STEP3 2022 -- Pure Mathematics

STEP3 2022 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2022  |  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	纯数	Standard	参数方程代入消元, 韦达定理, 对称多项式化简, 将四次方程倒数变换
2	纯数	Challenging	无穷递降法, 反证法, 模3/模2同余分析, 奇偶性分析, 平方剩余
3	纯数	Challenging	隐函数微分, 驻点条件推导, 四次方程根的分析, 联立方程消元
4	纯数	Challenging	数学归纳法, 麦克劳林展开求极限, 虚数代换, 几何级数指数化简
5	纯数	Standard	换元积分(u=-x), 奇偶函数分解, 微积分基本定理, 函数方程推理
6	纯数	Hard	三角函数小量展开, 参数方程构建, 周期性条件处理, 导数的几何解释
7	纯数	Challenging	向量叉积计算, 单位向量性质利用, Rodrigues旋转公式, 几何直觉与图示
8	纯数	Challenging	De Moivre定理, 二项式展开, sec²θ-1=tan²θ代换, 奇偶性分析, 反证法

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Question 1

Topic: 纯数  |  Difficulty: Standard  |  Marks: 20

Problem

1 Let $C_1$ be the curve given by the parametric equations

$x = ct, \quad y = \frac{c}{t}$

where $c > 0$ and $t \neq 0$, and let $C_2$ be the circle

$(x - a)^2 + (y - b)^2 = r^2 .$

$C_1$ and $C_2$ intersect at the four points $P_i$ ($i = 1, 2, 3, 4$), and the corresponding values of the parameter $t$ at these points are $t_i$.

(i) Show that $t_i$ are the roots of the equation

$c^2t^4 - 2act^3 + (a^2 + b^2 - r^2)t^2 - 2bct + c^2 = 0 . \quad (*)$

(ii) Show that

$\sum_{i=1}^4 t_i^2 = \frac{2}{c^2} (a^2 - b^2 + r^2)$

(iii) Show that

the center of the circle $C_2$ lies on the line $y = x$ if and only if $t_1t_2t_3t_4 = 1$ and $t_1 + t_2 + t_3 + t_4 = \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \frac{1}{t_4}$.

Hint

解题思路

Part (i)

交点：$(ct-a)^2+(\frac{c}{t}-b)^2=r^2$。 展开乘 $t^2$：$c^2t^4-2act^3+(a^2+b^2-r^2)t^2-2bct+c^2=0$。

Part (ii)

Vieta：$\sum t_i=\frac{2a}{c}$，$\sum_{i<j}t_it_j=\frac{a^2+b^2-r^2}{c^2}$。 $\sum t_i^2=(\frac{2a}{c})^2-2\cdot\frac{a^2+b^2-r^2}{c^2}=\frac{2}{c^2}(a^2-b^2+r^2)$。 交换 $a\leftrightarrow b$：$\sum\frac{1}{t_i^2}=\frac{2}{c^2}(b^2-a^2+r^2)$。

Part (iii)

$\sum OP_i^2=\sum(c^2t_i^2+\frac{c^2}{t_i^2})=c^2(\frac{2}{c^2}(a^2-b^2+r^2)+\frac{2}{c^2}(b^2-a^2+r^2))=4r^2$。

Part (iv)

两切点 $\implies$ 两对重根 $t_1=t_3,t_2=t_4$。 $t_1^2t_2^2=1 \implies t_1t_2=\pm 1$。 Vieta：$t_1+t_2=\frac{a}{c}$，$t_1t_2(t_1+t_2)=\frac{b}{c} \implies \frac{b}{a}=\pm 1 \implies a=\pm b$。 圆心 $(a,\pm a)$ 在 $y=\pm x$ 上。

Model Solution

Part (i)

At an intersection point of $C_1$ and $C_2$, the parametric coordinates satisfy the circle equation:

$(ct - a)^2 + \left(\frac{c}{t} - b\right)^2 = r^2$

Expanding each squared term:

$c^2t^2 - 2act + a^2 + \frac{c^2}{t^2} - \frac{2bc}{t} + b^2 = r^2$

Multiplying both sides by $t^2$ (valid since $t \neq 0$):

$c^2t^4 - 2act^3 + a^2t^2 + c^2 - 2bct + b^2t^2 = r^2t^2$

Collecting terms by powers of $t$:

$c^2t^4 - 2act^3 + (a^2 + b^2 - r^2)t^2 - 2bct + c^2 = 0 \qquad (*)$

Since the four intersection points $P_i$ correspond to parameter values $t_i$, each $t_i$ satisfies this equation, so the $t_i$ are roots of $(*)$.

Part (ii)

By Vieta’s formulas applied to the quartic $(*)$ with leading coefficient $c^2$:

$\sum_{i=1}^4 t_i = \frac{2a}{c}, \qquad \sum_{1 \leqslant i < j \leqslant 4} t_it_j = \frac{a^2 + b^2 - r^2}{c^2}$

Using the identity $\sum t_i^2 = \left(\sum t_i\right)^2 - 2\sum_{i<j} t_it_j$:

$\sum_{i=1}^4 t_i^2 = \left(\frac{2a}{c}\right)^2 - 2 \cdot \frac{a^2 + b^2 - r^2}{c^2} = \frac{4a^2}{c^2} - \frac{2(a^2 + b^2 - r^2)}{c^2}$

$= \frac{4a^2 - 2a^2 - 2b^2 + 2r^2}{c^2} = \frac{2a^2 - 2b^2 + 2r^2}{c^2} = \frac{2}{c^2}(a^2 - b^2 + r^2)$

Part (iii)

We need to show that $a = b$ if and only if both conditions hold.

Direction 1: Suppose $a = b$. Then the center of $C_2$ lies on $y = x$.

By Vieta’s formulas, the product of all four roots of $(*)$ is:

$t_1t_2t_3t_4 = \frac{c^2}{c^2} = 1$

The sum of the reciprocals is:

$\sum_{i=1}^4 \frac{1}{t_i} = \frac{\sum_{i<j<k} t_it_jt_k}{t_1t_2t_3t_4} = \frac{S_3}{S_4}$

By Vieta’s formulas, $S_3 = \frac{2bc}{c^2} = \frac{2b}{c}$ and $S_4 = 1$, so:

$\sum_{i=1}^4 \frac{1}{t_i} = \frac{2b}{c}$

Since $a = b$, we have $\sum t_i = \frac{2a}{c} = \frac{2b}{c} = \sum \frac{1}{t_i}$. Both conditions are satisfied.

Direction 2: Suppose $t_1t_2t_3t_4 = 1$ and $\sum t_i = \sum \frac{1}{t_i}$. We show $a = b$.

From Vieta’s formulas:

$t_1t_2t_3t_4 = \frac{c^2}{c^2} = 1$

This is automatically true (the constant term and leading coefficient of $(*)$ are both $c^2$), so this condition holds regardless of $a$ and $b$.

The condition $\sum t_i = \sum \frac{1}{t_i}$ gives:

$\frac{2a}{c} = \frac{2b}{c}$

Therefore $a = b$, so the center $(a, b)$ of $C_2$ lies on $y = x$.

Examiner Notes

最受欢迎题目(94%考生选择)，平均分约14/20。第(i)部分几乎全员正确。第(ii)部分求和符号书写不规范是主要扣分点，更高效的方法是将四次方程除以t⁴后与原方程比较。部分考生从(iii)反推，未获认可。第(iii)部分需注意利用(ii)的结果而非独立推导。

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Question 2

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

2 (i) Suppose that there are three non-zero integers $a$, $b$ and $c$ for which $a^3 + 2b^3 + 4c^3 = 0$. Explain why there must exist an integer $p$, with $|p| < |a|$, such that $4p^3 + b^3 + 2c^3 = 0$, and show further that there must exist integers $p$, $q$ and $r$, with $|p| < |a|$, $|q| < |b|$ and $|r| < |c|$, such that $p^3 + 2q^3 + 4r^3 = 0$. Deduce that no such integers $a$, $b$ and $c$ can exist.

(ii) Prove that there are no non-zero integers $a$, $b$ and $c$ for which $9a^3 + 10b^3 + 6c^3 = 0$.

(iii) By considering the expression $(3n \pm 1)^2$, prove that, unless an integer is a multiple of three, its square is one more than a multiple of 3. Deduce that the sum of the squares of two integers can only be a multiple of three if each of the integers is a multiple of three.

Hence prove that there are no non-zero integers $a$, $b$ and $c$ for which $a^2 + b^2 = 3c^2$.

(iv) Prove that there are no non-zero integers $a$, $b$ and $c$ for which $a^2 + b^2 + c^2 = 4abc$.

Hint

解题思路

Part (i)

$a^3+2b^3+4c^3=0 \implies a$ 偶数，设 $a=2p$，$|p|<|a|$。 $8p^3+2b^3+4c^3=0 \implies 4p^3+b^3+2c^3=0$。 同理 $b=2q,c=2r$，得 $p^3+2q^3+4r^3=0$。 无穷递降：不存在非零整数解。

Part (ii)

$9a^3+10b^3+6c^3=0 \implies 3|b$，设 $b=3q$。 $9a^3+270q^3+6c^3=0 \implies 3a^3+90q^3+2c^3=0 \implies 3|c$。 同理 $3|a$，设 $a=3p$：$9p^3+10q^3+6r^3=0$。无穷递降。

Part (iii)

平方模 3：$n\not\equiv 0 \implies n^2\equiv 1$。$a^2+b^2\equiv 0 \implies a,b$ 均为 3 的倍数。 设 $a=3p,b=3q$，则 $c=3r$，$p^2+q^2=3r^2$。无穷递降。

Part (iv)

$a^2+b^2+c^2=4abc$。RHS $\equiv 0 \pmod 4 \implies a,b,c$ 全偶。 设 $a=2p,b=2q,c=2r$：$p^2+q^2+r^2=8pqr$。无穷递降。

Model Solution

Part (i)

Suppose for contradiction that non-zero integers $a, b, c$ exist with $a^3 + 2b^3 + 4c^3 = 0$.

Step 1: Show $a$ is even and produce $p$ with $|p| < |a|$.

Since $a^3 + 2b^3 + 4c^3 = 0$, we have $a^3 = -2b^3 - 4c^3 = -2(b^3 + 2c^3)$. The right side is even, so $a^3$ is even, hence $a$ is even. Write $a = 2p$ for some integer $p$.

Substituting: $(2p)^3 + 2b^3 + 4c^3 = 0$, i.e., $8p^3 + 2b^3 + 4c^3 = 0$.

Dividing by 2:

$4p^3 + b^3 + 2c^3 = 0$

Since $a \neq 0$, we have $p \neq 0$, and $|p| = |a|/2 < |a|$ (as $|a| \geqslant 2$).

Step 2: Show $b$ and $c$ are also even, producing $q, r$ with $|q| < |b|$, $|r| < |c|$.

From $4p^3 + b^3 + 2c^3 = 0$, we get $b^3 = -4p^3 - 2c^3 = -2(2p^3 + c^3)$, so $b^3$ is even, hence $b$ is even. Write $b = 2q$.

Substituting: $4p^3 + 8q^3 + 2c^3 = 0$, dividing by 2: $2p^3 + 4q^3 + c^3 = 0$.

So $c^3 = -2p^3 - 4q^3 = -2(p^3 + 2q^3)$, hence $c$ is even. Write $c = 2r$.

Substituting: $2p^3 + 4q^3 + 8r^3 = 0$, dividing by 2:

$p^3 + 2q^3 + 4r^3 = 0$

We have $|p| < |a|$, $|q| < |b|$, $|r| < |c|$, and $(p, q, r)$ satisfies the same equation as $(a, b, c)$.

Step 3: Deduction by infinite descent.

Starting from any non-zero solution $(a, b, c)$, we produce a new non-zero solution $(p, q, r)$ with strictly smaller absolute values. Repeating this process produces an infinite strictly decreasing sequence of positive integers $|a| > |p| > \cdots > 0$. This is impossible since the positive integers are well-ordered (every non-empty set of positive integers has a least element). Therefore no non-zero integers $a, b, c$ with $a^3 + 2b^3 + 4c^3 = 0$ can exist.

Part (ii)

Suppose for contradiction that non-zero integers $a, b, c$ exist with $9a^3 + 10b^3 + 6c^3 = 0$.

Step 1: Show $3 \mid b$.

Reducing modulo 3: $9a^3 + 10b^3 + 6c^3 \equiv 0 + b^3 + 0 \equiv b^3 \pmod{3}$.

Since the left side equals 0, we need $b^3 \equiv 0 \pmod{3}$, so $3 \mid b$. Write $b = 3q$.

Substituting: $9a^3 + 10(27q^3) + 6c^3 = 0$, i.e., $9a^3 + 270q^3 + 6c^3 = 0$.

Dividing by 3: $3a^3 + 90q^3 + 2c^3 = 0$.

Step 2: Show $3 \mid c$.

Reducing modulo 3: $3a^3 + 90q^3 + 2c^3 \equiv 0 + 0 + 2c^3 \equiv 2c^3 \pmod{3}$.

Since the left side equals 0, we need $2c^3 \equiv 0 \pmod{3}$, hence $c^3 \equiv 0 \pmod{3}$ (as $\gcd(2, 3) = 1$), so $3 \mid c$. Write $c = 3r$.

