【数学I】例題3.3.27：方程式の解の存在範囲3（One More）★★

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% 例題I3.3.27：方程式の解の存在範囲3（One More）★★
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2次方程式$x^2-ax+2a^2-11=0$の異なる2つの実数解のうち，1つは3より大きく，他の1つは3より小さくなるような定数$a$の値の範囲を求めよ．

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$y=f(x)=x^2-ax+2a^2-11$とする．

$y=f(x)$のグラフは，下に凸の放物線である．

$f(x)=0$が異なる2つの実数解をもち，1つは3より大きく，他の1つは3より小さくなるのは，$y=f(x)$のグラフが右の図のようになるときである．

したがって，求める条件は，$f(3)<0$である．

$f(3)=3^2-a\cdot 3+2a^2-11=2a^2-3a-2=(2a+1)(a-2)$より，

\[
(2a+1)(a-2)<0
\]

よって，$-\frac{1}{2}<a<2$

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% 問題I3.3.27：方程式の解の存在範囲3（One More）★★
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2次方程式$x^2-ax+3a^2-20=0$の異なる2つの実数解のうち，1つは$4$より大きく，他の1つは$4$より小さくなるような定数$a$の値の範囲を求めよ．

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% 問題I3.3.27の解答（One More）★★
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$y=f(x)=x^2-ax+3a^2-20$とする．

$y=f(x)$のグラフは，下に凸の放物線である．

$f(x)=0$が異なる2つの実数解をもち，1つは$4$より大きく，他の1つは$4$より小さくなるのは，$y=f(x)$のグラフが右の図のようになるときである．

したがって，求める条件は，$f(4)<0$である．

$f(4)=4^2-a\cdot 4+3a^2-20=3a^2-4a-4=(3a+2)(a-2)$より，

\[
(3a+2)(a-2)<0
\]

よって，$-\frac{2}{3}<a<2$

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% 例題I3.3.27の側注：方程式の解の存在範囲3（One More）★★
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% 側注1
たすき掛けを用いる．

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% 問題I3.3.27の側注：方程式の解の存在範囲3（One More）★★
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% 側注1
たすき掛けを用いる．

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% 例題I3.3.27の考え方・補足など：方程式の解の存在範囲3（One More）★★
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$y=f(x)=x^2-ax+2a^2-11$とすると，$D>0$を示す必要はなく，条件は$f(3)<0$を考えるのみでよい．下に凸の放物線は，その関数が負の値をとるとき，$x$軸と異なる2点で交わることを利用する．

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異なる2つの実数解のうち，1つは$p$より大きく，他の1つは$p$より小さい場合は，

\begin{align*}
&\text{放物線が下に凸のとき，}f(p)<0\\
&\text{放物線が上に凸のとき，}f(p)>0
\end{align*}

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「条件は$f(3)<0$のみでよいのか，判別式や軸の条件は考えなくてもよいのか」と考える人もいるかもしれない．例題では，$f(3)<0$より，$x=3$のときの$y$座標は負である．このとき，右の図のように，下に凸の放物線は$x$軸と異なる2点で交わる．また，2つの異なる交点の$x$座標を$\alpha,\beta(\alpha<\beta)$とすると，$f(3)<0$のとき，軸の位置に関係なく$\alpha<3<\beta$となる．

よって，$f(3)<0$を満たすとき，判別式や軸の条件は考えなくてもよい．