STEP3 2008 -- Pure Mathematics

STEP3 2008 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2008  |  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	代数 (Algebra)	Standard	对称多项式, 联立方程消元, 二次方程求根
2	数列与级数 (Sequences and Series)	Challenging	裂项求和, 数学归纳法, 二项式定理展开
3	解析几何 (Coordinate Geometry)	Challenging	参数微分, 联立方程求交点, 消参求轨迹
4	双曲函数 (Hyperbolic Functions)	Challenging	双曲函数恒等式, 分部积分, 不等式积分
5	递推关系与多项式 (Recurrence Relations and Polynomials)	Challenging	数学归纳法, 递推关系, 特征方程求解, 等比数列
6	微分方程 (Differential Equations)	Challenging	Clairaut方程法,积分因子法,换元微分
7	复数 (Complex Numbers)	Challenging	复数旋转,几何变换,对称性分析
8	级数 (Series)	Standard	递推关系,乘因子消项法,部分分数分解,等比级数求和

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Question 1

Topic: 代数 (Algebra)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Find all values of $a, b, x$ and $y$ that satisfy the simultaneous equations

$$\begin{aligned} a + b &= 1 \\ ax + by &= \frac{1}{3} \\ ax^2 + by^2 &= \frac{1}{5} \\ ax^3 + by^3 &= \frac{1}{7} . \end{aligned}$$

[ Hint: you may wish to start by multiplying the second equation by $x + y$. ]

Hint

Following the hint yields

$ax^2 + by^2 + (a + b)xy = \frac{1}{3}(x + y)$

$which\ is\ \frac{1}{5} + xy = \frac{1}{3}(x + y)$

The same trick applied to the third equation gives $\frac{1}{7} + \frac{1}{3}xy = \frac{1}{5}(x + y)$.

The two equations can be solved simultaneously for $xy$ and $(x+y)$, giving

$xy = \frac{3}{35}\ and\ (x + y) = \frac{6}{7}$

Thus $x$ and $y$ are the roots of the quadratic equation $35z^2 - 30z + 3 = 0$ ($x$ and $y$ are interchangeable).

$a$ and $b$ are then found by substituting back into two of the original equations and the full solution is

$x = \frac{3}{7} \pm \frac{2}{35}\sqrt{30} = \frac{3}{7} \pm \frac{2}{7}\sqrt{\frac{6}{5}}$

$y = \frac{3}{7} \mp \frac{2}{35}\sqrt{30} = \frac{3}{7} \mp \frac{2}{7}\sqrt{\frac{6}{5}}$

$a = \frac{1}{2} \mp \frac{\sqrt{30}}{36} = \frac{1}{2} \mp \frac{1}{6}\sqrt{\frac{5}{6}}$

$b = \frac{1}{2} \pm \frac{\sqrt{30}}{36} = \frac{1}{2} \pm \frac{1}{6}\sqrt{\frac{5}{6}}$

Model Solution

Multiply the second equation by $(x + y)$:

$ax^2 + by^2 + (a + b)xy = \frac{1}{3}(x + y)$

Since $a + b = 1$ and $ax^2 + by^2 = \frac{1}{5}$:

$\frac{1}{5} + xy = \frac{1}{3}(x + y) \qquad \text{(i)}$

Multiply the third equation by $(x + y)$:

$ax^3 + by^3 + (ax^2 + by^2)(x + y) \cdot \frac{1}{x+y} \cdot (x+y)$

More precisely: $(ax^2 + by^2)(x + y) = ax^3 + by^3 + (ax^2 y + by^2 x)$, so

$(ax^3 + by^3) + xy(ax + by) = \frac{1}{5}(x + y)$

$\frac{1}{7} + \frac{1}{3}xy = \frac{1}{5}(x + y) \qquad \text{(ii)}$

From (i): $x + y = 3\left(\frac{1}{5} + xy\right) = \frac{3}{5} + 3xy$. Substitute into (ii):

$\frac{1}{7} + \frac{1}{3}xy = \frac{1}{5}\left(\frac{3}{5} + 3xy\right) = \frac{3}{25} + \frac{3}{5}xy$

$\frac{1}{7} - \frac{3}{25} = \left(\frac{3}{5} - \frac{1}{3}\right)xy$

$\frac{25 - 21}{175} = \frac{4}{15}xy$

$\frac{4}{175} = \frac{4}{15}xy \implies xy = \frac{3}{35}$

From (i): $x + y = \frac{3}{5} + 3 \cdot \frac{3}{35} = \frac{3}{5} + \frac{9}{35} = \frac{21 + 9}{35} = \frac{30}{35} = \frac{6}{7}$.

Thus $x$ and $y$ are roots of $z^2 - \frac{6}{7}z + \frac{3}{35} = 0$, i.e.\ $35z^2 - 30z + 3 = 0$.

$z = \frac{30 \pm \sqrt{900 - 420}}{70} = \frac{30 \pm \sqrt{480}}{70} = \frac{30 \pm 4\sqrt{30}}{70} = \frac{15 \pm 2\sqrt{30}}{35}$

So $x = \dfrac{15 + 2\sqrt{30}}{35}$ and $y = \dfrac{15 - 2\sqrt{30}}{35}$ (or vice versa, since $x$ and $y$ are interchangeable).

Finding $a$ and $b$. From $a + b = 1$ and $ax + by = \frac{1}{3}$, with $b = 1 - a$:

$ax + (1 - a)y = \frac{1}{3} \implies a(x - y) = \frac{1}{3} - y$

$a = \frac{1/3 - y}{x - y}$

With $x - y = \frac{4\sqrt{30}}{35}$ and $\frac{1}{3} - y = \frac{1}{3} - \frac{15 - 2\sqrt{30}}{35} = \frac{35 - 45 + 6\sqrt{30}}{105} = \frac{-10 + 6\sqrt{30}}{105}$:

$a = \frac{-10 + 6\sqrt{30}}{105} \cdot \frac{35}{4\sqrt{30}} = \frac{-10 + 6\sqrt{30}}{12\sqrt{30}} = \frac{-5 + 3\sqrt{30}}{6\sqrt{30}}$

Rationalizing: $a = \frac{(-5 + 3\sqrt{30})\sqrt{30}}{180} = \frac{-5\sqrt{30} + 90}{180} = \frac{18 - \sqrt{30}}{36}$.

Similarly $b = 1 - a = \frac{18 + \sqrt{30}}{36}$.

Verification: $ax + by = \frac{1}{3}$ can be checked by direct substitution (the algebra confirms this).

Summary: $x$ and $y$ are $\dfrac{15 \pm 2\sqrt{30}}{35}$, and with $x$ taking the $+$ sign, $a = \dfrac{18 - \sqrt{30}}{36}$, $b = \dfrac{18 + \sqrt{30}}{36}$ (with $a$ and $b$ swapped if $x$ and $y$ are interchanged). $\blacksquare$

Examiner Notes

This was the most popular question on the paper, and many earned good marks on it. Nearly all the candidates followed the hint, and most then applied the same trick with the third equation. Subsequent success depended on a candidate realising that they had simultaneous equations in $xy$ and $x + y$, although very rarely some managed to solve directly in $x$ and $y$.

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Question 2

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

Problem

2 Let $S_k(n) \equiv \sum_{r=0}^{n} r^k$, where $k$ is a positive integer, so that

$S_1(n) \equiv \frac{1}{2}n(n + 1) \quad \text{and} \quad S_2(n) \equiv \frac{1}{6}n(n + 1)(2n + 1) .$

(i) By considering $\sum_{r=0}^{n} \left[ (r + 1)^k - r^k \right]$, show that

$kS_{k-1}(n) = (n + 1)^k - (n + 1) - \binom{k}{2} S_{k-2}(n) - \binom{k}{3} S_{k-3}(n) - \dots - \binom{k}{k - 1} S_1(n) . \quad (*)$

Obtain simplified expressions for $S_3(n)$ and $S_4(n)$.

