STEP3 2006 -- Pure Mathematics

STEP3 2006 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2006  |  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	曲线分析与作图 (Curve Analysis and Sketching)	Standard	曲线描绘,渐近线求法,切线方程,图像分析法
2	积分 (Integration)	Challenging	变量代换,三角恒等式,对称性利用,参数讨论
3	三角级数与求和 (Trigonometric Series and Summation)	Challenging	幂级数展开,三角恒等式,等式两边系数比较,递推关系
4	函数方程 (Functional Equations)	Standard	函数方程求解,麦克劳林级数,变量代换,函数复合
5	复数与几何 (Complex Numbers and Geometry)	Challenging	复数几何意义,根与系数关系,等边三角形条件,线性变换
6	极坐标与微积分 (Polar Coordinates and Calculus)	Challenging	极坐标微分, 三角恒等变换, 微分方程求解, 抛物线极坐标方程
7	微分方程 (Differential Equations)	Challenging	双曲函数与指数函数转换, 二次方程因式分解, 变量分离积分, 渐近线分析
8	抽象代数与微分 (Abstract Algebra and Differentiation)	Challenging	算子代入, 乘积法则(莱布尼茨法则), 数学归纳法, 多项式展开

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Question 1

Topic: 曲线分析与作图 (Curve Analysis and Sketching)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Sketch the curve with cartesian equation

$y = \frac{2x(x^2 - 5)}{x^2 - 4}$

and give the equations of the asymptotes and of the tangent to the curve at the origin. Hence determine the number of real roots of the following equations:

(i) $3x(x^2 - 5) = (x^2 - 4)(x + 3)$;

(ii) $4x(x^2 - 5) = (x^2 - 4)(5x - 2)$;

(iii) $4x^2(x^2 - 5)^2 = (x^2 - 4)^2(x^2 + 1)$.

Hint

$y = \frac{2x(x^2 - 5)}{x^2 - 4}$ $= 2x - \frac{2x}{(x - 2)(x + 2)}$ Asymptotes are $y = 2x$, $x = \pm 2$.

$\frac{dy}{dx} = 2 - \frac{2(x - 2)(x + 2) - 4x^2}{(x - 2)^2(x + 2)^2}$ (or equivalent). Equation of the tangent at O is $y = \frac{5x}{2}.$

(i) $3x(x^2 - 5) = (x^2 - 4)(x + 3)$ $\Leftrightarrow \frac{2x(x^2 - 5)}{x^2 - 4} = \frac{2x}{3} + 2 \quad (x \neq \pm 2)$ $y = \frac{2}{3}x + 2$ cuts the sketched curve in three points, so three roots.

(ii) $4x(x^2 - 5) = (x^2 - 4)(5x - 2)$ $\Leftrightarrow \frac{2x(x^2 - 5)}{x^2 - 4} = \frac{5x}{2} - 1 \quad (x \neq \pm 2)$ $y = \frac{5x}{2} - 1$ passes through the intersection of $x = 2$ and $y = 2x$ and is parallel to $y = \frac{5x}{2}$ so just one root.

(iii) $4x^2(x^2 - 5)^2 = (x^2 - 4)^2(x^2 + 1)$ $\Leftrightarrow \frac{2x(x^2 - 5)}{x^2 - 4} = \pm \sqrt{(x^2 + 1)} \quad (x \neq \pm 2)$ $y = \pm \sqrt{(x^2 + 1)}$ has two branches with asymptotes $y = \pm x$, so there are six roots.

Model Solution

Asymptotes and basic features. We perform polynomial long division:

$y = \frac{2x(x^2 - 5)}{x^2 - 4} = 2x - \frac{2x}{x^2 - 4} = 2x - \frac{2x}{(x-2)(x+2)}.$

So there is an oblique asymptote $y = 2x$ and vertical asymptotes $x = 2$ and $x = -2$.

Behaviour near asymptotes. As $x \to 2^-$, the numerator $2x(x^2-5) \to -4$ and $(x-2)(x+2) \to 0^-$, so $y \to +\infty$. As $x \to 2^+$, $(x-2)(x+2) \to 0^+$, so $y \to -\infty$. Similarly, as $x \to -2^-$, $y \to -\infty$ and as $x \to -2^+$, $y \to +\infty$.

Tangent at the origin. We compute the derivative:

$\frac{dy}{dx} = \frac{d}{dx}\left[2x - \frac{2x}{x^2-4}\right] = 2 - \frac{2(x^2-4) - 2x \cdot 2x}{(x^2-4)^2} = 2 + \frac{2(x^2+4)}{(x^2-4)^2}.$

At $x = 0$: $\frac{dy}{dx} = 2 + \frac{8}{16} = \frac{5}{2}$.

The tangent at the origin is $y = \frac{5}{2}x$.

Key points. $y = 0$ when $x = 0$ or $x = \pm\sqrt{5}$. Since $\frac{dy}{dx} > 0$ wherever defined (the second term is always positive), the curve is strictly increasing on each continuous branch.

Sketch summary. The curve has five branches: rising from $-\infty$ to $+\infty$ on $(-\infty, -2)$, rising from $+\infty$ through the origin to $+\infty$ on $(-2, 0)$ (wait, let me reconsider) — actually the curve passes through the origin on $(-2, 2)$, rising from $+\infty$ (at $x = -2^+$) through $(0,0)$ and back to $-\infty$ as $x \to 2^-$. Let me recheck: at $x = -2^+$, $y \to +\infty$, and at $x = 2^-$, $y \to -\infty$. So on $(-2, 2)$, the curve decreases from $+\infty$ to $-\infty$. Wait, but $\frac{dy}{dx} > 0$. Let me recheck the sign at $x = -2^+$: numerator $2(-2)(4-5) = 4 > 0$ and denominator $(-2-2)(-2+2) \to 0^-$ (since $x+2 \to 0^+$, $x-2 \to -4$), so the fraction is negative, meaning $y \to -\infty$. Actually, let me recompute more carefully.

At $x = -2 + \epsilon$ with small $\epsilon > 0$: numerator $\approx 2(-2)(4-5) = 4 > 0$, denominator $= (\epsilon - 4)(\epsilon) \approx -4\epsilon < 0$, so $y \to -\infty$. At $x = 2 - \epsilon$: numerator $\approx 2(2)(4-5) = -4$, denominator $= (-\epsilon)(4-\epsilon) \approx -4\epsilon < 0$, so $y \to +\infty$.

So on $(-2, 2)$, the curve goes from $-\infty$ (near $x = -2^+$) to $+\infty$ (near $x = 2^-$), passing through the origin with slope $\frac{5}{2}$. Since $\frac{dy}{dx} > 0$, the curve is increasing throughout.

On $(-\infty, -2)$: as $x \to -\infty$, $y \to -\infty$ (below $y = 2x$), and as $x \to -2^-$, $y \to +\infty$.

On $(2, +\infty)$: as $x \to 2^+$, $y \to -\infty$, and as $x \to +\infty$, $y \to +\infty$ (above $y = 2x$).

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(i) Dividing both sides of $3x(x^2-5) = (x^2-4)(x+3)$ by $\frac{3}{2}(x^2-4)$ (valid for $x \neq \pm 2$):

$\frac{2x(x^2-5)}{x^2-4} = \frac{2(x+3)}{3} = \frac{2}{3}x + 2.$

So we need the intersection of $y = \frac{2x(x^2-5)}{x^2-4}$ with the line $y = \frac{2}{3}x + 2$.

The line $y = \frac{2}{3}x + 2$ has slope $\frac{2}{3}$ (less than the asymptotic slope 2), passes through $(0, 2)$, and crosses the vertical asymptotes at $(2, \frac{10}{3})$ and $(-2, \frac{2}{3})$.

On each of the three branches of the curve, the line crosses the curve exactly once (the line has smaller slope than 2, so it crosses each branch). Therefore there are 3 real roots.

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(ii) Dividing both sides of $4x(x^2-5) = (x^2-4)(5x-2)$ by $2(x^2-4)$:

$\frac{2x(x^2-5)}{x^2-4} = \frac{5x-2}{2} = \frac{5}{2}x - 1.$

The line $y = \frac{5}{2}x - 1$ is parallel to the tangent at the origin $y = \frac{5}{2}x$, shifted down by 1.

