STEP3 1999 -- Pure Mathematics

STEP3 1999 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 1999  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	代数 Algebra	Standard	韦达定理,多项式根与系数关系,等比数列性质
2	分析 Analysis	Challenging	求导判断单调性,函数图像分析,分段讨论参数范围
3	微积分 Calculus	Standard	黎曼和转化为定积分,极限与积分的联系
4	几何 Geometry	Standard	欧拉公式V-E+F=2,面与顶点的计数关系,整数约束分析
5	数列与递推 Sequences and Recurrence	Challenging	数学归纳法,递推关系代换,代数恒等式变形
6	参数方程与曲线面积 (Parametric Curves and Area)	Challenging	三角参数化,格林面积公式,贝塔函数积分
7	抽象代数/群论 (Abstract Algebra/Group Theory)	Challenging	结合律直接验证,单位元与逆元求解,子群构造
8	常微分方程 (Ordinary Differential Equations)	Challenging	二阶常系数ODE,分段拼接,连续性匹配,无穷级数求和

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Question 1

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

Problem

1 Consider the cubic equation $x^3 - px^2 + qx - r = 0 ,$ where $p \neq 0$ and $r \neq 0$.

(i) If the three roots can be written in the form $ak^{-1}$, $a$ and $ak$ for some constants $a$ and $k$, show that one root is $q/p$ and that $q^3 - rp^3 = 0$.

(ii) If $r = q^3/p^3$, show that $q/p$ is a root and that the product of the other two roots is $(q/p)^2$. Deduce that the roots are in geometric progression.

(iii) Find a necessary and sufficient condition involving $p$, $q$ and $r$ for the roots to be in arithmetic progression.

Model Solution

Part (i) Show that one root is $q/p$ and that $q^3 - rp^3 = 0$.

Let the three roots be $ak^{-1}$, $a$, and $ak$. By Vieta’s formulas for $x^3 - px^2 + qx - r = 0$:

$p = ak^{-1} + a + ak = a(k^{-1} + 1 + k), \qquad \text{(...1)}$

$q = (ak^{-1})(a) + (ak^{-1})(ak) + (a)(ak) = a^2(k^{-1} + 1 + k), \qquad \text{(...2)}$

$r = (ak^{-1})(a)(ak) = a^3. \qquad \text{(...3)}$

From (1) and (2), dividing $q$ by $p$:

$\frac{q}{p} = \frac{a^2(k^{-1} + 1 + k)}{a(k^{-1} + 1 + k)} = a.$

So $a = q/p$ is one of the three roots, as required.

From (3), $r = a^3 = (q/p)^3$, hence $rp^3 = q^3$, i.e.

$q^3 - rp^3 = 0. \qquad \blacksquare$

Part (ii) Show that $q/p$ is a root and the product of the other two roots is $(q/p)^2$; deduce the roots are in GP.

Suppose $r = q^3/p^3$. We check that $x = q/p$ is a root:

$\left(\frac{q}{p}\right)^3 - p\left(\frac{q}{p}\right)^2 + q\left(\frac{q}{p}\right) - r = \frac{q^3}{p^3} - \frac{q^2}{p} + \frac{q^2}{p} - \frac{q^3}{p^3} = 0. \checkmark$

So $q/p$ is a root. Let the other two roots be $\alpha$ and $\beta$. By Vieta’s formulas:

$\frac{q}{p} + \alpha + \beta = p, \qquad \frac{q}{p} \cdot \alpha\beta = r = \frac{q^3}{p^3}.$

From the second relation:

$\alpha\beta = \frac{q^3}{p^3} \cdot \frac{p}{q} = \frac{q^2}{p^2} = \left(\frac{q}{p}\right)^2.$

Now $\alpha + \beta = p - q/p$ and $\alpha\beta = (q/p)^2$. The roots $\alpha, \beta$ satisfy

$x^2 - \left(p - \frac{q}{p}\right)x + \left(\frac{q}{p}\right)^2 = 0.$

The three roots are $q/p$, $\alpha$, $\beta$ with $\alpha\beta = (q/p)^2$. Since the product of all three roots is $r = (q/p)^3$ and one root is $q/p$, the remaining product is $(q/p)^2$, meaning

$\alpha \cdot \beta = \left(\frac{q}{p}\right)^2.$

If $\beta = \frac{(q/p)^2}{\alpha}$, then the three roots $q/p$, $\alpha$, $\frac{(q/p)^2}{\alpha}$ are in geometric progression with common ratio $\frac{\alpha}{q/p} = \frac{p\alpha}{q}$. Indeed, $\frac{(q/p)^2}{\alpha} = \frac{q/p}{\alpha/(q/p)}$ confirms the GP structure. $\blacksquare$

Part (iii) Necessary and sufficient condition for roots in arithmetic progression.

Let the three roots be $a - d$, $a$, $a + d$ (common difference $d$). By Vieta’s:

$p = (a-d) + a + (a+d) = 3a, \qquad \text{so } a = \frac{p}{3}.$

$q = (a-d)a + (a-d)(a+d) + a(a+d) = a^2 - ad + a^2 - d^2 + a^2 + ad = 3a^2 - d^2.$

$r = (a-d) \cdot a \cdot (a+d) = a(a^2 - d^2).$

From $q = 3a^2 - d^2$ we get $d^2 = 3a^2 - q$. Substituting into $r$:

$r = a(3a^2 - d^2 - (3a^2 - d^2) + d^2) \quad \text{wait, let me redo this cleanly.}$

$r = a(a^2 - d^2) = a \cdot q - 2a^3 + a^3 \quad \text{no.}$

We have $d^2 = 3a^2 - q$ and $r = a(a^2 - d^2) = a(a^2 - 3a^2 + q) = a(q - 2a^2)$.

