STEP3 2016 -- Pure Mathematics

STEP3 2016 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2016  |  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	积分 (Integration)	Standard	三角换元 $x+a=\sqrt{b-a^2}\tan u$, 完全平方配方, 递推关系推导, 数学归纳法
2	坐标几何 (Coordinate Geometry)	Challenging	参数化表示 $(at^2, 2at)$, 隐函数求导求法线方程, 韦达定理 ($q+r=-p, qr=2$), 联立方程求交点
3	微分方程 (Differential Equations)	Challenging	对含指数函数的表达式求导, 因式定理/余式定理, 多项式次数比较, 反证法
4	级数与数列 (Series and Sequences)	Challenging	裂项法（望远镜求和）, 双曲函数定义 $\text{sech}(ry) = \frac{2e^{-ry}}{1+e^{-2ry}}$, 变量替换 $x=e^{-2y}$, 利用偶函数性质处理负下标
5	组合与数论 (Combinatorics and Number Theory)	Hard	二项式展开对称性, 素数整除性论证, 乘积分解 $P_{1,2m+1} = P_{1,m+1} \cdot P_{m+1,2m+1}$, 强归纳法（分奇偶讨论）
6	双曲函数 (Hyperbolic Functions)	Challenging	双曲函数加法公式,反双曲函数,分类讨论,必要充分条件论证
7	复数 (Complex Numbers)	Hard	单位根因式分解,因子定理,复数模的乘积,正多边形几何性质
8	函数方程 (Functional Equations)	Standard	代入-x建立方程组,函数迭代构造循环,代数消元法,验证解

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Question 1

Topic: 积分 (Integration)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Let $I_n = \int_{-\infty}^{\infty} \frac{1}{(x^2 + 2ax + b)^n} \, dx \, ,$ where $a$ and $b$ are constants with $b > a^2$, and $n$ is a positive integer.

(i) By using the substitution $x + a = \sqrt{b - a^2} \tan u$ , or otherwise, show that $I_1 = \frac{\pi}{\sqrt{b - a^2}} \, .$

(ii) Show that $2n(b - a^2) \, I_{n+1} = (2n - 1) \, I_n$ .

(iii) Hence prove by induction that $I_n = \frac{\pi}{2^{2n-2}(b - a^2)^{n-\frac{1}{2}}} \binom{2n-2}{n-1} \, .$

Hint

Part (i) is most simply dealt with by the suggested method, change of variable, and it is worth completing the square in the denominator to simplify the algebra leading to a trivial integral. Part (ii) can either be attempted immediately using integration by parts, starting from $I_n$ and obtaining $\int_{-\infty}^{\infty} \frac{2nx(x+a)}{(x^2+2ax+b)^{n+1}} dx$ and then writing the numerator as $2n(x^2 + 2ax + b) - na(2x + 2a) - 2n(b - a^2)$. Alternatively, use of the same substitution in $I_{n+1}$ as in part (i) leads to the need to integrate $\cos^{2n} u$, which in turn can be written as $\cos^{2n-2} u(1 - \sin^2 u) = \cos^{2n-2} u - \cos^{2n-2} u \sin u \cdot \sin u$, with the second term being susceptible to integration by parts. Part (iii) follows from the previous parts by induction using part (ii) to achieve the inductive step and (i) the base case.

Model Solution

Part (i)

Complete the square in the denominator:

$x^2 + 2ax + b = (x + a)^2 + (b - a^2)$

Let $x + a = \sqrt{b - a^2} \tan u$, so $dx = \sqrt{b - a^2} \sec^2 u \, du$. As $x \to \infty$, $u \to \frac{\pi}{2}$; as $x \to -\infty$, $u \to -\frac{\pi}{2}$.

The denominator becomes:

$(x + a)^2 + (b - a^2) = (b - a^2)\tan^2 u + (b - a^2) = (b - a^2)\sec^2 u$

Therefore:

$I_1 = \int_{-\pi/2}^{\pi/2} \frac{\sqrt{b - a^2} \sec^2 u}{(b - a^2)\sec^2 u} \, du = \frac{1}{\sqrt{b - a^2}} \int_{-\pi/2}^{\pi/2} du = \frac{\pi}{\sqrt{b - a^2}}$

Part (ii)

Let $c^2 = b - a^2$ and substitute $t = x + a$, so $dt = dx$ and $x^2 + 2ax + b = t^2 + c^2$:

$I_n = \int_{-\infty}^{\infty} \frac{dt}{(t^2 + c^2)^n}$

Write $1 = \frac{(t^2 + c^2) - t^2}{t^2 + c^2}$ in the numerator:

$I_n = \int_{-\infty}^{\infty} \frac{t^2 + c^2}{(t^2 + c^2)^{n+1}} \, dt - \int_{-\infty}^{\infty} \frac{t^2}{(t^2 + c^2)^{n+1}} \, dt$

$I_n = I_{n+1} + c^2 I_{n+1} - \int_{-\infty}^{\infty} \frac{t^2}{(t^2 + c^2)^{n+1}} \, dt \qquad \text{(*)}$

Evaluate the remaining integral by parts. Let $u = t$ and $dv = \frac{t}{(t^2 + c^2)^{n+1}} \, dt$, so $du = dt$ and $v = \frac{-1}{2n(t^2 + c^2)^n}$:

$\int_{-\infty}^{\infty} \frac{t^2}{(t^2 + c^2)^{n+1}} \, dt = \left[ \frac{-t}{2n(t^2 + c^2)^n} \right]_{-\infty}^{\infty} + \frac{1}{2n} \int_{-\infty}^{\infty} \frac{dt}{(t^2 + c^2)^n}$

The boundary term vanishes (the numerator is $O(t)$ while the denominator is $O(t^{2n})$ with $n \geq 1$), so:

$\int_{-\infty}^{\infty} \frac{t^2}{(t^2 + c^2)^{n+1}} \, dt = \frac{1}{2n} I_n$

Substituting back into $(*)$:

$I_n = (1 + c^2) I_{n+1} - \frac{1}{2n} I_n = c^2 I_{n+1} + \frac{2n - 1}{2n} I_n$

Rearranging:

$I_n - \frac{2n - 1}{2n} I_n = c^2 I_{n+1}$

$\frac{1}{2n} I_n = c^2 I_{n+1}$

$(2n - 1) I_n = 2n c^2 I_{n+1}$

That is, $2n(b - a^2) I_{n+1} = (2n - 1) I_n$.

Part (iii)

We prove by induction on $n$ that

$I_n = \frac{\pi}{2^{2n-2}(b - a^2)^{n - 1/2}} \binom{2n - 2}{n - 1}$

Base case ($n = 1$): The formula gives $\frac{\pi}{2^0 (b - a^2)^{1/2}} \binom{0}{0} = \frac{\pi}{\sqrt{b - a^2}}$, which matches part (i).

Inductive step: Assume the formula holds for some $n \geq 1$. By part (ii):

$I_{n+1} = \frac{2n - 1}{2n(b - a^2)} I_n = \frac{2n - 1}{2n(b - a^2)} \cdot \frac{\pi}{2^{2n-2}(b - a^2)^{n - 1/2}} \binom{2n - 2}{n - 1}$

Combine the powers of $(b - a^2)$:

$(b - a^2)^{-(n - 1/2) - 1} = (b - a^2)^{-(n + 1/2)}$

Combine the powers of $2$:

$\frac{1}{2n \cdot 2^{2n-2}} = \frac{1}{2^{2n-1} \cdot n}$

So:

$I_{n+1} = \frac{\pi (2n - 1)}{2^{2n-1} \cdot n \cdot (b - a^2)^{n + 1/2}} \binom{2n - 2}{n - 1}$

We need to show this equals $\frac{\pi}{2^{2n}(b - a^2)^{n+1/2}} \binom{2n}{n}$, i.e., we need:

$\frac{(2n - 1)}{2^{2n-1} \cdot n} \binom{2n - 2}{n - 1} = \frac{1}{2^{2n}} \binom{2n}{n}$

Multiply both sides by $2^{2n}$:

$\frac{2(2n - 1)}{n} \binom{2n - 2}{n - 1} = \binom{2n}{n}$

Expand the right side:

$\binom{2n}{n} = \frac{(2n)!}{n! \, n!} = \frac{2n(2n - 1)}{n^2} \cdot \frac{(2n - 2)!}{(n - 1)! \, (n - 1)!} = \frac{2(2n - 1)}{n} \binom{2n - 2}{n - 1}$

This confirms the identity. By induction, the formula holds for all positive integers $n$.

Examiner Notes

这是最常被尝试的题目，超过93%的考生作答，成功率也较高（第二高），平均得分约三分之二。多数考生顺利完成(i)和(iii)，但有各种代数错误，部分考生在(i)中忘记代入积分限。许多考生在(ii)中使用(i)的换元后开始，但面对所得积分时卡住。

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Question 2

Topic: 坐标几何 (Coordinate Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 The distinct points $P(ap^2, 2ap)$, $Q(aq^2, 2aq)$ and $R(ar^2, 2ar)$ lie on the parabola $y^2 = 4ax$, where $a > 0$. The points are such that the normal to the parabola at $Q$ and the normal to the parabola at $R$ both pass through $P$.