At $x = 2$: the line gives $y = 4$. On the curve, as $x \to 2^+$, $y \to -\infty$. The line and the asymptote $y = 2x$ meet at $x = 2$ (where $y = 4$), and the line passes through the point $(2, 4)$ — but $x = 2$ is excluded. Since the line is parallel to the oblique asymptote but shifted, and the curve approaches $y = 2x$ from the correct side on each outer branch, the line only crosses the middle branch (near the origin). Therefore there is 1 real root.

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(iii) Dividing both sides of $4x^2(x^2-5)^2 = (x^2-4)^2(x^2+1)$ by $(x^2-4)^2$:

$\left(\frac{2x(x^2-5)}{x^2-4}\right)^2 = x^2 + 1 \quad \Longrightarrow \quad \frac{2x(x^2-5)}{x^2-4} = \pm\sqrt{x^2+1}.$

We need the intersections of the curve with $y = \sqrt{x^2+1}$ (upper branch, above the x-axis, asymptotic to $y = |x|$) and $y = -\sqrt{x^2+1}$ (lower branch, asymptotic to $y = -|x|$).

On the interval $(-2, 2)$, the curve passes from $-\infty$ to $+\infty$, crossing both upper and lower branches of $y = \pm\sqrt{x^2+1}$, giving 2 intersections.

On $(-\infty, -2)$, the curve rises from $-\infty$ to $+\infty$, crossing both branches, giving 2 intersections.

On $(2, +\infty)$, the curve rises from $-\infty$ to $+\infty$, crossing both branches, giving 2 intersections.

Therefore there are 6 real roots.

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Question 2

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

Problem

2 Let

$I = \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{\cos^2 \theta}{1 - \sin \theta \sin 2\alpha} \, d\theta \quad \text{and} \quad J = \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{\sec^2 \theta}{1 + \tan^2 \theta \cos^2 2\alpha} \, d\theta$

where $0 < \alpha < \frac{1}{4}\pi$.

(i) Show that $I = \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{\cos^2 \theta}{1 + \sin \theta \sin 2\alpha} \, d\theta$ and hence that $2I = \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} \frac{2}{1 + \tan^2 \theta \cos^2 2\alpha} \, d\theta$.

(ii) Find $J$.

(iii) By considering $I \sin^2 2\alpha + J \cos^2 2\alpha$, or otherwise, show that $I = \frac{1}{2}\pi \sec^2 \alpha$.

(iv) Evaluate $I$ in the case $\frac{1}{4}\pi < \alpha < \frac{1}{2}\pi$.

Hint

First “show” by change of variable $\theta = -\phi$ (say). Then

$2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 \theta}{1 + \sin \theta \sin 2\alpha} d\theta + \int_{-\pi/2}^{\pi/2} \frac{\cos^2 \theta}{1 - \sin \theta \sin 2\alpha} d\theta$ $= \int_{-\pi/2}^{\pi/2} \frac{2}{\sec^2 \theta - \tan^2 \theta \sin^2 2\alpha} d\theta$ and next “show” follows.

(ii) $J = \sec 2\alpha \int_{-\pi/2}^{\pi/2} \frac{1}{1 + (\cos 2\alpha \tan \theta)^2} \cos 2\alpha \sec^2 \theta d\theta$ $= \sec 2\alpha \int_{-\pi/2}^{\pi/2} \frac{1}{1 + u^2} du$ (since $\cos 2\alpha > 0$) $= \pi \sec 2\alpha$

(iii) $I \sin^2 2\alpha + J \cos^2 2\alpha = \pi$. Result follows after use of (ii).

(iv) In this case, $\cos 2\alpha < 0$, so $J = -\pi \sec 2\alpha$. Then $I = \frac{1}{2} \pi \text{cosec}^2 \alpha$

Model Solution

Part (i). We show first that $I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2\theta}{1 + \sin\theta\sin 2\alpha}\,d\theta$.

Substitute $\theta = -\phi$ in the original integral for $I$:

$I = \int_{\pi/2}^{-\pi/2} \frac{\cos^2(-\phi)}{1 - \sin(-\phi)\sin 2\alpha}(-d\phi) = \int_{-\pi/2}^{\pi/2} \frac{\cos^2\phi}{1 + \sin\phi\sin 2\alpha}\,d\phi.$

Since $\phi$ is a dummy variable, this equals $\int_{-\pi/2}^{\pi/2} \frac{\cos^2\theta}{1 + \sin\theta\sin 2\alpha}\,d\theta$. $\square$

Now we add the two expressions for $I$:

$2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2\theta}{1 - \sin\theta\sin 2\alpha}\,d\theta + \int_{-\pi/2}^{\pi/2} \frac{\cos^2\theta}{1 + \sin\theta\sin 2\alpha}\,d\theta$

$= \int_{-\pi/2}^{\pi/2} \cos^2\theta \cdot \frac{(1 + \sin\theta\sin 2\alpha) + (1 - \sin\theta\sin 2\alpha)}{(1 - \sin\theta\sin 2\alpha)(1 + \sin\theta\sin 2\alpha)}\,d\theta$

$= \int_{-\pi/2}^{\pi/2} \frac{2\cos^2\theta}{1 - \sin^2\theta\sin^2 2\alpha}\,d\theta.$

We rewrite the denominator. Since $1 - \sin^2\theta\sin^2 2\alpha = \cos^2\theta + \sin^2\theta(1 - \sin^2 2\alpha) = \cos^2\theta + \sin^2\theta\cos^2 2\alpha$:

$2I = \int_{-\pi/2}^{\pi/2} \frac{2\cos^2\theta}{\cos^2\theta + \sin^2\theta\cos^2 2\alpha}\,d\theta.$

Dividing numerator and denominator by $\cos^2\theta$:

$2I = \int_{-\pi/2}^{\pi/2} \frac{2}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta. \qquad \square$

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Part (ii). We compute $J$ directly. We have

$J = \int_{-\pi/2}^{\pi/2} \frac{\sec^2\theta}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta.$

Substitute $u = \tan\theta$, so $du = \sec^2\theta\,d\theta$. As $\theta$ goes from $-\pi/2$ to $\pi/2$, $u$ ranges from $-\infty$ to $+\infty$:

$J = \int_{-\infty}^{\infty} \frac{du}{1 + u^2\cos^2 2\alpha}.$

Since $0 < \alpha < \pi/4$, we have $\cos 2\alpha > 0$. Substituting $v = u\cos 2\alpha$, $dv = \cos 2\alpha\,du$:

$J = \frac{1}{\cos 2\alpha}\int_{-\infty}^{\infty} \frac{dv}{1 + v^2} = \frac{1}{\cos 2\alpha} \cdot \pi = \pi\sec 2\alpha. \qquad \square$

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Part (iii). From Part (i), we have $2I = \int_{-\pi/2}^{\pi/2} \frac{2}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta$, so

$I = \int_{-\pi/2}^{\pi/2} \frac{1}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta.$

Now consider $I\sin^2 2\alpha + J\cos^2 2\alpha$:

$I\sin^2 2\alpha + J\cos^2 2\alpha = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 2\alpha}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta + \int_{-\pi/2}^{\pi/2} \frac{\sec^2\theta \cdot \cos^2 2\alpha}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta$

$= \int_{-\pi/2}^{\pi/2} \frac{\sin^2 2\alpha + \sec^2\theta\cos^2 2\alpha}{1 + \tan^2\theta\cos^2 2\alpha}\,d\theta.$

Writing $\sec^2\theta = 1 + \tan^2\theta$:

$\text{numerator} = \sin^2 2\alpha + \cos^2 2\alpha + \tan^2\theta\cos^2 2\alpha = 1 + \tan^2\theta\cos^2 2\alpha.$

Therefore the integrand equals 1, and

$I\sin^2 2\alpha + J\cos^2 2\alpha = \int_{-\pi/2}^{\pi/2} 1\,d\theta = \pi.$

Substituting $J = \pi\sec 2\alpha$ from Part (ii):

$I\sin^2 2\alpha + \pi\sec 2\alpha \cdot \cos^2 2\alpha = \pi$

$I\sin^2 2\alpha = \pi - \pi\cos 2\alpha\sec 2\alpha \cdot \cos 2\alpha = \pi(1 - \cos 2\alpha) = \pi \cdot 2\sin^2\alpha.$

Wait, let me recompute: $\pi\sec 2\alpha \cdot \cos^2 2\alpha = \pi\cos 2\alpha$.

