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IAL 2021 Jan Q4

A Level / Edexcel / P1

IAL 2021 Jan Paper · Question 4

题目

Problem

The points PP and QQ, as shown in Figure 2, have coordinates (−2,13)(-2,13) and (4,−5)(4,-5) respectively.

Figure 2

The straight line ll passes through PP and QQ.

(a) Find an equation for ll, writing your answer in the form y=mx+cy=mx+c, where mm and cc are integers to be found.

(3)

The quadratic curve CC passes through PP and has a minimum point at QQ.

(b) Find an equation for CC.

(3)

The region RR, shown shaded in Figure 2, lies in the second quadrant and is bounded by CC and ll only.

(c) Use inequalities to define region RR.

(2)

解答

(a)

解法一

思路

展开

用两点求斜率，再代入其中一个点求截距。

答题过程

展开 m= −5−134−(−2)= −186= −3.\begin{align*} m =&\,\frac{-5-13}{4-(-2)}\\ =&\,\frac{-18}{6}\\ =&\,-3. \end{align*}

Use P(−2,13)P(-2,13):

y−13= −3(x+2)y= −3x+7.\begin{align*} y-13=&\,-3(x+2)\\ y=&\,-3x+7. \end{align*}

(b)

解法一

思路

展开

最低点是 Q(4,−5)Q(4,-5)，所以曲线可写成 y=a(x−4)2−5y=a(x-4)^2-5。再代入 P(−2,13)P(-2,13) 求 aa。

答题过程

展开

Since the minimum point is (4,−5)(4,-5),

y=a(x−4)2−5.\begin{align*} y=a(x-4)^2-5. \end{align*}

The curve passes through P(−2,13)P(-2,13):

13= a(−2−4)2−518= 36aa= 12.\begin{align*} 13=&\,a(-2-4)^2-5\\ 18=&\,36a\\ a=&\,\frac12. \end{align*}

Therefore

y=12(x−4)2−5.\begin{align*} y=\frac12(x-4)^2-5. \end{align*}

(c)

解法一

思路

展开

区域在第二象限、在直线上方、在抛物线下方，并且位于左侧交点 PP 的左边。

答题过程

展开

The region is above the line:

y>−3x+7.\begin{align*} y>-3x+7. \end{align*}

It is below the curve:

y<12(x−4)2−5.\begin{align*} y<\frac12(x-4)^2-5. \end{align*}

It lies to the left of PP, so

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

Thus

y>−3x+7,y<12(x−4)2−5,x<−2.\begin{align*} y&>-3x+7,\\ y&<\frac12(x-4)^2-5,\\ x&<-2. \end{align*}