(ii) Explain, using $(*)$, why $S_k(n)$ is a polynomial of degree $k + 1$ in $n$. Show that in this polynomial the constant term is zero and the sum of the coefficients is 1.

Hint

2. (i) On the one hand

$\sum_{r=0}^{n} \left[ (r+1)^k - r^k \right] = \sum_{r=0}^{n} (r+1)^k - \sum_{r=0}^{n} r^k = \sum_{r=1}^{n+1} r^k - \sum_{r=0}^{n} r^k = (n+1)^k\ \text{whilst expanding}$

binomially yields

$k \sum_{r=0}^{n} r^{k-1} + \binom{k}{2} \sum_{r=0}^{n} r^{k-2} + \binom{k}{3} \sum_{r=0}^{n} r^{k-3} + \dots + \binom{k}{k-1} \sum_{r=0}^{n} r^{k-1} + \sum_{r=0}^{n} 1$

$= kS_{k-1}(n) + \binom{k}{2} S_{k-2}(n) + \binom{k}{3} S_{k-3}(n) + \dots + \binom{k}{k-1} S_1(n) + (n+1)$

and hence the required result.

Applying this in the case $k = 4$ gives

$4S_3(n) = (n+1)^4 - (n+1) - \binom{4}{2} S_2(n) - \binom{4}{3} S_1(n)$

which, after substitution of the two given results and factorization, yields the familiar

$S_3(n) = \frac{1}{4} n^2 (n + 1)^2$

The identical process with $k = 5$ results in

$S_4(n) = \frac{1}{30} n(n + 1)(6n^3 + 9n^2 + n - 1) = \frac{1}{30} n(n + 1)(2n + 1)(3n^2 + 3n - 1)$

(ii) Applying induction, with the assumption that $S_t(n)$ is a polynomial of degree $t + 1$ in $n$ for $t < r$ for some $r$, and then considering (*),

$(n + 1)^{r+1} - (n + 1)$ is a polynomial of degree $r + 1$ in $n$,

and each of the terms $-\binom{r + 1}{j} S_{r+1-j}(n)$ is a polynomial of degree $r + 2 - j$ in

$n$ where $j \ge 2$, i.e. the degree is $\le r$. A sum of polynomials of degree $\le r + 1$ in $n$, is a polynomial of degree $\le r + 1$ in $n$, and there is a single non-zero term in $n^{r+1}$ from just $(n + 1)^{r+1}$ so the degree of the polynomial is not reduced to $< r + 1$, i.e. it is $r + 1$. (The initial case is true to complete the proof.)

If, $S_k(n) = \sum_{i=0}^{k+1} a_i n^i = \sum_{r=0}^n r^k$ then $S_k(0) = a_0 + \sum_{i=1}^{k+1} a_i 0^i = \sum_{r=0}^0 r^k = 0$ and so $a_0 = 0$

$S_k(1) = \sum_{i=0}^{k+1} a_i 1^i = \sum_{r=0}^1 r^k = 1$ and so $\sum_{i=0}^{k+1} a_i = 1$ as required.

Model Solution

Part (i): Deriving the recurrence

Evaluate $\sum_{r=0}^{n} [(r+1)^k - r^k]$ in two ways.

Method 1 (telescoping):

$\sum_{r=0}^{n} (r+1)^k - \sum_{r=0}^{n} r^k = \sum_{s=1}^{n+1} s^k - \sum_{r=0}^{n} r^k = (n+1)^k$

Method 2 (binomial expansion):

$(r+1)^k - r^k = \binom{k}{1}r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k-1}r + 1$

Summing from $r = 0$ to $n$:

$= kS_{k-1}(n) + \binom{k}{2}S_{k-2}(n) + \binom{k}{3}S_{k-3}(n) + \cdots + \binom{k}{k-1}S_1(n) + (n+1)$

Equating the two methods:

$(n+1)^k = kS_{k-1}(n) + \binom{k}{2}S_{k-2}(n) + \cdots + \binom{k}{k-1}S_1(n) + (n+1)$

$kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2}S_{k-2}(n) - \cdots - \binom{k}{k-1}S_1(n) \qquad (*)$

Computing $S_3(n)$. Set $k = 4$ in $(*)$:

$4S_3(n) = (n+1)^4 - (n+1) - 6S_2(n) - 4S_1(n)$

Expanding: $(n+1)^4 - (n+1) = n^4 + 4n^3 + 6n^2 + 3n$.

Substituting $S_2(n) = \frac{n(n+1)(2n+1)}{6}$ and $S_1(n) = \frac{n(n+1)}{2}$:

$6S_2(n) = n(n+1)(2n+1) = 2n^3 + 3n^2 + n$

$4S_1(n) = 2n(n+1) = 2n^2 + 2n$

$4S_3(n) = (n^4 + 4n^3 + 6n^2 + 3n) - (2n^3 + 3n^2 + n) - (2n^2 + 2n) = n^4 + 2n^3 + n^2 = n^2(n+1)^2$

$S_3(n) = \frac{1}{4}n^2(n+1)^2$

Computing $S_4(n)$. Set $k = 5$ in $(*)$:

$5S_4(n) = (n+1)^5 - (n+1) - 10S_3(n) - 10S_2(n) - 5S_1(n)$

Expanding: $(n+1)^5 - (n+1) = n^5 + 5n^4 + 10n^3 + 10n^2 + 4n$.

$10S_3(n) = \frac{5}{2}n^2(n+1)^2 = \frac{5}{2}n^4 + 5n^3 + \frac{5}{2}n^2$

$10S_2(n) = \frac{5}{3}n(n+1)(2n+1) = \frac{10}{3}n^3 + 5n^2 + \frac{5}{3}n$

$5S_1(n) = \frac{5}{2}n^2 + \frac{5}{2}n$

Subtracting each:

$5S_4(n) = n^5 + \frac{5}{2}n^4 + 5n^3 + \frac{15}{2}n^2 + 4n - \frac{10}{3}n^3 - 5n^2 - \frac{5}{3}n - \frac{5}{2}n^2 - \frac{5}{2}n$

$= n^5 + \frac{5}{2}n^4 + \frac{5}{3}n^3 - \frac{n}{6}$

$S_4(n) = \frac{n}{30}\left(6n^4 + 15n^3 + 10n^2 - 1\right)$

Since $6(1)^4 + 15(1)^3 + 10(1)^2 - 1 = 30$, we have $n = 1$ as a root, so $(n - 1)$ is a factor:

$6n^4 + 15n^3 + 10n^2 - 1 = (n - 1)(6n^3 + 21n^2 + 31n + 1)$

Hmm, let me re-derive. Actually, dividing: $6n^4 + 15n^3 + 10n^2 - 1 = (n + 1)(6n^3 + 9n^2 + n - 1)$.

Verification: $(n+1)(6n^3 + 9n^2 + n - 1) = 6n^4 + 9n^3 + n^2 - n + 6n^3 + 9n^2 + n - 1 = 6n^4 + 15n^3 + 10n^2 - 1$. $\checkmark$

We can further factor: $6n^3 + 9n^2 + n - 1 = (3n^2 + 3n - 1)(2n + 1)$.

Verification: $(3n^2 + 3n - 1)(2n + 1) = 6n^3 + 3n^2 + 6n^2 + 3n - 2n - 1 = 6n^3 + 9n^2 + n - 1$. $\checkmark$

$S_4(n) = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)$

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Part (ii): $S_k(n)$ is a polynomial of degree $k+1$

We proceed by strong induction. $S_1(n) = \frac{1}{2}n^2 + \frac{1}{2}n$ is a polynomial of degree 2. Suppose $S_t(n)$ is a polynomial of degree $t + 1$ in $n$ for all $t < k$. From $(*)$:

$(k+1)S_k(n) = (n+1)^{k+1} - (n+1) - \binom{k+1}{2}S_{k-1}(n) - \cdots - \binom{k+1}{k}S_1(n)$

The term $(n+1)^{k+1} - (n+1)$ is a polynomial of degree $k+1$ in $n$. Each $S_{k+1-j}(n)$ for $j \geq 2$ is a polynomial of degree $k + 2 - j \leq k$ by the inductive hypothesis. Therefore $S_k(n)$ is a polynomial of degree $k + 1$ in $n$.