With $a = p/3$:

$r = \frac{p}{3}\left(q - \frac{2p^2}{9}\right) = \frac{pq}{3} - \frac{2p^3}{27}.$

Multiplying through by 27:

$27r = 9pq - 2p^3.$

Rearranging:

$2p^3 - 9pq + 27r = 0.$

This is the necessary and sufficient condition for the roots to be in arithmetic progression.

Verification of sufficiency: If $2p^3 - 9pq + 27r = 0$, then $r = \frac{9pq - 2p^3}{27} = \frac{p}{3}(q - \frac{2p^2}{9})$, and setting $a = p/3$, $d^2 = 3a^2 - q = \frac{p^2}{3} - q$, the cubic factors as $(x - a + d)(x - a)(x - a + d)$, giving three real roots in AP (when $d^2 \geq 0$, i.e. $q \leq p^2/3$, the roots are distinct or repeated).

$2p^3 - 9pq + 27r = 0 \qquad \blacksquare$

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Question 2

Topic: 分析 Analysis  |  Difficulty: Challenging  |  Marks: 20

Problem

2 (i) Let $f(x) = (1 + x^2)e^x$. Show that $f'(x) \geqslant 0$ and sketch the graph of $f(x)$. Hence, or otherwise, show that the equation $(1 + x^2)e^x = k,$ where $k$ is a constant, has exactly one real root if $k > 0$ and no real roots if $k \leqslant 0$.

(ii) Determine the number of real roots of the equation $(e^x - 1) - k \tan^{-1} x = 0$ in the cases (a) $0 < k \leqslant 2/\pi$ and (b) $2/\pi < k < 1$.

Model Solution

Part (i) Show $f'(x) \geq 0$, sketch $f$, and determine roots of $(1+x^2)e^x = k$.

Let $f(x) = (1 + x^2)e^x$. Differentiating:

$f'(x) = 2xe^x + (1 + x^2)e^x = e^x(1 + 2x + x^2) = e^x(1 + x)^2.$

Since $e^x > 0$ for all $x$ and $(1+x)^2 \geq 0$ for all $x$, we have $f'(x) \geq 0$ for all $x$, with equality only at $x = -1$.

Therefore $f$ is non-decreasing (strictly increasing except at $x = -1$ where it has a stationary inflection point).

Key features of the graph:

• $f(0) = 1$, $f(-1) = 2e^{-1} = 2/e$.
• As $x \to +\infty$: $f(x) \to +\infty$ (exponential growth dominates).
• As $x \to -\infty$: $f(x) \to 0^+$ (since $e^x \to 0$ and $1 + x^2$ grows polynomially, but $e^x$ decays faster).

Since $f$ is non-decreasing with $\lim_{x\to-\infty} f(x) = 0$ and $\lim_{x\to+\infty} f(x) = +\infty$, the range of $f$ is $(0, +\infty)$.

For $k > 0$: the horizontal line $y = k$ crosses the graph of $f$ exactly once (since $f$ is non-decreasing and its range is $(0, +\infty)$), so the equation has exactly one real root.

For $k \leq 0$: since $f(x) = (1+x^2)e^x > 0$ for all $x$ (as $1+x^2 > 0$ and $e^x > 0$), the equation has no real roots.

Part (ii) Determine the number of real roots of $(e^x - 1) - k\arctan x = 0$.

Let $g(x) = e^x - 1 - k\arctan x$. Note $g(0) = 0$, so $x = 0$ is always a root. We need to determine if there are other roots.

$g'(x) = e^x - \frac{k}{1 + x^2}.$

Case (a): $0 < k \leq 2/\pi$.

We claim $g(x) < 0$ for $x < 0$ and $g(x) > 0$ for $x > 0$, so $x = 0$ is the only root.

For $x > 0$: We show $e^x - 1 > k\arctan x$. Since $e^x > 1 + x$ for $x > 0$ (standard inequality from the Taylor series), and $\arctan x < x$ for $x > 0$:

$e^x - 1 > x > k \cdot \frac{x}{\text{(we need a tighter bound)}}.$

Better approach: consider $h(x) = \frac{e^x - 1}{\arctan x}$ for $x > 0$. We need $h(x) > k$. As $x \to 0^+$, by L’Hopital:

$\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \frac{e^x}{1/(1+x^2)} = 1.$

As $x \to +\infty$: $h(x) \approx \frac{e^x}{\pi/2} \to +\infty$.

We compute $h'(x)$ to check monotonicity. Actually, let us instead check $g'(x) > 0$ for all $x > 0$, which would mean $g$ is strictly increasing on $(0, \infty)$ and hence $g(x) > g(0) = 0$ for $x > 0$.

$g'(x) = e^x - \frac{k}{1+x^2}.$

For $x > 0$: $e^x > 1$ and $\frac{k}{1+x^2} < k \leq \frac{2}{\pi} < 1$. So $g'(x) > 0$ for all $x > 0$.

Hence $g$ is strictly increasing on $(0, \infty)$, giving $g(x) > 0$ for $x > 0$.

For $x < 0$: We need $g(x) < 0$, i.e. $e^x - 1 < k\arctan x$. Note both sides are negative for $x < 0$. Rearranging: $1 - e^x > k|\arctan x|$ (both sides positive).