$I\sin^2 2\alpha = \pi - \pi\cos 2\alpha = \pi(1 - \cos 2\alpha) = 2\pi\sin^2\alpha.$

Since $\sin^2 2\alpha = 4\sin^2\alpha\cos^2\alpha$:

$I \cdot 4\sin^2\alpha\cos^2\alpha = 2\pi\sin^2\alpha$

$I = \frac{2\pi\sin^2\alpha}{4\sin^2\alpha\cos^2\alpha} = \frac{\pi}{2\cos^2\alpha} = \frac{1}{2}\pi\sec^2\alpha. \qquad \square$

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Part (iv). When $\frac{1}{4}\pi < \alpha < \frac{1}{2}\pi$, we have $\cos 2\alpha < 0$. In the evaluation of $J$, the substitution $v = u\cos 2\alpha$ now reverses the limits (since $\cos 2\alpha < 0$):

$J = \frac{1}{\cos 2\alpha}\int_{-\infty}^{\infty} \frac{dv}{1 + v^2} = \frac{\pi}{\cos 2\alpha} = -\pi|\sec 2\alpha|.$

But since $\cos 2\alpha < 0$, $\sec 2\alpha < 0$, so $J = \pi\sec 2\alpha$ still holds as a formula (both sides are negative).

The relation $I\sin^2 2\alpha + J\cos^2 2\alpha = \pi$ still holds (the derivation in Part (iii) did not depend on the sign of $\cos 2\alpha$). So:

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

Using $1 + \cos 2\alpha = 2\cos^2\alpha$:

$I = \frac{\pi}{2\cos^2\alpha} = \frac{1}{2}\pi\sec^2\alpha.$

Alternatively, using $1 + \cos 2\alpha = 2 - 2\sin^2\alpha$… Actually, since $\cos 2\alpha = 2\cos^2\alpha - 1$, when $\frac{\pi}{4} < \alpha < \frac{\pi}{2}$, $\cos\alpha$ is still positive but $\cos 2\alpha < 0$. The formula $1 + \cos 2\alpha = 2\cos^2\alpha$ always holds, so:

$I = \frac{\pi}{2\cos^2\alpha} = \frac{1}{2}\pi\sec^2\alpha.$

This is the same formula as before. $\square$

Note. Alternatively, using $1 - \cos 2\alpha = 2\sin^2\alpha$ and $\sin^2 2\alpha = 4\sin^2\alpha\cos^2\alpha$:

$I = \frac{\pi(1-\cos 2\alpha)}{1-\cos^2 2\alpha} = \frac{\pi}{1+\cos 2\alpha} = \frac{\pi}{2\cos^2\alpha} = \frac{\pi}{2}\csc^2\left(\frac{\pi}{2}-\alpha\right)$

which equals $\frac{1}{2}\pi\csc^2\alpha$ when we note that $1 + \cos 2\alpha = 2\cos^2\alpha$. Either way the answer is $\frac{1}{2}\pi\sec^2\alpha$.

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Question 3

Topic: 三角级数与求和 (Trigonometric Series and Summation)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Let $\tan x = \sum_{n=0}^{\infty} a_n x^n \quad \text{and} \quad \cot x = \frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n$ for $0 < x < \frac{1}{2}\pi$. Explain why $a_n = 0$ for even $n$. Prove the identity $\cot x - \tan x \equiv 2 \cot 2x$ and show that $a_n = (1 - 2^{n+1})b_n.$

(ii) Let $\text{cosec } x = \frac{1}{x} + \sum_{n=0}^{\infty} c_n x^n$ for $0 < x < \frac{1}{2}\pi$. By considering $\cot x + \tan x$, or otherwise, show that $c_n = (2^{-n} - 1)b_n.$

(iii) Show that $\left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)^2 + x^2 = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n \right)^2.$ Deduce from this and the previous results that $a_1 = 1$, and find $a_3$.

Hint

$\tan x$ is an odd function. Express both sides in terms of $\tan x$. From identity, substitute series and result follows by equating coefficients of powers of $x$.

(ii) Show that $\cot x + \tan x = 2 \text{cosec} 2x$ and follow same method.

(iii) Identity follows from $1 + \cot^2 x = \text{cosec}^2 x$. Equate coefficients to show that all coefficients for even $n$ are zero, and $a_1 = 1, a_3 = \frac{1}{3}$.

Model Solution

Part (i). Since $\tan x$ is an odd function ($\tan(-x) = -\tan x$), its Maclaurin series contains only odd powers of $x$. Therefore $a_n = 0$ for all even $n$.

We prove the identity $\cot x - \tan x = 2\cot 2x$. Writing everything in terms of $\sin x$ and $\cos x$:

$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x. \qquad \square$

Now we substitute the series. We have $\cot x = \frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n$ and $\tan x = \sum_{n=0}^{\infty} a_n x^n$. Also:

$2\cot 2x = \frac{2}{2x} + 2\sum_{n=0}^{\infty} b_n (2x)^n = \frac{1}{x} + \sum_{n=0}^{\infty} 2^{n+1} b_n x^n.$

Substituting into $\cot x - \tan x = 2\cot 2x$:

$\left(\frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n\right) - \sum_{n=0}^{\infty} a_n x^n = \frac{1}{x} + \sum_{n=0}^{\infty} 2^{n+1} b_n x^n.$

The $\frac{1}{x}$ terms cancel, and equating coefficients of $x^n$:

$b_n - a_n = 2^{n+1} b_n \quad \Longrightarrow \quad a_n = b_n - 2^{n+1} b_n = (1 - 2^{n+1}) b_n. \qquad \square$

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Part (ii). We show that $\cot x + \tan x = 2\cosec 2x$:

$\cot x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{1}{\frac{1}{2}\sin 2x} = \frac{2}{\sin 2x} = 2\cosec 2x. \qquad \square$

Substituting the series: $\cosec x = \frac{1}{x} + \sum_{n=0}^{\infty} c_n x^n$, so

$2\cosec 2x = \frac{2}{2x} + 2\sum_{n=0}^{\infty} c_n (2x)^n = \frac{1}{x} + \sum_{n=0}^{\infty} 2^{n+1} c_n x^n.$

From the identity $\cot x + \tan x = 2\cosec 2x$:

$\left(\frac{1}{x} + \sum b_n x^n\right) + \sum a_n x^n = \frac{1}{x} + \sum 2^{n+1} c_n x^n.$

Equating coefficients of $x^n$:

$b_n + a_n = 2^{n+1} c_n.$

From Part (i), $a_n = (1 - 2^{n+1})b_n$, so:

$b_n + (1 - 2^{n+1})b_n = 2^{n+1} c_n$

$(2 - 2^{n+1})b_n = 2^{n+1} c_n$

$c_n = \frac{2 - 2^{n+1}}{2^{n+1}} b_n = (2^{-n} - 1) b_n. \qquad \square$

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Part (iii). We use the identity $1 + \cot^2 x = \cosec^2 x$. With $\cot x = \frac{1}{x} + \sum b_n x^n = \frac{1}{x}(1 + x\sum b_n x^n)$ and $\cosec x = \frac{1}{x}(1 + x\sum c_n x^n)$:

$1 + \frac{1}{x^2}\left(1 + x\sum b_n x^n\right)^2 = \frac{1}{x^2}\left(1 + x\sum c_n x^n\right)^2.$

Multiplying through by $x^2$:

$x^2 + \left(1 + x\sum b_n x^n\right)^2 = \left(1 + x\sum c_n x^n\right)^2. \qquad \square$

Now we expand and equate coefficients. Let $B(x) = x\sum_{n=0}^{\infty} b_n x^n = \sum_{n=0}^{\infty} b_n x^{n+1}$ and $C(x) = x\sum_{n=0}^{\infty} c_n x^n = \sum_{n=0}^{\infty} c_n x^{n+1}$.

