STEP3 2007 -- Pure Mathematics

STEP3 2007 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2007  |  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	三角函数与多项式根 (Trigonometry and Polynomial Roots)	Challenging	复合角公式展开,韦达定理（根与系数关系）,正切替换化为多项式
2	级数与二项式定理 (Series and Binomial Theorem)	Challenging	阶乘表达式化简,二项式级数展开,幂级数逐项微分与积分,代入特定值求和
3	斐波那契数列与数学归纳法 (Fibonacci Sequences and Mathematical Induction)	Challenging	数学归纳法,斐波那契递推关系,恒等式变形与化简
4	参数方程与微分几何 (Parametric Equations and Differential Geometry)	Challenging	参数方程求导,链式法则,曲率公式,曲率中心计算
5	双曲函数与高阶导数 (Hyperbolic Functions and Higher Derivatives)	Hard	双曲函数代换,隐函数求导,高阶导数计算,数学归纳法
6	复数 (Complex Numbers)	Hard	复数极坐标形式,复数共轭运算,垂直条件转化
7	积分 (Integration)	Challenging	变量代换,定积分拆分,函数方程,级数与积分联系
8	微分方程 (Differential Equations)	Challenging	线性ODE通解,Riccati方程代换,初值条件代入

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Question 1

Topic: 三角函数与多项式根 (Trigonometry and Polynomial Roots)  |  Difficulty: Challenging  |  Marks: 20

Problem

1 In this question, do not consider the special cases in which the denominators of any of your expressions are zero.

Express $\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4)$ in terms of $t_i$, where $t_1 = \tan \theta_1$, etc.

Given that $\tan \theta_1, \tan \theta_2, \tan \theta_3$ and $\tan \theta_4$ are the four roots of the equation

$at^4 + bt^3 + ct^2 + dt + e = 0$

(where $a \neq 0$), find an expression in terms of $a, b, c, d$ and $e$ for $\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4)$.

The four real numbers $\theta_1, \theta_2, \theta_3$ and $\theta_4$ lie in the range $0 \le \theta_i < 2\pi$ and satisfy the equation

$p \cos 2\theta + \cos(\theta - \alpha) + p = 0,$

where $p$ and $\alpha$ are independent of $\theta$. Show that $\theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi$ for some integer $n$.

Hint

The first result can be obtained by applying a compound angle formula to $\tan((\theta_1 + \theta_2) + (\theta_3 + \theta_4))$ and then repeating the application to each of $\tan(\theta_1 + \theta_2)$ and $\tan(\theta_3 + \theta_4)$ where they appear. On simplification, this gives

$\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{t_1 + t_2 + t_3 + t_4 - t_2 t_3 t_4 - t_3 t_4 t_1 - t_4 t_1 t_2 - t_1 t_2 t_3}{1 - t_1 t_2 - t_1 t_3 - t_1 t_4 - t_2 t_3 - t_2 t_4 - t_3 t_4 + t_1 t_2 t_3 t_4} .$

As $t_i$, etc are the roots of the equation $at^4 + bt^3 + ct^2 + dt + e = 0$, then $at^4 + bt^3 + ct^2 + dt + e = a(t - t_1)(t - t_2)(t - t_3)(t - t_4)$, which yields, from expansion and comparison of coefficients, the four results

$t_1 + t_2 + t_3 + t_4 = \frac{-b}{a}, \quad t_1 t_2 + t_1 t_3 + t_1 t_4 + t_2 t_3 + t_2 t_4 + t_3 t_4 = \frac{c}{a},$ $t_2 t_3 t_4 + t_3 t_4 t_1 + t_4 t_1 t_2 + t_1 t_2 t_3 = \frac{-d}{a}, \text{ and } t_1 t_2 t_3 t_4 = \frac{e}{a} .$

These substituted in the first result lead to $\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{-b + d}{a - c + e}$.

Applying double and compound angle formulae to $p \cos 2\theta + \cos(\theta - \alpha) + p = 0$ gives the equation $2p \cos^2 \theta + \cos \theta \cos \alpha + \sin \theta \sin \alpha = 0$, which can be rearranged as $\cos \alpha + \tan \theta \sin \alpha = \frac{-2p}{\sec \theta} .$

Squaring this and replacing $\tan \theta$ by $t$, $(\cos \alpha + t \sin \alpha)^2 = \frac{4p^2}{1 + t^2}$.

Rearranging this obtains the quartic equation $t^4 \sin^2 \alpha + t^3 \sin 2\alpha + t^2 + t \sin 2\alpha + (\cos^2 \alpha - 4p^2) = 0$, and so, from the second result $\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{0}{-4p^2} = 0$, and thus $\theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi$.

Model Solution

Expressing $\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4)$ in terms of $t_i$

We apply the compound angle formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ repeatedly.

First, $\tan(\theta_1 + \theta_2) = \frac{t_1 + t_2}{1 - t_1 t_2}$ and $\tan(\theta_3 + \theta_4) = \frac{t_3 + t_4}{1 - t_3 t_4}$.

Then

$\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\tan(\theta_1 + \theta_2) + \tan(\theta_3 + \theta_4)}{1 - \tan(\theta_1 + \theta_2)\tan(\theta_3 + \theta_4)}.$

Let $A = \tan(\theta_1 + \theta_2) = \frac{t_1 + t_2}{1 - t_1 t_2}$ and $B = \tan(\theta_3 + \theta_4) = \frac{t_3 + t_4}{1 - t_3 t_4}$. Then

$\frac{A + B}{1 - AB} = \frac{\frac{t_1 + t_2}{1 - t_1 t_2} + \frac{t_3 + t_4}{1 - t_3 t_4}}{1 - \frac{(t_1 + t_2)(t_3 + t_4)}{(1 - t_1 t_2)(1 - t_3 t_4)}}.$

Multiplying numerator and denominator by $(1 - t_1 t_2)(1 - t_3 t_4)$:

$= \frac{(t_1 + t_2)(1 - t_3 t_4) + (t_3 + t_4)(1 - t_1 t_2)}{(1 - t_1 t_2)(1 - t_3 t_4) - (t_1 + t_2)(t_3 + t_4)}.$

Expanding the numerator:

$(t_1 + t_2) - (t_1 + t_2)t_3 t_4 + (t_3 + t_4) - (t_3 + t_4)t_1 t_2$ $= t_1 + t_2 + t_3 + t_4 - t_1 t_3 t_4 - t_2 t_3 t_4 - t_1 t_2 t_3 - t_1 t_2 t_4.$

Expanding the denominator:

$1 - t_1 t_2 - t_3 t_4 + t_1 t_2 t_3 t_4 - t_1 t_3 - t_1 t_4 - t_2 t_3 - t_2 t_4.$

Therefore

$\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{S_1 - S_3}{1 - S_2 + S_4} \qquad \text{(i)}$

where $S_1 = \sum t_i$, $S_2 = \sum t_i t_j$, $S_3 = \sum t_i t_j t_k$, $S_4 = t_1 t_2 t_3 t_4$ are the elementary symmetric functions.

Using Vieta’s relations

Since $t_1, t_2, t_3, t_4$ are roots of $at^4 + bt^3 + ct^2 + dt + e = 0$, we have $at^4 + bt^3 + ct^2 + dt + e = a(t - t_1)(t - t_2)(t - t_3)(t - t_4)$. Expanding and comparing coefficients:

$S_1 = t_1 + t_2 + t_3 + t_4 = \frac{-b}{a}, \quad S_2 = \sum t_i t_j = \frac{c}{a},$ $S_3 = \sum t_i t_j t_k = \frac{-d}{a}, \quad S_4 = t_1 t_2 t_3 t_4 = \frac{e}{a}.$

Substituting into (i):

$\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\frac{-b}{a} - \frac{-d}{a}}{1 - \frac{c}{a} + \frac{e}{a}} = \frac{-b + d}{a - c + e}. \qquad \text{(ii)}$

Showing $\theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi$

Starting from $p\cos 2\theta + \cos(\theta - \alpha) + p = 0$, we use $\cos 2\theta = 2\cos^2\theta - 1$:

$p(2\cos^2\theta - 1) + \cos(\theta - \alpha) + p = 0,$ $2p\cos^2\theta + \cos(\theta - \alpha) = 0.$

Expanding $\cos(\theta - \alpha) = \cos\theta\cos\alpha + \sin\theta\sin\alpha$:

$2p\cos^2\theta + \cos\theta\cos\alpha + \sin\theta\sin\alpha = 0.$

Dividing by $\cos\theta$ (assuming $\cos\theta \neq 0$; if $\cos\theta = 0$ the equation becomes $\sin\alpha = 0$, a special case excluded by the problem):

$2p\cos\theta + \cos\alpha + \tan\theta\sin\alpha = 0.$

Since $\cos\theta = \frac{1}{\sec\theta} = \frac{1}{\sqrt{1 + t^2}}$ where $t = \tan\theta$, we rearrange:

$\cos\alpha + t\sin\alpha = \frac{-2p}{\sqrt{1 + t^2}}.$

Squaring both sides:

$(\cos\alpha + t\sin\alpha)^2 = \frac{4p^2}{1 + t^2}.$

Multiplying by $(1 + t^2)$:

$(\cos\alpha + t\sin\alpha)^2(1 + t^2) = 4p^2.$

Expanding the left side:

$(\cos^2\alpha + 2t\cos\alpha\sin\alpha + t^2\sin^2\alpha)(1 + t^2) = 4p^2,$

$\cos^2\alpha + t^2\cos^2\alpha + 2t\cos\alpha\sin\alpha + 2t^3\cos\alpha\sin\alpha + t^2\sin^2\alpha + t^4\sin^2\alpha = 4p^2.$

Collecting by powers of $t$:

$t^4\sin^2\alpha + t^3\sin 2\alpha + t^2 + t\sin 2\alpha + (\cos^2\alpha - 4p^2) = 0. \qquad \text{(iii)}$

This is a quartic in $t = \tan\theta$. The four roots are $t_1 = \tan\theta_1, \ldots, t_4 = \tan\theta_4$.