(i) Show that $q^2 + qp + 2 = 0$ .

(ii) Show that $QR$ passes through a certain point that is independent of the choice of $P$.

(iii) Let $T$ be the point of intersection of $OP$ and $QR$, where $O$ is the coordinate origin. Show that $T$ lies on a line that is independent of the choice of $P$.

Show further that the distance from the $x$-axis to $T$ is less than $\frac{a}{\sqrt{2}}$ .

Hint

Working parametrically with $x = at^2, y = 2at$ gives a normal as $tx + y = at^3 + 2at$ and imposing that this passes through $(ap^2, 2ap)$ yields $t^2 + tp + 2 = 0$ () which has roots $q$ and $r$, the former giving (i). As a consequence, $r + q = -p$ and $rq = 2$, so that QR, $2x - (r + q)y + 2aqr = 0$, simplifies to $2x + py + 4a = 0$, and thus passes through $(-2a, 0)$ for (ii). T can be shown to be $(-a, \frac{-2a}{p})$, which, of course, lies on $x = -a$, and as () had two real distinct roots, $q$ and $r$, $p^2 - 8 > 0$, which yields $|\frac{-2a}{p}| < \frac{a}{\sqrt{2}}$.

Model Solution

Part (i)

The parabola is $y^2 = 4ax$. Differentiating implicitly: $2y \frac{dy}{dx} = 4a$, so $\frac{dy}{dx} = \frac{2a}{y}$.

At the point $(at^2, 2at)$ on the parabola, the slope of the tangent is $\frac{2a}{2at} = \frac{1}{t}$, so the slope of the normal is $-t$.

The normal at $(at^2, 2at)$ is:

$y - 2at = -t(x - at^2)$

For the normal at $Q(aq^2, 2aq)$ to pass through $P(ap^2, 2ap)$, substitute $x = ap^2$, $y = 2ap$:

$2ap - 2aq = -q(ap^2 - aq^2)$

$2a(p - q) = -qa(p - q)(p + q)$

Since $P$ and $Q$ are distinct points, $p \neq q$, so we may divide both sides by $a(p - q)$:

$2 = -q(p + q)$

$q^2 + qp + 2 = 0$

Part (ii)

Since both $q$ and $r$ satisfy $t^2 + pt + 2 = 0$, by Vieta’s formulas:

$q + r = -p, \qquad qr = 2$

The line $QR$ passes through $Q(aq^2, 2aq)$ and $R(ar^2, 2ar)$. Its slope is:

$\frac{2aq - 2ar}{aq^2 - ar^2} = \frac{2a(q - r)}{a(q - r)(q + r)} = \frac{2}{q + r}$

The equation of $QR$ is:

$y - 2aq = \frac{2}{q + r}(x - aq^2)$

Substituting $q + r = -p$ and $qr = 2$:

$y - 2aq = \frac{2}{-p}(x - aq^2)$

$-p(y - 2aq) = 2(x - aq^2)$

$-py + 2apq = 2x - 2aq^2$

$2x + py = 2aq^2 + 2apq = 2aq(q + p)$

From $q^2 + pq + 2 = 0$, we get $q(q + p) = -2$, so:

$2x + py = 2a(-2) = -4a$

$2x + py + 4a = 0$

This line passes through $(-2a, 0)$ regardless of $P$, since $2(-2a) + p(0) + 4a = 0$.

Part (iii)

The line $OP$ passes through $O(0, 0)$ and $P(ap^2, 2ap)$, so its equation is $y = \frac{2}{p}x$, i.e., $py = 2x$.

To find the intersection $T$ with $QR$: $2x + py + 4a = 0$. Substitute $2x = py$:

$py + py + 4a = 0 \implies 2py = -4a \implies y = \frac{-2a}{p}$

Then $x = \frac{py}{2} = \frac{p \cdot (-2a/p)}{2} = -a$.

So $T = \left(-a, \frac{-2a}{p}\right)$, which lies on the line $x = -a$. This line is independent of $P$.

Showing the distance bound: The equation $t^2 + pt + 2 = 0$ must have two distinct real roots $q$ and $r$ (since $Q$ and $R$ are distinct points). The discriminant must be positive:

$p^2 - 8 > 0 \implies p^2 > 8 \implies |p| > 2\sqrt{2}$

The distance from $T$ to the $x$-axis is:

$\left| \frac{-2a}{p} \right| = \frac{2a}{|p|}$

Since $|p| > 2\sqrt{2}$:

$\frac{2a}{|p|} < \frac{2a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}$

Therefore the distance from $T$ to the $x$-axis is less than $\frac{a}{\sqrt{2}}$.

Examiner Notes

约四分之三的考生尝试此题，成功率中等。多数人使用隐函数求导而非参数方法，能找到切线、法线和弦的方程但不总是因式分解简化。那些能因式分解的考生进展良好，而不能的则在(i)上挣扎。其他弱点包括未能利用 $r+q$ 和 $rq$ 的关系，从而找不到(ii)中的固定点。在最后一部分，对不等式开方后只考虑正数情况并不罕见。

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Question 3

Topic: 微分方程 (Differential Equations)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Given that $\int \frac{x^3 - 2}{(x + 1)^2} \text{e}^x \text{d}x = \frac{P(x)}{Q(x)} \text{e}^x + \text{constant} ,$ where $P(x)$ and $Q(x)$ are polynomials, show that $Q(x)$ has a factor of $x + 1$.

Show also that the degree of $P(x)$ is exactly one more than the degree of $Q(x)$, and find $P(x)$ in the case $Q(x) = x + 1$.

(ii) Show that there are no polynomials $P(x)$ and $Q(x)$ such that $\int \frac{1}{x + 1} \text{e}^x \text{d}x = \frac{P(x)}{Q(x)} \text{e}^x + \text{constant} .$ You need consider only the case when $P(x)$ and $Q(x)$ have no common factors.

Hint

Differentiating, multiplying by denominators and dividing by the exponential function, gives $Q(P + P') - PQ'^2 = (x^3 - 2)Q^2$ which, invoking the factor theorem, gives the first required result. Denoting the degree of P by $p$ and that of Q by $q$ in this expression yields $p + q + 2 = 2q + 3$ and hence the desired result in (i). Furthermore, in the given case, substitution in the same result and postulating $P(x) = ax^2 + bx + c$ yields consistent equations for $a, b$ and $c$ and thus $P(x) = x^2 - 2x$.

For part (ii), commencing as in part (i) demonstrates again that Q has a factor $(x + 1)$ as $Q(P + P') - PQ' = Q^2$. Supposing $Q(x) = (x + 1)^n S(x)$, where $n \ge 2$ and $S(-1) \ne 0$, with $P(-1) \ne 0$ and substituting in the expression already derived leads to a contradiction.

Model Solution

Part (i)

We start by noting that if $\frac{P(x)}{Q(x)}e^x$ differentiates to $\frac{x^3 - 2}{(x+1)^2}e^x$, then by the product rule:

$\frac{d}{dx}\left[\frac{P}{Q}e^x\right] = \left(\frac{P}{Q} + \frac{P'Q - PQ'}{Q^2}\right)e^x = \frac{PQ + P'Q - PQ'}{Q^2}\,e^x$

Setting this equal to the integrand:

$\frac{PQ + P'Q - PQ'}{Q^2} = \frac{x^3 - 2}{(x+1)^2} \qquad \text{(...)}$

Cross-multiplying:

$(x^3 - 2)\,Q^2 = (x+1)^2\bigl[PQ + P'Q - PQ'\bigr] = (x+1)^2\,Q\bigl[P + P'\bigr] - (x+1)^2\,PQ' \qquad \text{(...)}$

To show $Q$ has a factor of $x + 1$: substitute $x = -1$ into the equation $(x^3 - 2)Q^2 = Q^2(P + P') - QQ'P$ (which follows from cross-multiplying the original identity differently — multiply both sides by $Q^2(x+1)^2$ and divide by $(x+1)^2$):

$(x^3 - 2)Q = Q(P + P') - Q'P \qquad \text{(...)}$

(This comes from dividing both sides of $(x^3-2)Q^2 = Q^2(P+P') - QQ'P$ by $Q$, noting $Q \not\equiv 0$.)

At $x = -1$:

$(-1 - 2)\,Q(-1) = Q(-1)\bigl[P(-1) + P'(-1)\bigr] - Q'(-1)\,P(-1)$

$-3\,Q(-1) = Q(-1)\bigl[P(-1) + P'(-1)\bigr] - Q'(-1)\,P(-1)$

This is a relation between values at $x = -1$ and does not immediately force $Q(-1) = 0$. Instead, we use equation (**) more carefully.

