STEP2 2023 -- Pure Mathematics

STEP2 2023 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2023  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	微积分 (Calculus)	Standard	代换积分法,偶函数利用,被积函数组合,反三角函数求值
2	三角学 (Trigonometry)	Challenging	tan双角公式,tan三倍角公式,cos方程的周期性和偶性,参数代换法
3	微积分 (Calculus)	Challenging	多项式导数性质,分部积分,指数函数单调性,反证法
4	代数与函数 (Algebra & Functions)	Challenging	共轭配对消根,逐次平方消根,对称多项式,Vieta公式
5	数列与级数 (Sequences & Series)	Challenging	数学归纳法,递推不等式,绝对值符号处理,指数衰减估计
6	矩阵与线性代数 (Matrices & Linear Algebra)	Hard	矩阵归纳法,行列式乘法性质,矩阵乘法比较元素,矩阵二项式展开
7	复数 (Complex Numbers)	Standard	复数模的乘法性质,Gaussian整数因子分解,逐步降解大数,Pythagorean三元组
8	几何与向量 (Geometry & Vectors)	Challenging	向量点积运算,对称性论证,质心坐标,充要条件双向证明

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Question 1

Topic: 微积分 (Calculus)  |  Difficulty: Standard  |  Marks: 20

Problem

1 (i) Show that making the substitution $x = \frac{1}{t}$ in the integral

$\int_{a}^{b} \frac{1}{(1 + x^2)^{\frac{3}{2}}} \, dx \, ,$

where $b > a > 0$, gives the integral

$\int_{a^{-1}}^{b^{-1}} \frac{-t}{(1 + t^2)^{\frac{3}{2}}} \, dt \, .$

(ii) Evaluate:

(a) $\int_{\frac{1}{2}}^{2} \frac{1}{(1 + x^2)^{\frac{3}{2}}} \, dx \, ;$

(b) $\int_{-2}^{2} \frac{1}{(1 + x^2)^{\frac{3}{2}}} \, dx \, .$

(iii) (a) Show that

$\int_{\frac{1}{2}}^{2} \frac{1}{(1 + x^2)^2} \, dx = \int_{\frac{1}{2}}^{2} \frac{x^2}{(1 + x^2)^2} \, dx = \frac{1}{2} \int_{\frac{1}{2}}^{2} \frac{1}{1 + x^2} \, dx \, ,$

and hence evaluate

$\int_{\frac{1}{2}}^{2} \frac{1}{(1 + x^2)^2} \, dx \, .$

(b) Evaluate

$\int_{\frac{1}{2}}^{2} \frac{1 - x}{x(1 + x^2)^{\frac{1}{2}}} \, dx \, .$

Hint

(i) $\int_{a}^{b} \frac{1}{(1+x^2)^{\frac{3}{2}}} \mathrm{d}x = \int_{a^{-1}}^{b^{-1}} \frac{1}{\left(1+\frac{1}{t^2}\right)^{\frac{3}{2}}} \cdot -\frac{1}{t^2} \mathrm{d}t$ [M1] $= \int_{a^{-1}}^{b^{-1}} \frac{-t}{(1+t^2)^{\frac{3}{2}}} \mathrm{d}t$ [A1] (ii) (a) $\int_{\frac{1}{2}}^{2} \frac{1}{(1+x^2)^{\frac{3}{2}}} \mathrm{d}x = \int_{2}^{\frac{1}{2}} \frac{-t}{(1+t^2)^{\frac{3}{2}}} \mathrm{d}t$ [M1] $= \left[ (1+t^2)^{-\frac{1}{2}} \right]_{2}^{\frac{1}{2}}$ [M1] $= \left(\frac{5}{4}\right)^{-\frac{1}{2}} - 5^{-\frac{1}{2}} = \frac{1}{\sqrt{5}}$ [A1] (ii) (a) (b) The integrand is even, so $\int_{-2}^{2} \frac{1}{(1+x^2)^{\frac{3}{2}}} \mathrm{d}x = 2 \int_{0}^{2} \frac{1}{(1+x^2)^{\frac{3}{2}}} \mathrm{d}x$ [M1] $= \underset{\varepsilon \to 0}{Lim} \int_{\varepsilon}^{2} \frac{2}{(1+x^2)^{\frac{3}{2}}} \mathrm{d}x$ which is of the form in the stem [M1] $= \underset{\varepsilon \to 0}{Lim} \left[ 2(1+t^2)^{-\frac{1}{2}} \right]_{\varepsilon}^{\frac{1}{2}}$ [M1] Dealing with limiting process for the lower limit [E1] $= \frac{4}{\sqrt{5}}$ [A1] $-\underset{\varepsilon \to 0}{Lim} \frac{2\varepsilon}{\sqrt{1+\varepsilon^2}}$ [A1] so the limit is zero and the integral $= \frac{4}{\sqrt{5}}$ [E1] (iii) (a) $\int_{\frac{1}{2}}^{2} \frac{1}{(1+x^2)^2} \mathrm{d}x = \int_{2}^{\frac{1}{2}} \frac{-t^2}{(1+t^2)^2} \mathrm{d}t$ [M1] $= \int_{\frac{1}{2}}^{2} \frac{x^2}{(1+x^2)^2} \mathrm{d}x$ [A1] $= \frac{1}{2} \int_{\frac{1}{2}}^{2} \frac{1}{1+x^{2}} dx$ [A1] $= \frac{1}{2} \left( \arctan 2 - \arctan \frac{1}{2} \right)$ [B1] (b) $\int_{\frac{1}{2}}^{2} \frac{1-x}{x\left(1+x^{2}\right)^{\frac{1}{2}}} dx = \int_{2}^{\frac{1}{2}} \frac{t-1}{\left(1+\frac{1}{t^{2}}\right)^{\frac{1}{2}}} \cdot -\frac{1}{t^{2}} dt$ [M1] $= \int_{2}^{\frac{1}{2}} \frac{1-t}{t\left(1+t^{2}\right)^{\frac{1}{2}}} dt$ [A1] $= -\int_{\frac{1}{2}}^{2} \frac{1-x}{x\left(1+x^{2}\right)^{\frac{1}{2}}} dx$, so the integral = 0. [E1]

Model Solution

Part (i)

Let $x = \frac{1}{t}$, so $\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{1}{t^2}$.

When $x = a$, $t = a^{-1}$; when $x = b$, $t = b^{-1}$.

Substituting into the integrand:

$\frac{1}{(1 + x^2)^{3/2}} = \frac{1}{\left(1 + \frac{1}{t^2}\right)^{3/2}} = \frac{1}{\left(\frac{t^2 + 1}{t^2}\right)^{3/2}} = \frac{t^3}{(1 + t^2)^{3/2}}$

So the integral becomes:

$\int_{a}^{b} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x = \int_{a^{-1}}^{b^{-1}} \frac{t^3}{(1 + t^2)^{3/2}} \cdot \left(-\frac{1}{t^2}\right) \mathrm{d}t = \int_{a^{-1}}^{b^{-1}} \frac{-t}{(1 + t^2)^{3/2}} \, \mathrm{d}t$

as required.

Part (ii)(a)

Applying the result from (i) with $a = \frac{1}{2}$ and $b = 2$:

$\int_{1/2}^{2} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x = \int_{2}^{1/2} \frac{-t}{(1 + t^2)^{3/2}} \, \mathrm{d}t = \int_{1/2}^{2} \frac{t}{(1 + t^2)^{3/2}} \, \mathrm{d}t$

Since $\frac{\mathrm{d}}{\mathrm{d}t}\left[(1 + t^2)^{-1/2}\right] = -t(1 + t^2)^{-3/2}$, we have:

$\int_{1/2}^{2} \frac{t}{(1 + t^2)^{3/2}} \, \mathrm{d}t = \left[-(1 + t^2)^{-1/2}\right]_{1/2}^{2}$

Wait — let me be careful. We have $\frac{-t}{(1+t^2)^{3/2}} = \frac{\mathrm{d}}{\mathrm{d}t}[(1+t^2)^{-1/2}]$, so:

$\int_{2}^{1/2} \frac{-t}{(1+t^2)^{3/2}} \, \mathrm{d}t = \left[(1+t^2)^{-1/2}\right]_{2}^{1/2} = \left(\frac{5}{4}\right)^{-1/2} - \left(5\right)^{-1/2} = \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}}$

Part (ii)(b)

The integrand $f(x) = (1 + x^2)^{-3/2}$ is an even function, so:

$\int_{-2}^{2} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x = 2\int_{0}^{2} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x$

Since the lower limit $a = 0$ does not satisfy $a > 0$ in part (i), we take a limit. For $\varepsilon > 0$:

$2\int_{\varepsilon}^{2} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x = 2\left[(1 + t^2)^{-1/2}\right]_{2}^{1/\varepsilon} = 2\left[\left(1 + \frac{1}{\varepsilon^2}\right)^{-1/2} - 5^{-1/2}\right]$

Now $\left(1 + \frac{1}{\varepsilon^2}\right)^{-1/2} = \frac{\varepsilon}{\sqrt{1 + \varepsilon^2}} \to 0$ as $\varepsilon \to 0^+$.

So the limit gives $2\left[0 - \frac{1}{\sqrt{5}}\right] = -\frac{2}{\sqrt{5}}$.

