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IAL 2024 Jan Q9

A Level / Edexcel / P1

IAL 2024 Jan Paper · Question 9

题目

Problem

The curve C1C_1 has equation y=f(x)y=f(x).

Given that

  • f(x)f(x) is a quadratic expression
  • C1C_1 has a maximum turning point at (2,20)(2,20)
  • C1C_1 passes through the origin

(a) sketch a graph of C1C_1 showing the coordinates of any points where C1C_1 cuts the coordinate axes.

(2)

(b) find an expression for f(x)f(x).

(3)

The curve C2C_2 has equation

y=x(x24).\begin{align*} y=x(x^2-4). \end{align*}

Curve C1C_1 and C2C_2 meet at the origin, and at the points PP and QQ.

Given that the xx coordinate of the point PP is negative,

(c) using algebra and showing all stages of your working, find the coordinates of PP.

(5)

解答

(a)

解法一

思路

展开

二次曲线有最大转折点,所以它开口向下。已知顶点是 (2,20)(2,20),并且经过原点。由对称轴 x=2x=2 可知另一个 xx 轴截距在 x=4x=4

答题过程

展开

The graph is a downward-opening parabola with maximum turning point (2,20)(2,20).

It passes through the origin, so one intercept is

(0,0).\begin{align*} (0,0). \end{align*}

Since the axis of symmetry is x=2x=2, the other xx-intercept is

(4,0).\begin{align*} (4,0). \end{align*}

The sketch should show a downward-opening parabola through (0,0)(0,0) and (4,0)(4,0), with maximum point (2,20)(2,20).

(b)

解法一

思路

展开

因为两个根是 0044,所以可设 f(x)=Ax(x4)f(x)=Ax(x-4)。再代入顶点 (2,20)(2,20)AA

答题过程

展开

Since the roots are 00 and 44, let

f(x)=Ax(x4).\begin{align*} f(x)=Ax(x-4). \end{align*}

Using the point (2,20)(2,20),

20=A(2)(24)20=4AA=5.\begin{align*} 20=&\,A(2)(2-4)\\ 20=&\,-4A\\ A=&\,-5. \end{align*}

Therefore

f(x)=5x(x4)=5x(4x).\begin{align*} f(x)=&\,-5x(x-4)\\ =&\,5x(4-x). \end{align*}

解法二

思路

展开

也可以直接用顶点式。最大点是 (2,20)(2,20),所以写成 f(x)=a(x2)2+20f(x)=a(x-2)^2+20。因为开口向下,所以 aa 会是负数。

答题过程

展开

Using vertex form,

f(x)=a(x2)2+20.\begin{align*} f(x)=a(x-2)^2+20. \end{align*}

Since the curve passes through (0,0)(0,0),

0=a(02)2+200=4a+20a=5.\begin{align*} 0=&\,a(0-2)^2+20\\ 0=&\,4a+20\\ a=&\,-5. \end{align*}

So

f(x)=5(x2)2+20=5(x24x+4)+20=5x2+20x=5x(4x).\begin{align*} f(x)=&\,-5(x-2)^2+20\\ =&\,-5(x^2-4x+4)+20\\ =&\,-5x^2+20x\\ =&\,5x(4-x). \end{align*}

(c)

解法一

思路

展开

交点满足两个 yy 值相等。把 C1C_1 的表达式与 x(x24)x(x^2-4) 联立,因式分解后得到三个交点的 xx 坐标。题目要 xx 坐标为负的点,所以选 x=8x=-8

答题过程

展开

At intersections of C1C_1 and C2C_2,

x(x24)=5x(4x).\begin{align*} x(x^2-4)=&\,5x(4-x). \end{align*}

Expand both sides:

x34x=20x5x2x3+5x224x=0x(x2+5x24)=0x(x+8)(x3)=0.\begin{align*} x^3-4x=&\,20x-5x^2\\ x^3+5x^2-24x=&\,0\\ x(x^2+5x-24)=&\,0\\ x(x+8)(x-3)=&\,0. \end{align*}

So

x=0,x=8,x=3.\begin{align*} x=0,\quad x=-8,\quad x=3. \end{align*}

Point PP has negative xx coordinate, so

x=8.\begin{align*} x=-8. \end{align*}

Substitute into C2C_2:

y=x(x24)=(8)(644)=480.\begin{align*} y =&\,x(x^2-4)\\ =&\,(-8)(64-4)\\ =&\,-480. \end{align*}

Therefore

P=(8,480).\begin{align*} P=(-8,-480). \end{align*}

解法二

思路

展开

直接除以 xx 的二次降维法。 由于交点 PPxx 坐标是负数(即 x0x \neq 0),在联立方程 x(x24)=5x(4x)x(x^2 - 4) = 5x(4-x) 时,我们可以直接在等号两边同除以 xx。 这使我们能够直接得到关于 xx 的一元二次方程 x24=5(4x)x^2 - 4 = 5(4-x),省去了展开和分解三次多项式的步骤,让求解过程更加直接、精炼。

答题过程

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At the intersection points of C1C_1 and C2C_2, we have:

x(x24)=5x(4x)\begin{align*} x(x^2 - 4) = 5x(4-x) \end{align*}

Since point PP is not the origin, its xx-coordinate is non-zero (x0x \neq 0). Thus, we can divide both sides by xx directly:

x24=5(4x)x24=205xx2+5x24=0.\begin{align*} x^2 - 4 =&\,\, 5(4-x)\\[3mm] x^2 - 4 =&\,\, 20 - 5x\\[3mm] x^2 + 5x - 24 =&\,\, 0. \end{align*}

Factorise the quadratic equation:

(x+8)(x3)=0.\begin{align*} (x + 8)(x - 3) =&\,\, 0. \end{align*}

So

x=8orx=3.x = -8 \quad\text{or}\quad x = 3.

\begin{align*} Since the xx-coordinate of PP is negative: \end{align*}

x=8.x = -8.

\begin{align*} Substituting x=8x = -8 into y=x(x24)y = x(x^2 - 4) gives: \end{align*}

y=8((8)24)=8(644)=480.\begin{align*} y =&\,\, -8((-8)^2 - 4)\\[3mm] =&\,\, -8(64 - 4)\\[3mm] =&\,\, -480. \end{align*}

Therefore, the coordinates of PP are:

P=(8,480).P = (-8, -480).