【数学I】3章：2次関数（節末問題・章末問題）

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% 節末問題I3.1.1：（One More）★
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関数$f(x)=3x^2-2x+2$について，$f(f(a))$の値を求めよ．

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% 節末問題I3.1.1の解答（One More）★
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$f(x)$の式に，$x=f(a)=3a^2-2a+2$を代入して，

\begin{align*}
f(f(a))&=3(3a^2-2a+2)^2-2(3a^2-2a+2)+2\\
&=27a^4-36a^3+42a^2-20a+10
\end{align*}

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% 節末問題I3.1.2：（One More）★★
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2つの放物線$y=3x^2-18x+25$と$y=ax^2+8x+b$の頂点が一致するように定数$a,b$の値を定めよ．

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% 節末問題I3.1.2の解答（One More）★★
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\[
y=3x^2-18x+25=3(x^2-6x)+25=3(x-3)^2-2,
\]

\begin{align*}
y&=ax^2+8x+b\\
&=a(x^2+\frac{8}{a}x)+b\\
&=a(x+\frac{4}{a})^2-a(\frac{4}{a})^2+b\\
&=a(x+\frac{4}{a})^2-\frac{16}{a}+b
\end{align*}

よって，2つの放物線の頂点の座標は，それぞれ，

\[
(3,-2),(-\frac{4}{a},-\frac{16}{a}+b)
\]

与えられた条件より，

\[
3=-\frac{4}{a},-2=-\frac{16}{a}+b
\]

これを解くと，$a=-\frac{4}{3},b=-14$

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% 節末問題I3.1.2の別解
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放物線$y=3x^2-18x+25$の頂点は，点$(3,-2)$

したがって，2つの放物線の頂点が一致するための条件は，$y=ax^2+8x+b$が$y=a(x-3)^2-2\cdots(\mathrm{i})$と表されることであるから，(i)を展開すると，

\[
y=ax^2-6ax+9a-2
\]

これと$y=ax^2+8x+b$の係数を比較すると，$8=-6a,b=9a-2$

これを解くと，$a=-\frac{4}{3},b=-14$

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% 節末問題I3.1.3：（One More）★★
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$y=ax^2+bx+c$で表される放物線が点$(1,-2)$に関して放物線$y=-3x^2$と点対称であるとき，定数$a,b,c$の値を求めよ．

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% 節末問題I3.1.3の解答（One More）★★
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放物線$y=-3x^2$の頂点は点$(0,0)$

この点を点$(1,-2)$に関して対称に移動すると，点$(2,-4)$となる．

もとのグラフの$x^2$の係数が$-3$であるから，移動したグラフは下に凸の放物線であり，$x^2$の係数は$3$である．

したがって，点$(1,-2)$に関して放物線$y=-3x^2$と点対称である放物線の方程式は

\[
y=3(x-2)^2-4=3x^2-12x+8
\]

よって，$a=3,b=-12,c=8$

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% 節末問題I3.1.4：（One More）★★
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放物線$y=x^2-4x+1$を$x$軸方向に3だけ平行移動し，さらに直線$y=2$に関して折り返してできる放物線の方程式を求めよ．

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% 節末問題I3.1.4の解答（One More）★★
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\[
y=x^2-4x+1=(x-2)^2-3
\]

したがって，もとの放物線$y=x^2-4x+1$の頂点は点$(2,-3)$である．

この頂点を$x$軸方向に3だけ平行移動すると，点$(5,-3)$となる．

さらに，直線$y=2$に関して折り返すと，頂点は点$(5,7)$となる．

もとのグラフの$x^2$の係数が$1$であるから，移動したグラフは上に凸の放物線であり，$x^2$の係数は$-1$である．

よって，$y=-(x-5)^2+7$

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% 節末問題I3.1.5：（One More）★★★
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次の関数$f(x)$の最小値とそのときの$x$の値を求めよ．

(1) $f(x)=|x-2|+|x-4|+|x-6|$

(2) $f(x)=|x+|2x-10||$

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% 節末問題I3.1.5の解答（One More）★★★
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(1) $x<2$のとき

\[
f(x)=-(x-2)-(x-4)-(x-6)=-3x+12
\]

