2019학년도 11월 가형 30번

30. 최고차항의 계수가 $6\pi$$6\pi$인 삼차함수 $f(x)$$f(x)$에 대하여 $g(x)=\frac{1}{2+\sin (f(x))}$$g(x)=\cfrac 1{2+\sin (f(x))}$이 $x=\alpha$$x=\alpha$에서 극대 또는 극소이고, $\alpha \ge 0$$\alpha \ge 0$인 모든 $\alpha$$\alpha$를 작은 수부터 크기순으로 나열한 것을 ${\alpha }_1,{\alpha }_2,{\alpha }_3,\cdots$$\alpha_1,~\alpha_2,~\alpha_3,~\cdots$라 할 때, $g(x)$$g(x)$는 다음 조건을 만족시킨다.

(가) ${\alpha }_1=0$$\alpha_1=0$이고 $g({\alpha }_1)=\frac{2}{5}$$g(\alpha_1)=\cfrac 25$이다.

(나) $\frac{1}{g({\alpha }_5)}=\frac{1}{g({\alpha }_2)}+\frac{1}{2}$$\cfrac 1{g(\alpha_5)}=\cfrac 1{g(\alpha_2)}+\cfrac 12$

$g^{\prime }(-\frac{1}{2})=a\pi$$g'\Big(-\cfrac 12\Big)=a\pi$라 할 때, $a^2$$a^2$의 값을 구하시오. (단, $0<f(0)<\frac{\pi }{2}$$0<f(0)<\cfrac {\pi}2$)

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i. 정리

• $f(x)=6\pi x^3+\sim$$f(x)=6\pi x^3 +\sim$
• $g(x)=\frac{1}{2+\sin f(x)}\alpha =$$g(x)=\cfrac 1{2+\sin f(x)} \quad \alpha=$극값
• $\alpha \ge 0⟶{\alpha }_1,{\alpha }_2,{\alpha }_3,\cdots$$\alpha\ge 0\quad \longrightarrow \quad \alpha_1,~\alpha_2,~\alpha_3,~\cdots$
• ${\alpha }_1=0⟶g({\alpha }_1)=\frac{2}{5}$$\alpha_1=0\quad \longrightarrow \quad g(\alpha_1)=\cfrac 25$
• $\frac{1}{g({\alpha }_5)}=\frac{1}{g({\alpha }_2)}+\frac{1}{2}$$\cfrac 1{g(\alpha_5)}=\cfrac 1{g(\alpha_2)}+\cfrac 12$

ii. 생각

• $g^{\prime }(x)$$g'(x)$를 계산하면,

  $g^{\prime }(x)=-\frac{\cos f(x)\times f^{\prime }(x)}{(2+\sin f(x){)}^2}$$g'(x)=-\cfrac {\cos f(x)\times f'(x)}{(2+\sin f(x))^2}$

  $\cos f(x)=0$$\cos f(x)=0$이거나 $f^{\prime }(x)=0$$f'(x)=0$

• $\{\begin{array}{l}{f}^{\prime }(0)=0\\ \\ \cos f(0)=0\end{array},g(0)=\frac{2}{5}$$\begin{cases} f'(0)=0\\ \\ \cos f(0)=0\end{cases}~~~ ,\quad ~ g(0)=\cfrac 25$

  • $\cos f(0)=0$$\cos f(0)=0$이라 가정하면, $\sin f(0)=\pm 1$$\sin f(0)=\pm 1$이면 $g(0)\ne 25$$g(0)\neq 25$

  $\therefore f^{\prime }(0)=0$$\therefore~ f'(0)=0$

  이를 이용하면,

  $\sin f(0)=\frac{1}{2}⟶f(0)=\frac{\pi }{6}$$\sin f(0)=\cfrac 12\quad\longrightarrow \quad f(0)=\cfrac {\pi}6$

  $f(x)$$f(x)$를 조금 더 특정시키자.

  $f(x)=6\pi x^3+k{x}^2+mx+\frac{\pi }{6}$$f(x)=6\pi x^3 +kx^2 +mx+\cfrac {\pi}6$이라 하면,

  $f^{\prime }(x)=18\pi x^2+2kx+m$$f'(x)=18\pi x^2 +2kx+m$

  그런데, $f^{\prime }(0)=0$$f'(0)=0$이므로

  $\{\begin{array}{l}f(x)=6\pi x^3+k{x}^2+\frac{\pi }{6}\\ \\ f^{\prime }(x)=2x(9\pi x+k)\end{array}$$\begin{cases} f(x)=6\pi x^3 +kx^2 +\cfrac {\pi}6\\ \\ f'(x)=2x(9\pi x+k)\end{cases}$

• $f(x)$$f(x)$의 개형을 그리고 조건 (나)를 생각하자.

  

  조건 (나)를 정리하면,

  $\sin f({\alpha }_5)=\sin f({\alpha }_2)+\frac{1}{2}$$\sin f(\alpha_5)=\sin f(\alpha _2)+\cfrac 12$

  $\{\begin{array}{l}\sin f({\alpha }_2)=-1⟶\sin f({\alpha }_5)=-\frac{1}{2}\\ \\ \sin f({\alpha }_5)=1⟶\sin f({\alpha }_2)=-\frac{1}{2}\end{array}$$\begin{cases} \sin f(\alpha_2 )=-1\quad \longrightarrow \quad \sin f(\alpha_5)=-\cfrac 12\\ \\ \sin f(\alpha_5)=1\quad \longrightarrow \quad \sin f(\alpha_2)=-\cfrac 12\end{cases}$

  

  가능한 경우는 두번째 그래프의 형태일 때이다.

  

  $\sin f({\alpha }_5)=-\frac{1}{2}$$\sin f(\alpha_5)=-\cfrac 12$이기 위해서는 $f({\alpha }_5)=-3\pi +\frac{\pi }{6}$$f(\alpha_5)=-3\pi +\cfrac{\pi}6$ 이고 $f^{\prime }({\alpha }_5)=0$$f'(\alpha_5)=0$

   

  위에서 구한 $f^{\prime }(x)$$f'(x)$를 보면, ${\alpha }_5=-\frac{k}{9\pi }$$\alpha_5=-\cfrac k{9\pi}$

  $f(-\frac{k}{9\pi })=-3\pi +\frac{\pi }{6}$$f\Big(-\cfrac k{9\pi}\Big)=-3\pi +\cfrac {\pi}6$을 풀면,

  $k=-9\pi$$k=-9\pi$

  $\begin{array}{c}\therefore \{\begin{array}{l}f(x)=6\pi x^3-9\pi x^2+\frac{\pi }{6}\\ \\ f^{\prime }(x)=18\pi x^2-19\pi x\end{array}\end{array}$$\therefore~ \begin{cases} f(x)=6\pi x^3 -9\pi x^2 +\cfrac {\pi}6\\ \\ f'(x)=18\pi x^2 -19\pi x\end{cases}$

• $g^{\prime }(-\frac{1}{2})$$g'(-\cfrac 12)$를 구하자.

  $f^{\prime }(-\frac{1}{2})=\frac{27}{2}\pi$$f'(-\cfrac 12)=\cfrac {27}2\pi$

  $f(-\frac{1}{2})=-3\pi +\frac{\pi }{6}$$f(-\cfrac 12)=-3\pi+\cfrac {\pi}6$

  $\therefore g^{\prime }(-\frac{1}{2})=3\sqrt{3}\pi$$\therefore~ g'(-\cfrac 12)=3\sqrt 3\pi$

  $\therefore a=3\sqrt{3}$$\therefore~ a=3\sqrt 3$

$\therefore a^2=27$$\therefore~ a^2 =27$