Applying Vieta’s relations to (iii) with leading coefficient $\sin^2\alpha$:

$S_1 = \frac{-\sin 2\alpha}{\sin^2\alpha}, \quad S_3 = \frac{-\sin 2\alpha}{\sin^2\alpha}.$

Using formula (ii), where the coefficients of the quartic are $a = \sin^2\alpha$, $b = \sin 2\alpha$, $c = 1$, $d = \sin 2\alpha$, $e = \cos^2\alpha - 4p^2$:

$\tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{-b + d}{a - c + e} = \frac{-\sin 2\alpha + \sin 2\alpha}{\sin^2\alpha - 1 + \cos^2\alpha - 4p^2} = \frac{0}{-4p^2} = 0.$

Therefore $\theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi$ for some integer $n$.

Examiner Notes

This question was popular. Many candidates did not simplify their first expression into the symmetrical form which made it harder for them to spot the use of the sums and products of roots results. A common slip was to make a 1 by default which also obscured what was going on. Most struggled to take the given equation requiring solution and produce a quartic equation in $t$ ($\tan \theta$), some producing a quartic equation in $\cos \theta$, and somehow expecting to use the earlier results.

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Question 2

Topic: 级数与二项式定理 (Series and Binomial Theorem)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 (i) Show that $1.3.5.7. \dots .(2n - 1) = \frac{(2n)!}{2^n n!}$ and that, for $|x| < \frac{1}{4}$,

$\frac{1}{\sqrt{1 - 4x}} = 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} x^n.$

(ii) By differentiating the above result, deduce that

$\sum_{n=1}^{\infty} \frac{(2n)!}{n! (n - 1)!} \left( \frac{6}{25} \right)^n = 60.$

(iii) Show that

$\sum_{n=1}^{\infty} \frac{2^{n+1}(2n)!}{3^{2n}(n + 1)! n!} = 1.$

Hint

(i) $1.3.5.7.....(2n - 1) = \frac{1.2.3.4.....2n}{2.4.6.8.....2n} = \frac{(2n)!}{2.1.2.2.2.3.2.4.....2.n} = \frac{(2n)!}{2^n.1.2.3.4.....n} = \frac{(2n)!}{2^n n!}$

Using the binomial theorem, which is valid given the condition $|x| < \frac{1}{4}$,

$(1-4x)^{-\frac{1}{2}} = 1 + \frac{-\frac{1}{2}}{2}(-4x) + \frac{\frac{-1}{2} \cdot \frac{-3}{2}}{2!}(-4x)^2 + ...$ $= 1 + 1.(2x) + \frac{1.3}{2!}(2x)^2 + ... + \frac{1.3.5.7...(2n - 1)}{n!}(2x)^n + ...$

So the first result of the question yields $(1 - 4x)^{-\frac{1}{2}} = 1 + \sum_{n=1}^{\infty} \frac{\frac{(2n)!}{2^n n!}(2x)^n}{n!}$ leading to the required expression.

(ii) Differentiating $(1 - 4x)^{-\frac{1}{2}} = 1 + \sum_{n=1}^{\infty} \frac{(2n)!x^n}{(n!)^2}$ with respect to $x$, and

multiplying the result by $x$ gives $\frac{2x}{(1 - 4x)^{\frac{3}{2}}} = \sum_{n=1}^{\infty} \frac{(2n)!x^n}{n!(n - 1)!}$ and substituting $x = \frac{6}{25} < \frac{1}{4}$, gives the desired result.

(iii) Integrating $(1 - 4x)^{-\frac{1}{2}} = 1 + \sum_{n=1}^{\infty} \frac{(2n)!x^n}{(n!)^2}$ with respect to $x$, gives

$-\frac{1}{2}(1 - 4x)^{\frac{1}{2}} = x + \sum_{n=1}^{\infty} \frac{(2n)!x^{n+1}}{(n + 1)!n!} + c$, and substituting $x = 0 < \frac{1}{4}$, gives $c = -\frac{1}{2}$.

Now substituting $x = \frac{2}{9} = \frac{2}{3^2} < \frac{1}{4}$ and simplifying, gives the desired result.

Model Solution

Part (i)

We show that $1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n-1) = \frac{(2n)!}{2^n n!}$.

$1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n)}{2 \cdot 4 \cdot 6 \cdots (2n)} = \frac{(2n)!}{2 \cdot 1 \cdot 2 \cdot 2 \cdot 3 \cdots 2 \cdot n} = \frac{(2n)!}{2^n(1 \cdot 2 \cdot 3 \cdots n)} = \frac{(2n)!}{2^n \, n!}.$

Now we prove the series expansion. By the binomial theorem, for $|4x| < 1$ (i.e. $|x| < \frac{1}{4}$):

$(1 - 4x)^{-1/2} = \sum_{n=0}^{\infty} \binom{-1/2}{n}(-4x)^n.$

The binomial coefficient is:

$\binom{-1/2}{n} = \frac{(-\tfrac{1}{2})(-\tfrac{3}{2}) \cdots (-\tfrac{2n-1}{2})}{n!} = \frac{(-1)^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^n \, n!}.$

Therefore the $n$-th term (for $n \ge 1$) is:

$\frac{(-1)^n \cdot 1 \cdot 3 \cdots (2n-1)}{2^n \, n!} \cdot (-4x)^n = \frac{(-1)^n \cdot 1 \cdot 3 \cdots (2n-1)}{2^n \, n!} \cdot (-1)^n \cdot 4^n \cdot x^n = \frac{1 \cdot 3 \cdots (2n-1) \cdot 2^{2n}}{2^n \, n!} x^n = \frac{1 \cdot 3 \cdots (2n-1) \cdot 2^n}{n!} x^n.$

Using the identity just proved, $1 \cdot 3 \cdots (2n-1) = \frac{(2n)!}{2^n \, n!}$, the $n$-th term becomes:

$\frac{(2n)!}{2^n \, n!} \cdot \frac{2^n}{n!} x^n = \frac{(2n)!}{(n!)^2} x^n.$

Hence

$\frac{1}{\sqrt{1 - 4x}} = 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} x^n. \qquad \text{(i)}$

Part (ii)

Differentiating (i) with respect to $x$:

$\frac{\mathrm{d}}{\mathrm{d}x}(1 - 4x)^{-1/2} = \frac{2}{(1 - 4x)^{3/2}}.$

On the right side, differentiating term by term:

$\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} \cdot n \, x^{n-1} = \sum_{n=1}^{\infty} \frac{(2n)!}{n! \, (n-1)!} x^{n-1}.$

Therefore

$\frac{2}{(1 - 4x)^{3/2}} = \sum_{n=1}^{\infty} \frac{(2n)!}{n! \, (n-1)!} x^{n-1}.$

Multiplying both sides by $x$:

$\frac{2x}{(1 - 4x)^{3/2}} = \sum_{n=1}^{\infty} \frac{(2n)!}{n! \, (n-1)!} x^n. \qquad \text{(ii)}$

Substituting $x = \frac{6}{25}$:

$1 - 4 \cdot \frac{6}{25} = 1 - \frac{24}{25} = \frac{1}{25},$

so $(1 - 4x)^{3/2} = \left(\frac{1}{25}\right)^{3/2} = \frac{1}{125}$.

$\sum_{n=1}^{\infty} \frac{(2n)!}{n!(n-1)!}\left(\frac{6}{25}\right)^n = \frac{2 \cdot \frac{6}{25}}{\frac{1}{125}} = \frac{12/25}{1/125} = \frac{12}{25} \cdot 125 = 60.$

Part (iii)

We integrate (i) with respect to $x$:

$\int \frac{\mathrm{d}x}{\sqrt{1 - 4x}} = \int \left(1 + \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} x^n\right) \mathrm{d}x.$

The left side: let $u = 1 - 4x$, so $\mathrm{d}u = -4\,\mathrm{d}x$:

$\int (1 - 4x)^{-1/2} \mathrm{d}x = -\frac{1}{4} \cdot 2(1 - 4x)^{1/2} = -\frac{1}{2}(1 - 4x)^{1/2}.$

The right side:

$x + \sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} \cdot \frac{x^{n+1}}{n+1} + c = x + \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} x^{n+1} + c.$

Setting $x = 0$: $-\frac{1}{2} = 0 + 0 + c$, so $c = -\frac{1}{2}$.

Therefore

$-\frac{1}{2}(1 - 4x)^{1/2} = x + \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} x^{n+1} - \frac{1}{2}. \qquad \text{(iii)}$

Substituting $x = \frac{2}{9}$:

$1 - 4 \cdot \frac{2}{9} = 1 - \frac{8}{9} = \frac{1}{9},$

so $(1 - 4x)^{1/2} = \frac{1}{3}$.

$-\frac{1}{2} \cdot \frac{1}{3} = \frac{2}{9} + \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^{n+1} - \frac{1}{2},$

$-\frac{1}{6} = \frac{2}{9} - \frac{1}{2} + \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^{n+1},$

$-\frac{1}{6} = -\frac{5}{18} + \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^{n+1},$

$\sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^{n+1} = -\frac{1}{6} + \frac{5}{18} = \frac{-3 + 5}{18} = \frac{2}{18} = \frac{1}{9}.$

Now observe that

$\sum_{n=1}^{\infty} \frac{2^{n+1}(2n)!}{3^{2n}(n+1)! \, n!} = \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \cdot \frac{2^{n+1}}{9^n} = 2\sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^{n+1} \cdot \frac{9}{2} \cdot \frac{1}{2^n} \cdot \frac{2^n}{9^n}.$

Actually, more directly: $\frac{2^{n+1}}{3^{2n}} = \frac{2 \cdot 2^n}{9^n} = 2 \cdot \left(\frac{2}{9}\right)^n$, so

$\sum_{n=1}^{\infty} \frac{2^{n+1}(2n)!}{3^{2n}(n+1)! \, n!} = 2\sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^n = 2 \cdot \frac{9}{2} \sum_{n=1}^{\infty} \frac{(2n)!}{(n+1)! \, n!} \left(\frac{2}{9}\right)^{n+1} = 9 \cdot \frac{1}{9} = 1.$

Examiner Notes

This question was popular though not well answered. Solutions to part (i) were frequently unconvincing, though to part (ii) were quite good if they avoided elementary errors in working. Part (iii) was less well attempted with some not spotting to use integration, some stumbling over "$+ c$" and some not spotting the value of $x$ to substitute.