Constant term is zero. Setting $n = 0$: $S_k(0) = \sum_{r=0}^{0} r^k = 0^k = 0$ (for $k \geq 1$). So the constant term is 0.

Sum of coefficients is 1. Setting $n = 1$: $S_k(1) = \sum_{r=0}^{1} r^k = 0 + 1 = 1$. Since $S_k(1)$ equals the sum of the coefficients of $S_k(n)$, this sum is 1. $\blacksquare$

Examiner Notes

About three fifths attempted this question, often obtaining the starred result and the familiar $S_3(n)$ successfully, but with $S_4(n)$ tripping up many. Any that made progress on part (ii) tended to be able to complete the whole question.

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Question 3

Topic: 解析几何 (Coordinate Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 The point $P(a \cos \theta, b \sin \theta)$, where $a > b > 0$, lies on the ellipse

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

The point $S(-ea, 0)$, where $b^2 = a^2(1 - e^2)$, is a focus of the ellipse. The point $N$ is the foot of the perpendicular from the origin, $O$, to the tangent to the ellipse at $P$. The lines $SP$ and $ON$ intersect at $T$. Show that the $y$-coordinate of $T$ is

$\frac{b \sin \theta}{1 + e \cos \theta} .$

Show that $T$ lies on the circle with centre $S$ and radius $a$.

Hint

3. $\frac{dy}{dx} = \frac{b \cos \theta}{-a \sin \theta}$

So the line $ON$ is $y = \frac{a \sin \theta}{b \cos \theta} x$

$SP$ is $y = \frac{b \sin \theta}{a(\cos \theta + e)} (x + ae)$

Solving simultaneously by substituting for $x$ to find the $y$ coordinate of $T$,

$y = \frac{b \sin \theta}{a(\cos \theta + e)} \left( \frac{b \cos \theta}{a \sin \theta} y + ae \right)$

and using $b^2 = a^2(1 - e^2)$ to eliminate $a^2$ gives the required result.

Then the $x$ coordinate of $T$ is $\frac{b^2 \cos \theta}{a(1 + e \cos \theta)}$.

Eliminating $\theta$ using $\sec \theta + e = \frac{b^2}{ax}$ and $\tan \theta = \frac{by}{ax}$,

$(x, y)$ satisfies $\left( \frac{b^2}{ax} - e \right)^2 = 1 + \left( \frac{by}{ax} \right)^2$

and again using $b^2 = a^2(1 - e^2)$, this time to eliminate $b^2$, gives, following simplifying algebra

$(x + ae)^2 + y^2 = a^2$, as required.

Model Solution

Finding the tangent at $P$ and the line $ON$.

The ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiating: $\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.

At $P(a\cos\theta, b\sin\theta)$: $\frac{dy}{dx} = -\frac{b^2 \cdot a\cos\theta}{a^2 \cdot b\sin\theta} = -\frac{b\cos\theta}{a\sin\theta}$.

The tangent at $P$ has gradient $-\frac{b\cos\theta}{a\sin\theta}$, so $ON$ (perpendicular from origin to tangent) has gradient $\frac{a\sin\theta}{b\cos\theta}$.

Therefore $ON$: $y = \frac{a\sin\theta}{b\cos\theta}\,x$, i.e.\ $x = \frac{b\cos\theta}{a\sin\theta}\,y$.

Finding the line $SP$.

$S = (-ea, 0)$ and $P = (a\cos\theta, b\sin\theta)$. The gradient of $SP$ is:

$\frac{b\sin\theta - 0}{a\cos\theta - (-ea)} = \frac{b\sin\theta}{a(\cos\theta + e)}$

So $SP$: $y = \frac{b\sin\theta}{a(\cos\theta + e)}(x + ea)$.

Finding the $y$-coordinate of $T$.

Substituting $x = \frac{b\cos\theta}{a\sin\theta}\,y$ from $ON$ into the equation of $SP$:

$y = \frac{b\sin\theta}{a(\cos\theta + e)}\left(\frac{b\cos\theta}{a\sin\theta}\,y + ea\right)$

$y = \frac{b^2\cos\theta}{a^2(\cos\theta + e)}\,y + \frac{bea\sin\theta}{a(\cos\theta + e)}$

$y\left(1 - \frac{b^2\cos\theta}{a^2(\cos\theta + e)}\right) = \frac{be\sin\theta}{\cos\theta + e}$

$y \cdot \frac{a^2(\cos\theta + e) - b^2\cos\theta}{a^2(\cos\theta + e)} = \frac{be\sin\theta}{\cos\theta + e}$

Using $b^2 = a^2(1 - e^2)$:

$a^2(\cos\theta + e) - a^2(1-e^2)\cos\theta = a^2[\cos\theta + e - \cos\theta + e^2\cos\theta] = a^2 e(1 + e\cos\theta)$

Therefore:

$y = \frac{be\sin\theta}{\cos\theta + e} \cdot \frac{a^2(\cos\theta + e)}{a^2 e(1 + e\cos\theta)} = \frac{b\sin\theta}{1 + e\cos\theta}$

Finding the $x$-coordinate of $T$.

From $ON$: $x_T = \frac{b\cos\theta}{a\sin\theta} \cdot y_T = \frac{b\cos\theta}{a\sin\theta} \cdot \frac{b\sin\theta}{1 + e\cos\theta} = \frac{b^2\cos\theta}{a(1 + e\cos\theta)}$.

Showing $T$ lies on the circle with centre $S$ and radius $a$.

We need $(x_T + ea)^2 + y_T^2 = a^2$.

$x_T + ea = \frac{b^2\cos\theta}{a(1+e\cos\theta)} + ea = \frac{b^2\cos\theta + ea^2(1+e\cos\theta)}{a(1+e\cos\theta)}$

Using $b^2 = a^2(1-e^2)$:

$= \frac{a^2(1-e^2)\cos\theta + a^2 e + a^2 e^2\cos\theta}{a(1+e\cos\theta)} = \frac{a^2(\cos\theta + e)}{a(1+e\cos\theta)} = \frac{a(\cos\theta + e)}{1 + e\cos\theta}$

$(x_T + ea)^2 + y_T^2 = \frac{a^2(\cos\theta + e)^2}{(1+e\cos\theta)^2} + \frac{b^2\sin^2\theta}{(1+e\cos\theta)^2}$

$= \frac{a^2(\cos\theta + e)^2 + a^2(1-e^2)\sin^2\theta}{(1+e\cos\theta)^2}$

$= \frac{a^2[(\cos\theta + e)^2 + (1-e^2)\sin^2\theta]}{(1+e\cos\theta)^2}$

Expanding the numerator bracket:

$\cos^2\theta + 2e\cos\theta + e^2 + \sin^2\theta - e^2\sin^2\theta$

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

Therefore $(x_T + ea)^2 + y_T^2 = \frac{a^2(1+e\cos\theta)^2}{(1+e\cos\theta)^2} = a^2$. $\blacksquare$

Examiner Notes

Just under half attempted this. Most were reluctant to use parametric differentiation. Some found $T$‘s coordinates successfully and got not further, but most either made very little progress on the whole question, or got right through it.

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Question 4

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

Problem

4 (i) Show, with the aid of a sketch, that $y > \tanh(y/2)$ for $y > 0$ and deduce that

$\text{arcosh } x > \frac{x - 1}{\sqrt{x^2 - 1}} \quad \text{for} \quad x > 1. \qquad \text{(*)}$

(ii) By integrating (*), show that $\text{arcosh } x > 2 \frac{x - 1}{\sqrt{x^2 - 1}}$ for $x > 1$.

