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IAL 2023 Oct Q11

A Level / Edexcel / P1

IAL 2023 Oct Paper · Question 11

题目

Problem

Figure 5 shows part of the curve CC with equation y=f(x)y=f(x) where

f(x)=2x2−12x+14.\begin{align*} f(x)=2x^2-12x+14. \end{align*}

Figure 5

(a) Write 2x2−12x+142x^2-12x+14 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)

Given that CC has a minimum at the point PP,

(b) state the coordinates of PP.

(1)

The line ll intersects CC at (−1,28)(-1,28) and at PP as shown in Figure 5.

(c) Find the equation of ll giving your answer in the form y=mx+cy=mx+c where mm and cc are constants to be found.

(3)

The finite region RR, shown shaded in Figure 5, is bounded by the xx-axis, ll, the yy-axis, and CC.

(d) Use inequalities to define the region RR.

(3)

解答

(a)

解法一

思路

展开

把二次项和一次项先提出 22，再完成平方。注意题目形式是 a(x+b)2+ca(x+b)^2+c，所以最后括号中会是 x−3x-3，也就是 b=−3b=-3。

答题过程

展开 2x2−12x+14= 2(x2−6x)+14= 2((x−3)2−9)+14= 2(x−3)2−18+14= 2(x−3)2−4.\begin{align*} 2x^2-12x+14 =&\,2(x^2-6x)+14\\ =&\,2\left((x-3)^2-9\right)+14\\ =&\,2(x-3)^2-18+14\\ =&\,2(x-3)^2-4. \end{align*}

Therefore

a=2,b=−3,c=−4.\begin{align*} a=2,\qquad b=-3,\qquad c=-4. \end{align*}

(b)

解法一

思路

展开

完成平方形式 2(x−3)2−42(x-3)^2-4 直接给出顶点。因为 2>02>0，这个顶点是最低点。

答题过程

展开

From part (a),

f(x)=2(x−3)2−4.\begin{align*} f(x)=2(x-3)^2-4. \end{align*}

The minimum occurs when

x=3.\begin{align*} x=3. \end{align*}

Then

y=−4.\begin{align*} y=-4. \end{align*}

Therefore

P=(3,−4).\begin{align*} P=(3,-4). \end{align*}

(c)

解法一

思路

展开

直线 ll 经过 (−1,28)(-1,28) 和 P(3,−4)P(3,-4)。先用两点求斜率，再代入一点求方程。

答题过程

展开

The gradient of ll is

m= 28−(−4)−1−3= 32−4= −8.\begin{align*} m =&\,\frac{28-(-4)}{-1-3}\\ =&\,\frac{32}{-4}\\ =&\,-8. \end{align*}

Using (−1,28)(-1,28),

y−28= −8(x+1)y−28= −8x−8y= −8x+20.\begin{align*} y-28=&\,-8(x+1)\\ y-28=&\,-8x-8\\ y=&\,-8x+20. \end{align*}

(d)

解法一

思路

展开

区域 RR 在 yy 轴右侧、xx 轴上方、直线下方，并且在曲线上方。把这些边界分别写成不等式即可。

答题过程

展开

The region is to the right of the yy-axis:

x≥0.\begin{align*} x\ge0. \end{align*}

It is above the xx-axis:

y≥0.\begin{align*} y\ge0. \end{align*}

It is below the line ll:

y≤−8x+20.\begin{align*} y\le -8x+20. \end{align*}

It is above the curve CC:

y≥2x2−12x+14.\begin{align*} y\ge 2x^2-12x+14. \end{align*}

Therefore the region RR is defined by

x≥0,y≥0,y≤−8x+20,y≥2x2−12x+14.\begin{align*} x&\ge0,\\ y&\ge0,\\ y&\le -8x+20,\\ y&\ge 2x^2-12x+14. \end{align*}