Substituting: $3a^3 + 90q^3 + 54r^3 = 0$. Dividing by 3: $a^3 + 30q^3 + 18r^3 = 0$.

Step 3: Show $3 \mid a$.

Reducing modulo 3: $a^3 + 30q^3 + 18r^3 \equiv a^3 \pmod{3}$.

Since the left side equals 0, $a^3 \equiv 0 \pmod{3}$, so $3 \mid a$. Write $a = 3p$.

Substituting: $27p^3 + 30q^3 + 18r^3 = 0$. Dividing by 3:

$9p^3 + 10q^3 + 6r^3 = 0$

Step 4: Infinite descent.

We have a new solution $(p, q, r)$ satisfying the same equation, with $|p| < |a|$, $|q| < |b|$, $|r| < |c|$. By the same infinite descent argument as part (i), no non-zero solution can exist.

Part (iii)

Step 1: Show that if $n$ is not a multiple of 3, then $n^2 \equiv 1 \pmod{3}$.

If $n$ is not a multiple of 3, then $n = 3k + 1$ or $n = 3k - 1$ for some integer $k$.

Case 1: $n = 3k + 1$. Then $n^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 \equiv 1 \pmod{3}$.

Case 2: $n = 3k - 1$. Then $n^2 = 9k^2 - 6k + 1 = 3(3k^2 - 2k) + 1 \equiv 1 \pmod{3}$.

In both cases, $n^2$ is one more than a multiple of 3.

Step 2: Deduce that $a^2 + b^2 \equiv 0 \pmod{3}$ implies $3 \mid a$ and $3 \mid b$.

The possible values of $a^2 \pmod{3}$ and $b^2 \pmod{3}$ are:

• If $3 \mid a$: $a^2 \equiv 0 \pmod{3}$.
• If $3 \nmid a$: $a^2 \equiv 1 \pmod{3}$ (by Step 1).

So $a^2 + b^2 \pmod{3}$ can be $0$, $1$, or $2$:

$a \pmod{3}$	$b \pmod{3}$	$a^2 + b^2 \pmod{3}$
$0$	$0$	$0$
$0$	$\neq 0$	$1$
$\neq 0$	$0$	$1$
$\neq 0$	$\neq 0$	$2$

Therefore $a^2 + b^2 \equiv 0 \pmod{3}$ only when $3 \mid a$ and $3 \mid b$.

Step 3: Prove no non-zero integers satisfy $a^2 + b^2 = 3c^2$.

Suppose for contradiction that non-zero integers $a, b, c$ exist with $a^2 + b^2 = 3c^2$.

Since $3c^2 \equiv 0 \pmod{3}$, by Step 2, $3 \mid a$ and $3 \mid b$. Write $a = 3p$, $b = 3q$.

Substituting: $9p^2 + 9q^2 = 3c^2$, so $3(p^2 + q^2) = c^2$.

This means $3 \mid c^2$, and since 3 is prime, $3 \mid c$. Write $c = 3r$.

Substituting: $3(p^2 + q^2) = 9r^2$, so $p^2 + q^2 = 3r^2$.

We have a new solution $(p, q, r)$ satisfying the same equation, with $|p| < |a|$, $|q| < |b|$, $|r| < |c|$. By infinite descent, no non-zero solution can exist.

Part (iv)

Suppose for contradiction that non-zero integers $a, b, c$ exist with $a^2 + b^2 + c^2 = 4abc$.

Step 1: Show $a, b, c$ are all even.

Since $4abc$ is divisible by 4, we need $a^2 + b^2 + c^2 \equiv 0 \pmod{4}$.

The square of any integer satisfies $(2k)^2 = 4k^2 \equiv 0 \pmod{4}$ and $(2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod{4}$. So $a^2 + b^2 + c^2 \pmod{4}$ equals the number of odd values among $a, b, c$.

We check each case:

• All three odd: $a^2 + b^2 + c^2 \equiv 3 \pmod{4}$. But $4abc \equiv 0 \pmod{4}$. So $3 \equiv 0 \pmod{4}$, contradiction.

• Exactly two odd: $a^2 + b^2 + c^2 \equiv 2 \pmod{4}$. But $4abc$ is divisible by 4. So $2 \equiv 0 \pmod{4}$, contradiction.

• Exactly one odd: $a^2 + b^2 + c^2 \equiv 1 \pmod{4}$. But $4abc \equiv 0 \pmod{4}$. So $1 \equiv 0 \pmod{4}$, contradiction.

• All three even: $a^2 + b^2 + c^2 \equiv 0 \pmod{4}$. This is consistent.

Therefore $a, b, c$ must all be even.

Step 2: Infinite descent.

Write $a = 2p$, $b = 2q$, $c = 2r$. Substituting:

$4p^2 + 4q^2 + 4r^2 = 4(2p)(2q)(2r) = 32pqr$

Dividing by 4:

$p^2 + q^2 + r^2 = 8pqr$

This equation has the same form as the original (a sum of three squares equals a constant times the product). We have $|p| < |a|$, $|q| < |b|$, $|r| < |c|$.

By the same infinite descent argument, no non-zero solution can exist. $\blacksquare$

Examiner Notes

仅约半数考生尝试，平均分9/20。主要问题在于逻辑表述不清——“可以无限继续”或”模不断减小”的论述不够严谨，必须明确解释为什么整数性导致矛盾。第(ii)部分关键是利用模3分析。第(iv)部分出乎意料地需要更精巧的论证（非简单复制(i)的结构），许多考生在此受阻。

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Question 3

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) The curve $C_1$ has equation $ax^2 + bxy + cy^2 = 1$ where $abc \neq 0$ and $a > 0$.

Show that, if the curve has two stationary points, then $b^2 < 4ac$.

(ii) The curve $C_2$ has equation $ay^3 + bx^2y + cx = 1$ where $abc \neq 0$ and $b > 0$.

Show that the $x$-coordinates of stationary points on this curve satisfy $4cb^3x^4 - 8b^3x^3 - ac^3 = 0.$ Show that, if the curve has two stationary points, then $4ac^6 + 27b^3 > 0$.

(iii) Consider the simultaneous equations $ay^3 + bx^2y + cx = 1$ $2bxy + c = 0$ $3ay^2 + bx^2 = 0$ where $abc \neq 0$ and $b > 0$.

Show that, if these simultaneous equations have a solution, then $4ac^6 + 27b^3 = 0$.

Hint

解题思路

Part (i)

$ax^2+bxy+cy^2=1$。隐函数求导：$2ax+by+bx y'+2cy y'=0$。 驻点 $y'=0 \implies by=-2ax$。 原方程乘 $b^2$：$ab^2x^2+b^3xy+b^2cy^2=b^2$。 代入 $by=-2ax$：$ab^2x^2-2ab^2x^2+4a^2cx^2=b^2 \implies a(4ac-b^2)x^2=b^2$。 $b^2>0,a>0 \implies 4ac-b^2>0 \implies b^2<4ac$。

Part (ii)

$ay^3+bx^2y+cx=1$。$3ay^2 y'+2bxy+bx^2 y'+c=0$。 驻点 $2bxy=-c$。 原方程乘 $8b^3x^3$：$8ab^3x^3y^3+8b^4x^5y+8b^3cx^4=8b^3x^3$。 $(2bxy)^3=-c^3 \implies -ac^3-4cb^3x^4+8cb^3x^4=8b^3x^3 \implies 4cb^3x^4-8b^3x^3-ac^3=0$。 $f'(x)=8b^3x^2(2cx-3)=0$，极值 $x=\frac{3}{2c}$。 $f(\frac{3}{2c})<0$（或 $>0$）$\implies 4ac^6+27b^3>0$。

Part (iii)

联立三方程。$3ay^2+bx^2=0 \implies a<0$。 $2bxy=-c$，乘 $4by^2$：$12aby^4+c^2=0 \implies y^4=\frac{-c^2}{12ab}$。 原方程乘 $y$：$ay^4+bx^2y^2+cxy=y$。 $ay^4=\frac{-c^2}{12b}$，$bx^2y^2=\frac{c^2}{4b}$，$cxy=\frac{-c^2}{2b}$。 $y=\frac{-c^2}{3b}$。$y^4=(\frac{-c^2}{3b})^4=\frac{-c^2}{12ab} \implies 4ac^6+27b^3=0$。

Model Solution

Part (i)

We differentiate $ax^2 + bxy + cy^2 = 1$ implicitly with respect to $x$:

$2ax + by + bx\frac{dy}{dx} + 2cy\frac{dy}{dx} = 0$

At a stationary point, $\frac{dy}{dx} = 0$, so:

$2ax + by = 0 \qquad \text{i.e.} \quad by = -2ax \qquad \cdots (*)$

We substitute $by = -2ax$ into the original equation. Multiply the original equation by $b^2$:

$ab^2x^2 + b^3xy + b^2cy^2 = b^2$

Using $by = -2ax$, we get $b^3xy = b^2(bxy) = b^2(-2ax^2) = -2ab^2x^2$ and $b^2cy^2 = c(by)^2 = c \cdot 4a^2x^2 = 4a^2cx^2$. Substituting:

$ab^2x^2 - 2ab^2x^2 + 4a^2cx^2 = b^2$

$a(4ac - b^2)x^2 = b^2 \qquad \cdots (**)$

Since $a > 0$ and $b^2 > 0$, equation $(**)$ requires $4ac - b^2 > 0$ for real solutions $x$ to exist. If $4ac - b^2 \leq 0$, the left-hand side is non-positive while the right-hand side is positive, a contradiction.

Therefore, if the curve has stationary points, $b^2 < 4ac$. $\blacksquare$

Part (ii)

We differentiate $ay^3 + bx^2y + cx = 1$ implicitly:

$3ay^2\frac{dy}{dx} + 2bxy + bx^2\frac{dy}{dx} + c = 0$

At a stationary point, $\frac{dy}{dx} = 0$:

$2bxy + c = 0 \qquad \cdots (*)$

From $(*)$: $y = \frac{-c}{2bx}$ (note $x \neq 0$ since $c \neq 0$). Substituting into the original equation $ay^3 + bx^2y + cx = 1$:

$a\left(\frac{-c}{2bx}\right)^3 + bx^2 \cdot \frac{-c}{2bx} + cx = 1$

$\frac{-ac^3}{8b^3x^3} - \frac{cx}{2} + cx = 1$

$\frac{-ac^3}{8b^3x^3} + \frac{cx}{2} = 1$

Multiply through by $8b^3x^3$:

$-ac^3 + 4cb^3x^4 = 8b^3x^3$

$4cb^3x^4 - 8b^3x^3 - ac^3 = 0 \qquad \blacksquare$

Now we show that if the curve has two stationary points, then $4ac^6 + 27b^3 > 0$.

Let $f(x) = 4cb^3x^4 - 8b^3x^3 - ac^3$. The stationary points of the curve correspond to roots of $f(x) = 0$.

Compute $f'(x) = 16cb^3x^3 - 24b^3x^2 = 8b^3x^2(2cx - 3)$.

Since $b > 0$, $f'(x) = 0$ at $x = 0$ (double root) and $x = \frac{3}{2c}$.

Because $x = 0$ is a double root of $f'$, it is an inflection point of $f$, not a local extremum. The only extremum of $f$ is at $x = \frac{3}{2c}$.

Evaluate:

$f\!\left(\frac{3}{2c}\right) = 4cb^3 \cdot \frac{81}{16c^4} - 8b^3 \cdot \frac{27}{8c^3} - ac^3 = \frac{81b^3}{4c^3} - \frac{27b^3}{c^3} - ac^3 = \frac{-(4ac^6 + 27b^3)}{4c^3}$

Also $f(0) = -ac^3$. The leading coefficient of $f$ is $4cb^3$, so $f(x) \to +\infty$ as $x \to \pm\infty$ when $c > 0$, and $f(x) \to -\infty$ as $x \to \pm\infty$ when $c < 0$.

For $f(x) = 0$ to have exactly two real roots (corresponding to two stationary points), the unique extremum must be on the opposite side of the $x$-axis from the end behaviour.

• If $c > 0$: $f \to +\infty$ at both ends. Two roots requires $f\!\left(\frac{3}{2c}\right) < 0$, i.e. $\frac{-(4ac^6 + 27b^3)}{4c^3} < 0$. Since $4c^3 > 0$, this gives $4ac^6 + 27b^3 > 0$.

• If $c < 0$: $f \to -\infty$ at both ends. Two roots requires $f\!\left(\frac{3}{2c}\right) > 0$, i.e. $\frac{-(4ac^6 + 27b^3)}{4c^3} > 0$. Since $4c^3 < 0$, we need $-(4ac^6 + 27b^3) < 0$, which again gives $4ac^6 + 27b^3 > 0$.

In both cases, two stationary points requires $4ac^6 + 27b^3 > 0$. $\blacksquare$

Part (iii)

We have the simultaneous equations:

$ay^3 + bx^2y + cx = 1 \qquad \cdots (1)$ $2bxy + c = 0 \qquad \cdots (2)$ $3ay^2 + bx^2 = 0 \qquad \cdots (3)$

From $(3)$: $bx^2 = -3ay^2$. Since $b > 0$, this requires $a < 0$.

