[문제 2-1]
주어진 f ( x ) \displaystyle f(x) f ( x ) f ′ ( x ) = 3 x 2 + 2 m x + n \displaystyle f^{\prime}(x)=3x^2+2mx+n f ′ ( x ) = 3 x 2 + 2 m x + n f ′ ( x ) = 0 \displaystyle f^{\prime}(x)=0 f ′ ( x ) = 0 D / 4 = m 2 − 3 n > 0 \displaystyle D/4=m^2-3n>0 D /4 = m 2 − 3 n > 0 α + β = − 2 m 3 \displaystyle \alpha+\beta=-\frac{2m}3 α + β = − 3 2 m α β = n 3 \displaystyle \alpha\beta=\frac n3 α β = 3 n
+4점 · 1 α + 1 β = α + β α β = − 2 m n \displaystyle \frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=-\frac{2m}n α 1 + β 1 = α β α + β = − n 2 m f ( 1 ) = 11 \displaystyle f(1)=11 f ( 1 ) = 11 m + n + k = 10 \displaystyle m+n+k=10 m + n + k = 10 m \displaystyle m m n \displaystyle n n k \displaystyle k k 1 ≤ m , n , k ≤ 8 \displaystyle 1\le m,n,k\le8 1 ≤ m , n , k ≤ 8 m + n ≤ 9 \displaystyle m+n\le9 m + n ≤ 9 1 α + 1 β = α + β α β = − 2 m n \displaystyle \frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=-\frac{2m}n α 1 + β 1 = α β α + β = − n 2 m n \displaystyle n n m \displaystyle m m
+4점 · m \displaystyle m m n \displaystyle n n m 2 − 3 n > 0 \displaystyle m^2-3n>0 m 2 − 3 n > 0 m + n ≤ 9 \displaystyle m+n\le9 m + n ≤ 9 m \displaystyle m m n \displaystyle n n (1) m = 1 \displaystyle m=1 m = 1 n \displaystyle n n m = 2 \displaystyle m=2 m = 2 n = 1 \displaystyle n=1 n = 1 m = 3 \displaystyle m=3 m = 3 n = 1 , 2 \displaystyle n=1,2 n = 1 , 2 (4) m = 4 \displaystyle m=4 m = 4 n = 1 , 2 , … , 5 \displaystyle n=1,2,\ldots,5 n = 1 , 2 , … , 5 m = 5 \displaystyle m=5 m = 5 n = 1 , … , 4 \displaystyle n=1,\ldots,4 n = 1 , … , 4 (6) m = 6 \displaystyle m=6 m = 6 n = 1 , 2 , 3 \displaystyle n=1,2,3 n = 1 , 2 , 3 m = 7 \displaystyle m=7 m = 7 n = 1 , 2 \displaystyle n=1,2 n = 1 , 2 m = 8 \displaystyle m=8 m = 8 n = 1 \displaystyle n=1 n = 1
+2점 · 모든 계산을 정확히 계산하여 m = 4 \displaystyle m=4 m = 4 n = 5 \displaystyle n=5 n = 5 k = 1 \displaystyle k=1 k = 1 이 중 m = 4 \displaystyle m=4 m = 4 n = 5 \displaystyle n=5 n = 5 − 2 m n = − 8 5 \displaystyle -\frac{2m}n=-\frac85 − n 2 m = − 5 8 k = 1 \displaystyle k=1 k = 1 따라서, f ( x ) = x 3 + 4 x 2 + 5 x + 1 \displaystyle f(x)=x^3+4x^2+5x+1 f ( x ) = x 3 + 4 x 2 + 5 x + 1
[문제 2-2]
+3점 · a = − 3 \displaystyle a=-3 a = − 3 x > 0 \displaystyle x>0 x > 0 f ′ ( x ) = 3 cos x ( 4 sin 2 x − 1 ) \displaystyle f^{\prime}(x)=3\cos x(4\sin^2x-1) f ′ ( x ) = 3 cos x ( 4 sin 2 x − 1 ) x < 0 \displaystyle x<0 x < 0 f ′ ( x ) = 9 π 2 x 2 + a \displaystyle f^{\prime}(x)=\frac9{\pi^2}x^2+a f ′ ( x ) = π 2 9 x 2 + a f ( x ) \displaystyle f(x) f ( x ) x = 0 \displaystyle x=0 x = 0 lim h → 0 + f ( h ) − f ( 0 ) h = lim h → 0 − f ( h ) − f ( 0 ) h \displaystyle \lim_{h\to0+}\frac{f(h)-f(0)}h=\lim_{h\to0-}\frac{f(h)-f(0)}h h → 0 + lim h f ( h ) − f ( 0 ) = h → 0 − lim h f ( h ) − f ( 0 ) 따라서, a = − 3 \displaystyle a=-3 a = − 3
(i) − π ≤ x < 0 \displaystyle -\pi\le x<0 − π ≤ x < 0 f ( x ) = 3 π 2 x ( x 2 − π 2 ) \displaystyle f(x)=\frac3{\pi^2}x(x^2-\pi^2) f ( x ) = π 2 3 x ( x 2 − π 2 ) f ( − π ) = 0 \displaystyle f(-\pi)=0 f ( − π ) = 0 f ′ ( x ) = 9 π 2 x 2 − 3 \displaystyle f^{\prime}(x)=\frac9{\pi^2}x^2-3 f ′ ( x ) = π 2 9 x 2 − 3 f ′ ( − π 3 ) = 0 \displaystyle f^{\prime}\left(-\frac\pi{\sqrt3}\right)=0 f ′ ( − 3 π ) = 0 f ( − π 3 ) = 2 π 3 \displaystyle f\left(-\frac\pi{\sqrt3}\right)=\frac{2\pi}{\sqrt3} f ( − 3 π ) = 3 2 π
(ii) 0 ≤ x ≤ π \displaystyle 0\le x\le\pi 0 ≤ x ≤ π f ( x ) = sin x ( 4 sin 2 x − 3 ) \displaystyle f(x)=\sin x(4\sin^2x-3) f ( x ) = sin x ( 4 sin 2 x − 3 ) f ( 0 ) = f ( π 3 ) = f ( 2 π 3 ) = f ( π ) = 0 \displaystyle f(0)=f\left(\frac\pi3\right)=f\left(\frac{2\pi}3\right)=f(\pi)=0 f ( 0 ) = f ( 3 π ) = f ( 3 2 π ) = f ( π ) = 0 f ′ ( x ) = 3 cos x ( 4 sin 2 x − 1 ) \displaystyle f^{\prime}(x)=3\cos x(4\sin^2x-1) f ′ ( x ) = 3 cos x ( 4 sin 2 x − 1 ) f ′ ( π 6 ) = f ′ ( π 2 ) = f ′ ( 5 π 6 ) = 0 \displaystyle f^{\prime}\left(\frac\pi6\right)=f^{\prime}\left(\frac\pi2\right)=f^{\prime}\left(\frac{5\pi}6\right)=0 f ′ ( 6 π ) = f ′ ( 2 π ) = f ′ ( 6 5 π ) = 0 이를 이용한 증가와 감소를 나타낸 표는 다음과 같다.