(iii) Show that $\text{arcosh } x > 3 \frac{\sqrt{x^2 - 1}}{x + 2}$ for $x > 1$.

[Note: $\text{arcosh } x$ is another notation for $\cosh^{-1} x$.]

Hint

4. (i)

y	z = y	z = tanh(y/2)
0	0	0
y > 0	y	tanh(y/2)

The graph of $z = y$ has gradient 1 and passes through the origin.

The graph of $z = \tanh\left(\frac{y}{2}\right)$ which has gradient $\frac{1}{2} \operatorname{sech}^2\left(\frac{y}{2}\right) \le \frac{1}{2}$ for $y \ge 0$ also passes though the origin and is asymptotic to $z = 1$.

Thus $y \ge \tanh\left(\frac{y}{2}\right)$ for $y \ge 0$.

If $x = \cosh y$, then $\sqrt{\frac{x-1}{x+1}} = \sqrt{\frac{\cosh y - 1}{\cosh y + 1}} = \sqrt{\frac{2 \sinh^2\left(\frac{y}{2}\right)}{2 \cosh^2\left(\frac{y}{2}\right)}} = \tanh\left(\frac{y}{2}\right)$

and as $y \ge \tanh\left(\frac{y}{2}\right)$ for $y \ge 0$, $\operatorname{arcosh} x \ge \sqrt{\frac{x-1}{x+1}}$ for $x \ge 1$.

$\sqrt{\frac{x-1}{x+1}} = \sqrt{\frac{x-1}{x+1}} \sqrt{\frac{x-1}{x-1}} = \frac{x-1}{\sqrt{x^2-1}}$ for $x > 1$, and (*) is obtained.

(ii) By parts $\int \operatorname{arcosh} x dx = x \operatorname{arcosh} x - \sqrt{x^2-1} + c$

and $\int \frac{x-1}{\sqrt{x^2-1}} dx = \sqrt{x^2-1} - \operatorname{arcosh} x + c'$

Thus $\int_1^x \operatorname{arcosh} x dx \ge \int_1^x \frac{x-1}{\sqrt{x^2-1}} dx$ for $x > 1$ gives

$x \operatorname{arcosh} x - \sqrt{x^2-1} \ge \sqrt{x^2-1} - \operatorname{arcosh} x$ for $x > 1$, which rearranges to give result

(iii) Integrating (ii) similarly gives $x \operatorname{arcosh} x - \sqrt{x^2-1} \ge 2\left(\sqrt{x^2-1} - \operatorname{arcosh} x\right)$ for $x > 1$, which also can be rearranged as desired.

Model Solution

Part (i): $y > \tanh(y/2)$ for $y > 0$

Consider the functions $f(y) = y$ and $g(y) = \tanh(y/2)$ for $y \geq 0$.

Both pass through the origin. The derivative of $g$ is $g'(y) = \frac{1}{2}\operatorname{sech}^2(y/2)$. Since $\operatorname{sech}^2(y/2) \leq 1$ for all $y$, we have $g'(y) \leq \frac{1}{2} < 1 = f'(y)$ for all $y \geq 0$.

Since $f(0) = g(0)$ and $f'(y) > g'(y)$ for all $y > 0$, it follows that $f(y) > g(y)$ for $y > 0$, i.e.\ $y > \tanh(y/2)$ for $y > 0$.

Deducing the inequality (*). Let $x = \cosh y$ with $y > 0$ (so $x > 1$). Then:

$\frac{x - 1}{x + 1} = \frac{\cosh y - 1}{\cosh y + 1} = \frac{2\sinh^2(y/2)}{2\cosh^2(y/2)} = \tanh^2(y/2)$

So $\sqrt{\frac{x-1}{x+1}} = \tanh(y/2)$ (taking the positive root since $y > 0$).

Since $y > \tanh(y/2)$:

$\operatorname{arcosh} x > \sqrt{\frac{x-1}{x+1}}$

For $x > 1$: $\sqrt{\frac{x-1}{x+1}} = \sqrt{\frac{(x-1)^2}{(x+1)(x-1)}} = \frac{x-1}{\sqrt{x^2-1}}$.

Therefore:

$\operatorname{arcosh} x > \frac{x-1}{\sqrt{x^2-1}} \quad \text{for } x > 1 \qquad (*)$

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Part (ii): $\operatorname{arcosh} x > 2\frac{x-1}{\sqrt{x^2-1}}$

Since both sides of $(*)$ are continuous on $[1, x]$ and the inequality is strict for $t > 1$, integrate from $1$ to $x$ (where $x > 1$):

$\int_1^x \operatorname{arcosh} t\, dt > \int_1^x \frac{t-1}{\sqrt{t^2-1}}\, dt$

Left side. Integration by parts with $u = \operatorname{arcosh} t$, $dv = dt$:

$\int_1^x \operatorname{arcosh} t\, dt = \left[t\operatorname{arcosh} t\right]_1^x - \int_1^x \frac{t}{\sqrt{t^2-1}}\, dt = x\operatorname{arcosh} x - \sqrt{x^2-1}$

(since $\operatorname{arcosh} 1 = 0$ and $\int \frac{t}{\sqrt{t^2-1}}\,dt = \sqrt{t^2-1}$).

Right side. $\frac{t-1}{\sqrt{t^2-1}} = \frac{t}{\sqrt{t^2-1}} - \frac{1}{\sqrt{t^2-1}}$

$\int_1^x \frac{t}{\sqrt{t^2-1}}\, dt = \sqrt{x^2-1}, \qquad \int_1^x \frac{1}{\sqrt{t^2-1}}\, dt = \operatorname{arcosh} x$

So $\int_1^x \frac{t-1}{\sqrt{t^2-1}}\, dt = \sqrt{x^2-1} - \operatorname{arcosh} x$.

Combining:

$x\operatorname{arcosh} x - \sqrt{x^2-1} > \sqrt{x^2-1} - \operatorname{arcosh} x$

$(x+1)\operatorname{arcosh} x > 2\sqrt{x^2-1} = 2(x-1) \cdot \frac{\sqrt{x^2-1}}{x-1} = 2\sqrt{x^2-1}$

Dividing by $(x+1) > 0$ and noting $\frac{2\sqrt{x^2-1}}{x+1} = \frac{2(x-1)}{\sqrt{x^2-1}}$:

$\operatorname{arcosh} x > 2\frac{x-1}{\sqrt{x^2-1}} \qquad \blacksquare$

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Part (iii): $\operatorname{arcosh} x > 3\frac{\sqrt{x^2-1}}{x+2}$

The result of part (ii) gives $\operatorname{arcosh} t > 2\frac{t-1}{\sqrt{t^2-1}}$ for $t > 1$. Integrate from $1$ to $x$:

$\int_1^x \operatorname{arcosh} t\, dt > 2\int_1^x \frac{t-1}{\sqrt{t^2-1}}\, dt$

From part (ii), the left side is $x\operatorname{arcosh} x - \sqrt{x^2-1}$ and the right side is $2(\sqrt{x^2-1} - \operatorname{arcosh} x)$:

$x\operatorname{arcosh} x - \sqrt{x^2-1} > 2\sqrt{x^2-1} - 2\operatorname{arcosh} x$

$(x+2)\operatorname{arcosh} x > 3\sqrt{x^2-1}$

Dividing by $(x+2) > 0$:

$\operatorname{arcosh} x > 3\frac{\sqrt{x^2-1}}{x+2} \qquad \blacksquare$

Examiner Notes

Almost exactly the same number attempted this as question 3, but with much less success. The initial inequality was frequently poorly justified, but some managed to apply it correctly to obtain the starred result, and went on to do part (ii) respectably. However, for most, it was a case of all or nothing.

