문제

Q3.svg

해설

경과된 시간이 tt 일 때 xx 만큼 반응하였다고 한다면 [A]t=[A]0−x[\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x, [B]t=[B]0−2x[\mathrm{B}]_{t}=[\mathrm{B}]_{0}-2x 이다.

v=dxdt=k[A]t[B]t  2=k([A]0−x)([B]0−2x)2v=\dfrac{\mathrm{d}x}{\mathrm{d}t}=k[\mathrm{A}]_{t}[\mathrm{B}]_{t}^{\;2}= k\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2}

양변에 dt\mathrm{d}t 를 곱하면 dx([A]0−x)([B]0−2x)2=k dt\dfrac{\mathrm{d}x}{\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2}}=k\:\mathrm{d}t

kk 를 구하기 위해서는 양변을 적분하여 ∫ 0 tk dt=kt\displaystyle\int_{\,0}^{\,t}{k\:\mathrm{d}t}=kt 를 도출해야 하므로, 좌변의 적분을 용이하게 하기 위해

dx([A]0−x)([B]0−2x)2=p[A]0−x+4px+q([B]0−2x)2\dfrac{\mathrm{d}x}{\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2}}=\dfrac{p}{[\mathrm{A}]_{0}-x}+\dfrac{4px+q}{\left([\mathrm{B}]_{0}-2x\right)^{2}} 형태로 변환하자.

이는 p[A]0−x\dfrac{p}{[\mathrm{A}]_0-x }를 우선 잡은 후에, 분모를 ([A]0−x)([B]0−2x)2\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2} 로 통분하였을 때 분자의 이차항이 00 이 되도록 4px+q([B]0−2x)2\dfrac{4px+q}{([\mathrm{B}]_0-2x)^{2}} 의 일차항을 잡은 것이다.

양변을 통분한 후 양변에 ([A]0−x)([B]0−2x)2\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2} 를 곱하면

1=p([B]0−2x)2+(4px+q)([A]0−x)=[B]0  2 p−4[B]0 px+4px2+4[A]0 px−4px2+[A]0 q−q\begin{aligned}1&=p\left([\mathrm{B}]_{0}-2x\right)^{2}+\left(4px+q\right)\left([\mathrm{A}]_{0}-x\right) \\ &=[\mathrm{B}]_{0}^{\;2}\,p-4[\mathrm{B}]_{0}\,px+4px^{2}+4[\mathrm{A}]_{0}\,px-4px^{2}+[\mathrm{A}]_{0}\,q-q \end{aligned}

(4[A]0 p−4[B]0 p−q)x+([B]0  2 p+[A]0 q−1)=0\left(4[\mathrm{A}]_{0}\,p-4[\mathrm{B}]_{0}\,p-q\right)x+\left([\mathrm{B}]_{0}^{\;2}\,p+[\mathrm{A}]_{0}\,q-1\right)=0

4([A]0−b)p=q4\left([\mathrm{A}]_{0}-b\right)p=q, [B]0 p+[A]0 q=1[\mathrm{B}]_{0}\,p+[\mathrm{A}]_{0}\,q=1

[B]0  2 p+4[A]0([A]0−[B]0)p=1[\mathrm{B}]_{0}^{\;2}\,p+4[\mathrm{A}]_{0}\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)p=1

(4[A]0  2−4[A]0[B]0+4[B]0  2)p=1\left(4[\mathrm{A}]_{0}^{\;2}-4[\mathrm{A}]_{0}[\mathrm{B}]_{0}+4[\mathrm{B}]_{0}^{\;2}\right)p=1

p=1(2[A]0−[B]0)2p=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}, q=4([A]0−[B]0)(2[A]0−[B]0)2q=\dfrac{4([\mathrm{A}]_{0}-[\mathrm{B}]_{0})}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}

∴1([A]0−x)([B]0−2x)2=1(2[A]0−[B]0)2×{1[A]0−x+4x+4([A]0−[B]0)([B]0−2x)2}\therefore \dfrac{1}{\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2}}=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\dfrac{1}{[\mathrm{A}]_{0}-x}+\dfrac{4x+4\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{\left([\mathrm{B}]_{0}-2x\right)^{2}}\right\} 이므로,

