【数学I】例題3.3.23：ある区間で常に成り立つ不等式（One More）★★★

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% 例題I3.3.23：ある区間で常に成り立つ不等式（One More）★★★
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$0\leqq x\leqq 8$のすべての$x$の値に対して，不等式$x^2-2ax+a+2>0$が成り立つような定数$a$の値の範囲を求めよ．

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% 例題I3.3.23の解答（One More）★★★
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$f(x)=(x-a)^2-a^2+a+2$であるから，放物線$y=f(x)$の軸は直線$x=a$

(i) $a<0$のとき

$f(x)$は$x=0$で最小となり，最小値は$f(0)=a+2$

したがって，$a+2>0$

ゆえに，$a>-2$

これと$a<0$より，$-2<a<0$

(ii) $0\leqq a\leqq 8$のとき

$f(x)$は$x=a$で最小となり，最小値は$f(a)=-a^2+a+2$

したがって，$-a^2+a+2>0$

ゆえに，$a^2-a-2<0$

したがって，$(a+1)(a-2)<0$であるから，$-1<a<2$

これと$0\leqq a\leqq 8$より，$0\leqq a<2$

(iii) $8<a$のとき

$f(x)$は$x=8$で最小となり，最小値は$f(8)=-15a+66$

したがって，$-15a+66>0$であるから，$a<\frac{22}{5}$

これは$8<a$を満たさない．

よって，(i)〜(iii)より，求める$a$の値の範囲は，$-2<a<2$

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% 問題I3.3.23：ある区間で常に成り立つ不等式（One More）★★★
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$-1\leqq x\leqq 9$のすべての$x$の値に対して，不等式$x^2-2ax+a+6>0$が成り立つような定数$a$の値の範囲を求めよ．

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% 問題I3.3.23の解答（One More）★★★
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$f(x)=(x-a)^2-a^2+a+6$であるから，放物線$y=f(x)$の軸は直線$x=a$

(i) $a<-1$のとき

$f(x)$は$x=-1$で最小となり，最小値は$f(-1)=3a+7$

したがって，$3a+7>0$

ゆえに，$a>-\frac{7}{3}$

これと$a<-1$より，$-\frac{7}{3}<a<-1$

(ii) $-1\leqq a\leqq 9$のとき

$f(x)$は$x=a$で最小となり，最小値は$f(a)=-a^2+a+6$

したがって，$-a^2+a+6>0$

ゆえに，$a^2-a-6<0$

これを解くと，$(a-3)(a+2)<0$であるから，$-2<a<3$

これと$-1\leqq a\leqq 9$より，$-1\leqq a<3$

(iii) $a>9$のとき

$f(x)$は$x=9$で最小となり，最小値は$f(9)=-17a+87$

したがって，$-17a+87>0$であるから，$a<\frac{87}{17}$

これは$a>9$を満たさない．

よって，(i)〜(iii)より，求める$a$の値の範囲は，$-\frac{7}{3}<a<3$

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% 例題I3.3.23の側注：ある区間で常に成り立つ不等式（One More）★★★
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% 側注1
軸が区間より左側か，区間内か，右側かで場合分けをする．全部で3通りの場合分けとなる．

% 側注2
場合分けの条件である，$0\leqq a\leqq 8$を満たすか否かの確認を忘れないように注意すること．

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% 問題I3.3.23の側注：ある区間で常に成り立つ不等式（One More）★★★
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% 側注1
軸が区間より左側か，区間内か，右側かで場合分けをする．全部で3通りの場合分けとなる．

% 側注2
場合分けの条件である，$-1\leqq a\leqq 9$を満たすか否かの確認を忘れないように注意すること．

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% 例題I3.3.23の考え方・補足など：ある区間で常に成り立つ不等式（One More）★★★
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% 考え方
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求める条件は，$0\leqq x\leqq 8$における$f(x)=x^2-2ax+a+2$の最小値が正となることである．

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% $f(x)$の符号がある区間内で一定である条件
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区間内で$f(x)>0\Longleftrightarrow($区間内の$f(x)$の最小値$)>0$

区間内で$f(x)<0\Longleftrightarrow($区間内の$f(x)$の最大値$)<0$