문제

Q4.svg

해설

Step 1. f(x)f(x) 구하기


∫ 0 x{f′(t)}2dt=ln⁡∣ 1+cos⁡f(x) ∣−ln⁡∣ sin⁡f(x) ∣  ⋯(가)\displaystyle\int_{\,0}^{\,x} \left\{ f^{\prime}(t) \right\}^{2}dt=\ln \big\lvert\, 1+\cos f(x) \,\big\rvert-\ln \big\lvert\, \sin f(x) \,\big\rvert \;\cdots \textrm{(가)}

{f′(x)}2={−sin⁡f(x)1+cos⁡f(x)−cos⁡f(x)sin⁡f(x)}⋅f′(x)\left\{ f^{\prime}(x) \right\}^{2}=\left\{ \dfrac{-\sin f(x)}{1+\cos f(x)} - \dfrac{\cos f(x)}{\sin f(x)} \right\} \cdot f^{\prime}(x)

f′(x)=−sin⁡2f(x)+cos⁡f(x)+cos⁡2f(x){1+cos⁡f(x)}⋅sin⁡f(x)=−1+cos⁡f(x){1+cos⁡f(x)}⋅sin⁡f(x)=−1sin⁡f(x)\begin{aligned} f^{\prime}(x)&= -\dfrac{\sin^{2} f(x) + \cos f(x) + \cos^{2} f(x)}{\left\{1+\cos f(x) \right\} \cdot \sin f(x)} \\ &= -\dfrac{1 + \cos f(x)}{\left\{1+\cos f(x) \right\} \cdot \sin f(x)} \\ &= -\dfrac{1}{\sin f(x)}\end{aligned}

f′(x)⋅sin⁡f(x)=−1f^{\prime}(x) \cdot \sin f(x) = -1

cos⁡f(x)=x+C\cos f(x) = x + C (CC 는 적분상수)

(가)에 x=0x=0 을 대입하면 0=ln⁡∣ 1+cos⁡f(0)sin⁡f(0) ∣0=\ln \bigg\lvert\,\dfrac{1+\cos f(0)}{\sin f(0)}\,\bigg\rvert

∴1+cos⁡f(0)=sin⁡f(0)\therefore 1 + \cos f(0) = \sin f(0) 또는 1+cos⁡f(0)=−sin⁡f(0)1 + \cos f(0) = -\sin f(0)

어느 경우든 간에 {1+cos⁡f(0)}2={sin⁡f(0)}2\left\{1 + \cos f(0)\right\}^{2}=\left\{\sin f(0) \right\}^{2} 이다.

그러므로 {sin⁡f(0)}2+{cos⁡f(0)}2=1\left\{\sin f(0) \right\}^{2} + \left\{\cos f(0) \right\}^{2} = 1 에서

{1+cos⁡f(0)}2+{cos⁡f(0)}2=1\left\{1 + \cos f(0) \right\}^{2} + \left\{\cos f(0) \right\}^{2} = 1

2{cos⁡f(0)}2+2cos⁡f(0)=02 \left\{\cos f(0) \right\}^{2} + 2 \cos f(0) = 0

cos⁡f(0)=0\cos f(0) = 0 또는 cos⁡f(0)=−1\cos f(0) = -1

문제 조건에서 ff 의 공역이 (0,π)(0, \pi) 이기 때문에, cos⁡f(0)≠−1\cos f(0) \neq -1 이다. 따라서 cos⁡f(0)=0\cos f(0) = 0 이고, C=0C=0 이다.

∴cos⁡f(x)=x\therefore \cos f(x) = x


Step 2. g(x)g(x) 구하기

∫ π2 2ysin⁡z dz=[−cos⁡z ]π22y=−cos⁡2y\displaystyle\int _{\,\frac{\pi}{2}}^{\,2y}\sin z \: \mathrm{d}z = \bigg[-\cos z \,\bigg]_{\frac{\pi}{2}}^{2y} = -\cos 2y

∫ 0 w2−cos⁡2y dy=[ −12sin⁡2y ]0w2=−12sin⁡w\displaystyle\int _{\,0}^{\,\frac{w}{2}}-\cos 2y \: \mathrm{d}y = \left[\,-\dfrac{1}{2}\sin 2y \,\right]_{0}^{\frac{w}{2}} = -\dfrac{1}{2} \sin w

