【数学I】例題3.3.25：方程式の解の存在範囲1（One More）★★★

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% 例題I3.3.25：方程式の解の存在範囲1（One More）★★★
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2次方程式$x^2-2ax+4a=0$の異なる2つの実数解が，ともに3より大きくなるような定数$a$の値の範囲を求めよ．

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% 例題I3.3.25の解答（One More）★★★
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$y=f(x)=x^2-2ax+4a$とし，2次方程式$f(x)=0$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，軸は直線$x=a$である．

$f(x)=0$の異なる2つの実数解が，ともに3より大きくなるのは，$y=f(x)$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，

(i) $D>0$，

(ii) 軸が$x>3$の範囲にある，

(iii) $f(3)>0$である．

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

\[
a(a-4)>0
\]

したがって，$a<0,4<a$

(ii) 軸は直線$x=a$であるから，$a>3$

(iii) $f(3)=3^2-2a\cdot 3+4a=-2a+9$であり，$f(3)>0$より，$a<\frac{9}{2}$よって，(i)〜(iii)より，求める$a$の値の範囲は，

\[
4<a<\frac{9}{2}
\]

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% 問題I3.3.25：方程式の解の存在範囲1（One More）★★★
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2次方程式$x^2-2ax-a+2=0$の異なる2つの実数解が，ともに$2$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I3.3.25の解答（One More）★★★
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$y=f(x)=x^2-2ax-a+2$とし，2次方程式$f(x)=0$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，軸は直線$x=a$である．

$f(x)=0$の異なる2つの実数解が，ともに$2$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，

(i) $D>0$，

(ii) 軸が$x<2$の範囲にある，

(iii) $f(2)>0$である．

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

したがって，$a<-2,1<a$

(ii) 軸は直線$x=a$であるから，$a<2$

(iii) $f(2)=2^2-2a\cdot 2-a+2=-5a+6$であり，$f(2)>0$より，$a<\frac{6}{5}$

よって，(i)〜(iii)より，求める$a$の値の範囲は，

\[
a<-2,1<a<\frac{6}{5}
\]

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% 例題I3.3.25の側注：方程式の解の存在範囲1（One More）★★★
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% 側注1
軸は$x=-\frac{b}{2a}$

% 側注2
判別式，軸，$f(3)$（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．

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% 問題I3.3.25の側注：方程式の解の存在範囲1（One More）★★★
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% 側注1
軸は$x=-\frac{b}{2a}$

% 側注2
判別式，軸，$f(2)$（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．

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% 例題I3.3.25の考え方・補足など：方程式の解の存在範囲1（One More）★★★
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% 考え方
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$f(x)=x^2-2ax+4a$とし，2次方程式$f(x)=0$の判別式を$D$としたとき，「異なる2つの実数解がともに3より大きくなる」を満たす条件を考える．ここではグラフより，

(i) $D>0$，

(ii) 軸が$x>3$の範囲にある，

(iii) $f(3)>0$を満たすように，定数$a$の範囲を定める．

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(i)〜(iii)は，3つのうち1つでも成り立たないと，次のように与えられた条件を満たさない例が挙げられる可能性があるので注意すること．(ii)，(iii)が成り立つが，(i)が成り立たない．

$x$軸と共有点をもたない，または$x$軸と接する例が挙げられる．

(i)，(iii)が成り立つが，(ii)が成り立たない．

異なる2つの実数解が，ともに3より小さくなる例が挙げられる．

(i)，(ii)が成り立つが，(iii)が成り立たない．

実数解が，3より小さくなる，または$3$となる例が挙げられる．