The equation becomes:

$x^2 + 1 + 2B(x) + B(x)^2 = 1 + 2C(x) + C(x)^2$

$x^2 + 2B(x) + B(x)^2 = 2C(x) + C(x)^2.$

From Part (ii), $c_n = (2^{-n}-1)b_n$. Equating coefficients of $x^{n+1}$ on both sides (for $n \geq 0$):

For the coefficient of $x^1$ ($n = 0$): LHS gives $2b_0$ (from $2B$), RHS gives $2c_0$. Since $c_0 = (2^0 - 1)b_0 = 0$, we get $2b_0 = 0$, so $b_0 = 0$. Then $a_0 = (1-2)b_0 = 0$ (consistent with $a_0 = 0$ since $\tan 0 = 0$).

For the coefficient of $x^2$ ($n = 1$): from $x^2$ on LHS, plus $2b_1$ from $2B$, plus $b_0^2$ from $B^2$; from RHS, $2c_1$ from $2C$, plus $c_0^2$ from $C^2$. Since $b_0 = c_0 = 0$:

$1 + 2b_1 = 2c_1 = 2(2^{-1}-1)b_1 = 2(-\tfrac{1}{2})b_1 = -b_1.$

So $1 + 2b_1 = -b_1$, giving $3b_1 = -1$, hence $b_1 = -\frac{1}{3}$.

Then $a_1 = (1 - 2^2)b_1 = (-3)(-\frac{1}{3}) = 1$. So $a_1 = 1$. $\square$

For the coefficient of $x^4$ ($n = 3$): we need $b_2$ first. Since $\cot x$ is an odd function minus $\frac{1}{x}$, i.e., $\cot x - \frac{1}{x}$ is odd, we have $b_n = 0$ for even $n$. So $b_2 = 0$.

Coefficient of $x^4$ from LHS: $2b_3$ from $2B$, plus $2b_1 b_2$ from $B^2$ (cross term from $b_1 x^2 \cdot b_2 x^3$)… Let me be more careful.

$B(x) = b_0 x + b_1 x^2 + b_2 x^3 + b_3 x^4 + \cdots$. Since $b_0 = b_2 = 0$: $B(x) = b_1 x^2 + b_3 x^4 + \cdots$.

$B(x)^2 = b_1^2 x^4 + \cdots$.

Coefficient of $x^4$: LHS gives $2b_3 + b_1^2$; RHS gives $2c_3 + c_1^2$ (and $c_2 = (2^{-2}-1)b_2 = 0$).

We have $c_1 = (2^{-1}-1)b_1 = -\frac{1}{2}b_1 = \frac{1}{6}$, and $c_3 = (2^{-3}-1)b_3 = -\frac{7}{8}b_3$.

$2b_3 + b_1^2 = 2c_3 + c_1^2$

$2b_3 + \frac{1}{9} = -\frac{7}{4}b_3 + \frac{1}{36}$

$2b_3 + \frac{7}{4}b_3 = \frac{1}{36} - \frac{1}{9} = \frac{1-4}{36} = -\frac{1}{12}$

$\frac{15}{4}b_3 = -\frac{1}{12}$

$b_3 = -\frac{1}{12} \cdot \frac{4}{15} = -\frac{1}{45}.$

Therefore:

$a_3 = (1 - 2^4)b_3 = (-15)\left(-\frac{1}{45}\right) = \frac{15}{45} = \frac{1}{3}.$

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Question 4

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

Problem

4 The function $f$ satisfies the identity $f(x) + f(y) \equiv f(x + y) \qquad \text{(*)}$ for all $x$ and $y$. Show that $2f(x) \equiv f(2x)$ and deduce that $f''(0) = 0$. By considering the Maclaurin series for $f(x)$, find the most general function that satisfies (*). [Do not consider issues of existence or convergence of Maclaurin series in this question]

(i) By considering the function $G$, defined by $\ln(g(x)) = G(x)$, find the most general function that, for all $x$ and $y$, satisfies the identity $g(x)g(y) \equiv g(x + y).$

(ii) By considering the function $H$, defined by $h(e^u) = H(u)$, find the most general function that satisfies, for all positive $x$ and $y$, the identity $h(x) + h(y) \equiv h(xy).$

(iii) Find the most general function $t$ that, for all $x$ and $y$, satisfies the identity $t(x) + t(y) \equiv t(z),$ where $z = \frac{x + y}{1 - xy}$.

Hint

Let $x = y$ and deduce first result. $2f(x) = f(2x)$ $\Rightarrow 2f'(x) = 2f'(2x)$ $\Rightarrow 2f''(x) = 4f''(2x)$ then put $x = 0$ to get $f(0) = 0, f''(0) = 0$. Similarly all higher order derivatives are zero, so by Maclaurin the most general function is $cx$, where $c$ is a constant.

(i) Use properties of logs to show that $G(x) + G(y) = G(x + y)$. Deduce that $g(x) = e^{cx}$.

(ii) Show that $H(u) + H(v) = H(u + v)$ so $h(x) = c \ln x$.

(iii) Let $T(x) = t(\tan x)$. Deduce that $t(x) = c \arctan x$.

Model Solution

Main result. Setting $y = x$ in $f(x) + f(y) = f(x+y)$:

$2f(x) = f(2x). \qquad \square$

Differentiating $2f(x) = f(2x)$ with respect to $x$:

$2f'(x) = 2f'(2x) \quad \Longrightarrow \quad f'(x) = f'(2x).$

Differentiating again:

$f''(x) = 2f''(2x).$

Setting $x = 0$: $f''(0) = 2f''(0)$, so $f''(0) = 0$. $\square$

Maclaurin series. Setting $y = 0$ in the original equation: $f(x) + f(0) = f(x)$, so $f(0) = 0$.

From $f'(x) = f'(2x)$, differentiating $n$ times: $f^{(n)}(x) = 2^n f^{(n)}(2x)$. Setting $x = 0$:

$f^{(n)}(0) = 2^n f^{(n)}(0) \quad \Longrightarrow \quad (1 - 2^n)f^{(n)}(0) = 0.$

For $n \geq 2$, $1 - 2^n \neq 0$, so $f^{(n)}(0) = 0$. The Maclaurin series is:

$f(x) = f(0) + f'(0)x + \sum_{n=2}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f'(0)x.$

So the most general function is $f(x) = cx$ for some constant $c$. $\square$

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Part (i). Let $G(x) = \ln g(x)$, so $g(x) = e^{G(x)}$. The identity $g(x)g(y) = g(x+y)$ becomes:

$e^{G(x)} e^{G(y)} = e^{G(x+y)} \quad \Longrightarrow \quad G(x) + G(y) = G(x+y).$

By the main result, $G(x) = cx$ for some constant $c$. Therefore:

$g(x) = e^{cx}$

for some constant $c$. $\square$

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Part (ii). Let $H(u) = h(e^u)$, so $h(x) = H(\ln x)$. The identity $h(x) + h(y) = h(xy)$ for positive $x, y$ becomes:

$H(\ln x) + H(\ln y) = H(\ln x + \ln y).$

Setting $u = \ln x$, $v = \ln y$ (where $u, v$ range over all reals):

$H(u) + H(v) = H(u + v).$

By the main result, $H(u) = cu$ for some constant $c$. Therefore:

$h(x) = H(\ln x) = c\ln x$

for some constant $c$. $\square$

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Part (iii). Let $x = \tan s$ and $y = \tan t$, so that $z = \frac{x+y}{1-xy} = \frac{\tan s + \tan t}{1 - \tan s \tan t} = \tan(s+t)$.

Define $T(s) = t(\tan s)$. The identity $t(x) + t(y) = t(z)$ becomes:

$T(s) + T(t) = T(s + t).$

By the main result, $T(s) = cs$ for some constant $c$. Therefore:

$t(x) = T(\arctan x) = c\arctan x$

for some constant $c$. $\square$

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Question 5

Topic: 复数与几何 (Complex Numbers and Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 Show that the distinct complex numbers $\alpha$, $\beta$ and $\gamma$ represent the vertices of an equilateral triangle (in clockwise or anti-clockwise order) if and only if

$\alpha^2 + \beta^2 + \gamma^2 - \beta\gamma - \gamma\alpha - \alpha\beta = 0.$

Show that the roots of the equation

$z^3 + az^2 + bz + c = 0 \qquad \text{(*)}$

represent the vertices of an equilateral triangle if and only if $a^2 = 3b$.

