【数学I】例題3.3.36：2次式の絶対値を含む方程式（定数分離）（One More）★★★

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% 例題I3.3.36：2次式の絶対値を含む方程式（定数分離）（One More）★★★
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方程式$|x^2+2x-3|=2x+a$の異なる実数解の個数を調べよ．ただし，$a$は定数とする．

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% 例題I3.3.36の解答（One More）★★★
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$|x^2+2x-3|=2x+a$より，$f(x)=|x^2+2x-3|-2x$とする．

(i) $x^2+2x-3\geqq 0$のとき，すなわち，$(x+3)(x-1)\geqq 0$より，$x\leqq-3,1\leqq x$のとき

\[
f(x)=x^2+2x-3-2x=x^2-3
\]

(ii) $x^2+2x-3<0$のとき，すなわち，$(x+3)(x-1)<0$より，$-3<x<1$のとき

\begin{align*}
f(x)&=-(x^2+2x-3)-2x\\
=&-x^2-4x+3\\
=&-(x+2)^2+7
\end{align*}

よって，(i)，(ii)より，$y=f(x)$のグラフは右の図のようになる．

求める実数解の個数は，$y=f(x)$と$y=a$のグラフの共有点の個数と一致するので，右の図より，

$a<-2$のとき，0個

$a=-2$のとき，1個

$-2<a<6$のとき，2個

$a=6$のとき，3個

$6<a<7$のとき，4個

$a=7$のとき，3個

$a>7$のとき，2個

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% 問題I3.3.36：2次式の絶対値を含む方程式（定数分離）（One More）★★★
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方程式$|x^2-4x+3|=x+a$の異なる実数解の個数を調べよ．ただし，$a$は定数とする．

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% 問題I3.3.36の解答（One More）★★★
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$|x^2-4x+3|=x+a$より，$f(x)=|x^2-4x+3|-x$とする．

(i) $x^2-4x+3\geqq 0$のとき，すなわち，$(x-1)(x-3)\geqq 0$より，$x\leqq 1$または$x\geqq 3$のとき

\begin{align*}
f(x)&=x^2-4x+3-x\\
=&x^2-5x+3\\
=&(x-\frac{5}{2})^2-\frac{13}{4}
\end{align*}

(ii) $x^2-4x+3<0$のとき，すなわち，$1<x<3$のとき

\begin{align*}
f(x)&=-(x^2-4x+3)-x\\
=&-x^2+3x-3\\
=&-(x-\frac{3}{2})^2-\frac{3}{4}
\end{align*}

よって，(i)，(ii)より，$y=f(x)$のグラフは右の図のようになる．

求める実数解の個数は，$y=f(x)$と$y=a$のグラフの共有点の個数と一致するので，右の図より，

$a<-3$のとき，0個

$a=-3$のとき，1個

$-3<a<-1$のとき，2個

$a=-1$のとき，3個

$-1<a<-\frac{3}{4}$のとき，4個

$a=-\frac{3}{4}$のとき，3個

$a>-\frac{3}{4}$のとき，2個

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% 例題I3.3.36の側注：2次式の絶対値を含む方程式（定数分離）（One More）★★★
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% 側注1
グラフをかき，$y=a$を動かすことをイメージして，共有点の個数を調べる．

% 側注2
$y=|x^2+2x-3|$のグラフは次のようになる．このグラフと直線$y=2x+a$の共有点を調べるよりも，解答のように定数分離した方が，共有点の個数を調べやすい．

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% 問題I3.3.36の側注：2次式の絶対値を含む方程式（定数分離）（One More）★★★
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% 側注1
グラフをかき，$y=a$を動かすことをイメージして，共有点の個数を調べる．

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% 例題I3.3.36の考え方・補足など：2次式の絶対値を含む方程式（定数分離）（One More）★★★
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% 考え方
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与えられた方程式は，$\{\begin{array}{l}y=|x^2+2x-3|\\y=2x+a\end{array}$の連立方程式の$y$を消去した形と見ることができるが，このままでは共有点の個数を調べにくい．そこで，与えられた方程式を次のように式変形するとよい．

\[
|x^2+2x-3|-2x=a(\text{定数}a\text{を分離した形})
\]

このように分ける方法を，定数分離という．

定数分離を用いて，$y=|x^2+2x-3|-2x$のグラフと直線$y=a$の共有点の個数を調べることで，$a$の範囲を求めることができる．