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Question 5

Topic: 递推关系与多项式 (Recurrence Relations and Polynomials)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The functions $\text{T}_n(x)$, for $n = 0, 1, 2, \dots$, satisfy the recurrence relation

$\text{T}_{n+1}(x) - 2x\text{T}_n(x) + \text{T}_{n-1}(x) = 0 \quad (n \geqslant 1). \qquad \text{(*)}$

Show by induction that

$(\text{T}_n(x))^2 - \text{T}_{n-1}(x)\text{T}_{n+1}(x) = \text{f}(x),$

where $\text{f}(x) = (\text{T}_1(x))^2 - \text{T}_0(x)\text{T}_2(x)$.

In the case $\text{f}(x) \equiv 0$, determine (with proof) an expression for $\text{T}_n(x)$ in terms of $\text{T}_0(x)$ (assumed to be non-zero) and $\text{r}(x)$, where $\text{r}(x) = \text{T}_1(x)/\text{T}_0(x)$. Find the two possible expressions for $\text{r}(x)$ in terms of $x$.

Hint

5. There are a number of correct routes to proving the induction, though the simplest is to consider $((T_{k+1}(x))^2 - T_k(x)T_{k+2}(x)) - ((T_k(x))^2 - T_{k-1}(x)T_{k+1}(x))$

For $f(x) = 0$, $(T_n(x))^2 - T_{n-1}(x)T_{n+1}(x) = 0$

and so $\frac{T_{n+1}(x)}{T_n(x)} = \frac{T_n(x)}{T_{n-1}(x)}$ provided that neither denominator is zero, leading to

$\frac{T_n(x)}{T_{n-1}(x)} = \frac{T_1(x)}{T_0(x)} = r(x)$,

and so $\frac{T_n(x)}{T_{n-1}(x)} \times \frac{T_{n-1}(x)}{T_{n-2}(x)} \times \dots \times \frac{T_1(x)}{T_0(x)} = (r(x))^n$

Thus $T_n(x) = (r(x))^n T_0(x)$

Substituting this result into (*) for $n = 1$, $((r(x))^2 - 2xr(x) + 1)T_0(x) = 0$, and as $T_0(x) \neq 0$, solving the quadratic gives

$r(x) = x \pm \sqrt{x^2 - 1}$

Model Solution

Induction proof that $(T_n(x))^2 - T_{n-1}(x)T_{n+1}(x) = f(x)$.

Define $g(n) = (T_n(x))^2 - T_{n-1}(x)T_{n+1}(x)$.

Base case ($n = 1$): $g(1) = (T_1(x))^2 - T_0(x)T_2(x) = f(x)$ by definition. $\checkmark$

Inductive step: Suppose $g(k) = f(x)$ for some $k \geqslant 1$. We show $g(k+1) = f(x)$.

Consider the difference:

$g(k+1) - g(k) = (T_{k+1})^2 - T_k T_{k+2} - (T_k)^2 + T_{k-1} T_{k+1}$

$= T_{k+1}(T_{k+1} + T_{k-1}) - T_k(T_{k+2} + T_k)$

From the recurrence $T_{n+1} + T_{n-1} = 2xT_n$, we have $T_{k+1} + T_{k-1} = 2xT_k$ and $T_{k+2} + T_k = 2xT_{k+1}$. Substituting:

$g(k+1) - g(k) = T_{k+1} \cdot 2xT_k - T_k \cdot 2xT_{k+1} = 0$

Therefore $g(k+1) = g(k) = f(x)$.

By the principle of mathematical induction, $g(n) = f(x)$ for all $n \geqslant 1$. $\blacksquare$

Expression for $T_n(x)$ when $f(x) \equiv 0$.

When $f(x) = 0$, we have $(T_n)^2 = T_{n-1}T_{n+1}$ for all $n \geqslant 1$.

Rearranging (and assuming $T_n \neq 0$):

$\frac{T_{n+1}}{T_n} = \frac{T_n}{T_{n-1}}$

This ratio is constant for all $n \geqslant 1$, so:

$\frac{T_n}{T_{n-1}} = \frac{T_{n-1}}{T_{n-2}} = \cdots = \frac{T_1}{T_0} = r(x)$

Telescoping:

$\frac{T_n}{T_0} = \frac{T_n}{T_{n-1}} \cdot \frac{T_{n-1}}{T_{n-2}} \cdots \frac{T_1}{T_0} = (r(x))^n$

Therefore:

$T_n(x) = (r(x))^n \, T_0(x) \qquad \text{(1)}$

Finding $r(x)$ in terms of $x$.

Substituting (1) into the recurrence $T_2 - 2xT_1 + T_0 = 0$ (i.e.\ the case $n = 1$):

$(r)^2 T_0 - 2x \cdot r \cdot T_0 + T_0 = 0$

Since $T_0 \neq 0$, divide through:

$r^2 - 2xr + 1 = 0$

By the quadratic formula:

$r(x) = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}$

Examiner Notes

In terms of attempts and success, this resembled question 2. Apart from some that made no progress at all, the induction was accessible to many, as was the expression for $T_n(x)$. In both of these there were frequent gaps or inaccuracies even though the solutions were understood in essence.

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Question 6

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

Problem

6 In this question, $p$ denotes $\frac{\text{d}y}{\text{d}x}$.

(i) Given that

$y = p^2 + 2xp,$

show by differentiating with respect to $x$ that

$\frac{\text{d}x}{\text{d}p} = -2 - \frac{2x}{p}.$

Hence show that $x = -\frac{2}{3}p + Ap^{-2}$, where $A$ is an arbitrary constant. Find $y$ in terms of $x$ if $p = -3$ when $x = 2$.

(ii) Given instead that

$y = 2xp + p \ln p,$

and that $p = 1$ when $x = -\frac{1}{4}$, show that $x = -\frac{1}{2} \ln p - \frac{1}{4}$ and find $y$ in terms of $x$.

Hint

6. (i) Differentiating $y = p^2 + 2xp$ with respect to $x$ gives

$p = 2p \frac{dp}{dx} + 2x \frac{dp}{dx} + 2p$ which can be rearranged suitably.

The differential equation $\frac{dx}{dp} + \frac{2}{p}x = -2$ has an integrating factor $p^2$ and integrating will give the required general solution.

Substituting $x = 2, p = -3$, leads to $A = 0$, i.e. $p = -\frac{3}{2}x$ which can be substituted in the original equation and so $y = -\frac{3}{4}x^2$.

(ii) The same approach as in part (i) generates $\frac{dx}{dp} + \frac{2}{p}x = -\frac{(\ln p + 1)}{p}$, which with the same integrating factor has general solution

$x = -\frac{1}{4} - \frac{1}{2} \ln p + Bp^{-2}$

and particular solution

$x = -\frac{1}{2} \ln p - \frac{1}{4}$

Again, substitution of $\ln p$ (and $p$) in the original equation leads to the solution which is $y = -\frac{1}{2} e^{-2x-\frac{1}{2}}$

Model Solution

Part (i)

We are given $y = p^2 + 2xp$ where $p = \dfrac{dy}{dx}$.

Differentiate both sides with respect to $x$:

$\frac{dy}{dx} = 2p\frac{dp}{dx} + 2x\frac{dp}{dx} + 2p$

Since $\dfrac{dy}{dx} = p$:

$p = 2p\frac{dp}{dx} + 2x\frac{dp}{dx} + 2p$

$0 = 2p\frac{dp}{dx} + 2x\frac{dp}{dx} + p$

$-p = \frac{dp}{dx}(2p + 2x)$

Provided $p \neq 0$, divide both sides by $p$:

$-1 = \frac{dp}{dx}\left(2 + \frac{2x}{p}\right)$

$\frac{dx}{dp} = -2 - \frac{2x}{p} \qquad \checkmark$

This is a linear first-order ODE in $x$ as a function of $p$:

$\frac{dx}{dp} + \frac{2}{p}x = -2$

The integrating factor is $e^{\int \frac{2}{p}\,dp} = p^2$. Multiplying through:

$p^2\frac{dx}{dp} + 2px = -2p^2$

$\frac{d}{dp}(p^2 x) = -2p^2$

Integrate with respect to $p$:

$p^2 x = -\frac{2}{3}p^3 + A$

$x = -\frac{2}{3}p + Ap^{-2} \qquad \blacksquare$

Finding $y$ in terms of $x$ when $p = -3$ at $x = 2$.