From $(2)$: $x = \frac{-c}{2by}$. Squaring: $x^2 = \frac{c^2}{4b^2y^2}$. Substituting into $(3)$:

$3ay^2 + \frac{c^2}{4by^2} = 0$

Multiply by $4by^2$:

$12aby^4 + c^2 = 0 \qquad \cdots (4)$

Now multiply equation $(1)$ by $y$:

$ay^4 + bx^2y^2 + cxy = y$

Using $bx^2 = -3ay^2$ (from $(3)$), we get $bx^2y^2 = -3ay^4$.

From $(2)$: $2bxy = -c$, so $cxy = c \cdot \frac{-c}{2b} = \frac{-c^2}{2b}$.

Substituting:

$ay^4 - 3ay^4 - \frac{c^2}{2b} = y$

$-2ay^4 - \frac{c^2}{2b} = y$

From $(4)$: $ay^4 = \frac{-c^2}{12b}$, so $-2ay^4 = \frac{c^2}{6b}$.

$\frac{c^2}{6b} - \frac{c^2}{2b} = y$

$y = \frac{c^2 - 3c^2}{6b} = \frac{-c^2}{3b} \qquad \cdots (5)$

Substitute $(5)$ into $(4)$:

$12ab \cdot \left(\frac{-c^2}{3b}\right)^4 + c^2 = 0$

$12ab \cdot \frac{c^8}{81b^4} + c^2 = 0$

$\frac{4ac^8}{27b^3} + c^2 = 0$

Divide by $c^2$ (valid since $c \neq 0$):

$\frac{4ac^6}{27b^3} + 1 = 0$

$4ac^6 + 27b^3 = 0 \qquad \blacksquare$

Examiner Notes

超过90%考生尝试但平均分仅约5.5/20，是高尝试率中得分最低的题目。常见错误：错误假设两个驻点是重根、未考虑c<0的情况、对四次方程在驻点处取值符号的错误断言。值得注意的是，第(iii)部分通过直接消元求解方程组的考生反而比试图联系(ii)结果的考生表现更好。

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Question 4

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

4 You may assume that all infinite sums and products in this question converge.

(i) Prove by induction that for all positive integers $n$,

$\sinh x = 2^n \cosh \left( \frac{x}{2} \right) \cosh \left( \frac{x}{4} \right) \cdots \cosh \left( \frac{x}{2^n} \right) \sinh \left( \frac{x}{2^n} \right)$

and deduce that, for $x \neq 0$,

$\frac{\sinh x}{x} \frac{\frac{x}{2^n}}{\sinh \left( \frac{x}{2^n} \right)} = \cosh \left( \frac{x}{2} \right) \cosh \left( \frac{x}{4} \right) \cdots \cosh \left( \frac{x}{2^n} \right) .$

(ii) You are given that the Maclaurin series for $\sinh x$ is

$\sinh x = \sum_{r=0}^{\infty} \frac{x^{2r+1}}{(2r+1)!} .$

Use this result to show that, as $y$ tends to 0, $\frac{y}{\sinh y}$ tends to 1.

Deduce that, for $x \neq 0$,

$\frac{\sinh x}{x} = \cosh \left( \frac{x}{2} \right) \cosh \left( \frac{x}{4} \right) \cdots \cosh \left( \frac{x}{2^n} \right) \cdots .$

(iii) Let $x = \ln 2$. Evaluate $\cosh \left( \frac{x}{2} \right)$ and show that

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

Use part (ii) to show that

$\frac{1}{\ln 2} = \frac{1 + 2^{\frac{1}{2}}}{2} \times \frac{1 + 2^{\frac{1}{4}}}{2} \times \frac{1 + 2^{\frac{1}{8}}}{2} \cdots .$

(iv) Show that

$\frac{2}{\pi} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2 + \sqrt{2}}}{2} \times \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \cdots .$

Hint

解题思路

Part (i)

$n=1$：$\sinh x=2\cosh\frac{x}{2}\sinh\frac{x}{2}$。 归纳：$2^{k+1}\cdots\sinh\frac{x}{2^{k+1}}=2\sinh x\cdot\frac{\cosh\frac{x}{2^{k+1}}\sinh\frac{x}{2^{k+1}}}{\sinh\frac{x}{2^k}}=\sinh x$。 推论：两边除 $x$ 整理即得。

Part (ii)

$\frac{y}{\sinh y}\to 1$（$y\to 0$）。令 $n\to\infty$：$\frac{\sinh x}{x}=\prod_{n=1}^{\infty}\cosh\frac{x}{2^n}$。

Part (iii)

$x=\ln 2$：$\sinh(\ln 2)=\frac{3}{4}$，$\cosh\frac{\ln 2}{2}=\frac{3}{2\sqrt{2}}$。 $\frac{1}{\ln 2}=\frac{1+\sqrt{2}}{2}\times\frac{1+\sqrt[4]{2}}{2}\times\cdots=\prod_{n=1}^{\infty}\frac{1+2^{1/2^n}}{2}$。

Part (iv)

$x=\frac{i\pi}{2}$：$\frac{\sinh(i\pi/2)}{i\pi/2}=\frac{2}{\pi}$。 $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$，$\cos\frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$。 $\frac{2}{\pi}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{2+\sqrt{2}}}{2}\times\cdots$（Vieta 公式）。

Model Solution

Part (i)

We prove by induction on $n$ that:

$\sinh x = 2^n \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \sinh\!\left(\frac{x}{2^n}\right) \qquad \text{P}(n)$

Base case ($n = 1$): We need $\sinh x = 2\cosh\!\left(\frac{x}{2}\right)\sinh\!\left(\frac{x}{2}\right)$.

Using the identity $\sinh x = \sinh\!\left(2 \cdot \frac{x}{2}\right) = 2\sinh\!\left(\frac{x}{2}\right)\cosh\!\left(\frac{x}{2}\right)$, the base case holds. (This follows from $\sinh 2\theta = 2\sinh\theta\cosh\theta$, which is the hyperbolic analogue of $\sin 2\theta = 2\sin\theta\cos\theta$.)

Inductive step: Assume P$(k)$ holds for some positive integer $k$, i.e.:

$\sinh x = 2^k \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^k}\right) \sinh\!\left(\frac{x}{2^k}\right)$

We need to show P$(k+1)$: $\sinh x = 2^{k+1} \cosh\!\left(\frac{x}{2}\right) \cdots \cosh\!\left(\frac{x}{2^{k+1}}\right) \sinh\!\left(\frac{x}{2^{k+1}}\right)$.

Applying $\sinh\theta = 2\sinh\!\left(\frac{\theta}{2}\right)\cosh\!\left(\frac{\theta}{2}\right)$ with $\theta = \frac{x}{2^k}$:

$\sinh\!\left(\frac{x}{2^k}\right) = 2\sinh\!\left(\frac{x}{2^{k+1}}\right)\cosh\!\left(\frac{x}{2^{k+1}}\right)$

Substituting into the inductive hypothesis:

$\sinh x = 2^k \cosh\!\left(\frac{x}{2}\right) \cdots \cosh\!\left(\frac{x}{2^k}\right) \cdot 2\sinh\!\left(\frac{x}{2^{k+1}}\right)\cosh\!\left(\frac{x}{2^{k+1}}\right)$

$= 2^{k+1} \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^{k+1}}\right) \sinh\!\left(\frac{x}{2^{k+1}}\right)$

This is P$(k+1)$. By induction, P$(n)$ holds for all positive integers $n$. $\blacksquare$

Deduction: For $x \neq 0$, divide both sides of P$(n)$ by $x$:

$\frac{\sinh x}{x} = 2^n \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdot \frac{\sinh\!\left(\frac{x}{2^n}\right)}{\frac{x}{2^n} \cdot \frac{2^n}{1}} \cdot \frac{1}{1}$

More directly, rewrite P$(n)$ as:

$\sinh x = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdot \frac{2^n \sinh\!\left(\frac{x}{2^n}\right)}{1}$

Divide both sides by $x$:

$\frac{\sinh x}{x} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdot \frac{2^n \sinh\!\left(\frac{x}{2^n}\right)}{x}$

Since $x = 2^n \cdot \frac{x}{2^n}$:

$\frac{\sinh x}{x} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \cdot \frac{\sinh\!\left(\frac{x}{2^n}\right)}{\frac{x}{2^n}}$

Rearranging:

$\frac{\sinh x}{x} \cdot \frac{\frac{x}{2^n}}{\sinh\!\left(\frac{x}{2^n}\right)} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) \qquad \blacksquare$

Part (ii)

Show that $\frac{y}{\sinh y} \to 1$ as $y \to 0$.

From the Maclaurin series:

$\sinh y = y + \frac{y^3}{3!} + \frac{y^5}{5!} + \cdots = y\left(1 + \frac{y^2}{6} + \frac{y^4}{120} + \cdots\right)$

So:

$\frac{y}{\sinh y} = \frac{1}{1 + \frac{y^2}{6} + \frac{y^4}{120} + \cdots}$

As $y \to 0$, the terms $\frac{y^2}{6}, \frac{y^4}{120}, \ldots$ all tend to $0$, so:

$\lim_{y \to 0} \frac{y}{\sinh y} = \frac{1}{1 + 0} = 1 \qquad \blacksquare$

Deduction: From Part (i), for any fixed $x \neq 0$ and all $n$:

$\frac{\sinh x}{x} \cdot \frac{\frac{x}{2^n}}{\sinh\!\left(\frac{x}{2^n}\right)} = \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right)$

As $n \to \infty$, $\frac{x}{2^n} \to 0$, so $\frac{\frac{x}{2^n}}{\sinh\!\left(\frac{x}{2^n}\right)} \to 1$. Therefore:

$\frac{\sinh x}{x} = \lim_{n \to \infty} \cosh\!\left(\frac{x}{2}\right) \cosh\!\left(\frac{x}{4}\right) \cdots \cosh\!\left(\frac{x}{2^n}\right) = \prod_{m=1}^{\infty} \cosh\!\left(\frac{x}{2^m}\right) \qquad \blacksquare$

Part (iii)

Let $x = \ln 2$.

Evaluate $\cosh\!\left(\frac{x}{2}\right)$:

$\sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4}$

Using $\sinh x = 2\sinh\!\left(\frac{x}{2}\right)\cosh\!\left(\frac{x}{2}\right)$ and $\cosh^2\!\left(\frac{x}{2}\right) - \sinh^2\!\left(\frac{x}{2}\right) = 1$:

$\sinh(\ln 2) = 2\sinh\!\left(\frac{\ln 2}{2}\right)\cosh\!\left(\frac{\ln 2}{2}\right) = \frac{3}{4}$

Alternatively, compute directly:

$\cosh\!\left(\frac{\ln 2}{2}\right) = \frac{e^{\ln 2/2} + e^{-\ln 2/2}}{2} = \frac{\sqrt{2} + \frac{1}{\sqrt{2}}}{2} = \frac{\sqrt{2} + \frac{\sqrt{2}}{2}}{2} = \frac{\frac{3\sqrt{2}}{2}}{2} = \frac{3}{2\sqrt{2}}$

So $\cosh\!\left(\frac{x}{2}\right) = \frac{3}{2\sqrt{2}}$. Note that $\frac{3}{2\sqrt{2}} = \frac{3}{2 \cdot 2^{1/2}} = \frac{1 + 2}{2 \cdot 2^{1/2}} = \frac{1 + 2^{1/2}}{2 \cdot 2^{1/2}}$ … Actually, let me write this differently.

$\frac{3}{2\sqrt{2}} = \frac{3}{2^{3/2}} = \frac{3}{2 \cdot 2^{1/2}}$

We can verify: $\frac{1 + 2^{1/2}}{2 \cdot 2^{1/4}}$ … hmm, let me check the pattern the question wants.

Actually, $\cosh\!\left(\frac{x}{2}\right) = \frac{3}{2\sqrt{2}}$, and note:

$\frac{3}{2\sqrt{2}} = \frac{3}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4}$

But let’s think about the pattern more carefully. We have $e^{x/2} = 2^{1/2}$, so:

$\cosh\!\left(\frac{x}{2}\right) = \frac{2^{1/2} + 2^{-1/2}}{2} = \frac{2^{1/2}(1 + 2^{-1})}{2} = \frac{2^{1/2} \cdot \frac{3}{2}}{2} = \frac{3}{2^{3/2}}$

More usefully: $\frac{2^{1/2} + 2^{-1/2}}{2} = \frac{1}{2}(2^{1/2} + 2^{-1/2})$.

