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IAL 2022 Oct Q9

A Level / Edexcel / P1

IAL 2022 Oct Paper · Question 9

题目

Problem

Figure 3 shows a sketch of the curve CC with equation

y=12x2−10x+22.\begin{align*} y=\frac12x^2-10x+22. \end{align*}

Figure 3

(a) Write 12x2−10x+22\frac12x^2-10x+22 in the form

a(x+b)2+c\begin{align*} a(x+b)^2+c \end{align*}

where aa, bb and cc are constants to be found.

(3)

The point MM is the minimum turning point of CC, as shown in Figure 3.

(b) Deduce the coordinates of MM.

(2)

The line ll is the normal to CC at the point PP, as shown in Figure 3.

Given that ll has equation y=k−18xy=k-\dfrac18x, where kk is a constant,

(c) (i) find the coordinates of PP

(ii) find the value of kk

(6)

Figure 4 is a copy of Figure 3. The finite region RR, shown shaded in Figure 4, is bounded by ll, CC and the line through MM parallel to the yy-axis.

Figure 4

(d) Identify the inequalities that define RR.

(3)

解答

(a)

解法一

思路

展开

先把 12\frac12 提出来，让括号里出现 x2−20xx^2-20x。完成平方时，x2−20x=(x−10)2−100x^2-20x=(x-10)^2-100。

答题过程

展开 12x2−10x+22= 12(x2−20x)+22= 12((x−10)2−100)+22= 12(x−10)2−50+22= 12(x−10)2−28.\begin{align*} \frac12x^2-10x+22 =&\,\frac12(x^2-20x)+22\\ =&\,\frac12\left((x-10)^2-100\right)+22\\ =&\,\frac12(x-10)^2-50+22\\ =&\,\frac12(x-10)^2-28. \end{align*}

So

a=12,b=−10,c=−28.\begin{align*} a=\frac12, \qquad b=-10, \qquad c=-28. \end{align*}

(b)

解法一

思路

展开

完成平方形式 12(x−10)2−28\frac12(x-10)^2-28 直接给出最低点。平方项最小时是 00，发生在 x=10x=10。

答题过程

展开

From

y=12(x−10)2−28,\begin{align*} y=\frac12(x-10)^2-28, \end{align*}

the minimum occurs when

x=10.\begin{align*} x=10. \end{align*}

Then

y=−28.\begin{align*} y=-28. \end{align*}

Therefore

M=(10,−28).\begin{align*} M=(10,-28). \end{align*}

(c)(i)

解法一

思路

展开

法线斜率是 −18-\frac18，所以切线斜率是它的负倒数 88。曲线的切线斜率来自导函数，令导函数等于 88 就能求出点 PP 的横坐标。

答题过程

展开

The gradient of the normal is

−18.\begin{align*} -\frac18. \end{align*}

Therefore the gradient of the tangent is

8.\begin{align*} 8. \end{align*}

Differentiate the curve:

dydx= x−10.\begin{align*} \frac{dy}{dx} =&\,x-10. \end{align*}

At PP,

x−10= 8x= 18.\begin{align*} x-10=&\,8\\ x=&\,18. \end{align*}

Find the yy coordinate:

y= 12(18)2−10(18)+22= 162−180+22= 4.\begin{align*} y =&\,\frac12(18)^2-10(18)+22\\ =&\,162-180+22\\ =&\,4. \end{align*}

Thus

P=(18,4).\begin{align*} P=(18,4). \end{align*}

(c)(ii)

解法一

思路

展开

点 PP 在法线上，所以把 P=(18,4)P=(18,4) 代入 y=k−18xy=k-\frac18x，即可求 kk。

答题过程

展开

Since P=(18,4)P=(18,4) lies on

y=k−18x,\begin{align*} y=k-\frac18x, \end{align*}

we have

4= k−18(18)4= k−94k= 254.\begin{align*} 4=&\,k-\frac18(18)\\ 4=&\,k-\frac94\\ k=&\,\frac{25}{4}. \end{align*}

(d)

解法一

思路

展开

区域左边界是过最低点的竖直线 x=10x=10。阴影区域在法线下方、曲线上方，所以用三个不等式描述：在竖线右侧、低于直线、高于曲线。

答题过程

展开

The vertical boundary through MM is

x=10.\begin{align*} x=10. \end{align*}

The region is to the right of this line, below the normal and above the curve. Therefore

x≥10,y≤254−18x,y≥12x2−10x+22.\begin{align*} x&\ge10,\\ y&\le \frac{25}{4}-\frac18x,\\ y&\ge \frac12x^2-10x+22. \end{align*}