题目

Problem

Figure 4 shows a sketch of part of the curve $C$ with equation $y=f(x)$ , where

\begin{align*} f(x)=(3x+20)(x+6)(2x-3). \end{align*}

Figure 4

(a) Use the given information to state the values of $x$ for which

\begin{align*} f(x)>0. \end{align*}

(2)

(b) Expand $(3x+20)(x+6)(2x-3)$ , writing your answer as a polynomial in simplest form.

(3)

The straight line $l$ is the tangent to $C$ at the point where $C$ cuts the $y$ -axis.

Given that $l$ cuts $C$ at the point $P$ , as shown in Figure 4,

(Solutions based on calculator technology are not acceptable.)

(5)

解答

(a)

解法一

思路

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三个因式给出三个 $x$ 轴截距： $-\frac{20}{3}$ 、 $-6$ 、 $\frac32$ 。结合三次曲线的图像，读出曲线在 $x$ 轴上方的区间。

答题过程

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The roots are found from

\begin{align*} 3x+20=&\,0, &x+6=&\,0, &2x-3=&\,0. \end{align*}

\begin{align*} x=-\frac{20}{3}, \qquad x=-6, \qquad x=\frac32. \end{align*}

From the sketch,

\begin{align*} f(x)>0 \end{align*}

for

\begin{align*} -\frac{20}{3}<x<-6 \quad\text{or}\quad x>\frac32. \end{align*}

(b)

解法一

思路

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先乘后两个因式会比较顺： $(x+6)(2x-3)$ 。得到二次式后，再乘以 $3x+20$ 。

答题过程

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First expand two brackets:

\begin{align*} (x+6)(2x-3) =&\,2x^2-3x+12x-18\\ =&\,2x^2+9x-18. \end{align*}

Then

\begin{align*} (3x+20)(x+6)(2x-3) =&\,(3x+20)(2x^2+9x-18)\\ =&\,3x(2x^2+9x-18)\\ &\quad +20(2x^2+9x-18)\\ =&\,6x^3+27x^2-54x\\ &\quad +40x^2+180x-360\\ =&\,6x^3+67x^2+126x-360. \end{align*}

(c)

解法一

思路

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先求 $C$ 在 $y$ 轴截点处的切线，也就是在 $x=0$ 处的切线。切线斜率来自 $f'(0)$ ，切点是 $(0,f(0))$ 。然后把切线方程和曲线方程联立，除了 $x=0$ 这个切点，另一个交点就是 $P$ 。

答题过程

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From part (b),

\begin{align*} f(x)=6x^3+67x^2+126x-360. \end{align*}

Differentiate:

\begin{align*} f'(x)=18x^2+134x+126. \end{align*}

At the point where $C$ cuts the $y$ -axis, $x=0$ , so

\begin{align*} f(0)=&\,-360,\\ f'(0)=&\,126. \end{align*}

Therefore the tangent is

\begin{align*} y+360=&\,126(x-0)\\ y=&\,126x-360. \end{align*}

Now find where this line meets the curve:

\begin{align*} 6x^3+67x^2+126x-360 =&\,126x-360. \end{align*}

Cancel common terms:

\begin{align*} 6x^3+67x^2=&\,0\\ x^2(6x+67)=&\,0. \end{align*}

Thus

\begin{align*} x=0 \quad\text{or}\quad x=-\frac{67}{6}. \end{align*}

The solution $x=0$ is the point of tangency, so the $x$ coordinate of $P$ is

\begin{align*} -\frac{67}{6}. \end{align*}

IAL 2023 Jan Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程