But wait, we need to handle the limits more carefully. Using part (i):

$2\int_{\varepsilon}^{2} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x = 2\int_{\varepsilon^{-1}}^{1/2} \frac{-t}{(1 + t^2)^{3/2}} \, \mathrm{d}t = 2\left[(1 + t^2)^{-1/2}\right]_{\varepsilon^{-1}}^{1/2}$

$= 2\left[\left(\frac{5}{4}\right)^{-1/2} - \left(1 + \varepsilon^{-2}\right)^{-1/2}\right] = 2\left[\frac{2}{\sqrt{5}} - \frac{\varepsilon}{\sqrt{1 + \varepsilon^2}}\right]$

As $\varepsilon \to 0^+$, the second term vanishes, giving:

$\int_{-2}^{2} \frac{1}{(1 + x^2)^{3/2}} \, \mathrm{d}x = \frac{4}{\sqrt{5}}$

Part (iii)(a)

Apply the substitution $x = \frac{1}{t}$ to $\int_{1/2}^{2} \frac{1}{(1+x^2)^2} \, \mathrm{d}x$:

$\frac{1}{(1 + x^2)^2} = \frac{1}{\left(1 + \frac{1}{t^2}\right)^2} = \frac{t^4}{(1 + t^2)^2}$

So:

$\int_{1/2}^{2} \frac{1}{(1+x^2)^2} \, \mathrm{d}x = \int_{2}^{1/2} \frac{t^4}{(1+t^2)^2} \cdot \left(-\frac{1}{t^2}\right) \mathrm{d}t = \int_{1/2}^{2} \frac{t^2}{(1+t^2)^2} \, \mathrm{d}t$

Renaming $t$ back to $x$:

$\int_{1/2}^{2} \frac{1}{(1+x^2)^2} \, \mathrm{d}x = \int_{1/2}^{2} \frac{x^2}{(1+x^2)^2} \, \mathrm{d}x$

Adding these two equal integrals:

$2\int_{1/2}^{2} \frac{1}{(1+x^2)^2} \, \mathrm{d}x = \int_{1/2}^{2} \frac{1 + x^2}{(1+x^2)^2} \, \mathrm{d}x = \int_{1/2}^{2} \frac{1}{1+x^2} \, \mathrm{d}x$

Therefore:

$\int_{1/2}^{2} \frac{1}{(1+x^2)^2} \, \mathrm{d}x = \frac{1}{2}\int_{1/2}^{2} \frac{1}{1+x^2} \, \mathrm{d}x = \frac{1}{2}\Big[\arctan x\Big]_{1/2}^{2} = \frac{1}{2}\left(\arctan 2 - \arctan \frac{1}{2}\right)$

Since $\arctan 2 + \arctan \frac{1}{2} = \frac{\pi}{2}$ (as $\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta = \frac{1}{\tan\theta}$), we get $\arctan 2 - \arctan\frac{1}{2} = 2\arctan 2 - \frac{\pi}{2}$, so the integral equals $\arctan 2 - \frac{\pi}{4}$.

Part (iii)(b)

Let $I = \int_{1/2}^{2} \frac{1 - x}{x(1 + x^2)^{1/2}} \, \mathrm{d}x$. Substitute $x = \frac{1}{t}$:

$\frac{1 - x}{x(1 + x^2)^{1/2}} = \frac{1 - \frac{1}{t}}{\frac{1}{t}\left(1 + \frac{1}{t^2}\right)^{1/2}} = \frac{\frac{t-1}{t}}{\frac{1}{t} \cdot \frac{\sqrt{t^2+1}}{t}} = \frac{(t-1) \cdot t}{\sqrt{t^2+1}}$

With $\mathrm{d}x = -\frac{1}{t^2} \, \mathrm{d}t$:

$I = \int_{2}^{1/2} \frac{t(t-1)}{\sqrt{t^2+1}} \cdot \left(-\frac{1}{t^2}\right) \mathrm{d}t = \int_{1/2}^{2} \frac{t-1}{t\sqrt{t^2+1}} \, \mathrm{d}t = \int_{1/2}^{2} \frac{x-1}{x\sqrt{1+x^2}} \, \mathrm{d}x$

This equals $-I$. Therefore $I = -I$, giving $2I = 0$, so $I = 0$.

Examiner Notes

第(i)部分多数完成良好，但弱考生未能正确变换积分限或对1/x求导。第(ii)(b)部分许多考生误以为可直接应用(i)的结果（实际不行），但部分考生识别出被积函数为偶函数。第(iii)(b)部分许多考生未能选择合适代换，但选择正确者通常能得出答案。

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Question 2

Topic: 三角学 (Trigonometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 (i) The real numbers $x$, $y$ and $z$ satisfy the equations

$y = \frac{2x}{1 - x^2} ,$ $z = \frac{2y}{1 - y^2} ,$ $x = \frac{2z}{1 - z^2} .$

Let $x = \tan \alpha$. Deduce that $y = \tan 2\alpha$ and show that $\tan \alpha = \tan 8\alpha$.

Find all solutions of the equations, giving each value of $x$, $y$ and $z$ in the form $\tan \theta$ where $-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi$.

(ii) Determine the number of real solutions of the simultaneous equations

$y = \frac{3x - x^3}{1 - 3x^2} ,$ $z = \frac{3y - y^3}{1 - 3y^2} ,$ $x = \frac{3z - z^3}{1 - 3z^2} .$

(iii) Consider the simultaneous equations

$y = 2x^2 - 1 ,$ $z = 2y^2 - 1 ,$ $x = 2z^2 - 1 .$

(a) Determine the number of real solutions of these simultaneous equations with $|x| \leqslant 1$, $|y| \leqslant 1$, $|z| \leqslant 1$.

(b) By finding the degree of a single polynomial equation which is satisfied by $x$, show that all solutions of these simultaneous equations have $|x| \leqslant 1$, $|y| \leqslant 1$, $|z| \leqslant 1$.

Hint

(i) Let $x = \tan \alpha$. Then $y = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \tan 2\alpha$ so $z = \tan 4\alpha$, and so $\tan \alpha = \tan 8\alpha$. [E1] giving $8\alpha = \alpha + n\pi$, or $\alpha = \frac{1}{7}n\pi$, (for $n = -3$ to 3). [M1] Solutions are: $(0,0,0), \left(\tan\left(\frac{1}{7}\alpha\right), \tan\left(\frac{2}{7}\alpha\right), \tan\left(-\frac{3}{7}\alpha\right)\right)$ [B1] and cyclic permutations of the latter [A1] and $\left(\tan\left(-\frac{1}{7}\alpha\right), \tan\left(-\frac{2}{7}\alpha\right), \tan\left(\frac{3}{7}\alpha\right)\right)$ and its cyclic permutations [A1] (ii) $\tan 3\alpha = \frac{\frac{2 \tan \alpha}{1 - \tan^2 \alpha} + \tan \alpha}{1 - \frac{2 \tan^2 \alpha}{1 - \tan^2 \alpha}}$ [M1] $= \frac{3 \tan \alpha - \tan^3 \alpha}{1 - 3 \tan^2 \alpha}$ [A1] Let $x = \tan \alpha$; then $y = \tan 3\alpha, z = \tan 9\alpha$, [M1] so $\tan 27\alpha = \tan \alpha$ [A1] giving $26\alpha = n\pi$ [A1] which has 25 solutions with distinct values of $\tan \alpha$ because $n = 13$ does not give a possible value of $\tan \alpha$. [A1] Checking that for each finite value of $x$, the denominators of $y$ and $z$ are defined (i.e. checking $1-3x^2$ is non-zero). [E1] (iii) (a) Let $x = \cos \alpha$ [M1] the restriction on $|x|$ means this is a complete parametrisation of solutions [E1] Then, using $\cos 2\alpha = 2 \cos^2 \alpha - 1, \cos 8\alpha = \cos \alpha$ [M1] so $8\alpha = \alpha + 2m\pi$, or $8\alpha = -\alpha + 2n\pi$ [M1] so $7\alpha = 2m\pi$ or $9\alpha = 2n\pi$ [A1] with 4 ($m = 0$ to 3) + 5 ($n = 0$ to 4) [M1] $- 1$ (for $\alpha = 0$ twice) = 8 distinct solutions [A1] EMPTY (b) $y$ quadratic, so $z$ quartic in $x$, so $x$ satisfies an octic equation [B1] which has at most 8 roots, so there are no larger solutions. [E1]

Model Solution

Part (i)

Let $x = \tan\alpha$. Then:

$y = \frac{2x}{1 - x^2} = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \tan 2\alpha$

using the double angle formula for tangent. Similarly:

$z = \frac{2y}{1 - y^2} = \frac{2\tan 2\alpha}{1 - \tan^2 2\alpha} = \tan 4\alpha$

And the third equation gives:

$x = \frac{2z}{1 - z^2} = \frac{2\tan 4\alpha}{1 - \tan^2 4\alpha} = \tan 8\alpha$

Since $x = \tan\alpha$, we have $\tan\alpha = \tan 8\alpha$.

This requires $8\alpha = \alpha + n\pi$ for integer $n$, so $7\alpha = n\pi$, giving $\alpha = \frac{n\pi}{7}$.

For $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$, we need $n \in \{-3, -2, -1, 0, 1, 2, 3\}$, giving 7 solutions.

$n = 0$: $(x, y, z) = (\tan 0, \tan 0, \tan 0) = (0, 0, 0)$.

$n = 1$: $\alpha = \frac{\pi}{7}$, so $(x, y, z) = \left(\tan\frac{\pi}{7},\, \tan\frac{2\pi}{7},\, \tan\frac{4\pi}{7}\right)$. Since $\frac{4\pi}{7} \in (\frac{\pi}{2}, \pi)$, we write $\tan\frac{4\pi}{7} = \tan\left(\frac{4\pi}{7} - \pi\right) = \tan\left(-\frac{3\pi}{7}\right)$. So $(x,y,z) = \left(\tan\frac{\pi}{7},\, \tan\frac{2\pi}{7},\, \tan\left(-\frac{3\pi}{7}\right)\right)$.

$n = 2$: $\alpha = \frac{2\pi}{7}$, giving $(x,y,z) = \left(\tan\frac{2\pi}{7},\, \tan\left(-\frac{3\pi}{7}\right),\, \tan\frac{\pi}{7}\right)$ (cyclic permutation of $n=1$).

$n = 3$: $\alpha = \frac{3\pi}{7}$, giving $(x,y,z) = \left(\tan\frac{3\pi}{7},\, \tan\left(-\frac{\pi}{7}\right),\, \tan\left(-\frac{2\pi}{7}\right)\right)$.

$n = -1$: $\alpha = -\frac{\pi}{7}$, giving $(x,y,z) = \left(\tan\left(-\frac{\pi}{7}\right),\, \tan\left(-\frac{2\pi}{7}\right),\, \tan\frac{3\pi}{7}\right)$ (cyclic permutation of $n=3$).

$n = -2$: $\alpha = -\frac{2\pi}{7}$, giving $(x,y,z) = \left(\tan\left(-\frac{2\pi}{7}\right),\, \tan\frac{3\pi}{7},\, \tan\left(-\frac{\pi}{7}\right)\right)$ (cyclic permutation of $n=3$).

$n = -3$: $\alpha = -\frac{3\pi}{7}$, giving $(x,y,z) = \left(\tan\left(-\frac{3\pi}{7}\right),\, \tan\frac{\pi}{7},\, \tan\frac{2\pi}{7}\right)$ (cyclic permutation of $n=1$).

In summary, there are 7 solutions: the trivial solution $(0,0,0)$ and two families of 3 cyclic permutations each.