Substitute $x = 2$, $p = -3$:

$2 = -\frac{2}{3}(-3) + \frac{A}{(-3)^2} = 2 + \frac{A}{9}$

So $A = 0$, giving $x = -\dfrac{2}{3}p$, hence $p = -\dfrac{3}{2}x$.

Substituting into $y = p^2 + 2xp$:

$y = \left(-\frac{3}{2}x\right)^2 + 2x\left(-\frac{3}{2}x\right) = \frac{9}{4}x^2 - 3x^2 = -\frac{3}{4}x^2$

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Part (ii)

We are given $y = 2xp + p\ln p$ where $p = \dfrac{dy}{dx}$.

Differentiate both sides with respect to $x$:

$\frac{dy}{dx} = 2p + 2x\frac{dp}{dx} + \frac{dp}{dx}\ln p + p \cdot \frac{1}{p}\frac{dp}{dx}$

Since $\dfrac{dy}{dx} = p$:

$p = 2p + 2x\frac{dp}{dx} + (\ln p)\frac{dp}{dx} + \frac{dp}{dx}$

$-p = \frac{dp}{dx}(2x + \ln p + 1)$

Divide through by $p$:

$-1 = \frac{dp}{dx}\left(\frac{2x}{p} + \frac{\ln p + 1}{p}\right)$

$\frac{dx}{dp} = -\frac{2x}{p} - \frac{\ln p + 1}{p}$

This is a linear ODE:

$\frac{dx}{dp} + \frac{2}{p}x = -\frac{\ln p + 1}{p}$

The integrating factor is again $p^2$. Multiplying through:

$\frac{d}{dp}(p^2 x) = -p(\ln p + 1)$

Integrate the right-hand side. Since $\dfrac{d}{dp}(p \ln p) = \ln p + 1$:

$\int p(\ln p + 1)\,dp = \int p \cdot \frac{d}{dp}(p\ln p)\,dp$

Using integration by parts with $u = p$, $dv = (\ln p + 1)\,dp$ (so $v = p\ln p$):

$\int p(\ln p + 1)\,dp = p \cdot p\ln p - \int p\ln p\,dp$

Alternatively, note that $\dfrac{d}{dp}\!\left(\frac{p^2}{2}\ln p\right) = p\ln p + \frac{p}{2}$, so

$\int p(\ln p + 1)\,dp = \int p\ln p\,dp + \frac{p^2}{2}$

and $\displaystyle\int p\ln p\,dp = \frac{p^2}{2}\ln p - \frac{p^2}{4}$ (by parts with $u = \ln p$, $dv = p\,dp$).

Therefore:

$\int p(\ln p + 1)\,dp = \frac{p^2}{2}\ln p - \frac{p^2}{4} + \frac{p^2}{2} = \frac{p^2}{2}\ln p + \frac{p^2}{4}$

So:

$p^2 x = -\frac{p^2}{2}\ln p - \frac{p^2}{4} + B$

$x = -\frac{1}{2}\ln p - \frac{1}{4} + Bp^{-2}$

Substitute the condition $p = 1$, $x = -\dfrac{1}{4}$:

$-\frac{1}{4} = -\frac{1}{2}\ln 1 - \frac{1}{4} + B = -\frac{1}{4} + B$

So $B = 0$, giving:

$x = -\frac{1}{2}\ln p - \frac{1}{4} \qquad \checkmark$

Finding $y$ in terms of $x$.

From $x = -\dfrac{1}{2}\ln p - \dfrac{1}{4}$, we get $\ln p = -2x - \dfrac{1}{2}$, so $p = e^{-2x - 1/2}$.

Substituting into $y = 2xp + p\ln p$:

$y = 2x \cdot e^{-2x-1/2} + e^{-2x-1/2}(-2x - \tfrac{1}{2})$

$= e^{-2x-1/2}\bigl[2x - 2x - \tfrac{1}{2}\bigr] = -\frac{1}{2}e^{-2x-1/2}$

Examiner Notes

More than 80% attempted this, and with more success than any other question. Having obtained the relation between $x$ and $p$ in each part, quite a few attempts then treated these as differential equations rather than merely substituting back to find expressions for $y$, and consequent inaccuracies lost marks.

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Question 7

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

Problem

7 The points $A$, $B$ and $C$ in the Argand diagram are the vertices of an equilateral triangle described anticlockwise. Show that the complex numbers $a$, $b$ and $c$ representing $A$, $B$ and $C$ satisfy $2c = (a + b) + i\sqrt{3}(b - a).$

Find a similar relation in the case that $A$, $B$ and $C$ are the vertices of an equilateral triangle described clockwise.

(i) The quadrilateral $DEFG$ lies in the Argand diagram. Show that points $P$, $Q$, $R$ and $S$ can be chosen so that $PDE$, $QEF$, $RFG$ and $SGD$ are equilateral triangles and $PQRS$ is a parallelogram.

(ii) The triangle $LMN$ lies in the Argand diagram. Show that the centroids $U$, $V$ and $W$ of the equilateral triangles drawn externally on the sides of $LMN$ are the vertices of an equilateral triangle.

[Note: The centroid of a triangle with vertices represented by the complex numbers $x$, $y$ and $z$ is the point represented by $\frac{1}{3}(x + y + z)$.]

Hint

7. The starting point $c - a = \frac{1}{2}(1 + i\sqrt{3})(b - a)$ leads to the given result.

Interchanging $a$ and $b$ gives $2c = (a + b) + i\sqrt{3}(a - b)$ if A, B, C are described clockwise.

(i) The clue to this is the phrase “can be chosen” and a sketch demonstrates that a pair of the equilateral triangles need to be clockwise, and the other pair anti-clockwise

Applying the results in the stem of the question to this configuration,

$2p = (d + e) + i\sqrt{3}(e - d)$ $2q = (e + f) + i\sqrt{3}(e - f)$ $2r = (f + g) + i\sqrt{3}(g - f)$ $2s = (g + d) + i\sqrt{3}(g - d)$

and so $2PS = (g - e) + i\sqrt{3}(g - e) = -2RQ$, PSQR is a parallelogram.

(The pairs could have been chosen with opposite parity leading to very similar working.)

(ii) Supposing LMN is clockwise, U is the centroid of equilateral triangle LMH, V of MNJ, and W of NLK, then

$3u = l + m + h$ where $2h = (l + m) + i\sqrt{3}(m - l)$ with similar results for $v$ and $w$.

Both $6w$, and $3[(u + v) + i\sqrt{3}(u - v)]$ can be shown to equal $3(n + l) + i\sqrt{3}(l - n)$ and so UVW is a clockwise equilateral triangle.

Model Solution

Establishing the relation for anticlockwise equilateral triangles

Let $A$, $B$, $C$ be vertices of an equilateral triangle described anticlockwise. The vector from $A$ to $B$ is $b - a$. Rotating this by $60°$ anticlockwise (multiplying by $e^{i\pi/3}$) gives the vector from $A$ to $C$:

$c - a = (b - a)e^{i\pi/3}$

Since $e^{i\pi/3} = \frac{1}{2} + \frac{i\sqrt{3}}{2}$:

$c - a = (b - a)\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)$

$2(c - a) = (b - a)(1 + i\sqrt{3})$

$2c = 2a + (b - a) + i\sqrt{3}(b - a) = (a + b) + i\sqrt{3}(b - a)$

For a clockwise equilateral triangle, we rotate by $-60°$ instead:

$c - a = (b - a)e^{-i\pi/3} = (b - a)\left(\frac{1}{2} - \frac{i\sqrt{3}}{2}\right)$

$2c = (a + b) - i\sqrt{3}(b - a)$

Part (i): Constructing the parallelogram PQRS

Given quadrilateral $DEFG$, we construct equilateral triangles on each side. For each side, we choose the triangle to be on the outside of the quadrilateral (consistent orientation).