Show $\cosh\!\left(\frac{x}{4}\right) = \frac{1 + 2^{1/2}}{2 \times 2^{1/4}}$:

$\cosh\!\left(\frac{x}{4}\right) = \frac{e^{x/4} + e^{-x/4}}{2} = \frac{2^{1/4} + 2^{-1/4}}{2}$

Now $2^{-1/4} = \frac{1}{2^{1/4}}$, so:

$\cosh\!\left(\frac{x}{4}\right) = \frac{2^{1/4} + \frac{1}{2^{1/4}}}{2} = \frac{2^{1/2} + 1}{2 \cdot 2^{1/4}} = \frac{1 + 2^{1/2}}{2 \times 2^{1/4}} \qquad \blacksquare$

General pattern: For any $m \geq 1$:

$\cosh\!\left(\frac{x}{2^m}\right) = \frac{2^{1/2^m} + 2^{-1/2^m}}{2} = \frac{2^{1/2^{m+1}} + 2^{-1/2^{m+1}}}{2} \cdot \frac{2^{1/2^{m+1}}}{2^{1/2^{m+1}}}$

Actually, let me just compute each factor directly. For $m \geq 1$, $e^{x/2^m} = 2^{1/2^m}$:

$\cosh\!\left(\frac{x}{2^m}\right) = \frac{2^{1/2^m} + 2^{-1/2^m}}{2}$

Note that $2^{1/2^m} + 2^{-1/2^m} = 2^{-1/2^m}(2^{2/2^m} + 1) = 2^{-1/2^m}(1 + 2^{1/2^{m-1}})$. So:

$\cosh\!\left(\frac{x}{2^m}\right) = \frac{1 + 2^{1/2^{m-1}}}{2 \cdot 2^{1/2^m}}$

For $m = 1$: $\cosh\!\left(\frac{x}{2}\right) = \frac{1 + 2^{1/1}}{2 \cdot 2^{1/2}} = \frac{3}{2\sqrt{2}}$.

For $m = 2$: $\cosh\!\left(\frac{x}{4}\right) = \frac{1 + 2^{1/2}}{2 \cdot 2^{1/4}}$.

For general $m$: $\cosh\!\left(\frac{x}{2^m}\right) = \frac{1 + 2^{1/2^{m-1}}}{2 \cdot 2^{1/2^m}}$.

Deduction using Part (ii): From Part (ii), $\frac{\sinh x}{x} = \prod_{m=1}^{\infty} \cosh\!\left(\frac{x}{2^m}\right)$. With $x = \ln 2$:

$\frac{\sinh(\ln 2)}{\ln 2} = \frac{3/4}{\ln 2} = \frac{3}{4\ln 2}$

And:

$\prod_{m=1}^{\infty} \cosh\!\left(\frac{\ln 2}{2^m}\right) = \prod_{m=1}^{\infty} \frac{1 + 2^{1/2^{m-1}}}{2 \cdot 2^{1/2^m}}$

So:

$\frac{3}{4\ln 2} = \prod_{m=1}^{\infty} \frac{1 + 2^{1/2^{m-1}}}{2 \cdot 2^{1/2^m}}$

We need to show this gives $\frac{1}{\ln 2} = \prod_{m=1}^{\infty} \frac{1 + 2^{1/2^m}}{2}$.

Let’s compute the partial products. Let $P_n = \prod_{m=1}^{n} \cosh\!\left(\frac{\ln 2}{2^m}\right)$.

From Part (i): $P_n = \frac{\sinh(\ln 2)}{\ln 2} \cdot \frac{\ln 2 / 2^n}{\sinh(\ln 2 / 2^n)} = \frac{3}{4\ln 2} \cdot \frac{\ln 2 / 2^n}{\sinh(\ln 2 / 2^n)}$.

As $n \to \infty$: $\frac{\ln 2 / 2^n}{\sinh(\ln 2 / 2^n)} \to 1$, so $P_n \to \frac{3}{4\ln 2}$.

Now compute $P_n$ directly:

$P_n = \prod_{m=1}^{n} \frac{2^{1/2^m} + 2^{-1/2^m}}{2}$

Let $a_m = 2^{1/2^m}$. Then $\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{a_m + a_m^{-1}}{2} = \frac{a_m^2 + 1}{2a_m}$.

Note $a_m^2 = 2^{1/2^{m-1}} = a_{m-1}$ (where $a_0 = 2$). So $\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{a_{m-1} + 1}{2a_m}$.

$P_n = \prod_{m=1}^{n} \frac{a_{m-1} + 1}{2a_m} = \frac{\prod_{m=1}^{n}(a_{m-1} + 1)}{2^n \prod_{m=1}^{n} a_m}$

The denominator: $\prod_{m=1}^{n} a_m = 2^{1/2} \cdot 2^{1/4} \cdots 2^{1/2^n} = 2^{\sum_{m=1}^{n} 1/2^m} = 2^{1 - 1/2^n}$.

So $2^n \cdot 2^{1-1/2^n} = 2^{n+1-1/2^n}$.

The numerator: $\prod_{m=1}^{n}(a_{m-1} + 1) = (a_0 + 1)(a_1 + 1) \cdots (a_{n-1} + 1) = 3 \cdot (1 + \sqrt{2}) \cdot (1 + 2^{1/4}) \cdots (1 + 2^{1/2^{n-1}})$.

Hmm, this is getting complicated. Let me try a telescoping approach instead.

Note that $\frac{a_{m-1}+1}{2a_m} = \frac{a_{m-1}+1}{2a_m}$. Since $a_{m-1} = a_m^2$:

$\frac{a_m^2 + 1}{2a_m} = \frac{(a_m+1)^2 - 2a_m}{2a_m}$

That doesn’t telescope nicely either. Let me try yet another approach.

Actually, let’s just use the result from Part (ii) directly. We have $\frac{\sinh x}{x} = \prod_{m=1}^{\infty} \cosh\!\left(\frac{x}{2^m}\right)$, so:

$\frac{1}{\ln 2} = \frac{1}{\sinh(\ln 2)} \cdot \frac{\sinh(\ln 2)}{\ln 2} = \frac{4}{3} \cdot \prod_{m=1}^{\infty} \cosh\!\left(\frac{\ln 2}{2^m}\right)$

Wait, $\sinh(\ln 2) = 3/4$, so $\frac{\sinh(\ln 2)}{\ln 2} = \frac{3}{4\ln 2}$, giving $\frac{1}{\ln 2} = \frac{4}{3} \cdot \frac{3}{4\ln 2} = \frac{1}{\ln 2}$. That’s circular.

Let me think about this differently. We need to show:

$\frac{1}{\ln 2} = \frac{1 + 2^{1/2}}{2} \cdot \frac{1 + 2^{1/4}}{2} \cdot \frac{1 + 2^{1/8}}{2} \cdots$

Let me define $b_m = \frac{1 + 2^{1/2^m}}{2}$ for $m = 1, 2, 3, \ldots$

So the right side is $\prod_{m=1}^{\infty} b_m$.

Now, $\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{2^{1/2^m} + 2^{-1/2^m}}{2}$. We showed this equals $\frac{1 + 2^{1/2^{m-1}}}{2 \cdot 2^{1/2^m}} = \frac{b_{m-1}}{2^{1/2^m}}$ (for $m \geq 2$, and $\cosh\!\left(\frac{\ln 2}{2}\right) = \frac{3}{2\sqrt{2}}$).

Hmm wait, let me be more careful. We have:

$\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{2^{1/2^m} + 2^{-1/2^m}}{2}$

Let $t = 2^{1/2^m}$. Then $\cosh = \frac{t + t^{-1}}{2} = \frac{t^2 + 1}{2t}$.

Now $t^2 = 2^{1/2^{m-1}}$, so $t^2 + 1 = 1 + 2^{1/2^{m-1}}$.

And $2t = 2 \cdot 2^{1/2^m} = 2^{1 + 1/2^m}$.

So $\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{1 + 2^{1/2^{m-1}}}{2^{1+1/2^m}}$.

For $m = 1$: $\cosh\!\left(\frac{\ln 2}{2}\right) = \frac{1 + 2}{2^{1+1/2}} = \frac{3}{2^{3/2}} = \frac{3}{2\sqrt{2}}$. Correct.

For $m \geq 2$: $\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{1 + 2^{1/2^{m-1}}}{2^{1+1/2^m}} = \frac{b_{m-1}}{2^{1/2^m}} \cdot \frac{1}{1} = \frac{2 \cdot b_{m-1}}{2 \cdot 2^{1/2^m}}$…

Actually $b_{m-1} = \frac{1 + 2^{1/2^{m-1}}}{2}$, so $1 + 2^{1/2^{m-1}} = 2b_{m-1}$.

Therefore $\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{2b_{m-1}}{2^{1+1/2^m}} = \frac{b_{m-1}}{2^{1/2^m}}$.

So for $m \geq 2$:

$\cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{b_{m-1}}{2^{1/2^m}}$

And $\cosh\!\left(\frac{\ln 2}{2}\right) = \frac{3}{2\sqrt{2}} = \frac{3}{2^{3/2}}$.

Now:

$\prod_{m=1}^{\infty} \cosh\!\left(\frac{\ln 2}{2^m}\right) = \frac{3}{2^{3/2}} \cdot \prod_{m=2}^{\infty} \frac{b_{m-1}}{2^{1/2^m}} = \frac{3}{2^{3/2}} \cdot \frac{\prod_{m=1}^{\infty} b_m}{\prod_{m=2}^{\infty} 2^{1/2^m}}$

Now $\prod_{m=2}^{\infty} 2^{1/2^m} = 2^{\sum_{m=2}^{\infty} 1/2^m} = 2^{1/2}$.

Also $b_1 = \frac{1 + \sqrt{2}}{2}$, and $\frac{3}{2^{3/2}} = \frac{3}{2\sqrt{2}}$.

So:

$\frac{3}{4\ln 2} = \frac{3}{2^{3/2}} \cdot \frac{\prod_{m=1}^{\infty} b_m}{2^{1/2}} = \frac{3}{2^{3/2} \cdot 2^{1/2}} \cdot \prod_{m=1}^{\infty} b_m = \frac{3}{4} \cdot \prod_{m=1}^{\infty} b_m$

Therefore:

$\frac{1}{\ln 2} = \prod_{m=1}^{\infty} b_m = \frac{1 + 2^{1/2}}{2} \cdot \frac{1 + 2^{1/4}}{2} \cdot \frac{1 + 2^{1/8}}{2} \cdots \qquad \blacksquare$

Part (iv)

We substitute $x = \frac{i\pi}{2}$ into the result of Part (ii): $\frac{\sinh x}{x} = \prod_{m=1}^{\infty} \cosh\!\left(\frac{x}{2^m}\right)$.

Left side: Using $\sinh(i\theta) = i\sin\theta$:

$\frac{\sinh\!\left(\frac{i\pi}{2}\right)}{\frac{i\pi}{2}} = \frac{i\sin\!\left(\frac{\pi}{2}\right)}{\frac{i\pi}{2}} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}$

Right side: Using $\cosh(i\theta) = \cos\theta$:

$\cosh\!\left(\frac{i\pi}{2 \cdot 2^m}\right) = \cos\!\left(\frac{\pi}{2^{m+1}}\right)$

So:

$\frac{2}{\pi} = \prod_{m=1}^{\infty} \cos\!\left(\frac{\pi}{2^{m+1}}\right) = \cos\!\left(\frac{\pi}{4}\right) \cos\!\left(\frac{\pi}{8}\right) \cos\!\left(\frac{\pi}{16}\right) \cdots$

Now we evaluate each factor.

$\cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

For $\cos\!\left(\frac{\pi}{8}\right)$, use the half-angle formula $\cos\!\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos\theta}{2}}$ with $\theta = \frac{\pi}{4}$:

$\cos\!\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos(\pi/4)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

For $\cos\!\left(\frac{\pi}{16}\right)$, apply the half-angle formula again with $\theta = \frac{\pi}{8}$:

$\cos\!\left(\frac{\pi}{16}\right) = \sqrt{\frac{1 + \cos(\pi/8)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2+\sqrt{2}}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2}$

In general, defining $c_1 = \sqrt{2}$ and $c_{k+1} = \sqrt{2 + c_k}$, we get $\cos\!\left(\frac{\pi}{2^{m+1}}\right) = \frac{c_m}{2}$.

Therefore:

$\frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \cdots \qquad \blacksquare$

Examiner Notes

第四受欢迎题目(80%+考生尝试)，平均分9/20。归纳法中逻辑流向不规范和未讨论除零问题是常见扣分点。第(ii)部分许多考生误用洛必达法则而非题目要求的麦克劳林级数。第(iv)部分最佳方法是虚数代换(令x=iπ/2)，但许多考生试图使用Osborn法则却未充分论证。

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Question 5

Topic: 纯数  |  Difficulty: Standard  |  Marks: 20

Problem

5 (i) Show that $\int_{-a}^{a} \frac{1}{1+e^{x}} dx = a \text{ for all } a \geqslant 0.$

(ii) Explain why, if $g$ is a continuous function and $\int_{0}^{a} g(x) dx = 0 \text{ for all } a \geqslant 0,$ then $g(x) = 0$ for all $x \geqslant 0$.

Let $f$ be a continuous function with $f(x) \geqslant 0$ for all $x$. Show that $\int_{-a}^{a} \frac{1}{1+f(x)} dx = a \text{ for all } a \geqslant 0$ if and only if $\frac{1}{1+f(x)} + \frac{1}{1+f(-x)} - 1 = 0 \text{ for all } x \geqslant 0,$ and hence if and only if $f(x)f(-x) = 1$ for all $x$.

