STEP3 1998 -- Pure Mathematics

STEP3 1998 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 1998  |  Questions: Q1—Q8  |  Total marks per question: 20

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

Overview

Q	Topic	Difficulty	Key Techniques
1	三角函数与函数变换 Trigonometric Functions and Transformations	Challenging	三角恒等变换,求导求驻点,复合函数求导,分式函数分析
2	积分与特殊函数 Integration and Special Functions	Challenging	分部积分,递推关系,对称性论证,Beta函数
3	递推关系与差分方程 Recurrence Relations	Standard	一阶线性递推,特解与齐次解,极限过渡
4	极坐标与面积计算 Polar Coordinates and Area	Challenging	极坐标转直角坐标,极坐标面积公式,三角函数周期性,对称性
5	矩阵指数与几何变换 Matrix Exponential and Geometric Transformations	Challenging	矩阵幂级数,泰勒展开,M^2=-I的利用,旋转与剪切变换
6	几何 Geometry	Challenging	坐标几何,体积公式,对称性分析,向量运算
7	微积分 Calculus	Hard	函数图像分析,求导找极值,换元积分,量纲分析,近似估算
8	向量与几何 Vectors and Geometry	Challenging	向量方程,点积运算,球面方程,正交条件,切面分析

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Question 1

Topic: 三角函数与函数变换 Trigonometric Functions and Transformations  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Let $f(x) = \sin^2 x + 2 \cos x + 1$ for $0 \leqslant x \leqslant 2\pi$. Sketch the curve $y = f(x)$, giving the coordinates of the stationary points. Now let $g(x) = \frac{af(x) + b}{cf(x) + d} \qquad ad \neq bc \, , \ d \neq -3c \, , \ d \neq c \, .$ Show that the stationary points of $y = g(x)$ occur at the same values of $x$ as those of $y = f(x)$, and find the corresponding values of $g(x)$.

Explain why, if $d/c < -3$ or $d/c > 1$, $|g(x)|$ cannot be arbitrarily large.

Model Solution

Stationary points of $f(x)$

Using $\sin^2 x = 1 - \cos^2 x$, we rewrite:

$f(x) = 1 - \cos^2 x + 2\cos x + 1 = 2 + 2\cos x - \cos^2 x = 3 - (\cos x - 1)^2.$

Differentiating:

$f'(x) = 2\sin x \cos x - 2\sin x = 2\sin x(\cos x - 1).$

Setting $f'(x) = 0$: either $\sin x = 0$ or $\cos x = 1$.

In $[0, 2\pi]$: $\sin x = 0$ at $x = 0, \pi, 2\pi$, and $\cos x = 1$ at $x = 0, 2\pi$. So the stationary points are $x = 0, \pi, 2\pi$.

Classifying the stationary points:

$f''(x) = 2\cos x(\cos x - 1) + 2\sin x(-\sin x) = 2\cos^2 x - 2\cos x - 2\sin^2 x.$

Using $\sin^2 x = 1 - \cos^2 x$:

$f''(x) = 2\cos^2 x - 2\cos x - 2(1 - \cos^2 x) = 4\cos^2 x - 2\cos x - 2.$

At $x = \pi$: $f''(\pi) = 4(1) - 2(-1) - 2 = 4 > 0$, so $x = \pi$ is a minimum with $f(\pi) = 3 - (-1-1)^2 = 3 - 4 = -1$.

At $x = 0$: $f''(0) = 4(1) - 2(1) - 2 = 0$, so the second derivative test is inconclusive. But since $f(x) = 3 - (\cos x - 1)^2$ and $(\cos x - 1)^2 \geqslant 0$ with equality at $x = 0, 2\pi$, these are maxima with $f = 3$.

Sketch: $f$ starts at $f(0) = 3$, decreases monotonically to $f(\pi) = -1$, then increases monotonically back to $f(2\pi) = 3$. The curve is symmetric about $x = \pi$.

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Stationary points of $g(x)$

Let $u = f(x)$. Then $g(x) = \dfrac{au + b}{cu + d}$, and by the chain rule:

$g'(x) = \frac{d}{du}\!\left(\frac{au + b}{cu + d}\right) \cdot f'(x) = \frac{(ad - bc)}{(cu + d)^2} \cdot f'(x).$

Since $ad \neq bc$, the factor $\dfrac{ad - bc}{(cu + d)^2} \neq 0$ wherever it is defined. Thus $g'(x) = 0$ if and only if $f'(x) = 0$, provided the denominator $cf(x) + d \neq 0$.

The stationary values of $f$ are $f = 3$ and $f = -1$. The constraints $d \neq -3c$ (i.e. $3c + d \neq 0$) and $d \neq c$ (i.e. $-c + d \neq 0$) ensure $cf(x) + d \neq 0$ at both stationary values. Therefore $g'(x) = 0$ at exactly the same $x$-values as $f'(x) = 0$. $\blacksquare$

Corresponding values of $g$:

At $x = 0$ and $x = 2\pi$ (where $f = 3$):

$g = \frac{3a + b}{3c + d}.$

At $x = \pi$ (where $f = -1$):

$g = \frac{-a + b}{-c + d} = \frac{b - a}{d - c}.$

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Why $|g(x)|$ cannot be arbitrarily large when $d/c < -3$ or $d/c > 1$

We write $cf(x) + d = c(f(x) + d/c)$. For $|g(x)|$ to be arbitrarily large, we need $cf(x) + d$ to approach zero, which requires $f(x) = -d/c$ for some $x \in [0, 2\pi]$.

The range of $f$ on $[0, 2\pi]$ is $[-1, 3]$.

