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P2 2024 Jan Q10

A Level / Edexcel / P2

IAL 2024 Jan Paper · Question 10

题目

Problem

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

Figure 2 shows a sketch of the curve with equation

y=12x2+1458x3−74x>0y=\frac12x^2+\frac{1458}{\sqrt{x^3}}-74 \qquad x>0

The point PP is the only stationary point on the curve.

(a) Use calculus to show that the xx coordinate of PP is 99

(4)

The line ll passes through the point PP and is parallel to the xx-axis.

The region RR, shown shaded in Figure 2, is bounded by the curve, the line ll and the line with equation x=4x=4

(b) Use algebraic integration to find the exact area of RR.

(5)
题目中文翻译

本题必须展示详细推理过程。不能完全依赖计算器技术。

图 2 显示曲线

y=12x2+1458x3−74,x>0y=\frac12x^2+\frac{1458}{\sqrt{x^3}}-74, \qquad x>0

的草图。PP 是曲线上唯一的驻点。

(a) 用微积分证明 PP 的 xx 坐标为 99。

（4分）

直线 ll 经过点 PP，并平行于 xx 轴。图 2 中阴影区域 RR 由曲线、直线 ll 和直线 x=4x=4 围成。

(b) 用代数积分求 RR 的精确面积。

（5分）

解答

(a)

解法一

思路

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先把 1458x3\frac{1458}{\sqrt{x^3}} 写成 1458x−321458x^{-\frac32}，再求导。驻点满足 dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0。

答题过程

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Rewrite the curve as

y=12x2+1458x−32−74.\begin{align*} y=\frac12x^2+1458x^{-\frac32}-74. \end{align*}

Differentiate:

dydx= x+1458(−32)x−52= x−2187x−52.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,x+1458\bigg(-\frac32\bigg)x^{-\frac52}\\ =&\,x-2187x^{-\frac52}. \end{align*}

At a stationary point,

dydx=0.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=0. \end{align*}

So

x−2187x−52= 0x= 2187x−52x72= 2187.\begin{align*} x-2187x^{-\frac52}=&\,0\\ x=&\,2187x^{-\frac52}\\ x^{\frac72}=&\,2187. \end{align*}

Since

2187=37,\begin{align*} 2187=3^7, \end{align*}

we have

x=(37)27=32=9.\begin{align*} x=(3^7)^{\frac27}=3^2=9. \end{align*}

Therefore, the xx coordinate of PP is 99.

还可将已经求出的 x=9x=9 代入导数作检查：

y′(9)=9−218795/2=9−2187243=0.y'(9)=9-\frac{2187}{9^{5/2}} =9-\frac{2187}{243}=0.

(b)

解法一

思路

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区域在曲线和水平线 ll 之间。先求 PP 的 yy 坐标，也就是直线 ll 的高度，再计算从 x=4x=4 到 x=9x=9 的「曲线面积减矩形面积」。

答题过程

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At PP, x=9x=9.

So

yP= 12(9)2+145893−74= 812+145827−74= 812+54−74= 412.\begin{align*} y_P=&\,\frac12(9)^2+\frac{1458}{\sqrt{9^3}}-74\\ =&\,\frac{81}{2}+\frac{1458}{27}-74\\ =&\,\frac{81}{2}+54-74\\ =&\,\frac{41}{2}. \end{align*}

Thus the line ll has equation

y=412.\begin{align*} y=\frac{41}{2}. \end{align*}

The required area is

∫49(12x2+1458x−32−74−412) dx.\int_4^9 \bigg( \frac12x^2+1458x^{-\frac32}-74-\frac{41}{2} \bigg)\,\mathrm{d}x.

Simplify the integrand:

12x2+1458x−32−1892.\begin{align*} \frac12x^2+1458x^{-\frac32}-\frac{189}{2}. \end{align*}

Therefore,

Area= ∫49(12x2+1458x−32−1892) dx= [16x3−2916x−12−1892x]49.\begin{align*} \text{Area} =&\,\int_4^9 \bigg( \frac12x^2+1458x^{-\frac32}-\frac{189}{2} \bigg)\,\mathrm{d}x\\ =&\,\bigg[ \frac16x^3-2916x^{-\frac12}-\frac{189}{2}x \bigg]_4^9. \end{align*}

Substitute the limits:

Area= (16(9)3−2916(9)−12−1892(9)) −(16(4)3−2916(4)−12−1892(4))= (7296−972−17012) −(646−1458−378)= 3733.\begin{align*} \text{Area} =&\,\bigg( \frac16(9)^3-2916(9)^{-\frac12} -\frac{189}{2}(9) \bigg)\\ &\,\hspace{2pt}- \bigg( \frac16(4)^3-2916(4)^{-\frac12} -\frac{189}{2}(4) \bigg)\\ =&\,\bigg(\frac{729}{6}-972-\frac{1701}{2}\bigg)\\ &\,\hspace{2pt}- \bigg(\frac{64}{6}-1458-378\bigg)\\ =&\,\frac{373}{3}. \end{align*}

Therefore, the exact area of RR is

3733.\begin{align*} \frac{373}{3}. \end{align*}

解法二

思路

展开

先积分曲线在 x=4x=4 到 x=9x=9 下方的面积，再减去高度为 412\frac{41}{2}、宽为 5 的矩形。

答题过程

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At x=9x=9,

yP=812+54−74=412.y_P=\frac{81}{2}+54-74=\frac{41}{2}.

Hence the line ll is y=412y=\dfrac{41}{2}.

Integrating the curve gives

AC= ∫49(12x2+1458x−3/2−74) dx= [16x3−2916x−1/2−74x]49.\begin{align*} A_C =&\,\int_4^9 \bigg(\frac12x^2+1458x^{-3/2}-74\bigg) \,\mathrm{d}x\\ =&\,\bigg[\frac16x^3-2916x^{-1/2}-74x\bigg]_4^9. \end{align*}

Evaluate the endpoints separately:

F(9)= 7296−972−666=−30332,F(4)= 646−1458−296=−52303,AC= F(9)−F(4)=13616.\begin{align*} F(9) =&\,\frac{729}{6}-972-666 =-\frac{3033}{2},\\ F(4) =&\,\frac{64}{6}-1458-296 =-\frac{5230}{3},\\ A_C =&\,F(9)-F(4)=\frac{1361}{6}. \end{align*}

The rectangle has area

(9−4)×412=2052.(9-4)\times\frac{41}{2}=\frac{205}{2}.

Therefore

AR= 13616−2052= 3733.\begin{align*} A_R =&\,\frac{1361}{6}-\frac{205}{2}\\ =&\,\boxed{\frac{373}{3}}. \end{align*}