$2\leqq x<4$のとき

\[
f(x)=(x-2)-(x-4)-(x-6)=-x+8
\]

$4\leqq x<6$のとき

\[
f(x)=(x-2)+(x-4)-(x-6)=x
\]

$6\leqq x$のとき

\[
f(x)=(x-2)+(x-4)+(x-6)=3x-12
\]

よって，$y=f(x)$のグラフは右の図のようになるから，$f(x)$は$x=4$で最小値4をとる．

(2) $2x-10\geqq 0$，すなわち，$x\geqq 5$のとき

\[
f(x)=|x+2x-10|=|3x-10|
\]

$x\geqq 5$より，$3x-10>0$であるから，$f(x)=3x-10$

$2x-10<0$，すなわち，$x<5$のとき

\[
f(x)=|x-(2x-10)|=|10-x|
\]

$x<5$より，$10-x>0$であるから，$f(x)=-x+10$

よって，$y=f(x)$のグラフは右の図のようになるから，$f(x)$は$x=5$で最小値5をとる．

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% 節末問題I3.2.1：（One More）★★
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関数$y=x^2-6x+k+3(-1\leqq x\leqq 3)$の最大値と最小値の和が$0$であるとき，定数$k$の値とそのときの最大値，最小値を求めよ．

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% 節末問題I3.2.1の解答（One More）★★
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\[
y=x^2-6x+k+3=(x-3)^2+k-6
\]

したがって，グラフは右の図のようになる．

グラフより，$x=-1$のとき，最大値$k+10$，$x=3$のとき，最小値$k-6$をとる．

最大値と最小値の和が$0$であるから，

\[
(k+10)+(k-6)=0
\]

よって，$k=-2$

また，$x=-1$のとき，最大値$8$，$x=3$のとき，最小値$-8$

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% 節末問題I3.2.2：（One More）★★★★
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2次関数$y=-2x^2+8x$について，次の問いに答えよ．

(1) この関数のグラフの頂点，$x$軸との共有点，$y$軸との共有点の座標を求め，グラフをかけ．

(2) $a\leqq x\leqq a+1$における関数の最大値が6であるような定数$a$の値を求めよ．

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% 節末問題I3.2.2の解答（One More）★★★★
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(1) $f(x)=-2x^2+8x$とすると，

\[
f(x)=-2x^2+8x=-2(x^2-4x)=-2(x-2)^2+8
\]

したがって，$y=f(x)$のグラフの頂点の座標は，$(2,8)$

$f(x)=0$とすると，$-2x^2+8x=0$

ゆえに，$-2x(x-4)=0$より，$x=0,4$

したがって，$y=f(x)$と$x$軸との共有点の座標は，$(0,0),(4,0)$

また，$f(0)=0$より，$y$軸との共有点の座標は，$(0,0)$

よって，$y=f(x)$のグラフは右の図のようになる．

(2) $y=f(x)$の最大値が6であるから，$a\leqq x\leqq a+1$に軸である直線$x=2$は含まれない．

(i) $a+1<2$，すなわち$a<1$のとき

$f(x)$は$x=a+1$のとき最大となるから，

\[
f(a+1)=-2(a+1)^2+8(a+1)=-2a^2+4a+6
\]

したがって，$f(a+1)=6$より，$-2a^2+4a+6=6$

ゆえに，$-2a(a-2)=0$であるから，$a=0,2$

したがって，$a<1$より，$a=0$

(ii) $2\leqq a$のとき

$f(x)$は$x=a$のとき最大となるから，

\[
f(a)=-2a^2+8a
\]

したがって，$f(a)=6$より，$-2a^2+8a=6$

ゆえに，$(a-1)(a-3)=0$であるから，$a=1,3$

したがって，$2\leqq a$より，$a=3$

よって，(i)，(ii)より，求める$a$の値は，$a=0,3$

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% 節末問題I3.2.3：（One More）★★
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$a$を定数として，関数$y=(x^2-4x)^2+2a(x^2-4x)+a+2$の最小値を$m$とする．このとき，次の問いに答えよ．