Consider $g'(x) = e^x - \frac{k}{1+x^2}$ for $x < 0$. We have $e^x < 1$ and $\frac{k}{1+x^2} \geq \frac{k}{1+x^2}$. At $x \leq -1$: $e^x \leq e^{-1} \approx 0.37$ and $\frac{k}{1+x^2} \geq \frac{k}{1+x^2}$; we need more care.

Let us check $g'(x)$ for $x < 0$. We have $g'(0) = 1 - k > 0$ (since $k \leq 2/\pi < 1$). Also $g'(x) \to 0 - 0 = 0$ as $x \to -\infty$ (wait, $e^x \to 0$ and $\frac{k}{1+x^2} \to 0$, so $g'(x) \to 0$).

Actually, let’s verify $g'(x) < 0$ is impossible for $x < 0$ when $k \leq 2/\pi$. We need $e^x < \frac{k}{1+x^2}$, i.e. $(1+x^2)e^x < k \leq 2/\pi$.

From Part (i), $f(x) = (1+x^2)e^x$ is non-decreasing with minimum value approaching 0. So for $x$ very negative, $f(x) < 2/\pi$, meaning $g'(x) < 0$ is possible there. But this just means $g$ decreases before it increases.

The key insight: $g(x) \to -1 - k(-\pi/2) = -1 + k\pi/2$ as $x \to -\infty$. For $k \leq 2/\pi$: $-1 + k\pi/2 \leq -1 + 1 = 0$. So $g(x) \leq 0$ as $x \to -\infty$, with equality only when $k = 2/\pi$.

For $k < 2/\pi$: $g(x) \to -1 + k\pi/2 < 0$ as $x \to -\infty$. Since $g(0) = 0$ and $g$ is continuous, if $g$ had a root at some $x_0 < 0$ with $g(x_0) = 0$, then between $-\infty$ and $x_0$ or between $x_0$ and $0$, $g$ would need to cross zero. Since $g'(x) < 0$ for very negative $x$ (where $(1+x^2)e^x$ is small), $g$ is decreasing there. Then $g'$ changes sign at most once (since $(1+x^2)e^x$ is non-decreasing), so $g$ has at most one local minimum. If $g$ goes from negative values, reaches a minimum, then increases to $g(0) = 0$, it stays negative for all $x < 0$.

For $k = 2/\pi$: $g(x) \to 0$ as $x \to -\infty$. Since $g$ is negative for large negative $x$ (it approaches 0 from below, because $e^x$ approaches 0 faster than $k\arctan x$ approaches $-k\pi/2$, making $e^x - 1 \approx -1$ while $k\arctan x \approx -1$, but $e^x - 1 < -1 + k\pi/2 = 0$ for any finite $x$), the same argument gives no additional roots.

Conclusion for case (a): The only real root is $x = 0$.

Case (b): $2/\pi < k < 1$.

Now $g(x) \to -1 + k\pi/2 > 0$ as $x \to -\infty$. Since $g(0) = 0$ and $g$ is continuous, $g$ must cross zero at least once for $x < 0$ (since it starts positive and reaches 0).

More precisely: $g(-\infty) = k\pi/2 - 1 > 0$ and $g(0) = 0$. If $g$ stayed positive on $(-\infty, 0)$, it would need $g'(0) \leq 0$, but $g'(0) = 1 - k > 0$. So $g$ must go below zero somewhere in $(-\infty, 0)$, meaning it crosses zero at least twice on $(-\infty, 0)$ (once going down, once coming back up to $g(0) = 0$).

To count precisely, consider $g'(x) = e^x - \frac{k}{1+x^2}$. Let $\phi(x) = (1+x^2)e^x$. Then $g'(x) = 0$ iff $\phi(x) = k$. From Part (i), $\phi(x) = k > 0$ has exactly one real root $x_0$. Since $k < 1 = \phi(0)$, we have $x_0 < 0$.

So $g'(x) < 0$ for $x < x_0$ and $g'(x) > 0$ for $x > x_0$. This means $g$ decreases on $(-\infty, x_0)$ and increases on $(x_0, \infty)$, with a unique minimum at $x_0$.

Since $g(-\infty) = k\pi/2 - 1 > 0$, $g$ decreases from positive values to $g(x_0)$, then increases to $g(0) = 0$ and beyond (since $g'(0) > 0$, $g$ is increasing at 0, so $g(x) > 0$ for $x > 0$).

If $g(x_0) < 0$: $g$ crosses zero once on $(-\infty, x_0)$ and once on $(x_0, 0)$, giving 2 negative roots plus $x = 0$, total 3 roots.

If $g(x_0) = 0$: $g$ touches zero at $x_0$ and at 0, total 2 roots.

If $g(x_0) > 0$: $g$ never crosses zero on $(-\infty, 0)$ except at $x = 0$ where it reaches 0 from above. But $g'(0) = 1 - k > 0$ means $g$ is increasing at 0, so $g(x) < 0$ just to the left of 0. This contradicts $g(x_0) > 0$ and $g$ being increasing on $(x_0, 0)$.

Wait, let me reconsider. $g$ is increasing on $(x_0, 0)$ with $g(0) = 0$. If $g(x_0) > 0$, then $g$ would decrease from $g(-\infty) > 0$ to $g(x_0) > 0$ (never crossing zero), then increase from $g(x_0) > 0$ to $g(0) = 0$… but $g$ increasing on $(x_0, 0)$ with $g(0) = 0$ requires $g(x_0) < 0$. Contradiction.

So $g(x_0) < 0$ is forced. Therefore $g$ crosses zero exactly once on $(-\infty, x_0)$ and exactly once on $(x_0, 0)$ (since $g$ is monotone on each interval).

