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FP1 2024 May Q2

A Level / Edexcel / FP1

IAL 2024 May Paper · Question 2

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

 f(z)= z3−13z2+59z+p\begin{align*} &\,f(z)\\ =&\,z^3-13z^2+59z+p \end{align*} p∈Zp\in\mathbb{Z}

Given that z=3z = 3 is a root of the equation f(z)=0f(z) = 0

(a) show that p=−87p = -87

(2)

(b) Use algebra to determine the other roots of f(z)=0f(z) = 0, giving your answers in simplest form.

(4)

On an Argand diagram

  • the root z=3z = 3 is represented by the point PP
  • the other roots of f(z)=0f(z) = 0 are represented by the points QQ and RR
  • the number z=−9z = -9 is represented by the point SS

(c) Show on a single Argand diagram the positions of PP, QQ, RR and SS

(1)

(d) Determine the perimeter of the quadrilateral PQSRPQSR, giving your answer as a simplified surd.

(2)
题目中文翻译

本题必须展示完整解题过程。

不允许完全依赖计算器技术。

已知

 f(z)= z3−13z2+59z+p\begin{align*} &\,f(z)\\ =&\,z^3-13z^2+59z+p \end{align*} p∈Zp\in\mathbb{Z}

已知 z=3z = 3 是方程 f(z)=0f(z) = 0 的一个根。

(a) 证明 p=−87p = -87。

(2)

(b) 用代数方法确定方程 f(z)=0f(z) = 0 的其余根，结果以最简形式给出。

(4)

在阿甘图（Argand diagram）上：

  • 根 z=3z = 3 用点 PP 表示
  • 方程 f(z)=0f(z) = 0 的其余根用点 QQ 和 RR 表示
  • 数 z=−9z = -9 用点 SS 表示

(c) 在同一个阿甘图上标出 PP、QQ、RR、SS 的位置。

(1)

(d) 确定四边形 PQSRPQSR 的周长，结果以最简根式给出。

(2)

解答

(a)

解法一

思路

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因为 z=3z=3 是方程的根，所以 f(3)=0f(3)=0。代入并整理即可自然推出题目给定的 pp 值。

答题过程

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Since z=3z=3 is a root,

 f(3)= 33−13(3)2+59(3)+p= 27−117+177+p= 87+p.\begin{align*} &\,f(3)\\ =&\,3^3-13(3)^2+59(3)+p\\ =&\,27-117+177+p\\ =&\,87+p. \end{align*}

Therefore,

87+p= 0p= −87.\begin{align*} 87+p=&\,0\\ p=&\,\boxed{-87}. \end{align*}

(b)

解法一

思路

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承接 (a)，先把 p=−87p=-87 代回多项式。已知 z=3z=3 是根，所以 z−3z-3 是因式；用代数除法得到二次因式，再用二次方程公式求其两个复根。

答题过程

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Using p=−87p=-87 from part (a),

 f(z)= z3−13z2+59z−87.\begin{align*} &\,f(z)\\ =&\,z^3-13z^2+59z-87. \end{align*}

For division by z−3z-3, the first quotient term is z2z^2. Subtracting its product with z−3z-3 gives

 (z3−13z2+59z−87) −(z3−3z2)= −10z2+59z−87.\begin{align*} &\,(z^3-13z^2+59z-87)\\ &\,\hspace{2pt}-(z^3-3z^2)\\ =&\,-10z^2+59z-87. \end{align*}

The next quotient term is −10z-10z:

 (−10z2+59z−87) −(−10z2+30z)= 29z−87.\begin{align*} &\,(-10z^2+59z-87)\\ &\,\hspace{2pt}-(-10z^2+30z)\\ =&\,29z-87. \end{align*}

The final quotient term is 2929, since

 (29z−87) −29(z−3)= 0.\begin{align*} &\,(29z-87)\\ &\,\hspace{2pt}-29(z-3)\\ =&\,0. \end{align*}

Thus the quotient is z2−10z+29z^2-10z+29 and the remainder is zero, giving

 f(z)= (z−3)(z2−10z+29).\begin{align*} &\,f(z)\\ =&\,(z-3)(z^2-10z+29). \end{align*}

Hence the other roots satisfy

z2−10z+29=0.z^2-10z+29=0.

