【数学I】例題1.3.8：絶対値記号を含む方程式・不等式2（One More）★★★

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% 例題I1.3.8：絶対値記号を含む方程式・不等式2（One More）★★★
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次の方程式，不等式を解け．

(1) $|x-3|=2x$

(2) $|x|+|x-3|<x+1$

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% 例題I1.3.8の解答（One More）★★★
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(1)

(i) $x-3\geqq 0$，すなわち，$x\geqq 3$のとき

$x-3=2x$より，$x=-3$

これは，$x\geqq 3$を満たさない．

(ii) $x-3<0$，すなわち，$x<3$のとき

$-(x-3)=2x$より，$x=1$

これは，$x<3$を満たす．

よって，(i)，(ii)より，$x=1$

(2)

(i) $x\geqq 3$のとき

$x+(x-3)<x+1$より，$x<4$

したがって，$x\geqq 3$より，$3\leqq x<4$

(ii) $0\leqq x<3$のとき

$x-(x-3)<x+1$より，$x>2$

したがって，$0\leqq x<3$より，$2<x<3$

(iii) $x<0$のとき

$-x-(x-3)<x+1$より，$x>\frac{2}{3}$

これは，$x<0$を満たさないので，解なし

よって，(i)〜(iii)より，$2<x<4$

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% 問題I1.3.8：絶対値記号を含む方程式・不等式2（One More）★★★
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次の方程式，不等式を解け．

(1) $|x+2|=3x$

(2) $|x+2|-|x-1|\geqq x$

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% 問題I1.3.8の解答（One More）★★★
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(1)

(i) $x+2\geqq 0$，すなわち，$x\geqq-2$のとき

$x+2=3x$より，$2x=2$

したがって，$x=1$

これは，$x\geqq-2$を満たす．

(ii) $x+2<0$，すなわち，$x<-2$のとき

$-(x+2)=3x$より，$-x-2=3x$

したがって，$-4x=2$より，$x=-\frac{1}{2}$

これは，$x<-2$を満たさない．

よって，(i)，(ii)より，$x=1$

(2)

(i) $x\geqq 1$のとき

$x+2-(x-1)\geqq x$より，$x\leqq 3$

したがって，$x\geqq 1$より，$1\leqq x\leqq 3$

(ii) $-2\leqq x<1$のとき

$(x+2)+(x-1)\geqq x$より，$x\geqq-1$

したがって，$-2\leqq x<1$より，$-1\leqq x<1$

(iii) $x<-2$のとき

$-(x+2)+(x-1)\geqq x$より，$x\leqq-3$

したがって，$x<-2$より，$x\leqq-3$

よって，(i)〜(iii)より，$x\leqq-3$または$-1\leqq x\leqq 3$

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% 例題I1.3.8の側注：絶対値記号を含む方程式・不等式2（One More）★★★
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% 側注1
$x-3=X$とおくと，
\[
|X|=\{\begin{array}{rr}X&(X\geqq 0)\\-X&(X<0)\end{array}
\]
であるので，$X$が$0$以上のときと負のときで場合分けをする．

% 側注2
求めた$x$の値が$x$の条件を満たすか否かを調べる（忘れないように注意すること）．

% 側注3
$x\geqq 3$のとき，$|x|=x,|x-3|=x-3$

% 側注4
$0\leqq x<3$のとき，$|x|=x,|x-3|=-(x-3)$

% 側注5
$x<0$のとき，$|x|=-x,|x-3|=-(x-3)$

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% 問題I1.3.8の側注：絶対値記号を含む方程式・不等式2（One More）★★★
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% 側注1
$x+2=X$とおくと，
\[
|X|=\{\begin{array}{rr}X&(X\geqq 0)\\-X&(X<0)\end{array}
\]
であるので，$X$が$0$以上のときと負のときで場合分けをする．

% 側注2
求めた$x$の値が$x$の条件を満たすか否かを調べる．

% 側注3
$x\geqq 1$のとき，$|x+2|=x+2,|x-1|=x-1$

% 側注4
$-2\leqq x<1$のとき，$|x+2|=x+2,|x-1|=-(x-1)$

% 側注5
$x<-2$のとき，$|x+2|=-(x+2),|x-1|=-(x-1)$

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% 例題I1.3.8の考え方・補足など：絶対値記号を含む方程式・不等式2（One More）★★★
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% 考え方
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絶対値記号の外に文字がある場合や，2つ以上の絶対値記号がある場合は，場合分けをするとよい．とくに，(2)は，

\[
|x|=\{\begin{array}{rl}
x&(x\geqq 0)\\
-x&(x<0)
\end{array},|x-3|=\begin{cases}x-3&(x\geqq 3)\\
-(x-3)&(x<3)\end{cases}
\]

となるので，右の図のように数直線を用いて，(i)$x\geqq 3$，(ii)$0\leqq x<3$，(iii)$x<0$の3つの部分に分けて考えるとよい．