[문제 3 − 1 \displaystyle 3- 1 3 − 1
+5점 · [문제 3] (1)
(1) 극한의 성질을 이용하여 k \displaystyle k k (1) 주어진 조건에 의해 g ( x ) \displaystyle g ( x ) g ( x ) x = 0 \displaystyle x = 0 x = 0
lim x → 0 − g ( x ) = lim x → 0 + g ( x ) , 2 f ( k ) = f ( 0 ) , 2 ∣ k 2 − k − 12 ∣ = 12 , ∣ k 2 − k − 12 ∣ = 6 \displaystyle \lim_{x \rightarrow 0 -} g ( x ) = \lim_{x \rightarrow 0 +} g ( x ) , 2 f ( k ) = f ( 0 ) , 2 \left| k^{2} - k - 12 \right| = 12 , \left| k^{2} - k - 12 \right| = 6 x → 0 − lim g ( x ) = x → 0 + lim g ( x ) , 2 f ( k ) = f ( 0 ) , 2 k 2 − k − 12 = 12 , k 2 − k − 12 = 6
+5점 · [문제 3] (1)
(2) (1) k < − 3 \displaystyle k < {} - 3 k < − 3 k > 4 \displaystyle k > 4 k > 4 k 2 − k − 18 = 0 \displaystyle k^{2} - k - 18 = 0 k 2 − k − 18 = 0 k \displaystyle k k − 3 ≤ k ≤ 4 \displaystyle - 3 \le k \le 4 − 3 ≤ k ≤ 4 k 2 − k − 6 = 0 , k = − 2 \displaystyle k^{2} - k - 6 = 0 , k = - 2 k 2 − k − 6 = 0 , k = − 2 k = 3 \displaystyle k = 3 k = 3 (2) ① k < − 3 \displaystyle k < {} - 3 k < − 3 k > 4 \displaystyle k > 4 k > 4 k 2 − k − 18 = 0 \displaystyle k^{2} - k - 18 = 0 k 2 − k − 18 = 0 k \displaystyle k k − 3 ≤ k ≤ 4 \displaystyle - 3 \le k \le 4 − 3 ≤ k ≤ 4 k 2 − k − 6 = 0 , k = − 2 \displaystyle k^{2} - k - 6 = 0 , k = - 2 k 2 − k − 6 = 0 , k = − 2 k = 3 \displaystyle k = 3 k = 3
+5점 · [문제 3] (1)
(3) (2) h ′ ( x ) \displaystyle h ^{\prime} ( x ) h ′ ( x ) h ( x ) \displaystyle h ( x ) h ( x ) (3) k = − 2 \displaystyle k = - 2 k = − 2
① 함수 g ( x ) \displaystyle g ( x ) g ( x ) x = − 1 , x = 0 , x = 4 \displaystyle x = - 1 , x = 0 , x = 4 x = − 1 , x = 0 , x = 4
g ( x ) h ( x ) \displaystyle g ( x ) h ( x ) g ( x ) h ( x ) h ( x ) = x ( x + 1 ) ( x − 4 ) = x 3 − 3 x 2 − 4 x \displaystyle h ( x ) = x ( x + 1 ) ( x - 4 ) = x^{3} - 3 x^{2} - 4 x h ( x ) = x ( x + 1 ) ( x − 4 ) = x 3 − 3 x 2 − 4 x
② h ′ ( x ) = 3 x 2 − 6 x − 4 \displaystyle h ^{\prime} ( x ) = 3 x^{2} - 6 x - 4 h ′ ( x ) = 3 x 2 − 6 x − 4 h ′ ( 4 ) = 20 \displaystyle h ^{\prime} ( 4 ) = 20 h ′ ( 4 ) = 20
+5점 · [문제 3] (1)
(4) (2) h ′ ( x ) \displaystyle h ^{\prime} ( x ) h ′ ( x ) h ( x ) \displaystyle h ( x ) h ( x ) (4) k = 3 \displaystyle k = 3 k = 3
① 함수 g ( x ) \displaystyle g ( x ) g ( x ) x = − 6 , x = 0 , x = 4 \displaystyle x = - 6 , x = 0 , x = 4 x = − 6 , x = 0 , x = 4
g ( x ) h ( x ) \displaystyle g ( x ) h ( x ) g ( x ) h ( x ) h ( x ) = x ( x + 6 ) ( x − 4 ) = x 3 + 2 x 2 − 24 x \displaystyle h ( x ) = x ( x + 6 ) ( x - 4 ) = x^{3} + 2 x^{2} - 24 x h ( x ) = x ( x + 6 ) ( x − 4 ) = x 3 + 2 x 2 − 24 x
