题目

Problem

A function is defined by $f(x) = \frac{4}{x^3} - \frac{3}{x} + 2$ for $x \ne 0$ . The graph of $y = f(x)$ is shown in the diagram.

(a) Find the set of values of $x$ for which $f(x)$ is decreasing.

[5]

(b) A triangle is bounded by the $y$ -axis, the normal to the curve at the point where $x = 1$ and the tangent to the curve at the point where $x = -1$ .

Find the area of the triangle. Give your answer correct to $3$ significant figures.

[8]

题目中文翻译

函数定义为 $f(x) = \frac{4}{x^3} - \frac{3}{x} + 2$ ，其中 $x \ne 0$ 。图中显示 $y = f(x)$ 的图像。

(a) 求使 $f(x)$ 递减的 $x$ 的取值范围。

[5]

(b) 一个三角形由 $y$ 轴、曲线上 $x = 1$ 处的法线，以及曲线上 $x = -1$ 处的切线围成。

求该三角形的面积。答案精确到 $3$ 位有效数字。

[8]

解答

(a)

解法一

思路

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把导数通分。分母在定义域内总为正，只需判断分子，并保留 $x=0$ 处的断点。

答题过程

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For $x\ne0$ ,

\begin{align*} f'(x)=&\,-\frac{12}{x^4}+\frac3{x^2}\\ =&\,\frac{3(x^2-4)}{x^4}. \end{align*}

The derivative is negative when $x^2<4$ and $x\ne0$ . Thus the decreasing intervals are $(-2,0)$ and $(0,2)$ .

(b)

解法一

思路

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先求两个接触点处的高度和斜率。用两条直线的纵截距差作底，交点到 $y$ 轴的距离作高。

答题过程

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We have $f(1)=3$ , $f(-1)=1$ , and $f'(1)=f'(-1)=-9$ . The normal at $x=1$ is

y-3=\frac19(x-1),\qquad y=\frac{x+26}{9}.

The tangent at $x=-1$ is $y-1=-9(x+1)$ , or $y=-9x-8$ . At their intersection, $(x+26)/9=-9x-8$ , so $x=-49/41$ .

The $y$ -axis intercepts differ by $26/9-(-8)=98/9$ . The perpendicular height is $49/41$ . Thus the area is

\begin{align*} &\,\frac12\cdot\frac{98}{9}\cdot\frac{49}{41}\\ =&\,\frac{2401}{369}\approx6.51 \end{align*}

square units to three significant figures.

CIE 9709 2024 June Paper 11 Q11

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程