+7점 · 증가와 감소 표 혹은 그래프 개형 혹은 극점과 극값을 바르게 구하면 x \displaystyle x x − π \displaystyle -\pi − π … − π 3 \displaystyle -\frac\pi{\sqrt3} − 3 π … 0 … π 6 \displaystyle \frac\pi6 6 π … π 3 \displaystyle \frac\pi3 3 π … π 2 \displaystyle \frac\pi2 2 π … 2 π 3 \displaystyle \frac{2\pi}3 3 2 π … 5 π 6 \displaystyle \frac{5\pi}6 6 5 π … π \displaystyle \pi π f ′ ( x ) \displaystyle f^{\prime}(x) f ′ ( x ) + 0 − − 0 + + 0 − − 0 + f ( x ) \displaystyle f(x) f ( x ) 0 ↗ 2 3 π \displaystyle \frac2{\sqrt3}\pi 3 2 π ↘ 0 ↘ −1 ↗ 0 ↗ 1 ↘ 0 ↘ −1 ↗ 0
부호 변화를 고려한 극솟값은 f ( π 6 ) = f ( 5 π 6 ) = − 1 \displaystyle f\left(\frac\pi6\right)=f\left(\frac{5\pi}6\right)=-1 f ( 6 π ) = f ( 6 5 π ) = − 1 f ( π 2 ) = 1 \displaystyle f\left(\frac\pi2\right)=1 f ( 2 π ) = 1 이를 그래프로 그리면
+5점 · 불연속인 k = − 1 , 0 , 1 , 2 3 π \displaystyle k=-1,0,1,\frac2{\sqrt3}\pi k = − 1 , 0 , 1 , 3 2 π 위 그래프의 개형으로부터 불연속인 k \displaystyle k k lim k → − 1 − 0 g ( k ) = 0 \displaystyle \lim_{k\to-1-0}g(k)=0 k → − 1 − 0 lim g ( k ) = 0 lim k → − 1 + 0 g ( k ) = 4 \displaystyle \lim_{k\to-1+0}g(k)=4 k → − 1 + 0 lim g ( k ) = 4 g ( − 1 ) = 2 \displaystyle g(-1)=2 g ( − 1 ) = 2 k = − 1 \displaystyle k=-1 k = − 1 lim k → 0 − g ( k ) = lim k → 0 + g ( k ) = 4 \displaystyle \lim_{k\to0-}g(k)=\lim_{k\to0+}g(k)=4 k → 0 − lim g ( k ) = k → 0 + lim g ( k ) = 4 g ( 0 ) = 5 \displaystyle g(0)=5 g ( 0 ) = 5 k = 0 \displaystyle k=0 k = 0 lim k → 1 − 0 g ( k ) = 4 \displaystyle \lim_{k\to1-0}g(k)=4 k → 1 − 0 lim g ( k ) = 4 lim k → 1 + 0 g ( k ) = 2 \displaystyle \lim_{k\to1+0}g(k)=2 k → 1 + 0 lim g ( k ) = 2 g ( 1 ) = 3 \displaystyle g(1)=3 g ( 1 ) = 3 k = 1 \displaystyle k=1 k = 1 lim k → 2 π 3 − 0 g ( k ) = 2 \displaystyle \lim_{k\to\frac{2\pi}{\sqrt3}-0}g(k)=2 k → 3 2 π − 0 lim g ( k ) = 2 lim k → 2 π 3 + 0 g ( k ) = 0 \displaystyle \lim_{k\to\frac{2\pi}{\sqrt3}+0}g(k)=0 k → 3 2 π + 0 lim g ( k ) = 0 g ( 2 π 3 ) = 1 \displaystyle g\left(\frac{2\pi}{\sqrt3}\right)=1 g ( 3 2 π ) = 1 k = 2 3 π \displaystyle k=\frac2{\sqrt3}\pi k = 3 2 π 그러므로, 불연속인 k \displaystyle k k − 1 , 0 , 1 , 2 3 π \displaystyle -1,0,1,\frac2{\sqrt3}\pi − 1 , 0 , 1 , 3 2 π
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