[문제 2-1]
+5점 · 영역을 정적분으로 표현한 후 정적분 값 − ln ( tan ( π 4 − θ 2 ) ) \displaystyle -\ln\left(\tan\left(\frac\pi4-\frac\theta2\right)\right) − ln ( tan ( 4 π − 2 θ ) ) 직선 l 1 \displaystyle l_1 l 1 x \displaystyle x x α \displaystyle \alpha α α = π 4 − θ 2 \displaystyle \alpha=\frac\pi4-\frac\theta2 α = 4 π − 2 θ y = ( tan α ) x \displaystyle y=(\tan\alpha)x y = ( tan α ) x y = 1 x \displaystyle y=\frac1x y = x 1 ( 1 tan α , tan α ) \displaystyle \left(\frac1{\sqrt{\tan\alpha}},\sqrt{\tan\alpha}\right) ( tan α 1 , tan α ) y = x \displaystyle y=x y = x
1 2 + ∫ 1 1 / tan α 1 x d x − 1 2 1 tan α tan α = − 1 2 ln ( tan α ) \displaystyle \frac12+\int_1^{1/\sqrt{\tan\alpha}}\frac1x\,dx-\frac12\frac1{\sqrt{\tan\alpha}}\sqrt{\tan\alpha}=-\frac12\ln(\tan\alpha) 2 1 + ∫ 1 1/ t a n α x 1 d x − 2 1 tan α 1 tan α = − 2 1 ln ( tan α )
를 구한다. 따라서 영역의 전체 넓이는
f ( θ ) = − ln ( tan α ) = − ln ( tan ( π 4 − θ 2 ) ) \displaystyle f(\theta)=-\ln(\tan\alpha)=-\ln\left(\tan\left(\frac\pi4-\frac\theta2\right)\right) f ( θ ) = − ln ( tan α ) = − ln ( tan ( 4 π − 2 θ ) )
+5점 · 합성함수의 미분법을 활용하여 θ = π 3 \displaystyle \theta=\frac\pi3 θ = 3 π 이다. 합성함수의 미분법을 이용하여 f \displaystyle f f
f ′ ( θ ) = − 1 tan ( π 4 − θ 2 ) sec 2 ( π 4 − θ 2 ) ( − 1 2 ) = 1 2 sin ( π 4 − θ 2 ) cos ( π 4 − θ 2 ) = 1 sin ( π 2 − θ ) = 1 cos θ \displaystyle \begin{aligned}f^{\prime}(\theta)&=-\frac1{\tan\left(\frac\pi4-\frac\theta2\right)}\sec^2\left(\frac\pi4-\frac\theta2\right)\left(-\frac12\right)\\&=\frac1{2\sin\left(\frac\pi4-\frac\theta2\right)\cos\left(\frac\pi4-\frac\theta2\right)}=\frac1{\sin\left(\frac\pi2-\theta\right)}=\frac1{\cos\theta}\end{aligned} f ′ ( θ ) = − tan ( 4 π − 2 θ ) 1 sec 2 ( 4 π − 2 θ ) ( − 2 1 ) = 2 sin ( 4 π − 2 θ ) cos ( 4 π − 2 θ ) 1 = sin ( 2 π − θ ) 1 = cos θ 1
계산하여 정답 θ = π 3 \displaystyle \theta=\frac\pi3 θ = 3 π
[문제 2-2]
+3점 · t = x + 1 − 1 \displaystyle t=\sqrt{x+1}-1 t = x + 1 − 1 t = x + 1 − 1 \displaystyle t=\sqrt{x+1}-1 t = x + 1 − 1 x + 1 + 1 \displaystyle \sqrt{x+1}+1 x + 1 + 1
x + 1 + 1 = x x + 1 − 1 = x t \displaystyle \sqrt{x+1}+1=\frac{x}{\sqrt{x+1}-1}=\frac xt x + 1 + 1 = x + 1 − 1 x = t x
