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IAL 2022 May Q6

A Level / Edexcel / P1

IAL 2022 May Paper · Question 6

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

(a) Given that

2xy−3x2=50\begin{align*} 2xy-3x^2=50 \end{align*}

and

y−x3+6x=0\begin{align*} y-x^3+6x=0 \end{align*}

show that

2x4−15x2−50=0.\begin{align*} 2x^4-15x^2-50=0. \end{align*}
(2)

(b) Hence solve the simultaneous equations

2xy−3x2= 50,y−x3+6x= 0.\begin{align*} 2xy-3x^2=&\,50,\\ y-x^3+6x=&\,0. \end{align*}

Give your answers in fully simplified surd form.

(5)

解答

(a)

解法一

思路

展开

从第二个方程把 yy 表示成 x3−6xx^3-6x，再代入第一个方程。Show that 题要把中间展开写清楚。

答题过程

展开

From

y−x3+6x=0,\begin{align*} y-x^3+6x=0, \end{align*}

we have

y=x3−6x.\begin{align*} y=x^3-6x. \end{align*}

Substitute into 2xy−3x2=502xy-3x^2=50:

2x(x3−6x)−3x2= 502x4−12x2−3x2= 502x4−15x2−50= 0.\begin{align*} 2x(x^3-6x)-3x^2=&\,50\\ 2x^4-12x^2-3x^2=&\,50\\ 2x^4-15x^2-50=&\,0. \end{align*}

(b)

解法一

思路

展开

把上一小题看成关于 x2x^2 的二次方程。解出 x2=10x^2=10 后，再用 y=x3−6xy=x^3-6x 求 yy，注意正负 xx 会配对出正负 yy。

答题过程

展开

From part (a),

2x4−15x2−50=0.\begin{align*} 2x^4-15x^2-50=0. \end{align*}

Factorise:

2x4−15x2−50= (2x2+5)(x2−10).\begin{align*} 2x^4-15x^2-50 =&\,(2x^2+5)(x^2-10). \end{align*}

So

(2x2+5)(x2−10)= 0.\begin{align*} (2x^2+5)(x^2-10)=&\,0. \end{align*}

Since 2x2+5=02x^2+5=0 gives no real value of xx,

x2=10.\begin{align*} x^2=10. \end{align*}

Thus

x=10orx=−10.\begin{align*} x=\sqrt{10} \quad\text{or}\quad x=-\sqrt{10}. \end{align*}

Using y=x3−6xy=x^3-6x:

when x=10,y= (10)3−610= 1010−610= 410.\begin{align*} \text{when }x=\sqrt{10},\quad y=&\,(\sqrt{10})^3-6\sqrt{10}\\ =&\,10\sqrt{10}-6\sqrt{10}\\ =&\,4\sqrt{10}. \end{align*}

Similarly, when x=−10x=-\sqrt{10},

y=−410.\begin{align*} y=-4\sqrt{10}. \end{align*}

Therefore the solutions are

(x,y)=(10,410)or(x,y)=(−10,−410).\begin{align*} (x,y)=\left(\sqrt{10},4\sqrt{10}\right) \quad\text{or}\quad (x,y)=\left(-\sqrt{10},-4\sqrt{10}\right). \end{align*}