② h ′ ( x ) = 3 x 2 + 4 x − 24 \displaystyle h ^{\prime} ( x ) = 3 x^{2} + 4 x - 24 h ′ ( x ) = 3 x 2 + 4 x − 24 h ′ ( 4 ) = 20 \displaystyle h ^{\prime} ( 4 ) = 20 h ′ ( 4 ) = 20
+5점 · [문제 3] (1)
(5)와 같이 함수 h ( x ) \displaystyle h ( x ) h ( x ) h ( k ) \displaystyle h ( k ) h ( k ) (5) 그러므로 h ( x ) = x 3 − 3 x 2 − 4 x \displaystyle h ( x ) = x^{3} - 3 x^{2} - 4 x h ( x ) = x 3 − 3 x 2 − 4 x h ( k ) = h ( − 2 ) = − 12 \displaystyle h ( k ) = h ( - 2 ) = - 12 h ( k ) = h ( − 2 ) = − 12
[문제 3 − 2 \displaystyle 3-2 3 − 2
+15점 · [문제 3] (2)
(1)과 같이 x = 2 \displaystyle x = 2 x = 2 j ( x ) = − 4 x − 4 \displaystyle j ( x ) = - 4 x - 4 j ( x ) = − 4 x − 4 (1) j ( x ) = a x + b \displaystyle j ( x ) = a x + b j ( x ) = a x + b
h ( 2 ) = j ( 2 ) , h ( 2 ) = − 12 , j ( 2 ) = 2 a + b = − 12 \displaystyle h ( 2 ) = j ( 2 ) , h ( 2 ) = - 12 , j ( 2 ) = 2 a + b = - 12 h ( 2 ) = j ( 2 ) , h ( 2 ) = − 12 , j ( 2 ) = 2 a + b = − 12
함수 y = h ( x ) \displaystyle y = h ( x ) y = h ( x ) y = j ( x ) \displaystyle y = j ( x ) y = j ( x ) x = 2 \displaystyle x = 2 x = 2 h ′ ( 2 ) = j ′ ( 2 ) \displaystyle h ^{\prime} ( 2 ) = j ^{\prime} ( 2 ) h ′ ( 2 ) = j ′ ( 2 )
h ′ ( x ) = 3 x 2 − 6 x − 4 \displaystyle h ^{\prime} ( x ) = 3 x^{2} - 6 x - 4 h ′ ( x ) = 3 x 2 − 6 x − 4 h ′ ( 2 ) = − 4 , j ′ ( 2 ) = a ∴ a = − 4 , b = − 4 \displaystyle h ^{\prime} ( 2 ) = - 4 , j ^{\prime} ( 2 ) = a \therefore a = - 4 , b = - 4 h ′ ( 2 ) = − 4 , j ′ ( 2 ) = a ∴ a = − 4 , b = − 4
j ( x ) = − 4 x − 4 \displaystyle j ( x ) = - 4 x - 4 j ( x ) = − 4 x − 4
+10점 · [문제 3] (2)
(2)와 같이 도함수를 이용하여 극값을 갖는 x \displaystyle x x (2) D ( x ) = x 3 − 3 x 2 − 4 x − ( − 4 x − 4 ) = x 3 − 3 x 2 + 4 \displaystyle D ( x ) = x^{3} - 3 x^{2} - 4 x - ( - 4 x - 4 ) = x^{3} - 3 x^{2} + 4 D ( x ) = x 3 − 3 x 2 − 4 x − ( − 4 x − 4 ) = x 3 − 3 x 2 + 4
D ′ ( x ) = 3 x 2 − 6 x = 0 \displaystyle D ^{\prime} ( x ) = 3 x^{2} - 6 x = 0 D ′ ( x ) = 3 x 2 − 6 x = 0 x \displaystyle x x 0 \displaystyle 0 0 2 \displaystyle 2 2
x \displaystyle x x ⋯ \displaystyle \cdots ⋯ 0 \displaystyle 0 0 ⋯ \displaystyle \cdots ⋯ 2 \displaystyle 2 2 ⋯ \displaystyle \cdots ⋯ D ′ ( x ) \displaystyle D ^{\prime} ( x ) D ′ ( x ) + 0 \displaystyle 0 0 - 0 \displaystyle 0 0 + D ( x ) \displaystyle D ( x ) D ( x ) 증가 4 \displaystyle 4 4 감소 0 \displaystyle 0 0 증가
x = 0 \displaystyle x = 0 x = 0 4 , x = 2 \displaystyle 4 , x = 2 4 , x = 2 0 \displaystyle 0 0
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수리논술, 이게 다다.