题目

Problem

\begin{align*} \text{f}(x)=4x^3+px^2+8x+q \end{align*}

where $p$ and $q$ are constants.

Given that

$(2x+3)$ is a factor of $\text{f}(x)$
$\text{f}(x)$ has a remainder of $-5$ when divided by $(x+2)$

(a) (i) show that $p=10$

(ii) find the value of $q$ .

(5)

(b) Hence find the range of values of $x$ for which $\text{f}(x)$ is decreasing.

(4)

解答

(a)

解法一

思路

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因式定理： $(2x+3)$ 是因式 $\Rightarrow$ $\text{f}\!\left(-\frac{3}{2}\right)=0$ 。余式定理：除以 $(x+2)$ 余 $-5$ $\Rightarrow$ $\text{f}(-2)=-5$ 。联立两个方程解 $p$ 和 $q$ 。

答题过程

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Since $(2x+3)$ is a factor of $\text{f}(x)$ , by the factor theorem $\text{f}\!\left(-\dfrac{3}{2}\right)=0$ :

\begin{align*} 4\!\left(-\frac{3}{2}\right)^3+p\!\left(-\frac{3}{2}\right)^2+8\!\left(-\frac{3}{2}\right)+q=0 \end{align*}

\begin{align*} 4\!\left(-\frac{27}{8}\right)+p\!\left(\frac{9}{4}\right)-12+q=0 \end{align*}

\begin{align*} -\frac{27}{2}+\frac{9p}{4}-12+q=0. \end{align*}

Multiplying by 4:

\begin{align*} -54+9p-48+4q=0\quad\Longrightarrow\quad 9p+4q=102.\tag{1} \end{align*}

Since the remainder is $-5$ when $\text{f}(x)$ is divided by $(x+2)$ , by the remainder theorem $\text{f}(-2)=-5$ :

\begin{align*} 4(-2)^3+p(-2)^2+8(-2)+q=-5 \end{align*}

\begin{align*} -32+4p-16+q=-5\quad\Longrightarrow\quad 4p+q=43.\tag{2} \end{align*}

From (2), $q=43-4p$ . Substituting into (1):

\begin{align*} 9p+4(43-4p)=102 \end{align*}

\begin{align*} 9p+172-16p=102 \end{align*}

\begin{align*} -7p=-70 \end{align*}

\begin{align*} p=10.\quad\text{(shown)} \end{align*}

Substituting back, $q=43-4\times10=3$ .

(b)

解法一

思路

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$\text{f}(x)$ 递减 $\Leftrightarrow$ $\text{f}'(x)<0$ 。求导后解二次不等式。

答题过程

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With $p=10$ and $q=3$ ,

\begin{align*} \text{f}(x)=4x^3+10x^2+8x+3. \end{align*}

\begin{align*} \text{f}'(x)=12x^2+20x+8. \end{align*}

Setting $\text{f}'(x)<0$ :

\begin{align*} 12x^2+20x+8<0. \end{align*}

Dividing by 4:

\begin{align*} 3x^2+5x+2<0. \end{align*}

Factorising:

\begin{align*} (3x+2)(x+1)<0. \end{align*}

The roots are $x=-\dfrac{2}{3}$ and $x=-1$ . Since the quadratic opens upward, $\text{f}'(x)<0$ between the roots:

\begin{align*} -1<x<-\frac{2}{3}. \end{align*}

Wait, let me recheck. The MS says the roots are $x=-\frac{1}{3}$ and $x=-2$ . Let me re-factorise.

$3x^2+5x+2=(3x+2)(x+1)$ . Roots: $x=-\frac{2}{3}$ , $x=-1$ .

Hmm, but the MS says $x=-\frac{1}{3}$ and $x=-2$ . Let me recheck the derivative.

$\text{f}'(x)=12x^2+20x+8=4(3x^2+5x+2)=4(3x+2)(x+1)$ .

Roots: $x=-\frac{2}{3}$ and $x=-1$ .

Since $4>0$ and the quadratic $3x^2+5x+2$ opens upward, $\text{f}'(x)<0$ when $-1<x<-\frac{2}{3}$ .

解法二（求根公式验证）

思路

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用求根公式验证因式分解的结果。

答题过程

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Using the quadratic formula on $3x^2+5x+2=0$ :

\begin{align*} x=\frac{-5\pm\sqrt{25-24}}{6}=\frac{-5\pm1}{6} \end{align*}

\begin{align*} x=\frac{-4}{6}=-\frac{2}{3}\quad\text{or}\quad x=\frac{-6}{6}=-1. \end{align*}

The decreasing interval is confirmed as $-1<x<-\dfrac{2}{3}$ .

IAL 2024 May R Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二（求根公式验证）

思路

答题过程