 2 4 t 2 − 4 d t 여기서 ∫ 3 5 { 3 4 ( x − 1 ) 2 − 3 } d x \displaystyle \int_3^5\left\{\sqrt{\frac{3}{4}(x-1)^2-3}\right\}dx ∫ 3 5 { 4 3 ( x − 1 ) 2 − 3 } d x x − 1 = t \displaystyle x-1=t x − 1 = t d t d x = 1 \displaystyle \frac{dt}{dx}=1 d x d t = 1 x = 3 \displaystyle x=3 x = 3 t = 2 \displaystyle t=2 t = 2 x = 5 \displaystyle x=5 x = 5 t = 4 \displaystyle t=4 t = 4 ∫ 3 5 { 3 4 ( x − 1 ) 2 − 3 } d x = ∫ 2 4 3 4 t 2 − 3 d t = 3 2 ∫ 2 4 t 2 − 4 d t \displaystyle \int_3^5\left\{\sqrt{\frac{3}{4}(x-1)^2-3}\right\}dx=\int_2^4\sqrt{\frac{3}{4}t^2-3}\,dt=\frac{\sqrt{3}}{2}\int_2^4\sqrt{t^2-4}\,dt ∫ 3 5 { 4 3 ( x − 1 ) 2 − 3 } d x = ∫ 2 4 4 3 t 2 − 3 d t = 2 3 ∫ 2 4 t 2 − 4 d t 이다.
+5점 · [1-1]의 결과를 이용하여 S = − 3 2 + 3 ln ( 2 + 3 ) \displaystyle S=-\frac{3}{2}+\sqrt{3}\ln\left(2+\sqrt{3}\right) S = − 2 3 + 3 ln ( 2 + 3 ) [1-1]의 결과를 이용하면 ∫ 2 4 3 2 t 2 − 4 d t = 3 4 [ t t 2 − 4 − 4 ln ∣ t + t 2 − 4 ∣ ] 2 4 = 3 4 { 8 3 − 4 ln ( 4 + 2 3 ) + 4 ln 2 } = 3 { 2 3 − ln ( 4 + 2 3 ) + ln 2 } = 3 { 2 3 − ln ( 2 + 3 ) } \displaystyle \begin{aligned}\int_2^4\frac{\sqrt{3}}{2}\sqrt{t^2-4}\,dt&=\frac{\sqrt{3}}{4}\Biggl[t\sqrt{t^2-4}-4\ln\left|t+\sqrt{t^2-4}\right|\Biggr]_2^4\\&=\frac{\sqrt{3}}{4}\left\{8\sqrt{3}-4\ln\left(4+2\sqrt{3}\right)+4\ln2\right\}\\&=\sqrt{3}\left\{2\sqrt{3}-\ln\left(4+2\sqrt{3}\right)+\ln2\right\}\\&=\sqrt{3}\left\{2\sqrt{3}-\ln\left(2+\sqrt{3}\right)\right\}\end{aligned} ∫ 2 4 2 3 t 2 − 4 d t = 4 3 [ t t 2 − 4 − 4 ln t + t 2 − 4 ] 2 4 = 4 3 { 8 3 − 4 ln ( 4 + 2 3 ) + 4 ln 2 } = 3 { 2 3 − ln ( 4 + 2 3 ) + ln 2 } = 3 { 2 3 − ln ( 2 + 3 ) } 이다. 따라서 S = 9 2 − 3 { 2 3 − ln ( 2 + 3 ) } = − 3 2 + 3 ln ( 2 + 3 ) \displaystyle S=\frac{9}{2}-\sqrt{3}\left\{2\sqrt{3}-\ln\left(2+\sqrt{3}\right)\right\}=-\frac{3}{2}+\sqrt{3}\ln\left(2+\sqrt{3}\right) S = 2 9 − 3 { 2 3 − ln ( 2 + 3 ) } = − 2 3 + 3 ln ( 2 + 3 ) 이다.
수리논술, 배움에서 논증의 완성까지.