From $(x^3 - 2)Q = Q(P + P') - Q'P$, rearrange:

$Q\bigl[(x^3 - 2) - P - P'\bigr] = -Q'P \qquad \text{(...*)}$

If $Q(-1) \neq 0$, then at $x = -1$:

$Q(-1)\bigl[-3 - P(-1) - P'(-1)\bigr] = -Q'(-1)\,P(-1)$

This is just a constraint on the values — it does not lead to a contradiction on its own. However, we can establish the result by a different route. Differentiate the given identity directly.

Suppose $\frac{P(x)}{Q(x)}e^x$ has derivative $\frac{x^3 - 2}{(x+1)^2}e^x$. Write $Q(x) = (x+1)^n S(x)$ where $S(-1) \neq 0$ and $n \geq 0$. We want to show $n \geq 1$.

From the equation $(x^3 - 2)Q = Q(P + P') - Q'P$:

The left side has a zero of order $n$ at $x = -1$ (from the factor $Q$).

On the right side:

• $Q(P + P')$ has a zero of order $\geq n$ at $x = -1$.
• $Q'P$: since $Q = (x+1)^n S$, we have $Q' = n(x+1)^{n-1}S + (x+1)^n S'$, so $Q'$ has a zero of order $n - 1$ at $x = -1$ (assuming $n \geq 1$; if $n = 0$, $Q' = S'$ has no guaranteed zero).

If $n = 0$: the left side is $(x^3 - 2)S(x)$, which at $x = -1$ gives $-3S(-1) \neq 0$. The right side is $S(P + P') - S'P$. At $x = -1$: $S(-1)[P(-1) + P'(-1)] - S'(-1)P(-1)$. Setting equal: $-3S(-1) = S(-1)[P(-1) + P'(-1)] - S'(-1)P(-1)$. This can be satisfied, so we need a different argument.

We use the degree argument instead. From $(x^3 - 2)Q = Q(P + P') - Q'P$, let $\deg P = p$, $\deg Q = q$. The left side has degree $q + 3$. On the right side, $Q(P + P')$ has degree $q + p$ (the leading terms of $P$ and $P'$ do not cancel since they have different degrees), and $Q'P$ has degree $(q - 1) + p = q + p - 1$. So the right side has degree $q + p$.

Therefore $q + 3 = q + p$, giving $p = 3$.

Now, from equation (*), cross-multiplying the original identity $\frac{PQ + P'Q - PQ'}{Q^2} = \frac{x^3 - 2}{(x+1)^2}$:

$(PQ + P'Q - PQ')(x+1)^2 = (x^3 - 2)Q^2$

At $x = -1$: $0 = (-3)Q(-1)^2$, so $Q(-1) = 0$.

This means $Q(-1) = 0$, so $(x + 1)$ divides $Q(x)$.

Showing $\deg P = \deg Q + 1$: From the degree analysis above, $p = 3$ regardless of $q$. Wait — let me redo this more carefully from the cross-multiplied form.

From $(PQ + P'Q - PQ')(x+1)^2 = (x^3 - 2)Q^2$:

• LHS degree: $\max(\deg(PQ), \deg(P'Q), \deg(PQ')) + 2 = \max(p+q, p+q-1, p+q-1) + 2 = p + q + 2$.
• RHS degree: $3 + 2q$.

So $p + q + 2 = 2q + 3$, giving $p = q + 1$.

Finding $P(x)$ when $Q(x) = x + 1$: Here $q = 1$, so $p = 2$. Write $P(x) = ax^2 + bx + c$.

From equation (**): $(x^3 - 2)(x+1) = (x+1)(P + P') - P$

where $P' = 2ax + b$, so $P + P' = ax^2 + (2a + b)x + (b + c)$.

Left side: $x^4 + x^3 - 2x - 2$.

Right side: $(x+1)[ax^2 + (2a+b)x + (b+c)] - (ax^2 + bx + c)$.

Expanding $(x+1)[ax^2 + (2a+b)x + (b+c)]$:

$= ax^3 + (2a+b)x^2 + (b+c)x + ax^2 + (2a+b)x + (b+c)$ $= ax^3 + (3a+b)x^2 + (2a+2b+c)x + (b+c)$

Subtracting $P = ax^2 + bx + c$:

$ax^3 + (3a+b)x^2 + (2a+2b+c)x + (b+c) - ax^2 - bx - c$ $= ax^3 + (2a+b)x^2 + (2a+b+c)x + b$

Matching coefficients with $x^4 + x^3 - 2x - 2$:

• $x^4$: $0 = 1$??

This is a contradiction, so let me re-examine. The issue is that the equation (**) came from dividing by $Q$, but I need to use the original cross-multiplied form.

From $(PQ + P'Q - PQ')(x+1)^2 = (x^3 - 2)Q^2$ with $Q = x + 1$:

LHS: $[(x+1)P + (x+1)P' - P](x+1)^2$

$= [xP + P + xP' + P' - P](x+1)^2$

$= [xP + xP' + P'](x+1)^2$

$= [x(P + P') + P'](x+1)^2$

With $P = ax^2 + bx + c$:

$P + P' = ax^2 + (2a+b)x + (b+c)$

$x(P + P') = ax^3 + (2a+b)x^2 + (b+c)x$

$P' = 2ax + b$

$x(P+P') + P' = ax^3 + (2a+b)x^2 + (b+c+2a)x + b$

Multiply by $(x+1)^2 = x^2 + 2x + 1$:

$[ax^3 + (2a+b)x^2 + (2a+b+c)x + b](x^2 + 2x + 1)$

This is getting unwieldy. Let me use a cleaner approach.

Cleaner approach for finding $P$: Since $\frac{d}{dx}\left[\frac{P}{Q}e^x\right] = \frac{x^3-2}{(x+1)^2}e^x$ with $Q = x+1$, we need:

$\frac{P}{x+1} + \frac{P'(x+1) - P}{(x+1)^2} = \frac{x^3 - 2}{(x+1)^2}$

Multiply through by $(x+1)^2$:

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

$xP + P'x + P' = x^3 - 2$

$x(P + P') + P' = x^3 - 2 \qquad \text{(***)}$

With $P = ax^2 + bx + c$, $P' = 2ax + b$:

$P + P' = ax^2 + (2a+b)x + (b+c)$

$x(P+P') = ax^3 + (2a+b)x^2 + (b+c)x$

$x(P+P') + P' = ax^3 + (2a+b)x^2 + (b+c+2a)x + b$

Matching with $x^3 - 2$:

• $x^3$: $a = 1$
• $x^2$: $2a + b = 0 \Rightarrow b = -2$
• $x^1$: $b + c + 2a = 0 \Rightarrow -2 + c + 2 = 0 \Rightarrow c = 0$
• $x^0$: $b = -2$ ✓

So $P(x) = x^2 - 2x$.

Verification: $\frac{d}{dx}\left[\frac{x^2 - 2x}{x+1}e^x\right]$.

Write $\frac{x^2 - 2x}{x+1} = x - 3 + \frac{3}{x+1}$ (by polynomial division). Then:

$\frac{d}{dx}\left[\left(x - 3 + \frac{3}{x+1}\right)e^x\right] = \left(1 - \frac{3}{(x+1)^2} + x - 3 + \frac{3}{x+1}\right)e^x$

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

$= \frac{(x-2)(x+1)^2 + 3(x+1) - 3}{(x+1)^2}\,e^x$

Numerator: $(x-2)(x^2+2x+1) + 3x + 3 - 3 = x^3 + 2x^2 + x - 2x^2 - 4x - 2 + 3x = x^3 - 2$. ✓

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Part (ii)

Suppose for contradiction that $\int \frac{1}{x+1}e^x\,dx = \frac{P}{Q}e^x + C$ where $P, Q$ are polynomials with no common factors.

Differentiating and clearing denominators as before:

$P(x+1) + P'(x+1) - P = Q \qquad \text{from } \frac{P}{Q} + \frac{P'Q - PQ'}{Q^2} = \frac{1}{x+1}$

More precisely, $\frac{PQ + P'Q - PQ'}{Q^2} = \frac{1}{x+1}$, so:

$(PQ + P'Q - PQ')(x+1) = Q^2 \qquad \text{(...)}$

At $x = -1$: $0 = Q(-1)^2$, so $Q(-1) = 0$. Write $Q(x) = (x+1)^n S(x)$ with $S(-1) \neq 0$, $n \geq 1$.

Since $P$ and $Q$ share no common factors, $P(-1) \neq 0$.

We compute the order of vanishing of each term in $PQ + P'Q - PQ'$ at $x = -1$:

• $PQ = P \cdot (x+1)^n S$ vanishes to order $n$.
• $P'Q = P' \cdot (x+1)^n S$ vanishes to order $\geq n$.
• $PQ' = P\bigl[n(x+1)^{n-1}S + (x+1)^n S'\bigr]$: since $P(-1) \neq 0$, this vanishes to order exactly $n - 1$.

