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IAL 2025 May Q9

A Level / Edexcel / P1

IAL 2025 May Paper · Question 9

题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Figure 4 shows a sketch of

Figure 4
  • the graph C1C_1 with equation y=2xy=\sqrt{2x}
  • the graph C2C_2 with equation y=12−xy=12-\sqrt{x}

(a) Describe fully the single transformation that would transform

(i) the graph with equation y=xy=\sqrt{x} onto C1C_1

(ii) the graph with equation y=−xy=-\sqrt{x} onto C2C_2

(4)

The graphs C1C_1 and C2C_2 meet at the point PP, as shown in Figure 4.

(b) (i) Show that the xx coordinate of PP satisfies

x=12(2−1)\begin{align*} \sqrt{x}=12(\sqrt2-1) \end{align*}

(ii) Hence find, in simplest form, the exact coordinates of PP.

(6)

解答

(a)(i)

解法一

思路

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2x\sqrt{2x} 可以看成把 xx 替换成 2x2x，这是水平方向压缩到原来的 12\frac12；也可以看成 2x\sqrt2\sqrt{x}，即竖直方向放大 2\sqrt2。题目只要求一个完整描述，写其中一种即可。

答题过程

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The graph of

y=x\begin{align*} y=\sqrt{x} \end{align*}

is transformed to

y=2x\begin{align*} y=\sqrt{2x} \end{align*}

by a stretch parallel to the xx-axis with scale factor

12.\begin{align*} \frac12. \end{align*}

解法二

思路

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因为 2x=2x\sqrt{2x}=\sqrt2\sqrt{x}，所以也可以描述为竖直方向的伸缩。

答题过程

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Since

2x=2x,\begin{align*} \sqrt{2x}=\sqrt2\sqrt{x}, \end{align*}

the transformation can also be described as a stretch parallel to the yy-axis with scale factor

2.\begin{align*} \sqrt2. \end{align*}

(a)(ii)

解法一

思路

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y=12−xy=12-\sqrt{x} 就是 y=−xy=-\sqrt{x} 整体向上平移 1212 个单位。

答题过程

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The graph of

y=−x\begin{align*} y=-\sqrt{x} \end{align*}

is transformed to

y=12−x\begin{align*} y=12-\sqrt{x} \end{align*}

by a translation by the vector

(012).\begin{align*} \begin{pmatrix}0\\12\end{pmatrix}. \end{align*}

(b)(i)

解法一

思路

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交点处两条曲线的 yy 值相等。把 2x\sqrt{2x} 写成 2x\sqrt2\sqrt{x}，再提出公因式 x\sqrt{x}，最后对分母有理化。

答题过程

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At PP,

2x= 12−x.\begin{align*} \sqrt{2x}=&\,12-\sqrt{x}. \end{align*}

So

2x+x= 122x+x= 12x(2+1)= 12.\begin{align*} \sqrt{2x}+\sqrt{x}=&\,12\\ \sqrt2\sqrt{x}+\sqrt{x}=&\,12\\ \sqrt{x}(\sqrt2+1)=&\,12. \end{align*}

Hence

x= 122+1= 122+1⋅2−12−1= 12(2−1).\begin{align*} \sqrt{x} =&\,\frac{12}{\sqrt2+1}\\ =&\,\frac{12}{\sqrt2+1}\cdot\frac{\sqrt2-1}{\sqrt2-1}\\ =&\,12(\sqrt2-1). \end{align*}

This is the required result.

解法二

思路

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也可以把方程两边平方，得到关于 x\sqrt{x} 的二次方程。最后要选正值，因为 x>0\sqrt{x}>0。

答题过程

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At PP,

2x= 12−x.\begin{align*} \sqrt{2x}=&\,12-\sqrt{x}. \end{align*}

Square both sides:

2x= (12−x)22x= 144−24x+xx+24x−144= 0.\begin{align*} 2x=&\,(12-\sqrt{x})^2\\ 2x=&\,144-24\sqrt{x}+x\\ x+24\sqrt{x}-144=&\,0. \end{align*}

Let

u=x.\begin{align*} u=\sqrt{x}. \end{align*}

Then

u2+24u−144= 0.\begin{align*} u^2+24u-144=&\,0. \end{align*}

Using the quadratic formula,

u= −24±242−4(1)(−144)2= −24±11522= −12±122.\begin{align*} u=&\,\frac{-24\pm\sqrt{24^2-4(1)(-144)}}{2}\\ =&\,\frac{-24\pm\sqrt{1152}}{2}\\ =&\,-12\pm12\sqrt2. \end{align*}

Since u=x>0u=\sqrt{x}>0,

x=122−12=12(2−1).\begin{align*} \sqrt{x}=12\sqrt2-12=12(\sqrt2-1). \end{align*}

(b)(ii)

解法一

思路

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由 (b)(i) 得到 x=12(2−1)\sqrt{x}=12(\sqrt2-1)，所以平方可得 xx。再把 x\sqrt{x} 代入 y=12−xy=12-\sqrt{x} 求 yy。

答题过程

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From part (b)(i),

x=12(2−1).\begin{align*} \sqrt{x}=12(\sqrt2-1). \end{align*}

So

x= (12(2−1))2= 144(2−1)2= 144(2−22+1)= 144(3−22).\begin{align*} x =&\,\left(12(\sqrt2-1)\right)^2\\ =&\,144(\sqrt2-1)^2\\ =&\,144(2-2\sqrt2+1)\\ =&\,144(3-2\sqrt2). \end{align*}

Also,

y= 12−x= 12−12(2−1)= 24−122.\begin{align*} y =&\,12-\sqrt{x}\\ =&\,12-12(\sqrt2-1)\\ =&\,24-12\sqrt2. \end{align*}

Therefore

P=(144(3−22), 24−122).\begin{align*} P=\bigl(144(3-2\sqrt2),\,24-12\sqrt2\bigr). \end{align*}