∫ v f(x)−12sin⁡w dw=[ 12cos⁡w ]vf(x)=12{cos⁡f(x)−cos⁡v}=12(x−cos⁡v)\displaystyle\int _{\,v}^{\,f(x)}-\dfrac{1}{2} \sin w \: \mathrm{d}w = \left[\,\dfrac{1}{2}\cos w \,\right]_{v}^{f(x)} = \dfrac{1}{2} \left\{\cos f(x) - \cos v \right\}=\dfrac{1}{2} \left(x - \cos v \right)

∫ 0 u212(x−cos⁡v)dv=[ sin⁡v−12xv ]0u2=12{sin⁡(u2)−xu2}\displaystyle\int _{\,0}^{\,u^2}\dfrac{1}{2} \left(x - \cos v \right) \mathrm{d}v = \left[\,\sin v - \dfrac{1}{2} xv\,\right]_{0}^{u^2}=\dfrac{1}{2} \left\{\sin(u^2) - xu^2\right\}

h(t)=∫π t{sin⁡(u2)−xu2}duh(t)=\displaystyle\int _{\sqrt{\pi}} ^{\,t} \left\{\sin\left(u^2\right) - xu^2\right\}\mathrm{d}u 라 하자.

h′(t)=−xt2+sin⁡(t2)h^{\prime}(t)=-x t^{2} + \sin \left(t ^2 \right)

∫ 0 th(s) ds=[ sh(s) ]0t−∫ 0 tsh′(s) ds=th(t)−∫ 0 ts⋅{−xs2+sin⁡(s2)} ds\begin{aligned} \displaystyle\int_{\,0}^{\,t} h(s) \: \mathrm{d}s &= \bigg[\,sh(s)\,\bigg]_{0}^{t} - \displaystyle\int_{\,0}^{\,t} s h^{\prime}(s) \: \mathrm{d}s \\ &= th(t) - \displaystyle\int_{\,0}^{\,t}s \cdot \left\{ -xs^2 + \sin\left(s^2\right)\right\}\, \mathrm{d} s \end{aligned}

k=s2k=s^2 라 하면 dk=2s ds\mathrm{d} k = 2s\: \mathrm{d} s

∫ 0 th(s) ds=th(t)−12∫ 0 t2(−xk+sin⁡k) dk=th(t)−12[−x2v2−cos⁡v ]0t2=th(t)−12{−x2t4−cos⁡(t2)+1}\begin{aligned} \displaystyle\int_{\,0}^{\,t} h(s) \: \mathrm{d}s &= th(t) - \dfrac{1}{2}\displaystyle\int_{\,0}^{\,t^2}\left(-xk + \sin k \right)\, \mathrm{d}k \\ &= th(t) - \dfrac{1}{2} \bigg[-\dfrac{x}{2}v^{2} - \cos v \,\bigg]_{0}^{t^2} \\ &= th(t) - \dfrac{1}{2} \left\{-\dfrac{x}{2} t^{4} - \cos \left( t^2 \right) + 1 \right\}\end{aligned}

g(x)=lim⁡t→012∫ 0 th(s) ds12 h(t)sin⁡t=lim⁡t→0th(t)−12{−x2t4−cos⁡(t2)+1}sin⁡t⋅h(t)=lim⁡t→0tsin⁡t+14lim⁡t→0xt4sin⁡t⋅h(t)−12lim⁡t→01−cos⁡(t2)sin⁡t⋅h(t)=1+14lim⁡t→0(x−1)t3h(t)\begin{aligned} g(x)&=\lim\limits_{t\rightarrow0}\dfrac{\dfrac{1}{2}\displaystyle\int_{\,0}^{\,t}{h(s) \: \mathrm{d} s}}{\dfrac{1}{2}\,h(t) \sin t} \\ &= \lim\limits_{t\rightarrow0}\dfrac{th(t) - \dfrac{1}{2} \left\{-\dfrac{x}{2} t^4 - \cos \left(t^2 \right) + 1 \right\}}{\sin t \cdot h(t)} \\ &= \lim\limits_{t\rightarrow0}\dfrac{t}{\sin t} + \dfrac{1}{4} \lim\limits_{t\rightarrow0}\dfrac{xt^4}{\sin t \cdot h(t)}-\dfrac{1}{2} \lim\limits_{t\rightarrow0}\dfrac{1-\cos\left(t^2\right)}{\sin t \cdot h(t)} \\ &= 1 + \dfrac{1}{4} \lim\limits_{t \rightarrow 0}\dfrac{(x-1) t^3}{h(t)}\end{aligned}


