2018년 10월 가형 21번

21. 함수 $f(x)=-\frac{k{x}^3}{x^2+1}(k>1)$$f(x)=-\cfrac {kx^3}{x^2+1}~(k>1)$에 대하여 곡선 $y=f(x)$$y=f(x)$와 곡선 $y=f^{-1}(x)$$y=f^{-1}(x)$가 만나는 점의 $x$$x$ 좌표 중 가장 작은 값을 $\alpha$$\alpha$, 가장 큰 값을 $\beta$$\beta$라 하자. 함수 $y=f(x-2\beta )+2\alpha$$y=f(x-2\beta)+2\alpha$의 역함수 $g(x)$$g(x)$에 대하여 $f^{\prime }(\beta )=2g^{\prime }(\alpha )$$f'(\beta)=2g'(\alpha)$일 때, 상수 $k$$k$의 값을 구하시오.

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i. 정리

• $f(x)=-\frac{k{x}^3}{x^2+1}(k>1)$$f(x)=-\cfrac {kx^3}{x^2+1} \quad (k>1)$
• $f(x)=f^{-1}(x)⟶\alpha <\beta$$f(x)=f^{-1} (x)\quad \longrightarrow \quad \alpha<\beta$
• $h(x)=f(x-2\beta )+2\alpha$$h(x)=f(x-2\beta )+2\alpha$
• $g(x)=h^{-1}(x)$$g(x)=h^{-1}(x)$
• $f^{\prime }(\beta )=2g^{\prime }(\alpha )$$f'(\beta)=2g'(\alpha)$

ii. 생각

• $y=f(x)$$y=f(x)$의 개형을 살펴보자.

  • $x^2+1>0$$x^2 +1 >0$

    $\therefore y=f(x)$$\therefore~ y=f(x)$는 증감은 $y=x^3$$y=x^3$과 대충 비슷할 것이다

  어? 그러면, $f(x)=x$$f(x)=x$로 접근하면 안된다!

  • $f^{\prime }(x)$$f'(x)$를 구하자.

    $f^{\prime }(x)=-\frac{k{x}^2(x^2+3k)}{(x+1{)}^2}$$f'(x)=-\cfrac {kx^2 (x^2+3k)}{(x+1)^2}$

• $f(x)=-x$$f(x)=-x$의 근을 혹시 구할 수 있는가?

  $-\frac{x^2}{x^2+1}=-x$$-\cfrac {x^2}{x^2+1} =-x$

  $x=\pm \frac{1}{\sqrt{k-1}}$$x=\pm \cfrac 1{\sqrt{k-1}}$

  $\therefore \alpha =-\frac{1}{\sqrt{k-1}},\beta =\frac{1}{\sqrt{k-1}}$$\therefore~ \alpha =-\cfrac 1{\sqrt{k-1}},~\beta =\cfrac 1{\sqrt {k-1}}$

  $\alpha +\beta =0$$\alpha +\beta =0$

• $g^{\prime }(x)$$g'(x)$를 구하자.

  $h(g(x))=x⟶h^{\prime }(g(x))=\frac{1}{g^{\prime }(x)}$$h(g(x))=x\quad \longrightarrow \quad h'(g(x))=\cfrac 1{g'(x)}$

  $\therefore g^{\prime }(x)=\frac{1}{h^{\prime }(g(x))}$$\therefore~ g'(x)=\cfrac 1{h'(g(x))}$

• $f^{\prime }(\beta )=2g^{\prime }(\alpha )$$f'(\beta )=2g'(\alpha)$를 이용하자.

  • $g(\alpha )=x$$g(\alpha )=x$가 되는 $x$$x$의 값을 찾도록 하자.

    $h(x)=\alpha$$h(x)=\alpha$

    $f(x-2\beta )+2\alpha =\alpha$$f(x-2\beta)+2\alpha =\alpha$

    $f(x+2\alpha )=-\alpha$$f(x+2\alpha )=-\alpha$

    $\therefore x+2\alpha =\alpha ((\alpha ,-\alpha ),(\beta ,-\beta )$$\therefore~ x+2\alpha =\alpha \quad \Big((\alpha ,~-\alpha ),~(\beta,~-\beta)$을 지난다.$)$$\Big)$

    $x=-\alpha$$x=-\alpha$

  $\therefore g^{\prime }(\alpha )=\frac{1}{h^{\prime }(g(x))}=\frac{1}{h^{\prime }(-\alpha )}$$\therefore~ g'(\alpha )=\cfrac 1{h'(g(x))}=\cfrac 1{h'(-\alpha )}$

  • $h^{\prime }(-\alpha )=f^{\prime }(\beta -2\beta )=f^{\prime }(-\beta )$$h'(-\alpha )=f'(\beta -2\beta )=f'(-\beta)$

    $f^{\prime }(\beta )\cdot f^{\prime }(-\beta )=2$$f'(\beta)\cdot f'(-\beta)=2$

    그런데, $y=f^{\prime }(x)$$y=f'(x)$는 $y$$y$ 축 대칭이다.

    $\therefore f^{\prime }(\beta {)}^2=2⟶f^{\prime }(\beta )=\pm \sqrt{2}$$\therefore~ f'(\beta )^2 =2\quad \longrightarrow \quad f'(\beta )=\pm \sqrt 2$

    그런데, $f^{\prime }(x)<0$$f'(x)<0$이다.

    $\therefore f^{\prime }(\beta )=-\sqrt{2}$$\therefore~ f'(\beta )=-\sqrt 2$

• $f^{\prime }(\beta )=-\sqrt{2}$$f'(\beta)=-\sqrt 2$를 계산하자.

  $\begin{array}{rl}\frac{k{\beta }^2({\beta }^2+3k)}{({\beta }^2+1{)}^2}& =\sqrt{2}\\ \\ & \leftarrow {\beta }^2=\frac{1}{k-1}\\ \\ & =\frac{k\cdot \frac{1}{k-1}\cdot (\frac{1}{k-1}+3k)}{(\frac{1}{k-1}+1{)}^2}\\ \\ & =\frac{\frac{k}{k-1}\cdot (\frac{3k-2}{k-1})}{\frac{k^2}{(k-1{)}^2}}\\ \\ & =\frac{3k-2}{k}=\sqrt{2}\end{array}$$\begin{aligned} \cfrac {k\beta ^2 (\beta^2 +3k)}{(\beta ^2 +1)^2}&=\sqrt 2\\ \\ &\leftarrow \beta^2 =\cfrac 1{k-1} \\ \\ &=\cfrac {k\cdot \cfrac 1{k-1}\cdot \big(\cfrac 1{k-1}+3k\big)}{(\cfrac 1{k-1}+1)^2}\\ \\ &= \cfrac {\cfrac k{k-1}\cdot \big(\cfrac {3k-2}{k-1}\big)}{\cfrac {k^2}{(k-1)^2}}\\ \\ &=\cfrac {3k-2}k=\sqrt 2 \end{aligned}$

  $\therefore k=\frac{2}{3-\sqrt{2}}=\frac{6+2\sqrt{2}}{7}$$\therefore~ k=\cfrac 2{3-\sqrt 2}=\cfrac {6+2\sqrt 2}{7}$