(1) $m$を$a$の式で表せ．

(2) $m$を最大にする$a$の値を求めよ．

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% 節末問題I3.2.3の解答（One More）★★
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(1)

\[
y=(x^2-4x)^2+2a(x^2-4x)+a+2
\]

$t=x^2-4x$とおくと，$t=(x-2)^2-4$より，$t\geqq-4$

したがって，

\begin{align*}
y&=t^2+2at+a+2\\
&=(t+a)^2-a^2+a+2(t\geqq-4)
\end{align*}

(i) $-a<-4$，すなわち，$a>4$のとき

$y$は$t=-4$のとき，最小値をとる．

したがって，$m=(-4)^2+2a\cdot(-4)+a+2=-7a+18$

(ii) $-a\geqq-4$，すなわち，$a\leqq 4$のとき

$y$は$t=-a$のとき，最小値をとる．

したがって，$m=-a^2+a+2$

よって，(i)，(ii)より，

\[
m=\begin{cases}-7a+18&(a>4)\\
-a^2+a+2&(a\leqq 4)\end{cases}
\]

(2) $a>4$のとき

\[
m=-7a+18
\]

$a\leqq 4$のとき

\[
m=-a^2+a+2=-(a-\frac{1}{2})^2+\frac{9}{4}
\]

よって，グラフは右の図のようになり，$m$は$a=\frac{1}{2}$のとき，最大値$\frac{9}{4}$

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% 節末問題I3.2.4：（One More）★★★
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放物線$y=ax^2+bx+c$は，頂点の座標が$(3,7)$で，点$(6,-5)$を通る．このとき，定数$a,b,c$の値を求めよ．

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% 節末問題I3.2.4の解答（One More）★★★
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頂点が点$(3,7)$であるから，求める放物線は，

\[
y=a(x-3)^2+7
\]

と表される．

この放物線が，点$(6,-5)$を通るから，

\[
-5=a(6-3)^2+7
\]

したがって$a=-\frac{4}{3}$

このとき，$y=-\frac{4}{3}(x-3)^2+7=-\frac{4}{3}x^2+8x-5$

これが，$y=ax^2+bx+c$と一致するので，$b=8,c=-5$

よって，$a=-\frac{4}{3},b=8,c=-5$

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% 節末問題I3.2.5：（One More）★★★
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$a>0,b>0,a+b=1$のとき，$a^3+b^3$の最小値を求めよ．

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% 節末問題I3.2.5の解答（One More）★★★
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$a+b=1$から，$b=1-a\cdots(\mathrm{i})$

$b>0$であるから，$1-a>0$

したがって，$a<1$

$a>0$との共通範囲は，$0<a<1\cdots(\mathrm{ii})$

$a^3+b^3=t$とすると，

\[
t=a^3+(1-a)^3=3a^2-3a+1=3(a-\frac{1}{2})^2+\frac{1}{4}
\]

(ii) において，$t$は$a=\frac{1}{2}$のとき，最小値$\frac{1}{4}$

$a=\frac{1}{2}$のとき，(i)から$b=\frac{1}{2}$

よって，$a=\frac{1}{2},b=\frac{1}{2}$のとき，最小値$\frac{1}{4}$

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% 節末問題I3.2.6：（One More）★★★
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2次関数$f(x)=ax^2+bx+c$が，$f(-2)=f(4)=0$を満たし，その最大値が9であるとき，定数$a,b,c$の値を求めよ．

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% 節末問題I3.2.6の解答（One More）★★★
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$f(-2)=f(4)=0$であるから，放物線$y=f(x)$の軸は，2点$(-2,0),(4,0)$を結ぶ線分の中点$(1,0)$を通る．

したがって，$f(x)$は$x=1$で最大値9をとり，$f(x)$は$f(x)=a(x-1)^2+9(a<0)$と表される．

$f(-2)=0$から，$9a+9=0$

ゆえに，$a=-1$

これは，$a<0$を満たす．

したがって，$f(x)=-(x-1)^2+9$

展開すると，$f(x)=-x^2+2x+8$である．

よって，$a=-1,b=2,c=8$となる．

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% 節末問題I3.2.6の別解
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$f(-2)=f(4)=0$であるから，$f(x)=a(x+2)(x-4)$と表される．

