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IAL 2021 Oct Q9

A Level / Edexcel / P1

IAL 2021 Oct 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 5 shows a sketch of the curve with equation y=f(x)y=f(x) where

f(x)=x,x>0.\begin{align*} f(x)=\sqrt{x},\qquad x>0. \end{align*}

Figure 5

The point P(9,3)P(9,3) lies on the curve and is shown in Figure 5.

On the next page there is a copy of Figure 5 called Diagram 1.

Diagram 1

(a) On Diagram 1, sketch and clearly label the graphs of

y=f(2x)andy=f(x)+3.\begin{align*} y=f(2x)\quad\text{and}\quad y=f(x)+3. \end{align*}

Show on each graph the coordinates of the point to which PP is transformed.

(3)

The graph of y=f(2x)y=f(2x) meets the graph of y=f(x)+3y=f(x)+3 at the point QQ.

(b) Show that the xx coordinate of QQ is the solution of

x=3(2+1).\begin{align*} \sqrt{x}=3(\sqrt2+1). \end{align*}
(3)

(c) Hence find, in simplest form, the coordinates of QQ.

(3)

解答

(a)

解法一

思路

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y=f(2x)y=f(2x) 是水平压缩，点 P(9,3)P(9,3) 变为 (92,3)\left(\frac92,3\right)。y=f(x)+3y=f(x)+3 是向上平移 33，点 P(9,3)P(9,3) 变为 (9,6)(9,6)。

答题过程

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For y=f(2x)y=f(2x), the xx coordinate is halved, while the yy coordinate is unchanged:

P(9,3)↦(92,3).\begin{align*} P(9,3)\mapsto \left(\frac92,3\right). \end{align*}

For y=f(x)+3y=f(x)+3, the graph is translated 33 units upwards:

P(9,3)↦(9,6).\begin{align*} P(9,3)\mapsto (9,6). \end{align*}

The sketch should show y=f(2x)y=f(2x) as a horizontally compressed square-root curve, and y=f(x)+3y=f(x)+3 as the original square-root curve shifted up by 33.

(b)

解法一

思路

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交点处两条图像的 yy 值相等。由于 f(x)=xf(x)=\sqrt{x}，所以 f(2x)=2xf(2x)=\sqrt{2x}，而 f(x)+3=x+3f(x)+3=\sqrt{x}+3。

答题过程

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At QQ,

f(2x)= f(x)+32x= x+3.\begin{align*} f(2x)=&\,f(x)+3\\ \sqrt{2x}=&\,\sqrt{x}+3. \end{align*}

Since 2x=2x\sqrt{2x}=\sqrt2\sqrt{x},

2x= x+3(2−1)x= 3x= 32−1.\begin{align*} \sqrt2\sqrt{x}=&\,\sqrt{x}+3\\ (\sqrt2-1)\sqrt{x}=&\,3\\ \sqrt{x}=&\,\frac3{\sqrt2-1}. \end{align*}

Rationalise the denominator:

x= 32−1⋅2+12+1= 3(2+1)2−1= 3(2+1).\begin{align*} \sqrt{x} =&\,\frac3{\sqrt2-1}\cdot\frac{\sqrt2+1}{\sqrt2+1}\\ =&\,\frac{3(\sqrt2+1)}{2-1}\\ =&\,3(\sqrt2+1). \end{align*}

This is the required result.

(c)

解法一

思路

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由 (b) 直接平方求 xx。再代入较简单的 y=f(x)+3=x+3y=f(x)+3=\sqrt{x}+3 求 yy。

答题过程

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From part (b),

x=3(2+1).\begin{align*} \sqrt{x}=3(\sqrt2+1). \end{align*}

Square both sides:

x= (3(2+1))2= 9(2+1)2= 9(2+22+1)= 9(3+22).\begin{align*} x =&\,\bigl(3(\sqrt2+1)\bigr)^2\\ =&\,9(\sqrt2+1)^2\\ =&\,9(2+2\sqrt2+1)\\ =&\,9(3+2\sqrt2). \end{align*}

Then

y= x+3= 3(2+1)+3= 32+6.\begin{align*} y =&\,\sqrt{x}+3\\ =&\,3(\sqrt2+1)+3\\ =&\,3\sqrt2+6. \end{align*}

Therefore

Q=(9(3+22), 32+6).\begin{align*} Q=\bigl(9(3+2\sqrt2),\,3\sqrt2+6\bigr). \end{align*}