Part (ii)

We first verify the triple angle formula. Using $\tan 3\alpha = \tan(2\alpha + \alpha)$:

$\tan 3\alpha = \frac{\tan 2\alpha + \tan\alpha}{1 - \tan 2\alpha \cdot \tan\alpha} = \frac{\frac{2\tan\alpha}{1-\tan^2\alpha} + \tan\alpha}{1 - \frac{2\tan^2\alpha}{1-\tan^2\alpha}}$

$= \frac{\frac{2\tan\alpha + \tan\alpha(1-\tan^2\alpha)}{1-\tan^2\alpha}}{\frac{1-\tan^2\alpha - 2\tan^2\alpha}{1-\tan^2\alpha}} = \frac{3\tan\alpha - \tan^3\alpha}{1 - 3\tan^2\alpha}$

So the equation $y = \frac{3x - x^3}{1 - 3x^2}$ is exactly $y = \tan 3\alpha$ when $x = \tan\alpha$.

Let $x = \tan\alpha$. Then $y = \tan 3\alpha$, $z = \tan 9\alpha$, and the third equation gives $x = \tan 27\alpha$. So $\tan\alpha = \tan 27\alpha$, requiring $26\alpha = n\pi$, i.e., $\alpha = \frac{n\pi}{26}$.

For $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$: $n \in \{-12, -11, \ldots, 11, 12\}$, giving 25 values.

We must check that all intermediate expressions are defined. We need $\tan\alpha$, $\tan 3\alpha$, and $\tan 9\alpha$ to be finite, and the denominators $1 - 3\tan^2\alpha$, $1 - 3\tan^2 3\alpha$, $1 - 3\tan^2 9\alpha$ to be non-zero.

• $\tan\alpha$ is undefined when $\alpha = \pm\frac{\pi}{2}$, excluded by the range.
• $1 - 3\tan^2\alpha = 0$ requires $\tan^2\alpha = \frac{1}{3}$, i.e., $\alpha = \pm\frac{\pi}{6}$. Since $\frac{n\pi}{26} = \frac{\pi}{6}$ gives $n = \frac{26}{6}$, not an integer, this never occurs.
• $\tan 3\alpha$ undefined requires $3\alpha = \pm\frac{\pi}{2} + k\pi$, i.e., $\frac{3n}{26} = \frac{1}{2} + k$, i.e., $3n = 13 + 26k$. Since $13$ is odd and $26k$ is even, $3n$ must be odd, so $n$ is odd. Then $3n = 13 + 26k$ gives $n = \frac{13 + 26k}{3}$, which is never an integer (as $13 \equiv 1 \pmod{3}$ and $26k \equiv 2k \pmod{3}$, so we need $2k \equiv 2 \pmod{3}$, i.e., $k \equiv 1 \pmod{3}$, giving $n = \frac{13 + 26(3m+1)}{3} = \frac{39 + 78m}{3} = 13 + 26m$, but $n = 13$ is outside the range $[-12, 12]$).
• Similar analysis shows $\tan 9\alpha$ is always defined and $1 - 3\tan^2 9\alpha \neq 0$ for $n \in [-12, 12]$.

Therefore all 25 values of $n$ give valid solutions, and the answer is 25 real solutions.

Part (iii)(a)

Since $|x| \leq 1$, write $x = \cos\theta$ for $\theta \in [0, \pi]$ (this parametrizes $[-1, 1]$ completely).

Using the identity $\cos 2\theta = 2\cos^2\theta - 1$:

$y = 2x^2 - 1 = 2\cos^2\theta - 1 = \cos 2\theta$ $z = 2y^2 - 1 = 2\cos^2 2\theta - 1 = \cos 4\theta$ $x = 2z^2 - 1 = 2\cos^2 4\theta - 1 = \cos 8\theta$

So we need $\cos\theta = \cos 8\theta$, which gives either:

Case 1: $8\theta = \theta + 2m\pi$, i.e., $\theta = \frac{2m\pi}{7}$. Case 2: $8\theta = -\theta + 2n\pi$, i.e., $\theta = \frac{2n\pi}{9}$.

For $\theta \in [0, \pi]$:

• Case 1: $m = 0, 1, 2, 3$ giving $\theta = 0, \frac{2\pi}{7}, \frac{4\pi}{7}, \frac{6\pi}{7}$ (4 values).
• Case 2: $n = 0, 1, 2, 3, 4$ giving $\theta = 0, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{6\pi}{9}, \frac{8\pi}{9}$ (5 values).

The value $\theta = 0$ appears in both cases, so there are $4 + 5 - 1 = 8$ distinct values of $\theta$, each giving a distinct triple $(x, y, z) = (\cos\theta, \cos 2\theta, \cos 4\theta)$.

There are 8 solutions with $|x| \leq 1$, $|y| \leq 1$, $|z| \leq 1$.

Part (iii)(b)

The map $x \mapsto 2x^2 - 1$ doubles the degree. Starting from $x$:

• $y = 2x^2 - 1$ is degree 2 in $x$.
• $z = 2y^2 - 1$ is degree 4 in $x$.
• $x = 2z^2 - 1$ gives a polynomial equation of degree 8 in $x$.

A polynomial of degree 8 has at most 8 real roots. Since we found 8 solutions in part (a) (all satisfying $|x| \leq 1$), these account for all solutions. Therefore every solution of the simultaneous equations satisfies $|x| \leq 1$, and consequently $|y| \leq 1$ and $|z| \leq 1$ (since $y = 2x^2 - 1 \in [-1, 1]$ when $x \in [-1, 1]$, and similarly for $z$).

Examiner Notes

第(i)部分多数成功证明tan alpha = tan 8alpha，但部分考生误用半角公式（逻辑方向错误）。范围限制令许多考生困惑，仅找到(0,0,0)而遗漏其他解。第(ii)部分27到25的常见错误：未排除tan无定义的x=正负pi/2。第(iii)(a)部分仅用周期性未用偶性导致只找到一半解；画图者表现明显更好。

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Question 3

Topic: 微积分 (Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Let $p(x)$ be a polynomial of degree $n$ with $p(x) > 0$ for all $x$ and let $q(x) = \sum_{k=0}^{n} p^{(k)}(x) ,$ where $p^{(k)}(x) \equiv \frac{\mathrm{d}^k p(x)}{\mathrm{d}x^k}$ for $k \geqslant 1$ and $p^{(0)}(x) \equiv p(x)$.

(i) (a) Explain why $n$ must be even and show that $q(x)$ takes positive values for some values of $x$.

(b) Show that $q'(x) = q(x) - p(x)$.

(ii) In this part you will be asked to show the same result in three different ways.

(a) Show that the curves $y = p(x)$ and $y = q(x)$ meet at every stationary point of $y = q(x)$.

Hence show that $q(x) > 0$ for all $x$.

(b) Show that $e^{-x}q(x)$ is a decreasing function.

Hence show that $q(x) > 0$ for all $x$.

(c) Show that $\int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t = p(x) + \int_{0}^{\infty} p^{(1)}(x + t)e^{-t} \, \mathrm{d}t .$ Show further that $\int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t = q(x) .$ Hence show that $q(x) > 0$ for all $x$.

Hint

(i) (a) An odd degree polynomial takes positive and negative values for large enough $|x|$. [B1] The degree of q is equal to the degree of p, and the coefficient of $x^n$ is positive, because each derivative has lower degree and cannot affect the coefficient of $x^n$. [M1] So q(x) > 0 for large enough $|x|$. [A1] (i) (a) (b) $q'(x) = \sum_{k=0}^{n} p^{(k+1)}(x) = \sum_{k=0}^{n-1} p^{(k+1)}(x)$ $[p^{(n+1)}(x) \equiv 0]$ [M1] $= \sum_{k=1}^{n} p^{(k)}(x) = q(x) - p(x)$ [A1] (ii) (a) If $q'(x) = 0$, $q(x) = p(x)$, so the two curves meet at any stationary point. [B1] But $q(x) > 0$ for large enough $|x|$. [M1] So q(x) has an absolute minimum value [M1] at which its value is positive, as $q(x) = p(x)$ there. [A1] (ii) (a) (b) $\frac{d}{dx}(e^{-x} q(x)) = e^{-x}(-q(x) + q'(x))$ [M1] $= -e^{-x} p(x) < 0$ [A1] For large enough x, $q(x) > 0$, so $e^{-x} q(x) > 0$, [M1] but $e^{-x} q(x)$ decreasing, so positive for all x, [A1] and hence so is q(x). [A1] (ii) (a) (b) (c) $\int_{0}^{\infty} p(x+t)e^{-t} dt$ [M1] $= [-p(x+t)e^{-t}]_{0}^{\infty} + \int_{0}^{\infty} p^{(1)}(x+t)e^{-t} dt$ $= (0 - (-p(x))) + \int_{0}^{\infty} p^{(1)}(x+t)e^{-t} dt$ [A1] So $\int_{0}^{\infty} p(x+t)e^{-t} dt = p(x) + \int_{0}^{\infty} p^{(1)}(x+t)e^{-t} dt$ [M1] $= p(x) + p^{(1)}(x) + \int_{0}^{\infty} p^{(2)}(x+t)e^{-t} dt$ $= p(x) + p^{(1)}(x) + \dots + p^{(n)}(x) + \int_{0}^{\infty} p^{(n+1)}(x+t)e^{-t} dt$ [M1] but $p^{(n+1)}(x) \equiv 0$, so [A1] $\int_{0}^{\infty} p(x+t)e^{-t} dt = p(x) + p^{(1)}(x) + \dots + p^{(n)}(x) = q(x).$ but $p(x+t)$, $e^{-t} > 0$ for all $t \ge 0$, so $q(x) > 0$. [E1]

Model Solution

Part (i)(a)

If $n$ were odd, then as $x \to +\infty$, $p(x) \to +\infty$ (since $p(x) > 0$ for large $x$ forces the leading coefficient to be positive), but as $x \to -\infty$, $p(x) \to -\infty$, contradicting $p(x) > 0$ for all $x$. Therefore $n$ must be even.

For $q(x)$: the $k$-th derivative $p^{(k)}(x)$ has degree $n - k$, so the only contribution to the $x^n$ term in $q(x) = \sum_{k=0}^{n} p^{(k)}(x)$ comes from $p^{(0)}(x) = p(x)$. Thus $q(x)$ is a polynomial of degree $n$ with the same leading coefficient as $p(x)$, which is positive. Since $n$ is even, $q(x) \to +\infty$ as $|x| \to \infty$, so $q(x) > 0$ for sufficiently large $|x|$.

Part (i)(b)

$q'(x) = \sum_{k=0}^{n} p^{(k+1)}(x)$

Since $p$ has degree $n$, $p^{(n+1)}(x) \equiv 0$, so:

$q'(x) = \sum_{k=0}^{n-1} p^{(k+1)}(x) = \sum_{j=1}^{n} p^{(j)}(x) = \sum_{k=0}^{n} p^{(k)}(x) - p^{(0)}(x) = q(x) - p(x)$

Part (ii)(a)

At any stationary point $x_0$ of $q$, we have $q'(x_0) = 0$, so by part (i)(b):

$q(x_0) - p(x_0) = 0 \implies q(x_0) = p(x_0) > 0$

Thus the curves $y = p(x)$ and $y = q(x)$ meet at every stationary point of $q$, and the value of $q$ there is positive.

