【数学I】例題3.3.28：方程式の解の存在範囲4（One More）★★★

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% 例題I3.3.28：方程式の解の存在範囲4（One More）★★★
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2次方程式$ax^2-(a+2)x-a-5=0$が，$-1<x<0$の範囲に1つの解があり，$1<x<3$の範囲に他の解があるような定数$a$の値の範囲を定めよ．

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% 例題I3.3.28の解答（One More）★★★
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$f(x)=ax^2-(a+2)x-a-5$とする．

与えられた方程式は2次方程式であるから，$a\neq 0$

求める条件は，$y=f(x)$のグラフが$-1<x<0,1<x<3$の範囲でそれぞれ$x$軸と1点で交わることである．

すなわち，$f(-1)\cdot f(0)<0$かつ$f(1)\cdot f(3)<0$

ここで，$f(-1)=a-3,f(0)=-a-5,f(1)=-a-7,f(3)=5a-11$

$f(-1)\cdot f(0)<0$より，$(a-3)(-a-5)<0$

すなわち，$(a+5)(a-3)>0$

したがって，$a<-5,3<a\cdots(\mathrm{i})$

$f(1)\cdot f(3)<0$より，$(-a-7)(5a-11)<0$

すなわち，$(a+7)(5a-11)>0$

したがって，$a<-7,\frac{11}{5}<a\cdots(\mathrm{ii})$

よって，(i)，(ii)より，求める$a$の値の範囲は，

\[
a<-7,3<a
\]

これは，$a\neq 0$を満たす．

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% 問題I3.3.28：方程式の解の存在範囲4（One More）★★★
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2次方程式$ax^2-(a+1)x-2=0$が，$-1<x<1$の範囲に1つの解があり，$3<x<5$の範囲に他の解があるような定数$a$の値の範囲を定めよ．

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% 問題I3.3.28の解答（One More）★★★
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$f(x)=ax^2-(a+1)x-2$とする．

与えられた方程式は2次方程式であるから，$a\neq 0$

求める条件は，$y=f(x)$のグラフが$-1<x<1,3<x<5$の範囲でそれぞれ$x$軸と1点で交わることである．

すなわち，$f(-1)\cdot f(1)<0$かつ$f(3)\cdot f(5)<0$

ここで，$f(-1)=2a-1,f(1)=-3,f(3)=6a-5,f(5)=20a-7$

$f(-1)\cdot f(1)<0$より，$(2a-1)\cdot(-3)<0$

したがって，$a>\frac{1}{2}\cdots(\mathrm{i})$

$f(3)\cdot f(5)<0$より，$(6a-5)(20a-7)<0$

したがって，$\frac{7}{20}<a<\frac{5}{6}\cdots(\mathrm{ii})$

よって，(i)，(ii)より，求める$a$の値の範囲は，

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

これは，$a\neq 0$を満たす．

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% 例題I3.3.28の側注：方程式の解の存在範囲4（One More）★★★
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% 側注1
グラフの凹凸に関係なく，$f(-1)\cdot f(0)<0$かつ$f(1)\cdot f(3)<0$が求める条件である．$a>0,a<0$のときで場合分けをする必要はないので注意すること．

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% 問題I3.3.28の側注：方程式の解の存在範囲4（One More）★★★
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% 側注1
グラフの凹凸に関係なく，$f(-1)\cdot f(1)<0$かつ$f(3)\cdot f(5)<0$が求める条件である．$a>0,a<0$のときで場合分けをする必要はないので注意すること．

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% 例題I3.3.28の考え方・補足など：方程式の解の存在範囲4（One More）★★★
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$y=ax^2-(a+2)x-a-5$とすると，求める条件を満たすためには，$y=f(x)$のグラフが右の図のようになればよい．

よって，グラフの凹凸に関係なく，

$f(-1)\cdot f(0)<0$かつ$f(1)\cdot f(3)<0$

（$f(-1)$と$f(0)$が異符号かつ$f(1)$と$f(3)$が異符号）

となるときを求めればよい．

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$f(p)\cdot f(q)<0$のとき，$p$と$q$の間に解（交点）がある．