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P2 2025 May A Q10

A Level / Edexcel / P2

IAL 2025 May A Paper · Question 10

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 the curve CC with equation

y=2x+64x2−3,x>0y=2x+\frac{64}{x^2}-3,\qquad x>0

The point PP, shown in Figure 3, is the stationary point on CC.

(a) Show, using calculus, that the xx coordinate of PP is 44

(4)

The point QQ lies on CC and has xx coordinate 22

The line segments MQMQ and NPNP, shown in Figure 3, are parallel to the xx-axis. The region RR, shown shaded in Figure 3, is bounded by CC, the yy-axis and line segments MQMQ and NPNP.

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

(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3是曲线 CC 的示意图，其方程为

y=2x+64x2−3,x>0y=2x+\frac{64}{x^2}-3,\qquad x>0

图3中的点 PP 是 CC 上的驻点。

(a) 使用微积分证明 PP 的 xx 坐标为 44。

(4)

点 QQ 在 CC 上，其 xx 坐标为 22。

图3中的线段 MQMQ 和 NPNP 均平行于 xx 轴。

图3中阴影区域 RR 由 CC、yy 轴和线段 MQMQ、NPNP 围成。

(b) 使用代数积分求 RR 的精确面积。

必须清楚展示所用方法。

(5)

解答

(a)

解法一

思路

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

答题过程

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Differentiate

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

Then

dydx=2−128x−3.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=2-128x^{-3}. \end{align*}

At a stationary point,

2−128x−3= 02−128x3= 02x3= 128x3= 64x= 4.\begin{align*} 2-128x^{-3}=&\,0\\ 2-\frac{128}{x^3}=&\,0\\ 2x^3=&\,128\\ x^3=&\,64\\ x=&\,4. \end{align*}

Therefore, the xx coordinate of PP is 44.

(b)

解法一

思路

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先求 QQ 和 PP 的高度。区域可以看成：从 x=2x=2 到 x=4x=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=2x=2,

y= 2(2)+6422−3= 4+16−3= 17.\begin{align*} y=&\,2(2)+\frac{64}{2^2}-3\\ =&\,4+16-3\\ =&\,17. \end{align*}

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

At x=4x=4,

y= 2(4)+6442−3= 8+4−3= 9.\begin{align*} y=&\,2(4)+\frac{64}{4^2}-3\\ =&\,8+4-3\\ =&\,9. \end{align*}

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

Now integrate the curve:

 ∫(2x+64x−2−3) dx= x2−64x−1−3x= x2−64x−3x.\begin{align*} &\,\int(2x+64x^{-2}-3)\,\mathrm{d}x\\ =&\,x^2-64x^{-1}-3x\\ =&\,x^2-\frac{64}{x}-3x. \end{align*}

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

 [x2−64x−3x]24= (16−16−12) −(4−32−6)= −12−(−34)= 22.\begin{align*} &\,\bigg[x^2-\frac{64}{x}-3x\bigg]_2^4\\ =&\,(16-16-12)\\ &\,\hspace{2pt}-(4-32-6)\\ =&\,-12-(-34)\\ =&\,22. \end{align*}

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

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

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

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

Therefore,

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

Hence the exact area of RR is 2020.

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=9y=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), PP has xx coordinate 44.

At x=4x=4,

y= 2(4)+6442−3= 9.\begin{align*} y=&\,2(4)+\frac{64}{4^2}-3\\ =&\,9. \end{align*}

At x=2x=2,

y= 2(2)+6422−3= 17.\begin{align*} y=&\,2(2)+\frac{64}{2^2}-3\\ =&\,17. \end{align*}

So the left rectangular part has area

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

For 2⩽x⩽42\leqslant x\leqslant4, the lower boundary is the horizontal line y=9y=9.

Therefore the curved part has area

 ∫24(2x+64x−2−12) dx= [x2−64x−12x]24.\begin{align*} &\,\int_2^4(2x+64x^{-2}-12)\,\mathrm{d}x\\ =&\,\bigg[x^2-\frac{64}{x}-12x\bigg]_2^4. \end{align*}

Evaluate:

 [x2−64x−12x]24= (16−16−48) −(4−32−24)= −48−(−52)= 4.\begin{align*} &\,\bigg[x^2-\frac{64}{x}-12x\bigg]_2^4\\ =&\,(16-16-48)\\ &\,\hspace{2pt}-(4-32-24)\\ =&\,-48-(-52)\\ =&\,4. \end{align*}

Hence

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

Therefore, the exact area of RR is 2020.