2020년 10월 나형 21번

21. 두 곡선 $y=2^{-x}$$y=2^{-x}$과 $y=|{\log }_{2}x|$$y=|\log_2 x|$가 만나는 두 점을 $(x_1,y_1),(x_2,y_2)$$(x_1,~y_1),~(x_2,~y_2)$라 하자. $x_1<x_2$$x_1<x_2$일 때, <보기>에서 옳은 것을 모두 고르시오.

ㄱ. $\frac{1}{2}<x_1<\frac{\sqrt{2}}{2}$$\cfrac 12<x_1<\cfrac {\sqrt2}2$

ㄴ. $\sqrt[3]{2}<x_2<\sqrt{2}$$\sqrt[3]2<x_2<\sqrt 2$

ㄷ. $y_1-y_2<\frac{3\sqrt{2}-2}{6}$$y_1-y_2<\cfrac {3\sqrt 2-2}6$

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i. 생각

• $f(x)=2^{-x},g(x)=|{\log }_{2}x|$$f(x)=2^{-x},~g(x)=|\log_2x|$라 하자.

• ㄱ

  $f(\frac{1}{2})<g(\frac{1}{2})⟶\frac{1}{2}<x_1$$f(\cfrac 12) < g(\cfrac 12)\quad \longrightarrow \quad \cfrac 12 <x_1$

  $f(\frac{\sqrt{2}}{2})<g(\frac{\sqrt{2}}{2})⟶x_1<\frac{\sqrt{2}}{2}$$f(\cfrac {\sqrt 2}2)<g(\cfrac {\sqrt 2}2)\quad \longrightarrow \quad x_1<\cfrac {\sqrt 2}2$

  $\therefore \frac{1}{2}<x_1<\frac{\sqrt{2}}{2}$$\therefore~ \cfrac 12<x_1<\cfrac {\sqrt 2}2$

  True

• ㄴ

  $f(\sqrt[3]{2})>g(\sqrt[3]{2})$$f(\sqrt[3]2)>g(\sqrt[3]2)$

  $f(\sqrt{2})<g(\sqrt{2})$$f(\sqrt 2)<g(\sqrt2)$

  $\sqrt[3]{2}<x_2<\sqrt{2}$$\sqrt[3]2<x_2<\sqrt 2$

  True

• ㄷ

  $y_1-y_2<\frac{3\sqrt{2}-2}{6}=\frac{\sqrt{2}}{2}-\frac{1}{3}$$y_1-y_2<\cfrac {3\sqrt2 -2}6=\cfrac {\sqrt 2}2-\cfrac 13$

  가장 깔끔한 건 $y_1$$y_1$에서 $\frac{\sqrt{2}}{2}$$\cfrac {\sqrt2 }2$가 나오고 $y_2$$y_2$에서 $\frac{1}{3}$$\cfrac 13$을 끄집어내면 되겠다?

  • ㄴ에 ${\log }_2$$\log_2$를 씌우면,

    $\sqrt[3]{2}<{\log }_2{x}_2=y_2<\frac{1}{2}$$\sqrt[3]2<\log_2 x_2=y_2<\cfrac 12$

    $-\frac{1}{2}<-y_2<-\frac{1}{3}$$-\cfrac 12 <-y_2<-\cfrac 13$

  • $y_1$$y_1$을 생각하자.

    $y_1=2^{-x}=-{\log }_2{x}_1$$y_1=2^{-x}=-\log_2 x_1$

    어? 역함수인가?

    $y=2^{-x}$$y=2^{-x}$의 역함수를 구하면,

    $x=2^{-y}⟶-y={\log }_{2}x$$x=2^{-y} \quad\longrightarrow \quad -y=\log_2x$

    $y=-{\log }_{2}x$$y=-\log_2x$

    오호라...

    $x_1=y_1$$x_1=y_1$이다

    $\frac{1}{2}<y_1<\frac{\sqrt{2}}{2}$$\cfrac 12 <y_1 <\cfrac {\sqrt 2}2$

  • $y_1-y_2$$y_1-y_2$의 범위를 구하자.

    $0<y_1-y_2<\frac{\sqrt{2}}{2}-\frac{1}{3}=\frac{3\sqrt{2}-2}{6}$$0<y_1-y_2<\cfrac {\sqrt 2}2 -\cfrac 13 =\cfrac {3\sqrt 2-2}6$

  True