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Question 3

Topic: 斐波那契数列与数学归纳法 (Fibonacci Sequences and Mathematical Induction)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 A sequence of numbers, $F_1, F_2, \dots$, is defined by $F_1 = 1, F_2 = 1$, and

$F_n = F_{n-1} + F_{n-2} \quad \text{for } n \ge 3.$

(i) Write down the values of $F_3, F_4, \dots, F_8$.

(ii) Prove that $F_{2k+3}F_{2k+1} - F_{2k+2}^2 = -F_{2k+2}F_{2k} + F_{2k+1}^2$.

(iii) Prove by induction or otherwise that $F_{2n+1}F_{2n-1} - F_{2n}^2 = 1$ and deduce that $F_{2n}^2 + 1$ is divisible by $F_{2n+1}$.

(iv) Prove that $F_{2n-1}^2 + 1$ is divisible by $F_{2n+1}$.

Hint

(i) $F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21$

(ii) The result requires no term beyond $F_{2k+2}$ should appear on the RHS so the first strategy is to replace $F_{2k+3}$ and hence

$F_{2k+3}F_{2k+1} - {F_{2k+2}}^2 = (F_{2k+2} + F_{2k+1})F_{2k+1} - {F_{2k+2}}^2 = (F_{2k+1} - F_{2k+2})F_{2k+2} + {F_{2k+1}}^2 = -F_{2k}F_{2k+2} + {F_{2k+1}}^2$ as required.

(iii) The initial case is trivial to demonstrate, and so the induction runs from assuming that $F_{2k+1}F_{2k-1} - {F_{2k}}^2 = 1$, and attempting to prove that

$F_{2(k+1)+1}F_{2(k+1)-1} - {F_{2(k+1)}}^2 = 1$. $F_{2(k+1)+1}F_{2(k+1)-1} - {F_{2(k+1)}}^2 = F_{2k+3}F_{2k+1} - {F_{2k+2}}^2 = -F_{2k}F_{2k+2} + {F_{2k+1}}^2$ from (ii)

$= -(F_{2k-1}F_{2k+1} + {F_{2k}}^2)$ by a similar argument to (ii) $= -(-1)$ by inductive hypothesis.

The deduction follows from adding ${F_{2n}}^2$ to both sides of the result just proved.

(iv) This result cannot be deduced directly from (iii) as the nature of the expression differs in the type of subscript. Thus consider

${F_{2n-1}}^2 + 1 = \left( F_{2n+1} - F_{2n} \right)^2 + 1 = {F_{2n+1}}^2 - 2F_{2n+1}F_{2n} + {F_{2n}}^2 + 1 = {F_{2n+1}}^2 - 2F_{2n+1}F_{2n} + F_{2n-1}F_{2n+1}$ from (iii) and hence the desired result is obtained.

Model Solution

Part (i)

$F_3 = F_2 + F_1 = 2, \quad F_4 = F_3 + F_2 = 3, \quad F_5 = F_4 + F_3 = 5,$ $F_6 = F_5 + F_4 = 8, \quad F_7 = F_6 + F_5 = 13, \quad F_8 = F_7 + F_6 = 21.$

Part (ii)

Using $F_{2k+3} = F_{2k+2} + F_{2k+1}$ (the Fibonacci recurrence):

$F_{2k+3}F_{2k+1} - F_{2k+2}^2 = (F_{2k+2} + F_{2k+1})F_{2k+1} - F_{2k+2}^2$ $= F_{2k+2}F_{2k+1} + F_{2k+1}^2 - F_{2k+2}^2$ $= F_{2k+1}^2 + F_{2k+2}(F_{2k+1} - F_{2k+2}).$

Since $F_{2k+2} = F_{2k+1} + F_{2k}$, we have $F_{2k+1} - F_{2k+2} = -F_{2k}$. Therefore

$F_{2k+3}F_{2k+1} - F_{2k+2}^2 = F_{2k+1}^2 - F_{2k+2}F_{2k}. \qquad \text{(ii)}$

Part (iii)

We prove by induction that $F_{2n+1}F_{2n-1} - F_{2n}^2 = 1$ for all $n \ge 1$.

Base case ($n = 1$): $F_3 F_1 - F_2^2 = 2 \cdot 1 - 1^2 = 1$. True.

Inductive step: Assume $F_{2k+1}F_{2k-1} - F_{2k}^2 = 1$ for some $k \ge 1$. We wish to show $F_{2k+3}F_{2k+1} - F_{2k+2}^2 = 1$.

By result (ii):

$F_{2k+3}F_{2k+1} - F_{2k+2}^2 = F_{2k+1}^2 - F_{2k+2}F_{2k}.$

Now we evaluate $F_{2k+2}F_{2k}$ using a similar trick. Since $F_{2k+2} = F_{2k+1} + F_{2k}$:

$F_{2k+2}F_{2k} = (F_{2k+1} + F_{2k})F_{2k} = F_{2k+1}F_{2k} + F_{2k}^2.$

So

$F_{2k+1}^2 - F_{2k+2}F_{2k} = F_{2k+1}^2 - F_{2k+1}F_{2k} - F_{2k}^2 = F_{2k+1}(F_{2k+1} - F_{2k}) - F_{2k}^2.$

Since $F_{2k+1} - F_{2k} = F_{2k-1}$:

$= F_{2k+1}F_{2k-1} - F_{2k}^2 = 1$

by the inductive hypothesis. This completes the induction.

Deduction: From $F_{2n+1}F_{2n-1} - F_{2n}^2 = 1$, adding $F_{2n}^2$ to both sides:

$F_{2n}^2 + 1 = F_{2n+1}F_{2n-1}.$

Since $F_{2n-1}$ is an integer, $F_{2n}^2 + 1$ is divisible by $F_{2n+1}$.

Part (iv)

We need to show $F_{2n+1} \mid F_{2n-1}^2 + 1$.

Since $F_{2n+1} = F_{2n} + F_{2n-1}$, we have $F_{2n-1} = F_{2n+1} - F_{2n}$. Therefore

$F_{2n-1}^2 + 1 = (F_{2n+1} - F_{2n})^2 + 1 = F_{2n+1}^2 - 2F_{2n+1}F_{2n} + F_{2n}^2 + 1.$

From part (iii), $F_{2n}^2 + 1 = F_{2n+1}F_{2n-1}$. Substituting:

$F_{2n-1}^2 + 1 = F_{2n+1}^2 - 2F_{2n+1}F_{2n} + F_{2n+1}F_{2n-1} = F_{2n+1}(F_{2n+1} - 2F_{2n} + F_{2n-1}).$

Since $F_{2n+1} - 2F_{2n} + F_{2n-1} = (F_{2n} + F_{2n-1}) - 2F_{2n} + F_{2n-1} = 2F_{2n-1} - F_{2n}$, which is an integer, we conclude $F_{2n+1}$ divides $F_{2n-1}^2 + 1$.

Examiner Notes

This question was popular. Many solutions to part (ii) were rambling and lacked a sense of direction, even if correct. The induction in (iii) was frequently incorrectly handled and a common error was to replace $n$ by $k/2$. Part (iv) caused difficulties.

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Question 4

Topic: 参数方程与微分几何 (Parametric Equations and Differential Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 A curve is given parametrically by

$x = a(\cos t + \ln \tan \tfrac{1}{2}t),$ $y = a \sin t,$

where $0 < t < \frac{1}{2}\pi$ and $a$ is a positive constant. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \tan t$ and sketch the curve.

Let $P$ be the point with parameter $t$ and let $Q$ be the point where the tangent to the curve at $P$ meets the $x$-axis. Show that $PQ = a$.

The radius of curvature, $\rho$, at $P$ is defined by

$\rho = \frac{(\dot{x}^2 + \dot{y}^2)^{\frac{3}{2}}}{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|},$

where the dots denote differentiation with respect to $t$. Show that $\rho = a \cot t$.

The point $C$ lies on the normal to the curve at $P$, a distance $\rho$ from $P$ and above the curve. Show that $CQ$ is parallel to the $y$-axis.

Hint

$y = a \sin t \Rightarrow \dot{y} = a \cos t$

$x = a \left( \cos t + \ln \tan \frac{t}{2} \right) \Rightarrow \dot{x} = a \left( -\sin t + \frac{\frac{1}{2} \sec^2 \frac{t}{2}}{\tan \frac{t}{2}} \right) = a(-\sin t + \csc t) = a \cos t \cot t$

giving $\frac{dy}{dx} = \tan t$.

(y intercept $a$, y axis tangential to curve, x axis asymptote)

Tangent is $y - a \sin t = \tan t \left( x - a \left( \cos t + \ln \tan \frac{t}{2} \right) \right)$ giving $Q$ as $\left( a \ln \tan \frac{t}{2}, 0 \right)$ and

thus $PQ = \sqrt{(a \cos t)^2 + (a \sin t)^2} = a$

$\dot{y} = a \cos t \Rightarrow \ddot{y} = -a \sin t$

$\dot{x} = a(-\sin t + \csc t) \Rightarrow \ddot{x} = a(-\cos t - \csc t \cot t)$

$\dot{x}^2 + \dot{y}^2 = (a \cos t \cot t)^2 + (a \cos t)^2 = a^2 \cot^2 t$

$\dot{x} \ddot{y} - \dot{y} \ddot{x} = a \cos t \cot t \times -a \sin t - a \cos t \times a(-\cos t - \csc t \cot t)$

$= a^2 (-\cos^2 t + \cos^2 t + \cot^2 t) = a^2 \cot^2 t$

giving $\rho = a \cot t$.