Conclusion for case (b): There are exactly 3 real roots: one negative (left of $x_0$), one negative (between $x_0$ and $0$), and $x = 0$.

Hmm, wait. Let me recount. $g$ starts at $k\pi/2 - 1 > 0$ as $x \to -\infty$, decreases to $g(x_0) < 0$, then increases to $g(0) = 0$. So $g$ crosses zero exactly once on $(-\infty, x_0)$ (from positive to negative). On $(x_0, 0)$, $g$ increases from negative to 0, so $g(x) < 0$ on $(x_0, 0)$ and $g(0) = 0$. So the only crossing is one root in $(-\infty, x_0)$, plus $x = 0$.

For $x > 0$: $g'(x) > 0$ and $g(0) = 0$, so $g(x) > 0$ for $x > 0$, no additional roots.

Conclusion for case (b): There are exactly 2 real roots: one negative root and $x = 0$.

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Question 3

Topic: 微积分 Calculus  |  Difficulty: Standard  |  Marks: 20

Problem

3 Justify, by means of a sketch, the formula $\lim_{n \to \infty} \left\{ \frac{1}{n} \sum_{m=1}^{n} f(1 + m/n) \right\} = \int_{1}^{2} f(x) \, dx .$ Show that $\lim_{n \to \infty} \left\{ \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + n} \right\} = \ln 2 .$ Evaluate $\lim_{n \to \infty} \left\{ \frac{n}{n^2 + 1} + \frac{n}{n^2 + 4} + \dots + \frac{n}{n^2 + n^2} \right\} .$

Model Solution

Part (i): Justifying the Riemann sum formula

Consider the integral $\int_1^2 f(x) \, dx$. We partition the interval $[1, 2]$ into $n$ equal sub-intervals, each of width $\Delta x = \frac{2 - 1}{n} = \frac{1}{n}$.

The partition points are $x_0 = 1, \; x_1 = 1 + \tfrac{1}{n}, \; x_2 = 1 + \tfrac{2}{n}, \; \ldots, \; x_n = 2$.

The right-hand Riemann sum uses the value of $f$ at the right endpoint of each sub-interval:

$\sum_{m=1}^{n} f(x_m) \, \Delta x = \sum_{m=1}^{n} f\!\left(1 + \tfrac{m}{n}\right) \cdot \frac{1}{n} = \frac{1}{n} \sum_{m=1}^{n} f\!\left(1 + \tfrac{m}{n}\right).$

Sketch: For a continuous function $f$ on $[1, 2]$, the Riemann sum represents the total area of $n$ rectangles of width $\frac{1}{n}$, where the $m$-th rectangle has height $f(1 + m/n)$. As $n \to \infty$, these rectangles fill the region under the curve $y = f(x)$ from $x = 1$ to $x = 2$ more and more precisely, so the sum converges to the definite integral. Hence

$\lim_{n \to \infty} \left\{ \frac{1}{n} \sum_{m=1}^{n} f\!\left(1 + \tfrac{m}{n}\right) \right\} = \int_1^2 f(x) \, dx.$

Part (ii): Showing the limit equals $\ln 2$

We rewrite the sum by dividing numerator and denominator of each term by $n$:

$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{n+n} = \sum_{m=1}^{n} \frac{1}{n+m} = \sum_{m=1}^{n} \frac{1}{n\!\left(1 + m/n\right)} = \frac{1}{n} \sum_{m=1}^{n} \frac{1}{1 + m/n}.$

This is exactly the Riemann sum from Part (i) with $f(x) = \frac{1}{x}$. Therefore

$\lim_{n \to \infty} \left\{ \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right\} = \int_1^2 \frac{1}{x} \, dx = \Big[\ln x\Big]_1^2 = \ln 2 - \ln 1 = \ln 2.$

Part (iii): Evaluating the second limit

We rewrite each term by dividing numerator and denominator by $n^2$:

$\frac{n}{n^2 + m^2} = \frac{n}{n^2\!\left(1 + m^2/n^2\right)} = \frac{1}{n} \cdot \frac{1}{1 + (m/n)^2}.$

So the full sum becomes

$\sum_{m=1}^{n} \frac{n}{n^2 + m^2} = \frac{1}{n} \sum_{m=1}^{n} \frac{1}{1 + (m/n)^2}.$

This is the Riemann sum from Part (i) with $f(x) = \frac{1}{1 + x^2}$. Therefore

$\lim_{n \to \infty} \left\{ \frac{n}{n^2+1} + \frac{n}{n^2+4} + \cdots + \frac{n}{n^2+n^2} \right\} = \int_1^2 \frac{1}{1+x^2} \, dx = \Big[\arctan x\Big]_1^2 = \arctan 2 - \arctan 1.$

Since $\arctan 1 = \frac{\pi}{4}$, the limit is

$\arctan 2 - \frac{\pi}{4}.$

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Question 4

Topic: 几何 Geometry  |  Difficulty: Standard  |  Marks: 20

Problem

4 A polyhedron is a solid bounded by $F$ plane faces, which meet in $E$ edges and $V$ vertices. You may assume Euler’s formula, that $V - E + F = 2$.

In a regular polyhedron the faces are equal regular $m$-sided polygons, $n$ of which meet at each vertex. Show that $F = \frac{4n}{h},$ where $h = 4 - (n - 2)(m - 2)$.

By considering the possible values of $h$, or otherwise, prove that there are only five regular polyhedra, and find $V$, $E$ and $F$ for each.