Therefore $PQ + P'Q - PQ'$ vanishes to order exactly $n - 1$ (the $PQ'$ term dominates).

The left side of (*), $(PQ + P'Q - PQ')(x+1)$, vanishes to order $n$.

The right side, $Q^2 = (x+1)^{2n}S^2$, vanishes to order $2n$.

For the equation to hold: $n = 2n$, giving $n = 0$. But $n \geq 1$, a contradiction.

Therefore no such polynomials $P$ and $Q$ exist.

Examiner Notes

比 Q2 稍少人尝试但成功率略高。两部分中，考生成功将右侧表达式的导数等于被积函数（消去指数函数）。第一部分中，许多人求出了给定 $Q(x)$ 情况下的 $P(x)$，但证明 $\deg P$ 比 $\deg Q$ 大一以及(ii)中不存在这样的多项式时出现许多不合逻辑的步骤。许多人如果乘开分母并使用余式/因式定理会更好，而非尝试基于有理表达式次数的论证。

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Question 4

Topic: 级数与数列 (Series and Sequences)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 (i) By considering $\frac{1}{1 + x^r} - \frac{1}{1 + x^{r+1}}$ for $|x| \neq 1$, simplify $\sum_{r=1}^{N} \frac{x^r}{(1 + x^r)(1 + x^{r+1})} .$ Show that, for $|x| < 1$, $\sum_{r=1}^{\infty} \frac{x^r}{(1 + x^r)(1 + x^{r+1})} = \frac{x}{1 - x^2} .$ (ii) Deduce that $\sum_{r=1}^{\infty} \text{sech}(ry) \text{sech}((r + 1)y) = 2\text{e}^{-y} \text{cosech}(2y)$ for $y > 0$.

Hence simplify $\sum_{r=-\infty}^{\infty} \text{sech}(ry) \text{sech}((r + 1)y) ,$ for $y > 0$.

Hint

The considered expression equates to $\frac{(x-1)x^r}{(1+x^r)(1+x^{r+1})}$ and so, by the method of differences, $\sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac{1}{(x-1)} \left[ \frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right]$, and letting $N \to \infty$, the desired result is obtained. Writing $\text{sech}(ry)$ as $\frac{2e^{-ry}}{1+e^{-2ry}}$ and similarly $\text{sech}((r + 1)y)$, the result of part (i) with $x = e^{-2y}$ can be used to obtain the result. Care needs to be taken to write $\sum_{r=-\infty}^{\infty} \text{sech}(ry) \text{sech}((r + 1)y)$ as $2[\sum_{r=1}^{\infty} \text{sech}(ry) \text{sech}((r + 1)y) + \text{sech } y]$ which with the previous deduction of (ii) can be simplified to $2 \text{ cosech } y$.

Model Solution

Part (i)

Consider the expression suggested in the problem:

$\frac{1}{1 + x^r} - \frac{1}{1 + x^{r+1}}$

Combining over a common denominator:

$= \frac{(1 + x^{r+1}) - (1 + x^r)}{(1 + x^r)(1 + x^{r+1})} = \frac{x^{r+1} - x^r}{(1 + x^r)(1 + x^{r+1})} = \frac{x^r(x - 1)}{(1 + x^r)(1 + x^{r+1})}$

Therefore:

$\frac{x^r}{(1 + x^r)(1 + x^{r+1})} = \frac{1}{x - 1}\left(\frac{1}{1 + x^r} - \frac{1}{1 + x^{r+1}}\right)$

Summing from $r = 1$ to $N$ (telescoping):

$\sum_{r=1}^{N} \frac{x^r}{(1 + x^r)(1 + x^{r+1})} = \frac{1}{x - 1}\sum_{r=1}^{N}\left(\frac{1}{1 + x^r} - \frac{1}{1 + x^{r+1}}\right)$

$= \frac{1}{x - 1}\left(\frac{1}{1 + x} - \frac{1}{1 + x^{N+1}}\right)$

For $|x| < 1$: as $N \to \infty$, $x^{N+1} \to 0$, so $\frac{1}{1 + x^{N+1}} \to 1$.

$\sum_{r=1}^{\infty} \frac{x^r}{(1 + x^r)(1 + x^{r+1})} = \frac{1}{x - 1}\left(\frac{1}{1 + x} - 1\right) = \frac{1}{x - 1} \cdot \frac{1 - (1 + x)}{1 + x} = \frac{1}{x - 1} \cdot \frac{-x}{1 + x}$

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

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Part (ii)

Using the definition $\cosh \theta = \frac{e^\theta + e^{-\theta}}{2}$, we write:

$\text{sech}(ry) = \frac{2}{e^{ry} + e^{-ry}} = \frac{2e^{-ry}}{1 + e^{-2ry}}$

$\text{sech}((r+1)y) = \frac{2e^{-(r+1)y}}{1 + e^{-2(r+1)y}}$

Therefore:

$\text{sech}(ry)\,\text{sech}((r+1)y) = \frac{4e^{-(2r+1)y}}{(1 + e^{-2ry})(1 + e^{-2(r+1)y})}$

Set $x = e^{-2y}$. Since $y > 0$, we have $0 < x < 1$, so $|x| < 1$. Then:

$e^{-2ry} = x^r, \qquad e^{-(2r+1)y} = e^{-y} \cdot x^r$

$\text{sech}(ry)\,\text{sech}((r+1)y) = \frac{4e^{-y}\,x^r}{(1 + x^r)(1 + x^{r+1})}$

Summing from $r = 1$ to $\infty$ and using part (i):

$\sum_{r=1}^{\infty} \text{sech}(ry)\,\text{sech}((r+1)y) = 4e^{-y}\sum_{r=1}^{\infty} \frac{x^r}{(1 + x^r)(1 + x^{r+1})} = 4e^{-y} \cdot \frac{x}{1 - x^2}$

Substituting $x = e^{-2y}$:

$= \frac{4e^{-y} \cdot e^{-2y}}{1 - e^{-4y}} = \frac{4e^{-3y}}{1 - e^{-4y}}$

Multiply numerator and denominator by $e^{4y}$:

$= \frac{4e^{y}}{e^{4y} - 1} = \frac{4e^{y}}{(e^{2y} - 1)(e^{2y} + 1)}$

Now we express the right side $2e^{-y}\,\text{cosech}(2y)$ in the same form:

$2e^{-y}\,\text{cosech}(2y) = \frac{2e^{-y}}{\sinh(2y)} = \frac{2e^{-y}}{\frac{e^{2y} - e^{-2y}}{2}} = \frac{4e^{-y}}{e^{2y} - e^{-2y}}$

Multiply numerator and denominator by $e^{2y}$:

$= \frac{4e^{y}}{e^{4y} - 1}$

These are identical, so:

$\sum_{r=1}^{\infty} \text{sech}(ry)\,\text{sech}((r+1)y) = 2e^{-y}\,\text{cosech}(2y)$

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Hence result:

We evaluate $\sum_{r=-\infty}^{\infty} \text{sech}(ry)\,\text{sech}((r+1)y)$.

Since $\text{sech}$ is an even function ($\text{sech}(-\theta) = \text{sech}(\theta)$):

For $r \leq -2$, set $k = -(r+1) \geq 1$:

$\text{sech}(ry)\,\text{sech}((r+1)y) = \text{sech}((k+1)y)\,\text{sech}(ky)$

So the terms for $r = -2, -3, -4, \ldots$ match the terms for $r = 1, 2, 3, \ldots$.

The remaining terms are $r = -1$ and $r = 0$:

• $r = -1$: $\text{sech}(-y)\,\text{sech}(0) = \text{sech}(y) \cdot 1 = \text{sech}(y)$
• $r = 0$: $\text{sech}(0)\,\text{sech}(y) = 1 \cdot \text{sech}(y) = \text{sech}(y)$

Therefore:

$\sum_{r=-\infty}^{\infty} \text{sech}(ry)\,\text{sech}((r+1)y) = 2\sum_{r=1}^{\infty} \text{sech}(ry)\,\text{sech}((r+1)y) + 2\,\text{sech}(y)$

$= 2 \cdot 2e^{-y}\,\text{cosech}(2y) + 2\,\text{sech}(y)$

$= 4e^{-y}\,\text{cosech}(2y) + 2\,\text{sech}(y)$

We simplify. Using $\text{cosech}(2y) = \frac{1}{\sinh(2y)} = \frac{1}{2\sinh y\,\cosh y}$:

$4e^{-y}\,\text{cosech}(2y) = \frac{4e^{-y}}{2\sinh y\,\cosh y} = \frac{2e^{-y}}{\sinh y\,\cosh y}$

And $2\,\text{sech}(y) = \frac{2}{\cosh y}$.

