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【数学I】例題1.3.6:文字を含む1次不等式(One More)★★★

【数学I】例題1.3.6:文字を含む1次不等式(One More)
【数学I】例題1.3.6:文字を含む1次不等式の例題ページ
問題の解答

【数学I】問題1.3.6:文字を含む1次不等式の解答
検索用コード(LaTeX)
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$a$を定数とするとき,次の問いに答えよ.

(1) $x$の不等式$ax-3<0$を解け.

(2) $x$の不等式$(a+1)x<a^2+a$を解け.

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(1) $ax-3<0$より,$ax<3$

(i) $a>0$のとき,$x<\frac{3}{a}$

(ii) $a=0$のとき,不等式は,$0\cdot x<3$

したがって,解はすべての実数

(iii) $a<0$のとき,不等式は,$x>\frac{3}{a}$

よって,(i)〜(iii)より,求める解は,

\[
\{
\begin{array}{lll}
a>0&\text{のとき,}&x<\frac{3}{a}\\
a=0&\text{のとき,}&\text{すべての実数}\\
a<0&\text{のとき,}&x>\frac{3}{a}
\end{array}
\]

(2) $(a+1)x<a^2+a$より,$(a+1)x<a(a+1)$

(i) $a+1>0$,すなわち,$a>-1$のとき,$x<a$

(ii) $a+1=0$,すなわち,$a=-1$のとき,$0\cdot x<0$

これを満たす$x$の値はない.したがって,解なし

(iii) $a+1<0$,すなわち,$a<-1$のとき,$x>a$

よって,(i)〜(iii)より,求める解は,

\[
\{
\begin{array}{lll}
a>-1&\text{のとき,}&x<a\\
a=-1&\text{のとき,}&\text{解なし}\\
a<-1&\text{のとき,}&x>a
\end{array}
\]

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$a$を定数とするとき,次の問いに答えよ.

(1) $x$の不等式$ax+2>0$を解け.

(2) $x$の不等式$(a-1)x\leqq a^2-a$を解け.

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% 問題I1.3.6の解答(One More)★★★
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(1) $ax+2>0$より,$ax>-2$

(i) $a>0$のとき,$x>-\frac{2}{a}$

(ii) $a=0$のとき,不等式は$0\cdot x>-2$

したがって,解はすべての実数

(iii) $a<0$のとき,不等式は$x<-\frac{2}{a}$

よって,(i)〜(iii)より,求める解は,

\[
\{
\begin{array}{lll}
a>0&\text{のとき,}&x>-\frac{2}{a}\\
a=0&\text{のとき,}&\text{すべての実数}\\
a<0&\text{のとき,}&x<-\frac{2}{a}
\end{array}
\]

(2) $(a-1)x\leqq a^2-a$より,$(a-1)x\leqq a(a-1)$

(i) $a-1>0$すなわち,$a>1$のとき,$x\leqq a$

(ii) $a-1=0$すなわち,$a=1$のとき,$0\cdot x\leqq 0$

これを満たす$x$の値はすべての実数.したがって,解はすべての実数

(iii) $a-1<0$すなわち,$a<1$のとき,$x\geqq a$

よって,(i)〜(iii)より,求める解は,

\[
\{
\begin{array}{lll}
a>1&\text{のとき,}&x\leqq a\\
a=1&\text{のとき,}&\text{すべての実数}\\
a<1&\text{のとき,}&x\geqq a
\end{array}
\]

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