(iii) Let $f$ be a continuous function such that, for all $x$, $f(x) \geqslant 0$ and $f(x)f(-x) = 1$. Show that, if $h$ is a continuous function with $h(x) = h(-x)$ for all $x$, then $\int_{-a}^{a} \frac{h(x)}{1+f(x)} dx = \int_{0}^{a} h(x) dx .$

(iv) Hence find the exact value of $\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{e^{-x} \cos x}{\cosh x} dx .$

Hint

解题思路

本题围绕积分对称性展开，四个部分层层递进：

• Part (i) 通过具体函数 1/(1+e^x) 建立一个积分恒等式
• Part (ii) 利用微积分基本定理推广到一般函数，建立充要条件
• Part (iii) 将结论推广到含偶函数 h(x) 的被积函数
• Part (iv) 将理论应用于具体积分求值

Part (i)：证明 ∫_{-a}^{a} 1/(1+e^x) dx = a（a ≥ 0）

方法一（分子分母同乘 e^{-x}）： ∫{-a}^{a} 1/(1+e^x) dx = ∫{-a}^{a} e^{-x}/(e^{-x}+1) dx = [-ln(e^{-x}+1)]_{-a}^{a} = ln((e^a+1)/(e^{-a}+1)) = ln(e^a) = a

方法二（拆项法）： 1/(1+e^x) + 1/(1+e^{-x}) = 1/(1+e^x) + e^x/(1+e^x) = 1 因此 ∫_{-a}^{a} 1/(1+e^x) dx = ∫_0^a 1 dx = a。

Part (ii) 第一部分：若 g 连续且 ∫_0^a g(x) dx = 0 对所有 a ≥ 0，则 g(x) = 0（x ≥ 0）

由微积分基本定理（FTC），设 G(a) = ∫_0^a g(x) dx，则 G(a) = 0 对所有 a ≥ 0，故 G’(a) = g(a) = 0 对所有 a ≥ 0。

Part (ii) 第二部分：证明充要条件

必要性（⇒）： ∫{-a}^{a} 1/(1+f(x)) dx = a ⇔ ∫{-a}^{0} 1/(1+f(x)) dx + ∫{0}^{a} 1/(1+f(x)) dx = a 对第一项作 x → -x 代换： ⇔ ∫{0}^{a} 1/(1+f(-x)) dx + ∫{0}^{a} 1/(1+f(x)) dx = a ⇔ ∫{0}^{a} [1/(1+f(-x)) + 1/(1+f(x)) - 1] dx = 0 由第一部分结论，被积函数恒为零： 1/(1+f(-x)) + 1/(1+f(x)) - 1 = 0 ∀ x ≥ 0

充分性（⇐）：反向推导同样成立。

化简得 f(x)f(-x) = 1： 1/(1+f(x)) + 1/(1+f(-x)) = 1 ⇔ 1+f(-x) + 1+f(x) = (1+f(x))(1+f(-x)) ⇔ 2 + f(x) + f(-x) = 1 + f(x) + f(-x) + f(x)f(-x) ⇔ f(x)f(-x) = 1

Part (iii)：设 f(x) ≥ 0，f(x)f(-x) = 1，h 连续且 h(x) = h(-x)，证明 ∫_{-a}^{a} h(x)/(1+f(x)) dx = ∫_0^a h(x) dx

∫{-a}^{a} h(x)/(1+f(x)) dx = ∫{-a}^{0} h(x)/(1+f(x)) dx + ∫_{0}^{a} h(x)/(1+f(x)) dx

对第一项作 x → -x 代换，利用 h(-x) = h(x)： = ∫{0}^{a} h(x)/(1+f(-x)) dx + ∫{0}^{a} h(x)/(1+f(x)) dx = ∫_0^a h(x)[1/(1+f(-x)) + 1/(1+f(x))] dx

由 Part (ii) 的结论 1/(1+f(-x)) + 1/(1+f(x)) = 1： = ∫_0^a h(x) dx

Part (iv)：求 ∫_{-π/2}^{π/2} (e^{-x} cos x)/cosh x dx

第一步：化简被积函数 (e^{-x} cos x)/cosh x = (e^{-x} cos x)/((e^x + e^{-x})/2) = (2 cos x)/(e^{2x} + 1) = (2 cos x)/(1 + e^{2x})

第二步：验证条件

• h(x) = cos x：连续且 h(-x) = cos(-x) = cos x = h(x)（偶函数）✓
• f(x) = e^{2x}：连续，f(x) > 0，f(x)f(-x) = e^{2x} · e^{-2x} = 1 ✓

第三步：应用 Part (iii) 结论 ∫_{-π/2}^{π/2} (2 cos x)/(1+e^{2x}) dx = 2∫_0^{π/2} cos x dx = 2[sin x]_0^{π/2} = 2(1 - 0) = 2

常见错误（Examiner Notes）

1. Part (i)：积分 ∫ 1/(1+e^x) dx 错误地算成 ln(1+e^x)
2. Part (ii)：直接从 ∫_{-a}^{a} g(x) dx = 0 推出 g(x) = 0，这是错误推理
3. Part (ii)：使用 u = -x 代换时未展示充分的推导步骤
4. Part (iii)：未意识到 h(x) = h(-x) 即使题目已给出此条件
5. Part (iv)：未验证 cos x 和 e^{2x} 满足 Part (iii) 的条件；遗漏因子 2

Model Solution

Part (i)

Let $I = \int_{-a}^{a} \frac{1}{1+e^{x}} dx$.

Using the substitution $u = -x$ (so $du = -dx$), when $x = -a$, $u = a$; when $x = a$, $u = -a$. Thus

$I = \int_{a}^{-a} \frac{1}{1+e^{-u}} (-du) = \int_{-a}^{a} \frac{1}{1+e^{-u}} du.$

Since $\frac{1}{1+e^{-u}} = \frac{e^{u}}{e^{u}+1}$, renaming the dummy variable back to $x$:

$I = \int_{-a}^{a} \frac{e^{x}}{1+e^{x}} dx.$

Adding the two expressions for $I$:

$2I = \int_{-a}^{a} \frac{1}{1+e^{x}} dx + \int_{-a}^{a} \frac{e^{x}}{1+e^{x}} dx = \int_{-a}^{a} \frac{1+e^{x}}{1+e^{x}} dx = \int_{-a}^{a} 1 \, dx = 2a.$

Therefore $I = a$.

Part (ii)

First we show that if $g$ is continuous and $\int_{0}^{a} g(x) dx = 0$ for all $a \geqslant 0$, then $g(x) = 0$ for all $x \geqslant 0$.

Define $G(a) = \int_{0}^{a} g(x) dx$. By hypothesis, $G(a) = 0$ for all $a \geqslant 0$. Since $g$ is continuous, by the Fundamental Theorem of Calculus, $G$ is differentiable and $G'(a) = g(a)$ for all $a > 0$. But $G(a) = 0$ for all $a \geqslant 0$ implies $G'(a) = 0$ for all $a > 0$. Hence $g(x) = 0$ for all $x > 0$. By continuity of $g$, $g(0) = \lim_{x \to 0^+} g(x) = 0$. So $g(x) = 0$ for all $x \geqslant 0$.

Now let $f$ be continuous with $f(x) \geqslant 0$ for all $x$. We show that $\int_{-a}^{a} \frac{1}{1+f(x)} dx = a$ for all $a \geqslant 0$ if and only if $\frac{1}{1+f(x)} + \frac{1}{1+f(-x)} - 1 = 0$ for all $x \geqslant 0$.

Forward direction ($\Rightarrow$): Suppose $\int_{-a}^{a} \frac{1}{1+f(x)} dx = a$ for all $a \geqslant 0$. Split the integral:

$\int_{-a}^{0} \frac{1}{1+f(x)} dx + \int_{0}^{a} \frac{1}{1+f(x)} dx = a.$

In the first integral, substitute $u = -x$ (so $du = -dx$):

$\int_{-a}^{0} \frac{1}{1+f(x)} dx = \int_{a}^{0} \frac{1}{1+f(-u)} (-du) = \int_{0}^{a} \frac{1}{1+f(-u)} du.$

Renaming $u$ to $x$:

$\int_{0}^{a} \frac{1}{1+f(-x)} dx + \int_{0}^{a} \frac{1}{1+f(x)} dx = a.$

Combining:

$\int_{0}^{a} \left[ \frac{1}{1+f(-x)} + \frac{1}{1+f(x)} \right] dx = a = \int_{0}^{a} 1 \, dx.$

Therefore:

$\int_{0}^{a} \left[ \frac{1}{1+f(-x)} + \frac{1}{1+f(x)} - 1 \right] dx = 0 \quad \text{for all } a \geqslant 0.$

Since $f$ is continuous, the integrand $g(x) = \frac{1}{1+f(-x)} + \frac{1}{1+f(x)} - 1$ is continuous. By the result just proved, $g(x) = 0$ for all $x \geqslant 0$, i.e.,

$\frac{1}{1+f(x)} + \frac{1}{1+f(-x)} - 1 = 0 \quad \text{for all } x \geqslant 0.$

Backward direction ($\Leftarrow$): Suppose $\frac{1}{1+f(x)} + \frac{1}{1+f(-x)} - 1 = 0$ for all $x \geqslant 0$. Then for all $a \geqslant 0$:

$\int_{0}^{a} \left[ \frac{1}{1+f(x)} + \frac{1}{1+f(-x)} - 1 \right] dx = 0,$

so $\int_{0}^{a} \left[ \frac{1}{1+f(x)} + \frac{1}{1+f(-x)} \right] dx = a$. Reversing the substitution $u = -x$ on the second integral:

$\int_{0}^{a} \frac{1}{1+f(-x)} dx = \int_{-a}^{0} \frac{1}{1+f(x)} dx.$

Therefore $\int_{-a}^{a} \frac{1}{1+f(x)} dx = a$.

Showing the equivalence with $f(x)f(-x) = 1$: Starting from $\frac{1}{1+f(x)} + \frac{1}{1+f(-x)} = 1$, multiply both sides by $(1+f(x))(1+f(-x))$:

$(1+f(-x)) + (1+f(x)) = (1+f(x))(1+f(-x)).$

Expanding the right side:

$2 + f(x) + f(-x) = 1 + f(x) + f(-x) + f(x)f(-x).$

Cancelling $f(x) + f(-x)$ from both sides:

$2 = 1 + f(x)f(-x),$

hence $f(x)f(-x) = 1$ for all $x$.

Part (iii)

Let $f$ be continuous with $f(x) \geqslant 0$ and $f(x)f(-x) = 1$ for all $x$, and let $h$ be continuous with $h(x) = h(-x)$ for all $x$. We show $\int_{-a}^{a} \frac{h(x)}{1+f(x)} dx = \int_{0}^{a} h(x) dx$.

Split the integral:

$\int_{-a}^{a} \frac{h(x)}{1+f(x)} dx = \int_{-a}^{0} \frac{h(x)}{1+f(x)} dx + \int_{0}^{a} \frac{h(x)}{1+f(x)} dx.$

In the first integral, substitute $u = -x$:

$\int_{-a}^{0} \frac{h(x)}{1+f(x)} dx = \int_{a}^{0} \frac{h(-u)}{1+f(-u)} (-du) = \int_{0}^{a} \frac{h(-u)}{1+f(-u)} du.$

Since $h(-u) = h(u)$, this equals $\int_{0}^{a} \frac{h(u)}{1+f(-u)} du$. Renaming $u$ to $x$ and combining:

$\int_{-a}^{a} \frac{h(x)}{1+f(x)} dx = \int_{0}^{a} h(x) \left[ \frac{1}{1+f(-x)} + \frac{1}{1+f(x)} \right] dx.$

By Part (ii), since $f(x)f(-x) = 1$ for all $x$ (and $f(x) \geqslant 0$), we have $\frac{1}{1+f(-x)} + \frac{1}{1+f(x)} = 1$ for all $x \geqslant 0$. Therefore:

$\int_{-a}^{a} \frac{h(x)}{1+f(x)} dx = \int_{0}^{a} h(x) \cdot 1 \, dx = \int_{0}^{a} h(x) dx.$

Part (iv)

We evaluate $\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{e^{-x} \cos x}{\cosh x} dx$.

First, rewrite the integrand using $\cosh x = \frac{e^{x} + e^{-x}}{2}$:

$\frac{e^{-x} \cos x}{\cosh x} = \frac{e^{-x} \cos x}{\frac{e^{x} + e^{-x}}{2}} = \frac{2e^{-x} \cos x}{e^{x} + e^{-x}}.$

Multiplying numerator and denominator by $e^{x}$:

$\frac{2e^{-x} \cos x}{e^{x} + e^{-x}} \cdot \frac{e^{x}}{e^{x}} = \frac{2 \cos x}{e^{2x} + 1} = \frac{2\cos x}{1 + e^{2x}}.$

We identify $h(x) = 2\cos x$ and $f(x) = e^{2x}$, and verify the conditions of Part (iii):

• $f(x) = e^{2x} > 0$ for all $x$, so $f(x) \geqslant 0$.
• $f(x)f(-x) = e^{2x} \cdot e^{-2x} = 1$ for all $x$.
• $h(x) = 2\cos x$ is continuous and $h(-x) = 2\cos(-x) = 2\cos x = h(x)$, so $h$ is even.

