题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1 shows a sketch of part of the curve $C$ with equation

y=\frac{9x-x^2}{2\sqrt{x}},\quad x>0

The point $P$ is the stationary point on $C$ .

(a) Find, using calculus, the $x$ coordinate of $P$ .

(4)

The finite region $R$ , shown shaded in Figure 1, is bounded by the curve $C$ , the $x$ -axis and the line with equation $x=1$

(b) Using calculus, calculate the exact area of $R$ .

(5)

题目中文翻译

本题必须展示计算过程的所有步骤。

完全依赖计算器技术的解答不予接受。

图1给出曲线 $C$ 的一部分的草图，其方程为

y=\frac{9x-x^2}{2\sqrt{x}},\quad x>0

点 $P$ 是 $C$ 上的驻点。

(a) 用微积分求 $P$ 的 $x$ 坐标。(4)

图1中阴影所示的有限区域 $R$ 由曲线 $C$ 、 $x$ 轴和方程为 $x=1$ 的直线围成。

(b) 用微积分计算 $R$ 的精确面积。(5)

解答

(a)

解法一

思路

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把函数拆成两个幂函数，逐项求导后令导数为零。

答题过程

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Rewrite the equation as

y=\frac92x^{1/2}-\frac12x^{3/2}.

Differentiating,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac94x^{-1/2}\\ &\,\hspace{2pt}-\frac34x^{1/2}. \end{align*}

At $P$ , the derivative is zero. Multiplying by $4\sqrt{x}$ , which is positive since $x>0$ , gives

\begin{align*} 9-3x=&\,0\\ x=&\,3. \end{align*}

The $x$ coordinate of $P$ is $3$ .

解法二

思路

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保留原来的商式，用商法求导。分母为正，因此驻点条件等价于导数分子为零。

答题过程

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Let $u=9x-x^2$ and $v=2\sqrt{x}$ . Then

u'=9-2x,\qquad v'=\frac1{\sqrt{x}}.

By the quotient rule,

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{N}{4x},

where

\begin{align*} N=&\,2\sqrt{x}(9-2x)\\ &\,\hspace{2pt}-\frac{9x-x^2}{\sqrt{x}}. \end{align*}

Since $x>0$ , a stationary point requires $N=0$ . Multiplying by $\sqrt{x}$ ,

\begin{align*} 2x(9-2x)-(9x-x^2)=&\,0\\ 9x-3x^2=&\,0\\ 3x(3-x)=&\,0. \end{align*}

Reject $x=0$ since $x>0$ . Hence the $x$ coordinate of $P$ is $3$ .

(b)

解法一

思路

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题目只给出左侧边界 $x=1$ ，右侧边界要从曲线与 $x$ 轴的交点求出。确认曲线在两边界之间位于轴上方，再积分求面积。

答题过程

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To find the intersection with the $x$ -axis, set $y=0$ . Since $x>0$ ,

\begin{align*} 9x-x^2=&\,0\\ x(9-x)=&\,0\\ x=&\,9. \end{align*}

For $1<x<9$ , both $x(9-x)$ and $2\sqrt{x}$ are positive, so the curve is above the $x$ -axis. Therefore

A=\int_1^9\frac{9x-x^2}{2\sqrt{x}}\,\mathrm{d}x.

Using the power form,

\begin{align*} A=&\,\frac92\int_1^9x^{1/2}\,\mathrm{d}x\\ &\,\hspace{2pt}-\frac12\int_1^9x^{3/2}\,\mathrm{d}x\\ =&\,\bigg[3x^{3/2}-\frac15x^{5/2}\bigg]_1^9. \end{align*}

At the upper limit,

\begin{align*} 3(9)^{3/2}-\frac15(9)^{5/2} =&\,81-\frac{243}{5}\\ =&\,\frac{162}{5}. \end{align*}

At the lower limit,

3(1)^{3/2}-\frac15(1)^{5/2}=\frac{14}{5}.

Hence the exact area is

A=\frac{162}{5}-\frac{14}{5}=\frac{148}{5}.

P2 2024 May R Q10

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程