题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 3 shows a sketch of part of the curve $C$ with equation

y = \frac{2}{3}x^2 - 9\sqrt{x} + 13 \qquad x \geqslant 0

(a) Find, using calculus, the range of values of $x$ for which $y$ is increasing.

(4)

The point $P$ lies on $C$ and has coordinates $(9, 40)$ .

The line $l$ is the tangent to $C$ at the point $P$ .

The finite region $R$ , shown shaded in Figure 3, is bounded by the curve $C$ , the line $l$ , the $x$ -axis and the $y$ -axis.

(b) Find, using calculus, the exact area of $R$ .

(8)

题目中文翻译

本题必须展示所有解题步骤，不能完全依赖计算器技术。

图3显示了曲线 $C$ 的一部分的草图，曲线方程为：

$y = \dfrac{2}{3}x^2 - 9\sqrt{x} + 13 \quad (x \geqslant 0)$

(a) 用微积分求 $y$ 递增时 $x$ 的取值范围。

（4分）

点 $P$ 在 $C$ 上，坐标为 $(9, 40)$ 。

直线 $l$ 是 $C$ 在点 $P$ 处的切线。

图3中阴影所示的有限区域 $R$ 由曲线 $C$ 、直线 $l$ 、 $x$ 轴和 $y$ 轴围成。

(b) 用微积分求 $R$ 的精确面积。

（8分）

解答

(a)

解法一

思路

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先求导，再解导数大于零的不等式。乘以 $6\sqrt{x}$ 前说明它在 $x>0$ 时为正。

答题过程

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Write $y=\dfrac{2}{3}x^2-9x^{1/2}+13$ . Then

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{4}{3}x-\frac{9}{2\sqrt{x}}.

For $x>0$ , $6\sqrt{x}>0$ , so

\begin{align*} \frac{4}{3}x-\frac{9}{2\sqrt{x}}>0 &\,\iff 8x^{3/2}-27>0\\ &\,\iff x^{3/2}>\frac{27}{8}\\ &\,\iff x>\bigg(\frac{27}{8}\bigg)^{2/3}. \end{align*}

Therefore $y$ is increasing for

\boxed{x>\frac{9}{4}}.

(b)

解法一

思路

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曲线在 $0\leq x\leq9$ 下方的面积，减去切线在其横截距到 $P$ 之间下方的面积。先求切线与横截距，再分别积分。

答题过程

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Let

\begin{align*} f(x) =&\,\frac{2}{3}x^2-9\sqrt{x}+13,\\ F(x) =&\,\frac{2}{9}x^3-6x^{3/2}+13x. \end{align*}

The area under the curve from $0$ to $9$ is

\begin{align*} A_C =&\,\int_0^9 f(x)\,\mathrm{d}x\\ =&\,[F(x)]_0^9\\ =&\,162-162+117\\ =&\,117. \end{align*}

At $P(9,40)$ , the gradient is

\begin{align*} f'(9) =&\,\frac{4}{3}\times9-\frac{9}{2\sqrt{9}}\\ =&\,12-\frac{3}{2}=\frac{21}{2}. \end{align*}

Hence the tangent has equation

\begin{align*} y-40 =&\,\frac{21}{2}(x-9)\\ y =&\,\frac{21}{2}x-\frac{109}{2}. \end{align*}

Its $x$ -intercept is $c=\dfrac{109}{21}$ , obtained by setting $y=0$ .

For the tangent, let $G(x)=\dfrac{21}{4}x^2-\dfrac{109}{2}x$ . Then

\begin{align*} G(9) =&\,\frac{1701}{4}-\frac{981}{2} =-\frac{261}{4},\\ G(c) =&\,\frac{11881}{84}-\frac{11881}{42} =-\frac{11881}{84}. \end{align*}

Therefore the area under the tangent between $c$ and $9$ is

\begin{align*} A_l =&\,\int_c^9 \bigg(\frac{21}{2}x-\frac{109}{2}\bigg)\,\mathrm{d}x\\ =&\,G(9)-G(c)\\ =&\,-\frac{261}{4}+\frac{11881}{84}\\ =&\,\frac{6400}{84}=\frac{1600}{21}. \end{align*}

Thus

\begin{align*} A_R =&\,117-\frac{1600}{21}\\ =&\,\frac{2457-1600}{21}\\ =&\,\boxed{\frac{857}{21}}. \end{align*}

也可把上述切线下方区域看作由切线、 $x$ 轴及 $x=9$ 围成的三角形：

\begin{align*} A_l =&\,\frac12(9-c)\times40\\ =&\,\frac12\times\frac{80}{21}\times40\\ =&\,\frac{1600}{21}. \end{align*}

解法二

思路

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在切线横截距 $c$ 处分段： $0$ 到 $c$ 的上边界是曲线、下边界是 $x$ 轴； $c$ 到 $9$ 的上边界是曲线、下边界是切线。两个积分合并时 $F(c)$ 抵消。

答题过程

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At $P$ , $f'(9)=\dfrac{21}{2}$ , so the tangent is

l(x)=\frac{21}{2}x-\frac{109}{2}, \qquad c=\frac{109}{21}.

Use the antiderivatives

\begin{align*} F(x) =&\,\frac{2}{9}x^3-6x^{3/2}+13x,\\ G(x) =&\,\frac{21}{4}x^2-\frac{109}{2}x. \end{align*}

The required area is

\begin{align*} A_R =&\,\int_0^c f(x)\,\mathrm{d}x\\ &\,\hspace{2pt}+\int_c^9\big(f(x)-l(x)\big)\,\mathrm{d}x\\ =&\,F(c)-F(0)\\ &\,\hspace{2pt}+F(9)-F(c)-G(9)+G(c)\\ =&\,117-\bigg(-\frac{261}{4}\bigg) -\frac{11881}{84}\\ =&\,\frac{9828+5481-11881}{84}\\ =&\,\boxed{\frac{857}{21}}. \end{align*}

解法三

思路

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先在 $0$ 到 $9$ 全段积分“曲线减切线”。切线在 $0$ 到 $c$ 位于 $x$ 轴下方，这样会多算一个三角形，最后减去它。

答题过程

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The tangent gradient at $P$ is $\dfrac{21}{2}$ , giving

l(x)=\frac{21}{2}x-\frac{109}{2}.

Its intercepts are $\big(\dfrac{109}{21},0\big)$ and $\big(0,-\dfrac{109}{2}\big)$ .

The area between the curve and the tangent over $0\leq x\leq9$ is

\begin{align*} A_* =&\,\int_0^9\big(f(x)-l(x)\big)\,\mathrm{d}x\\ =&\,\bigg[\frac{2}{9}x^3-6x^{3/2}+13x\\ &\,\hspace{2pt}-\frac{21}{4}x^2+\frac{109}{2}x\bigg]_0^9\\ =&\,117-\frac{1701}{4}+\frac{981}{2}\\ =&\,\frac{729}{4}. \end{align*}

This includes the extra triangle below the $x$ -axis:

A_{\triangle} =\frac12\times\frac{109}{21}\times\frac{109}{2} =\frac{11881}{84}.

Subtract it:

\begin{align*} A_R =&\,\frac{729}{4}-\frac{11881}{84}\\ =&\,\frac{15309-11881}{84}\\ =&\,\boxed{\frac{857}{21}}. \end{align*}

P2 2023 Oct Q9

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

思路

答题过程

解法三

思路

答题过程