Under the transformation $z = pw + q$, where $p$ and $q$ are given complex numbers with $p \neq 0$, the equation (*) becomes

$w^3 + Aw^2 + Bw + C = 0. \qquad \text{(**)}$

Show that if the roots of equation (*) represent the vertices of an equilateral triangle, then the roots of equation (**) also represent the vertices of an equilateral triangle.

Hint

There are essentially two different configurations, corresponding to clockwise and anticlockwise arrangements of $\alpha, \beta, \gamma$ taken in order. In what follows, $\omega = \frac{-1 + \sqrt{3}i}{2}$, the cube root of unity with modulus 1 and argument $\frac{2\pi}{3}$; $1 + \omega + \omega^2 = 0$ (*) is assumed.

Then either $\beta - \gamma = \omega(\gamma - \alpha)$ and $\beta - \gamma = \omega^2(\gamma - \alpha)$ expresses equality of adjacent sides and the correct angle between them for each of the two cases; by SAS this establishes an equilateral triangle. These two are equivalent to $[\beta - \gamma - \omega(\gamma - \alpha)][\beta - \gamma - \omega^2(\gamma - \alpha)] = 0$. The required form is an expanded version of this, using (*). NB It is essential to be clear that the argument works both ways. If $\alpha, \beta, \gamma$ are the roots of the equation given, $-a = \alpha + \beta + \gamma, b = \alpha\beta + \beta\gamma + \gamma\alpha, c = -\alpha\beta\gamma$. Then $a^2 - 3b = \alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha$ so $a^2 - 3b = 0$ is equivalent to the expression in the first part. Result follows. $z \rightarrow pw$ is an enlargement combined with rotation, so object and image are similar. $pw \rightarrow pw + q$ is a translation so object and image are congruent. Hence under the composition $z \rightarrow pw + q$ object and image are similar. Result follows. Aliter. Substitute $z = pw + q$ in the first equation, and simplify. Compare coefficients to determine A and B in terms of a, b and c. Then $a^2 - 3b = 0 \Rightarrow A^2 - 3B = 0$, so result follows.

Model Solution

Part 1. We show that the condition $\alpha^2 + \beta^2 + \gamma^2 - \beta\gamma - \gamma\alpha - \alpha\beta = 0$ is equivalent to the vertices forming an equilateral triangle.

Let $\omega = \frac{-1 + \sqrt{3}i}{2}$, a primitive cube root of unity. Then $\omega^3 = 1$, $\omega^2 + \omega + 1 = 0$, and $|\omega| = 1$.

The three vertices form an equilateral triangle (in some cyclic order) if and only if two adjacent sides are equal in length and meet at angle $\frac{\pi}{3}$. In the complex plane, rotating a vector by $\frac{2\pi}{3}$ is multiplication by $\omega$. So the vertices in anti-clockwise order $\alpha, \beta, \gamma$ form an equilateral triangle if and only if

$\beta - \alpha = \omega(\gamma - \beta),$

and in clockwise order if and only if

$\beta - \alpha = \omega^2(\gamma - \beta).$

Combining both cases: the triangle is equilateral if and only if

$[\beta - \alpha - \omega(\gamma - \beta)][\beta - \alpha - \omega^2(\gamma - \beta)] = 0.$

We expand this product. Writing $u = \beta - \alpha$ and $v = \gamma - \beta$:

$[u - \omega v][u - \omega^2 v] = u^2 - (\omega + \omega^2)uv + \omega^3 v^2.$

Since $\omega + \omega^2 = -1$ and $\omega^3 = 1$:

$= u^2 + uv + v^2.$

Now substituting back $u = \beta - \alpha$, $v = \gamma - \beta$:

$u^2 = (\beta - \alpha)^2 = \alpha^2 - 2\alpha\beta + \beta^2$

$uv = (\beta - \alpha)(\gamma - \beta) = \beta\gamma - \beta^2 - \alpha\gamma + \alpha\beta$

$v^2 = (\gamma - \beta)^2 = \beta^2 - 2\beta\gamma + \gamma^2$

Summing:

$u^2 + uv + v^2 = \alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha.$

Hence the triangle is equilateral if and only if

$\alpha^2 + \beta^2 + \gamma^2 - \beta\gamma - \gamma\alpha - \alpha\beta = 0$

as required. $\square$

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Part 2. Let the roots of $z^3 + az^2 + bz + c = 0$ \qquad \text{(*)} be $\alpha, \beta, \gamma$. By Vieta’s relations:

$\alpha + \beta + \gamma = -a, \qquad \alpha\beta + \beta\gamma + \gamma\alpha = b.$

We compute $\alpha^2 + \beta^2 + \gamma^2$. From $(\alpha + \beta + \gamma)^2$:

$(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)$

$a^2 = \alpha^2 + \beta^2 + \gamma^2 + 2b$

$\alpha^2 + \beta^2 + \gamma^2 = a^2 - 2b.$

Therefore:

$\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha = (a^2 - 2b) - b = a^2 - 3b.$

By Part 1, the roots form an equilateral triangle if and only if this equals zero, i.e., $a^2 = 3b$. $\square$

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Part 3. We substitute $z = pw + q$ into equation \qquad \text{(*)}:

$(pw + q)^3 + a(pw + q)^2 + b(pw + q) + c = 0.$

Expanding:

$p^3w^3 + 3p^2qw^2 + 3pq^2w + q^3 + a(p^2w^2 + 2pqw + q^2) + b(pw + q) + c = 0.$

Collecting by powers of $w$:

$p^3w^3 + (3p^2q + ap^2)w^2 + (3pq^2 + 2apq + bp)w + (q^3 + aq^2 + bq + c) = 0.$

Dividing through by $p^3$ (since $p \neq 0$):

$w^3 + \frac{3q + a}{p}w^2 + \frac{3q^2 + 2aq + b}{p^2}w + \frac{q^3 + aq^2 + bq + c}{p^3} = 0. \qquad \text{(**)}$

So the coefficients of \qquad \text{(**)} are:

$A = \frac{3q + a}{p}, \qquad B = \frac{3q^2 + 2aq + b}{p^2}.$

We compute $A^2 - 3B$:

$A^2 = \frac{(3q + a)^2}{p^2} = \frac{9q^2 + 6aq + a^2}{p^2}$

$3B = \frac{3(3q^2 + 2aq + b)}{p^2} = \frac{9q^2 + 6aq + 3b}{p^2}$

$A^2 - 3B = \frac{(9q^2 + 6aq + a^2) - (9q^2 + 6aq + 3b)}{p^2} = \frac{a^2 - 3b}{p^2}.$

If the roots of \qquad \text{(*)} form an equilateral triangle, then by Part 2 we have $a^2 - 3b = 0$, so $A^2 - 3B = 0$, i.e., $A^2 = 3B$. By Part 2 applied to equation \qquad \text{()}, the roots of \qquad \text{()} also form an equilateral triangle.

Remark. Geometrically, the transformation $z \mapsto pw + q$ is a rotation-scaling (multiplication by $p$) followed by a translation (addition of $q$). Both are similarity transformations, so the image of an equilateral triangle is again equilateral. $\square$

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Question 6

Topic: 极坐标与微积分 (Polar Coordinates and Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Show that in polar coordinates the gradient of any curve at the point $(r, \theta)$ is

$\frac{\frac{\mathrm{d}r}{\mathrm{d}\theta} \tan \theta + r}{\frac{\mathrm{d}r}{\mathrm{d}\theta} - r \tan \theta}.$

A mirror is designed so that if an incident ray of light is parallel to a fixed line $L$ the reflected ray passes through a fixed point $O$ on $L$. Prove that the mirror intersects any plane containing $L$ in a parabola. You should assume that the angle between the incident ray and the normal to the mirror is the same as the angle between the reflected ray and the normal.

Hint

6 $x = r \cos \theta, y = r \sin \theta, r = r(\theta)$

$\Rightarrow \frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}$

and result follows.

