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IAL 2019 Jan Q11

A Level / Edexcel / P1

IAL 2019 Jan Paper · Question 11

题目

Problem

(a) On Diagram 1 sketch the graphs of

(i) y=x(3−x)y=x(3-x),

(ii) y=x(x−2)(5−x)y=x(x-2)(5-x),

showing clearly the coordinates of the points where the curves cross the coordinate axes.

(4)

Diagram 1

(b) Show that the xx coordinates of the points of intersection of

y=x(3−x)andy=x(x−2)(5−x)\begin{align*} y=x(3-x) \quad\text{and}\quad y=x(x-2)(5-x) \end{align*}

are given by the solutions to the equation

x(x2−8x+13)=0.\begin{align*} x(x^2-8x+13)=0. \end{align*}
(3)

The point PP lies on both curves. Given that PP lies in the first quadrant,

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

(5)

解答

(a)

解法一

思路

展开

第一条是开口向下的二次曲线，根是 00 和 33。第二条是负三次曲线，根是 0,2,50,2,5。

答题过程

展开

For y=x(3−x)y=x(3-x), the graph is a downward-opening quadratic crossing the xx-axis at

(0,0)and(3,0).\begin{align*} (0,0)\quad\text{and}\quad(3,0). \end{align*}

For y=x(x−2)(5−x)y=x(x-2)(5-x), the graph is a negative cubic crossing the xx-axis at

(0,0),(2,0),(5,0).\begin{align*} (0,0),\quad(2,0),\quad(5,0). \end{align*}

(b)

解法一

思路

展开

交点满足两个 yy 值相等。把两式联立后，把所有项移到一边并提出 xx。

答题过程

展开

At the points of intersection,

x(3−x)= x(x−2)(5−x).\begin{align*} x(3-x)=&\,x(x-2)(5-x). \end{align*}

So

x(x−2)(5−x)−x(3−x)= 0x{(x−2)(5−x)−(3−x)}= 0.\begin{align*} x(x-2)(5-x)-x(3-x)=&\,0\\ x\{(x-2)(5-x)-(3-x)\}=&\,0. \end{align*}

Now

(x−2)(5−x)= −x2+7x−10.\begin{align*} (x-2)(5-x) =&\,-x^2+7x-10. \end{align*}

Therefore

x(−x2+7x−10−3+x)= 0x(−x2+8x−13)= 0.\begin{align*} x(-x^2+7x-10-3+x)=&\,0\\ x(-x^2+8x-13)=&\,0. \end{align*}

Multiplying by −1-1 gives

x(x2−8x+13)=0.\begin{align*} x(x^2-8x+13)=0. \end{align*}

解法二

思路

展开

两边完全展开后合并同类项求解的方法。 我们也可以将等式两边完全展开，再合并同类项。 方程的左边展开为：

y=3x−x2\begin{align*} y = 3x - x^2 \end{align*}

方程的右边展开为：

y=x(x2−7x+10)=x3−7x2+10x\begin{align*} y = x(x^2 - 7x + 10) = x^3 - 7x^2 + 10x \end{align*}

令两边相等：

3x−x2=x3−7x2+10x\begin{align*} 3x - x^2 = x^3 - 7x^2 + 10x \end{align*}

将所有项移到等号同一边并提取公因式 xx，即可得出 x(x2−8x+13)=0x(x^2 - 8x + 13) = 0。该方法极为直观，非常符合常规的代数展开和整理习惯。

答题过程

展开

Expand the LHS of the intersection equation y=x(3−x)y = x(3-x):

LHS=3x−x2.\begin{align*} \text{LHS} = 3x - x^2. \end{align*}

Expand the RHS of the intersection equation y=x(x−2)(5−x)y = x(x-2)(5-x):

RHS=  x(5x−x2−10+2x)=  x(−x2+7x−10)=  −x3+7x2−10x.\begin{align*} \text{RHS} =&\,\, x(5x - x^2 - 10 + 2x)\\[3mm] =&\,\, x(-x^2 + 7x - 10)\\[3mm] =&\,\, -x^3 + 7x^2 - 10x. \end{align*}

Set LHS equal to RHS:

3x−x2=−x3+7x2−10x3x - x^2 = -x^3 + 7x^2 - 10x

\begin{align*} Move all terms to one side (e.g. to the left-hand side): \end{align*}

x3−x2−7x2+3x+10x=  0x3−8x2+13x=  0.\begin{align*} x^3 - x^2 - 7x^2 + 3x + 10x =&\,\, 0\\[3mm] x^3 - 8x^2 + 13x =&\,\, 0. \end{align*}

Factor out the common term xx:

x(x2−8x+13)=0.x(x^2 - 8x + 13) = 0.

This is the required equation.

(c)

解法一

思路

展开

第一象限的交点不是原点，所以解 x2−8x+13=0x^2-8x+13=0。两个解中，较小的那个给出第一象限交点。

答题过程

展开 x2−8x+13= 0x= 8±64−522= 4±3.\begin{align*} x^2-8x+13=&\,0\\ x=&\,\frac{8\pm\sqrt{64-52}}{2}\\ =&\,4\pm\sqrt3. \end{align*}

For the first quadrant point,

x=4−3.\begin{align*} x=4-\sqrt3. \end{align*}

Then

y= x(3−x)= (4−3)(3−4+3)= (4−3)(−1+3)= −4+43+3−3= −7+53.\begin{align*} y=&\,x(3-x)\\ =&\,(4-\sqrt3)(3-4+\sqrt3)\\ =&\,(4-\sqrt3)(-1+\sqrt3)\\ =&\,-4+4\sqrt3+\sqrt3-3\\ =&\,-7+5\sqrt3. \end{align*}

Therefore

P=(4−3, −7+53).\begin{align*} P=\left(4-\sqrt3,\,-7+5\sqrt3\right). \end{align*}