東京大学(理系) 2022年　問題1


$次の関数 \ f(x)\ を考える。$
\[\quad f(x)=(\cos x)\log(\cos x) - \cos x + \int_0^x (\cos t)\log(\cos t)dt \ \ (0 \leqq x < \cfrac{\pi}{2})\] $(1)\ \ f(x)\ は区間 \ 0 \leqq x < \cfrac{\pi}{2} \ において最小値を持つことを示せ。$
$(2)\ \ f(x)\ の区間 \ 0 \leqq x < \cfrac{\pi}{2} \ における最小値を求めよ。$


$(解説)$

\[(1)\ \ \cfrac{d}{dt}\int _a^x g(t)dt=g(x) \ \ です。f(x)の増減表をつくって調べます。\] $(2)\ \ いくつかの定積分を順次求めることになります。$


(1)


$f'(x)=-\sin x\log \cos x - \sin x + \sin x + \cos x \log \cos x =(\cos x - \sin x)\log \cos x$

$ 0 \leqq x < \cfrac{\pi}{2} \quad より \quad 0 < \cos x \leqq 1 \quad \therefore \ \ \log \cos x \leqq 0$

$f'(x)=0 \quad より \quad \cos x=\sin x \qquad \tan x=1 \qquad \therefore x=\cfrac{\pi}{4}$

\[ f(x)\ の増減表は \qquad \begin{array}{c||c|c|c|c|c} x& 0 & \cdots & \small{\cfrac{\pi}{4}} & \cdots & \small{\cfrac{\pi}{2}}\\ \hline f'(x) & 0 & - & 0 & + & \\ \hline f(x)& & \searrow & 極小 & \nearrow & \\ \end{array} \]
$f(x)\ は \ \ x=\cfrac{\pi}{4} \ \ で極小かつ最小となる。$


(2)


\[最小値は \quad f(\cfrac{\pi}{4})=\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} -\cfrac{1}{\sqrt{2}}+ \int _0^{\scriptsize{\cfrac{\pi}{4}}} \cos t \log \cos t dt\] $第 \ 3\ 項目を \ I\ とおくと$
\[\quad I=\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cos t \log \cos t dt =\big[\sin t \log \cos t \big]_0^{\scriptsize{\cfrac{\pi}{4}}} - \int _0^{\scriptsize{\cfrac{\pi}{4}}} \sin t \times \cfrac{-\sin t}{\cos t} dt =\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} + \int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{\sin ^2t}{\cos t} dt\] $第 \ 2\ 項目を \ J\ とおくと$
\[\quad J=\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{\sin ^2t}{\cos t} dt =\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{1-\cos ^2t}{\cos t} dt =\int _0^{\scriptsize{\cfrac{\pi}{4}}} \big(\cfrac{1}{\cos t} -\cos t\big) dt =\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{dt}{\cos t} - \big[\sin t\big] _0^{\scriptsize{\cfrac{\pi}{4}}} =\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{dt}{\cos t} - \cfrac{1}{\sqrt{2}}\] $第 \ 1\ 項目を \ K\ とおくと$
\[\quad K=\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{1}{\cos t}dt =\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{\cos t}{\cos ^2t}dt =\int _0^{\scriptsize{\cfrac{\pi}{4}}} \cfrac{\cos t}{1-\sin ^2t}dt\] \[\quad \sin t=u \quad とおくと \quad \cos t dt=du \qquad \quad \begin{array}{c|c} t & \ 0 \ \rightarrow \small{\cfrac{\pi}{4}} \quad \\ \hline u & \ 0 \ \rightarrow \small{\cfrac{1}{\sqrt{2}}} \\ \end{array} \] \[\quad K=\int _0^{\scriptsize{\cfrac{1}{\sqrt{2}}}} \cfrac{du}{1-u^2} =\cfrac{1}{2} \int _0^{\scriptsize{\cfrac{1}{\sqrt{2}}}} \big(\cfrac{1}{1+u}+\cfrac{1}{1-u}\big)du =\cfrac{1}{2} \big[\log (1+u) - \log (1-u) \big]_0^{\scriptsize{\cfrac{1}{\sqrt{2}}}} =\cfrac{1}{2} \big[\log \cfrac{1+u}{1-u} \big]_0^{\scriptsize{\cfrac{1}{\sqrt{2}}}}\] \[ \hspace{2em} =\cfrac{1}{2} \log \cfrac{ 1+ \scriptsize{\cfrac{1}{\sqrt{2}}} }{ 1- \scriptsize{\cfrac{1}{\sqrt{2}}}} =\cfrac{1}{2} \log \cfrac{ \sqrt{2}+1 }{ \sqrt{2}-1 } =\cfrac{1}{2} \log (\sqrt{2}+1)^2 =\log (\sqrt{2}+1)\] $よって$

$\quad J=K - \cfrac{1}{\sqrt{2}}= \log (\sqrt{2}+1)- \cfrac{1}{\sqrt{2}} $

$\quad I=\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} +J =\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} + \log (\sqrt{2}+1)- \cfrac{1}{\sqrt{2}} $

$したがって　最小値は$

\begin{eqnarray*} \quad f(\cfrac{\pi}{4}) &=&\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} -\cfrac{1}{\sqrt{2}}+ I\\ \\ &=&\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} -\cfrac{1}{\sqrt{2}}+\cfrac{1}{\sqrt{2}}\log \cfrac{1}{\sqrt{2}} + \log (\sqrt{2}+1)- \cfrac{1}{\sqrt{2}}\\ \\ &=&\sqrt{2}\log \cfrac{1}{\sqrt{2}} -\sqrt{2} + \log (\sqrt{2}+1)\\ \\ &=&-\cfrac{\sqrt{2}}{2}\log 2 -\sqrt{2} + \log (\sqrt{2}+1)\\ \end{eqnarray*}

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