All conditions are satisfied. Applying Part (iii) with $a = \frac{\pi}{2}$:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2\cos x}{1 + e^{2x}} dx = \int_{0}^{\frac{\pi}{2}} 2\cos x \, dx = 2[\sin x]_{0}^{\frac{\pi}{2}} = 2(1 - 0) = 2.$

Therefore:

$\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{e^{-x} \cos x}{\cosh x} dx = 2.$

Examiner Notes

第三成功题目，平均分接近满分的一半。第(i)部分常见错误是积分∫1/(1+eˣ)dx直接得到ln(1+eˣ)而非正确简化，以及推导步骤不充分。第(ii)部分最难，关键错误警示：∫₋ₐᵃg(x)dx=0不能直接推出g(x)=0（g(x)=sin x是反例），必须使用微积分基本定理。第(iv)部分必须使用前面部分的结果（题目要求”hence”），否则不给分。

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Question 6

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

6 (i) Show that when $\alpha$ is small, $\cos(\theta + \alpha) - \cos \theta \approx -\alpha \sin \theta - \frac{1}{2} \alpha^2 \cos \theta$.

Find the limit as $\alpha \to 0$ of $\frac{\sin(\theta + \alpha) - \sin \theta}{\cos(\theta + \alpha) - \cos \theta} \qquad \text{(*)}$

in the case $\sin \theta \neq 0$.

In the case $\sin \theta = 0$, what happens to the value of expression ($*$) when $\alpha \to 0$?

(ii) A circle $C_1$ of radius $a$ rolls without slipping in an anti-clockwise direction on a fixed circle $C_2$ with centre at the origin $O$ and radius $(n - 1)a$, where $n$ is an integer greater than 2. The point $P$ is fixed on $C_1$. Initially the centre of $C_1$ is at $(na, 0)$ and $P$ is at $((n + 1)a, 0)$.

(a) Let $Q$ be the point of contact of $C_1$ and $C_2$ at any time in the rolling motion. Show that when $OQ$ makes an angle $\theta$, measured anticlockwise, with the positive $x$-axis, the $x$-coordinate of $P$ is $x(\theta) = a(n \cos \theta + \cos n\theta)$, and find the corresponding expression for the $y$-coordinate, $y(\theta)$, of $P$.

(b) Find the values of $\theta$ for which the distance $OP$ is $(n - 1)a$.

(c) Let $\theta_0 = \frac{1}{n - 1} \pi$. Find the limit as $\alpha \to 0$ of $\frac{y(\theta_0 + \alpha) - y(\theta_0)}{x(\theta_0 + \alpha) - x(\theta_0)}.$

Hence show that, at the point $(x(\theta_0), y(\theta_0))$, the tangent to the curve traced out by $P$ is parallel to $OP$.

Hint

解题思路概述： 本题分为两部分：(i) 小量近似与三角函数极限；(ii) 滚动圆的参数方程（外摆线），涉及几何推理、代数化简和极限计算。

Part (i) 第一问（Show that）： 利用展开 cos(θ + α) = cos θ cos α - sin θ sin α，取小量近似 cos α ≈ 1 - α²/2，sin α ≈ α： cos(θ + α) - cos θ ≈ cos θ(1 - α²/2) - sin θ · α - cos θ = -α sin θ - (α²/2) cos θ

Part (i) 第二问（sin θ ≠ 0 的情况）： 类似地，sin(θ + α) - sin θ ≈ α cos θ - (α²/2) sin θ。 lim_{α→0} [sin(θ+α)-sinθ]/[cos(θ+α)-cosθ] = lim_{α→0} [α cos θ - (α²/2) sin θ]/[-α sin θ - (α²/2) cos θ] = lim_{α→0} [cos θ - (α/2) sin θ]/[-sin θ - (α/2) cos θ] = -cot θ

（替代方法：L’Hôpital 法则直接求导得 cos(θ+α)/(-sin(θ+α)) → -cot θ）

Part (i) 第三问（sin θ = 0 的情况）： 当 sin θ = 0 时，分子的 α cos θ 项不为零，分母 -α sin θ 为零： lim_{α→0} (α cos θ)/(-(α²/2) cos θ) = lim_{α→0} (-2)/α 因此当 α → +0 时趋向 -∞，当 α → -0 时趋向 +∞。

Part (ii)(a)： C₂ 上初始接触点 Q₀，当 OQ 转过角度 θ 时，弧 Q₀Q 在 C₂ 上的长度为 (n-1)aθ。由于无滑动滚动，C₁ 上滚过的弧长也等于 (n-1)aθ。在 C₁ 上这对应角度 (n-1)aθ/a = (n-1)θ。

C₁ 的中心 T 的位置为 T = (na cos θ, na sin θ)。P 相对于 T 的方向角为 θ + (n-1)θ = nθ。 x(θ) = na cos θ + a cos nθ = a(n cos θ + cos nθ) y(θ) = na sin θ + a sin nθ = a(n sin θ + sin nθ)

Part (ii)(b)： OP = (n-1)a 要求： (n cos θ + cos nθ)² + (n sin θ + sin nθ)² = (n-1)² 展开得 n² + 1 + 2n cos(n-1)θ = n² - 2n + 1，即 cos(n-1)θ = -1。

因此 (n-1)θ = (2r+1)π，r = 0, 1, 2, … θ = (2r+1)π/(n-1)，r = 0, 1, 2, …

Part (ii)(c)： 设 θ₀ = π/(n-1)，需要求： lim_{α→0} [y(θ₀+α) - y(θ₀)]/[x(θ₀+α) - x(θ₀)]

利用小量展开，分子和分母分别化为： [cos θ₀ + cos nθ₀ - (α/2)(sin θ₀ + n sin nθ₀)]/[-(sin θ₀ + sin nθ₀) - (α/2)(cos θ₀ + n cos nθ₀)]

关键观察：因为 (n-1)θ₀ = π，利用和差化积：

• cos θ₀ + cos nθ₀ = 2cos((n+1)θ₀/2)cos((n-1)θ₀/2) = 2cos((n+1)θ₀/2)cos(π/2) = 0
• sin θ₀ + sin nθ₀ = 2sin((n+1)θ₀/2)cos((n-1)θ₀/2) = 0

利用加法公式 cos nθ₀ = cos((n-1)θ₀ + θ₀) = -cos θ₀，sin nθ₀ = -sin θ₀：

• sin θ₀ + n sin nθ₀ = (1-n) sin θ₀
• cos θ₀ + n cos nθ₀ = (1-n) cos θ₀

因此极限为：[(1-n) sin θ₀]/[(1-n) cos θ₀] = tan θ₀

这表明在点 (x(θ₀), y(θ₀)) 处，曲线的切线与 OP 平行。

常见错误（来自 Examiner Report）：

• Part (i)：除以一个未必非零的量。
• Part (ii)：缺少图示或图示过小；未说明 x(θ) 中第二项 cos nθ 的来源。
• Part (ii)(b)：得出 cos(n-1)θ = -1 后忘记考虑周期性。
• Part (ii)(c)：错误地将极限当作 0/0 直接计算 cotangent。

Model Solution

Part (i)

Show that $\cos(\theta + \alpha) - \cos\theta \approx -\alpha\sin\theta - \frac{1}{2}\alpha^2\cos\theta$ when $\alpha$ is small.

Using the addition formula $\cos(\theta + \alpha) = \cos\theta\cos\alpha - \sin\theta\sin\alpha$, and the small-angle approximations $\cos\alpha \approx 1 - \frac{\alpha^2}{2}$, $\sin\alpha \approx \alpha$:

$\cos(\theta + \alpha) - \cos\theta \approx \cos\theta\left(1 - \frac{\alpha^2}{2}\right) - \sin\theta \cdot \alpha - \cos\theta = -\alpha\sin\theta - \frac{1}{2}\alpha^2\cos\theta.$

Find the limit when $\sin\theta \neq 0$.

Similarly, $\sin(\theta + \alpha) = \sin\theta\cos\alpha + \cos\theta\sin\alpha$, so:

$\sin(\theta + \alpha) - \sin\theta \approx \sin\theta\left(1 - \frac{\alpha^2}{2}\right) + \cos\theta \cdot \alpha - \sin\theta = \alpha\cos\theta - \frac{1}{2}\alpha^2\sin\theta.$

Therefore:

$\frac{\sin(\theta+\alpha) - \sin\theta}{\cos(\theta+\alpha) - \cos\theta} \approx \frac{\alpha\cos\theta - \frac{1}{2}\alpha^2\sin\theta}{-\alpha\sin\theta - \frac{1}{2}\alpha^2\cos\theta}.$

Dividing numerator and denominator by $\alpha$ (valid since $\alpha \neq 0$):

$\frac{\cos\theta - \frac{\alpha}{2}\sin\theta}{-\sin\theta - \frac{\alpha}{2}\cos\theta}.$

Taking $\alpha \to 0$:

$\lim_{\alpha \to 0} \frac{\sin(\theta+\alpha) - \sin\theta}{\cos(\theta+\alpha) - \cos\theta} = \frac{\cos\theta}{-\sin\theta} = -\cot\theta.$

When $\sin\theta = 0$:

In this case $\cos\theta = \pm 1$ (so $\cos\theta \neq 0$). The numerator is $\alpha\cos\theta - \frac{1}{2}\alpha^2\sin\theta = \alpha\cos\theta$ (first-order in $\alpha$), while the denominator is $-\alpha\sin\theta - \frac{1}{2}\alpha^2\cos\theta = -\frac{1}{2}\alpha^2\cos\theta$ (second-order in $\alpha$). Thus:

$\frac{\sin(\theta+\alpha) - \sin\theta}{\cos(\theta+\alpha) - \cos\theta} \approx \frac{\alpha\cos\theta}{-\frac{1}{2}\alpha^2\cos\theta} = \frac{-2}{\alpha}.$

As $\alpha \to 0^+$, this tends to $-\infty$; as $\alpha \to 0^-$, this tends to $+\infty$. The expression has no finite limit.

Part (ii)(a)

When the rolling circle $C_1$ has rolled so that the contact point $Q$ on $C_2$ is at angle $\theta$ from the positive $x$-axis, the arc from the initial contact point to $Q$ on $C_2$ has length $(n-1)a\theta$. By the no-slip condition, this equals the arc rolled on $C_1$, which subtends angle $\frac{(n-1)a\theta}{a} = (n-1)\theta$ at the centre of $C_1$.

The centre $T$ of $C_1$ is at distance $na$ from $O$ (since the radii of $C_1$ and $C_2$ are $a$ and $(n-1)a$ respectively), so $T = (na\cos\theta, \, na\sin\theta)$.

Initially, $P$ was on the positive $x$-side of $T$. After rolling, $P$ has rotated by angle $(n-1)\theta$ relative to $T$ in the same direction as the rolling. The direction from $T$ to $P$ relative to the positive $x$-axis is $\theta + (n-1)\theta = n\theta$. Therefore:

$x(\theta) = na\cos\theta + a\cos(n\theta) = a(n\cos\theta + \cos n\theta),$

$y(\theta) = na\sin\theta + a\sin(n\theta) = a(n\sin\theta + \sin n\theta).$

Part (ii)(b)

We need $OP = (n-1)a$, i.e., $x(\theta)^2 + y(\theta)^2 = (n-1)^2 a^2$. Dividing by $a^2$:

$(n\cos\theta + \cos n\theta)^2 + (n\sin\theta + \sin n\theta)^2 = (n-1)^2.$

Expanding the left side:

$n^2\cos^2\theta + 2n\cos\theta\cos n\theta + \cos^2 n\theta + n^2\sin^2\theta + 2n\sin\theta\sin n\theta + \sin^2 n\theta$

$= n^2 + 1 + 2n(\cos\theta\cos n\theta + \sin\theta\sin n\theta) = n^2 + 1 + 2n\cos(n-1)\theta.$

Setting this equal to $(n-1)^2 = n^2 - 2n + 1$:

$n^2 + 1 + 2n\cos(n-1)\theta = n^2 - 2n + 1,$

$2n\cos(n-1)\theta = -2n,$

$\cos(n-1)\theta = -1.$

Therefore $(n-1)\theta = (2r+1)\pi$ for integer $r$, giving:

$\theta = \frac{(2r+1)\pi}{n-1}, \qquad r = 0, 1, 2, \ldots$

Part (ii)(c)

Let $\theta_0 = \frac{\pi}{n-1}$, so $(n-1)\theta_0 = \pi$. We first compute $x(\theta_0)$ and $y(\theta_0)$.

Since $(n-1)\theta_0 = \pi$, we have $n\theta_0 = \pi + \theta_0$, so $\cos n\theta_0 = \cos(\pi + \theta_0) = -\cos\theta_0$ and $\sin n\theta_0 = \sin(\pi + \theta_0) = -\sin\theta_0$. Therefore:

$x(\theta_0) = a(n\cos\theta_0 - \cos\theta_0) = a(n-1)\cos\theta_0,$

$y(\theta_0) = a(n\sin\theta_0 - \sin\theta_0) = a(n-1)\sin\theta_0.$

Now we expand $x(\theta_0 + \alpha)$ and $y(\theta_0 + \alpha)$ for small $\alpha$.

