[문제 3-1]
+3점 · 교점 x = 0 , e − 1 , e 5 − 1 \displaystyle x=0,\sqrt{e-1},\sqrt{e^5-1} x = 0 , e − 1 , e 5 − 1 f ( x ) = x x 2 + 1 ( ln ( x 2 + 1 ) − 1 ) ( ln ( x 2 + 1 ) − 5 ) \displaystyle f(x)=\frac x{x^2+1}(\ln(x^2+1)-1)(\ln(x^2+1)-5) f ( x ) = x 2 + 1 x ( ln ( x 2 + 1 ) − 1 ) ( ln ( x 2 + 1 ) − 5 ) x = 0 , e − 1 , e 5 − 1 \displaystyle x=0,\sqrt{e-1},\sqrt{e^5-1} x = 0 , e − 1 , e 5 − 1 f ( x ) = 0 \displaystyle f(x)=0 f ( x ) = 0
+3점 · 그래프의 개형을 이해하고 ∫ 0 e − 1 f ( x ) d x − ∫ e − 1 e 5 − 1 f ( x ) d x \displaystyle \int_0^{\sqrt{e-1}}f(x)dx-\int_{\sqrt{e-1}}^{\sqrt{e^5-1}}f(x)dx ∫ 0 e − 1 f ( x ) d x − ∫ e − 1 e 5 − 1 f ( x ) d x 0 ≤ x ≤ e − 1 \displaystyle 0\le x\le\sqrt{e-1} 0 ≤ x ≤ e − 1 f ( x ) ≥ 0 \displaystyle f(x)\ge0 f ( x ) ≥ 0 e − 1 ≤ x ≤ e 5 − 1 \displaystyle \sqrt{e-1}\le x\le\sqrt{e^5-1} e − 1 ≤ x ≤ e 5 − 1 f ( x ) ≤ 0 \displaystyle f(x)\le0 f ( x ) ≤ 0 ∫ 0 e − 1 f ( x ) d x − ∫ e − 1 e 5 − 1 f ( x ) d x \displaystyle \int_0^{\sqrt{e-1}}f(x)dx-\int_{\sqrt{e-1}}^{\sqrt{e^5-1}}f(x)dx ∫ 0 e − 1 f ( x ) d x − ∫ e − 1 e 5 − 1 f ( x ) d x
+4점 · 치환 적분을 하여 13 2 \displaystyle \frac{13}2 2 13 t = ln ( x 2 + 1 ) \displaystyle t=\ln(x^2+1) t = ln ( x 2 + 1 ) 1 2 ∫ 0 1 ( t 2 − 6 t + 5 ) d t − 1 2 ∫ 1 5 ( t 2 − 6 t + 5 ) d t = 7 6 + 16 3 = 13 2 \displaystyle \frac12\int_0^1(t^2-6t+5)dt-\frac12\int_1^5(t^2-6t+5)dt=\frac76+\frac{16}3=\frac{13}2 2 1 ∫ 0 1 ( t 2 − 6 t + 5 ) d t − 2 1 ∫ 1 5 ( t 2 − 6 t + 5 ) d t = 6 7 + 3 16 = 2 13 이다.
[문제 3-2]
+8점 · f ( t ) = 2 t + 2 t 2 + t + 1 \displaystyle f(t)=\frac{2\sqrt{t+2}}{t^2+t+1} f ( t ) = t 2 + t + 1 2 t + 2 f ′ ( t ) = − 3 ( t 2 + 3 t + 1 ) ( t 2 + t + 1 ) 2 t + 2 \displaystyle f^{\prime}(t)=-\frac{3(t^2+3t+1)}{(t^2+t+1)^2\sqrt{t+2}} f ′ ( t ) = − ( t 2 + t + 1 ) 2 t + 2 3 ( t 2 + 3 t + 1 ) 점 P의 좌표는 ( t , t + 2 ) \displaystyle (t,\sqrt{t+2}) ( t , t + 2 ) ∠ P A B = α \displaystyle \angle\mathrm{PAB}=\alpha ∠ PAB = α ∠ P B A = β \displaystyle \angle\mathrm{PBA}=\beta ∠ PBA = β tan α = t + 2 1 + t \displaystyle \tan\alpha=\frac{\sqrt{t+2}}{1+t} tan α = 1 + t t + 2 tan β = t + 2 1 − t \displaystyle \tan\beta=\frac{\sqrt{t+2}}{1-t} tan β = 1 − t t + 2 θ = π − α − β \displaystyle \theta=\pi-\alpha-\beta θ = π − α − β tan ( θ ) = f ( t ) \displaystyle \tan(\theta)=f(t) tan ( θ ) = f ( t ) f ( t ) = 2 t + 2 t 2 + t + 1 \displaystyle f(t)=\frac{2\sqrt{t+2}}{t^2+t+1} f ( t ) = t 2 + t + 1 2 t + 2 f ′ ( t ) = − 3 ( t 2 + 3 t + 1 ) ( t 2 + t + 1 ) 2 t + 2 \displaystyle f^{\prime}(t)=-\frac{3(t^2+3t+1)}{(t^2+t+1)^2\sqrt{t+2}} f ′ ( t ) = − ( t 2 + t + 1 ) 2 t + 2 3 ( t 2 + 3 t + 1 )
+4점 · t = − 3 ± 5 2 \displaystyle t=\frac{-3\pm\sqrt5}2 t = 2 − 3 ± 5 t = − 3 + 5 2 \displaystyle t=\frac{-3+\sqrt5}2 t = 2 − 3 + 5 t = − 3 ± 5 2 \displaystyle t=\frac{-3\pm\sqrt5}2 t = 2 − 3 ± 5 − 3 − 5 2 < − 1 < − 3 + 5 2 < 1 \displaystyle \frac{-3-\sqrt5}2<-1<\frac{-3+\sqrt5}2<1 2 − 3 − 5 < − 1 < 2 − 3 + 5 < 1 t = − 3 + 5 2 \displaystyle t=\frac{-3+\sqrt5}2 t = 2 − 3 + 5
+3점 · 최댓값 11 + 5 5 2 \displaystyle \frac{11+5\sqrt5}2 2 11 + 5 5 tan 2 θ \displaystyle \tan^2\theta tan 2 θ { f ( − 3 + 5 2 ) } 2 = 11 + 5 5 2 \displaystyle \left\{f\left(\frac{-3+\sqrt5}2\right)\right\}^2=\frac{11+5\sqrt5}2 { f ( 2 − 3 + 5 ) } 2 = 2 11 + 5 5
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