$\sum = \frac{2e^{-y}}{\sinh y\,\cosh y} + \frac{2}{\cosh y} = \frac{2e^{-y} + 2\sinh y}{\sinh y\,\cosh y}$

Now simplify $2e^{-y} + 2\sinh y$:

$2e^{-y} + 2 \cdot \frac{e^y - e^{-y}}{2} = 2e^{-y} + e^y - e^{-y} = e^y + e^{-y} = 2\cosh y$

Therefore:

$\sum_{r=-\infty}^{\infty} \text{sech}(ry)\,\text{sech}((r+1)y) = \frac{2\cosh y}{\sinh y\,\cosh y} = \frac{2}{\sinh y} = 2\,\text{cosech}(y)$

Examiner Notes

比 Q2 稍多人尝试（约五分之四），但成功率略低。第一部分（有明确提示）完成得不错，但 $N$ 趋于无穷的极限论证中记号很差。在(ii)中，相当多考生用正指数幂写出 sech(ry) 的表达式，导致无法应用(i)的结果。很少有考生完全简化最终结果，且许多人未能正确处理正负部分的求和，有些人只是简单地将从1到无穷的求和乘以2。

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Question 5

Topic: 组合与数论 (Combinatorics and Number Theory)  |  Difficulty: Hard  |  Marks: 20

Problem

5 (i) By considering the binomial expansion of $(1 + x)^{2m+1}$, prove that

$\binom{2m+1}{m} < 2^{2m},$

for any positive integer $m$.

(ii) For any positive integers $r$ and $s$ with $r < s$, $P_{r,s}$ is defined as follows: $P_{r,s}$ is the product of all the prime numbers greater than $r$ and less than or equal to $s$, if there are any such primes numbers; if there are no such primes numbers, then $P_{r,s} = 1$.

For example, $P_{3,7} = 35$, $P_{7,10} = 1$ and $P_{14,18} = 17$.

Show that, for any positive integer $m$, $P_{m+1, 2m+1}$ divides $\binom{2m+1}{m}$, and deduce that

$P_{m+1, 2m+1} < 2^{2m}.$

(iii) Show that, if $P_{1,k} < 4^k$ for $k = 2, 3, \dots, 2m$, then $P_{1,2m+1} < 4^{2m+1}$.

(iv) Prove that $P_{1,n} < 4^n$ for $n \geqslant 2$.

Hint

The binomial expansion, evaluated for $x = 1$, appreciating that terms are symmetrical contains two terms equal to the LHS of the inequality, and so truncating to them gives double the required result in (i). Appreciating that $\frac{(2m+1)!}{(m+1)!m!}$ is an integer and that if $m + 1 < p \le 2m + 1$, with p a prime, implies p divides the numerator and not the denominator of this expression and hence divides the integer then can be extended for all such primes yielding the result, with the deduction following from (i). For (iii), it can be shown that $m + 1 \le 2m$ and writing $P_{1,2m+1}$ as $P_{1,m+1} P_{m+1,2m+1}$, combining the given result and (ii), the desired result is obtained. Part (iv) is obtained by use of strong induction with the supposition, $P_{1,m} < 4^m$ for all $m \le k$ for some particular $k \ge 2$, and considering the cases k even and odd separately and making use of (iii).

Model Solution

Part (i)

Consider the binomial expansion of $(1+x)^{2m+1}$:

$(1+x)^{2m+1} = \sum_{k=0}^{2m+1} \binom{2m+1}{k} x^k.$

Setting $x = 1$:

$2^{2m+1} = \sum_{k=0}^{2m+1} \binom{2m+1}{k}.$

The binomial coefficients satisfy the symmetry $\binom{2m+1}{k} = \binom{2m+1}{2m+1-k}$. We pair the terms: the $k$-th term with the $(2m+1-k)$-th term. Since $2m+1$ is odd, every term has a distinct partner, and there are $m+1$ pairs:

$2^{2m+1} = 2\left[\binom{2m+1}{0} + \binom{2m+1}{1} + \cdots + \binom{2m+1}{m}\right].$

Since all binomial coefficients are strictly positive:

$\binom{2m+1}{0} + \binom{2m+1}{1} + \cdots + \binom{2m+1}{m} > \binom{2m+1}{m}.$

Therefore:

$2^{2m+1} = 2\left[\binom{2m+1}{0} + \cdots + \binom{2m+1}{m}\right] > 2\binom{2m+1}{m},$

which gives $\binom{2m+1}{m} < 2^{2m}$. $\qquad \text{(proved)}$

Part (ii)

Recall that $\binom{2m+1}{m} = \frac{(2m+1)!}{m!\,(m+1)!}$.

Let $p$ be any prime with $m + 1 < p \leq 2m + 1$. We show that $p \mid \binom{2m+1}{m}$ by counting the multiplicity of $p$ in the numerator and denominator.

• Since $m + 1 < p \leq 2m+1$ and $2p > 2(m+1) > 2m+1$, the number $p$ appears exactly once in $1, 2, \ldots, 2m+1$. So $p$ has multiplicity exactly 1 in $(2m+1)!$.
• Since $p > m + 1 > m$, the prime $p$ does not divide any of $1, 2, \ldots, m$, so $p$ has multiplicity 0 in $m!$.
• Since $p > m + 1$, the prime $p$ does not divide any of $1, 2, \ldots, m+1$, so $p$ has multiplicity 0 in $(m+1)!$.

Therefore $p$ has multiplicity $1 - 0 - 0 = 1$ in $\binom{2m+1}{m}$, so $p \mid \binom{2m+1}{m}$.

Since this holds for every prime $p$ with $m + 1 < p \leq 2m+1$, and these primes are distinct, their product $P_{m+1, 2m+1}$ divides $\binom{2m+1}{m}$.

From part (i), $\binom{2m+1}{m} < 2^{2m}$. Since $P_{m+1, 2m+1}$ is a positive divisor of $\binom{2m+1}{m}$:

$P_{m+1, 2m+1} \leq \binom{2m+1}{m} < 2^{2m}. \qquad \text{(deduced)}$

Part (iii)

We decompose the product of all primes up to $2m+1$:

$P_{1,2m+1} = P_{1,m+1} \cdot P_{m+1,2m+1}.$

This is valid because every prime $\leq 2m+1$ is either $\leq m+1$ or lies in $(m+1, 2m+1]$, and the two sets are disjoint.

Since $m \geq 1$, we have $m + 1 \geq 2$ and $m + 1 \leq 2m$, so $m+1$ lies in the range $\{2, 3, \ldots, 2m\}$. By hypothesis:

$P_{1,m+1} < 4^{m+1}.$

By part (ii):

$P_{m+1,2m+1} < 2^{2m} = 4^m.$

Multiplying:

$P_{1,2m+1} = P_{1,m+1} \cdot P_{m+1,2m+1} < 4^{m+1} \cdot 4^m = 4^{2m+1}. \qquad \text{(proved)}$

Part (iv)

We prove $P_{1,n} < 4^n$ for all $n \geq 2$ by strong induction.

Base cases.

$P_{1,2} = 2 < 16 = 4^2$.

$P_{1,3} = 2 \times 3 = 6 < 64 = 4^3$.

Inductive step. Suppose $P_{1,k} < 4^k$ for all $k$ with $2 \leq k \leq n$, where $n \geq 3$. We show $P_{1,n+1} < 4^{n+1}$.

Case 1: $n + 1$ is even. Write $n + 1 = 2m$ with $m \geq 2$ (since $n + 1 \geq 4$).

Since $P_{1,s}$ is the product of primes $\leq s$, increasing the upper bound from $2m - 1$ to $2m$ either multiplies by a new prime (if $2m$ is prime) or leaves the product unchanged (if $2m$ is composite). In either case:

$P_{1,2m} \leq P_{1,2m-1}.$

Since $2m - 1 \geq 3 \geq 2$, the inductive hypothesis gives $P_{1,2m-1} < 4^{2m-1}$. Therefore:

$P_{1,2m} \leq P_{1,2m-1} < 4^{2m-1} < 4^{2m}.$

Case 2: $n + 1$ is odd. Write $n + 1 = 2m + 1$ with $m \geq 2$ (since $n + 1 \geq 5$).

We decompose as in part (iii):

$P_{1,2m+1} = P_{1,m+1} \cdot P_{m+1,2m+1}.$

Since $m \geq 2$, we have $3 \leq m + 1 \leq 2m \leq n$, so $m + 1$ lies in the range $\{2, 3, \ldots, n\}$. By the inductive hypothesis:

$P_{1,m+1} < 4^{m+1}.$

By part (ii):

$P_{m+1,2m+1} < 2^{2m} = 4^m.$

Therefore:

$P_{1,2m+1} < 4^{m+1} \cdot 4^m = 4^{2m+1}.$

In both cases, $P_{1,n+1} < 4^{n+1}$. By strong induction, $P_{1,n} < 4^n$ for all $n \geq 2$. $\qquad \text{(proved)}$

Examiner Notes

约一半考生尝试此题，得分与 Q2 相似。二项展开及其对称性在第一部分处理得不错，(iii)的结果在(iv)中的应用也可以。但(ii)回答得不好，因为关于整除性的论证往往过于随意；在(iii)中，很少有人注意到需要 $m+1 \le 2m$ 的条件才能使用(ii)的结果。

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Question 6

Topic: 双曲函数 (Hyperbolic Functions)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Show, by finding $R$ and $\gamma$, that $A \sinh x + B \cosh x$ can be written in the form $R \cosh(x + \gamma)$ if $B > A > 0$. Determine the corresponding forms in the other cases that arise, for $A > 0$, according to the value of $B$.