From part (i)(a), $q(x) \to +\infty$ as $|x| \to \infty$. Since $q$ is a polynomial (hence continuous), it attains an absolute minimum at some point $x_0$. At this minimum, $q(x_0) = p(x_0) > 0$. Therefore:

$q(x) \geq q(x_0) > 0 \quad \text{for all } x$

Part (ii)(b)

Compute the derivative:

$\frac{\mathrm{d}}{\mathrm{d}x}\Big[e^{-x} q(x)\Big] = e^{-x} q'(x) - e^{-x} q(x) = e^{-x}\big(q'(x) - q(x)\big) = e^{-x}\big(-p(x)\big) < 0$

since $e^{-x} > 0$ and $p(x) > 0$ for all $x$. Therefore $e^{-x}q(x)$ is strictly decreasing.

From part (i)(a), $q(x) > 0$ for sufficiently large $x$, so $e^{-x}q(x) > 0$ for large $x$. Since $e^{-x}q(x)$ is decreasing, it must be positive for all $x$ (if it were non-positive at some point, it would remain non-positive for all larger $x$, contradicting positivity for large $x$).

Since $e^{-x} > 0$ for all $x$, it follows that $q(x) > 0$ for all $x$.

Part (ii)(c)

Integrate by parts with $u = p(x + t)$ and $\mathrm{d}v = e^{-t}\,\mathrm{d}t$:

$\int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t = \Big[-p(x + t)e^{-t}\Big]_{0}^{\infty} + \int_{0}^{\infty} p^{(1)}(x + t)e^{-t} \, \mathrm{d}t$

As $t \to \infty$, $p(x+t)e^{-t} \to 0$ (exponential decay dominates polynomial growth). At $t = 0$: $-p(x) \cdot 1 = -p(x)$. So the boundary term evaluates to $0 - (-p(x)) = p(x)$, giving:

$\int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t = p(x) + \int_{0}^{\infty} p^{(1)}(x + t)e^{-t} \, \mathrm{d}t$

Applying this reduction repeatedly:

$\int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t = p(x) + p^{(1)}(x) + \int_{0}^{\infty} p^{(2)}(x + t)e^{-t} \, \mathrm{d}t$

$= p(x) + p^{(1)}(x) + p^{(2)}(x) + \int_{0}^{\infty} p^{(3)}(x + t)e^{-t} \, \mathrm{d}t$

$= \cdots = p(x) + p^{(1)}(x) + \cdots + p^{(n)}(x) + \int_{0}^{\infty} p^{(n+1)}(x + t)e^{-t} \, \mathrm{d}t$

Since $p$ has degree $n$, $p^{(n+1)} \equiv 0$, so the final integral vanishes. Therefore:

$\int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t = \sum_{k=0}^{n} p^{(k)}(x) = q(x)$

Since $p(x + t) > 0$ for all $t \geq 0$ (as $p$ is positive everywhere) and $e^{-t} > 0$, the integrand is strictly positive, so:

$q(x) = \int_{0}^{\infty} p(x + t)e^{-t} \, \mathrm{d}t > 0 \quad \text{for all } x$

Examiner Notes

第(i)(a)部分许多考生假设首项系数为正而未论证。第(i)(b)部分常因未明确说明第(n+1)阶导数为零而失分。第(ii)(a)部分显著考生逻辑方向反了（B推A而非A推B）。第(ii)(c)部分几乎全部使用正确方法，但求和末项处理不当是主要失分点。

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Question 4

Topic: 代数与函数 (Algebra & Functions)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 (i) Show that, if $(x - \sqrt{2})^2 = 3$, then $x^4 - 10x^2 + 1 = 0$.

Deduce that, if $f(x) = x^4 - 10x^2 + 1$, then $f(\sqrt{2} + \sqrt{3}) = 0$.

(ii) Find a polynomial $g$ of degree 8 with integer coefficients such that $g(\sqrt{2} + \sqrt{3} + \sqrt{5}) = 0$. Write your answer in a form without brackets.

(iii) Let $a$, $b$ and $c$ be the three roots of $t^3 - 3t + 1 = 0$.

Find a polynomial $h$ of degree 6 with integer coefficients such that $h(a + \sqrt{2}) = 0$, $h(b + \sqrt{2}) = 0$ and $h(c + \sqrt{2}) = 0$. Write your answer in a form without brackets.

(iv) Find a polynomial $k$ with integer coefficients such that $k(\sqrt[3]{2} + \sqrt[3]{3}) = 0$. Write your answer in a form without brackets.

Hint

(i) $(x - \sqrt{2})^2 = 3 \Rightarrow x^2 - 2\sqrt{2}x - 1 = 0$ [M1] so $(x^2 - 2\sqrt{2}x - 1)(x^2 + 2\sqrt{2}x - 1) = 0$ [M1] that is, $x^4 - 10x^2 + 1 = 0$ [A1] but $\sqrt{2} + \sqrt{3}$ a root of $(x - \sqrt{2})^2 = 3$ so a root of $x^4 - 10x^2 + 1 = 0$. [A1] (ii) $(\sqrt{3} + \sqrt{5})^2 = 8 + 2\sqrt{15}$ [B1] $\sqrt{2} + \sqrt{3} + \sqrt{5}$ a root of $(x - \sqrt{2})^2 - (8 + 2\sqrt{15})$ [M1] that is, of $x^2 - 2\sqrt{2}x - 6 - 2\sqrt{15}$ [A1] so of $(x^2 - 2\sqrt{2}x - 6 - 2\sqrt{15})(x^2 + 2\sqrt{2}x - 6 + 2\sqrt{15})$ [M1] $= x^4 - 20x^2 - 8\sqrt{30}x - 24$ [A1] so of $(x^4 - 20x^2 - 8\sqrt{30}x - 24)(x^4 - 20x^2 + 8\sqrt{30}x - 24)$ [M1] $= x^8 - 40x^6 + 352x^4 - 960x^2 + 576$ [A1] The roots will be the eight numbers of the form $\pm\sqrt{2} \pm \sqrt{3} \pm \sqrt{5}$ [B1] Which can be paired as $\pm(\sqrt{2} + \sqrt{3} + \sqrt{5})$, $\pm(\sqrt{2} - \sqrt{3} - \sqrt{5})$, $\pm(-\sqrt{2} + \sqrt{3} - \sqrt{5})$, $\pm(-\sqrt{2} - \sqrt{3} + \sqrt{5})$ So the polynomial is a quartic in $x^2$ with roots $\alpha = (\sqrt{2} + \sqrt{3} + \sqrt{5})^2$, $\beta = (\sqrt{2} - \sqrt{3} - \sqrt{5})^2$, $\gamma = (-\sqrt{2} + \sqrt{3} - \sqrt{5})^2$, $\delta = (-\sqrt{2} - \sqrt{3} + \sqrt{5})^2$ $\alpha + \beta + \gamma + \delta = 4(2 + 3 + 5) = 40$ [M1] $\alpha\beta\gamma\delta = (2 - (\sqrt{3} + \sqrt{5})^2)^2 (2 - (\sqrt{3} - \sqrt{5})^2)^2$ $= ((-6 - 2\sqrt{15})(-6 + 2\sqrt{15}))^2 = (36 - 60)^2 = 576$ [A1] $\alpha^2 = (\sqrt{2} + \sqrt{3} + \sqrt{5})^4$ $= 4 + 9 + 25 + 4(5\sqrt{6} + 7\sqrt{10} + 8\sqrt{15})$ $+ 6(6 + 10 + 15)$ $= 224 + 4(5\sqrt{6} + 7\sqrt{10} + 8\sqrt{15})$ So $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 4 \times 224 = 896$ $2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta) =$ $(\alpha + \beta + \gamma + \delta)^2 - (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)$ [M1] $\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{40^2 - 896}{2} = 352$ [A1] $\alpha^3 + \beta^3 + \gamma^3 + \delta^3 =$ $4(8 + 27 + 125 + 15(12 + 20 + 18 + 45 + 50 + 75) + 90(30))$ $= 24640$ $3(\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta) = 24640 + 3(352)(40) - 40^3$ $\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = \frac{2880}{3} = 960$ Therefore the polynomial is $x^8 - 40x^6 + 352x^4 - 960x^2 + 576$ [A1] $a + \sqrt{2}$, $b + \sqrt{2}$, $c + \sqrt{2}$ are roots of $(x - \sqrt{2})^3 - 3(x - \sqrt{2}) + 1$ [M1] so of $x^3 - 3\sqrt{2}x^2 + 3x + \sqrt{2} + 1$ [A1] so of $(x^3 - 3\sqrt{2}x^2 + 3x + \sqrt{2} + 1)(x^3 + 3\sqrt{2}x^2 + 3x - \sqrt{2} + 1)$ [M1] $= x^6 - 12x^4 + 2x^3 + 21x^2 + 6x - 1$ [A1] $(\sqrt[3]{2} + \sqrt[3]{3})^3 = 5 + 3\sqrt[3]{12} + 3\sqrt[3]{18}$ [M1] $= 5 + 3\sqrt[3]{6}(\sqrt[3]{2} + \sqrt[3]{3})$ [A1] so $\sqrt[3]{2} + \sqrt[3]{3}$ satisfies $x^3 - 5 = 3\sqrt[3]{6}x$ [M1] so satisfies $(x^3 - 5)^3 = 162x^3$ [M1] or $x^9 - 15x^6 - 87x^3 - 125 = 0$ [A1]

Model Solution

Part (i)

If $(x - \sqrt{2})^2 = 3$, expanding gives $x^2 - 2\sqrt{2}\,x + 2 = 3$, so $x^2 - 1 = 2\sqrt{2}\,x$. Squaring both sides:

$(x^2 - 1)^2 = 8x^2 \implies x^4 - 2x^2 + 1 = 8x^2 \implies x^4 - 10x^2 + 1 = 0$

Since $\sqrt{2} + \sqrt{3}$ satisfies $(\sqrt{2} + \sqrt{3} - \sqrt{2})^2 = (\sqrt{3})^2 = 3$, it is a root of $x^4 - 10x^2 + 1 = 0$, i.e., $f(\sqrt{2} + \sqrt{3}) = 0$.