Gradient of the normal is $\tan \frac{\theta}{2} = t$, say. Then we have

$t = -\frac{\frac{dr}{d\theta} - r \tan \theta}{\frac{dr}{d\theta} \tan \theta + r}, \tan \theta = \frac{2t}{1 - t^2}$

This reduces to

$\frac{dr}{d\theta} = rt$

$\Rightarrow \ln r = \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} d\theta$ $= -2 \ln \left[ c \cos \frac{\theta}{2} \right]$

$\Rightarrow \frac{2}{c^2 r} = 1 + r \cos \theta \text{ (using } 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \text{)}$

This corresponds to the standard equation of a parabola in polars.

Model Solution

Polar gradient formula. Using $x = r\cos\theta$ and $y = r\sin\theta$ with $r = r(\theta)$:

$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta.$

Therefore:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}.$

Dividing numerator and denominator by $\cos\theta$:

$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\tan\theta + r}{\frac{dr}{d\theta} - r\tan\theta}. \qquad \square$

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Mirror problem. Place the origin at $O$ and let $L$ be a ray from $O$. We use polar coordinates $(r, \theta)$ centred at $O$. An incident ray parallel to $L$ travels in the direction of $L$ (the direction $\theta = 0$). The reflected ray passes through $O$.

At the point $P = (r, \theta)$ on the mirror, the reflected ray points from $P$ toward $O$, i.e., in the direction $\theta + \pi$. The incident ray is horizontal (direction 0). By the law of reflection, the angles of incidence and reflection with the normal are equal.

Let $\psi$ be the angle the normal at $P$ makes with the horizontal. The angle of incidence is $|\psi|$ and the angle of reflection is $|\psi - (\theta + \pi)|$. Setting these equal (with appropriate sign conventions), the bisector of the angle between the incident direction (0) and reflected direction ($\theta + \pi$) must be the normal direction.

The angle bisector of directions $0$ and $\theta + \pi$ is at angle $\frac{0 + (\theta + \pi)}{2} = \frac{\theta}{2} + \frac{\pi}{2}$ (or its supplement). So the normal to the curve at $P$ has direction $\frac{\theta}{2} + \frac{\pi}{2}$.

The tangent to the curve at $P$ has direction $\frac{\theta}{2}$ (perpendicular to the normal). Using the polar gradient formula, the tangent direction at $(r, \theta)$ satisfies:

$\tan\left(\frac{\theta}{2}\right) = \frac{\frac{dr}{d\theta}\tan\theta + r}{\frac{dr}{d\theta} - r\tan\theta}.$

Let $t = \tan\frac{\theta}{2}$. Using $\tan\theta = \frac{2t}{1-t^2}$:

$t = \frac{\frac{dr}{d\theta} \cdot \frac{2t}{1-t^2} + r}{\frac{dr}{d\theta} - r \cdot \frac{2t}{1-t^2}}.$

Cross-multiplying:

$t\left(\frac{dr}{d\theta} - \frac{2rt}{1-t^2}\right) = \frac{2t}{1-t^2}\frac{dr}{d\theta} + r$

$t\frac{dr}{d\theta} - \frac{2rt^2}{1-t^2} = \frac{2t}{1-t^2}\frac{dr}{d\theta} + r.$

Multiplying through by $1 - t^2$:

$t(1-t^2)\frac{dr}{d\theta} - 2rt^2 = 2t\frac{dr}{d\theta} + r(1-t^2)$

$t\frac{dr}{d\theta} - t^3\frac{dr}{d\theta} - 2t\frac{dr}{d\theta} = r(1-t^2) + 2rt^2$

$(-t - t^3)\frac{dr}{d\theta} = r(1 + t^2)$

$-t(1+t^2)\frac{dr}{d\theta} = r(1+t^2).$

Since $1 + t^2 \neq 0$:

$-t\frac{dr}{d\theta} = r \quad \Longrightarrow \quad \frac{1}{r}\frac{dr}{d\theta} = -t = -\tan\frac{\theta}{2}.$

Wait, let me recheck the angle bisector argument. The incident ray travels in direction $\theta = 0$ (rightward). The reflected ray at point $P$ goes from $P$ toward $O$, which is in the direction from $P$ to the origin, i.e., direction $\theta + \pi$.

The bisector of angle between directions $0$ and $\theta + \pi$ should be at $\frac{\theta + \pi}{2}$. The normal is at this angle, and the tangent is perpendicular to it, at angle $\frac{\theta + \pi}{2} - \frac{\pi}{2} = \frac{\theta}{2}$.

But I need to be more careful. Let me use the gradient formula differently. The slope of the normal is $-\frac{1}{dy/dx}$, and the tangent has slope $dy/dx$. The direction of the normal should make angle $\frac{\theta + \pi}{2}$ with the positive x-axis.

The slope of a line at angle $\frac{\theta + \pi}{2}$ is $\tan\frac{\theta+\pi}{2} = -\cot\frac{\theta}{2} = -\frac{1}{t}$. So the normal slope is $-\frac{1}{t}$, meaning the tangent slope is $t = \tan\frac{\theta}{2}$.

Using the polar gradient formula:

$\tan\frac{\theta}{2} = \frac{\frac{dr}{d\theta}\tan\theta + r}{\frac{dr}{d\theta} - r\tan\theta}.$

Let me set $t = \tan\frac{\theta}{2}$ and $\tan\theta = \frac{2t}{1-t^2}$. Then:

$t = \frac{\frac{2t}{1-t^2}\frac{dr}{d\theta} + r}{\frac{dr}{d\theta} - \frac{2rt}{1-t^2}}.$

Multiply numerator and denominator by $1-t^2$:

$t = \frac{2t\frac{dr}{d\theta} + r(1-t^2)}{(1-t^2)\frac{dr}{d\theta} - 2rt}.$

Cross-multiply:

$t[(1-t^2)\frac{dr}{d\theta} - 2rt] = 2t\frac{dr}{d\theta} + r(1-t^2)$

$t(1-t^2)\frac{dr}{d\theta} - 2rt^2 = 2t\frac{dr}{d\theta} + r - rt^2$

$[t(1-t^2) - 2t]\frac{dr}{d\theta} = r - rt^2 + 2rt^2 = r(1+t^2)$

$t(1-t^2-2)\frac{dr}{d\theta} = r(1+t^2)$

$-t(1+t^2)\frac{dr}{d\theta} = r(1+t^2).$

Dividing by $(1+t^2) \neq 0$:

$\frac{dr}{d\theta} = -\frac{r}{t} = -\frac{r}{\tan(\theta/2)}.$

Hmm, this gives $\frac{dr}{r} = -\frac{d\theta}{\tan(\theta/2)} = -\frac{\cos(\theta/2)}{\sin(\theta/2)}d\theta$.

Integrating:

$\ln r = -2\ln\sin\frac{\theta}{2} + \text{const} = \ln\left(\frac{1}{\sin^2(\theta/2)}\right) + \text{const}.$

Wait, $\int \frac{\cos(\theta/2)}{\sin(\theta/2)}d\theta = 2\ln|\sin(\theta/2)|$, so $\ln r = -2\ln|\sin(\theta/2)| + c'$, giving:

$r = \frac{A}{\sin^2(\theta/2)}$

for some constant $A > 0$. Using $\sin^2(\theta/2) = \frac{1-\cos\theta}{2}$:

$r = \frac{2A}{1-\cos\theta}.$

Hmm, this is a conic with $e = 1$ and directrix to the left. Let me recheck my angle bisector.

Actually, I think there’s a sign issue. Let me reconsider. The reflected ray goes FROM P TO O. The direction from P toward O is the direction of the vector $O - P = -P$, which has angle $\theta + \pi$. But the incident ray comes FROM the direction of L (from the left, say), traveling rightward (direction 0).

For the reflection law, the angle between the incident ray and the normal equals the angle between the reflected ray and the normal. Both on opposite sides of the normal.

Let me use a cleaner approach. At point $P = (r, \theta)$, the tangent has angle $\psi$ with the x-axis, where $\tan\psi = \frac{dy}{dx}$ (the gradient formula).

The incident ray has direction angle 0. The angle of incidence is $\psi - 0 = \psi$.

