CalcGospel 國際數學圖譜
返回

IAL 2022 May Q5

A Level / Edexcel / P1

IAL 2022 May Paper · Question 5

题目

Problem

The curve CC has equation y=f(x)y=f(x).

Given that

  • f(x)f(x) is a quadratic expression
  • the maximum turning point on CC has coordinates (−2,12)(-2,12)
  • CC cuts the negative xx-axis at −5-5

(a) find f(x)f(x).

(4)

The line l1l_1 has equation y=45xy=\dfrac45x.

Given that the line l2l_2 is perpendicular to l1l_1 and passes through (−5,0)(-5,0),

(b) find an equation for l2l_2, writing your answer in the form y=mx+cy=mx+c where mm and cc are constants to be found.

(3)

Figure 2 shows a sketch of the curve CC and the lines l1l_1 and l2l_2.

Figure 2

(c) Define the region RR, shown shaded in Figure 2, using inequalities.

(2)

解答

(a)

解法一

思路

展开

已知最高点是 (−2,12)(-2,12)，所以用顶点式最方便：f(x)=12−a(x+2)2f(x)=12-a(x+2)^2。再用截距 (−5,0)(-5,0) 求 aa。

答题过程

展开

Since the maximum turning point is (−2,12)(-2,12), write

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

Use the point (−5,0)(-5,0):

0= 12−a(−5+2)20= 12−9aa= 43.\begin{align*} 0=&\,12-a(-5+2)^2\\ 0=&\,12-9a\\ a=&\,\frac43. \end{align*}

Therefore

f(x)=12−43(x+2)2.\begin{align*} f(x)=12-\frac43(x+2)^2. \end{align*}

解法二

思路

展开

二次曲线关于 x=−2x=-2 对称。既然一个根是 −5-5，另一个根应是 11，因为 −5-5 和 11 的中点是 −2-2。

答题过程

展开

The axis of symmetry is

x=−2.\begin{align*} x=-2. \end{align*}

Since one root is x=−5x=-5, the other root is x=1x=1.

So

f(x)=a(x+5)(x−1).\begin{align*} f(x)=a(x+5)(x-1). \end{align*}

Use the maximum point (−2,12)(-2,12):

12= a(−2+5)(−2−1)12= a(3)(−3)a= −43.\begin{align*} 12=&\,a(-2+5)(-2-1)\\ 12=&\,a(3)(-3)\\ a=&\,-\frac43. \end{align*}

Therefore

f(x)=−43(x−1)(x+5).\begin{align*} f(x)=-\frac43(x-1)(x+5). \end{align*}

(b)

解法一

思路

展开

l1l_1 的斜率是 45\frac45，所以垂线 l2l_2 的斜率是负倒数 −54-\frac54。再代入点 (−5,0)(-5,0) 求截距。

答题过程

展开

The gradient of l1l_1 is 45\frac45, so the gradient of l2l_2 is

−54.\begin{align*} -\frac54. \end{align*}

Using the point (−5,0)(-5,0),

y−0= −54(x+5)y= −54x−254.\begin{align*} y-0=&\,-\frac54(x+5)\\ y=&\,-\frac54x-\frac{25}{4}. \end{align*}

(c)

解法一

思路

展开

从图像看，区域在 l1l_1 上方、l2l_2 上方、二次曲线下方。用三个不等式分别描述边界。

答题过程

展开

Using

f(x)=−43x2−163x+203,\begin{align*} f(x)=-\frac43x^2-\frac{16}{3}x+\frac{20}{3}, \end{align*}

the region RR is defined by

y≥45x,y≥−54x−254,y≤−43x2−163x+203.\begin{align*} y&\ge \frac45x,\\ y&\ge -\frac54x-\frac{25}{4},\\ y&\le -\frac43x^2-\frac{16}{3}x+\frac{20}{3}. \end{align*}