展開すると，$a(x+2)(x-4)=a(x^2-2x-8)=a(x-1)^2-9a$であるから，$f(x)=a(x-1)^2-9a$

最大値が9であるから，$a<0$かつ$-9a=9$

したがって，$a=-1$

これは$a<0$を満たす．

ゆえに，$f(x)=-(x+2)(x-4)=-x^2+2x+8$

よって，$a=-1,b=2,c=8$

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% 節末問題I3.3.1：（One More）★★
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2次方程式$x^2+4x+7a=0$，$2x^2-6x-a=0$がともに実数解をもつ整数$a$の個数を求めよ．

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% 節末問題I3.3.1の解答（One More）★★
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2次方程式$x^2+4x+7a=0$の判別式を$D_1$とすると，$\frac{D_1}{4}=2^2-1\cdot 7a=4-7a$

2次方程式が実数解をもつから，$D_1\geqq 0$

したがって，$4-7a\geqq 0$

ゆえに，$a\leqq\frac{4}{7}\cdots(\mathrm{i})$

2次方程式$2x^2-6x-a=0$の判別式を$D_2$とすると，$\frac{D_2}{4}=(-3)^2-2\cdot(-a)=9+2a$

2次方程式が実数解をもつから，$D_2\geqq 0$

したがって，$9+2a\geqq 0$

ゆえに，$a\geqq-\frac{9}{2}\cdots(\mathrm{ii})$

したがって，(i)と(ii)の共通範囲を求めると，
\[
-\frac{9}{2}\leqq a\leqq\frac{4}{7}
\]

この不等式を満たす整数$a$は，$a=-4,-3,-2,-1,0$

よって，求める整数$a$の個数は，5個

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% 節末問題I3.3.2：（One More）★★★
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実数$x,y$が$x^2+y^2=13$のもとで，$x-ay$の最大値が7となるとき，定数$a$の値を求めよ．

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% 節末問題I3.3.2の解答（One More）★★★
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$x-ay=k$とおくと，$x=k+ay$

これを$x^2+y^2=13$に代入すると，$(k+ay)^2+y^2=13$

したがって，$(a^2+1)y^2+2aky+k^2-13=0$

$y$は実数であるから，この2次方程式の判別式を$D$とすると，$D\geqq 0$

\[
\frac{D}{4}=(ak)^2-(a^2+1)(k^2-13)=-k^2+13(a^2+1)
\]

これより，$-k^2+13(a^2+1)\geqq 0$

ゆえに，$k^2\leqq 13(a^2+1)$

したがって，$-\sqrt{13(a^2+1)}\leqq k\leqq\sqrt{13(a^2+1)}$

$k$の最大値が7であるから，$\sqrt{13(a^2+1)}=7$

よって，$a^2=\frac{36}{13}$

これを解いて，$a=\pm\frac{6}{\sqrt{13}}$

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% 節末問題I3.3.3：（One More）★★★
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(1) 不等式$3x^4-7x^2+2>0$を解け．

(2) 不等式$(x^2-3x+2)^2-4(x^2-3x+2)+3\leqq 0$を解け．

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% 節末問題I3.3.3の解答（One More）★★★
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(1) $x^2=t$とおくと，$t\geqq 0$

与えられた不等式は，$3t^2-7t+2>0$

したがって，$(3t-1)(t-2)>0$

これを解くと，$0\leqq t<\frac{1}{3},2<t$

ゆえに，$0\leqq x^2<\frac{1}{3},2<x^2$

ここで，$x^2\geqq 0$は常に成り立つので，

$x^2<\frac{1}{3}$より，$-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

$2<x^2$より，$x<-\sqrt{2},\sqrt{2}<x$

よって，$x<-\sqrt{2},-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}},\sqrt{2}<x$