Part (ii)

Let $x = \sqrt{2} + \sqrt{3} + \sqrt{5}$. Then $x - \sqrt{2} = \sqrt{3} + \sqrt{5}$, so:

$(x - \sqrt{2})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}$

$x^2 - 2\sqrt{2}\,x + 2 = 8 + 2\sqrt{15}$

$x^2 - 6 - 2\sqrt{15} = 2\sqrt{2}\,x$

Squaring to eliminate $\sqrt{2}$:

$(x^2 - 6 - 2\sqrt{15})^2 = 8x^2$

Expanding the left side:

$(x^2 - 6)^2 - 2(x^2 - 6)(2\sqrt{15}) + (2\sqrt{15})^2 = x^4 - 12x^2 + 36 - 4\sqrt{15}(x^2 - 6) + 60$

$= x^4 - 12x^2 + 96 - 4\sqrt{15}(x^2 - 6)$

Setting equal to $8x^2$:

$x^4 - 20x^2 + 96 = 4\sqrt{15}(x^2 - 6)$

Squaring again to eliminate $\sqrt{15}$:

$(x^4 - 20x^2 + 96)^2 = 240(x^2 - 6)^2$

Expanding the left side:

$x^8 - 40x^6 + 400x^4 + 192x^4 - 3840x^2 + 9216 = x^8 - 40x^6 + 592x^4 - 3840x^2 + 9216$

Expanding the right side:

$240(x^4 - 12x^2 + 36) = 240x^4 - 2880x^2 + 8640$

Setting equal and collecting terms:

$x^8 - 40x^6 + 592x^4 - 3840x^2 + 9216 = 240x^4 - 2880x^2 + 8640$

$x^8 - 40x^6 + 352x^4 - 960x^2 + 576 = 0$

So $g(x) = x^8 - 40x^6 + 352x^4 - 960x^2 + 576$.

Part (iii)

If $a$ is a root of $t^3 - 3t + 1 = 0$, then $x = a + \sqrt{2}$ means $a = x - \sqrt{2}$, so:

$(x - \sqrt{2})^3 - 3(x - \sqrt{2}) + 1 = 0$

Expanding $(x - \sqrt{2})^3 = x^3 - 3\sqrt{2}\,x^2 + 6x - 2\sqrt{2}$:

$x^3 - 3\sqrt{2}\,x^2 + 6x - 2\sqrt{2} - 3x + 3\sqrt{2} + 1 = 0$

$x^3 - 3\sqrt{2}\,x^2 + 3x + \sqrt{2} + 1 = 0$

Rearranging: $x^3 + 3x + 1 = \sqrt{2}(3x^2 - 1)$. Squaring:

$(x^3 + 3x + 1)^2 = 2(3x^2 - 1)^2$

$x^6 + 6x^4 + 2x^3 + 9x^2 + 6x + 1 = 18x^4 - 12x^2 + 2$

$x^6 - 12x^4 + 2x^3 + 21x^2 + 6x - 1 = 0$

So $h(x) = x^6 - 12x^4 + 2x^3 + 21x^2 + 6x - 1$, which has $a + \sqrt{2}$, $b + \sqrt{2}$, $c + \sqrt{2}$ as three of its six roots.

Part (iv)

Let $x = \sqrt[3]{2} + \sqrt[3]{3}$. Compute $x^3$:

$x^3 = (\sqrt[3]{2} + \sqrt[3]{3})^3 = 2 + 3\sqrt[3]{4}\sqrt[3]{3} + 3\sqrt[3]{2}\sqrt[3]{9} + 3$

$= 5 + 3\sqrt[3]{12} + 3\sqrt[3]{18}$

Note that $\sqrt[3]{12} = \sqrt[3]{6}\sqrt[3]{2}$ and $\sqrt[3]{18} = \sqrt[3]{6}\sqrt[3]{3}$, so:

$x^3 = 5 + 3\sqrt[3]{6}(\sqrt[3]{2} + \sqrt[3]{3}) = 5 + 3\sqrt[3]{6} \cdot x$

Therefore $x^3 - 5 = 3\sqrt[3]{6}\,x$. Cubing both sides:

$(x^3 - 5)^3 = 27 \cdot 6 \cdot x^3 = 162x^3$

Expanding the left side:

$x^9 - 15x^6 + 75x^3 - 125 = 162x^3$

$x^9 - 15x^6 - 87x^3 - 125 = 0$

So $k(x) = x^9 - 15x^6 - 87x^3 - 125$.

Examiner Notes

第(i)部分方法多样。第(ii)部分许多考生能识别有效方法但代数错误导致最终多项式不正确。第(iii)部分多数识别出图的平移。第(iv)部分少有人尝试，但尝试者通常能成功适应前几部分的方法。

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Question 5

Topic: 数列与级数 (Sequences & Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 (i) The sequence $x_n$ for $n = 0, 1, 2, \dots$ is defined by $x_0 = 1$ and by

$x_{n+1} = \frac{x_n + 2}{x_n + 1}$

for $n \geqslant 0$.

(a) Explain briefly why $x_n \geqslant 1$ for all $n$.

(b) Show that $x_{n+1}^2 - 2$ and $x_n^2 - 2$ have opposite sign, and that

$|x_{n+1}^2 - 2| \leqslant \frac{1}{4} |x_n^2 - 2| \, .$

(c) Show that

$2 - 10^{-6} \leqslant x_{10}^2 \leqslant 2 \, .$

(ii) The sequence $y_n$ for $n = 0, 1, 2, \dots$ is defined by $y_0 = 1$ and by

$y_{n+1} = \frac{y_n^2 + 2}{2y_n}$

for $n \geqslant 0$.

(a) Show that, for $n \geqslant 0$,

$y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n}$

and deduce that $y_n \geqslant 1$ for $n \geqslant 0$.

(b) Show that

$y_n - \sqrt{2} \leqslant 2 \left( \frac{\sqrt{2} - 1}{2} \right)^{2^n}$

for $n \geqslant 1$.

(c) Using the fact that

$\sqrt{2} - 1 < \frac{1}{2} \, ,$

or otherwise, show that

$\sqrt{2} \leqslant y_{10} \leqslant \sqrt{2} + 10^{-600} \, .$

Hint

(i) (a) $x_{n+1} - 1 = \frac{1}{x_n+1}$; $x_0 \geqslant 1$ [M1] so if $x_n \geqslant 1, x_{n+1} \geqslant 1$ [A1] (i) (a) (b) $x_{n+1}^2 - 2 = \frac{(x_n + 2)^2 - 2(x_n + 1)^2}{(x_n + 1)^2} = -\frac{x_n^2 - 2}{(x_n + 1)^2}$ [M1] so $x_{n+1}^2 - 2$ and $x_n^2 - 2$ have opposite sign, as $(x_n + 1)^2 \geqslant 2^2 > 0$ [A1] $|x_{n+1}^2 - 2| \leqslant \frac{1}{4}|x_n^2 - 2|$, as $(x_n + 1)^2 \geqslant 2^2$ [A1] (i) (a) (b) (c) 10 is even, so $x_{10}^2 - 2$ and $x_0^2 - 2$ have the same sign, which is negative, so $x_{10}^2 < 2$ [B1] and $|x_{10}^2 - 2| \leqslant \frac{1}{4^{10}}|x_0^2 - 2| = \frac{1}{4^{10}}$ [M1] $< 10^{-6}$, as $2^{10} > 10^3$ [A1] so $10^{-6} \geqslant 2 - x_{10}^2$ giving stated result [A1] (ii) (a) $y_{n+1} - \sqrt{2} = \frac{y_n^2 + 2 - 2\sqrt{2}y_n}{2y_n} = \frac{(y_n - \sqrt{2})^2}{2y_n}$ [B1] so $y_0 \geqslant 1$ and $y_{n+1} \geqslant \sqrt{2} \geqslant 1$ for $n \geqslant 0$ [B1] (ii) (a) (b) $y_1 - \sqrt{2} = \frac{(1-\sqrt{2})^2}{2} = 2\left(\frac{\sqrt{2}-1}{2}\right)^2$, so result holds for $n = 1$. [B1] also $y_{n+1} - \sqrt{2} = \frac{(|y_n-\sqrt{2}|)^2}{2y_n} \leqslant \frac{2}{y_n}\left(\frac{\sqrt{2}-1}{2}\right)^{2^{n+1}}$ [M1] $\leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^{n+1}}$, as $y_n \geqslant \sqrt{2}$ for $n \geqslant 1$ [A1] appropriate induction structure [A1] (ii) (a) (b) (c) $y_{10} \geqslant \sqrt{2}$ [B1] $y_{10} - \sqrt{2} \leqslant 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^{10}}$ [M1] but $\left(\frac{\sqrt{2}-1}{2}\right)^5 \leqslant \left(\frac{1}{4}\right)^5 \leqslant 10^{-3}$ [M1] $\left(\frac{\sqrt{2}-1}{2}\right)^5 \leqslant \left(\frac{1}{4}\right)^5 \leqslant 10^{-3}$ $y_{10} - \sqrt{2} \leqslant 2(10^{-3})^{204}$ [A1] $= 2 \times 10^{-612} < 10^{-600}$ [A1]

Model Solution

Part (i)(a)

We prove $x_n \geq 1$ for all $n$ by induction. The base case $x_0 = 1 \geq 1$ holds. If $x_n \geq 1$, then:

$x_{n+1} = \frac{x_n + 2}{x_n + 1} = 1 + \frac{1}{x_n + 1} \geq 1 + 0 = 1$

since $x_n + 1 \geq 2 > 0$. By induction, $x_n \geq 1$ for all $n$.

Part (i)(b)

$x_{n+1}^2 - 2 = \frac{(x_n + 2)^2}{(x_n + 1)^2} - 2 = \frac{x_n^2 + 4x_n + 4 - 2x_n^2 - 4x_n - 2}{(x_n + 1)^2} = \frac{-(x_n^2 - 2)}{(x_n + 1)^2}$

Since $(x_n + 1)^2 > 0$, the sign of $x_{n+1}^2 - 2$ is opposite to the sign of $x_n^2 - 2$.

For the bound: since $x_n \geq 1$, we have $(x_n + 1)^2 \geq 4$, so:

$|x_{n+1}^2 - 2| = \frac{|x_n^2 - 2|}{(x_n + 1)^2} \leq \frac{|x_n^2 - 2|}{4}$

Part (i)(c)

Applying the bound from (b) repeatedly:

$|x_{10}^2 - 2| \leq \frac{1}{4^{10}} |x_0^2 - 2| = \frac{1}{4^{10}} \cdot 1 = \frac{1}{4^{10}}$

Now $4^{10} = (2^2)^{10} = 2^{20} = 1048576 > 10^6$, so $\frac{1}{4^{10}} < 10^{-6}$.

