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1b:I[3866,["51","static/chunks/795d4814-710d199cb1f51304.js","183","static/chunks/183-141f39fb5cb93742.js","23","static/chunks/app/advice/%5Bid%5D/page-26686d89cc5a4cf8.js"],"AdOnAdviceList1"] 1c:I[3866,["51","static/chunks/795d4814-710d199cb1f51304.js","183","static/chunks/183-141f39fb5cb93742.js","23","static/chunks/app/advice/%5Bid%5D/page-26686d89cc5a4cf8.js"],"AdOnAdviceList2"] 1d:I[3866,["51","static/chunks/795d4814-710d199cb1f51304.js","183","static/chunks/183-141f39fb5cb93742.js","23","static/chunks/app/advice/%5Bid%5D/page-26686d89cc5a4cf8.js"],"AdOnAdviceList3"] 1a:Tb60,一言でいえば「経験」です。 必要条件の利用は青チャートなどのいわゆる網羅系参考書などでは得られない少し発展的な技術ですが、数学がある程度得意な人は過去にやったことがあったり、習ったことがあったりとそれを利用した経験があるのです。 なので質問者さんももう少し経験を積めば普通に利用できるようになると思います。 ここで必要条件の利用に至るまでの思考回路の簡単な例を紹介しておきます。 問題 「k を正の整数とする. 5n^2 − 2kn + 1 < 0 ー①を満たす整数 n が,ちょうど 1 個であるような k をすべて求めよ.」 これは2008年の一橋の問題です。下にプロセスを書いてますが良問であり考えがいがあるので一回自力で考えてみてください。 まず大前提として数学における問題と解は全て必要十分性を保っている必要があります、従ってこの問題を解く際必要十分を保ちながら解く(同値変形)のと必要、十分を分けて解く2通りに分かれます。この問題では実数でなく整数の2次方程式であり同値変形で解くのはややこしい(できることはできます)と判断しまず必要条件から絞ろうと考えます。 この問題を考える際式①が成り立つとき5x^2-2kx+1=0(xは実数)が二つの異なる実数解を持つー②「必要」がある(つまり②は①の必要条件である)ことを利用します、そうすることによってkの条件が分かります。そのもとで①を満たす整数nがちょうど1個である条件は5x^2-2kx+1=0の二つの解(α、βとおく)の差が2未満ー③であればいいということがわかります。②と③から得られるkの範囲が5=