(2) $x^2-3x+2=t$とおくと，$t=(x-\frac{3}{2})^2-\frac{1}{4}$である．

$t$のとりうる値の範囲は，$t\geqq-\frac{1}{4}$

不等式を$t$で表すと，$t^2-4t+3\leqq 0$

したがって，$(t-1)(t-3)\leqq 0$より，$1\leqq t\leqq 3$

これは$t\geqq-\frac{1}{4}$を満たす．

ゆえに，$1\leqq x^2-3x+2\leqq 3$

$1\leqq x^2-3x+2$より，$x^2-3x+1\geqq 0$

$x^2-3x+1=0$を解くと，$x=\frac{3\pm\sqrt{5}}{2}$

このとき，$x^2-3x+1\geqq 0$の解は，

\[
x\leqq\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}\leqq x\cdots(\mathrm{i})
\]

$x^2-3x+2\leqq 3$より，$x^2-3x-1\leqq 0$

$x^2-3x-1=0$を解くと，$x=\frac{3\pm\sqrt{13}}{2}$

このとき，$x^2-3x-1\leqq 0$の解は，

\[
\frac{3-\sqrt{13}}{2}\leqq x\leqq\frac{3+\sqrt{13}}{2}\cdots(\mathrm{ii})
\]

よって，(i)，(ii)の共通範囲を求めると，

\[
\frac{3-\sqrt{13}}{2}\leqq x\leqq\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}\leqq x\leqq\frac{3+\sqrt{13}}{2}
\]

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% 節末問題I3.3.4：（One More）★★★★
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$a$を定数とする．$x$についての方程式$|(x-3)(x-5)|=ax-2a+\frac{1}{2}$が異なる4つの実数解をもつとき，$a$の値の範囲を求めよ．

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% 節末問題I3.3.4の解答（One More）★★★★
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\[
y=|(x-3)(x-5)|\cdots(\mathrm{i}),y=ax-2a+\frac{1}{2}\cdots(\mathrm{ii})
\]

とする．

$(x-3)(x-5)\geqq 0$の解は，$x\leqq 3,5\leqq x$

$(x-3)(x-5)<0$の解は，$3<x<5$

したがって，(i)は，

$x\leqq 3,5\leqq x$のとき，$y=(x-3)(x-5)=(x-4)^2-1$

$3<x<5$のとき，$y=-(x-3)(x-5)=-(x-4)^2+1$

よって，(i)のグラフは右の図のようになる．

(ii) は，$y=a(x-2)+\frac{1}{2}$と変形できるから，(ii)のグラフは定点$(2,\frac{1}{2})$を通る傾き$a$の直線である．

（ア）(ii)のグラフが(i)のグラフの$3\leqq x\leqq 5$の部分と接するとき

このとき，2次方程式$-(x-3)(x-5)=ax-2a+\frac{1}{2}$，すなわち，$x^2+(a-8)x-2a+\frac{31}{2}=0$の判別式を$D$とすると，

\[
D=(a-8)^2-4\cdot 1\cdot(-2a+\frac{31}{2})=a^2-8a+2
\]

$D=0$より，$a^2-8a+2=0$

これを解いて，$a=4\pm\sqrt{14}$

このとき，$3\leqq x\leqq 5$の部分と接するのは，$a=4-\sqrt{14}$のときである．

（イ）(ii)のグラフが点$(5,0)$を通るとき

$0=5a-2a+\frac{1}{2}$より，$a=-\frac{1}{6}$

よって，（ア），（イ）より，方程式$|(x-3)(x-5)|=ax-2a+\frac{1}{2}$が相異なる4つの実数解をもつとき，$a$の値の範囲は，

\[
-\frac{1}{6}<a<4-\sqrt{14}
\]

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% 節末問題I3.3.5：（One More）★★★
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2次不等式$x^2-3x-4<|x-2|$を満たす$x$の値の範囲を求めよ．

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\[
|x-2|=\begin{cases}x-2&(x\geqq 2)\\
-x+2&(x<2)\end{cases}
\]

$x^2-3x-4=(x-\frac{3}{2})^2-\frac{25}{4}$であるから，$y=x^2-3x-4$のグラフと$y=|x-2|$のグラフは，右の図のようになる．