The discriminant is

Δ= (−10)2−4(1)(29)= 100−116= −16.\begin{align*} \Delta =&\,(-10)^2-4(1)(29)\\ =&\,100-116\\ =&\,-16. \end{align*}

Using the quadratic formula,

 z= 10±Δ2= 10±−162= 5±2i.\begin{align*} &\,z\\ =&\,\frac{10\pm\sqrt{\Delta}}{2}\\ =&\,\frac{10\pm\sqrt{-16}}{2}\\ =&\,5\pm2\mathrm{i}. \end{align*}

Therefore, the other roots are

z=5+2iandz=5−2i.\begin{gathered} \boxed{z=5+2\mathrm{i}}\\ \text{and}\quad\boxed{z=5-2\mathrm{i}}. \end{gathered}

解法二

思路

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由上面的代数除法，另外两个根满足二次方程。也可以把二次式配方，得到复数的平方等于负数，再求出两个根。

答题过程

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Using the quadratic factor obtained above,

z2−10z+29=0.z^2-10z+29=0.

Completing the square,

(z−5)2−25+29= 0(z−5)2+4= 0(z−5)2= −4.\begin{align*} (z-5)^2-25+29=&\,0\\ (z-5)^2+4=&\,0\\ (z-5)^2=&\,-4. \end{align*}

Hence

z−5=±2i.z-5=\pm2\mathrm{i}.

The other roots are therefore

z=5+2iandz=5−2i.\begin{gathered} \boxed{z=5+2\mathrm{i}}\\ \text{and}\quad\boxed{z=5-2\mathrm{i}}. \end{gathered}

(c)

解法一

思路

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在 Argand 图上，横坐标表示实部，纵坐标表示虚部。两个非实根关于实轴对称；PP 与 SS 都位于实轴上。

答题过程

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The four points are

P= (3,0),Q= (5,2),R= (5,−2),S= (−9,0).\begin{align*} P=&\,(3,0),\\ Q=&\,(5,2),\\ R=&\,(5,-2),\\ S=&\,(-9,0). \end{align*}

The points QQ and RR are vertically aligned and symmetric about the real axis.

(d)

解法一

思路

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四边形关于实轴对称，因此 PQ=PRPQ=PR、QS=RSQS=RS。分别用两点间距离公式求一条短边与一条长边，再把它们各取两倍。

答题过程

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By symmetry,

PQ=PRPQ=PR

and

QS=RS.QS=RS.

Now,

 PQ= (5−3)2+(2−0)2= 8= 22,\begin{align*} &\,PQ\\ =&\,\sqrt{(5-3)^2+(2-0)^2}\\ =&\,\sqrt8\\ =&\,2\sqrt2, \end{align*}

For QSQS, the coordinate differences are

Δx= 5−(−9)=14,Δy= 2−0=2.\begin{align*} \Delta x=&\,5-(-9)=14,\\ \Delta y=&\,2-0=2. \end{align*}

Hence,

 QS= (Δx)2+(Δy)2= 142+22= 200= 102.\begin{align*} &\,QS\\ =&\,\sqrt{(\Delta x)^2+(\Delta y)^2}\\ =&\,\sqrt{14^2+2^2}\\ =&\,\sqrt{200}\\ =&\,10\sqrt2. \end{align*}

Therefore, the perimeter of PQSRPQSR is

 2PQ+2QS= 2(22)+2(102)= 242.\begin{align*} &\,2PQ+2QS\\ =&\,2(2\sqrt2)+2(10\sqrt2)\\ =&\,\boxed{24\sqrt2}. \end{align*}