【数学I】例題3.3.20：不等式の係数決定（One More）★★

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% 例題I3.3.20：不等式の係数決定（One More）★★
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2次不等式$ax^2+2x+b\geqq 0$の解が$-2\leqq x\leqq 3$となるとき，定数$a,b$の値を求めよ．

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% 例題I3.3.20の解答（One More）★★
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$y=ax^2+2x+b\cdots(\mathrm{i})$とする．

$ax^2+2x+b\geqq 0$の解が$-2\leqq x\leqq 3$となるのは，(i)のグラフが右の図のようになるときであるから，$a<0$

このとき，解が$-2\leqq x\leqq 3$であるから，$y=ax^2+2x+b$のグラフは$x$軸と$x=-2,3$で交わる．

したがって，2次方程式$ax^2+2x+b=0$の解は，$x=-2,3$となる．

ゆえに，$ax^2+2x+b=0$に$x=-2,3$をそれぞれ代入すると，

\[
\{\begin{array}{l}
4a-4+b=0\\
9a+6+b=0
\end{array}
\]

これを解くと，$a=-2,b=12$となり，$a<0$を満たす．

よって，$a=-2,b=12$

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% 例題I3.3.20の別解
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$-2\leqq x\leqq 3$を解にもつ2次不等式のうち，$x^2$の係数が1のものは，$(x+2)(x-3)\leqq 0$と表される．

したがって，$x^2-x-6\leqq 0\cdots(\mathrm{i})$

$ax^2+2x+b\geqq 0\cdots(\mathrm{ii})$の$x$の係数が$2$であるから，(i)の両辺に$-2$を掛けて，

\[
-2x^2+2x+12\geqq 0
\]

よって，(ii)と係数を比較すると，$a=-2,b=12$

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% 問題I3.3.20：不等式の係数決定（One More）★★
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2次不等式$2ax^2+bx+1\leqq 0$の解が$x\leqq-\frac{1}{2},3\leqq x$となるとき，定数$a,b$の値を求めよ．

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% 問題I3.3.20の解答（One More）★★
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$y=2ax^2+bx+1\cdots(\mathrm{i})$とする．

$2ax^2+bx+1\leqq 0$の解が$x\leqq-\frac{1}{2},3\leqq x$となるのは，(i)のグラフが右の図のようになるときであるから，$a<0$

このとき，解が$x\leqq-\frac{1}{2},3\leqq x$であるから，$y=2ax^2+bx+1$のグラフは$x$軸と$x=-\frac{1}{2},3$で交わる．

したがって，2次方程式$2ax^2+bx+1=0$の解は，$x=-\frac{1}{2},3$となる．

ゆえに，$2ax^2+bx+1=0$に$x=-\frac{1}{2},3$をそれぞれ代入すると，

\[
\{\begin{array}{l}
\frac{1}{2}a-\frac{1}{2}b+1=0\\
18a+3b+1=0
\end{array}
\]

これを解くと，$a=-\frac{1}{3},b=\frac{5}{3}$となり，$a<0$を満たす．

よって，$a=-\frac{1}{3},b=\frac{5}{3}$

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% 問題I3.3.20の別解
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$x\leqq-\frac{1}{2},3\leqq x$を解にもつ2次不等式のうち，$x^2$の係数が1のものは，$(x+\frac{1}{2})(x-3)\geqq 0$と表される．

したがって，$x^2-\frac{5}{2}x-\frac{3}{2}\geqq 0\cdots(\mathrm{i})$

$2ax^2+bx+1\leqq 0\cdots(\mathrm{ii})$の定数項が1であるから，(i)の両辺に$-\frac{2}{3}$を掛けて，

\[
-\frac{2}{3}x^2+\frac{5}{3}x+1\leqq 0
\]

よって，(ii)と係数を比較すると，$a=-\frac{1}{3},b=\frac{5}{3}$

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% 例題I3.3.20の側注：不等式の係数決定（One More）★★
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% 側注1
$a>0$のときは，解は$x\leqq-2,3\leqq x$の形になるので不適である．

% 側注2
$\alpha<\beta$のとき，$\alpha\leqq x\leqq\beta\Longleftrightarrow(x-\alpha)(x-\beta)\leqq 0$

% 側注3
両辺に$-2$を掛けたので，(i)の不等号の向きが変わり，(ii)の不等号の向きと一致する．

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% 問題I3.3.20の側注：不等式の係数決定（One More）★★
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% 側注1
$a>0$のときは，解は$-\frac{1}{2}\leqq x\leqq 3$の形になるので不適である．

% 側注2
$\alpha<\beta$のとき，$x\leqq\alpha,\beta\leqq x\Longleftrightarrow(x-\alpha)(x-\beta)\geqq 0$

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% 例題I3.3.20の考え方・補足など：不等式の係数決定（One More）★★
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% 考え方
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$y=ax^2+2x+b$として，2次不等式の解を2次関数のグラフで考える．$y=ax^2+2x+b$のグラフの上側にある部分の解は，$a>0$のときと，$a<0$のときで変わるので注意すること（ここでは，$a>0$のときは不適である）．

$a>0$のとき

$a<0$のとき