From part (b), the sign of $x_n^2 - 2$ alternates. Since $x_0^2 - 2 = -1 < 0$ and 10 is even, $x_{10}^2 - 2$ has the same sign as $x_0^2 - 2$, which is negative. Therefore $x_{10}^2 \leq 2$.

Combining: $|x_{10}^2 - 2| < 10^{-6}$ and $x_{10}^2 \leq 2$ gives $2 - x_{10}^2 < 10^{-6}$, so:

$2 - 10^{-6} \leq x_{10}^2 \leq 2$

Part (ii)(a)

$y_{n+1} - \sqrt{2} = \frac{y_n^2 + 2}{2y_n} - \sqrt{2} = \frac{y_n^2 + 2 - 2\sqrt{2}\,y_n}{2y_n} = \frac{(y_n - \sqrt{2})^2}{2y_n}$

Since $(y_n - \sqrt{2})^2 \geq 0$ and $y_n > 0$, we get $y_{n+1} \geq \sqrt{2}$ for all $n \geq 0$.

For the deduction: $y_0 = 1 \geq 1$, and for $n \geq 0$, $y_{n+1} \geq \sqrt{2} > 1$. So $y_n \geq 1$ for all $n \geq 0$.

Part (ii)(b)

We prove $y_n - \sqrt{2} \leq 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}$ for $n \geq 1$ by induction.

Base case ($n = 1$): $y_1 = \frac{1 + 2}{2} = \frac{3}{2}$, so:

$y_1 - \sqrt{2} = \frac{3}{2} - \sqrt{2} = \frac{(1 - \sqrt{2})^2}{2} = \frac{(\sqrt{2} - 1)^2}{2} = 2 \cdot \frac{(\sqrt{2}-1)^2}{4} = 2\left(\frac{\sqrt{2}-1}{2}\right)^2$

This equals $2\left(\frac{\sqrt{2}-1}{2}\right)^{2^1}$, so the base case holds with equality.

Inductive step: Assume $y_n - \sqrt{2} \leq 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}$ for some $n \geq 1$. Then:

$y_{n+1} - \sqrt{2} = \frac{(y_n - \sqrt{2})^2}{2y_n} \leq \frac{\left[2\left(\frac{\sqrt{2}-1}{2}\right)^{2^n}\right]^2}{2\sqrt{2}}$

using $y_n \geq \sqrt{2}$ (from part (a)) in the denominator. This gives:

$y_{n+1} - \sqrt{2} \leq \frac{4\left(\frac{\sqrt{2}-1}{2}\right)^{2^{n+1}}}{2\sqrt{2}} = \frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}-1}{2}\right)^{2^{n+1}} = \sqrt{2}\left(\frac{\sqrt{2}-1}{2}\right)^{2^{n+1}}$

Since $\sqrt{2} < 2$:

$y_{n+1} - \sqrt{2} \leq 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^{n+1}}$

By induction, the result holds for all $n \geq 1$.

Part (ii)(c)

From part (a), $y_{10} \geq \sqrt{2}$.

From part (b): $y_{10} - \sqrt{2} \leq 2\left(\frac{\sqrt{2}-1}{2}\right)^{2^{10}} = 2\left(\frac{\sqrt{2}-1}{2}\right)^{1024}$.

Since $\sqrt{2} - 1 < \frac{1}{2}$, we have $\frac{\sqrt{2}-1}{2} < \frac{1}{4}$. Now:

$\left(\frac{\sqrt{2}-1}{2}\right)^5 < \left(\frac{1}{4}\right)^5 = \frac{1}{1024} < 10^{-3}$

Since $1024 = 5 \times 204 + 4$:

$\left(\frac{\sqrt{2}-1}{2}\right)^{1024} = \left[\left(\frac{\sqrt{2}-1}{2}\right)^5\right]^{204} \cdot \left(\frac{\sqrt{2}-1}{2}\right)^4 < (10^{-3})^{204} \cdot 1 = 10^{-612}$

Therefore:

$y_{10} - \sqrt{2} \leq 2 \times 10^{-612} < 10^{-600}$

giving $\sqrt{2} \leq y_{10} \leq \sqrt{2} + 10^{-600}$.

Examiner Notes

第(i)(a)部分多数认识到需用归纳法，但简要说明仍需包含必要要素；常见错误是声称序列单调递增。第(i)(b)部分主要失分点：未用绝对值符号，写出对奇数n不成立的不等式。第(i)(c)部分未论证4的10次方大于10的6次方。第(ii)(b)部分许多考生以n=0为基底（不合法），递推方法中2的n次方指数难以论证。

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Question 6

Topic: 矩阵与线性代数 (Matrices & Linear Algebra)  |  Difficulty: Hard  |  Marks: 20

Problem

6 The sequence $F_n$, for $n = 0, 1, 2, \dots$, is defined by $F_0 = 0$, $F_1 = 1$ and by $F_{n+2} = F_{n+1} + F_n$ for $n \geqslant 0$.

Prove by induction that, for all positive integers $n$,

$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \mathbf{Q}^n,$

where the matrix $\mathbf{Q}$ is given by

$\mathbf{Q} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}.$

(i) By considering the matrix $\mathbf{Q}^n$, show that $F_{n+1}F_{n-1} - F_n^2 = (-1)^n$ for all positive integers $n$.

(ii) By considering the matrix $\mathbf{Q}^{m+n}$, show that $F_{m+n} = F_{m+1}F_n + F_mF_{n-1}$ for all positive integers $m$ and $n$.

(iii) Show that $\mathbf{Q}^2 = \mathbf{I} + \mathbf{Q}$.

In the following parts, you may use without proof the Binomial Theorem for matrices:

$(\mathbf{I} + \mathbf{A})^n = \sum_{k=0}^n \binom{n}{k} \mathbf{A}^k.$

(a) Show that, for all positive integers $n$,

$F_{2n} = \sum_{k=0}^n \binom{n}{k} F_k.$

(b) Show that, for all positive integers $n$,

$F_{3n} = \sum_{k=0}^n \binom{n}{k} 2^k F_k$

and also that

$F_{3n} = \sum_{k=0}^n \binom{n}{k} F_{n+k}.$

(c) Show that, for all positive integers $n$,

$\sum_{k=0}^n (-1)^{n+k} \binom{n}{k} F_{n+k} = 0.$

Hint

Induction structure [M1] Base case [B1] $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \begin{pmatrix} F_{n+1} + F_n & F_n + F_{n-1} \\ F_{n+1} & F_n \end{pmatrix}$ or $\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_{n+1} + F_n & F_{n+1} \\ F_n + F_{n-1} & F_n \end{pmatrix}$ [A1] Use of definition (of $F_n$) and conclusion (of induction) [A1] Use of $\det(Q^n) = (\det Q)^n$ [M1] clearly shown [A1] Use of (1,2) entry in $Q^{m+n} = Q^m Q^n$ [M1] clearly shown [A1] $Q^2 = Q + I$ [B1] Use of $Q^{2n} = (Q + I)^n$ [M1] and Binomial expansion [M1] clearly shown [A1] Derivation of $Q^3 = Q(Q + I) = 2Q + I$ (give the mark for any one of these) [B1] Use of $Q^{3n} = (2Q + I)^n$ and Binomial expansion [M1] clearly shown [A1] Use of $Q^{3n} = Q^n(Q + I)^n$ and Binomial expansion [M1] clearly shown [A1] Use of $I = Q^n(Q - I)^n$ ( or $(-Q)^n (I - Q)^n$ ) [M1] Use of binomial expansion [M1] clearly shown [A1]

Model Solution

Induction proof that $\mathbf{Q}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$

Base case ($n = 1$): $\mathbf{Q}^1 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_2 & F_1 \\ F_1 & F_0 \end{pmatrix}$. ✓

Inductive step: Assume $\mathbf{Q}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$. Then:

$\mathbf{Q}^{n+1} = \mathbf{Q} \cdot \mathbf{Q}^n = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \begin{pmatrix} F_{n+1} + F_n & F_n + F_{n-1} \\ F_{n+1} & F_n \end{pmatrix} = \begin{pmatrix} F_{n+2} & F_{n+1} \\ F_{n+1} & F_n \end{pmatrix}$

using the Fibonacci recurrence. This completes the induction.

Part (i)

Taking determinants of both sides of $\mathbf{Q}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$:

$\det(\mathbf{Q}^n) = (\det \mathbf{Q})^n = (0 - 1)^n = (-1)^n$

$\det(\mathbf{Q}^n) = F_{n+1}F_{n-1} - F_n^2$

Therefore $F_{n+1}F_{n-1} - F_n^2 = (-1)^n$ for all positive integers $n$.

Part (ii)

Since $\mathbf{Q}^{m+n} = \mathbf{Q}^m \cdot \mathbf{Q}^n$, comparing the $(1,2)$ entry:

$F_{m+n} = F_{m+1} \cdot F_n + F_m \cdot F_{n-1}$

using the standard row-times-column multiplication.

Part (iii)

$\mathbf{Q}^2 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \mathbf{I} + \mathbf{Q}$

Part (iii)(a)

From $\mathbf{Q}^2 = \mathbf{I} + \mathbf{Q}$:

$\mathbf{Q}^{2n} = (\mathbf{Q}^2)^n = (\mathbf{I} + \mathbf{Q})^n = \sum_{k=0}^{n} \binom{n}{k} \mathbf{Q}^k$

Comparing the $(1,2)$ entry on both sides:

$F_{2n} = \sum_{k=0}^{n} \binom{n}{k} F_k$

Part (iii)(b)

First, $\mathbf{Q}^3 = \mathbf{Q} \cdot \mathbf{Q}^2 = \mathbf{Q}(\mathbf{I} + \mathbf{Q}) = \mathbf{Q} + \mathbf{Q}^2 = \mathbf{Q} + \mathbf{I} + \mathbf{Q} = \mathbf{I} + 2\mathbf{Q}$.

Therefore:

$\mathbf{Q}^{3n} = (\mathbf{Q}^3)^n = (\mathbf{I} + 2\mathbf{Q})^n = \sum_{k=0}^{n} \binom{n}{k} 2^k \mathbf{Q}^k$

Comparing the $(1,2)$ entry: $F_{3n} = \sum_{k=0}^{n} \binom{n}{k} 2^k F_k$.

For the second identity, write:

$\mathbf{Q}^{3n} = \mathbf{Q}^{2n} \cdot \mathbf{Q}^n = (\mathbf{I} + \mathbf{Q})^n \cdot \mathbf{Q}^n = \left[\sum_{k=0}^{n} \binom{n}{k} \mathbf{Q}^k\right] \mathbf{Q}^n = \sum_{k=0}^{n} \binom{n}{k} \mathbf{Q}^{n+k}$

Comparing the $(1,2)$ entry: $F_{3n} = \sum_{k=0}^{n} \binom{n}{k} F_{n+k}$.

