【数学I】例題3.3.26：方程式の解の存在範囲2（One More）★★★

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% 例題I3.3.26：方程式の解の存在範囲2（One More）★★★
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2次方程式$x^2-3ax+2=0$が，$0<x<4$の範囲に異なる2つの実数解をもつような定数$a$の値の範囲を求めよ．

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% 例題I3.3.26の解答（One More）★★★
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$y=f(x)=x^2-3ax+2$とし，2次方程式$x^2-3ax+2=0$の判別式を$D$とする．

$y=f(x)$のグラフは，下に凸の放物線で，軸は$x=\frac{3a}{2}$である．

$f(x)=0$が$0<x<4$に異なる2つの実数解をもつのは，$y=f(x)$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，

(i) $D>0$，

(ii) 軸が$0<x<4$の間にある，

(iii) $f(0)>0,f(4)>0$である．

(i) $D=(-3a)^2-4\cdot 1\cdot 2=9a^2-8$であり，$D>0$であるから，

\[
9a^2-8>0
\]

したがって，$a<-\frac{2\sqrt{2}}{3},\frac{2\sqrt{2}}{3}<a$

(ii) 軸は直線$x=\frac{3a}{2}$であるから，$0<\frac{3a}{2}<4$より，$0<a<\frac{8}{3}$

(iii) $f(0)=0^2-3a\cdot 0+2=2,f(4)=4^2-3a\cdot 4+2=-12a+18$であり，$f(0)>0$かつ$f(4)>0$より，$-12a+18>0$

したがって，$a<\frac{3}{2}$

よって，(i)〜(iii)より，求める$a$の値の範囲は，

\[
\frac{2\sqrt{2}}{3}<a<\frac{3}{2}
\]

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% 問題I3.3.26：方程式の解の存在範囲2（One More）★★★
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2次方程式$x^2-4ax+3=0$が，$1<x<5$の範囲に異なる2つの実数解をもつような定数$a$の値の範囲を求めよ．

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% 問題I3.3.26の解答（One More）★★★
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$y=f(x)=x^2-4ax+3$とし，2次方程式$x^2-4ax+3=0$の判別式を$D$とする．

$y=f(x)$のグラフは，下に凸の放物線で，軸は$x=2a$である．

$f(x)=0$が$1<x<5$に異なる2つの実数解をもつのは，$y=f(x)$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，

(i) $D>0$，

(ii) 軸が$1<x<5$の間にある，

(iii) $f(1)>0,f(5)>0$である．

(i) $\frac{D}{4}=(-2a)^2-1\cdot 3=4a^2-3$であり，$D>0$であるから，

\[
4a^2-3>0
\]

したがって，$a<-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}<a$

(ii) 軸は直線$x=2a$であるから，$1<2a<5$より，$\frac{1}{2}<a<\frac{5}{2}$

(iii) $f(1)=1^2-4a\cdot 1+3=-4a+4,f(5)=5^2-4a\cdot 5+3=-20a+28$であり，$f(1)>0$かつ$f(5)>0$より，$4-4a>0,-20a+28>0$

したがって，$a<1$

よって，(i)〜(iii)より，求める$a$の値の範囲は，

\[
\frac{\sqrt{3}}{2}<a<1
\]

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% 例題I3.3.26の側注：方程式の解の存在範囲2（One More）★★★
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% 側注1
軸は$x=-\frac{b}{2a}$

% 側注2
判別式，軸，$f(0),f(4)$（端点）に注目する．

% 側注3
判別式$D$を用いる代わりに，「頂点の$y$座標が$0$より小さくなる」という条件を用いても同じ結果が得られる．

% 側注4
$(a+\frac{2\sqrt{2}}{3})(a-\frac{2\sqrt{2}}{3})>0$より，$a<-\frac{2\sqrt{2}}{3},\frac{2\sqrt{2}}{3}<a$

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% 問題I3.3.26の側注：方程式の解の存在範囲2（One More）★★★
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% 側注1
軸は$x=-\frac{b}{2a}$

% 側注2
判別式，軸，$f(1),f(5)$（端点）に注目する．

% 側注3
判別式$D$を用いる代わりに，「頂点の$y$座標が$0$より小さくなる」という条件を用いても同じ結果が得られる．

% 側注4
$(a+\frac{\sqrt{3}}{2})(a-\frac{\sqrt{3}}{2})>0$より，$a<-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}<a$

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% 例題I3.3.26の考え方・補足など：方程式の解の存在範囲2（One More）★★★
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% 考え方
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$f(x)=x^2-3ax+2$とし，グラフをかいて考える．

$f(x)$のグラフが右の図のようになればよいので，

(i) $D>0$，

(ii) 軸が$0<x<4$の間にある，

(iii) $f(0)>0,f(4)>0$を満たすように，定数$a$の範囲を定める．

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方程式の解の存在範囲は，判別式（頂点），軸，端点に注目する．