+3점 · 함수를 정리하여 g ( t ) = t + 2 cos t \displaystyle g(t)=t+2\cos t g ( t ) = t + 2 cos t 를 대입한다. 그 결과는
f ( x ) = x + 1 + 1 x + 2 ( x + 1 + 1 ) cos ( x + 1 − 1 ) = 1 t + 2 cos t \displaystyle f(x)=\frac{\sqrt{x+1}+1}{x+2(\sqrt{x+1}+1)\cos(\sqrt{x+1}-1)}=\frac1{t+2\cos t} f ( x ) = x + 2 ( x + 1 + 1 ) cos ( x + 1 − 1 ) x + 1 + 1 = t + 2 cos t 1
이다. 따라서 g ( t ) = t + 2 cos t \displaystyle g(t)=t+2\cos t g ( t ) = t + 2 cos t 0 ≤ t ≤ π \displaystyle 0\le t\le\pi 0 ≤ t ≤ π t \displaystyle t t
+6점 · g ( x ) \displaystyle g(x) g ( x ) 함수 g ( t ) \displaystyle g(t) g ( t ) g ′ ( t ) = 1 − 2 sin t \displaystyle g^{\prime}(t)=1-2\sin t g ′ ( t ) = 1 − 2 sin t t = π 6 , 5 π 6 \displaystyle t=\frac\pi6,\frac{5\pi}6 t = 6 π , 6 5 π g ( t ) \displaystyle g(t) g ( t ) ( 0 , π 6 ) \displaystyle \left(0,\frac\pi6\right) ( 0 , 6 π ) ( π 6 , 5 π 6 ) \displaystyle \left(\frac\pi6,\frac{5\pi}6\right) ( 6 π , 6 5 π ) ( 5 π 6 , π ) \displaystyle \left(\frac{5\pi}6,\pi\right) ( 6 5 π , π ) g ( t ) \displaystyle g(t) g ( t ) t = 0 \displaystyle t=0 t = 0 t = 5 π 6 \displaystyle t=\frac{5\pi}6 t = 6 5 π
+3점 · 정답 x = ( 5 π 6 + 1 ) 2 − 1 \displaystyle x=\left(\frac{5\pi}6+1\right)^2-1 x = ( 6 5 π + 1 ) 2 − 1 실제 g ( 0 ) = 1 \displaystyle g(0)=1 g ( 0 ) = 1 g ( 5 π 6 ) = 5 π 6 − 3 2 \displaystyle g\left(\frac{5\pi}6\right)=\frac{5\pi}6-\frac{\sqrt3}2 g ( 6 5 π ) = 6 5 π − 2 3 g ( t ) \displaystyle g(t) g ( t ) t = 5 π 6 \displaystyle t=\frac{5\pi}6 t = 6 5 π 5 π 12 − 3 2 \displaystyle \frac{5\pi}{12}-\frac{\sqrt3}2 12 5 π − 2 3 π ≤ 3.5 \displaystyle \pi\le3.5 π ≤ 3.5 3 ≥ 1.5 \displaystyle \sqrt3\ge1.5 3 ≥ 1.5 x = ( t + 1 ) 2 − 1 = ( 5 π 6 + 1 ) 2 − 1 \displaystyle x=(t+1)^2-1=\left(\frac{5\pi}6+1\right)^2-1 x = ( t + 1 ) 2 − 1 = ( 6 5 π + 1 ) 2 − 1 Correction. [2-2] 대학 답안에서 g ( t ) \displaystyle g(t) g ( t ) g ( t ) = t + 2 cos t \displaystyle g(t)=t+2\cos t g ( t ) = t + 2 cos t g ( 0 ) = 2 \displaystyle g(0)=2 g ( 0 ) = 2 g ( 5 π / 6 ) = 5 π / 6 − 3 \displaystyle g(5\pi/6)=5\pi/6-\sqrt3 g ( 5 π /6 ) = 5 π /6 − 3 g ( t ) / 2 \displaystyle g(t)/2 g ( t ) /2
수리논술, 배움에서 논증의 완성까지.