1京都大学 2007年 数学 第1問解答・解説

問1 積分を二つに分け、$\displaystyle \frac{d}{dx}\sqrt{x^2+4}=x/\sqrt{x^2+4}$ と $\displaystyle \frac{d}{dx}\log(x+\sqrt{x^2+4})=1/\sqrt{x^2+4}$ を用いると $$\displaystyle \int_0^2\frac{2x+1}{\sqrt{x^2+4}}dx =\left[2\sqrt{x^2+4}+\log(x+\sqrt{x^2+4})\right]_0^2 =\boxed{4\sqrt2-4+\log(1+\sqrt2)}.$$

問2 2段昇る歩数が $k$ 回なら、1段昇る歩数は $15-2k$ 回で、歩数の合計は $15-k$ 回となる。2段昇る $k$ 回は隣り合ってはならない。$m$ 個の位置から隣り合わない $k$ 個の位置を選ぶ方法は $\displaystyle \binom{m-k+1}{k}$ 通りである（選んだ位置の間に必要な空きを取り除いて数える）。ここでは $k\le5$ なので、求める総数は $$\displaystyle \sum\limits _{k=0}^5\binom{16-2k}{k}=1+14+66+120+70+6=\boxed{277}.$$

Tags. Unit: definite integrals and counting. Method: elementary antiderivatives, nonadjacent-position bijection and binomial coefficients. Theme: exact integral and constrained compositions of 15.

Quality comparison and disposition Independent derivation was saved before consulting the prior answer. The prior final result agrees: 問1=4sqrt(2)-4+ln(1+sqrt(2)); 問2=277.. The old integral value and nonadjacent-move count agree. The original accepted result files and their evidence metadata remain untouched. Prior actual-result evidence was not independently validated, so actual model status stays UNVERIFIED.　 Disposition: mathematics PASS; source eligible; display matched; adopt this independent high-school solution.