Let $P$ be the third vertex of the equilateral triangle on $DE$ (described anticlockwise), $Q$ on $EF$, $R$ on $FG$, $S$ on $GD$.

Using the relation from above:

$2p = (d + e) + i\sqrt{3}(e - d)$ $2q = (e + f) + i\sqrt{3}(f - e)$ $2r = (f + g) + i\sqrt{3}(g - f)$ $2s = (g + d) + i\sqrt{3}(d - g)$

To show $PQRS$ is a parallelogram, we need $q - p = s - r$ (or equivalently $p + r = q + s$).

$2(q - p) = (e + f) + i\sqrt{3}(f - e) - (d + e) - i\sqrt{3}(e - d)$ $= (f - d) + i\sqrt{3}(f - 2e + d)$

$2(s - r) = (g + d) + i\sqrt{3}(d - g) - (f + g) - i\sqrt{3}(g - f)$ $= (d - f) + i\sqrt{3}(d - 2g + f)$

These are equal if and only if $f - 2e + d = d - 2g + f$, i.e., $e = g$. That’s not generally true, so let me reconsider the orientation choices.

Actually, we can choose the equilateral triangles with alternating orientations. Let $P$ on $DE$ be clockwise, $Q$ on $EF$ anticlockwise, $R$ on $FG$ clockwise, $S$ on $GD$ anticlockwise:

$2p = (d + e) - i\sqrt{3}(e - d)$ $2q = (e + f) + i\sqrt{3}(f - e)$ $2r = (f + g) - i\sqrt{3}(g - f)$ $2s = (g + d) + i\sqrt{3}(d - g)$

Then:

$2(q - p) = (f - d) + i\sqrt{3}(f + e - d - e) + i\sqrt{3}(e - d) = (f - d) + i\sqrt{3}(f - d) + i\sqrt{3}(e - d) + i\sqrt{3}(e - d)$

Let me compute this more carefully:

$2q - 2p = [(e+f) - (d+e)] + i\sqrt{3}[(f-e) + (e-d)] = (f - d) + i\sqrt{3}(f - d)$

$2s - 2r = [(g+d) - (f+g)] + i\sqrt{3}[(d-g) + (g-f)] = (d - f) + i\sqrt{3}(d - f)$

So $2(q-p) = (f-d)(1 + i\sqrt{3})$ and $2(s-r) = (d-f)(1 + i\sqrt{3}) = -(f-d)(1+i\sqrt{3})$.

This gives $q - p = -(s - r)$, so $PSRQ$ is a parallelogram (with $PQRS$ having $PQ \parallel SR$ and $PQ = SR$). More precisely, $p + r = q + s$ confirms $PQRS$ is a parallelogram.

Part (ii): Napoleon’s theorem (centroids of external equilateral triangles)

Let triangle $LMN$ be given. Construct equilateral triangles externally on sides $LM$, $MN$, $NL$, with third vertices $H$, $J$, $K$ respectively. Their centroids are:

$u = \frac{l + m + h}{3}, \quad v = \frac{m + n + j}{3}, \quad w = \frac{n + l + k}{3}$

Using the anticlockwise relation (external to triangle $LMN$):

$2h = (l + m) + i\sqrt{3}(m - l)$ $2j = (m + n) + i\sqrt{3}(n - m)$ $2k = (n + l) + i\sqrt{3}(l - n)$

So:

$6u = 2l + 2m + (l+m) + i\sqrt{3}(m-l) = 3(l+m) + i\sqrt{3}(m-l)$ $6v = 3(m+n) + i\sqrt{3}(n-m)$ $6w = 3(n+l) + i\sqrt{3}(l-n)$

To show $UVW$ is equilateral, we verify $6w = 6u + 6v + i\sqrt{3}(6v - 6u)$… no, we use the relation $2w' = (u' + v') + i\sqrt{3}(v' - u')$ where $u', v', w'$ are the centroids. Let me check:

$6v - 6u = 3(n - l) + i\sqrt{3}[(n-m) - (m-l)] = 3(n-l) + i\sqrt{3}(n - 2m + l)$

$6u + 6v = 3(l + 2m + n) + i\sqrt{3}[(m-l) + (n-m)] = 3(l+2m+n) + i\sqrt{3}(n-l)$

$6u + 6v + i\sqrt{3}(6v - 6u) = 3(l+2m+n) + i\sqrt{3}(n-l) + i\sqrt{3}[3(n-l) + i\sqrt{3}(n-2m+l)]$

$= 3(l+2m+n) + i\sqrt{3}(n-l) + 3i\sqrt{3}(n-l) - 3(n-2m+l)$

$= 3[(l+2m+n) - (n-2m+l)] + i\sqrt{3}[(n-l) + 3(n-l)]$

$= 3(4m) + i\sqrt{3} \cdot 4(n-l) = 12m + 4i\sqrt{3}(n-l)$

And $12w = 6(n+l) + 2i\sqrt{3}(l-n) = 6(n+l) - 2i\sqrt{3}(n-l)$.

These aren’t equal, so let me recheck. Actually, the correct relation to verify is:

$2 \cdot 6w = (6u + 6v) + i\sqrt{3}(6v - 6u)$

LHS: $12w = 6(n+l) + 2i\sqrt{3}(l-n)$

RHS: $3(l+2m+n) + i\sqrt{3}(n-l) + i\sqrt{3}[3(n-l) + i\sqrt{3}(n-2m+l)]$

$= 3(l+2m+n) + i\sqrt{3}(n-l) + 3i\sqrt{3}(n-l) - 3(n-2m+l)$

$= 3l + 6m + 3n - 3n + 6m - 3l + 4i\sqrt{3}(n-l)$

$= 12m + 4i\sqrt{3}(n-l)$

So RHS $= 12m + 4i\sqrt{3}(n-l)$ while LHS $= 6n + 6l - 2i\sqrt{3}(n-l)$. These are not equal in general.

Let me try the clockwise relation instead. Using $2h = (l+m) - i\sqrt{3}(m-l)$ etc.:

$6u = 3(l+m) - i\sqrt{3}(m-l)$ $6v = 3(m+n) - i\sqrt{3}(n-m)$ $6w = 3(n+l) - i\sqrt{3}(l-n)$

$6v - 6u = 3(n-l) - i\sqrt{3}[(n-m)-(m-l)] = 3(n-l) - i\sqrt{3}(n-2m+l)$

$6u + 6v = 3(l+2m+n) - i\sqrt{3}(n-l)$

RHS: $(6u+6v) + i\sqrt{3}(6v-6u) = 3(l+2m+n) - i\sqrt{3}(n-l) + i\sqrt{3}[3(n-l) - i\sqrt{3}(n-2m+l)]$

$= 3(l+2m+n) - i\sqrt{3}(n-l) + 3i\sqrt{3}(n-l) + 3(n-2m+l)$

$= 3[(l+2m+n)+(n-2m+l)] + 2i\sqrt{3}(n-l)$

$= 3(2l+2n) + 2i\sqrt{3}(n-l) = 6(l+n) + 2i\sqrt{3}(n-l)$

And $12w = 6(n+l) - 2i\sqrt{3}(l-n) = 6(n+l) + 2i\sqrt{3}(n-l)$.

So RHS $= 12w =$ LHS. Therefore $UVW$ is an equilateral triangle (described clockwise). $\blacksquare$

Examiner Notes

Less than a fifth attempted this and frequently with little success except for obtaining the initial result. The configuration for part (i) tripped up many, although some skipped that to do part (ii) successfully.

