[문제 2-1]
+3점 · x = 1 , 5 \displaystyle x=1,5 x = 1 , 5 f ′ ( x ) = 0 \displaystyle f^{\prime}(x)=0 f ′ ( x ) = 0 f ′ ( x ) = ( x 2 + 5 ) { 1 − 2 sin ( π x x 2 + 5 ) } \displaystyle f^{\prime}(x)=(x^2+5)\left\{1-2\sin\left(\frac{\pi x}{x^2+5}\right)\right\} f ′ ( x ) = ( x 2 + 5 ) { 1 − 2 sin ( x 2 + 5 π x ) } f ( x ) \displaystyle f(x) f ( x ) π x x 2 + 5 = π 6 + 2 n π \displaystyle \frac{\pi x}{x^2+5}=\frac\pi6+2n\pi x 2 + 5 π x = 6 π + 2 nπ π x x 2 + 5 = 5 π 6 + 2 n π \displaystyle \frac{\pi x}{x^2+5}=\frac{5\pi}6+2n\pi x 2 + 5 π x = 6 5 π + 2 nπ f ′ ( 0 ) ≠ 0 \displaystyle f^{\prime}(0)\ne0 f ′ ( 0 ) = 0 x ≠ 0 \displaystyle x\ne0 x = 0 ∣ π x x 2 + 5 ∣ = π ∣ x ∣ + 5 ∣ x ∣ ≤ π 2 5 < π 4 \displaystyle \left|\frac{\pi x}{x^2+5}\right|=\frac\pi{|x|+\frac5{|x|}}\le\frac\pi{2\sqrt5}<\frac\pi4 x 2 + 5 π x = ∣ x ∣ + ∣ x ∣ 5 π ≤ 2 5 π < 4 π 이므로, π x 5 + x 2 = π 6 \displaystyle \frac{\pi x}{5+x^2}=\frac\pi6 5 + x 2 π x = 6 π x = 1 , 5 \displaystyle x=1,5 x = 1 , 5
+4점 · f ( x ) \displaystyle f(x) f ( x ) f ( 1 ) \displaystyle f(1) f ( 1 ) 도함수 f ′ ( x ) \displaystyle f^{\prime}(x) f ′ ( x ) g ( x ) = 1 − 2 sin ( π x x 2 + 5 ) \displaystyle g(x)=1-2\sin\left(\frac{\pi x}{x^2+5}\right) g ( x ) = 1 − 2 sin ( x 2 + 5 π x ) g ′ ( x ) = 2 cos ( π x x 2 + 5 ) π ( x 2 − 5 ) ( x 2 + 5 ) 2 \displaystyle g^{\prime}(x)=2\cos\left(\frac{\pi x}{x^2+5}\right)\frac{\pi(x^2-5)}{(x^2+5)^2} g ′ ( x ) = 2 cos ( x 2 + 5 π x ) ( x 2 + 5 ) 2 π ( x 2 − 5 ) g ( x ) \displaystyle g(x) g ( x ) ( − 1 , 5 ) \displaystyle (-1,\sqrt5) ( − 1 , 5 ) ( 5 , 5 ) \displaystyle (\sqrt5,5) ( 5 , 5 ) g ( 1 ) = g ( 5 ) = 0 \displaystyle g(1)=g(5)=0 g ( 1 ) = g ( 5 ) = 0 ( − 1 , 1 ) \displaystyle (-1,1) ( − 1 , 1 ) g ( x ) > 0 \displaystyle g(x)>0 g ( x ) > 0 f ( x ) \displaystyle f(x) f ( x ) ( 1 , 5 ) \displaystyle (1,5) ( 1 , 5 ) g ( x ) < 0 \displaystyle g(x)<0 g ( x ) < 0 f ( x ) \displaystyle f(x) f ( x ) x = 1 \displaystyle x=1 x = 1 f ( x ) \displaystyle f(x) f ( x )
+3점 · f ( 1 ) = 32 3 \displaystyle f(1)=\frac{32}3 f ( 1 ) = 3 32 마지막으로 ( x 2 + 5 ) sin ( π x x 2 + 5 ) \displaystyle (x^2+5)\sin\left(\frac{\pi x}{x^2+5}\right) ( x 2 + 5 ) sin ( x 2 + 5 π x ) ∫ − 1 1 ( x 2 + 5 ) sin ( π x x 2 + 5 ) d x = 0 \displaystyle \int_{-1}^1(x^2+5)\sin\left(\frac{\pi x}{x^2+5}\right)dx=0 ∫ − 1 1 ( x 2 + 5 ) sin ( x 2 + 5 π x ) d x = 0 f ( 1 ) = ∫ − 1 1 ( t 2 + 5 ) d t = 32 3 \displaystyle f(1)=\int_{-1}^1(t^2+5)dt=\frac{32}3 f ( 1 ) = ∫ − 1 1 ( t 2 + 5 ) d t = 3 32