The reflected ray goes in direction $\theta + \pi$ (toward origin). The angle of reflection is $\psi - (\theta + \pi)$.

Setting $|\psi| = |\psi - \theta - \pi|$ means either $\psi = \theta + \pi - \psi$ (i.e., $2\psi = \theta + \pi$, so $\psi = \frac{\theta + \pi}{2}$) or $\psi = -(\psi - \theta - \pi)$ (i.e., $2\psi = -\theta - \pi + 2\psi$, which is trivial).

Wait, the condition is: the angle of incidence equals the angle of reflection. If the normal has angle $\psi + \frac{\pi}{2}$ (perpendicular to tangent), then:

Angle of incidence = $(\psi + \frac{\pi}{2}) - 0 = \psi + \frac{\pi}{2}$

Angle of reflection = $(\theta + \pi) - (\psi + \frac{\pi}{2}) = \theta + \frac{\pi}{2} - \psi$

Setting equal: $\psi + \frac{\pi}{2} = \theta + \frac{\pi}{2} - \psi$, so $2\psi = \theta$, hence $\psi = \frac{\theta}{2}$.

So the tangent makes angle $\frac{\theta}{2}$ with the x-axis, and $\tan\psi = \tan\frac{\theta}{2}$.

Using the gradient formula: $\frac{dr}{d\theta} \tan\theta + r = \tan\frac{\theta}{2} \left(\frac{dr}{d\theta} - r\tan\theta\right)$

Let me use $t = \tan\frac{\theta}{2}$, $\tan\theta = \frac{2t}{1-t^2}$:

$\frac{dr}{d\theta} \cdot \frac{2t}{1-t^2} + r = t \cdot \frac{dr}{d\theta} - t \cdot r \cdot \frac{2t}{1-t^2}$

$\frac{dr}{d\theta}\left(\frac{2t}{1-t^2} - t\right) = -r - \frac{2rt^2}{1-t^2}$

$\frac{dr}{d\theta} \cdot t \cdot \frac{2 - (1-t^2)}{1-t^2} = -r \cdot \frac{1-t^2+2t^2}{1-t^2}$

$\frac{dr}{d\theta} \cdot t \cdot \frac{1+t^2}{1-t^2} = -r \cdot \frac{1+t^2}{1-t^2}$

Dividing by $\frac{1+t^2}{1-t^2} \neq 0$:

$t \cdot \frac{dr}{d\theta} = -r \quad \Longrightarrow \quad \frac{1}{r}\frac{dr}{d\theta} = -\frac{1}{t} = -\frac{1}{\tan(\theta/2)} = -\cot\frac{\theta}{2}.$

Hmm wait, let me redo this more carefully.

$\frac{dr}{d\theta}\left(\frac{2t}{1-t^2} - t\right) = -r\left(1 + \frac{2t^2}{1-t^2}\right)$

LHS: $\frac{2t - t(1-t^2)}{1-t^2} = \frac{2t - t + t^3}{1-t^2} = \frac{t + t^3}{1-t^2} = \frac{t(1+t^2)}{1-t^2}$

RHS: $-r\cdot\frac{1-t^2+2t^2}{1-t^2} = -r\cdot\frac{1+t^2}{1-t^2}$

So $\frac{dr}{d\theta} \cdot \frac{t(1+t^2)}{1-t^2} = -r \cdot \frac{1+t^2}{1-t^2}$

Dividing both sides by $\frac{1+t^2}{1-t^2}$:

$t \frac{dr}{d\theta} = -r$

$\frac{dr}{d\theta} = -\frac{r}{t} = -\frac{r}{\tan(\theta/2)}$

$\frac{dr}{r} = -\cot\frac{\theta}{2}\,d\theta = -\frac{\cos(\theta/2)}{\sin(\theta/2)}\,d\theta.$

Integrating:

$\ln r = -2\ln\left|\sin\frac{\theta}{2}\right| + c = \ln\frac{1}{\sin^2(\theta/2)} + c.$

Exponentiating:

$r = \frac{A}{\sin^2(\theta/2)} = \frac{2A}{1 - \cos\theta}$

where $A = e^c > 0$. Setting $\ell = 2A$:

$r = \frac{\ell}{1 - \cos\theta}.$

Converting to Cartesian: $r - r\cos\theta = \ell$, so $\sqrt{x^2+y^2} = \ell + x$. Squaring: $x^2 + y^2 = \ell^2 + 2\ell x + x^2$, hence $y^2 = \ell^2 + 2\ell x = 2\ell(x + \ell/2)$.

This is the equation of a parabola with focus at the origin and directrix $x = -\ell$ (the fixed line $L$), confirming that the mirror has parabolic cross-section. $\square$

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Question 7

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

Problem

7 (i) Solve the equation $u^2 + 2u \sinh x - 1 = 0$ giving $u$ in terms of $x$.

Find the solution of the differential equation

$\left( \frac{dy}{dx} \right)^2 + 2 \frac{dy}{dx} \sinh x - 1 = 0$

that satisfies $y = 0$ and $\frac{dy}{dx} > 0$ at $x = 0$.

(ii) Find the solution, not identically zero, of the differential equation

$\sinh y \left( \frac{dy}{dx} \right)^2 + 2 \frac{dy}{dx} - \sinh y = 0$

that satisfies $y = 0$ at $x = 0$, expressing your solution in the form $\cosh y = f(x)$. Show that the asymptotes to the solution curve are $y = \pm(-x + \ln 4)$.

Hint

7 (i) Express $\sinh x$ in terms of exponentials, factorise and solve to get

$u = -e^x$ or $u = e^{-x}$ (or $-\cosh x \pm \sinh x$).

Use both of these as equal to $\frac{dy}{dx}$ and integration to get alternative solutions

$y = -e^{\pm x} + c$.

From the given conditions the particular integral is

$y = 1 - e^{-x}$.

(ii) Solve the quadratic as before to get either

$u = \frac{-1 \pm \cosh y}{\sinh y} \text{ (or equivalent)}$

$\Rightarrow \frac{dx}{dy} = \frac{\sinh y}{-1 \pm \cosh y}$

$\Rightarrow x = \ln(\cosh y - 1) + c_1$ or $x = -\ln(\cosh y + 1) + c_2$

Only the first can satisfy the conditions $x = 0, y = 0$; then we have

$x = \ln \frac{2}{1 + \cosh y}$

$\Rightarrow \cosh y = 2e^{-x} - 1$

This is undefined for $x > 0$.

For $x \to -\infty \Rightarrow \cosh y \to \infty$, and there will be two branches, corresponding to $y \to \pm \infty$, as $\cosh$ is an even function.

So $x \to -\infty \Rightarrow \cosh y \to \infty \Rightarrow y \to \infty \Rightarrow e^y \approx 4e^{-x} \Rightarrow y = -x + \ln 4$ in one case, and similarly $y = x - \ln 4$ in the other.

Model Solution

Part (i). We solve $u^2 + 2u\sinh x - 1 = 0$. Using the quadratic formula:

$u = \frac{-2\sinh x \pm \sqrt{4\sinh^2 x + 4}}{2} = -\sinh x \pm \sqrt{\sinh^2 x + 1} = -\sinh x \pm \cosh x.$

Since $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$:

$u = -\sinh x + \cosh x = e^{-x} \qquad \text{or} \qquad u = -\sinh x - \cosh x = -e^x.$

For the differential equation $\left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx}\sinh x - 1 = 0$, we set $u = \frac{dy}{dx}$, giving:

$\frac{dy}{dx} = e^{-x} \qquad \text{or} \qquad \frac{dy}{dx} = -e^x.$

Case 1: $\frac{dy}{dx} = e^{-x}$. Integrating: $y = -e^{-x} + C$. At $x = 0$: $y = -1 + C = 0$, so $C = 1$. Also $\frac{dy}{dx}\big|_{x=0} = 1 > 0$. So $y = 1 - e^{-x}$.

Case 2: $\frac{dy}{dx} = -e^x$. At $x = 0$: $\frac{dy}{dx} = -1 < 0$, violating $\frac{dy}{dx} > 0$. Rejected.

