题目

Problem

Figure 3

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 the curve $C$ with equation

y=2x+\frac{64}{x^2}-3,\qquad x>0

The point $P$ , shown in Figure 3, is the stationary point on $C$ .

(a) Show, using calculus, that the $x$ coordinate of $P$ is $4$

(4)

The point $Q$ lies on $C$ and has $x$ coordinate $2$

The line segments $MQ$ and $NP$ , shown in Figure 3, are parallel to the $x$ -axis. The region $R$ , shown shaded in Figure 3, is bounded by $C$ , the $y$ -axis and line segments $MQ$ and $NP$ .

(b) Use algebraic integration to find the exact area of $R$ . You must make your method clear.

(5)

解答

(a)

解法一

思路

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stationary point 要令 $\frac{dy}{dx}=0$ 。题目要求 show that，所以不能只代 $x=4$ ，要从求导和解方程自然推出 $x=4$ 。

答题过程

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Differentiate

\begin{align*} y=2x+64x^{-2}-3. \end{align*}

Then

\begin{align*} \frac{dy}{dx}=2-128x^{-3}. \end{align*}

At a stationary point,

\begin{align*} 2-128x^{-3}=&\,0\\ 2-\frac{128}{x^3}=&\,0\\ 2x^3=&\,128\\ x^3=&\,64\\ x=&\,4. \end{align*}

Therefore, the $x$ coordinate of $P$ is $4$ .

(b)

解法一

思路

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先求 $Q$ 和 $P$ 的高度。区域可以看成：从 $x=2$ 到 $x=4$ 的曲线下面积，减去底部矩形，再加上左边的矩形。

答题过程

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At $x=2$ ,

\begin{align*} y=&\,2(2)+\frac{64}{2^2}-3\\ =&\,4+16-3\\ =&\,17. \end{align*}

So $Q=(2,17)$ and $M=(0,17)$ .

At $x=4$ ,

\begin{align*} y=&\,2(4)+\frac{64}{4^2}-3\\ =&\,8+4-3\\ =&\,9. \end{align*}

So $P=(4,9)$ and $N=(0,9)$ .

Now integrate the curve:

\begin{align*} \int\left(2x+\frac{64}{x^2}-3\right)\,dx =&\,\int\left(2x+64x^{-2}-3\right)\,dx\\ =&\,x^2-64x^{-1}-3x\\ =&\,x^2-\frac{64}{x}-3x. \end{align*}

The area under the curve from $x=2$ to $x=4$ is

\begin{align*} \left[x^2-\frac{64}{x}-3x\right]_2^4 =&\,\left(16-16-12\right) -\left(4-32-6\right)\\ =&\,-12-(-34)\\ =&\,22. \end{align*}

The rectangle below $y=9$ from $x=2$ to $x=4$ has area

\begin{align*} 9(4-2)=18. \end{align*}

The rectangle from $x=0$ to $x=2$ , between $y=9$ and $y=17$ , has area

\begin{align*} 2(17-9)=16. \end{align*}

Therefore,

\begin{align*} \text{area of }R=&\,22-18+16\\ =&\,20. \end{align*}

Hence the exact area of $R$ is $20$ .

解法二

思路

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也可以把右边那块直接看成曲线与直线 $y=9$ 之间的面积，再加左边矩形。这样图形意义更直接。

答题过程

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From part (a), $P$ has $x$ coordinate $4$ .

At $x=4$ ,

\begin{align*} y=2(4)+\frac{64}{4^2}-3=9. \end{align*}

At $x=2$ ,

\begin{align*} y=2(2)+\frac{64}{2^2}-3=17. \end{align*}

So the left rectangular part has area

\begin{align*} 2(17-9)=16. \end{align*}

For $2\leqslant x\leqslant4$ , the lower boundary is the horizontal line $y=9$ .

Therefore the curved part has area

\begin{align*} \int_2^4\left(2x+\frac{64}{x^2}-3-9\right)\,dx =&\,\int_2^4\left(2x+64x^{-2}-12\right)\,dx\\ =&\,\left[x^2-\frac{64}{x}-12x\right]_2^4. \end{align*}

Evaluate:

\begin{align*} \left[x^2-\frac{64}{x}-12x\right]_2^4 =&\,\left(16-16-48\right) -\left(4-32-24\right)\\ =&\,-48-(-52)\\ =&\,4. \end{align*}

Hence

\begin{align*} \text{area of }R=&\,4+16\\ =&\,20. \end{align*}

Therefore, the exact area of $R$ is $20$ .

IAL 2025 May A 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

解法二

思路

答题过程