Step 3. 답 구하기

(ⅰ) xx가 h(0)=0h(0) = 0 을 만족할 때

i(t)=(x−1)t3i(t) = (x-1)t^3 로 두면, i′(t)=3(x−1)t2i^{\prime}(t) = 3(x-1)t^2 이므로

lim⁡t→0i(t)h(t)=lim⁡t→0i(t)−i(0)th(t)−h(0)t=lim⁡t→0i′(t)h′(t)=lim⁡t→03(x−1)t2−xt2+sin⁡(t2)=−3\lim\limits_{t \rightarrow 0} \dfrac{i(t)}{h(t)} = \lim\limits_{t \rightarrow 0}\cfrac{\frac{i(t)-i(0)}{t}}{\frac{h(t)-h(0)}{t}}= \lim\limits_{t \rightarrow 0} \dfrac{i^{\prime}(t)}{h^{\prime}(t)} = \lim\limits_{t \rightarrow 0} \dfrac{3(x-1)t^2}{-x t^{2} + \sin \left(t ^2 \right)}=-3

g(x)=1−34=14g(x)=1-\dfrac{3}{4}=\dfrac{1}{4}

h(0)=∫π 0{sin⁡(l2)−xl2}dl=[−x3l3 ]0π−2∫ 0 πsin⁡(l2)dl=0\begin{aligned} h(0)&=\displaystyle\int _{\sqrt{\pi}} ^{\,0} \left\{\sin\left(l^2\right) - xl^2\right\}\mathrm{d}l \\ &= \bigg[ -\dfrac{x}{3} l^{3} \, \bigg] _{0}^{\sqrt{\pi}} - 2 \displaystyle\int_{\,0}^{\,\sqrt{\pi}} \sin \left( l^2 \right) \mathrm{d} l = 0\end{aligned}

∫ 0 πsin⁡(l2)dl=−x6ππ\displaystyle\int_{\,0}^{\,\sqrt{\pi}} \sin \left( l^2 \right) \mathrm{d} l = - \dfrac{x}{6} \pi \sqrt{\pi}

∫ 0 πsin⁡(l2)dl\displaystyle\int_{\,0}^{\,\sqrt{\pi}} \sin \left( l^2 \right) \mathrm{d} l 은 상수이므로, 위 식을 만족시키는 xx 는 α\alpha 로 유일하다.

(다시 말해 g(m)=14=kg(m)=\dfrac{1}{4} = k 를 만족하는 mm 은 α=−6∫0πsin⁡(l2)dlππ\alpha = -\dfrac{6\int_{0}^{\sqrt{\pi}} \sin \left( l^2 \right) \mathrm{d} l}{\pi \sqrt{\pi}} 로 유일하다.)

l=tl=\sqrt{t} 라 하면 dt=2l dl\mathrm{d} t = 2l \: \mathrm{d} l 이므로,

∫ 0 πsin⁡tt dt=∫ 0 πsin⁡(l2)l⋅2l dl=2∫ 0 πsin⁡(l2)dl=−x3π32\displaystyle\int_{\,0}^{\,\pi}\dfrac{\sin t}{\sqrt t} \: \mathrm{d} t = \displaystyle\int_{\,0}^{\,\sqrt{\pi}}\dfrac{\sin \left(l^2 \right)}{l} \cdot 2l \: \mathrm{d} l = 2 \displaystyle\int_{\,0}^{\,\sqrt{\pi}}\sin \left(l^2 \right)\mathrm{d} l = -\dfrac{x}{3} \pi^{\frac{3}{2}}

pk=3pk=3 이므로 p=12p=12, q=2q=2, r=3r=3

∴p×(q+r)=60\therefore p \times (q + r) = 60


(ⅱ) xx가 h(0)≠0h(0) \neq 0 을 만족할 때

lim⁡t→0(x−1)t3h(t)=0\lim\limits_{t \rightarrow 0}\dfrac{(x-1) t^3}{h(t)} = 0 이므로 g(x)=1g(x)=1 이다. 그런데 앞에서 h(0)=0h(0) = 0 을 만족하는 xx 는 단 한 개임을 보였으므로, h(0)≠0h(0) \neq 0 을 만족하는 xx 는 무수히 많다. 고로 이 경우에서 g(m)=1g(m)=1 을 만족하는 mm 이 α\alpha ‘뿐’인 경우는 없다.

정답

60

Comment

문제의 생김새에 비해서는 깔끔하게 풀린다고 볼 수 있을 것 같다. 식을 잘 정리하고 관찰하면 풀리기 때문에 가형 킬러보다는 다소 쉬운 난이도라고 할 수 있을 것 같다.

History

  • 최초 출제일: 2022.08.14.
  • 해설 작성일: 2022.08.14.
  • 웹 업로드일: 2023.02.07.