+20점 · f ( x ) = { x − ( 2 a − 4 ) } ( x − a ) ( x − 4 ) \displaystyle f(x)=\{x-(2a-4)\}(x-a)(x-4) f ( x ) = { x − ( 2 a − 4 )} ( x − a ) ( x − 4 ) 【2-1】 조건 (ㄱ)에서 x = 0 \displaystyle x=0 x = 0 f ( a ) ≠ 0 \displaystyle f(a)\ne0 f ( a ) = 0 { f ( a ) } 2 < 0 \displaystyle \{f(a)\}^2<0 { f ( a ) } 2 < 0 f ( a ) = 0 \displaystyle f(a)=0 f ( a ) = 0 조건 (ㄱ)에서 x = a − 4 \displaystyle x=a-4 x = a − 4 f ( a + x ) f ( a − x ) = f ( 2 a − 4 ) f ( 4 ) = 0 \displaystyle f(a+x)f(a-x)=f(2a-4)f(4)=0 f ( a + x ) f ( a − x ) = f ( 2 a − 4 ) f ( 4 ) = 0 f ( 2 a − 4 ) = f ( a + x ) = 0 \displaystyle f(2a-4)=f(a+x)=0 f ( 2 a − 4 ) = f ( a + x ) = 0 따라서 f ( x ) = { x − ( 2 a − 4 ) } ( x − a ) ( x − 4 ) \displaystyle f(x)=\{x-(2a-4)\}(x-a)(x-4) f ( x ) = { x − ( 2 a − 4 )} ( x − a ) ( x − 4 ) 2 a − 4 < a < 4 \displaystyle 2a-4<a<4 2 a − 4 < a < 4
+14점 · n ( a ) \displaystyle n(a) n ( a ) n ( a ) = { 1 ( a ≤ 0 ) 2 ( 0 < a ≤ 2 ) 3 ( 2 < a < 4 ) \displaystyle n(a)=\begin{cases}1&(a\le0)\\2&(0<a\le2)\\3&(2<a<4)\end{cases} n ( a ) = ⎩ ⎨ ⎧ 1 2 3 ( a ≤ 0 ) ( 0 < a ≤ 2 ) ( 2 < a < 4 ) 이므로
+6점 · lim a → 2 + n ( a ) = 3 , lim a → 2 − n ( a ) = 2 \displaystyle \lim_{a\to2^+}n(a)=3,\ \lim_{a\to2^-}n(a)=2 a → 2 + lim n ( a ) = 3 , a → 2 − lim n ( a ) = 2 lim a → 2 + n ( a ) = 3 , lim a → 2 − n ( a ) = 2 \displaystyle \lim_{a\to2^+}n(a)=3,\ \lim_{a\to2^-}n(a)=2 a → 2 + lim n ( a ) = 3 , a → 2 − lim n ( a ) = 2 이다.
+25점 · ∫ 0 2 f ( x ) d x = − 12 ( a − 5 3 ) 2 + 16 3 \displaystyle \int_0^2f(x)\,dx=-12\left(a-\frac53\right)^2+\frac{16}3 ∫ 0 2 f ( x ) d x = − 12 ( a − 3 5 ) 2 + 3 16 【2-2】 ∫ 0 2 f ( x ) d x = ∫ 0 2 { x − ( 2 a − 4 ) } ( x − a ) ( x − 4 ) d x = ∫ 0 2 { x − ( 2 a − 4 ) } { x 2 − ( a + 4 ) x + 4 a } d x = ∫ 0 2 { x 3 − ( a + 4 ) x 2 + 4 a x − ( 2 a − 4 ) x 2 + ( 2 a − 4 ) ( a + 4 ) x − 4 a ( 2 a − 4 ) } d x = 4 − 8 3 ( a + 4 ) + 8 a − 8 3 ( 2 a − 4 ) + 2 ( 2 a − 4 ) ( a + 4 ) − 8 a ( 2 a − 4 ) = − 12 a 2 + 40 a − 28 = − 12 ( a − 5 3 ) 2 + 16 3 \displaystyle \begin{aligned}\int_0^2f(x)\,dx&=\int_0^2\{x-(2a-4)\}(x-a)(x-4)\,dx\\&=\int_0^2\{x-(2a-4)\}\{x^2-(a+4)x+4a\}\,dx\\&=\int_0^2\{x^3-(a+4)x^2+4ax-(2a-4)x^2+(2a-4)(a+4)x-4a(2a-4)\}\,dx\\&=4-\frac83(a+4)+8a-\frac83(2a-4)+2(2a-4)(a+4)-8a(2a-4)\\&=-12a^2+40a-28=-12\left(a-\frac53\right)^2+\frac{16}3\end{aligned} ∫ 0 2 f ( x ) d x = ∫ 0 2 { x − ( 2 a − 4 )} ( x − a ) ( x − 4 ) d x = ∫ 0 2 { x − ( 2 a − 4 )} { x 2 − ( a + 4 ) x + 4 a } d x = ∫ 0 2 { x 3 − ( a + 4 ) x 2 + 4 a x − ( 2 a − 4 ) x 2 + ( 2 a − 4 ) ( a + 4 ) x − 4 a ( 2 a − 4 )} d x = 4 − 3 8 ( a + 4 ) + 8 a − 3 8 ( 2 a − 4 ) + 2 ( 2 a − 4 ) ( a + 4 ) − 8 a ( 2 a − 4 ) = − 12 a 2 + 40 a − 28 = − 12 ( a − 3 5 ) 2 + 3 16
+5점 · 3 a = 5 \displaystyle 3a=5 3 a = 5 이므로 a = 5 3 \displaystyle a=\frac53 a = 3 5 3 a = 5 \displaystyle 3a=5 3 a = 5
【2-3】 g ( x ) = ∣ f ( x ) f ( − x ) ∣ = ∣ { x − ( 2 a − 4 ) } ( x − a ) ( x − 4 ) { x + ( 2 a − 4 ) } ( x + a ) ( x + 4 ) ∣ \displaystyle g(x)=|f(x)f(-x)|=|\{x-(2a-4)\}(x-a)(x-4)\{x+(2a-4)\}(x+a)(x+4)| g ( x ) = ∣ f ( x ) f ( − x ) ∣ = ∣ { x − ( 2 a − 4 )} ( x − a ) ( x − 4 ) { x + ( 2 a − 4 )} ( x + a ) ( x + 4 ) ∣ 2 a − 4 < a < 4 , − 4 < − a < − ( 2 a − 4 ) \displaystyle 2a-4<a<4,\ -4<-a<-(2a-4) 2 a − 4 < a < 4 , − 4 < − a < − ( 2 a − 4 ) 이다.