2つのグラフの交点の$x$座標は，

(i) $x\geqq 2$のとき

$x^2-3x-4=x-2$より，$x^2-4x-2=0$

したがって，$x=2\pm\sqrt{6}$

ゆえに，$x\geqq 2$より，$x=2+\sqrt{6}$

(ii) $x<2$のとき

$x^2-3x-4=-(x-2)$より，$x^2-2x-6=0$

したがって，$x=1\pm\sqrt{7}$

ゆえに，$x<2$より，$x=1-\sqrt{7}$

よって，(i)，(ii)とグラフより，求める$x$の値の範囲は，$1-\sqrt{7}<x<2+\sqrt{6}$

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(i) $x\geqq 2$のとき

$x^2-3x-4<x-2$より，$x^2-4x-2<0$

ゆえに，$2-\sqrt{6}<x<2+\sqrt{6}$

これと$x\geqq 2$より，$2\leqq x<2+\sqrt{6}$

(ii) $x<2$のとき

$x^2-3x-4<-(x-2)$より，$x^2-2x-6<0$

ゆえに，$1-\sqrt{7}<x<1+\sqrt{7}$

これと$x<2$より，$1-\sqrt{7}<x<2$

よって，(i)，(ii)より，求める$x$の値の範囲は，$1-\sqrt{7}<x<2+\sqrt{6}$

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% 章末問題I3.1：（One More）★★★
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実数$x,y$が$x^2-xy+y^2+y-5=0$を満たすとき，$y$の最大値は$\boxed{\text{}}$であり，最小値は$\boxed{\text{}}$である．

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$y$が$k$という値をとるとすると，$x^2-kx+k^2+k-5=0\cdots(\mathrm{i})$を満たす実数$x$が存在する．

したがって，この条件は，(i)の判別式を$D$とすると，

\[
D=(-k)^2-4(k^2+k-5)
\]

$D\geqq 0$であるから，$k^2-4k^2-4k+20\geqq 0$

ゆえに，$-3k^2-4k+20\geqq 0$より，$(3k+10)(k-2)\leqq 0$

したがって，$-\frac{10}{3}\leqq k\leqq 2$

よって，$y$の最大値は$2$，最小値は$-\frac{10}{3}$

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% 章末問題I3.2：（One More）★★
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2つの放物線$y=x^2+4$，$y=-x^2+4x$の両方に接する直線の方程式を求めよ．

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求める直線の方程式を$y=ax+b\cdots(\mathrm{i})$とおく．

直線(i)と放物線$y=x^2+4$が接するから，$x^2+4=ax+b$

すなわち，$x^2-ax+(4-b)=0$

この2次方程式の判別式を$D_1$とすると，$D_1=(-a)^2-4\cdot 1\cdot(4-b)$

したがって，$D_1=0$より，$a^2+4b-16=0\cdots(\mathrm{ii})$

直線(i)と放物線$y=-x^2+4x$が接するから，$-x^2+4x=ax+b$

すなわち，$x^2+(a-4)x+b=0$

この2次方程式の判別式を$D_2$とすると，$D_2=(a-4)^2-4\cdot 1\cdot b$

したがって，$D_2=0$より，$a^2-8a+16-4b=0$

ゆえに，$4b=a^2-8a+16\cdots(\mathrm{iii})$

(iii) を(ii)に代入して整理すると，

\[
2a^2-8a=0
\]

したがって，$2a(a-4)=0$より，$a=0,4$

$a=0$のとき，(ii)より，$b=4$

$a=4$のとき，(ii)より，$b=0$

よって，求める直線の方程式は，$y=4,y=4x$

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% 章末問題I3.3：（One More）★★★
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$a,b$は自然数で，2次方程式$x^2+2ax+4a-4b=0$が重解$\alpha$をもつとき，$a,b,\alpha$の値を求めよ．

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2次方程式$x^2+2ax+4a-4b=0$の判別式を$D$とすると，

\[
\frac{D}{4}=a^2-1\cdot(4a-4b)=a^2-4a+4b
\]