From the results for $\frac{dy}{dx}$ and $\rho$, $C$ is

$\left( a \left( \cos t + \ln \tan \frac{t}{2} \right) - \rho \sin t, a \sin t + \rho \cos t \right) = \left( a \ln \tan \frac{t}{2}, a \operatorname{cosec} t \right)$

Which has the same $x$ coordinate as $Q$.

Model Solution

Showing $\frac{\mathrm{d}y}{\mathrm{d}x} = \tan t$

We have $x = a(\cos t + \ln \tan \frac{t}{2})$ and $y = a\sin t$.

$\dot{y} = a\cos t.$

For $\dot{x}$:

$\dot{x} = a\left(-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \sec^2\frac{t}{2} \cdot \frac{1}{2}\right) = a\left(-\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right) = a\left(-\sin t + \frac{1}{\sin t}\right) = a \cdot \frac{-\sin^2 t + 1}{\sin t} = \frac{a\cos^2 t}{\sin t}.$

Therefore

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\dot{y}}{\dot{x}} = \frac{a\cos t}{\frac{a\cos^2 t}{\sin t}} = \frac{\sin t}{\cos t} = \tan t.$

Sketch of the curve

At $t \to 0^+$: $y \to 0$ and $x \to a(-1 + \ln \tan 0^+) \to -\infty$ (since $\ln \tan \frac{t}{2} \to -\infty$). At $t = \frac{\pi}{2}$: $y = a$ and $x = a(0 + \ln \tan \frac{\pi}{4}) = a \cdot 0 = 0$. The curve approaches $y = a$ asymptotically as $t \to \frac{\pi}{2}$, with the $y$-axis tangent to the curve at $(0, a)$. The $x$-axis is a horizontal asymptote.

Showing $PQ = a$

The tangent at $P$ (with parameter $t$) has gradient $\tan t$. The equation of the tangent is:

$y - a\sin t = \tan t \left(x - a\cos t - a\ln\tan\tfrac{t}{2}\right).$

Setting $y = 0$ to find $Q$:

$-a\sin t = \tan t \left(x_Q - a\cos t - a\ln\tan\tfrac{t}{2}\right),$

$x_Q = a\cos t + a\ln\tan\tfrac{t}{2} - \frac{a\sin t}{\tan t} = a\cos t + a\ln\tan\tfrac{t}{2} - a\cos t = a\ln\tan\tfrac{t}{2}.$

So $Q = (a\ln\tan\frac{t}{2}, 0)$. Now $P = (a\cos t + a\ln\tan\frac{t}{2}, a\sin t)$, so

$PQ = \sqrt{(a\cos t)^2 + (a\sin t)^2} = \sqrt{a^2(\cos^2 t + \sin^2 t)} = a.$

Showing $\rho = a\cot t$

We already have $\dot{x} = \frac{a\cos^2 t}{\sin t} = a\cos t\cot t$ and $\dot{y} = a\cos t$.

Second derivatives:

$\ddot{y} = -a\sin t.$

$\ddot{x} = a(-\sin t \cdot \cot t + \cos t \cdot (-\csc^2 t)) = a\left(\frac{-\sin^2 t}{\sin t} - \frac{\cos t}{\sin^2 t}\right) = a\left(-\cos t - \frac{\cos t}{\sin^2 t}\right) = -a\cos t\left(1 + \csc^2 t\right).$

Actually, let me compute more carefully. $\dot{x} = a(-\sin t + \csc t)$, so

$\ddot{x} = a(-\cos t - \csc t\cot t).$

Now:

$\dot{x}^2 + \dot{y}^2 = a^2\cos^2 t\cot^2 t + a^2\cos^2 t = a^2\cos^2 t(\cot^2 t + 1) = a^2\cos^2 t \cdot \csc^2 t = a^2\cot^2 t.$

So $(\dot{x}^2 + \dot{y}^2)^{3/2} = a^3\cot^3 t$ (taking absolute value, noting $\cot t > 0$ for $0 < t < \frac{\pi}{2}$).

For the numerator of $\rho$:

$\dot{x}\ddot{y} - \dot{y}\ddot{x} = a\cos t\cot t \cdot (-a\sin t) - a\cos t \cdot a(-\cos t - \csc t\cot t)$

$= -a^2\cos^2 t - a^2\cos t(-\cos t - \csc t\cot t)$

$= -a^2\cos^2 t + a^2\cos^2 t + a^2\cos t \cdot \csc t \cot t = a^2 \cdot \frac{\cos t}{\sin t} \cdot \frac{\cos t}{\sin t} = a^2\cot^2 t.$

Therefore

$\rho = \frac{a^3\cot^3 t}{a^2\cot^2 t} = a\cot t.$

Showing $CQ$ is parallel to the $y$-axis

The normal at $P$ has gradient $-\cot t$ (negative reciprocal of $\tan t$). The centre of curvature $C$ lies on the normal at distance $\rho = a\cot t$ from $P$, on the side towards which the curve is concave (above the curve, i.e. the side of increasing $y$).

The unit normal pointing towards the centre of curvature has components $\frac{(-\sin t, \cos t)}{1}$ (rotating the tangent direction $(\cos t, \sin t)$ by $90^\circ$). So

$C = P + \rho(-\sin t, \cos t) = (a\cos t + a\ln\tan\tfrac{t}{2} - a\cot t \cdot \sin t, \; a\sin t + a\cot t \cdot \cos t).$

The $x$-coordinate of $C$:

$x_C = a\cos t + a\ln\tan\tfrac{t}{2} - a\cos t = a\ln\tan\tfrac{t}{2}.$

The $y$-coordinate of $C$:

$y_C = a\sin t + \frac{a\cos^2 t}{\sin t} = \frac{a\sin^2 t + a\cos^2 t}{\sin t} = \frac{a}{\sin t} = a\csc t.$

So $C = (a\ln\tan\frac{t}{2}, a\csc t)$. Since $Q = (a\ln\tan\frac{t}{2}, 0)$, both $C$ and $Q$ have $x$-coordinate $a\ln\tan\frac{t}{2}$. Therefore $CQ$ is parallel to the $y$-axis.

Examiner Notes

This question was quite popular. A lot of attempts involved rambling trigonometrical manipulations, and few spotted the standard differential of $\ln \tan \frac{t}{2}$. The curve sketch was often omitted or incorrect, and there was a lot of complicated working using e.g. the equation of the normal etc. to find the centre of curvature.

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Question 5

Topic: 双曲函数与高阶导数 (Hyperbolic Functions and Higher Derivatives)  |  Difficulty: Hard  |  Marks: 20

Problem

5 Let $y = \ln(x^2 - 1)$, where $x > 1$, and let $r$ and $\theta$ be functions of $x$ determined by $r = \sqrt{x^2 - 1}$ and $\coth \theta = x$. Show that

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2 \cosh \theta}{r} \quad \text{and} \quad \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{2 \cosh 2\theta}{r^2},$

and find an expression in terms of $r$ and $\theta$ for $\frac{\mathrm{d}^3y}{\mathrm{d}x^3}$.

Find, with proof, a similar formula for $\frac{\mathrm{d}^ny}{\mathrm{d}x^n}$ in terms of $r$ and $\theta$.

Hint

$\frac{dr}{dx} = x(x^2 - 1)^{-\frac{1}{2}} = \cosh \theta$

$y = \ln r^2 = 2 \ln r$

So $\frac{dy}{dx} = \frac{2}{r} \frac{dr}{dx} = \frac{2 \cosh \theta}{r}$

$\frac{dx}{d\theta} = -\operatorname{cosech}^2 \theta$ and $r = \operatorname{cosech} \theta$,

So differentiating the previous result and substituting,

$\frac{d^2 y}{dx^2} = \frac{2r \sinh \theta \frac{d\theta}{dx} - 2 \cosh \theta \frac{dr}{dx}}{r^2} = \frac{2(\operatorname{cosech} \theta \sinh \theta \times -\sinh^2 \theta - \cosh \theta \cosh \theta)}{r^2} = -\frac{2 \cosh 2\theta}{r^2}$

Similarly,

$\frac{d^3 y}{dx^3} = -\frac{2r^2 2 \sinh 2\theta \frac{d\theta}{dx} - 2 \cosh 2\theta \times 2r \frac{dr}{dx}}{r^4} = \frac{4}{r^4} (\sinh 2\theta + \cosh 2\theta \coth \theta) = \frac{4}{r^3} \cosh 3\theta$

In order to hypothesise a result for $\frac{d^n y}{dx^n}$, the important thing is to appreciate that the 4 has come from 2 times exponent of $r$ and multiple of $\theta$.

So $\frac{d^n y}{dx^n} = 2 \times (-1)^{n-1} \frac{(n-1)!}{r^n} \cosh n\theta$ which may be proved by induction, the

inductive differentiation step following the same pattern of working as used for $\frac{d^2 y}{dx^2}$ and $\frac{d^3 y}{dx^3}$.

Model Solution

First derivative

Given $y = \ln(x^2 - 1) = 2\ln r$ where $r = \sqrt{x^2 - 1}$, and $\coth\theta = x$.

We have $r = \sqrt{\coth^2\theta - 1} = \operatorname{cosech}\theta$ (since $x > 1$ implies $\theta > 0$, so $\operatorname{cosech}\theta > 0$).

Differentiating $r = \sqrt{x^2 - 1}$:

$\frac{\mathrm{d}r}{\mathrm{d}x} = \frac{x}{\sqrt{x^2 - 1}} = \frac{\coth\theta}{\operatorname{cosech}\theta} = \cosh\theta.$

Since $y = 2\ln r$:

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{r}\frac{\mathrm{d}r}{\mathrm{d}x} = \frac{2\cosh\theta}{r}. \qquad \text{(i)}$

Second derivative

Differentiating (i) with respect to $x$:

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 \cdot \frac{r\sinh\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} - \cosh\theta \cdot \frac{\mathrm{d}r}{\mathrm{d}x}}{r^2}.$

Since $\coth\theta = x$, we have $-\operatorname{cosech}^2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} = 1$, so $\frac{\mathrm{d}\theta}{\mathrm{d}x} = -\sinh^2\theta$.

