STEP3 2006 -- Pure Mathematics

Exam: STEP3  |  Year: 2006  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions are included below; examiner notes appear only where available.

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.

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 $y \to +\infty$.

Sketch summary. 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 this branch.

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

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):

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

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

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 integral remains positive, so

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

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 $\cos 2\alpha$). Thus

$I\sin^2 2\alpha-\pi\cos 2\alpha=\pi,$

so

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

Using $1-\cos 2\alpha=2\sin^2\alpha$ gives

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

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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 $b_2 = 0$.

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

The tangent makes angle $\frac{\theta}{2}$ with the fixed line. Put $t=\tan\frac{\theta}{2}$. Then

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

This reduces to

$\frac{dr}{d\theta}=-\frac{r}{t}=-r\cot\frac{\theta}{2}.$

Hence

$\ln r=-2\ln\sin\frac{\theta}{2}+C,$

so $r=\ell/(1-\cos\theta)$, the polar equation of a parabola with focus at $O$.

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 $\cos\theta$:

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

---

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

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

Let $t=\tan\frac{\theta}{2}$. Since $\tan\theta=\frac{2t}{1-t^2}$, this gives

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

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 $h(x)$. $\square$