Two curves have equations $y = \operatorname{sech} x$ and $y = a \tanh x + b$, where $a > 0$.

(i) In the case $b > a$, show that if the curves intersect then the $x$-coordinates of the points of intersection can be written in the form

$\pm \operatorname{arcosh} \left( \frac{1}{\sqrt{b^2 - a^2}} \right) - \operatorname{artanh} \frac{a}{b}.$

(ii) Find the corresponding result in the case $a > b > 0$.

(iii) Find necessary and sufficient conditions on $a$ and $b$ for the curves to intersect at two distinct points.

(iv) Find necessary and sufficient conditions on $a$ and $b$ for the curves to touch and, given that they touch, express the $y$-coordinate of the point of contact in terms of $a$.

Hint

Using $R \cosh(x + \gamma) = R(\cosh x \cosh \gamma + \sinh x \sinh \gamma)$, $R = \sqrt{B^2 - A^2}$ and $\gamma = \tanh^{-1} \frac{A}{B}$ if $B > A > 0$. If $B = A$, then $A \sinh x + B \cosh x = Ae^x$. If $-A < B < A$, the expression can be written as $R \sinh(x + \gamma)$ with $R = \sqrt{A^2 - B^2}$ and $\gamma = \tanh^{-1} \frac{B}{A}$. If $B = -A$, then $A \sinh x + B \cosh x = -Ae^{-x}$, and if $B < -A$, the expression can be written as $R \cosh(x + \gamma)$ with $R = -\sqrt{B^2 - A^2}$ and $\gamma = \tanh^{-1} \frac{A}{B}$.

For part (i), solving simultaneously gives $a \sinh x + b \cosh x = 1$, which gives the desired solutions using the first result of the question.

Similarly for part (ii) using the appropriate result, $x = \sinh^{-1} \left( \frac{1}{\sqrt{a^2-b^2}} \right) - \tanh^{-1} \frac{b}{a}$.

For (iii), we require that the conditions for (i) give two solutions, i.e. that $b > a$ and $\left( \frac{1}{\sqrt{b^2-a^2}} \right) > 1$, and so $a < b < \sqrt{a^2 + 1}$, and vice versa, if this applies there are indeed two solutions.

For (iv), we require case (i) to give coincident solutions, i.e. $b = \sqrt{a^2 + 1}$ and hence $x = -\tanh^{-1} \frac{a}{\sqrt{a^2+1}}$, and so $y = \frac{1}{\sqrt{a^2+1}}$. The reverse argument also applies.

Model Solution

Preamble: Combining $A \sinh x + B \cosh x$

We expand $R \cosh(x + \gamma) = R \cosh x \cosh \gamma + R \sinh x \sinh \gamma$ and match coefficients:

$A = R \sinh \gamma, \qquad B = R \cosh \gamma.$

Dividing: $\tanh \gamma = \frac{A}{B}$, so $\gamma = \operatorname{artanh} \frac{A}{B}$ (valid when $|A/B| < 1$).

Using $\cosh^2 \gamma - \sinh^2 \gamma = 1$:

$\frac{B^2}{R^2} - \frac{A^2}{R^2} = 1 \implies R^2 = B^2 - A^2.$

Case 1: $B > A > 0$. Then $B^2 - A^2 > 0$, so $R = \sqrt{B^2 - A^2} > 0$ and $\gamma = \operatorname{artanh} \frac{A}{B}$.

$A \sinh x + B \cosh x = \sqrt{B^2 - A^2} \, \cosh\!\left(x + \operatorname{artanh} \frac{A}{B}\right).$

Case 2: $B = A > 0$. Then $A \sinh x + A \cosh x = A(\sinh x + \cosh x) = Ae^x$.

Case 3: $0 < B < A$ (i.e., $-A < B < A$ with $B > 0$). We try $R \sinh(x + \gamma) = R \sinh x \cosh \gamma + R \cosh x \sinh \gamma$, matching $A = R \cosh \gamma$ and $B = R \sinh \gamma$. Then $\tanh \gamma = B/A$, so $\gamma = \operatorname{artanh} \frac{B}{A}$, and $R^2 = A^2 - B^2 > 0$, giving $R = \sqrt{A^2 - B^2}$.

$A \sinh x + B \cosh x = \sqrt{A^2 - B^2} \, \sinh\!\left(x + \operatorname{artanh} \frac{B}{A}\right).$

Case 4: $B = -A < 0$. Then $A \sinh x - A \cosh x = A(\sinh x - \cosh x) = -Ae^{-x}$.

Case 5: $B < -A < 0$. We return to the $\cosh$ form. $R^2 = B^2 - A^2 > 0$, but $R \cosh \gamma = B < 0$, so we take $R = -\sqrt{B^2 - A^2} < 0$. Then $\gamma = \operatorname{artanh} \frac{A}{B}$ (note $|A/B| < 1$ since $|B| > A$).

$A \sinh x + B \cosh x = -\sqrt{B^2 - A^2} \, \cosh\!\left(x + \operatorname{artanh} \frac{A}{B}\right).$

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Part (i)

Setting $\operatorname{sech} x = a \tanh x + b$ and multiplying both sides by $\cosh x$:

$1 = a \sinh x + b \cosh x.$

In the case $b > a > 0$, we apply Case 1 with $A = a$, $B = b$:

$\sqrt{b^2 - a^2} \, \cosh\!\left(x + \operatorname{artanh} \frac{a}{b}\right) = 1$

$\cosh\!\left(x + \operatorname{artanh} \frac{a}{b}\right) = \frac{1}{\sqrt{b^2 - a^2}}.$

For this to have a solution, we need $\frac{1}{\sqrt{b^2 - a^2}} \geq 1$, i.e., $b^2 - a^2 \leq 1$.

Since $\cosh$ is an even function, $\cosh u = \alpha$ gives $u = \pm \operatorname{arcosh} \alpha$ for $\alpha \geq 1$. Therefore:

$x + \operatorname{artanh} \frac{a}{b} = \pm \operatorname{arcosh}\!\left(\frac{1}{\sqrt{b^2 - a^2}}\right)$

$x = \pm \operatorname{arcosh}\!\left(\frac{1}{\sqrt{b^2 - a^2}}\right) - \operatorname{artanh} \frac{a}{b}. \qquad \text{(shown)}$

Part (ii)

Again the intersection condition is $a \sinh x + b \cosh x = 1$. In the case $a > b > 0$, we apply Case 3 with $A = a$, $B = b$:

$\sqrt{a^2 - b^2} \, \sinh\!\left(x + \operatorname{artanh} \frac{b}{a}\right) = 1$

$\sinh\!\left(x + \operatorname{artanh} \frac{b}{a}\right) = \frac{1}{\sqrt{a^2 - b^2}}.$

Since $\sinh$ is a strictly increasing function with range $(-\infty, \infty)$, the equation $\sinh u = \alpha$ has the unique solution $u = \operatorname{arsinh} \alpha$ for any real $\alpha$. Therefore:

$x + \operatorname{artanh} \frac{b}{a} = \operatorname{arsinh}\!\left(\frac{1}{\sqrt{a^2 - b^2}}\right)$

$x = \operatorname{arsinh}\!\left(\frac{1}{\sqrt{a^2 - b^2}}\right) - \operatorname{artanh} \frac{b}{a}. \qquad \text{(shown)}$

Part (iii)

From part (i), when $b > a$, the intersection $x$-coordinates satisfy $\cosh(x + \gamma) = \frac{1}{\sqrt{b^2 - a^2}}$. This equation has two distinct solutions (the $\pm$ branch) if and only if $\frac{1}{\sqrt{b^2 - a^2}} > 1$, i.e., $b^2 - a^2 < 1$, i.e., $b < \sqrt{a^2 + 1}$.

Combining with $b > a$: the curves intersect at two distinct points when $a < b < \sqrt{a^2 + 1}$.

From part (ii), when $a > b > 0$, the equation $\sinh(x + \gamma) = \frac{1}{\sqrt{a^2 - b^2}}$ has exactly one solution (since $\sinh$ is strictly monotone), so there is only one intersection point.

Therefore the necessary and sufficient condition for two distinct intersection points is:

$a < b < \sqrt{a^2 + 1}. \qquad \text{(shown)}$

Part (iv)

“Touching” means the curves meet at exactly one point with a common tangent, i.e., both the function values and derivatives agree.