Substituting $r = \operatorname{cosech}\theta$:

$r\sinh\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} = \operatorname{cosech}\theta \cdot \sinh\theta \cdot (-\sinh^2\theta) = -\sinh^2\theta.$

And $\cosh\theta \cdot \frac{\mathrm{d}r}{\mathrm{d}x} = \cosh\theta \cdot \cosh\theta = \cosh^2\theta$.

Therefore

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{2(-\sinh^2\theta - \cosh^2\theta)}{r^2} = -\frac{2\cosh 2\theta}{r^2}, \qquad \text{(ii)}$

using $\cosh 2\theta = \cosh^2\theta + \sinh^2\theta$.

Third derivative

Differentiating (ii) with respect to $x$:

$\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -2 \cdot \frac{r^2 \cdot 2\sinh 2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} - \cosh 2\theta \cdot 2r\frac{\mathrm{d}r}{\mathrm{d}x}}{r^4}.$

Substituting $\frac{\mathrm{d}\theta}{\mathrm{d}x} = -\sinh^2\theta$ and $\frac{\mathrm{d}r}{\mathrm{d}x} = \cosh\theta$:

Numerator: $r^2 \cdot 2\sinh 2\theta \cdot (-\sinh^2\theta) - \cosh 2\theta \cdot 2r\cosh\theta$

$= -2r^2\sinh 2\theta\sinh^2\theta - 2r\cosh 2\theta\cosh\theta$.

With $r = \operatorname{cosech}\theta$ and $r^2 = \operatorname{cosech}^2\theta$:

$= -2\operatorname{cosech}^2\theta \cdot 2\sinh\theta\cosh\theta \cdot \sinh^2\theta - 2\operatorname{cosech}\theta \cdot \cosh 2\theta \cdot \cosh\theta$ $= -4\cosh\theta\sinh\theta - 2\frac{\cosh 2\theta\cosh\theta}{\sinh\theta}.$

Actually, let me compute the numerator of $\frac{\mathrm{d}^3y}{\mathrm{d}x^3}$ more carefully. We have:

$\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{\mathrm{d}}{\mathrm{d}x}\left(-\frac{2\cosh 2\theta}{r^2}\right) = \frac{-2 \cdot 2\sinh 2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} \cdot r^2 - (-2\cosh 2\theta) \cdot 2r\frac{\mathrm{d}r}{\mathrm{d}x}}{r^4}$

$= \frac{-4r^2\sinh 2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} + 4r\cosh 2\theta \cdot \frac{\mathrm{d}r}{\mathrm{d}x}}{r^4}.$

Now $\sinh 2\theta = 2\sinh\theta\cosh\theta$, and with $r = \operatorname{cosech}\theta$:

$-4r^2\sinh 2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} = -4\operatorname{cosech}^2\theta \cdot 2\sinh\theta\cosh\theta \cdot (-\sinh^2\theta) = 8\cosh\theta\sinh\theta.$

$4r\cosh 2\theta \cdot \frac{\mathrm{d}r}{\mathrm{d}x} = 4\operatorname{cosech}\theta \cdot \cosh 2\theta \cdot \cosh\theta = \frac{4\cosh 2\theta\cosh\theta}{\sinh\theta}.$

So the numerator is $8\cosh\theta\sinh\theta + \frac{4\cosh 2\theta\cosh\theta}{\sinh\theta}$.

$= \frac{8\cosh\theta\sinh^2\theta + 4\cosh 2\theta\cosh\theta}{\sinh\theta} = \frac{4\cosh\theta(2\sinh^2\theta + \cosh 2\theta)}{\sinh\theta}.$

Since $\cosh 2\theta = 2\sinh^2\theta + 1$, we get $2\sinh^2\theta + \cosh 2\theta = 4\sinh^2\theta + 1$. Hmm, let me use $\cosh 2\theta = 2\cosh^2\theta - 1$ instead: $2\sinh^2\theta + 2\cosh^2\theta - 1 = 2(\sinh^2\theta + \cosh^2\theta) - 1 = 2\cosh 2\theta$.

Wait: $\sinh^2\theta + \cosh^2\theta = \cosh 2\theta$. So $2\sinh^2\theta + \cosh 2\theta = 2\sinh^2\theta + \sinh^2\theta + \cosh^2\theta = 3\sinh^2\theta + \cosh^2\theta$. That doesn’t simplify nicely. Let me try differently.

$2\sinh^2\theta + \cosh 2\theta = 2\sinh^2\theta + 2\sinh^2\theta + 1 = 4\sinh^2\theta + 1$… no, $\cosh 2\theta = 1 + 2\sinh^2\theta$.

So $2\sinh^2\theta + 1 + 2\sinh^2\theta = 4\sinh^2\theta + 1$. Hmm.

Let me try: $\frac{4\cosh\theta}{\sinh\theta}(2\sinh^2\theta + \cosh 2\theta) = \frac{4\cosh\theta}{\sinh\theta}(2\sinh^2\theta + 2\sinh^2\theta + 1) = \frac{4\cosh\theta(4\sinh^2\theta + 1)}{\sinh\theta}$.

And $r^4 = \operatorname{cosech}^4\theta$. So

$\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{4\cosh\theta(4\sinh^2\theta + 1)}{r^4\sinh\theta} = \frac{4\cosh\theta(4\sinh^2\theta + 1)\sinh^3\theta}{\sinh\theta} = 4\cosh\theta\sinh^2\theta(4\sinh^2\theta + 1).$

Hmm, this is getting messy. Let me use a cleaner approach.

We hypothesise from the pattern $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\cosh\theta}{r}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{-2\cosh 2\theta}{r^2}$ that $\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{4\cosh 3\theta}{r^3}$ and prove it.

From (ii), $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2r^{-2}\cosh 2\theta$. Differentiating:

$\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -2\left(-2r^{-3}\frac{\mathrm{d}r}{\mathrm{d}x}\cosh 2\theta + r^{-2} \cdot 2\sinh 2\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x}\right)$

$= -2\left(-2r^{-3}\cosh\theta\cosh 2\theta + r^{-2} \cdot 2\sinh 2\theta \cdot (-\sinh^2\theta)\right)$

$= -2\left(-\frac{2\cosh\theta\cosh 2\theta}{r^3} - \frac{2\sinh 2\theta\sinh^2\theta}{r^2}\right)$

$= \frac{4\cosh\theta\cosh 2\theta}{r^3} + \frac{4\sinh 2\theta\sinh^2\theta}{r^2}.$

With $r = \operatorname{cosech}\theta$, the second term is $4\sinh 2\theta\sinh^2\theta \cdot \sinh^2\theta = 4 \cdot 2\sinh\theta\cosh\theta \cdot \sinh^2\theta \cdot \sinh^2\theta$. Hmm, this is still messy.

Let me try differently. $\frac{4\sinh 2\theta\sinh^2\theta}{r^2} = \frac{4\sinh 2\theta\sinh^2\theta}{\operatorname{cosech}^2\theta} = 4\sinh 2\theta\sinh^4\theta$.

And $\frac{4\cosh\theta\cosh 2\theta}{r^3} = 4\cosh\theta\cosh 2\theta\sinh^3\theta$.

So $\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 4\cosh\theta\cosh 2\theta\sinh^3\theta + 4 \cdot 2\sinh\theta\cosh\theta\sinh^4\theta = 4\cosh\theta\sinh^3\theta(\cosh 2\theta + 2\sinh^2\theta)$.

$= 4\cosh\theta\sinh^3\theta(\cosh 2\theta + 2\sinh^2\theta)$.

Now $\cosh 2\theta + 2\sinh^2\theta = \cosh 2\theta + \cosh 2\theta - 1 = 2\cosh 2\theta - 1$… no.

$\cosh 2\theta = 2\cosh^2\theta - 1 = 1 + 2\sinh^2\theta$. So $\cosh 2\theta + 2\sinh^2\theta = 1 + 2\sinh^2\theta + 2\sinh^2\theta = 1 + 4\sinh^2\theta$.

Also $\cosh 3\theta = \cosh\theta(4\cosh^2\theta - 3) = \cosh\theta(1 + 4\sinh^2\theta - 3 + 3) = \cosh\theta(4\cosh^2\theta - 3)$. Hmm.

$\cosh 3\theta = 4\cosh^3\theta - 3\cosh\theta = \cosh\theta(4\cosh^2\theta - 3) = \cosh\theta(4(1+\sinh^2\theta) - 3) = \cosh\theta(1 + 4\sinh^2\theta)$.

So $4\cosh\theta\sinh^3\theta(1 + 4\sinh^2\theta) = 4\sinh^3\theta\cosh 3\theta$.

And $\frac{4\cosh 3\theta}{r^3} = 4\cosh 3\theta \cdot \sinh^3\theta$.

Therefore $\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \frac{4\cosh 3\theta}{r^3}$.

General formula and proof by induction

We conjecture

$\frac{\mathrm{d}^ny}{\mathrm{d}x^n} = \frac{2(-1)^{n-1}(n-1)!\cosh n\theta}{r^n}. \qquad \text{(iii)}$

Base cases: $n = 1$: $\frac{2\cosh\theta}{r}$. True by (i). $n = 2$: $\frac{-2\cosh 2\theta}{r^2}$. True by (ii). $n = 3$: $\frac{4\cosh 3\theta}{r^3}$. True by the computation above.