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Question 8

Topic: 级数 (Series)  |  Difficulty: Standard  |  Marks: 20

Problem

8 (i) The coefficients in the series $S = \frac{1}{3}x + \frac{1}{6}x^2 + \frac{1}{12}x^3 + \dots + a_rx^r + \dots$ satisfy a recurrence relation of the form $a_{r+1} + pa_r = 0$. Write down the value of $p$. By considering $(1 + px)S$, find an expression for the sum to infinity of $S$ (assuming that it exists). Find also an expression for the sum of the first $n + 1$ terms of $S$.

(ii) The coefficients in the series $T = 2 + 8x + 18x^2 + 37x^3 + \dots + a_rx^r + \dots$ satisfy a recurrence relation of the form $a_{r+2} + pa_{r+1} + qa_r = 0$. Find an expression for the sum to infinity of $T$ (assuming that it exists). By expressing $T$ in partial fractions, or otherwise, find an expression for the sum of the first $n + 1$ terms of $T$.

Hint

8. (i) $p = -\frac{1}{2}$ $(1 + px)S = \frac{1}{3}x$ with all other terms cancelling and so $S = \frac{1}{3}x / \left(1 - \frac{1}{2}x\right) = \frac{2x}{3(2 - x)}$

Using the sum of a GP

$S_{n+1} = \frac{1}{3}x + \frac{1}{6}x^2 + \frac{1}{12}x^3 + ... + \frac{1}{3 \times 2^n}x^{n+1} = \frac{\frac{1}{3}x\left(1 - \frac{x^{n+1}}{2^{n+1}}\right)}{\left(1 - \frac{x}{2}\right)}$

Alternatively $S_{n+1} = S - (a_{n+2}x^{n+2} + ...) = S - \frac{1}{2^{n+1}}x^{n+1}S$

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

(ii) Using similar working to part (i) $18 + 8p + 2q = 0$ $37 + 18p + 8q = 0$ so $p = -\frac{5}{2}, q = 1$ and so $(1 + px + qx^2)T = 2 + 3x$ giving $T = (2 + 3x) / \left(1 - \frac{5}{2}x + x^2\right) = \frac{4 + 6x}{2 - 5x + 2x^2} = \frac{4 + 6x}{(2 - x)(1 - 2x)}$ By partial fractions $T = \frac{14}{3}(1 - 2x)^{-1} - \frac{8}{3}\left(1 - \frac{x}{2}\right)^{-1}$ and so $T_{n+1} = \frac{14}{3}(1 + 2x + (2x)^2 + ... + (2x)^n) - \frac{8}{3}\left(1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + ... + \left(\frac{x}{2}\right)^n\right)$ $= \frac{14}{3} \frac{(1 - (2x)^{n+1})}{1 - 2x} - \frac{8}{3} \frac{\left(1 - \left(\frac{x}{2}\right)^{n+1}\right)}{1 - \frac{x}{2}}$

Model Solution

Part (i)

The series is $S = \frac{1}{3}x + \frac{1}{6}x^2 + \frac{1}{12}x^3 + \cdots$, so $a_1 = \frac{1}{3}$, $a_2 = \frac{1}{6}$, $a_3 = \frac{1}{12}$.

Checking: $\frac{a_2}{a_1} = \frac{1/6}{1/3} = \frac{1}{2}$ and $\frac{a_3}{a_2} = \frac{1/12}{1/6} = \frac{1}{2}$.

So $a_{r+1} = \frac{1}{2}a_r$, giving $p = -\frac{1}{2}$.

Sum to infinity: Consider $(1 - \frac{1}{2}x)S$:

$\left(1 - \frac{1}{2}x\right)S = \frac{1}{3}x + \left(\frac{1}{6} - \frac{1}{6}\right)x^2 + \left(\frac{1}{12} - \frac{1}{12}\right)x^3 + \cdots = \frac{1}{3}x$

since each coefficient of $x^r$ for $r \geq 2$ is $a_r - \frac{1}{2}a_{r-1} = 0$. Therefore:

$S = \frac{\frac{1}{3}x}{1 - \frac{1}{2}x} = \frac{2x}{3(2 - x)}$

Sum of first $n+1$ terms: The partial sum $S_{n+1} = \sum_{r=1}^{n+1} a_r x^r$ where $a_r = \frac{1}{3 \cdot 2^{r-1}}$.

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

Alternatively, $S_{n+1} = S - \sum_{r=n+2}^{\infty} a_r x^r = S - a_{n+2}x^{n+2} \cdot \frac{1}{1-x/2}$. Since $a_{n+2} = \frac{1}{3 \cdot 2^{n+1}}$:

$S_{n+1} = S - \frac{x^{n+1}}{2^{n+1}} \cdot S = \left(1 - \frac{x^{n+1}}{2^{n+1}}\right)\frac{2x}{3(2-x)}$

Part (ii)

The series is $T = 2 + 8x + 18x^2 + 37x^3 + \cdots$, so $a_0 = 2$, $a_1 = 8$, $a_2 = 18$, $a_3 = 37$.

From $a_{r+2} + pa_{r+1} + qa_r = 0$:

$18 + 8p + 2q = 0 \quad \text{and} \quad 37 + 18p + 8q = 0$

From the first: $q = -9 - 4p$. Substituting: $37 + 18p + 8(-9-4p) = 0 \Rightarrow 37 + 18p - 72 - 32p = 0 \Rightarrow -35 - 14p = 0$, so $p = -\frac{5}{2}$ and $q = -9 + 10 = 1$.

Sum to infinity: Consider $(1 - \frac{5}{2}x + x^2)T$. The coefficient of $x^r$ for $r \geq 2$ is $a_r - \frac{5}{2}a_{r-1} + a_{r-2} = 0$ by the recurrence. So:

$(1 - \tfrac{5}{2}x + x^2)T = a_0 + (a_1 - \tfrac{5}{2}a_0)x = 2 + 3x$

$T = \frac{2 + 3x}{1 - \frac{5}{2}x + x^2} = \frac{2(2 + 3x)}{(2 - x)(1 - 2x)}$

For the sum to exist, we need $|x| < \frac{1}{2}$ (so that both geometric series converge).

Partial fraction decomposition:

$\frac{2(2+3x)}{(2-x)(1-2x)} = \frac{A}{2-x} + \frac{B}{1-2x}$

Setting $x = 2$: $B(1-4) = 2(2+6) = 16$, so $B = -\frac{16}{3}$.

Setting $x = \frac{1}{2}$: $A(1 - 1)$… let me redo. Setting $x = \frac{1}{2}$: $A(\frac{1}{2} \cdot 2 - \frac{1}{4}) =$… Actually:

$2(2+3x) = A(1-2x) + B(2-x)$

$x = \frac{1}{2}$: $2 \cdot \frac{7}{2} = A \cdot 0 + B \cdot \frac{3}{2}$, so $B = \frac{14}{3}$.

$x = 2$: $2 \cdot 8 = A(-3) + B \cdot 0$, so $A = -\frac{16}{3}$.

Therefore:

$T = \frac{-\frac{16}{3}}{2-x} + \frac{\frac{14}{3}}{1-2x} = -\frac{8}{3}\cdot\frac{1}{1-x/2} + \frac{14}{3}\cdot\frac{1}{1-2x}$

Sum of first $n+1$ terms: Expanding as geometric series:

$T_{n+1} = -\frac{8}{3}\sum_{r=0}^{n}\left(\frac{x}{2}\right)^r + \frac{14}{3}\sum_{r=0}^{n}(2x)^r = -\frac{8}{3}\cdot\frac{1-(x/2)^{n+1}}{1-x/2} + \frac{14}{3}\cdot\frac{1-(2x)^{n+1}}{1-2x}$

Examiner Notes

Three fifths attempted this with most scoring about two thirds of the marks. Apart from minor errors, the last part (expressing $T$ in partial fractions etc.) was the pitfall for most.