Using $\cos(\theta_0 + \alpha) = \cos\theta_0\cos\alpha - \sin\theta_0\sin\alpha$ and $\cos(n\theta_0 + n\alpha) = \cos n\theta_0\cos n\alpha - \sin n\theta_0\sin n\alpha = -\cos\theta_0\cos n\alpha + \sin\theta_0\sin n\alpha$:

$x(\theta_0 + \alpha) = a\bigl[n(\cos\theta_0\cos\alpha - \sin\theta_0\sin\alpha) + (-\cos\theta_0\cos n\alpha + \sin\theta_0\sin n\alpha)\bigr]$

$= a\bigl[\cos\theta_0(n\cos\alpha - \cos n\alpha) - \sin\theta_0(n\sin\alpha - \sin n\alpha)\bigr].$

Similarly, using $\sin(\theta_0 + \alpha) = \sin\theta_0\cos\alpha + \cos\theta_0\sin\alpha$ and $\sin n\theta_0 = -\sin\theta_0$:

$y(\theta_0 + \alpha) = a\bigl[\sin\theta_0(n\cos\alpha - \cos n\alpha) + \cos\theta_0(n\sin\alpha - \sin n\alpha)\bigr].$

The differences from the values at $\theta_0$ are:

$x(\theta_0 + \alpha) - x(\theta_0) = a\bigl[\cos\theta_0(n\cos\alpha - \cos n\alpha - (n-1)) - \sin\theta_0(n\sin\alpha - \sin n\alpha)\bigr],$

$y(\theta_0 + \alpha) - y(\theta_0) = a\bigl[\sin\theta_0(n\cos\alpha - \cos n\alpha - (n-1)) + \cos\theta_0(n\sin\alpha - \sin n\alpha)\bigr].$

For small $\alpha$: $\cos\alpha \approx 1 - \frac{\alpha^2}{2}$, $\sin\alpha \approx \alpha$, $\cos n\alpha \approx 1 - \frac{n^2\alpha^2}{2}$, $\sin n\alpha \approx n\alpha$. So:

$n\cos\alpha - \cos n\alpha - (n-1) \approx n\left(1 - \frac{\alpha^2}{2}\right) - \left(1 - \frac{n^2\alpha^2}{2}\right) - (n-1) = \frac{n(n-1)\alpha^2}{2},$

$n\sin\alpha - \sin n\alpha \approx n\alpha - n\alpha = 0.$

The second expression vanishes to first order; more precisely, $n\sin\alpha - \sin n\alpha = n\alpha - \frac{n\alpha^3}{6} - n\alpha + \frac{n^3\alpha^3}{6} + \cdots = \frac{n(n^2-1)\alpha^3}{6} + \cdots$, which is third-order. So the dominant contributions come from the $\alpha^2$ terms:

$x(\theta_0 + \alpha) - x(\theta_0) \approx a \cdot \cos\theta_0 \cdot \frac{n(n-1)\alpha^2}{2},$

$y(\theta_0 + \alpha) - y(\theta_0) \approx a \cdot \sin\theta_0 \cdot \frac{n(n-1)\alpha^2}{2}.$

Taking the ratio:

$\lim_{\alpha \to 0} \frac{y(\theta_0 + \alpha) - y(\theta_0)}{x(\theta_0 + \alpha) - x(\theta_0)} = \frac{\sin\theta_0}{\cos\theta_0} = \tan\theta_0.$

Showing the tangent is parallel to $OP$:

At $\theta = \theta_0$, the point on the curve is $P = (a(n-1)\cos\theta_0, \, a(n-1)\sin\theta_0)$. The vector $OP$ has direction $(\cos\theta_0, \sin\theta_0)$, so the slope of $OP$ is $\frac{\sin\theta_0}{\cos\theta_0} = \tan\theta_0$.

The limit computed above is the slope of the tangent to the curve at this point, which also equals $\tan\theta_0$. Since both slopes are equal, the tangent to the curve at $(x(\theta_0), y(\theta_0))$ is parallel to $OP$.

Examiner Notes

约半数考生尝试，是得分最低的题目之一（平均分约1/4）。常见问题：图形缺失或过小且遗漏关键细节、未说明参数方程第二项的来源、第(b)部分忘记处理周期性（加2kπ）、第(c)部分错误地对0/0型极限直接求值后声称余切是答案。部分考生能正确识别零点并有效处理三角函数。

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Question 7

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Let n be a vector of unit length in three dimensions. For each vector r, f(r) is defined by f(r) = n × r.

(i) Given that

$\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \text{ and } \mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix},$

show that the $x$-component of f(f(r)) is $-x(b^2 + c^2) + aby + acz$. Show further that

$\text{f(f}(\mathbf{r})) = (\mathbf{n} \cdot \mathbf{r})\mathbf{n} - \mathbf{r}.$

Explain, by means of a diagram, how f(f(r)) is related to n and r.

(ii) Let $R$ be the point with position vector r and $P$ be the point with position vector g(r), where g is defined by

$\text{g}(\mathbf{s}) = \mathbf{s} + \sin \theta \, \text{f}(\mathbf{s}) + (1 - \cos \theta) \, \text{f(f}(\mathbf{s})).$

By considering g(n) and g(r) when r is perpendicular to n, state, with justification, the geometric transformation which maps $R$ onto $P$.

(iii) Let $R$ be the point with position vector r and $Q$ be the point with position vector h(r), where h is defined by

$\text{h}(\mathbf{s}) = -\mathbf{s} - 2 \, \text{f(f}(\mathbf{s})).$

State, with justification, the geometric transformation which maps $R$ onto $Q$.

Hint

解题思路： 本题考察向量叉积的性质、Rodrigues 旋转公式和反射变换。核心恒等式是 n × (n × r) = (n·r)n − r（当 |n|=1 时）。

Part (i)：证明 f(f(r)) = (n·r)n − r 并用图解释

步骤1：计算 f(r) = n × r f(r) = (a, b, c)ᵀ × (x, y, z)ᵀ = (bz − cy, cx − az, ay − bx)ᵀ

步骤2：计算 f(f(r)) 的 x 分量 f(f(r)) 的 x 分量 = b(ay − bx) − c(cx − az) = −x(b² + c²) + aby + acz

步骤3：利用 |n|=1 简化 由于 a² + b² + c² = 1： −x(b² + c²) + aby + acz = −x(a² + b² + c²) + a(ax + by + cz) = −x + a(n·r)

同理 y 分量为 −y + b(n·r)，z 分量为 −z + c(n·r)。

因此 f(f(r)) = −r + (n·r)n = (n·r)n − r ✓

几何解释：r 可分解为平行于 n 的分量 (n·r)n 和垂直于 n 的分量 r − (n·r)n。f(f(r)) = (n·r)n − r = −(r − (n·r)n)，即 r 的垂直分量取反。

Part (ii)：旋转

步骤1：计算 g(n) g(n) = n + sin θ(n × n) + (1 − cos θ)((n·n)n − n) = n + 0 + (1 − cos θ)(n − n) = n → n 方向的向量在变换下不变。

步骤2：计算 g(r)，其中 r ⊥ n 当 r ⊥ n 时，n·r = 0，所以 f(f(r)) = −r。 g(r) = r + sin θ(n × r) + (1 − cos θ)(−r) = r cos θ + sin θ(n × r)

步骤3：判断几何变换 r、n×r 都垂直于 n，且 |r| = |n×r|（因为 |n|=1），两者夹角为 θ。因此 g(r) 是将 r 绕 n 轴逆时针旋转角度 θ。

结论：g 表示绕 n 轴逆时针旋转角度 θ 的旋转变换。

Part (iii)：反射

步骤1：化简 h(s) h(s) = −s − 2f(f(s)) = −s − 2((n·s)n − s) = s − 2(n·s)n

步骤2：验证变换性质

• h(n) = n − 2(n·n)n = n − 2n = −n（n 方向反向）
• 若 r ⊥ n，则 h(r) = r − 2(n·r)n = r（垂直分量不变）
• h(n × r) = n × r − 2(n·(n × r))n = n × r（不变）

结论：h 表示关于过原点且垂直于 n 的平面的反射变换。

常见错误（来自 Examiner Report）：

1. Part (i)：向量叉积计算错误，尤其是符号。
2. Part (ii)：未能正确解释 g 的几何意义。
3. Part (iii)：未能将 h 与反射联系起来。

Model Solution

Part (i)

We compute f(r) = n x r using the determinant formula:

$\mathbf{n} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & b & c \\ x & y & z \end{vmatrix} = \begin{pmatrix} bz - cy \\ cx - az \\ ay - bx \end{pmatrix}$

Now we compute f(f(r)) = n x (f(r)). The $x$-component is:

$b(ay - bx) - c(cx - az) = aby - b^2 x - c^2 x + acz = -x(b^2 + c^2) + aby + acz$

This establishes the first result. Now we use $|\mathbf{n}| = 1$, i.e., $a^2 + b^2 + c^2 = 1$, so $b^2 + c^2 = 1 - a^2$. The $x$-component becomes:

$-x(1 - a^2) + a(by + cz) = -x + a^2 x + aby + acz = -x + a(ax + by + cz) = -x + a(\mathbf{n} \cdot \mathbf{r})$

By identical reasoning (cyclic permutation of components), the $y$-component is $-y + b(\mathbf{n} \cdot \mathbf{r})$ and the $z$-component is $-z + c(\mathbf{n} \cdot \mathbf{r})$. Therefore:

$\text{f(f}(\mathbf{r})) = -\mathbf{r} + (\mathbf{n} \cdot \mathbf{r})\mathbf{n} = (\mathbf{n} \cdot \mathbf{r})\mathbf{n} - \mathbf{r} \qquad \text{(*)}$

Diagram and geometric interpretation:

Any vector r can be decomposed into a component parallel to n and a component perpendicular to n:

$\mathbf{r} = \underbrace{(\mathbf{n} \cdot \mathbf{r})\mathbf{n}}_{\text{component parallel to } \mathbf{n}} + \underbrace{(\mathbf{r} - (\mathbf{n} \cdot \mathbf{r})\mathbf{n})}_{\text{component perpendicular to } \mathbf{n}}$

From (*), f(f(r)) = (n . r)n - r = -(r - (n . r)n). This means f(f(r)) is the vector obtained by negating the perpendicular component of r while leaving the parallel component unchanged. In other words, f(f(r)) is the reflection of r through the line spanned by n (equivalently, the reflection in the direction perpendicular to n).

Diagrammatically, if we draw r, its component $(\mathbf{n} \cdot \mathbf{r})\mathbf{n}$ along n, and the perpendicular residual $\mathbf{r} - (\mathbf{n} \cdot \mathbf{r})\mathbf{n}$, then f(f(r)) has the same component along n but the perpendicular residual is flipped.

Part (ii)

We examine g in two special cases.

Case 1: s = n. Since n x n = 0, we have f(n) = 0 and f(f(n)) = (n . n)n - n = n - n = 0. Therefore:

$\text{g}(\mathbf{n}) = \mathbf{n} + \mathbf{0} + \mathbf{0} = \mathbf{n}$

So points along the n-axis are unchanged by g.

Case 2: s = r where r is perpendicular to n (i.e., n . r = 0).

Since n . r = 0, we get f(f(r)) = -r from (*). Therefore:

$\text{g}(\mathbf{r}) = \mathbf{r} + \sin\theta \, (\mathbf{n} \times \mathbf{r}) + (1 - \cos\theta)(-\mathbf{r}) = \mathbf{r}\cos\theta + \sin\theta \, (\mathbf{n} \times \mathbf{r})$

Since |n| = 1 and r is perpendicular to n, we have |n x r| = |r|. The vectors r and n x r are both perpendicular to n and perpendicular to each other, with equal magnitudes. The expression $\mathbf{r}\cos\theta + (\mathbf{n} \times \mathbf{r})\sin\theta$ is a rotation of r by angle $\theta$ within the plane perpendicular to n, in the direction from r towards n x r (which is the right-hand rule / anticlockwise direction about n).

Since any vector s can be decomposed as $\mathbf{s} = (\mathbf{n} \cdot \mathbf{s})\mathbf{n} + (\mathbf{s} - (\mathbf{n} \cdot \mathbf{s})\mathbf{n})$, and g fixes the parallel component (by Case 1) and rotates the perpendicular component by $\theta$ (by Case 2), the transformation g is a rotation through angle $\theta$ about the axis through the origin in the direction of n, in the anticlockwise sense (right-hand rule). This is precisely the Rodrigues rotation formula.

Part (iii)

We simplify h(s):

$\text{h}(\mathbf{s}) = -\mathbf{s} - 2\,\text{f(f}(\mathbf{s})) = -\mathbf{s} - 2((\mathbf{n} \cdot \mathbf{s})\mathbf{n} - \mathbf{s}) = \mathbf{s} - 2(\mathbf{n} \cdot \mathbf{s})\mathbf{n}$

We verify the key properties:

• h(n) = n - 2(n . n)n = n - 2n = -n. So the component along n is negated.
• If r is perpendicular to n (i.e., n . r = 0), then h(r) = r. So the perpendicular component is unchanged.

Therefore h reflects each vector through the plane through the origin that is perpendicular to n: the component along n is negated, while all components in the plane perpendicular to n are preserved. This is precisely the reflection in the plane through $O$ perpendicular to n.

Examiner Notes

约三分之一考生尝试，略优于Q3和Q6。第(i)部分第一问通常正确，但许多考生忽略了n是单位向量的条件导致第二问推导困难。图形绘制是主要失分点——未能理解三个向量的大小关系和垂直关系。第(ii)(iii)部分少数考生能准确识别并论证变换，但多数虽有大致想法却无法精确定义和论证结论。

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Question 8

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

8 (i) Use De Moivre’s theorem to prove that for any positive integer $k > 1$,

$\sin(k\theta) = \sin \theta \cos^{k-1} \theta \left( k - \binom{k}{3}(\sec^2 \theta - 1) + \binom{k}{5}(\sec^2 \theta - 1)^2 - \cdots \right)$

and find a similar expression for $\cos(k\theta)$.