From $y = \operatorname{sech} x$ and $y = a \tanh x + b$:

$y' = -\operatorname{sech} x \tanh x \qquad \text{and} \qquad y' = a \operatorname{sech}^2 x.$

At a touching point, $-\operatorname{sech} x \tanh x = a \operatorname{sech}^2 x$. Since $\operatorname{sech} x > 0$, divide by $\operatorname{sech} x$:

$-\tanh x = a \operatorname{sech} x$

$-\frac{\sinh x}{\cosh x} = \frac{a}{\cosh x}$

$\sinh x = -a.$

Substituting into the intersection condition $a \sinh x + b \cosh x = 1$:

$a(-a) + b \cosh x = 1 \implies b \cosh x = 1 + a^2 \implies \cosh x = \frac{1 + a^2}{b}.$

Using $\cosh^2 x - \sinh^2 x = 1$:

$\left(\frac{1 + a^2}{b}\right)^2 - a^2 = 1$

$\frac{(1 + a^2)^2}{b^2} = 1 + a^2$

$\frac{1 + a^2}{b^2} = 1$

$b^2 = 1 + a^2$

$b = \sqrt{a^2 + 1} \quad (\text{since } b > 0).$

Conversely, if $b = \sqrt{a^2 + 1}$, then from part (i) (since $b = \sqrt{a^2+1} > a$), the $\cosh$ equation becomes $\cosh(x + \gamma) = \frac{1}{\sqrt{b^2 - a^2}} = \frac{1}{1} = 1$, which has the unique solution $x + \gamma = 0$, confirming exactly one intersection point.

The necessary and sufficient condition for the curves to touch is:

$b = \sqrt{a^2 + 1}. \qquad \text{(shown)}$

$y$-coordinate of the point of contact. At the touching point, $\cosh x = \frac{1+a^2}{b} = \frac{1+a^2}{\sqrt{a^2+1}} = \sqrt{a^2+1}$, so:

$y = \operatorname{sech} x = \frac{1}{\cosh x} = \frac{1}{\sqrt{a^2 + 1}}.$

Examiner Notes

约40%考生尝试此题，得分约三分之一。多数知道 cosh 加法公式，但 A=B 的情况处理很弱。(i) 部分解释不清，± 号理解有误，导致 (ii) 出现虚假的 ±。(iii)(iv) 必要充分条件论证薄弱。仅少数优秀考生完整解答。

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Question 7

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

Problem

7 Let $\omega = e^{2\pi i/n}$, where $n$ is a positive integer. Show that, for any complex number $z$,

$(z - 1)(z - \omega) \cdots (z - \omega^{n-1}) = z^n - 1 .$

The points $X_0, X_1, \dots, X_{n-1}$ lie on a circle with centre $O$ and radius 1, and are the vertices of a regular polygon.

(i) The point $P$ is equidistant from $X_0$ and $X_1$. Show that, if $n$ is even,

$|PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| = |OP|^n + 1 ,$

where $|PX_k|$ denotes the distance from $P$ to $X_k$.

Give the corresponding result when $n$ is odd. (There are two cases to consider.)

(ii) Show that

$|X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| = n .$

Hint

Considering $(\omega^r)^n$ establishes by the factor theorem that each factor on the LHS is a factor of the RHS, and comparing coefficients of $z^n$ between the two sides establishes that no numerical factor is required.

For part (i), representing $X_r$ by $\omega^r$, then there are two cases to consider, $P$ will be represented either by $re^{\frac{\pi i}{n}}$, or $re^{(\frac{\pi}{n}+\pi)i}$. The product of moduli is the moduli of the product of factors, and the product of the factors can be simplified using the stem and choosing $z$ in turn as the representations of $P$ to give the required result in both cases.

Proceeding similarly for $n$ odd, the first case yields $|OP|^n + 1$, and the second, $|OP|^n - 1$, if $|OP| \ge 1$, and $1 - |OP|^n$ if $|OP| < 1$.

Using the same representations for the $X_r$ in part (ii), and the same technique with the moduli, the stem can be divided by $(z - 1)$ to give $(z - \omega)(z - \omega^2) \dots (z - \omega^{n-1}) = z^{n-1} + z^{n-2} + \dots + 1$ which then gives the desired result when $z = 1$.

Model Solution

Proof of the stem result.

The equation $z^n = 1$ has exactly $n$ roots: $z = 1, \omega, \omega^2, \ldots, \omega^{n-1}$, where $\omega = e^{2\pi i/n}$.

To verify these are distinct: if $\omega^j = \omega^k$ for $0 \le j, k \le n-1$, then $e^{2\pi i(j-k)/n} = 1$, requiring $n \mid (j-k)$. Since $|j-k| < n$, we must have $j = k$.

By the Factor Theorem, each root $\omega^k$ means $(z - \omega^k)$ divides $z^n - 1$. Since $\omega^0, \omega^1, \ldots, \omega^{n-1}$ are $n$ distinct roots, the product

$(z - 1)(z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1})$

divides $z^n - 1$. Both sides are monic polynomials of degree $n$, so they must be equal:

$(z - 1)(z - \omega) \cdots (z - \omega^{n-1}) = z^n - 1. \qquad \text{(*)}$

Part (i)

We represent each vertex $X_k$ by $\omega^k$, so $X_0 = 1$ and $X_1 = \omega = e^{2\pi i/n}$.

Since $P$ is equidistant from $X_0$ and $X_1$, it lies on the perpendicular bisector of the chord $X_0 X_1$. This bisector passes through the center $O$ and makes an angle $\pi/n$ with the positive real axis. We can write

$P = r e^{i\pi/n}$

for some real $r$, where $|OP| = |r|$.

Taking the modulus of both sides of $(*)$ with $z = P$:

$\prod_{k=0}^{n-1} |P - \omega^k| = |P^n - 1|.$

Computing $P^n = r^n e^{i\pi} = -r^n$, so

$|P^n - 1| = |-r^n - 1| = |r^n + 1|.$

When $n$ is even: $r^n = |r|^n \ge 0$, so $r^n + 1 > 0$, giving

$\prod_{k=0}^{n-1} |PX_k| = r^n + 1 = |OP|^n + 1.$

This holds for all real $r$ (both sides of $O$), since $r^n = |r|^n$ when $n$ is even.

When $n$ is odd, there are two cases depending on the sign of $r$ (i.e., which side of $O$ the point $P$ lies):

Case 1: $r > 0$ (same side as the arc midpoint $e^{i\pi/n}$). Then $r^n > 0$, so

$|r^n + 1| = r^n + 1 = |OP|^n + 1.$

Case 2: $r < 0$ (opposite side of $O$). Write $r = -s$ where $s = |OP| > 0$. Since $n$ is odd, $r^n = -s^n$, so

$|r^n + 1| = |1 - s^n| = |1 - |OP|^n|.$

Concretely:

• If $|OP| > 1$: the product equals $|OP|^n - 1$.
• If $|OP| < 1$: the product equals $1 - |OP|^n$.
• If $|OP| = 1$: the product equals $0$ (here $P$ coincides with the vertex $X_{(n+1)/2}$).

Part (ii)

The distance $|X_0 X_k| = |1 - \omega^k|$, so the required product is

$\prod_{k=1}^{n-1} |1 - \omega^k| = \left| \prod_{k=1}^{n-1} (1 - \omega^k) \right|.$

From $(*)$, dividing both sides by $(z - 1)$:

$(z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1}) = \frac{z^n - 1}{z - 1} = z^{n-1} + z^{n-2} + \cdots + z + 1.$

Setting $z = 1$:

$(1 - \omega)(1 - \omega^2) \cdots (1 - \omega^{n-1}) = 1 + 1 + \cdots + 1 = n.$

Taking moduli:

$\prod_{k=1}^{n-1} |1 - \omega^k| = |n| = n.$

Examiner Notes

纯数部分最不受欢迎的题目，不到30%尝试，整体表现不佳。多数能完成 stem 部分，但经常忽略数值因子为1的验证。尽管有 stem，多数不知道如何处理 (i)，因此进展甚少。

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Question 8

Topic: 函数方程 (Functional Equations)  |  Difficulty: Standard  |  Marks: 20

Problem

8 (i) The function $f$ satisfies, for all $x$, the equation

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

Show that $f(-x) + (1 + x)f(x) = x^2$. Hence find $f(x)$ in terms of $x$. You should verify that your function satisfies the original equation.

(ii) The function $K$ is defined, for $x \neq 1$, by

$K(x) = \frac{x + 1}{x - 1} .$

Show that, for $x \neq 1$, $K(K(x)) = x$.

The function $g$ satisfies the equation

$g(x) + x g \left( \frac{x + 1}{x - 1} \right) = x \quad (x \neq 1) .$

Show that, for $x \neq 1$, $g(x) = \frac{2x}{x^2 + 1}$.