[1(2[A]0−[B]0)2×{1[A]0−x+4x+4([A]0−[B]0)([B]0−2x)2}]dx=k dt\left[\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\dfrac{1}{[\mathrm{A}]_{0}-x}+\dfrac{4x+4\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{\left([\mathrm{B}]_{0}-2x\right)^{2}}\right\}\right]\mathrm{d}x=k\,\mathrm{d}t

∫ 0 tk dt=∫ 0 xdx([A]0−x)([B]0−2x)2=1(2[A]0−[B]0)2×∫ 0 x{1[A]0−x+4x+4([A]0−[B]0)([B]0−2x)2}dx=1(2[A]0−[B]0)2×{∫ 0 x1[A]0−x dx+∫ 0 x4x([B]0−2x)2 dx+∫ 0 x4([A]0−[B]0)([B]0−2x)2 dx}\begin{aligned}\displaystyle\int_{\,0}^{\,t}{k\:\mathrm{d}t}&=\displaystyle\int_{\,0}^{\,x}{\dfrac{\mathrm{d}x}{\left([\mathrm{A}]_{0}-x\right)\left([\mathrm{B}]_{0}-2x\right)^{2}}}=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\displaystyle\int_{\,0}^{\,x}{\left\{\dfrac{1}{[\mathrm{A}]_{0}-x}+\dfrac{4x+4\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{\left([\mathrm{B}]_{0}-2x\right)^{2}}\right\}\mathrm{d}x } \\ &=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\displaystyle\int_{\,0}^{\,x}{\dfrac{1}{[\mathrm{A}]_{0}-x}\:\mathrm{d}x}+\displaystyle\int_{\,0}^{\,x}{\dfrac{4x}{\left([\mathrm{B}]_{0}-2x\right)^{2}}\:\mathrm{d}x+\displaystyle\int_{\,0}^{\,x}{\dfrac{4\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{\left([\mathrm{B}]_{0}-2x\right)^{2}}\:\mathrm{d}x}}\right\}\end{aligned}

∫ 0 x4x([B]0−2x)2 dx\displaystyle\int_{\,0}^{\,x}{\dfrac{4x}{\left([\mathrm{B}]_{0}-2x\right)^{2}}\:\mathrm{d}x } 를 구하기 위해 치환적분하자.

앞에서 [B]0−2x=[B]t[\mathrm{B}]_{0}-2x=[\mathrm{B}]_{t} 였다. x=[B]0−[B]t2x=\dfrac{[\mathrm{B}]_{0}-[\mathrm{B}]_{t}}{2}, −2 dx=d[B]t-2\:\mathrm{d}x=\mathrm{d}[\mathrm{B}]_{t} 이므로,

∫ 0 x4x[B]t  2 dx=∫ [B]0 [B]0−2x−[B]0−[B]t([B]t)2 d[B]t=∫ [B]0−2x [B]0([B]0[B]t  2−1[B]t)d[B]t=[ −[B]0[B]t−ln⁡[B]t ][B]0−2x[B]0\displaystyle\int_{\,0}^{\,x}{\dfrac{4x}{[\mathrm{B}]_{t}^{\;2}}\:\mathrm{d}x}=\displaystyle\int_{\,[\mathrm{B}]_{0}}^{\,[\mathrm{B}]_{0}-2x}{-\dfrac{[\mathrm{B}]_{0}-[\mathrm{B}]_{t}}{\left([\mathrm{B}]_{t}\right)^{2}}\:\mathrm{d}[\mathrm{B}]_{t}}=\displaystyle\int_{\,[\mathrm{B}]_{0}-2x}^{\,[\mathrm{B}]_{0}}{\left(\dfrac{[\mathrm{B}]_{0}}{[\mathrm{B}]_{t}^{\;2}}-\dfrac{1}{[\mathrm{B}]_{t}}\right)\mathrm{d}[\mathrm{B}]_{t}}=\left[\,-\dfrac{[\mathrm{B}]_{0}}{[\mathrm{B}]_{t}}-\ln [\mathrm{B}]_{t}\,\right]_{[\mathrm{B}]_{0}-2x}^{[\mathrm{B}]_{0}}