(ii) Let $\theta = \cos^{-1}(\frac{1}{a})$, where $\theta$ is measured in degrees, and $a$ is an odd integer greater than 1.

Suppose that there is a positive integer $k$ such that $\sin(k\theta) = 0$ and $\sin(m\theta) \neq 0$ for all integers $m$ with $0 < m < k$.

Show that it would be necessary to have $k$ even and $\cos(\frac{1}{2}k\theta) = 0$.

Deduce that $\theta$ is irrational.

(iii) Show that if $\phi = \cot^{-1}(\frac{1}{b})$, where $\phi$ is measured in degrees, and $b$ is an even integer greater than 1, then $\phi$ is irrational.

Hint

解题思路： 本题以 De Moivre 定理为核心工具，通过分析 sin(kθ) 展开式中各项的奇偶性来证明角度的无理性。核心技巧是：将 sec²θ - 1 替换为 tan²θ = (a²-1)，利用 a 为奇数时 a²-1 为偶数这一数论性质，证明展开式中括号内的值恒为奇数（从而非零），导出矛盾。

Part (i)：用 De Moivre 定理展开 sin(kθ) 和 cos(kθ)

由 De Moivre 定理：cos(kθ) + i sin(kθ) = (cos θ + i sin θ)^k

用二项式展开右边，分别取虚部和实部：

sin(kθ) 的展开： sin(kθ) = C(k,1)cos^{k-1}θ sin θ - C(k,3)cos^{k-3}θ sin³θ + C(k,5)cos^{k-5}θ sin⁵θ - …

提取公因子 sin θ cos^{k-1}θ，并用 tan²θ = sec²θ - 1： sin(kθ) = sin θ cos^{k-1}θ [k - C(k,3)(sec²θ - 1) + C(k,5)(sec²θ - 1)² - …]

cos(kθ) 的类似表达式： cos(kθ) = cos^k θ [1 - C(k,2)(sec²θ - 1) + C(k,4)(sec²θ - 1)² - …]

Part (ii)：证明 k 必须为偶数，并推导 θ 为无理数

设 θ = cos⁻¹(1/a)，其中 a 为大于 1 的奇数。

第一步：证明 k 必须为偶数。

假设 k 为奇数。代入 sec θ = a，sin(kθ) = 0 意味着： sin θ · (1/a^{k-1}) [k - C(k,3)(a²-1) + C(k,5)(a²-1)² - … + (-1)^{(k-1)/2}(a²-1)^{(k-1)/2}] = 0

由于 a 为奇数，a² - 1 为偶数。括号内各项中：

• 第一项 k 为奇数
• 其余各项都包含 (a²-1) 的幂次，因此都是偶数 所以括号内 = 奇数 + 偶数之和 = 奇数 ≠ 0。

又 sin θ ≠ 0，1/a^{k-1} ≠ 0。三个因子均非零，矛盾！故 k 必须为偶数。

第二步：证明 cos(kθ/2) = 0。

k 为偶数时，sin(kθ) = 2 sin(kθ/2) cos(kθ/2) = 0。由于 k/2 < k，由题设 sin(kθ/2) ≠ 0，故必有 cos(kθ/2) = 0。

第三步：导出矛盾，证明 θ 为无理数。

设 n = k/2。由 Part (i) 的 cos 展开： cos(nθ) = (1/a^n)[1 - C(n,2)(a²-1) + C(n,4)(a²-1)² - …]

括号内第一项为 1（奇数），其余各项含 (a²-1) 的幂次（偶数），故括号内为奇数 ≠ 0。因此 cos(nθ) ≠ 0，与 cos(nθ) = 0 矛盾！

这说明不存在最小的正整数 k 使得 sin(kθ) = 0。若 θ 为有理数，则 θ = 180p/k，此时 sin(kθ) = sin(180p) = 0，矛盾。故 θ 为无理数。

Part (iii)：证明 φ = cot⁻¹(1/b) 为无理数

设 b 为大于 1 的偶数，φ = cot⁻¹(1/b)，即 tan φ = b。

假设存在最小的奇数 k 使得 sin(kφ) = 0。由 Part (i)： sin(kφ) = sin φ cos^{k-1}φ [k - C(k,3)b² + C(k,5)b⁴ - …] = 0

括号内第一项 k 为奇数，其余各项含 b²（偶数，因为 b 为偶数），故括号内为奇数 ≠ 0。又 sin φ ≠ 0，cos φ ≠ 0，矛盾。故 k 不能为奇数。

若 k 为偶数，同 Part (ii) 论证，cos(kφ/2) = 0。设 n = k/2： cos(nφ) = cos^n φ [1 - C(n,2)b² + C(n,4)b⁴ - …]

括号内为奇数 ≠ 0，cos φ ≠ 0，故 cos(nφ) ≠ 0，矛盾。

因此不存在 k 使得 sin(kφ) = 0，故 φ 为无理数。

常见错误（来自 Examiner Report）：

1. 混淆度与弧度：题目明确说 θ 以度为单位，但有考生错误地引用 π 的无理性来论证。
2. Part (ii) 不完整：大多数考生能代入 sec θ = a 得到展开式，但难以完成”k 必须为偶数”的论证。
3. 忽视结果的意义：部分考生能证明 cos(kθ/2) = 0，但未能认识到它与无理性论证的关系。
4. Part (i) 工作不足：少数考生未给出充分的推导过程。

Model Solution

Part (i)

By De Moivre’s theorem, $\cos(k\theta) + i\sin(k\theta) = (\cos\theta + i\sin\theta)^k$. We expand the right side using the binomial theorem:

$(\cos\theta + i\sin\theta)^k = \sum_{r=0}^{k} \binom{k}{r} \cos^{k-r}\theta \, (i\sin\theta)^r$

Taking the imaginary part to obtain $\sin(k\theta)$, only odd powers of $i\sin\theta$ contribute (since $i^{2m+1}$ is purely imaginary):

$\sin(k\theta) = \binom{k}{1}\cos^{k-1}\theta\sin\theta - \binom{k}{3}\cos^{k-3}\theta\sin^3\theta + \binom{k}{5}\cos^{k-5}\theta\sin^5\theta - \cdots$

We factor out $\sin\theta\cos^{k-1}\theta$ from every term. The general term $\binom{k}{2j+1}\cos^{k-(2j+1)}\theta\sin^{2j+1}\theta$ becomes $\binom{k}{2j+1}\cos^{k-1}\theta\sin\theta \cdot \frac{\sin^{2j}\theta}{\cos^{2j}\theta} = \binom{k}{2j+1}\cos^{k-1}\theta\sin\theta \cdot \tan^{2j}\theta$.

Since $\tan^2\theta = \sec^2\theta - 1$, we obtain:

$\sin(k\theta) = \sin\theta\cos^{k-1}\theta\left(k - \binom{k}{3}(\sec^2\theta - 1) + \binom{k}{5}(\sec^2\theta - 1)^2 - \cdots\right) \qquad \text{(*)}$

where the signs alternate and the powers of $(\sec^2\theta - 1)$ increase by 1 at each step.

Taking the real part to obtain $\cos(k\theta)$, only even powers contribute:

$\cos(k\theta) = \binom{k}{0}\cos^k\theta - \binom{k}{2}\cos^{k-2}\theta\sin^2\theta + \binom{k}{4}\cos^{k-4}\theta\sin^4\theta - \cdots$

Factoring out $\cos^k\theta$ and using $\sin^{2j}\theta/\cos^{2j}\theta = \tan^{2j}\theta = (\sec^2\theta - 1)^j$:

$\cos(k\theta) = \cos^k\theta\left(1 - \binom{k}{2}(\sec^2\theta - 1) + \binom{k}{4}(\sec^2\theta - 1)^2 - \cdots\right) \qquad \text{(**)}$

Part (ii)

Let $\theta = \cos^{-1}(1/a)$ where $a$ is an odd integer greater than 1, so $\sec\theta = a$. Suppose $k$ is the smallest positive integer with $\sin(k\theta) = 0$.

Step 1: $k$ must be even.

Suppose for contradiction that $k$ is odd. Substituting $\sec\theta = a$ into (*):

$\sin(k\theta) = \sin\theta \cdot \frac{1}{a^{k-1}} \left[k - \binom{k}{3}(a^2 - 1) + \binom{k}{5}(a^2 - 1)^2 - \cdots + (-1)^{(k-1)/2}\binom{k}{k}(a^2 - 1)^{(k-1)/2}\right]$

Since $a$ is an odd integer, $a^2 - 1$ is even. In the bracket:

• The first term $k$ is odd (since $k$ is odd).
• Every other term contains a positive power of $(a^2 - 1)$, so each is even.

Therefore the bracket equals (odd) + (sum of evens) = odd, and in particular the bracket is non-zero.

Since $\theta = \cos^{-1}(1/a)$ with $a > 1$, we have $\sin\theta \neq 0$ and $1/a^{k-1} \neq 0$. The product of three non-zero factors is non-zero, contradicting $\sin(k\theta) = 0$. Hence $k$ must be even.

Step 2: $\cos(\tfrac{1}{2}k\theta) = 0$.

Since $k$ is even, write $k = 2n$ where $n$ is a positive integer. Then:

$\sin(k\theta) = \sin(2n\theta) = 2\sin(n\theta)\cos(n\theta) = 0$

Since $n = k/2 < k$, by the minimality of $k$ we have $\sin(n\theta) \neq 0$. Therefore $\cos(n\theta) = 0$, i.e., $\cos(\tfrac{1}{2}k\theta) = 0$.

Step 3: Deduce that $\theta$ is irrational.

We apply (**) with $n = k/2$ and $\sec\theta = a$:

$\cos(n\theta) = \frac{1}{a^n}\left[1 - \binom{n}{2}(a^2 - 1) + \binom{n}{4}(a^2 - 1)^2 - \cdots\right]$

In the bracket, the first term is 1 (odd), and every other term contains a positive power of $(a^2-1)$ (which is even), so each is even. The bracket is (odd) + (sum of evens) = odd, hence non-zero. Since $1/a^n \neq 0$, we conclude $\cos(n\theta) \neq 0$, contradicting Step 2.

This contradiction shows that no such smallest positive integer $k$ can exist. Now suppose $\theta$ is rational, say $\theta = 180p/q$ degrees where $p, q$ are positive integers with $\gcd(p, q) = 1$. Then $\sin(q\theta) = \sin(180p) = 0$, so there exists a positive integer $k \leq q$ with $\sin(k\theta) = 0$. By the argument above, no such $k$ exists. This is a contradiction, so $\theta$ must be irrational.

Part (iii)

Let $\phi = \cot^{-1}(1/b)$ where $b$ is an even integer greater than 1, so $\tan\phi = b$ (equivalently $\sec^2\phi = 1 + b^2$). We show $\phi$ is irrational by the same strategy.

Suppose for contradiction that there exists a smallest positive integer $k$ with $\sin(k\phi) = 0$. We use the expansion (*) with $\sec^2\phi - 1 = \tan^2\phi = b^2$:

$\sin(k\phi) = \sin\phi\cos^{k-1}\phi\left(k - \binom{k}{3}b^2 + \binom{k}{5}b^4 - \cdots\right)$

Since $b$ is even, $b^2$ is even, and every positive power of $b^2$ is even.

Case 1: $k$ is odd. The bracket is $k - (\text{even}) + (\text{even}) - \cdots = k + (\text{even sum})$. Since $k$ is odd, the bracket is odd, hence non-zero. Since $\phi = \cot^{-1}(1/b)$ with $b > 1$, we have $\sin\phi \neq 0$ and $\cos\phi \neq 0$. Three non-zero factors give $\sin(k\phi) \neq 0$, a contradiction.

Case 2: $k$ is even. Write $k = 2n$. As before, $\sin(2n\phi) = 2\sin(n\phi)\cos(n\phi) = 0$ with $\sin(n\phi) \neq 0$ (by minimality of $k$), so $\cos(n\phi) = 0$.

Now apply (**) with $\sec^2\phi - 1 = b^2$:

$\cos(n\phi) = \cos^n\phi\left(1 - \binom{n}{2}b^2 + \binom{n}{4}b^4 - \cdots\right)$

The bracket is $1 + (\text{sum of evens}) = \text{odd} \neq 0$, and $\cos\phi \neq 0$, so $\cos(n\phi) \neq 0$. This contradicts $\cos(n\phi) = 0$.

In both cases we reach a contradiction, so no such $k$ exists. If $\phi$ were rational, say $\phi = 180p/q$ degrees, then $\sin(q\phi) = 0$, giving such a $k$. This is impossible, so $\phi$ is irrational.

Examiner Notes

最不受欢迎的纯数题（略少于Q7），平均分6/20。第(i)部分通常正确但推导过程常不充分。第(ii)部分超过半数考生止步于此，主要困难在于证明k必须为偶数。严重错误警示：部分考生引用π的无理性来证明（题目明确θ以度为单位，π的无理性与此无关）。能完成(ii)的考生通常也能完成(iii)。