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IAL 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

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)

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\left(-\frac32\right)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.

(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 \left( \frac12x^2+1458x^{-\frac32}-74-\frac{41}{2} \right)\,\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 \left( \frac12x^2+1458x^{-\frac32}-\frac{189}{2} \right)\,\mathrm{d}x\\ =&\,\left[ \frac16x^3-2916x^{-\frac12}-\frac{189}{2}x \right]_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} =&\,\left( \frac16(9)^3-2916(9)^{-\frac12} -\frac{189}{2}(9) \right)\\ &\,\hspace{2pt}- \left( \frac16(4)^3-2916(4)^{-\frac12} -\frac{189}{2}(4) \right)\\ =&\,\left(\frac{729}{6}-972-\frac{1701}{2}\right)\\ &\,\hspace{2pt}- \left(\frac{64}{6}-1458-378\right)\\ =&\,\frac{373}{3}. \end{align*}

Therefore, the exact area of RR is

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