∫ 0 tk dt=1(2[A]0−[B]0)2×{[−ln⁡([A]0−x)]0x+[ −[B]0[B]t−ln⁡[B]t ][B]0−2x[B]0+[ 4([A]0−[B]0)2([B]0−2x) ]0x}=1(2[A]0−[B]0)2×[{ ⁣−ln⁡([A]0−x)+ln⁡[A]0}+{−1+[B]0[B]0−2x+ln⁡[B]0−2x[B]0}+{2([A]0−[B]0)[B]0−2x−2([A]0−[B]0)[B]0}]=1(2[A]0−[B]0)2×{2[A]0−[B]0[B]0−2x−2[A]0−[B]0[B]0+ln⁡[A]0([B]0−2x)[B]0([A]0−x)}=1(2[A]0−[B]0)2×{2(2[A]0−[B]0)x[B]0([B]0−2x)+ln⁡[A]0([B]0−2x)[B]0([A]0−x)}=1(2[A]0−[B]0)2×{2(2[A]0−[B]0)x[B]0[B]t+ln⁡[A]0[B]t[B]0[A]t}=[B]0−[B]t(2[A]0−[B]0)[B]0[B]t+1(2[A]0−[B]0)2ln⁡[A]0[B]t[B]0[A]t(∵2x=[B]0−[B]t)=1(2[A]0−[B]0)×(1[B]t−1[B]0)+1(2[A]0−[B]0)2ln⁡[A]0[B]t[B]0[A]t\begin{aligned}\displaystyle\int_{\,0}^{\,t}{k\:\mathrm{d}t} &=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\bigg[-\ln\left([\mathrm{A}]_{0}-x\right)\bigg]_{0}^{x}+\left[\,-\dfrac{[\mathrm{B}]_{0}}{[\mathrm{B}]_{t}}-\ln[\mathrm{B}]_{t}\,\right]_{[\mathrm{B}]_{0}-2x}^{[\mathrm{B}]_{0}}+\left[\,\dfrac{4\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{2\left([\mathrm{B}]_{0}-2x\right)}\,\right]_{0}^{x}\right\} \\ &=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left[\bigg\{\!-\ln\left([\mathrm{A}]_{0}-x\right)+\ln[\mathrm{A}]_{0}\bigg\}+\left\{-1+\dfrac{[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}-2x}+\ln\dfrac{[\mathrm{B}]_{0}-2x}{[\mathrm{B}]_{0}}\right\}+\left\{\dfrac{2\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{[\mathrm{B}]_{0}-2x }-\dfrac{2\left([\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}{[\mathrm{B}]_{0}}\right\}\right] \\ &=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}-2x }-\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}}+\ln\dfrac{[\mathrm{A}]_{0}\left([\mathrm{B}]_{0}-2x\right)}{[\mathrm{B}]_{0}\left([\mathrm{A}]_{0}-x\right)}\right\} \\ &=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\dfrac{2\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)x}{[\mathrm{B}]_{0}\left([\mathrm{B}]_{0}-2x\right)}+\ln\dfrac{[\mathrm{A}]_{0}\left([\mathrm{B}]_{0}-2x\right)}{[\mathrm{B}]_{0}\left([\mathrm{A}]_{0}-x\right)}\right\} \\ &=\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\times\left\{\dfrac{2\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)x}{[\mathrm{B}]_{0}[\mathrm{B}]_{t}}+\ln\dfrac{[\mathrm{A}]_{0}[\mathrm{B}]_{t}}{[\mathrm{B}]_{0}[\mathrm{A}]_{t}}\right\} \\&=\dfrac{[\mathrm{B}]_{0}-[\mathrm{B}]_{t}}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)[\mathrm{B}]_{0}[\mathrm{B}]_{t}}+\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\ln\dfrac{[\mathrm{A}]_{0}[\mathrm{B}]_{t}}{[\mathrm{B}]_{0}[\mathrm{A}]_{t}} \quad (\because 2x=[\mathrm{B}]_{0}-[\mathrm{B}]_{t}) \\ &= \dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)}\times\left(\dfrac{1}{[\mathrm{B}]_{t}}-\dfrac{1}{[\mathrm{B}]_{0}}\right)+\dfrac{1}{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}}\ln\dfrac{[\mathrm{A}]_{0}[\mathrm{B}]_{t}}{[\mathrm{B}]_{0}[\mathrm{A}]_{t}}\end{aligned}