The solution is $y = 1 - e^{-x}$. $\square$

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Part (ii). We solve $\sinh y\left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx} - \sinh y = 0$ as a quadratic in $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2 \pm \sqrt{4 + 4\sinh^2 y}}{2\sinh y} = \frac{-2 \pm 2\cosh y}{2\sinh y} = \frac{-1 \pm \cosh y}{\sinh y}.$

Since $\frac{dy}{dx}$ must be real and we need the reciprocal $\frac{dx}{dy}$ to integrate:

Case 1: $\frac{dx}{dy} = \frac{\sinh y}{\cosh y - 1}$. Integrating: substitute $w = \cosh y - 1$, $dw = \sinh y\,dy$:

$x = \int \frac{dw}{w} = \ln|\cosh y - 1| + C_1.$

At $(x, y) = (0, 0)$: $\cosh 0 - 1 = 0$, so $\ln 0$ is undefined. This case cannot satisfy the initial condition (except via a limiting argument; the solution is only valid for $y \neq 0$).

Case 2: $\frac{dx}{dy} = \frac{\sinh y}{-1 - \cosh y} = -\frac{\sinh y}{1 + \cosh y}$. Using $1 + \cosh y = 2\cosh^2(y/2)$ and $\sinh y = 2\sinh(y/2)\cosh(y/2)$:

$\frac{dx}{dy} = -\frac{2\sinh(y/2)\cosh(y/2)}{2\cosh^2(y/2)} = -\tanh\frac{y}{2}.$

Integrating:

$x = -2\ln\left|\cosh\frac{y}{2}\right| + C_2.$

At $(0, 0)$: $0 = -2\ln 1 + C_2$, so $C_2 = 0$. Therefore $x = -2\ln\cosh\frac{y}{2}$.

This gives $\cosh\frac{y}{2} = e^{-x/2}$, so $\cosh y = 2\cosh^2\frac{y}{2} - 1 = 2e^{-x} - 1$.

$\cosh y = 2e^{-x} - 1. \qquad \square$

Domain. For $\cosh y \geq 1$, we need $2e^{-x} - 1 \geq 1$, i.e., $e^{-x} \geq 1$, so $x \leq 0$. The solution exists only for $x \leq 0$.

Asymptotic behaviour. As $x \to -\infty$: $\cosh y = 2e^{-x} - 1 \to \infty$, so $|y| \to \infty$.

For $y > 0$: $\cosh y \approx \frac{e^y}{2}$ for large $y$. So $\frac{e^y}{2} \approx 2e^{-x}$, giving $e^y \approx 4e^{-x}$, hence $y \approx -x + \ln 4$.

For $y < 0$: $\cosh y = \cosh(-y) \approx \frac{e^{-y}}{2}$, so $\frac{e^{-y}}{2} \approx 2e^{-x}$, giving $e^{-y} \approx 4e^{-x}$, hence $-y \approx -x + \ln 4$, i.e., $y \approx x - \ln 4$.

The asymptotes are $y = -x + \ln 4$ and $y = x - \ln 4$. $\square$

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Question 8

Topic: 抽象代数与微分 (Abstract Algebra and Differentiation)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 $\Delta$ is an operation that takes polynomials in $x$ to polynomials in $x$; that is, given any polynomial $h(x)$, there is a polynomial called $\Delta h(x)$ which is obtained from $h(x)$ using the rules that define $\Delta$. These rules are as follows:

(i) $\Delta x = 1$;

(ii) $\Delta(f(x) + g(x)) = \Delta f(x) + \Delta g(x)$ for any polynomials $f(x)$ and $g(x)$;

(iii) $\Delta(\lambda f(x)) = \lambda \Delta f(x)$ for any constant $\lambda$ and any polynomial $f(x)$;

(iv) $\Delta(f(x)g(x)) = f(x)\Delta g(x) + g(x)\Delta f(x)$ for any polynomials $f(x)$ and $g(x)$.

Using these rules show that, if $f(x)$ is a polynomial of degree zero (that is, a constant), then $\Delta f(x) = 0$. Calculate $\Delta x^2$ and $\Delta x^3$.

Prove that $\Delta h(x) \equiv \frac{dh(x)}{dx}$ for any polynomial $h(x)$. You should make it clear whenever you use one of the above rules in your proof.

Hint

8 Use (iv) with $f(x) \equiv 1$, $g(x) \equiv 1$ to show that $\Delta 1 = 0$. Use (iii) with $\lambda \equiv k$, $f(x) \equiv 1$ to show that $\Delta k = 0$. By (iv), (i) $\Delta x^2 = 2x$; ditto $\Delta x^3 = 3x^2$. Now show $\Delta kx^n = knx^{n-1}$ by induction. Initial step is $\Delta k = 0$; inductive hypothesis is that $\Delta kx^N = kNx^{N-1}$. Use (iii) and (iv) with hypothesis to show that $\Delta kx^{N+1} = k(N+1)x^N$. Now express any $P_k(x)$, a polynomial of degree k, as a sum of such powers, and so use (ii) to establish required result.

Model Solution

Constants have zero image. We first show $\Delta(1) = 0$. By rule (iv) with $f(x) = 1$ and $g(x) = 1$:

$\Delta(1 \cdot 1) = 1 \cdot \Delta(1) + 1 \cdot \Delta(1) = 2\Delta(1).$

But $\Delta(1 \cdot 1) = \Delta(1)$, so $\Delta(1) = 2\Delta(1)$, hence $\Delta(1) = 0$.

For a general constant $k$, by rule (iii) with $\lambda = k$ and $f(x) = 1$:

$\Delta(k) = \Delta(k \cdot 1) = k\Delta(1) = k \cdot 0 = 0. \qquad \square$

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Computing $\Delta x^2$ and $\Delta x^3$. By rule (iv) with $f(x) = x$ and $g(x) = x$:

$\Delta(x^2) = \Delta(x \cdot x) = x\Delta(x) + x\Delta(x) = 2x\Delta(x).$

By rule (i), $\Delta x = 1$, so $\Delta(x^2) = 2x$. $\square$

By rule (iv) with $f(x) = x$ and $g(x) = x^2$:

$\Delta(x^3) = \Delta(x \cdot x^2) = x\Delta(x^2) + x^2\Delta(x) = x \cdot 2x + x^2 \cdot 1 = 3x^2. \qquad \square$

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General proof by induction. We prove that $\Delta h(x) = \frac{dh(x)}{dx}$ for every polynomial $h(x)$.

Claim: For all $n \geq 0$ and all constants $k$, $\Delta(kx^n) = knx^{n-1}$.

Base case ($n = 0$): $\Delta(k) = 0 = k \cdot 0 \cdot x^{-1}$. (Already proved above.) $\checkmark$

Inductive step: Assume $\Delta(kx^N) = kNx^{N-1}$ for some $N \geq 0$. We show $\Delta(kx^{N+1}) = k(N+1)x^N$.

By rule (iv) with $f(x) = x$ and $g(x) = kx^N$:

$\Delta(kx^{N+1}) = \Delta(x \cdot kx^N) = x \cdot \Delta(kx^N) + kx^N \cdot \Delta(x).$

Applying rule (i): $\Delta(x) = 1$. Applying the inductive hypothesis: $\Delta(kx^N) = kNx^{N-1}$.

$\Delta(kx^{N+1}) = x \cdot kNx^{N-1} + kx^N \cdot 1 = kNx^N + kx^N = k(N+1)x^N. \qquad \checkmark$

By induction, $\Delta(kx^n) = knx^{n-1}$ for all $n \geq 0$ and all constants $k$.

Extension to general polynomials. Let $h(x) = \sum_{i=0}^{n} k_i x^i$ be any polynomial. By rule (ii) (additivity) applied repeatedly:

$\Delta h(x) = \Delta\left(\sum_{i=0}^{n} k_i x^i\right) = \sum_{i=0}^{n} \Delta(k_i x^i) = \sum_{i=0}^{n} k_i i x^{i-1}.$

By rule (iii) (linearity), each $\Delta(k_i x^i) = k_i \Delta(x^i)$, and we have shown $\Delta(x^i) = ix^{i-1}$.

The right-hand side is exactly $\frac{dh}{dx}$. Therefore $\Delta h(x) \equiv \frac{dh(x)}{dx}$ for every polynomial $h(x)$. $\square$