Inductive step: Assume (iii) holds for some $n \ge 1$. Differentiating with respect to $x$:

$\frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} = 2(-1)^{n-1}(n-1)!\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\cosh n\theta}{r^n}\right).$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\cosh n\theta}{r^n}\right) = \frac{n\sinh n\theta \cdot \frac{\mathrm{d}\theta}{\mathrm{d}x} \cdot r^n - \cosh n\theta \cdot nr^{n-1}\frac{\mathrm{d}r}{\mathrm{d}x}}{r^{2n}}$

$= \frac{n\sinh n\theta(-\sinh^2\theta)}{r^n} - \frac{n\cosh n\theta\cosh\theta}{r^{n+1}}$

$= \frac{-n\sinh n\theta\sinh^2\theta \cdot r - n\cosh n\theta\cosh\theta}{r^{n+1}}.$

With $r = \operatorname{cosech}\theta$:

$= \frac{-n\sinh n\theta\sinh^2\theta \cdot \operatorname{cosech}\theta - n\cosh n\theta\cosh\theta}{r^{n+1}} = \frac{-n(\sinh n\theta\sinh\theta + \cosh n\theta\cosh\theta)}{r^{n+1}} = \frac{-n\cosh(n+1)\theta}{r^{n+1}},$

using the identity $\cosh A\cosh B + \sinh A\sinh B = \cosh(A+B)$.

Therefore

$\frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} = 2(-1)^{n-1}(n-1)! \cdot \frac{-n\cosh(n+1)\theta}{r^{n+1}} = \frac{2(-1)^n \cdot n! \cdot \cosh(n+1)\theta}{r^{n+1}}.$

This is exactly (iii) with $n$ replaced by $n+1$. The induction is complete.

Examiner Notes

This was frequently attempted, though lack of facility with hyperbolic functions meant that few progressed beyond the first two differentials, and for those going further, the working was not methodical enough to spot the factorial that would emerge in the general result.

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Question 6

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

Problem

6 The distinct points $P, Q, R$ and $S$ in the Argand diagram lie on a circle of radius $a$ centred at the origin and are represented by the complex numbers $p, q, r$ and $s$, respectively. Show that

$pq = -a^2 \frac{p - q}{p^* - q^*} .$

Deduce that, if the chords $PQ$ and $RS$ are perpendicular, then $pq + rs = 0$.

The distinct points $A_1, A_2, \dots, A_n$ (where $n \ge 3$) lie on a circle. The points $B_1, B_2, \dots, B_n$ lie on the same circle and are chosen so that the chords $B_1 B_2, B_2 B_3, \dots, B_n B_1$ are perpendicular, respectively, to the chords $A_1 A_2, A_2 A_3, \dots, A_n A_1$. Show that, for $n = 3$, there are only two choices of $B_1$ for which this is possible. What is the corresponding result for $n = 4$? State the corresponding results for values of $n$ greater than 4.

Hint

$pp^* = qq^* = a^2$ and so $a^2(p - q) = qq^* p - pp^* q = -pq(p^* - q^*)$ and hence the required result. If $PQ$ and $RS$ are perpendicular then $p - q = ki(r - s)$ for some real $k$, and thus

$p^* - q^* = -ki(r^* - s^*), \text{ and so } pq = -a^2 \frac{p - q}{p^* - q^*} = a^2 \frac{r - s}{r^* - s^*} = -rs$

For $n = 3$, $B_1 B_2 \perp A_1 A_2$ etc. $\Rightarrow a_1 a_2 + b_1 b_2 = 0$ etc.

Thus $b_1^2 = \frac{b_1 b_2 \times b_1 b_3}{b_2 b_3} = \frac{-a_1 a_2 \times -a_1 a_3}{-a_2 a_3} = -a_1^2$ and so $b_1 = \pm i a_1$

i.e. two choices of $B_1$.

For $n = 4$, $B_1 B_2 \perp A_1 A_2$ etc. $\Rightarrow a_1 a_2 + b_1 b_2 = 0$ etc. but this only yields 3 independent equations as e.g. $a_3 a_4 + b_3 b_4 = 0$ can be obtained from the other three equations by

$a_3 a_4 = \frac{a_2 a_3 \times a_4 a_1}{a_2 a_1} \text{ etc. Hence there are arbitrarily many possible choices for } B_i.$

For $n > 4$, the corresponding results are as for $n = 3$ or $n = 4$ depending on whether $n$ is odd or even.

Model Solution

Showing $pq = -a^2 \frac{p-q}{p^*-q^*}$

Since $P$ and $Q$ lie on a circle of radius $a$ centred at the origin, $|p| = |q| = a$, so $pp^* = qq^* = a^2$.

Therefore

$a^2(p - q) = pp^*q - qq^*p \quad \text{(using } pp^* = a^2 \text{ for the first term, } qq^* = a^2 \text{ for the second)}$

Wait, let me be more careful. $a^2(p-q) = qq^*p - pp^*q = pq(q^* - p^*) = -pq(p^* - q^*)$.

Hence

$pq = \frac{a^2(p-q)}{-(p^*-q^*)} = -a^2\frac{p-q}{p^*-q^*}.$

Deduction: if $PQ \perp RS$ then $pq + rs = 0$

If $PQ$ and $RS$ are perpendicular, then the direction from $P$ to $Q$ is perpendicular to the direction from $R$ to $S$. In the Argand diagram, perpendicularity means the ratio of the directions is purely imaginary: $p - q = ki(r - s)$ for some real $k \neq 0$.

Taking conjugates: $p^* - q^* = -ki(r^* - s^*)$ (since $k$ is real and $\bar{i} = -i$).

Therefore

$pq = -a^2\frac{p-q}{p^*-q^*} = -a^2\frac{ki(r-s)}{-ki(r^*-s^*)} = a^2\frac{r-s}{r^*-s^*} = -rs,$

where the last step uses the same identity $rs = -a^2\frac{r-s}{r^*-s^*}$.

Hence $pq + rs = 0$.

The case $n = 3$

We have points $A_1, A_2, A_3$ and $B_1, B_2, B_3$ on the same circle. The conditions $B_1B_2 \perp A_1A_2$, $B_2B_3 \perp A_2A_3$, $B_3B_1 \perp A_3A_1$ give, by the perpendicularity result:

$b_1 b_2 + a_1 a_2 = 0, \quad b_2 b_3 + a_2 a_3 = 0, \quad b_3 b_1 + a_3 a_1 = 0. \qquad \text{(i)}$

From these:

$b_1 b_2 = -a_1 a_2, \quad b_1 b_3 = -a_1 a_3, \quad b_2 b_3 = -a_2 a_3.$

Therefore

$b_1^2 = \frac{b_1 b_2 \cdot b_1 b_3}{b_2 b_3} = \frac{(-a_1 a_2)(-a_1 a_3)}{-a_2 a_3} = \frac{a_1^2 a_2 a_3}{-a_2 a_3} = -a_1^2.$

So $b_1^2 = -a_1^2 = (ia_1)^2$, giving $b_1 = \pm ia_1$. This means $B_1$ is one of two points: obtained by rotating $A_1$ by $\pm 90^\circ$ about the centre of the circle. Once $b_1$ is chosen, $b_2$ and $b_3$ are uniquely determined by (i). So there are exactly two choices of $B_1$.

The case $n = 4$

The conditions give:

$b_1 b_2 + a_1 a_2 = 0, \quad b_2 b_3 + a_2 a_3 = 0, \quad b_3 b_4 + a_3 a_4 = 0, \quad b_4 b_1 + a_4 a_1 = 0. \qquad \text{(ii)}$

From the first three: $b_1 b_2 = -a_1 a_2$, $b_2 b_3 = -a_2 a_3$, $b_3 b_4 = -a_3 a_4$.

Multiplying the first and third: $b_1 b_2 b_3 b_4 = a_1 a_2 a_3 a_4$. Also from the second and fourth: $b_2 b_3 b_4 b_1 = a_2 a_3 a_4 a_1 = a_1 a_2 a_3 a_4$.

So the four equations are not independent: the fourth equation is automatically satisfied once the first three hold (since $b_4 b_1 = \frac{b_1 b_2 b_3 b_4}{b_2 b_3} = \frac{a_1 a_2 a_3 a_4}{-a_2 a_3} = -a_1 a_4$).

More explicitly: from the first two equations, we get $\frac{b_1}{b_3} = \frac{a_1}{a_3}$, so $b_1 = \frac{a_1 b_3}{a_3}$. The third equation determines $b_4$ in terms of $b_3$: $b_4 = \frac{-a_3 a_4}{b_3}$. The first equation determines $b_2 = \frac{-a_1 a_2}{b_1}$. So $b_3$ is a free parameter (any point on the circle that avoids zero denominators), and the other three $B$-points are determined. This gives infinitely many (arbitrarily many) valid configurations.

General $n > 4$

For odd $n$: the same algebraic manipulation as for $n = 3$ yields $b_1^2 = (-1)^n a_1^2 = -a_1^2$ (since $n$ is odd), giving exactly two choices for $B_1$.

For even $n$: by the same dependency argument as for $n = 4$, the $n$ equations reduce to $n - 1$ independent ones, leaving one degree of freedom. So there are infinitely many valid configurations.

The corresponding results are: for odd $n \ge 3$, exactly two choices of $B_1$; for even $n \ge 4$, infinitely many configurations.

Examiner Notes

This was the least popular Pure question and very little success was achieved by the few that attempted it. The first result was often obtained correctly by expressing each of the four complex numbers in modulus-exponential form, but then the perpendicularity was the stumbling block.

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Question 7

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

Problem

7 The functions $s(x)$ ($0 \le x < 1$) and $t(x)$ ($x \ge 0$), and the real number $p$, are defined by

$s(x) = \int_0^x \frac{1}{\sqrt{1 - u^2}} \, \mathrm{d}u \ , \quad t(x) = \int_0^x \frac{1}{1 + u^2} \, \mathrm{d}u \ , \quad p = 2 \int_0^\infty \frac{1}{1 + u^2} \, \mathrm{d}u \ .$

For this question, do not evaluate any of the above integrals explicitly in terms of inverse trigonometric functions or the number $\pi$.