2次方程式が重解をもつから，$D=0$

整理すると，$4b=a(4-a)\cdots(\mathrm{i})$

ここで，$a,b$は自然数より，$4b>0$かつ$a>0$

したがって，$4-a>0$より，$0<a<4$

$a$は自然数であるから，$a=1,2,3$

これらのうち，(i)より，$b$が自然数となるのは，$a=2$のときである．

ゆえに，$b=1$

このとき，$x^2+2ax+4a-4b=0$の重解は，$x=-\frac{2a}{2}=-a$

したがって，$a=2$より，重解は$-2$となる．

よって，$a=2,b=1,\alpha=-2$

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% 章末問題I3.4：（One More）★★★
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$a,b$を異なる実数とするとき，$x$に関する方程式$(x-3a)(x-3b)-(3x-4a-5b)=0$は異なる2つの実数解をもつことを証明せよ．

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与えられた方程式を整理すると，

\[
x^2-3(a+b+1)x+9ab+4a+5b=0
\]

この2次方程式の判別式を$D$とすると，

\[
D=\{3(a+b+1)\}^2-4\cdot 1\cdot\{9ab+4a+5b\}=9(a-b)^2+2(a-b)+9
\]

ここで，$a-b=c$とおくと，

\[
D=9c^2+2c+9=(3c+\frac{1}{3})^2+\frac{80}{9}
\]

よって，この方程式は$D>0$であるから，相異なる2つの実数解をもつ．$\blacksquare$

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$f(x)=(x-3a)(x-3b)-(3x-4a-5b)$とする．方程式が異なる2つの実数解をもつことを示すには，$y=f(x)$が$x$軸と異なる2点で交わることを示せばよい．

\[
f(3a)=-5(a-b),f(3b)=4(a-b)
\]

より，$a\neq b$であるから，$f(3a)$と$f(3b)$は異符号であり，一方は負である．

よって，$y=f(x)$は$x$軸と異なる2点で交わる．$\blacksquare$

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% 章末問題I3.5：（One More）★★★
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方程式$4x^2+7xy+4y^2=15$を満たす$x,y$に対して，$u=x+y,v=xy$とおく．

(1) $u^2-4v\geqq 0$を示せ．

(2) $u,v$の間に成り立つ等式を求めよ．

(3) $k=u+v$がとる値の範囲を求めよ．

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(1)
\[
u^2-4v=(x+y)^2-4xy=x^2-2xy+y^2=(x-y)^2\geqq 0
\]

よって，$u^2-4v\geqq 0$が成り立つ．

(2) $x^2+y^2=(x+y)^2-2xy$であるから，与えられた方程式は，

\[
4\{(x+y)^2-2xy\}+7xy=15
\]

したがって，$4(x+y)^2-xy=15$

よって，$4u^2-v=15$

(3) $u^2-4v\geqq 0\cdots(\mathrm{i}),4u^2-v=15\cdots(\mathrm{ii})$とする．

(ii) より，$v=4u^2-15$

これを(i)に代入すると，$u^2-4(4u^2-15)\geqq 0$

したがって，$u^2-4\leqq 0$

ゆえに，$-2\leqq u\leqq 2\cdots(\mathrm{iii})$

$k$を$u$の式で表すと，(ii)より，

\[
k=u+v=u+4u^2-15=4(u+\frac{1}{8})^2-\frac{241}{16}
\]

(iii)より，$k$は$u=2$で最大値$3$，$u=-\frac{1}{8}$で最小値$-\frac{241}{16}$

よって，
\[
-\frac{241}{16}\leqq k\leqq 3
\]

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% 側注1
\begin{align*}
&(x+y+z)^2\\
&=x^2+y^2+z^2\\
&+2xy+2yz+2zx
\end{align*}

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% 側注1
$y=ax^2+8x+b$は放物線であるから，$a\neq 0$

% 側注2
連立方程式を解く．

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% 側注1
対称に移動した点を$(x,y)$とすると，

\[
\frac{x+0}{2}=1,\frac{y+0}{2}=-2
\]

より，$x=2,y=-4$と求められる．

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% 側注1
$\frac{y+(-3)}{2}=2$より，折り返したグラフの頂点の$y$座標は$7$である．

