45308 學測試題一個選項的討論

\begin{align*} \Delta=&\left|\begin{array}{ccccc} 4-r&~~&2-r&~~&1-r\\[4pt] 9-4r&&3-2r&&1-r\\[4pt] 16-9r&&4-3r&&1-r\end{array}\right|=\left|\begin{array}{ccccc} 4-r&~~&2-r&~~&1-r\\[4pt] 5-3r&&1-r&&0\\[4pt] 12-8r&&2-2r&&0\end{array}\right|\\[6pt] =&(1-r)\left|\begin{array}{ccc} 5-3r&~~&1-r\\[4pt] 12-8r&&2-2r\end{array}\right| =(1-r)^2\times \left|\begin{array}{ccc} 5-3r&~~&1\\[4pt] 12-8r&&2\end{array}\right|\\[6pt] =&(1-r)^2(-2+2r)=-2(1-r)^3, \end{align*}

\begin{align*} &\hskip 3cm\left\{\begin{array}{cccccl} (4-r)a&+&(2-r)b&+&(1-r)c&=r-8\\[4pt] (9-4r)a&+&(3-2r)b&+&(1-r)c&=8r-27\\[4pt] (16-9r)a&+&(4-3r)b&+&(1-r)c&=27r-64\\[4pt]\end{array}\right.,\tag*{$(*)$}\\ %\eqno{(*)}\\ &\left[\begin{array}{ccccccc} 4-r&~~&2-r&~~&1-r&~~&r-8\\[4pt] 9-4r&&3-2r&&1-r&~~&8r-27\\[4pt] 16-9r&&4-3r&&1-r&~~&27r-64\end{array}\right]\to \left[\begin{array}{ccccccc} 4-r&~~&2-r&~~&1-r&~~&r-8\\[4pt] 5-3r&&1-r&&0&~~&7r-19\\[4pt] 12-8r&&2-2r&&0&~~&26r-56\end{array}\right]\\ &\to\left[\begin{array}{ccccccc} 4-r&~~&2-r&~~&1-r&~~&r-8\\[4pt] 5-3r&&1-r&&0&~~&7r-19\\[4pt] 2-2r&&0&&0&~~&12r\!-\!18\end{array}\right]\to \left[\begin{array}{ccccccc} 4-r&~~&2-r&~~&1-r&~~&r-8\\[4pt] 5-3r&&1-r&&0&~~&7r-19\\[4pt] 1-r&&0&&0&~~&6r-9\end{array}\right] \end{align*}

\begin{align*} g(x)=&\, k\frac{(x -2)(x - 3)(x - 4)}{(1 - 2)(1 - 3)(1 - 4)} + rk\frac{(x -1)(x - 3)(x - 4)}{(2 - 1)(2 - 3)(2 - 4)}\\ & + r^2k\frac{(x -1)(x - 2)(x - 4)}{(3 - 1)(3 - 2)(3 - 4)} + r^3k\frac{(x -1)(x - 2)(x - 3)}{(4 - 1)(4 - 2)(4 - 3)}\\ =&-\frac k6(x -2)(x - 3)(x - 4)+\frac {rk}2(x -1)(x - 3)(x - 4)\\ &-\frac {r^2k}2(x -1)(x - 2)(x - 4)+\frac {r^3k}6(x -1)(x - 2)(x - 3) \end{align*}

$g(x)$ 的三次項係數為 $\dfrac k6(-1+3r-3r^2+r^3)=\dfrac{k(r-1)^3}{6}\not=0$,

\begin{align*} c\!=&-\dfrac{4\!-\!r}{1\!-\!r}a\!-\!\dfrac{2\!-\!r}{1\!-\!r}b+\dfrac{r\!-\!8}{1\!-\!r}\\ =&\Big(\!-1\!-\!\dfrac 3{1\!-\!r}\Big)\Big(\!-6\!-\!\dfrac 3{1\!-\!r}\Big) \!+\!\Big(\!-1\!-\!\dfrac 1{1\!-\!r}\Big)\Big(11\!+\!\dfrac 9{1\!-\!r}+\dfrac 6{(1\!-\!r)^2}\Big)\!-\!1\!-\!\dfrac 7{1\!-\!r}\\ =&-6-\dfrac 6{1\!-\!r}-\dfrac 6{(1\!-\!r)^2}-\dfrac 6{(1\!-\!r)^3}. \end{align*}

\begin{align*} f(x\!+\!1)\!-\!rf(x)=&(x\!+\!1)^3\!+\!a(x\!+\!1)^2\!+\!b(x\!+\!1)\!+\!c\!-\!rx^3\!-\!rax^2\!-\!rbx\!-\!rc\\ =&(1\!-\!r)x^3\!+\!(a\!+\!3\!-\!ra)x^2\!+\!(2a\!+\!b\!+\!3\!-\!rb)x\!+\!(a\!+\!b\!+\!c\!+\!1\!-\!rc) \end{align*}

$\therefore$ $f(x+1)-rf(x)=(1-r)(x-1)(x-2)(x-3)$, 即 $(1\!-\!r)x^3\!+\!(a\!+\!3\!-\!ra)x^2\!+\!(2a\!+\!b\!+\!3\!-\!rb)x\!+\!(a\!+\!b\!+\!c\!+\!1\!-\!rc)\!=\!(1\!-\!r)(x^3\!-\!6x^2\!+\!11x\!-\!6)$

\begin{align*} \therefore \qquad \frac{(1-r)a+3}{1-r}=-6\Rightarrow&\ a=-6-\frac 3{1-r},\\ \frac{(1-r)b+2a+3}{1-r}=11\Rightarrow&\ b=11-\frac {2a+3}{1-r},\\ \frac{(1-r)c+a+b+1}{1-r}=-6\Rightarrow&\ c=-6-\frac {a+b+1}{1-r}, \end{align*} \begin{align*} \therefore\ a=&-6-\frac 3{1-r},\\ b=&11-\frac 1{1-r}\Big[2\Big(\!-6-\frac 3{1-r}\Big)+3\Big]=11+\frac 9{1-r}+\frac 6{(1-r)^2},\\ c=&-6-\frac 1{1-r}\Big[\Big(\!-6-\frac 3{1-r}\Big)+\Big(11+\frac 9{1-r}+\frac 6{(1-r)^2}+1\Big)\Big]\\ =&-6-\frac 6{1-r}-\frac 6{(1-r)^2}-\frac 6{(1-r)^3}, \end{align*}

$f(1)=6$, $f(2)=12$, $f(3)=24$, $f(4)=48$,

### 參考文獻

110 年大學學科能力測驗數學考科試題卷。台北市。大學入學考試中心, 2021。

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