Model Solution

Part (i): Deriving the formula $F = \frac{4n}{h}$

Each face is a regular $m$-gon, so each face has $m$ edges. Since every edge is shared by exactly two faces, counting edge-face incidences gives

$2E = mF, \qquad \text{so} \qquad E = \frac{mF}{2}. \qquad \text{...(1)}$

At each vertex exactly $n$ faces meet, and each face has $m$ vertices. Counting vertex-face incidences gives

$nV = mF, \qquad \text{so} \qquad V = \frac{mF}{n}. \qquad \text{...(2)}$

Substituting (1) and (2) into Euler’s formula $V - E + F = 2$:

$\frac{mF}{n} - \frac{mF}{2} + F = 2.$

Factor out $F$:

$F\!\left(\frac{m}{n} - \frac{m}{2} + 1\right) = 2.$

Multiply through by $2n$:

$F(2m - mn + 2n) = 4n.$

Now expand $h = 4 - (n-2)(m-2)$:

$h = 4 - (nm - 2n - 2m + 4) = 4 - nm + 2n + 2m - 4 = 2m + 2n - mn.$

So $2m - mn + 2n = h$, and therefore

$Fh = 4n, \qquad \text{i.e.} \qquad F = \frac{4n}{h}. \qquad \blacksquare$

Part (ii): Proving there are exactly five regular polyhedra

Since $F \geqslant 4$ (a polyhedron has at least 4 faces), we need $F = \frac{4n}{h} > 0$. With $n \geqslant 3$ (at least 3 faces meet at each vertex) and $m \geqslant 3$ (each face is at least a triangle), we need $h > 0$, i.e.

$4 - (n-2)(m-2) > 0 \qquad \Longrightarrow \qquad (n-2)(m-2) < 4.$

Since $n \geqslant 3$ and $m \geqslant 3$, we have $n - 2 \geqslant 1$ and $m - 2 \geqslant 1$. The product of two positive integers being less than 4 gives only these cases:

$m$	$n$	$(n-2)(m-2)$	$h$	$F = \frac{4n}{h}$	$E = \frac{mF}{2}$	$V = \frac{mF}{n}$	Polyhedron
3	3	1	3	4	6	4	Tetrahedron
3	4	2	2	8	12	6	Octahedron
3	5	3	1	20	30	12	Icosahedron
4	3	2	2	6	12	8	Cube
5	3	3	1	12	30	20	Dodecahedron

For $m \geqslant 6$, we have $m - 2 \geqslant 4$, so $(n-2)(m-2) \geqslant 4$ for any $n \geqslant 3$, which violates $h > 0$. Similarly for $n \geqslant 6$.

Therefore only five pairs $(m, n)$ are possible, corresponding to exactly five regular polyhedra. $\blacksquare$

Summary of the five regular polyhedra:

Polyhedron	$m$	$n$	$F$	$E$	$V$
Tetrahedron	3	3	4	6	4
Cube	4	3	6	12	8
Octahedron	3	4	8	12	6
Dodecahedron	5	3	12	30	20
Icosahedron	3	5	20	30	12

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Question 5

Topic: 数列与递推 Sequences and Recurrence  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The sequence $u_0, u_1, u_2, \dots$ is defined by $u_0 = 1, \quad u_1 = 1, \quad u_{n+1} = u_n + u_{n-1} \quad \text{for } n \geqslant 1.$

Prove that $u_{n+2}^2 + u_{n-1}^2 = 2(u_{n+1}^2 + u_n^2).$

Using induction, or otherwise, prove the following result: $u_{2n} = u_n^2 + u_{n-1}^2 \quad \text{and} \quad u_{2n+1} = u_{n+1}^2 - u_{n-1}^2$ for any positive integer $n$.

Model Solution

Part (i): Prove that $u_{n+2}^2 + u_{n-1}^2 = 2(u_{n+1}^2 + u_n^2)$

We start by expanding $u_{n+2}^2$ using the recurrence $u_{n+2} = u_{n+1} + u_n$:

$u_{n+2}^2 = (u_{n+1} + u_n)^2 = u_{n+1}^2 + 2u_{n+1}u_n + u_n^2.$

Therefore

$u_{n+2}^2 + u_{n-1}^2 = u_{n+1}^2 + 2u_{n+1}u_n + u_n^2 + u_{n-1}^2. \qquad \text{(...)}$

From the recurrence $u_{n+1} = u_n + u_{n-1}$, we have $u_{n-1} = u_{n+1} - u_n$. Substituting into $u_{n-1}^2$:

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

Substituting this back into $(...)$:

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

This completes the proof. $\blacksquare$

Part (ii): Prove by induction that $u_{2n} = u_n^2 + u_{n-1}^2$ and $u_{2n+1} = u_{n+1}^2 - u_{n-1}^2$ for all positive integers $n$

Base case $(n = 1)$: We have $u_0 = 1$, $u_1 = 1$, $u_2 = u_1 + u_0 = 2$, $u_3 = u_2 + u_1 = 3$.

First identity: $u_2 = 2 = 1 + 1 = u_1^2 + u_0^2$. Confirmed.

Second identity: $u_3 = 3 = 4 - 1 = u_2^2 - u_0^2$. Confirmed.

Inductive step: Suppose both identities hold for some positive integer $n$, i.e.

$u_{2n} = u_n^2 + u_{n-1}^2, \qquad u_{2n+1} = u_{n+1}^2 - u_{n-1}^2. \qquad \text{(**)}$

We prove both identities for $n + 1$.

First identity for $n + 1$: We need $u_{2(n+1)} = u_{n+1}^2 + u_n^2$.

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

This confirms the first identity for $n + 1$.

Second identity for $n + 1$: We need $u_{2n+3} = u_{n+2}^2 - u_n^2$.