[문제 2-2]
+5점 · ( d x d t ) 2 + ( d y d t ) 2 = sin t ( 10 − sin 2 t ) 3 ( 8 + sin 2 t ) \displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\frac{\sin t(10-\sin^2t)}{\sqrt3(8+\sin^2t)} ( d t d x ) 2 + ( d t d y ) 2 = 3 ( 8 + sin 2 t ) sin t ( 10 − sin 2 t ) x \displaystyle x x y \displaystyle y y t \displaystyle t t ( d x d t ) 2 + ( d y d t ) 2 = ( − sin t 3 ) 2 + ( 2 3 sin t cos t 8 + sin 2 t ) 2 = sin t 3 ( 8 + sin 2 t ) ( 8 + sin 2 t ) 2 + 36 cos 2 t = sin t ( 10 − sin 2 t ) 3 ( 8 + sin 2 t ) \displaystyle \begin{aligned}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}&=\sqrt{\left(-\frac{\sin t}{\sqrt3}\right)^2+\left(\frac{2\sqrt3\sin t\cos t}{8+\sin^2t}\right)^2}\\&=\frac{\sin t}{\sqrt3(8+\sin^2t)}\sqrt{(8+\sin^2t)^2+36\cos^2t}=\frac{\sin t(10-\sin^2t)}{\sqrt3(8+\sin^2t)}\end{aligned} ( d t d x ) 2 + ( d t d y ) 2 = ( − 3 sin t ) 2 + ( 8 + sin 2 t 2 3 sin t cos t ) 2 = 3 ( 8 + sin 2 t ) sin t ( 8 + sin 2 t ) 2 + 36 cos 2 t = 3 ( 8 + sin 2 t ) sin t ( 10 − sin 2 t ) 를 구한다.
+5점 · u = cos t \displaystyle u=\cos t u = cos t ∫ − 1 1 9 + u 2 3 ( 9 − u 2 ) d u \displaystyle \int_{-1}^1\frac{9+u^2}{\sqrt3(9-u^2)}du ∫ − 1 1 3 ( 9 − u 2 ) 9 + u 2 d u 따라서 u = cos t \displaystyle u=\cos t u = cos t ∫ 0 π ( d x d t ) 2 + ( d y d t ) 2 d t = ∫ 0 π sin t ( 9 + cos 2 t ) 3 ( 9 − cos 2 t ) d t = ∫ − 1 1 9 + u 2 3 ( 9 − u 2 ) d u = 2 3 ∫ 0 1 9 + u 2 9 − u 2 d u \displaystyle \begin{aligned}\int_0^\pi\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt&=\int_0^\pi\frac{\sin t(9+\cos^2t)}{\sqrt3(9-\cos^2t)}dt=\int_{-1}^1\frac{9+u^2}{\sqrt3(9-u^2)}du\\&=\frac2{\sqrt3}\int_0^1\frac{9+u^2}{9-u^2}du\end{aligned} ∫ 0 π ( d t d x ) 2 + ( d t d y ) 2 d t = ∫ 0 π 3 ( 9 − cos 2 t ) sin t ( 9 + cos 2 t ) d t = ∫ − 1 1 3 ( 9 − u 2 ) 9 + u 2 d u = 3 2 ∫ 0 1 9 − u 2 9 + u 2 d u
+5점 · 정적분을 2 3 ∫ 0 1 ( − 1 + 3 3 − u + 3 3 + u ) d u \displaystyle \frac2{\sqrt3}\int_0^1\left(-1+\frac3{3-u}+\frac3{3+u}\right)du 3 2 ∫ 0 1 ( − 1 + 3 − u 3 + 3 + u 3 ) d u 2 3 ( − 1 + 3 ln 2 ) \displaystyle \frac2{\sqrt3}(-1+3\ln2) 3 2 ( − 1 + 3 ln 2 ) = 2 3 ∫ 0 1 ( − 1 + 3 3 − u + 3 3 + u ) d u = 2 3 ( − 1 + 3 ln 2 ) = − 2 3 + 2 3 ln 2 \displaystyle =\frac2{\sqrt3}\int_0^1\left(-1+\frac3{3-u}+\frac3{3+u}\right)du=\frac2{\sqrt3}(-1+3\ln2)=-\frac2{\sqrt3}+2\sqrt3\ln2 = 3 2 ∫ 0 1 ( − 1 + 3 − u 3 + 3 + u 3 ) d u = 3 2 ( − 1 + 3 ln 2 ) = − 3 2 + 2 3 ln 2 를 구한다.
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