(iii) Find $h(x)$, for $x \neq 0, x \neq 1$, given that

$h(x) + h \left( \frac{1}{1 - x} \right) = 1 - x - \frac{1}{1 - x} \quad (x \neq 0, \enspace x \neq 1) .$

Hint

The first result in (i) is obtained by the substitution $x = -u$ (followed by a second $u = x$!). Substituting for $f(-x)$ in the initial statement using the result obtained readily leads to $f(x) = x$ which is simply verified. Alternatively, subtracting the result from the initial equation leads to $f(x) = f(-x)$ which substituting gives the required result again.

In part (ii), substituting $K(x)$ for $x$ in the equation for $g(x)$ gives an equation for $g\left(\frac{x+1}{x-1}\right)$ which can be substituted in the equation to be solved to give the desired result.

Similarly, in part (iii), substituting $\frac{1}{1-x}$ for $x$ gives an equation for $h\left(\frac{1}{1-x}\right)$ and $h\left(\frac{x-1}{x}\right)$, and then repeating this substitution in the equation just obtained gives an equation for $h\left(\frac{x-1}{x}\right)$ and $h(x)$. Adding the given and last equations and subtracting that first found leads to $h(x) = \frac{1}{2} - x$.

Model Solution

Part (i)

The given equation is

$f(x) + (1 - x)f(-x) = x^2. \qquad \text{(1)}$

Replace $x$ by $-x$ throughout:

$f(-x) + (1 + x)f(x) = (-x)^2 = x^2. \qquad \text{(2)}$

This is the first required result.

From (2), solve for $f(-x)$:

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

Substitute into (1):

$f(x) + (1 - x)\bigl[x^2 - (1 + x)f(x)\bigr] = x^2.$

Expand:

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

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

Collect the $f(x)$ terms:

$f(x)\bigl[1 - (1 - x^2)\bigr] = x^2 - x^2 + x^3.$

$f(x) \cdot x^2 = x^3.$

$f(x) = x. \qquad \text{(3)}$

Verification: Substitute $f(x) = x$ into (1):

$x + (1 - x)(-x) = x - x + x^2 = x^2. \checkmark$

Part (ii)

Showing $K(K(x)) = x$:

$K(K(x)) = K\!\left(\frac{x+1}{x-1}\right) = \frac{\dfrac{x+1}{x-1} + 1}{\dfrac{x+1}{x-1} - 1} = \frac{\dfrac{x+1+x-1}{x-1}}{\dfrac{x+1-x+1}{x-1}} = \frac{\dfrac{2x}{x-1}}{\dfrac{2}{x-1}} = x. \qquad \text{(4)}$

Finding $g(x)$:

The equation for $g$ is

$g(x) + x \, g\!\left(\frac{x+1}{x-1}\right) = x. \qquad \text{(5)}$

Replace $x$ by $\frac{x+1}{x-1}$ in (5). Using $(4)$, $K(K(x)) = x$, so

$g\!\left(\frac{x+1}{x-1}\right) + \frac{x+1}{x-1} \, g(x) = \frac{x+1}{x-1}. \qquad \text{(6)}$

From (6), solve for $g\!\bigl(\frac{x+1}{x-1}\bigr)$:

$g\!\left(\frac{x+1}{x-1}\right) = \frac{x+1}{x-1} - \frac{x+1}{x-1} \, g(x) = \frac{x+1}{x-1}\bigl(1 - g(x)\bigr). \qquad \text{(7)}$

Substitute (7) into (5):

$g(x) + x \cdot \frac{x+1}{x-1}\bigl(1 - g(x)\bigr) = x.$

Multiply through by $(x - 1)$:

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

Expand the left side:

$(x - 1)g(x) + x(x + 1) - x(x + 1)g(x) = x^2 - x.$

Collect the $g(x)$ terms:

$g(x)\bigl[(x - 1) - x(x + 1)\bigr] = x^2 - x - x(x + 1).$

$g(x)\bigl[x - 1 - x^2 - x\bigr] = x^2 - x - x^2 - x.$

$g(x)(-1 - x^2) = -2x.$

$g(x) = \frac{2x}{x^2 + 1}. \qquad \text{(8)}$

Verification: We need to check that $(8)$ satisfies (5). First compute $g\!\bigl(\frac{x+1}{x-1}\bigr)$:

$g\!\left(\frac{x+1}{x-1}\right) = \frac{2 \cdot \frac{x+1}{x-1}}{\left(\frac{x+1}{x-1}\right)^2 + 1} = \frac{\frac{2(x+1)}{x-1}}{\frac{(x+1)^2 + (x-1)^2}{(x-1)^2}} = \frac{2(x+1)}{x-1} \cdot \frac{(x-1)^2}{(x+1)^2 + (x-1)^2}.$

The denominator simplifies: $(x+1)^2 + (x-1)^2 = x^2 + 2x + 1 + x^2 - 2x + 1 = 2x^2 + 2 = 2(x^2 + 1)$. So

$g\!\left(\frac{x+1}{x-1}\right) = \frac{2(x+1)(x-1)}{2(x^2+1)} = \frac{x^2 - 1}{x^2 + 1}.$

Now check (5):

$g(x) + x \, g\!\left(\frac{x+1}{x-1}\right) = \frac{2x}{x^2+1} + x \cdot \frac{x^2-1}{x^2+1} = \frac{2x + x^3 - x}{x^2+1} = \frac{x^3 + x}{x^2+1} = \frac{x(x^2+1)}{x^2+1} = x. \checkmark$

Part (iii)

Define $T(x) = \frac{1}{1 - x}$. We first check that $T$ has order 3 (i.e., applying $T$ three times returns to $x$):

$T^2(x) = T\!\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{1-x-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x}.$

$T^3(x) = T\!\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} = \frac{1}{\frac{x - (x-1)}{x}} = \frac{1}{\frac{1}{x}} = x. \qquad \text{(9)}$

The given equation is

$h(x) + h\!\left(\frac{1}{1-x}\right) = 1 - x - \frac{1}{1-x}. \qquad \text{(I)}$

Replace $x$ by $T(x) = \frac{1}{1-x}$ in (I):

$h\!\left(\frac{1}{1-x}\right) + h\!\left(\frac{x-1}{x}\right) = 1 - \frac{1}{1-x} - \frac{1}{1 - \frac{1}{1-x}}.$

The right side simplifies:

$1 - \frac{1}{1-x} - \frac{1-x}{-x} = 1 - \frac{1}{1-x} + \frac{1-x}{x}.$

So:

$h\!\left(\frac{1}{1-x}\right) + h\!\left(\frac{x-1}{x}\right) = 1 - \frac{1}{1-x} + \frac{1-x}{x}. \qquad \text{(II)}$

Replace $x$ by $T^2(x) = \frac{x-1}{x}$ in (I). Using $(9)$, $T^3(x) = x$:

$h\!\left(\frac{x-1}{x}\right) + h(x) = 1 - \frac{x-1}{x} - \frac{1}{1 - \frac{x-1}{x}}.$

The right side simplifies:

$1 - \frac{x-1}{x} - \frac{1}{\frac{1}{x}} = 1 - \frac{x-1}{x} - x = 1 - 1 + \frac{1}{x} - x = \frac{1}{x} - x.$

So:

$h\!\left(\frac{x-1}{x}\right) + h(x) = \frac{1}{x} - x. \qquad \text{(III)}$

Now compute $\text{(I)} + \text{(III)} - \text{(II)}$:

Left side: $[h(x) + h(T(x))] + [h(T^2(x)) + h(x)] - [h(T(x)) + h(T^2(x))] = 2h(x)$.

Right side:

$\left(1 - x - \frac{1}{1-x}\right) + \left(\frac{1}{x} - x\right) - \left(1 - \frac{1}{1-x} + \frac{1-x}{x}\right).$

Group terms:

$= 1 - x - \frac{1}{1-x} + \frac{1}{x} - x - 1 + \frac{1}{1-x} - \frac{1-x}{x}.$

The $\frac{1}{1-x}$ terms cancel:

$= -2x + \frac{1}{x} - \frac{1-x}{x} = -2x + \frac{1 - (1-x)}{x} = -2x + \frac{x}{x} = -2x + 1.$

Therefore $2h(x) = 1 - 2x$, giving

$h(x) = \frac{1}{2} - x.$

Verification:

$h(x) + h\!\left(\frac{1}{1-x}\right) = \left(\frac{1}{2} - x\right) + \left(\frac{1}{2} - \frac{1}{1-x}\right) = 1 - x - \frac{1}{1-x}. \checkmark$

Examiner Notes

尝试人数接近 Q1（最多），且略为更成功，不少人得满分。函数迭代构造循环的想法普遍被发现。(ii) 中找 g(x) 时有困难，部分考生代入已知的 g(x) 而非求解它。(iii) 中部分考生只做了第一次代换就停止，无法找到解答。也有部分猜出答案但未完整论证。