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IAL 2025 May A Q8

A Level / Edexcel / P1

IAL 2025 May A Paper · Question 8

题目

Problem

Figure 3 shows a sketch of a line ll and a quadratic curve CC.

Figure 3

Given that ll passes through (−3,0)(-3,0) and (2,18)(2,18),

(a) find an equation for ll in the form y=mx+cy=mx+c, where mm and cc are constants.

(3)

Given that

  • CC and ll intersect at the points (−3,0)(-3,0) and (2,18)(2,18)
  • CC crosses the yy-axis at (0,−6)(0,-6)

(b) find an equation for CC.

(4)

The region RR, shown shaded in Figure 3, is bounded by CC, ll and the yy-axis.

(c) Use inequalities to define RR.

(2)

解答

(a)

解法一

思路

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先用两点求直线斜率，再代入其中一点求截距。

答题过程

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The gradient of ll is

m= 18−02−(−3)= 185.\begin{align*} m=&\,\frac{18-0}{2-(-3)}\\ =&\,\frac{18}{5}. \end{align*}

So

y=185x+c.\begin{align*} y=\frac{18}{5}x+c. \end{align*}

Use the point (−3,0)(-3,0):

0= 185(−3)+cc= 545.\begin{align*} 0=&\,\frac{18}{5}(-3)+c\\ c=&\,\frac{54}{5}. \end{align*}

Therefore

y=185x+545.\begin{align*} y=\frac{18}{5}x+\frac{54}{5}. \end{align*}

(b)

解法一

思路

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二次曲线经过 (0,−6)(0,-6)，所以常数项是 −6-6。设

y=ax2+bx−6.\begin{align*} y=ax^2+bx-6. \end{align*}

再代入另外两个点求 a,ba,b。

答题过程

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Let the equation of CC be

y=ax2+bx−6.\begin{align*} y=ax^2+bx-6. \end{align*}

Using (−3,0)(-3,0):

0= 9a−3b−69a−3b= 6.\begin{align*} 0=&\,9a-3b-6\\ 9a-3b=&\,6. \end{align*}

Using (2,18)(2,18):

18= 4a+2b−64a+2b= 24.\begin{align*} 18=&\,4a+2b-6\\ 4a+2b=&\,24. \end{align*}

Simplify the two equations:

3a−b= 2,2a+b= 12.\begin{align*} 3a-b=&\,2,\\ 2a+b=&\,12. \end{align*}

Add them:

5a= 14a= 145.\begin{align*} 5a=&\,14\\ a=&\,\frac{14}{5}. \end{align*}

Then

2a+b= 122(145)+b= 12b= 325.\begin{align*} 2a+b=&\,12\\ 2\left(\frac{14}{5}\right)+b=&\,12\\ b=&\,\frac{32}{5}. \end{align*}

Therefore

y=145x2+325x−6.\begin{align*} y=\frac{14}{5}x^2+\frac{32}{5}x-6. \end{align*}

(c)

解法一

思路

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阴影区域在 yy 轴左侧，所以 x≤0x\leq0。在这段区域中，直线在上方，二次曲线在下方，所以 yy 夹在两者之间。

答题过程

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The region is to the left of the yy-axis, so

x≤0.\begin{align*} x\leq0. \end{align*}

It lies above the curve CC and below the line ll. Therefore

145x2+325x−6≤yy≤185x+545.\begin{align*} \frac{14}{5}x^2+\frac{32}{5}x-6 &\leq y\\ y&\leq \frac{18}{5}x+\frac{54}{5}. \end{align*}

So RR is defined by

145x2+325x−6≤y≤185x+545,x≤0.\begin{align*} \frac{14}{5}x^2+\frac{32}{5}x-6 &\leq y \leq \frac{18}{5}x+\frac{54}{5},\\ x&\leq0. \end{align*}