【数学I】例題3.3.24：2次不等式が整数解をもつ条件（One More）★★★

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% 例題I3.3.24：2次不等式が整数解をもつ条件（One More）★★★
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$x$についての不等式$x^2-(a+1)x+a<0,x^2+x-6>0$を満たす整数$x$がちょうど3個存在するような定数$a$の値の範囲を求めよ．

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% 例題I3.3.24の解答（One More）★★★
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$x^2+x-6>0$より，$(x+3)(x-2)>0$

したがって，$x<-3,2<x\cdots(\mathrm{i})$

$x^2-(a+1)x+a<0$より，$(x-1)(x-a)<0\cdots(\mathrm{ii})$

（ア）$a<1$のとき，(ii)より，$a<x<1$

これと(i)より，不等式を満たす整数$x$がちょうど3個となるのは，右の図のようなときである．

したがって，$-7\leqq a<-6$

（イ）$a=1$のとき，(ii)は解なしで不適である．

（ウ）$a>1$のとき，(ii)より，$1<x<a$

これと(i)より，不等式を満たす整数$x$がちょうど3個となるのは，右の図のようなときである．

したがって，$5<a\leqq 6$

よって，（ア）〜（ウ）より，求める$a$の値の範囲は$-7\leqq a<-6,5<a\leqq 6$

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% 問題I3.3.24：2次不等式が整数解をもつ条件（One More）★★★
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$x$についての不等式$x^2-(a+2)x+2a<0,x^2-x-12>0$を満たす整数$x$がちょうど3個存在するような定数$a$の値の範囲を求めよ．

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% 問題I3.3.24の解答（One More）★★★
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$x^2-x-12>0$より，$(x+3)(x-4)>0$

したがって，$x<-3,4<x\cdots(\mathrm{i})$

$x^2-(a+2)x+2a<0$より，$(x-2)(x-a)<0\cdots(\mathrm{ii})$

（ア）$a<2$のとき，(ii)より，$a<x<2$

これと(i)より，不等式を満たす整数$x$がちょうど3個となるのは，右の図のようなときである．

したがって，$-7\leqq a<-6$

（イ）$a=2$のとき，(ii)は解なしで不適である．

（ウ）$a>2$のとき，(ii)より，$2<x<a$

これと(i)より，不等式を満たす整数$x$がちょうど3個となるのは，右の図のようなときである．

したがって，$7<a\leqq 8$

よって，（ア）〜（ウ）より，求める$a$の値の範囲は$-7\leqq a<-6,7<a\leqq 8$

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% 例題I3.3.24の側注：2次不等式が整数解をもつ条件（One More）★★★
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% 側注1
等号を含むか否かに注意すること．

% 側注2
$5<a<6$としないように注意すること．$a=6$のとき，$2<x<6$となり，解はちょうど$3,4,5$の3個となるから条件を満たす．

% 側注3
$a=1$のとき，不等式は$(x-1)^2<0$となり，これを満たす実数$x$は存在しない．

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% 問題I3.3.24の側注：2次不等式が整数解をもつ条件（One More）★★★
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% 側注1
等号を含むか否かに注意すること．

% 側注2
$a=2$のとき，不等式は$(x-2)^2<0$となり，これを満たす実数$x$は存在しない．

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% 例題I3.3.24の考え方・補足など：2次不等式が整数解をもつ条件（One More）★★★
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% 考え方
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$a$と$1$の大小関係に注目して，$a$の値によって場合分けをする．数直線を利用して，整数がちょうど3個存在するような$a$の範囲を考えるとよい．なお，$a$の範囲は等号を含むか否かをよく考えて注意すること．

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$a$の値によって場合分けをして，等号を含むか否かをよく考える．