$u_{2n+3} = u_{2n+2} + u_{2n+1} = (u_{n+1}^2 + u_n^2) + (u_{n+1}^2 - u_{n-1}^2) = 2u_{n+1}^2 + u_n^2 - u_{n-1}^2.$

Using $u_{n-1} = u_{n+1} - u_n$:

$u_{n-1}^2 = u_{n+1}^2 - 2u_{n+1}u_n + u_n^2.$

Substituting:

$u_{2n+3} = 2u_{n+1}^2 + u_n^2 - u_{n+1}^2 + 2u_{n+1}u_n - u_n^2 = u_{n+1}^2 + 2u_{n+1}u_n = u_{n+1}(u_{n+1} + 2u_n).$

Now, $u_{n+2} - u_n = (u_{n+1} + u_n) - u_n = u_{n+1}$, so by the difference of squares:

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

It remains to show $u_{n+1} + 2u_n = u_{n+2} + u_n$. Since $u_{n+2} = u_{n+1} + u_n$:

$u_{n+2} + u_n = u_{n+1} + u_n + u_n = u_{n+1} + 2u_n. \checkmark$

Therefore $u_{2n+3} = u_{n+1}(u_{n+2} + u_n) = u_{n+2}^2 - u_n^2$, confirming the second identity for $n + 1$.

By induction, both identities hold for all positive integers $n$. $\blacksquare$

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Question 6

Topic: 参数方程与曲线面积 (Parametric Curves and Area)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 A closed curve is given by the equation $x^{2/n} + y^{2/n} = a^{2/n} \qquad \text{(*)}$ where $n$ is an odd integer and $a$ is a positive constant. Find a parametrization $x = x(t)$, $y = y(t)$ which describes the curve anticlockwise as $t$ ranges from $0$ to $2\pi$.

Sketch the curve in the case $n = 3$, justifying the main features of your sketch.

The area $A$ enclosed by such a curve is given by the formula $A = \frac{1}{2} \int_0^{2\pi} \left[ x(t) \frac{dy(t)}{dt} - y(t) \frac{dx(t)}{dt} \right] \, dt.$

Use this result to find the area enclosed by ($*$) for $n = 3$.

Model Solution

Part (i): Parametrization

The curve is $x^{2/n} + y^{2/n} = a^{2/n}$ where $n$ is odd. We seek a parametrization that traces the curve anticlockwise for $t \in [0, 2\pi]$.

Set

$x = a\cos^n t, \qquad y = a\sin^n t.$

Since $n$ is odd, the $n$-th root of a negative number is well-defined and real, so $x$ and $y$ take all real values as required. We verify:

$x^{2/n} + y^{2/n} = (a\cos^n t)^{2/n} + (a\sin^n t)^{2/n} = a^{2/n}\cos^2 t + a^{2/n}\sin^2 t = a^{2/n}(\cos^2 t + \sin^2 t) = a^{2/n}. \checkmark$

As $t$ increases from $0$ to $2\pi$, the point $(x,y)$ traces anticlockwise: at $t = 0$ we have $(a, 0)$; at $t = \pi/2$ we have $(0, a)$; at $t = \pi$ we have $(-a, 0)$; at $t = 3\pi/2$ we have $(0, -a)$.

Part (ii): Sketch for $n = 3$

The curve $x^{2/3} + y^{2/3} = a^{2/3}$ is called an astroid (or tetracuspid hypocycloid). Its key features are:

• Intercepts: $(\pm a, 0)$ and $(0, \pm a)$.
• Symmetry: symmetric about both axes (replacing $x \to -x$ or $y \to -y$ leaves the equation unchanged).
• Cusps: at each intercept, the curve has a cusp (pointed corner). To see why, note that $\frac{dy}{dx}$ at, say, $(a, 0)$ corresponds to $t = 0$, where $\frac{dy/dt}{dx/dt} = \frac{3a\sin^2 t\cos t}{-3a\cos^2 t\sin t} = \frac{-\sin t}{\cos t}\big|_{t=0} = 0$ from the $y$-side, but the tangent direction is degenerate at the cusp.
• Shape: the curve bulges inward between the axes, forming a concave shape somewhat like a rounded square. This is because when, say, $t = \pi/4$, we have $x = y = a(\frac{\sqrt{2}}{2})^3 = \frac{a}{2\sqrt{2}}$, which lies well inside the square with vertices $(\pm a, \pm a)$.

Part (iii): Area for $n = 3$

With $x = a\cos^3 t$ and $y = a\sin^3 t$:

$\frac{dx}{dt} = -3a\cos^2 t \sin t, \qquad \frac{dy}{dt} = 3a\sin^2 t \cos t.$

Compute the integrand $x\frac{dy}{dt} - y\frac{dx}{dt}$:

$x\frac{dy}{dt} = a\cos^3 t \cdot 3a\sin^2 t\cos t = 3a^2\sin^2 t\cos^4 t,$

$y\frac{dx}{dt} = a\sin^3 t \cdot (-3a\cos^2 t\sin t) = -3a^2\sin^4 t\cos^2 t.$

Therefore

$x\frac{dy}{dt} - y\frac{dx}{dt} = 3a^2\sin^2 t\cos^4 t + 3a^2\sin^4 t\cos^2 t = 3a^2\sin^2 t\cos^2 t(\cos^2 t + \sin^2 t) = 3a^2\sin^2 t\cos^2 t.$

Using the double angle identity $\sin t\cos t = \frac{1}{2}\sin 2t$:

$\sin^2 t\cos^2 t = \frac{1}{4}\sin^2 2t.$

So the area is

$A = \frac{1}{2}\int_0^{2\pi} 3a^2 \cdot \frac{1}{4}\sin^2 2t \, dt = \frac{3a^2}{8}\int_0^{2\pi}\sin^2 2t \, dt.$

Using $\sin^2 2t = \frac{1 - \cos 4t}{2}$:

$\int_0^{2\pi}\sin^2 2t \, dt = \int_0^{2\pi}\frac{1 - \cos 4t}{2} \, dt = \frac{1}{2}\left[t - \frac{\sin 4t}{4}\right]_0^{2\pi} = \frac{1}{2}(2\pi - 0 - 0 + 0) = \pi.$

Therefore

$A = \frac{3a^2}{8} \cdot \pi = \frac{3\pi a^2}{8}.$

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Question 7

Topic: 抽象代数/群论 (Abstract Algebra/Group Theory)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Let $a$ be a non-zero real number and define a binary operation on the set of real numbers by

$x * y = x + y + axy .$

Show that the operation $*$ is associative.

Show that $(G, *)$ is a group, where $G$ is the set of all real numbers except for one number which you should identify.

Find a subgroup of $(G, *)$ which has exactly 2 elements.

Model Solution

Associativity

We compute $(x * y) * z$ and $x * (y * z)$ and show they are equal.

$(x * y) * z = (x + y + axy) * z = (x + y + axy) + z + a(x + y + axy)z$

$= x + y + axy + z + axz + ayz + a^2xyz$

$= x + y + z + axy + axz + ayz + a^2xyz \qquad \text{(... 1)}$

$x * (y * z) = x * (y + z + ayz) = x + (y + z + ayz) + ax(y + z + ayz)$

$= x + y + z + ayz + axy + axz + a^2xyz$

$= x + y + z + axy + axz + ayz + a^2xyz \qquad \text{(... 2)}$

Since (1) and (2) are identical, $(x * y) * z = x * (y * z)$ for all real $x, y, z$. The operation $*$ is associative.

Identifying $G$ and showing it is a group

Identity element. We need $e$ such that $x * e = x$ for all $x \in G$:

$x + e + axe = x \implies e(1 + ax) = 0.$

For this to hold for all $x \in G$, we need $e = 0$. Check: $x * 0 = x + 0 + 0 = x$ and $0 * x = 0 + x + 0 = x$. So the identity is $0$.

Inverse element. Given $x \in G$, we need $x' \in G$ with $x * x' = 0$:

$x + x' + axx' = 0 \implies x'(1 + ax) = -x \implies x' = \frac{-x}{1 + ax}.$

This expression is defined provided $1 + ax \neq 0$, i.e. $x \neq -\frac{1}{a}$. Therefore the element that must be excluded from $G$ is $\boxed{-\dfrac{1}{a}}$, giving

$G = \mathbb{R} \setminus \left\{ -\frac{1}{a} \right\}.$

We verify that $x' \in G$: if $x' = -\frac{1}{a}$, then $\frac{-x}{1+ax} = -\frac{1}{a}$, giving $ax = 1 + ax$, hence $0 = 1$, a contradiction. So $x' \in G$ whenever $x \in G$.

Closure. We must check that $x * y \in G$ whenever $x, y \in G$. Suppose for contradiction that $x * y = -\frac{1}{a}$:

$x + y + axy = -\frac{1}{a} \implies a(x + y + axy) = -1 \implies ax + ay + a^2xy + 1 = 0.$

Factorising: $(1 + ax)(1 + ay) = 0$, so $x = -\frac{1}{a}$ or $y = -\frac{1}{a}$. This contradicts $x, y \in G$. Hence $x * y \in G$.

Summary. The set $G = \mathbb{R} \setminus \{-1/a\}$ with the operation $*$ satisfies all four group axioms: closure, associativity (proved above), identity ($e = 0$), and inverses $\left(x^{-1} = -\frac{x}{1+ax}\right)$. Therefore $(G, *)$ is a group.

A subgroup with exactly 2 elements

A subgroup of order 2 has the form $\{0, g\}$ where $g \neq 0$ and $g * g = 0$ (the identity). We solve:

$g + g + ag^2 = 0 \implies g(2 + ag) = 0.$

Since $g \neq 0$: $2 + ag = 0$, giving $g = -\frac{2}{a}$.

Check: $g = -\frac{2}{a} \neq -\frac{1}{a}$ (since $a \neq 0$), so $g \in G$.

Verify closure: $g * g = -\frac{2}{a} + \left(-\frac{2}{a}\right) + a \cdot \frac{4}{a^2} = -\frac{4}{a} + \frac{4}{a} = 0 \in G$.

The subgroup is $\left\{ 0, \, -\dfrac{2}{a} \right\}$.

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Question 8

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

Problem

8 The function $y(x)$ is defined for $x \geqslant 0$ and satisfies the conditions

$y = 0 \quad \text{and} \quad \frac{dy}{dx} = 1 \quad \text{at } x = 0.$

When $x$ is in the range $2(n - 1)\pi < x < 2n\pi$, where $n$ is a positive integer, $y(t)$ satisfies the differential equation

$\frac{d^2y}{dx^2} + n^2y = 0.$

Both $y$ and $\frac{dy}{dx}$ are continuous at $x = 2n\pi$ for $n = 0, 1, 2, \dots$.

(i) Find $y(x)$ for $0 \leqslant x \leqslant 2\pi$.

(ii) Show that $y(x) = \frac{1}{2} \sin 2x$ for $2\pi \leqslant x \leqslant 4\pi$, and find $y(x)$ for all $x \geqslant 0$.