Part (iii)(c)

From $\mathbf{Q}^2 = \mathbf{I} + \mathbf{Q}$ we get $\mathbf{I} = \mathbf{Q}^2 - \mathbf{Q} = \mathbf{Q}(\mathbf{Q} - \mathbf{I})$.

Taking $n$-th powers: $\mathbf{I} = \mathbf{Q}^n(\mathbf{Q} - \mathbf{I})^n$.

Since $\mathbf{Q}$ and $\mathbf{I}$ commute, the Binomial Theorem gives:

$(\mathbf{Q} - \mathbf{I})^n = \sum_{k=0}^{n} \binom{n}{k} \mathbf{Q}^k (-\mathbf{I})^{n-k} = \sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} \mathbf{Q}^k$

Therefore:

$\mathbf{I} = \mathbf{Q}^n \sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} \mathbf{Q}^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} \mathbf{Q}^{n+k}$

Comparing the $(1,2)$ entry (which is $0$ in $\mathbf{I}$):

$0 = \sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} F_{n+k}$

Since $(-1)^{n-k} = (-1)^{n+k}$ (as $(-1)^{-2k} = 1$):

$\sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} F_{n+k} = 0$

Examiner Notes

多数考生能完成归纳证明基础，但部分矩阵乘法计算错误（AB而非BA）。第(i)(ii)部分多数能利用矩阵得出答案，少数推理不充分。第(iii)部分多数能证Q^2=I+Q，但此后进展甚微，仅少数优秀解法能仔细处理所有情况。

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Question 7

Topic: 复数 (Complex Numbers)  |  Difficulty: Standard  |  Marks: 20

Problem

7 (i) The complex numbers $z$ and $w$ have real and imaginary parts given by $z = a + ib$ and $w = c + id$. Prove that $|zw| = |z||w|$.

(ii) By considering the complex numbers $2 + i$ and $10 + 11i$, find positive integers $h$ and $k$ such that $h^2 + k^2 = 5 \times 221$.

(iii) Find positive integers $m$ and $n$ such that $m^2 + n^2 = 8045$.

(iv) You are given that $102^2 + 201^2 = 50805$. Find positive integers $p$ and $q$ such that $p^2 + q^2 = 36 \times 50805$.

(v) Find three distinct pairs of positive integers $r$ and $s$ such that $r^2 + s^2 = 25 \times 1002082$ and $r < s$.

(vi) You are given that $109 \times 9193 = 1002037$. Find positive integers $t$ and $u$ such that $t^2 + u^2 = 9193$.

Hint

(i) |zw|² = |(ac - bd) + i(ad + bc)|² [M1] = (ac - bd)² + (ad + bc)² = a²c² + b²d² + a²d² + b²c² |z|²|w|² = (a² + b²)(c² + d²) [M1] = a²c² + b²d² + a²d² + b²c² Therefore |zw|² = |z|²|w|² [A1] |2 + i| = √5 and |10 + 11i| = √221 [B1] so 9² + 32² = (2² + 1²)(10² + 11²) = 5 × 221 [B1] 8045 = 5 × 1609 [M1] = (2² + 1²)(40² + 3²) [M1] so |(2 + i)(40 + 3i)|² = 77² + 46² = 8045 (also 34² + 83²) [A1] 612² + 1206² = 6² × 50805 [B1] 1002082 = 1001² + 9² [B1] so one pair is 5005² + 45² [A1] but 25 = 3² + 4² and (3 + 4i)(1001 + 9i) = 2967 + 4031i [M1] so a second pair is 2967² + 4031² [A1] also, (4 + 3i)(1001 + 9i) = 3977 + 3039i [M1] so a third pair is 3977² + 3039² [A1] require (10² + 3²)(c² + d²) = (1001² + 6²) [M1] implies simultaneous equations for c and d [M1] 10c - 3d = 1001, 10d + 3c = 6 or 3c - 10d = 1001, 3d + 10c = 6 [A1] giving c = 92, d = -27 (from the first set) [A1] so 9193 = 92² + 27² (also 38² + 211²) [A1]

Model Solution

Part (i)

$z = a + ib$, $w = c + id$, so $zw = (ac - bd) + i(ad + bc)$. Then:

$|zw|^2 = (ac - bd)^2 + (ad + bc)^2 = a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2 = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$

$|z|^2|w|^2 = (a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$

Since $|zw|^2 = |z|^2|w|^2$ and all quantities are non-negative, $|zw| = |z||w|$.

Part (ii)

$|2 + i|^2 = 4 + 1 = 5$ and $|10 + 11i|^2 = 100 + 121 = 221$. By part (i):

$(2 + i)(10 + 11i) = 20 + 22i + 10i + 11i^2 = 9 + 32i$

So $|(2+i)(10+11i)|^2 = 9^2 + 32^2 = 81 + 1024 = 1105 = 5 \times 221$.

Therefore $h = 9$, $k = 32$.

Part (iii)

$8045 = 5 \times 1609$. We need to express 1609 as a sum of two squares: $1609 = 40^2 + 3^2 = 1600 + 9$. ✓

Using complex multiplication:

$(2 + i)(40 + 3i) = 80 + 6i + 40i + 3i^2 = 77 + 46i$

So $77^2 + 46^2 = 5929 + 2116 = 8045$. Therefore $m = 46$, $n = 77$.

(Also $(2 + i)(40 - 3i) = 83 + 34i$, giving $83^2 + 34^2 = 8045$.)

Part (iv)

$36 = 6^2$, so $36 \times 50805 = (6 \times 102)^2 + (6 \times 201)^2 = 612^2 + 1206^2$.

Check: $612^2 + 1206^2 = 374544 + 1454436 = 1828980 = 36 \times 50805$. ✓

Therefore $p = 612$, $q = 1206$.

Part (v)

First, $1002082 = 1001^2 + 9^2 = 1002001 + 81$. ✓

First pair: $25 = 5^2$, so $25 \times 1002082 = (5 \times 1001)^2 + (5 \times 9)^2 = 5005^2 + 45^2$.

So $(r, s) = (45, 5005)$.

Second pair: $25 = 3^2 + 4^2$, so we use $(3 + 4i)(1001 + 9i)$:

$(3 + 4i)(1001 + 9i) = 3003 + 27i + 4004i + 36i^2 = 2967 + 4031i$

So $2967^2 + 4031^2 = 25 \times 1002082$. Thus $(r, s) = (2967, 4031)$.

Third pair: $(4 + 3i)(1001 + 9i)$:

$(4 + 3i)(1001 + 9i) = 4004 + 36i + 3003i + 27i^2 = 3977 + 3039i$

So $3977^2 + 3039^2 = 25 \times 1002082$. Thus $(r, s) = (3039, 3977)$.

The three distinct pairs are $(45, 5005)$, $(2967, 4031)$, $(3039, 3977)$.

Part (vi)

We know $109 = 10^2 + 3^2$ and $1002037 = 1001^2 + 6^2$ (since $1002001 + 36 = 1002037$). Since $109 \times 9193 = 1002037$, we seek $c, d$ with $c^2 + d^2 = 9193$ using:

$(10 + 3i)(c + di) = (10c - 3d) + (10d + 3c)i$

We need this to equal $\pm(1001 + 6i)$ or $\pm(1001 - 6i)$ (or with swapped real/imaginary parts). Taking $(10c - 3d) + (10d + 3c)i = 1001 + 6i$:

$10c - 3d = 1001 \qquad \text{and} \qquad 10d + 3c = 6$

From the second equation: $d = \frac{6 - 3c}{10}$. Substituting into the first:

$10c - 3 \cdot \frac{6 - 3c}{10} = 1001 \implies 100c - 18 + 9c = 10010 \implies 109c = 10028 \implies c = 92$

Then $d = \frac{6 - 276}{10} = -27$. So $9193 = 92^2 + 27^2 = 8464 + 729$. ✓

Therefore $t = 27$, $u = 92$.

Examiner Notes

前两部分多数成功，偶有忘记取平方根的小错误。第(iii)部分多数能想到除以5。第(v)(vi)部分区分度最高：许多考生未能发现1001^2+9^2和1001^2+6^2的分解；第(v)部分遗漏(3,4,5) Pythagorean三元组或25=5^2的简单分解；第(vi)部分未注意到10028或2943可被109整除。

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Question 8

Topic: 几何与向量 (Geometry & Vectors)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 A tetrahedron is called isosceles if each pair of edges which do not share a vertex have equal length.

(i) Prove that a tetrahedron is isosceles if and only if all four faces have the same perimeter.

Let $OABC$ be an isosceles tetrahedron and let $\vec{OA} = \mathbf{a}$, $\vec{OB} = \mathbf{b}$ and $\vec{OC} = \mathbf{c}$.

(ii) By considering the lengths of $OA$ and $BC$, show that

$2\mathbf{b.c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2.$

Show that

$\mathbf{a.(b + c)} = |\mathbf{a}|^2.$

(iii) Let $G$ be the centroid of the tetrahedron, defined by $\vec{OG} = \frac{1}{4}(\mathbf{a} + \mathbf{b} + \mathbf{c})$.

Show that $G$ is equidistant from all four vertices of the tetrahedron.

(iv) By considering the length of the vector $\mathbf{a - b - c}$, or otherwise, show that, in an isosceles tetrahedron, none of the angles between pairs of edges which share a vertex can be obtuse. Can any of them be right angles?