(i) Use the substitution $u = v^{-1}$ to show that $t(x) = \int_{1/x}^\infty \frac{1}{1 + v^2} \, \mathrm{d}v$. Hence evaluate $t(1/x) + t(x)$ in terms of $p$ and deduce that $2t(1) = \frac{1}{2}p$.

(ii) Let $y = \frac{u}{\sqrt{1 + u^2}}$. Express $u$ in terms of $y$, and show that $\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{1}{\sqrt{(1 - y^2)^3}}$.

By making a substitution in the integral for $t(x)$, show that

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

Deduce that $s\left( \frac{1}{\sqrt{2}} \right) = \frac{1}{4}p$.

(iii) Let $z = \frac{u + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}u}$. Show that $t\left( \frac{1}{\sqrt{3}} \right) = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1 + z^2} \, \mathrm{d}z$ , and hence that $3t\left( \frac{1}{\sqrt{3}} \right) = \frac{1}{2}p$.

Hint

7. (i) $u = v^{-1} \Rightarrow \frac{du}{dv} = -v^{-2}$ so $t(x) = \int_{\infty}^{\frac{1}{x}} \frac{1}{1 + v^{-2}} \times -v^{-2} dv = \int_{\frac{1}{x}}^{\infty} \frac{1}{v^2 + 1} dv$

so $t\left(\frac{1}{x}\right) + t(x) = \int_{0}^{\frac{1}{x}} \frac{1}{1 + u^2} du + \int_{\frac{1}{x}}^{\infty} \frac{1}{v^2 + 1} dv = \int_{0}^{\infty} \frac{1}{1 + u^2} du = \frac{1}{2} \pi$

Letting $x = 1$ gives the desired result.

(ii) $y = \frac{u}{\sqrt{1 + u^2}} \Rightarrow u = \frac{y}{\sqrt{1 - y^2}}$

so $\frac{du}{dy} = \frac{(1 - y^2)^{\frac{1}{2}} - y \times -y(1 - y^2)^{-\frac{1}{2}}}{1 - y^2} = \frac{(1 - y^2) + y^2}{(1 - y^2)^{\frac{3}{2}}}$ and hence the result.

Using the given substitution for $u$,

$t(x) = \int_{0}^{\frac{x}{\sqrt{1+x^2}}} \frac{1}{1 + \frac{y^2}{1 - y^2}} \times \frac{1}{(1 - y^2)^{\frac{3}{2}}} dy = \int_{0}^{\frac{x}{\sqrt{1+x^2}}} \frac{1}{(1 - y^2)^{\frac{1}{2}}} dy = s\left(\frac{x}{\sqrt{1+x^2}}\right)$

Again letting $x = 1$, and using the result from part (i) gives the desired result.

(iii) $z = \frac{u + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}u} \Rightarrow u = \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}z} \Rightarrow \frac{du}{dz} = \frac{\frac{4}{3}}{\left(1 + \frac{1}{\sqrt{3}}z\right)^2}$

Using this substitution,

$t(x) = \int_{\frac{1}{\sqrt{3}}}^{\frac{x + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}x}} \frac{1}{1 + \left(\frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}z}\right)^2} \times \frac{\frac{4}{3}}{\left(1 + \frac{1}{\sqrt{3}}z\right)^2} dz = \int_{\frac{1}{\sqrt{3}}}^{\frac{x + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}x}} \frac{\frac{4}{3}}{\left(1 + \frac{1}{\sqrt{3}}z\right)^2 + \left(z - \frac{1}{\sqrt{3}}\right)^2} dz = \int_{\frac{1}{\sqrt{3}}}^{\frac{x + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}x}} \frac{1}{1 + z^2} dz$

Letting $x = \frac{1}{\sqrt{3}}$ gives the required result.

By definition $t\left(\frac{1}{\sqrt{3}}\right) = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{1+u^2} \, du$, by the previous result just obtained

$t\left(\frac{1}{\sqrt{3}}\right) = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1+z^2} \, dz$, and from part (i) $t\left(\frac{1}{\sqrt{3}}\right) = \int_{\sqrt{3}}^{\infty} \frac{1}{1+v^2} \, dv$ and so adding these three

results gives $3t\left(\frac{1}{\sqrt{3}}\right) = \int_{0}^{\infty} \frac{1}{1+u^2} \, du = \frac{1}{2} \pi$

Model Solution

Part (i)

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

When $u = 0$, $v \to \infty$; when $u = x$, $v = 1/x$. Therefore

$t(x) = \int_0^x \frac{1}{1 + u^2} \, \mathrm{d}u = \int_\infty^{1/x} \frac{1}{1 + v^{-2}} \left(-v^{-2}\right) \mathrm{d}v = \int_{1/x}^{\infty} \frac{v^{-2}}{1 + v^{-2}} \, \mathrm{d}v = \int_{1/x}^{\infty} \frac{1}{v^2 + 1} \, \mathrm{d}v.$

Now we compute $t(1/x) + t(x)$:

$t(1/x) + t(x) = \int_0^{1/x} \frac{1}{1 + u^2} \, \mathrm{d}u + \int_{1/x}^{\infty} \frac{1}{1 + v^2} \, \mathrm{d}v = \int_0^{\infty} \frac{1}{1 + u^2} \, \mathrm{d}u = \frac{1}{2} p.$

Setting $x = 1$: $2t(1) = t(1) + t(1) = \frac{1}{2}p$, so $t(1) = \frac{1}{4}p$.

Part (ii)

We have $y = \frac{u}{\sqrt{1 + u^2}}$. Squaring both sides: $y^2 = \frac{u^2}{1 + u^2}$, so $y^2(1 + u^2) = u^2$, giving $u^2(1 - y^2) = y^2$, hence $u^2 = \frac{y^2}{1 - y^2}$ and

$u = \frac{y}{\sqrt{1 - y^2}}.$

Differentiating by the quotient rule:

$\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{\sqrt{1 - y^2} - y \cdot \frac{-y}{\sqrt{1 - y^2}}}{1 - y^2} = \frac{\frac{1 - y^2 + y^2}{\sqrt{1 - y^2}}}{1 - y^2} = \frac{1}{(1 - y^2)^{3/2}}.$

Now substitute $y = \frac{u}{\sqrt{1 + u^2}}$ into the integral for $t(x)$. When $u = 0$, $y = 0$; when $u = x$, $y = \frac{x}{\sqrt{1 + x^2}}$. Using $\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{1}{(1 - y^2)^{3/2}}$:

$t(x) = \int_0^{x/\sqrt{1+x^2}} \frac{1}{1 + u^2} \cdot \frac{\mathrm{d}u}{\mathrm{d}y} \, \mathrm{d}y.$

Since $u = \frac{y}{\sqrt{1 - y^2}}$, we have $1 + u^2 = 1 + \frac{y^2}{1 - y^2} = \frac{1}{1 - y^2}$. Therefore

$t(x) = \int_0^{x/\sqrt{1+x^2}} \frac{1}{\frac{1}{1 - y^2}} \cdot \frac{1}{(1 - y^2)^{3/2}} \, \mathrm{d}y = \int_0^{x/\sqrt{1+x^2}} \frac{1 - y^2}{(1 - y^2)^{3/2}} \, \mathrm{d}y = \int_0^{x/\sqrt{1+x^2}} \frac{1}{\sqrt{1 - y^2}} \, \mathrm{d}y.$

This is exactly $s\!\left(\frac{x}{\sqrt{1 + x^2}}\right)$.

Setting $x = 1$ and using $t(1) = \frac{1}{4}p$ from part (i):

$\frac{1}{4}p = t(1) = s\!\left(\frac{1}{\sqrt{2}}\right).$

Part (iii)

Let $z = \frac{u + 1/\sqrt{3}}{1 - u/\sqrt{3}}$. Solving for $u$: $z(1 - u/\sqrt{3}) = u + 1/\sqrt{3}$, so $z - zu/\sqrt{3} = u + 1/\sqrt{3}$, giving $u(1 + z/\sqrt{3}) = z - 1/\sqrt{3}$, hence

$u = \frac{z - 1/\sqrt{3}}{1 + z/\sqrt{3}}.$

Differentiating:

$\frac{\mathrm{d}u}{\mathrm{d}z} = \frac{(1 + z/\sqrt{3}) - (z - 1/\sqrt{3}) \cdot \frac{1}{\sqrt{3}}}{(1 + z/\sqrt{3})^2} = \frac{1 + \frac{z}{\sqrt{3}} - \frac{z}{\sqrt{3}} + \frac{1}{3}}{(1 + z/\sqrt{3})^2} = \frac{4/3}{(1 + z/\sqrt{3})^2}.$

When $u = 0$, $z = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$; when $u = \frac{1}{\sqrt{3}}$, $z = \frac{2/\sqrt{3}}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3} = \sqrt{3}$.