(iii) Show that

$\int_{0}^{\infty} y^2 dx = \pi \sum_{n=1}^{\infty} \frac{1}{n^2} .$

Model Solution

Part (i) Find $y(x)$ for $0 \leqslant x \leqslant 2\pi$.

On the interval $0 < x < 2\pi$ (corresponding to $n = 1$), the ODE is

$\frac{d^2y}{dx^2} + y = 0.$

The general solution is $y = A\cos x + B\sin x$.

Applying the initial conditions at $x = 0$:

$y(0) = 0 \implies A = 0.$

$y'(0) = 1 \implies B = 1.$

Therefore $y(x) = \sin x$ for $0 \leqslant x \leqslant 2\pi$.

We note the boundary values at $x = 2\pi$ that will be needed for continuity matching:

$y(2\pi) = \sin 2\pi = 0, \qquad y'(2\pi) = \cos 2\pi = 1.$

Part (ii) Show that $y(x) = \frac{1}{2}\sin 2x$ for $2\pi \leqslant x \leqslant 4\pi$, and find $y(x)$ for all $x \geqslant 0$.

On the interval $2\pi < x < 4\pi$ (corresponding to $n = 2$), the ODE is

$\frac{d^2y}{dx^2} + 4y = 0.$

The general solution is $y = C\cos 2x + D\sin 2x$.

Using continuity at $x = 2\pi$:

$y(2\pi) = 0 \implies C\cos 4\pi + D\sin 4\pi = 0 \implies C = 0.$

$y'(2\pi) = 1 \implies 2D\cos 4\pi = 1 \implies 2D = 1 \implies D = \frac{1}{2}.$

Therefore $y(x) = \frac{1}{2}\sin 2x$ for $2\pi \leqslant x \leqslant 4\pi$.

General formula. On the interval $2(n-1)\pi < x < 2n\pi$, the ODE is $y'' + n^2y = 0$ with general solution

$y = A_n\cos(nx) + B_n\sin(nx).$

We claim $y(x) = \frac{1}{n}\sin(nx)$, i.e. $A_n = 0$ and $B_n = \frac{1}{n}$ for all $n \geqslant 1$, and prove this by induction.

Base case $n = 1$: confirmed above, $y = \sin x$.

Inductive step. Suppose $y = \frac{1}{n}\sin(nx)$ on $[2(n-1)\pi, 2n\pi]$. At $x = 2n\pi$:

$y(2n\pi) = \frac{1}{n}\sin(2n^2\pi) = 0,$

$y'(2n\pi) = \cos(2n^2\pi) = 1.$

On the next interval $[2n\pi, 2(n+1)\pi]$ with the ODE $y'' + (n+1)^2y = 0$, the general solution is $y = A_{n+1}\cos((n+1)x) + B_{n+1}\sin((n+1)x)$.

$y(2n\pi) = 0 \implies A_{n+1}\cos(2n(n+1)\pi) + B_{n+1}\sin(2n(n+1)\pi) = 0.$

Since $2n(n+1)$ is an even integer, $\cos(2n(n+1)\pi) = 1$ and $\sin(2n(n+1)\pi) = 0$, so $A_{n+1} = 0$.

$y'(2n\pi) = 1 \implies (n+1)B_{n+1}\cos(2n(n+1)\pi) = 1 \implies (n+1)B_{n+1} = 1 \implies B_{n+1} = \frac{1}{n+1}.$

This completes the induction. Therefore for all $x \geqslant 0$:

$y(x) = \frac{1}{n}\sin(nx) \qquad \text{for } 2(n-1)\pi \leqslant x \leqslant 2n\pi, \quad n = 1, 2, 3, \dots$

Part (iii) Show that $\int_{0}^{\infty} y^2 \, dx = \pi \sum_{n=1}^{\infty} \frac{1}{n^2}$.

We split the integral over the intervals $[2(n-1)\pi, 2n\pi]$:

$\int_{0}^{\infty} y^2 \, dx = \sum_{n=1}^{\infty} \int_{2(n-1)\pi}^{2n\pi} \frac{1}{n^2}\sin^2(nx) \, dx.$

For each term, substitute $u = nx$ (so $dx = \frac{du}{n}$):

$\int_{2(n-1)\pi}^{2n\pi} \frac{1}{n^2}\sin^2(nx) \, dx = \frac{1}{n^3} \int_{2n(n-1)\pi}^{2n^2\pi} \sin^2 u \, du.$

The interval of integration has length $2n^2\pi - 2n(n-1)\pi = 2n\pi$. Since $\sin^2 u$ has period $\pi$, this interval contains exactly $2n$ complete periods. Over each period $[k\pi, (k+1)\pi]$:

$\int_{k\pi}^{(k+1)\pi} \sin^2 u \, du = \frac{\pi}{2}.$

Therefore:

$\int_{2n(n-1)\pi}^{2n^2\pi} \sin^2 u \, du = 2n \cdot \frac{\pi}{2} = n\pi.$

Substituting back:

$\int_{2(n-1)\pi}^{2n\pi} \frac{1}{n^2}\sin^2(nx) \, dx = \frac{1}{n^3} \cdot n\pi = \frac{\pi}{n^2}.$

Summing over all $n$:

$\int_{0}^{\infty} y^2 \, dx = \sum_{n=1}^{\infty} \frac{\pi}{n^2} = \pi \sum_{n=1}^{\infty} \frac{1}{n^2}. \qquad \blacksquare$

(Using the Basel series $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$, the integral evaluates to $\frac{\pi^3}{6}$.)