+12점 · (1) g ( x ) = ∣ f ( x ) f ( − x ) ∣ = x 2 ∣ x − 4 ∣ ∣ x − 2 ∣ ∣ x + 2 ∣ ∣ x + 4 ∣ \displaystyle g(x)=|f(x)f(-x)|=x^2|x-4||x-2||x+2||x+4| g ( x ) = ∣ f ( x ) f ( − x ) ∣ = x 2 ∣ x − 4∣∣ x − 2∣∣ x + 2∣∣ x + 4∣ +8점 · (1) b = 4 , 2 , − 2 , − 4 \displaystyle b=4,2,-2,-4 b = 4 , 2 , − 2 , − 4 (1) a = 2 \displaystyle a=2 a = 2 g ( x ) = ∣ f ( x ) f ( − x ) ∣ = x 2 ∣ x − 4 ∣ ∣ x − 2 ∣ ∣ x + 2 ∣ ∣ x + 4 ∣ \displaystyle g(x)=|f(x)f(-x)|=x^2|x-4||x-2||x+2||x+4| g ( x ) = ∣ f ( x ) f ( − x ) ∣ = x 2 ∣ x − 4∣∣ x − 2∣∣ x + 2∣∣ x + 4∣ g ( x ) \displaystyle g(x) g ( x ) x = b = 4 , 2 , − 2 , − 4 \displaystyle x=b=4,2,-2,-4 x = b = 4 , 2 , − 2 , − 4
(2) a = 0 \displaystyle a=0 a = 0 g ( x ) = x 2 ( x − 4 ) 2 ( x + 4 ) 2 \displaystyle g(x)=x^2(x-4)^2(x+4)^2 g ( x ) = x 2 ( x − 4 ) 2 ( x + 4 ) 2 이다.
+5점 · (2) h ( x 0 ) = x 0 ( x 0 − 4 ) 2 − k = 0 \displaystyle h(x_0)=x_0(x_0-4)^2-k=0 h ( x 0 ) = x 0 ( x 0 − 4 ) 2 − k = 0 방정식 g ( x ) ( x + 4 ) 2 = k x ( x ≠ − 4 ) \displaystyle \frac{g(x)}{(x+4)^2}=kx\ (x\ne-4) ( x + 4 ) 2 g ( x ) = k x ( x = − 4 ) x = 0 \displaystyle x=0 x = 0 k ≠ 0 \displaystyle k\ne0 k = 0 x ( x − 4 ) 2 = k \displaystyle x(x-4)^2=k x ( x − 4 ) 2 = k 0 , − 4 \displaystyle 0,-4 0 , − 4 h ( x ) = x ( x − 4 ) 2 \displaystyle h(x)=x(x-4)^2 h ( x ) = x ( x − 4 ) 2 y = h ( x ) \displaystyle y=h(x) y = h ( x ) y = k \displaystyle y=k y = k x \displaystyle x x 0 , − 4 \displaystyle 0,-4 0 , − 4
+5점 · (2) h ′ ( x 0 ) = 2 x 0 ( x 0 − 4 ) + ( x 0 − 4 ) 2 = 0 \displaystyle h^{\prime}(x_0)=2x_0(x_0-4)+(x_0-4)^2=0 h ′ ( x 0 ) = 2 x 0 ( x 0 − 4 ) + ( x 0 − 4 ) 2 = 0 +5점 · (2) x 0 = 4 3 \displaystyle x_0=\frac43 x 0 = 3 4 h ′ ( x ) = ( x − 4 ) ( 3 x − 4 ) \displaystyle h^{\prime}(x)=(x-4)(3x-4) h ′ ( x ) = ( x − 4 ) ( 3 x − 4 ) h ( x ) \displaystyle h(x) h ( x ) x = 4 3 \displaystyle x=\frac43 x = 3 4 256 27 \displaystyle \frac{256}{27} 27 256 x = 4 \displaystyle x=4 x = 4
+5점 · (2) 27 16 k \displaystyle \frac{27}{16}k 16 27 k 따라서 k = 256 27 \displaystyle k=\frac{256}{27} k = 27 256 27 16 k = 16 \displaystyle \frac{27}{16}k=16 16 27 k = 16 이다.
수리논술, 배움에서 논증의 완성까지.