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% 側注1
$x-2>0,x-4\geqq 0,x-6<0$

% 側注2
$x-2>0,x-4>0,x-6\geqq 0$

% 側注3
$x-2<0,x-4<0,x-6<0$

% 側注4
$x-2\geqq 0,x-4<0,x-6<0$

% 側注5
先に，$2x-10$の絶対値記号を外し，$2x-10\geqq 0,2x-10<0$で場合分けをする．

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% 側注1
平方完成する．

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% 側注1
軸が$a\leqq x\leqq a+1$より，右側にあるときを考える．

% 側注2
軸が$a\leqq x\leqq a+1$より，左側にあるときを考える．

% 側注3
$a\leqq x\leqq a+1$に軸が含まれる場合は，最大値が$8$となり不適である．

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% 側注1
グラフより，$t$のとりうる値の範囲は$t\geqq-4$

% 側注2
$y$の最小値の最大値が$\frac{9}{4}$である．

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% 側注1
問題文で$y=ax^2+bx+c$と与えられていても，$y=a(x-p)^2+q$などの形に設定し直した方が計算が楽になることもある．

% 側注2
係数を比較する．

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% 側注1
$a$の値の範囲を求めてから，最小値を考えるように注意すること．

% 側注2
$b$を代入して，文字を消去する．

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% 側注1
グラフは軸$x=1$に関して対称である．

% 側注2
$x$軸との交点が2つ与えられたときは，$y=a(x-\alpha)(x-\beta)$を用いて考えるとよい．

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% 側注1
2つの判別式を考えるので，$D_1,D_2$とする．

% 側注2
実数解をもつ条件は，$D>0,D=0$を合わせた，$D\geqq 0$である．

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% 側注1
$(\text{実数})^2\geqq 0$より，
\[
a^2+1>0
\]

% 側注2
$\sqrt{13(a^2+1)}=7$の両辺を2乗して，$13(a^2+1)=49$

これを整理すると，$a^2=\frac{36}{13}$が得られる．

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% 側注1
$0\leqq x^2<\frac{1}{3}$または$2<x^2$であり，解を合わせた範囲が求める解となるので注意すること．

% 側注2
$(\text{実数})^2\geqq 0$

% 側注3
たすき掛けを用いる．

% 側注4
$t$の変域を考える必要があるので注意すること．

% 側注5
解の公式を用いる．

% 側注6
解の公式を用いる．

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% 側注1
$y=(x-3)(x-5)$のグラフで，$x$軸より下側の部分を$x$軸に関して対称に折り返したものである．

% 側注2
$y=|(x-3)(x-5)|$のグラフと直線$y=ax-2a+\frac{1}{2}$の共有点について調べる．$a$を含む式を直線の形に分離して，2つのグラフが4つの共有点をもつ条件を考える．

% 側注3
2つのグラフが接するとき，共有点（実数解）の個数は3個である．

% 側注4
2つのグラフが接するとき，共有点（実数解）の個数は3個である．

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% 側注1
$y=|x-2|$のグラフが$y=x^2-3x-4$のグラフより上側にある$x$の範囲である．

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% 側注1
$y$が$k$(実数)という値をとりうる．$\Longleftrightarrow x^2-kx+k^2+k-5=0$を満たす実数$x$が存在する．$\Longleftrightarrow x^2-kx+k^2+k-5=0$の判別式$D\geqq 0$

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2つの放物線の両方に接する直線は，$y$軸に平行ではない．

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% 側注1
$a=1$のとき，$4b=3$より，$b=\frac{3}{4}$となり，$a=3$のとき，$4b=3$より，$b=\frac{3}{4}$となり不適である．

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% 側注1
下に凸の放物線は，その関数が負の値をとるとき，$x$軸と異なる2点で交わることを利用する．

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% 章末問題I3.5の側注：（One More）★★★
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% 側注1
$(\text{実数})^2\geqq 0$

% 側注2
与えられた方程式は，$x,y$の対称式であるので，$x+y,xy$で表す．

% 側注3
平方完成する．