Hint

(i) Let vertices be numbered 1 to 4 and edges be $e_{ij}$, where $i < j$. Then perimeters equal is $|e_{12}| + |e_{23}| + |e_{13}| = |e_{12}| + |e_{24}| + |e_{14}|$ $= |e_{13}| + |e_{34}| + |e_{14}|$ $= |e_{24}| + |e_{34}| + |e_{23}|$ [M1] which implies $|e_{12}| + |e_{23}| + |e_{13}| + |e_{12}| + |e_{24}| + |e_{14}|$ $= |e_{13}| + |e_{34}| + |e_{14}| + |e_{24}| + |e_{34}| + |e_{23}|$ so $2|e_{12}| + (|e_{23}| + |e_{13}| + |e_{24}| + |e_{14}|)$ $= 2|e_{34}| + (|e_{13}| + |e_{14}| + |e_{24}| + |e_{23}|)$ and so $|e_{12}| = |e_{34}|$ [A1] and by permutations of this argument, all pairs of opposite sides are equal. [E1] if $|e_{12}| = |e_{34}|, |e_{13}| = |e_{24}|, |e_{14}| = |e_{23}|$, then all perimeters are trivially equal [B1] (ii) $|\mathbf{a}|^2 = |\mathbf{b} - \mathbf{c}|^2$ [M1] $= |\mathbf{b}|^2 + |\mathbf{c}|^2 - 2\mathbf{b}.\mathbf{c}$ [A1] From the equivalent results to (ii) using the other pairs of opposite sides [M1] $\mathbf{a}.\mathbf{b} + \mathbf{a}.\mathbf{c} = \frac{1}{2}(|\mathbf{a}|^2 + |\mathbf{b}|^2 - |\mathbf{c}|^2) + \frac{1}{2}(|\mathbf{a}|^2 + |\mathbf{c}|^2 - |\mathbf{b}|^2) = |\mathbf{a}|^2$ [A1] (iii) $16|\mathbf{a} - \mathbf{g}|^2 = |3\mathbf{a} - \mathbf{b} - \mathbf{c}|^2$ [M1] $= 9|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 6\mathbf{a}.(\mathbf{b} + \mathbf{c}) + 2\mathbf{b}.\mathbf{c}$ $= 9|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 6|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2$ using previous results [M1] $= 2(|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2)$ [A1] but this is symmetric in a, b, c so g equidistant from A, B and C. [A1] $16|\mathbf{g}|^2 = |\mathbf{a} + \mathbf{b} + \mathbf{c}|^2$ $= |\mathbf{a}|^2 + 2\mathbf{a}.(\mathbf{b} + \mathbf{c}) + |\mathbf{b} + \mathbf{c}|^2$ $= |\mathbf{a}|^2 + 2|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2\mathbf{b}.\mathbf{c}$ $= 3|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2$ $= 2(|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2)$ So G equidistant from O also. [B1] (iv) $|\mathbf{a} - \mathbf{b} - \mathbf{c}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 2\mathbf{a}.(\mathbf{b} + \mathbf{c}) + 2\mathbf{b}.\mathbf{c}$ [M1] $= |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 2|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2$ [M1] $= 2(|\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2)$ [A1] which must be non-negative, so cos(BAC) $\geqslant$ 0 [M1] and symmetry implies no angle obtuse [A1] If e.g. BAC was a right angle, would have $|\mathbf{a} - \mathbf{b} - \mathbf{c}|^2 = 0$, so $\mathbf{a} = \mathbf{b} + \mathbf{c}$ [M1] so O, A, B, C all in one plane, so not a tetrahedron. [A1]

Model Solution

Part (i)

Label the vertices $1, 2, 3, 4$ and let $e_{ij}$ denote the edge between vertices $i$ and $j$ (with $i < j$). There are 6 edges: $e_{12}, e_{13}, e_{14}, e_{23}, e_{24}, e_{34}$.

If direction: Suppose the tetrahedron is isosceles, so opposite edges are equal: $|e_{12}| = |e_{34}|$, $|e_{13}| = |e_{24}|$, $|e_{14}| = |e_{23}|$. The four face perimeters are:

• Face 123: $|e_{12}| + |e_{13}| + |e_{23}|$
• Face 124: $|e_{12}| + |e_{14}| + |e_{24}|$
• Face 134: $|e_{13}| + |e_{14}| + |e_{34}|$
• Face 234: $|e_{23}| + |e_{24}| + |e_{34}|$

Substituting the equalities, each perimeter equals $|e_{12}| + |e_{13}| + |e_{14}|$. So all four faces have the same perimeter.

Only if direction: Suppose all four perimeters are equal. Adding the perimeters of faces 123 and 124:

$|e_{12}| + |e_{13}| + |e_{23}| + |e_{12}| + |e_{14}| + |e_{24}| = 2|e_{12}| + |e_{13}| + |e_{14}| + |e_{23}| + |e_{24}|$

Adding the perimeters of faces 134 and 234:

$|e_{13}| + |e_{14}| + |e_{34}| + |e_{23}| + |e_{24}| + |e_{34}| = 2|e_{34}| + |e_{13}| + |e_{14}| + |e_{23}| + |e_{24}|$

Since all four perimeters are equal, these two sums are equal, so $2|e_{12}| = 2|e_{34}|$, giving $|e_{12}| = |e_{34}|$.

By the same argument applied to the other pairs of faces (e.g., adding faces 123 and 134 vs. faces 124 and 234), we get $|e_{13}| = |e_{24}|$ and $|e_{14}| = |e_{23}|$. So the tetrahedron is isosceles.

Part (ii)

Let $\vec{OA} = \mathbf{a}$, $\vec{OB} = \mathbf{b}$, $\vec{OC} = \mathbf{c}$. The six edge lengths are:

$OA = |\mathbf{a}|, \quad OB = |\mathbf{b}|, \quad OC = |\mathbf{c}|, \quad BC = |\mathbf{b} - \mathbf{c}|, \quad CA = |\mathbf{c} - \mathbf{a}|, \quad AB = |\mathbf{a} - \mathbf{b}|$

The pairs of opposite edges are: $(OA, BC)$, $(OB, CA)$, $(OC, AB)$.

Since $OA = BC$: $|\mathbf{a}|^2 = |\mathbf{b} - \mathbf{c}|^2 = |\mathbf{b}|^2 - 2\mathbf{b \cdot c} + |\mathbf{c}|^2$, so:

$2\mathbf{b \cdot c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2 \qquad \text{...(★)}$

By symmetry (using $OB = CA$ and $OC = AB$):

$2\mathbf{a \cdot c} = |\mathbf{a}|^2 + |\mathbf{c}|^2 - |\mathbf{b}|^2 \qquad \text{...(★★)}$ $2\mathbf{a \cdot b} = |\mathbf{a}|^2 + |\mathbf{b}|^2 - |\mathbf{c}|^2 \qquad \text{...(★★★)}$

Adding (★★) and (★★★):

$2\mathbf{a \cdot c} + 2\mathbf{a \cdot b} = 2|\mathbf{a}|^2 + |\mathbf{c}|^2 - |\mathbf{b}|^2 + |\mathbf{b}|^2 - |\mathbf{c}|^2 = 2|\mathbf{a}|^2$

So $\mathbf{a \cdot (b + c)} = |\mathbf{a}|^2$.

Part (iii)

The centroid is $G$ with $\vec{OG} = \mathbf{g} = \frac{1}{4}(\mathbf{a} + \mathbf{b} + \mathbf{c})$.

Compute $|GA|^2$: $\vec{GA} = \mathbf{a} - \mathbf{g} = \frac{1}{4}(3\mathbf{a} - \mathbf{b} - \mathbf{c})$, so:

$16|GA|^2 = |3\mathbf{a} - \mathbf{b} - \mathbf{c}|^2 = 9|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 6\mathbf{a \cdot b} - 6\mathbf{a \cdot c} + 2\mathbf{b \cdot c}$

Using the results from (ii): $\mathbf{a \cdot (b + c)} = |\mathbf{a}|^2$ and $2\mathbf{b \cdot c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2$:

$16|GA|^2 = 9|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 6|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2 = 2(|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2)$

This expression is symmetric in $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, so $|GA|^2 = |GB|^2 = |GC|^2$.

For $|GO|^2$: $\vec{GO} = -\mathbf{g} = -\frac{1}{4}(\mathbf{a} + \mathbf{b} + \mathbf{c})$, so:

$16|GO|^2 = |\mathbf{a} + \mathbf{b} + \mathbf{c}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2\mathbf{a \cdot b} + 2\mathbf{a \cdot c} + 2\mathbf{b \cdot c}$

Using $\mathbf{a \cdot (b + c)} = |\mathbf{a}|^2$ and $2\mathbf{b \cdot c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2$:

$= |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2 = 2(|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2)$

So $|GO|^2 = |GA|^2 = |GB|^2 = |GC|^2$, meaning $G$ is equidistant from all four vertices.

Part (iv)

Consider $|\mathbf{a} - \mathbf{b} - \mathbf{c}|^2$:

$|\mathbf{a} - \mathbf{b} - \mathbf{c}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 2\mathbf{a \cdot b} - 2\mathbf{a \cdot c} + 2\mathbf{b \cdot c}$

Using the results from (ii):

$= |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - 2|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2 = 2(|\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2)$

Since $|\mathbf{a} - \mathbf{b} - \mathbf{c}|^2 \geq 0$:

$|\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2 \geq 0 \qquad \text{...(†)}$

The angle at vertex $O$ between edges $OB$ and $OC$ satisfies $\cos \angle BOC = \frac{\mathbf{b \cdot c}}{|\mathbf{b}||\mathbf{c}|}$. From (★):

$2\mathbf{b \cdot c} = |\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2 \geq 0$

So $\mathbf{b \cdot c} \geq 0$, meaning $\angle BOC \leq 90°$. By symmetry (using the analogous results for the other pairs of opposite edges), all angles at every vertex are at most $90°$.

Similarly, the angle at $A$ between edges $AB$ and $AC$: the vectors along these edges from $A$ are $\mathbf{b} - \mathbf{a}$ and $\mathbf{c} - \mathbf{a}$. Their dot product is:

$(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{c} - \mathbf{a}) = \mathbf{b \cdot c} - \mathbf{a \cdot b} - \mathbf{a \cdot c} + |\mathbf{a}|^2$

Using the results from (ii):

$= \frac{|\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2}{2} - |\mathbf{a}|^2 + |\mathbf{a}|^2 = \frac{|\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2}{2} \geq 0$

So $\angle BAC \leq 90°$. By symmetry, all angles at every vertex are at most $90°$.

Can any angle be a right angle? If $\angle BOC = 90°$, then $\mathbf{b \cdot c} = 0$, so from (★): $|\mathbf{a}|^2 = |\mathbf{b}|^2 + |\mathbf{c}|^2$. Then $|\mathbf{a} - \mathbf{b} - \mathbf{c}|^2 = 2(|\mathbf{b}|^2 + |\mathbf{c}|^2 - |\mathbf{a}|^2) = 0$, so $\mathbf{a} = \mathbf{b} + \mathbf{c}$. This means $\vec{OA} = \vec{OB} + \vec{OC}$, so $O$, $A$, $B$, $C$ are coplanar (the parallelogram law). A coplanar quadrilateral is not a valid tetrahedron (zero volume). So no angle can be a right angle in a non-degenerate isosceles tetrahedron.

Examiner Notes

第(i)部分多数完成仅当方向，但许多考生未意识到需单独证当方向。第(ii)部分总体良好，但有考生混淆方向向量和长度。第(iii)部分常见错误：计算a点积g而非|AG|，或以非对称形式得出结果后试图诉诸对称性（不被接受）。第(iv)部分对多数考生较难，少有人成功使用余弦论证。