Now we compute $1 + u^2$ in terms of $z$:

$1 + u^2 = 1 + \frac{(z - 1/\sqrt{3})^2}{(1 + z/\sqrt{3})^2} = \frac{(1 + z/\sqrt{3})^2 + (z - 1/\sqrt{3})^2}{(1 + z/\sqrt{3})^2}.$

Expanding the numerator:

$(1 + z/\sqrt{3})^2 + (z - 1/\sqrt{3})^2 = 1 + \frac{2z}{\sqrt{3}} + \frac{z^2}{3} + z^2 - \frac{2z}{\sqrt{3}} + \frac{1}{3} = \frac{4}{3} + \frac{4z^2}{3} = \frac{4}{3}(1 + z^2).$

Therefore

$\frac{1}{1 + u^2} \cdot \frac{\mathrm{d}u}{\mathrm{d}z} = \frac{(1 + z/\sqrt{3})^2}{\frac{4}{3}(1 + z^2)} \cdot \frac{4/3}{(1 + z/\sqrt{3})^2} = \frac{1}{1 + z^2}.$

Hence

$t\!\left(\frac{1}{\sqrt{3}}\right) = \int_0^{1/\sqrt{3}} \frac{1}{1 + u^2} \, \mathrm{d}u = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1 + z^2} \, \mathrm{d}z.$

From part (i), with $x = \sqrt{3}$:

$t(\sqrt{3}) = \int_{1/\sqrt{3}}^{\infty} \frac{1}{1 + v^2} \, \mathrm{d}v.$

Therefore

$t\!\left(\frac{1}{\sqrt{3}}\right) + t(\sqrt{3}) = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1 + z^2} \, \mathrm{d}z + \int_{\sqrt{3}}^{\infty} \frac{1}{1 + v^2} \, \mathrm{d}v = \int_{1/\sqrt{3}}^{\infty} \frac{1}{1 + z^2} \, \mathrm{d}z = t(\sqrt{3}).$

Wait, let me recombine more carefully. We have three equal integrals for $t\!\left(\frac{1}{\sqrt{3}}\right)$:

$t\!\left(\frac{1}{\sqrt{3}}\right) = \int_0^{1/\sqrt{3}} \frac{1}{1 + u^2} \, \mathrm{d}u, \qquad t\!\left(\frac{1}{\sqrt{3}}\right) = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1 + z^2} \, \mathrm{d}z, \qquad t\!\left(\frac{1}{\sqrt{3}}\right) = \int_{\sqrt{3}}^{\infty} \frac{1}{1 + v^2} \, \mathrm{d}v,$

where the third follows from part (i) with $x = 1/\sqrt{3}$: $t(1/\sqrt{3}) = \int_{\sqrt{3}}^{\infty} \frac{1}{1+v^2} \mathrm{d}v$.

Adding all three:

$3t\!\left(\frac{1}{\sqrt{3}}\right) = \int_0^{\infty} \frac{1}{1 + u^2} \, \mathrm{d}u = \frac{1}{2}p.$

Examiner Notes

This was a very popular question. As the question led the candidates through there were a number of unconvincing solutions to parts of the question, but overall it was reasonably well handled.

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Question 8

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

Problem

8 (i) Find functions $a(x)$ and $b(x)$ such that $u = x$ and $u = e^{-x}$ both satisfy the equation

$\frac{d^2u}{dx^2} + a(x)\frac{du}{dx} + b(x)u = 0.$

For these functions $a(x)$ and $b(x)$, write down the general solution of the equation.

Show that the substitution $y = \frac{1}{3u} \frac{du}{dx}$ transforms the equation

$\frac{dy}{dx} + 3y^2 + \frac{x}{1+x}y = \frac{1}{3(1+x)} \qquad \text{(*)}$

into

$\frac{d^2u}{dx^2} + \frac{x}{1+x} \frac{du}{dx} - \frac{1}{1+x}u = 0$

and hence show that the solution of equation (*) that satisfies $y = 0$ at $x = 0$ is given by

$y = \frac{1 - e^{-x}}{3(x + e^{-x})}.$

(ii) Find the solution of the equation

$\frac{dy}{dx} + y^2 + \frac{x}{1-x}y = \frac{1}{1-x}$

that satisfies $y = 2$ at $x = 0$.

Hint

8. (i) Substituting each $u$ into the differential equation yields simultaneous equations $a(x) + xb(x) = 0$ and $e^{-x}(1 - a(x) + b(x)) = 0$ which solve to give

$a(x) = \frac{x}{1+x}$ and $b(x) = \frac{-1}{1+x}$

The general solution is $u = Ax + Be^{-x}$.

$y = \frac{1}{3u} \frac{du}{dx} \Rightarrow \frac{dy}{dx} = \frac{-1}{3u^2} \left(\frac{du}{dx}\right)^2 + \frac{1}{3u} \frac{d^2u}{dx^2}$ which when substituted into equation (*),

multiplied by $3u$, and collected on one side gives the required result.

$u = Ax + Be^{-x} \Rightarrow \frac{du}{dx} = A - Be^{-x} \Rightarrow y = \frac{A - Be^{-x}}{3(Ax + Be^{-x})}$,

and substitution of $x = 0, y = 0$ gives $A = B$ and hence $y = \frac{1 - e^{-x}}{3(x + e^{-x})}$.

(ii) Substituting $y = \frac{1}{u} \frac{du}{dx}$ into the given equation yields

$\frac{d^2u}{dx^2} + \frac{x}{1-x} \frac{du}{dx} - \frac{1}{1-x} u = 0$ which is the equation in the first part with $x$ replaced by $-x$

So the general solution is $u = Cx + De^x$

Substitution of $x = 0, y = 2$ again gives $A = B$, and hence $y = \frac{1 + e^x}{x + e^x}$

Model Solution

Part (i)

Finding $a(x)$ and $b(x)$.

Substituting $u = x$ (so $u' = 1$, $u'' = 0$) into $u'' + a(x)u' + b(x)u = 0$:

$0 + a(x) \cdot 1 + b(x) \cdot x = 0 \implies a(x) + xb(x) = 0. \qquad \text{(i)}$

Substituting $u = e^{-x}$ (so $u' = -e^{-x}$, $u'' = e^{-x}$):

$e^{-x} + a(x)(-e^{-x}) + b(x)e^{-x} = 0 \implies 1 - a(x) + b(x) = 0. \qquad \text{(ii)}$

From (i): $a(x) = -xb(x)$. Substituting into (ii):

$1 + xb(x) + b(x) = 0 \implies b(x)(1 + x) = -1 \implies b(x) = \frac{-1}{1+x}.$

Therefore $a(x) = \frac{x}{1+x}$.

General solution. Since $u_1 = x$ and $u_2 = e^{-x}$ are two linearly independent solutions:

$u = Ax + Be^{-x}.$

Verifying the substitution $y = \frac{1}{3u}\frac{\mathrm{d}u}{\mathrm{d}x}$.

If $y = \frac{u'}{3u}$, then $3uy = u'$, so $u = \frac{u'}{3y}$.

Differentiating $3uy = u'$ with respect to $x$: $3u'y + 3uy' = u''$, so $u'' = 3u'y + 3uy'$.

Substituting into $u'' + \frac{x}{1+x}u' - \frac{1}{1+x}u = 0$:

$3u'y + 3uy' + \frac{x}{1+x}u' - \frac{1}{1+x}u = 0.$

Dividing by $3u$:

$\frac{u'}{u}y + y' + \frac{x}{3(1+x)}\frac{u'}{u} - \frac{1}{3(1+x)} = 0.$

Since $\frac{u'}{u} = 3y$:

$3y \cdot y + y' + \frac{x}{3(1+x)} \cdot 3y - \frac{1}{3(1+x)} = 0,$

$y' + 3y^2 + \frac{xy}{1+x} - \frac{1}{3(1+x)} = 0,$

which is exactly the equation (*).

Solving () with $y(0) = 0$.*

The general solution of the ODE for $u$ is $u = Ax + Be^{-x}$, so

$y = \frac{1}{3u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{A - Be^{-x}}{3(Ax + Be^{-x})}.$

Applying $y = 0$ at $x = 0$: $\frac{A - B}{3B} = 0$, so $A = B$. Taking $A = B = 1$:

$y = \frac{1 - e^{-x}}{3(x + e^{-x})}.$

Part (ii)

We need to solve $\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 + \frac{x}{1-x}y = \frac{1}{1-x}$ with $y(0) = 2$.

We use the substitution $y = \frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}x}$ (i.e. $y = \frac{u'}{u}$). Then $u' = uy$ and $u'' = u'y + uy' = uy^2 + uy'$, so $y' = \frac{u''}{u} - y^2 = \frac{u''}{u} - \frac{(u')^2}{u^2}$.

Substituting into the equation:

$\frac{u''}{u} - y^2 + y^2 + \frac{x}{1-x}y = \frac{1}{1-x},$

$\frac{u''}{u} + \frac{x}{1-x} \cdot \frac{u'}{u} = \frac{1}{1-x}.$

Multiplying by $u$:

$u'' + \frac{x}{1-x}u' - \frac{1}{1-x}u = 0. \qquad \text{(iii)}$

Comparing with the equation from part (i): $u'' + \frac{x}{1+x}u' - \frac{1}{1+x}u = 0$.

Replacing $x$ by $-x$ in (iii): let $v(x) = u(-x)$, then $v' = -u'(-x)$ and $v'' = u''(-x)$. Substituting into the part (i) equation with $x \to -x$:

$u''(-x) + \frac{-x}{1+(-x)}u'(-x) - \frac{1}{1+(-x)}u(-x) = 0,$

$u''(-x) + \frac{-x}{1-x}u'(-x) - \frac{1}{1-x}u(-x) = 0.$

But (iii) requires $u''(x) + \frac{x}{1-x}u'(x) - \frac{1}{1-x}u(x) = 0$. Replacing $x$ by $-x$ in (iii):

$u''(-x) + \frac{-x}{1-(-x)}u'(-x) - \frac{1}{1-(-x)}u(-x) = u''(-x) + \frac{-x}{1+x}u'(-x) - \frac{1}{1+x}u(-x) = 0.$

This is the part (i) equation with argument $-x$. So if $f(x)$ solves (iii), then $f(-x)$ satisfies the part (i) ODE.

The part (i) ODE has general solution $u = Ax + Be^{-x}$. Therefore the general solution of (iii) is:

$u(x) = A(-x) + Be^{-(-x)} = -Ax + Be^x,$

which we can write as $u = Cx + De^x$ (absorbing the sign into the constants).

Therefore

$y = \frac{u'}{u} = \frac{C + De^x}{Cx + De^x}.$

Applying $y(0) = 2$: $\frac{C + D}{D} = 2$, so $C + D = 2D$, giving $C = D$. Taking $C = D = 1$:

$y = \frac{1 + e^x}{x + e^x}.$

Examiner Notes

This ranked alongside question 5 in popularity and success. Frequently, it was calculation errors that obscured the path through part (i) and the two differences between part (i) and part (ii) were enough to put most off the track for part (ii), even if they had completed (i) successfully.