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IAL 2023 Oct Q8

A Level / Edexcel / P1

IAL 2023 Oct Paper · Question 8

题目

Problem

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

Solutions relying on calculator technology are not acceptable.

The curve C1C_1 has equation

xy=152−5x,x≠0.\begin{align*} xy=\frac{15}{2}-5x,\qquad x\ne0. \end{align*}

The curve C2C_2 has equation

y=x3−72x−5.\begin{align*} y=x^3-\frac72x-5. \end{align*}

(a) Show that C1C_1 and C2C_2 meet when

2x4−7x2−15=0.\begin{align*} 2x^4-7x^2-15=0. \end{align*}
(2)

Given that C1C_1 and C2C_2 meet at points PP and QQ,

(b) find, using algebra, the exact distance PQPQ.

(5)

解答

(a)

解法一

思路

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交点满足两个曲线的 yy 值相同。由 C1C_1 可写出 xy=152−5xxy=\frac{15}{2}-5x，再把 C2C_2 的 yy 代入即可。

答题过程

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At an intersection,

y=x3−72x−5.\begin{align*} y=x^3-\frac72x-5. \end{align*}

Since C1C_1 has equation

xy=152−5x,\begin{align*} xy=\frac{15}{2}-5x, \end{align*}

substitute the expression for yy:

x(x3−72x−5)= 152−5xx4−72x2−5x= 152−5x.\begin{align*} x\left(x^3-\frac72x-5\right) =&\,\frac{15}{2}-5x\\ x^4-\frac72x^2-5x =&\,\frac{15}{2}-5x. \end{align*}

The −5x-5x terms cancel:

x4−72x2−152= 0.\begin{align*} x^4-\frac72x^2-\frac{15}{2}=&\,0. \end{align*}

Multiplying by 22 gives

2x4−7x2−15=0.\begin{align*} 2x^4-7x^2-15=0. \end{align*}

(b)

解法一

思路

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把 x2x^2 看成一个整体来解四次方程。解出两个交点的坐标后，用距离公式求 PQPQ。

答题过程

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Let

u=x2.\begin{align*} u=x^2. \end{align*}

Then

2u2−7u−15= 0(2u+3)(u−5)= 0.\begin{align*} 2u^2-7u-15=&\,0\\ (2u+3)(u-5)=&\,0. \end{align*}

So

u=5oru=−32.\begin{align*} u=5 \quad\text{or}\quad u=-\frac32. \end{align*}

Since u=x2u=x^2, reject u=−32u=-\frac32. Hence

x2=5,x=±5.\begin{align*} x^2=5, \qquad x=\pm\sqrt5. \end{align*}

Use C2C_2 to find the yy coordinates.

For x=5x=\sqrt5,

y= (5)3−725−5= 55−725−5= 325−5.\begin{align*} y =&\,(\sqrt5)^3-\frac72\sqrt5-5\\ =&\,5\sqrt5-\frac72\sqrt5-5\\ =&\,\frac32\sqrt5-5. \end{align*}

For x=−5x=-\sqrt5,

y= (−5)3−72(−5)−5= −55+725−5= −325−5.\begin{align*} y =&\,(-\sqrt5)^3-\frac72(-\sqrt5)-5\\ =&\,-5\sqrt5+\frac72\sqrt5-5\\ =&\,-\frac32\sqrt5-5. \end{align*}

So the two points are

(5,325−5)and(−5,−325−5).\begin{align*} \left(\sqrt5,\frac32\sqrt5-5\right) \quad\text{and}\quad \left(-\sqrt5,-\frac32\sqrt5-5\right). \end{align*}

Therefore

PQ= (25)2+(35)2= 20+45= 65.\begin{align*} PQ =&\,\sqrt{\left(2\sqrt5\right)^2+\left(3\sqrt5\right)^2}\\ =&\,\sqrt{20+45}\\ =&\,\sqrt{65}. \end{align*}