(2[A]0−[B]0)2×kt=(2[A]0−[B]0)×(1[B]t−1[B]0)+ln⁡[A]0[B]t[B]0[A]t\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}\times kt=\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)\times\left(\dfrac{1}{[\mathrm{B}]_{t}}-\dfrac{1}{[\mathrm{B}]_{0}}\right)+\ln\dfrac{[\mathrm{A}]_{0}[\mathrm{B}]_{t}}{[\mathrm{B}]_{0}[\mathrm{A}]_{t}}

ln⁡[B]t[A]t+2[A]0−[B]0[B]t=(2[A]0−[B]0)2×kt+2[A]0−[B]0[B]0−ln⁡[A]0[B]0\ln\dfrac{[\mathrm{B}]_{t}}{[\mathrm{A}]_{t}}+\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{t}}=\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2}\times kt+\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}}-\ln\dfrac{[\mathrm{A}]_{0}}{[\mathrm{B}]_{0}}

{ f([A]0,[B]0)=2[A]0−[B]0g([A]0,[B]0)=(2[A]0−[B]0)2h([A]0,[B]0)=2[A]0−[B]0[B]0−ln⁡[A]0[B]0\begin{cases} \begin{aligned} \,f\left([\mathrm{A}]_{0}, [\mathrm{B}]_{0}\right)&=2[\mathrm{A}]_{0}-[\mathrm{B}]_{0} \\[1ex] g\left([\mathrm{A}]_{0}, [\mathrm{B}]_{0}\right)&=\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{2} \\[1ex] h\left([\mathrm{A}]_{0}, [\mathrm{B}]_{0}\right)&=\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}}-\ln\dfrac{[\mathrm{A}]_{0}}{[\mathrm{B}]_{0}} \end{aligned} \end{cases}

f([A]0,[B]0)a×g([A]0,[B]0)bp[A]0+q[B]0=2[A]0−[B]0[B]0\dfrac{f\left([\mathrm{A}]_{0}, [\mathrm{B}]_{0}\right)^{a}\times g\left([\mathrm{A}]_{0}, [\mathrm{B}]_{0}\right)^{b}}{p[\mathrm{A}]_{0}+q[\mathrm{B}]_{0}}=\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}}

(2[A]0−[B]0)a+2bp[A]0+q[B]0=2[A]0−[B]0[B]0\dfrac{\left(2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}\right)^{a+2b}}{p[\mathrm{A}]_{0}+q[\mathrm{B}]_{0}}=\dfrac{2[\mathrm{A}]_{0}-[\mathrm{B}]_{0}}{[\mathrm{B}]_{0}} 이 항등식이 되기 위해서는 a+2b=1a+2b=1, p=0p=0, q=1q=1 이 되어야 한다.

a+2b=1a+2b=1 을 만족하는 실수 aa, bb 에 대하여 abab 의 최댓값을 구하기 위해, ab=αab=\alpha 라고 두자.

우리가 구하고자 하는 것은 ‘최댓값’이므로, α≥0\alpha \geq 0 일 때를 우선적으로 검토할 것이다. 만약 α>0\alpha>0 일 때 a+2b=1a+2b=1 을 만족하는 실수 aa, bb 가 없다면, α<0\alpha<0 일 때를 검토하면 된다.

b=−12a+12b=-\dfrac{1}{2}a+\dfrac{1}{2} 와 b=αa(α>0)b=\dfrac{\alpha}{a}(\alpha >0) 를 abab 좌표평면에 나타낼 때, 두 그래프는 오직 제11사분면에서만 만난다.

이때 a≥0a\geq 0, b≥0b\geq 0 이며, 산술-기하평균을 사용하면 a+2b=1≥22aba+2b=1\geq 2\sqrt{2ab}

2ab≤12\sqrt{2ab}\leq\dfrac{1}{2}; ab≤18ab\leq\dfrac{1}{8}

ab=18ab=\dfrac{1}{8} 일 때 a=2ba=2b 이므로, a=12=ma=\dfrac{1}{2}=m, b=14=nb=\dfrac{1}{4}=n

∴m+n+p+q≤12+14+0+1=74=M\therefore m+n+p+q\leq\dfrac{1}{2}+\dfrac{1}{4}+0+1=\dfrac{7}{4}=M

16M=2816M=28

정답

28

Comment

풀이를 막상 까보면 단순 계산이 위주인 문제이지만, 비주얼로 압도하는 문제이다. [A]0[\mathrm{A}]_0, [B]0[\mathrm{B}]_0 문자를 aa, bb로 간단하게 표현한 다음, 식을 하나하나씩 정리해나가다보면 수월하게 풀릴 것이다.

History

  • 최초 출제일: 2022.10.26.
  • 해설 작성일: 2022.10.26.
  • 웹 업로드일: 2023.02.06.