• If $d/c > 1$: then $-d/c < -1$, which is below the range of $f$. So $f(x) + d/c > 0$ for all $x$, and in fact $f(x) + d/c \geqslant -1 + d/c > 0$.
• If $d/c < -3$: then $-d/c > 3$, which is above the range of $f$. So $f(x) + d/c < 0$ for all $x$, and in fact $f(x) + d/c \leqslant 3 + d/c < 0$.

In both cases, $|cf(x) + d| = |c| \cdot |f(x) + d/c|$ is bounded below by a positive constant. Since $|af(x) + b|$ is bounded (as $f$ is continuous on the closed interval $[0, 2\pi]$), it follows that $|g(x)|$ is bounded and cannot be arbitrarily large. $\blacksquare$

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Question 2

Topic: 积分与特殊函数 Integration and Special Functions  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Let $I(a, b) = \int_0^1 t^a (1 - t)^b \, dt \qquad (a \geqslant 0, \ b \geqslant 0).$ (i) Show that $I(a, b) = I(b, a)$,

(ii) Show that $I(a, b) = I(a + 1, b) + I(a, b + 1)$.

(iii) Show that $(a+1)I(a, b) = bI(a+1, b-1)$ when $a$ and $b$ are positive and hence calculate $I(a, b)$ when $a$ and $b$ are positive integers.

Model Solution

Part (i): Show that $I(a, b) = I(b, a)$

Substitute $u = 1 - t$, so $du = -dt$. When $t = 0$, $u = 1$; when $t = 1$, $u = 0$.

$I(a, b) = \int_0^1 t^a (1-t)^b \, dt = \int_1^0 (1-u)^a u^b (-du) = \int_0^1 u^b (1-u)^a \, du = I(b, a). \qquad \blacksquare$

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Part (ii): Show that $I(a, b) = I(a+1, b) + I(a, b+1)$

Since $1 = t + (1 - t)$, we multiply the integrand by $1$:

$I(a, b) = \int_0^1 t^a(1-t)^b \cdot 1 \, dt = \int_0^1 t^a(1-t)^b \bigl[t + (1-t)\bigr] \, dt$

$= \int_0^1 t^{a+1}(1-t)^b \, dt + \int_0^1 t^a(1-t)^{b+1} \, dt = I(a+1, b) + I(a, b+1). \qquad \blacksquare$

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Part (iii): Show $(a+1)I(a, b) = bI(a+1, b-1)$ and calculate $I(a,b)$ for positive integers

Deriving the recurrence:

Use integration by parts on $I(a, b) = \int_0^1 t^a(1-t)^b \, dt$ with:

$u = t^a, \quad dv = (1-t)^b \, dt \qquad \Longrightarrow \qquad du = at^{a-1} \, dt, \quad v = -\frac{(1-t)^{b+1}}{b+1}.$

$I(a,b) = \left[-\frac{t^a(1-t)^{b+1}}{b+1}\right]_0^1 + \frac{a}{b+1}\int_0^1 t^{a-1}(1-t)^{b+1} \, dt.$

The boundary term vanishes (at $t=0$ the factor $t^a = 0$ for $a > 0$; at $t=1$ the factor $(1-t)^{b+1} = 0$). So:

$I(a,b) = \frac{a}{b+1}\,I(a-1, b+1). \qquad (\star)$

Now apply the symmetry from Part (i) to $(\star)$ with $(a,b)$ replaced by $(b,a)$:

$I(b,a) = \frac{b}{a+1}\,I(b-1, a+1).$

Using $I(b,a) = I(a,b)$ and $I(b-1, a+1) = I(a+1, b-1)$ (by symmetry again):

$I(a,b) = \frac{b}{a+1}\,I(a+1, b-1),$

which gives $(a+1)I(a,b) = bI(a+1, b-1)$. $\qquad \blacksquare$

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Calculating $I(a, b)$ for positive integers:

The recurrence $(a+1)I(a,b) = bI(a+1, b-1)$ can be rearranged as:

$I(a+1, b-1) = \frac{a+1}{b}\,I(a,b).$

Applying this repeatedly to reduce the second argument:

$I(a,b) = \frac{a+1}{b}\,I(a+1, b-1) = \frac{(a+1)(a+2)}{b(b-1)}\,I(a+2, b-2) = \cdots$

After $b$ applications:

$I(a,b) = \frac{(a+1)(a+2)\cdots(a+b)}{b!}\,I(a+b, 0).$

We compute the base case directly:

$I(a+b, 0) = \int_0^1 t^{a+b} \, dt = \frac{1}{a+b+1}.$

Therefore:

$I(a,b) = \frac{(a+1)(a+2)\cdots(a+b)}{b!} \cdot \frac{1}{a+b+1} = \frac{(a+b)!}{a!\,b!} \cdot \frac{1}{a+b+1} = \frac{a!\,b!}{(a+b+1)!}.$

Hence for positive integers $a$ and $b$:

$I(a,b) = \frac{a!\,b!}{(a+b+1)!}. \qquad \blacksquare$

Verification: $I(1,2) = \int_0^1 t(1-t)^2\,dt = \int_0^1 (t - 2t^2 + t^3)\,dt = \frac{1}{2} - \frac{2}{3} + \frac{1}{4} = \frac{1}{12}$, and $\frac{1!\,2!}{4!} = \frac{2}{24} = \frac{1}{12}$. $\checkmark$

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Question 3

Topic: 递推关系与差分方程 Recurrence Relations  |  Difficulty: Standard  |  Marks: 20

Problem

3 The value $V_N$ of a bond after $N$ days is determined by the equation $V_{N+1} = (1 + c)V_N - d \qquad (c > 0, \ d > 0),$ where $c$ and $d$ are given constants. By looking for solutions of the form $V_T = Ak^T + B$ for some constants $A, B$ and $k$, or otherwise, find $V_N$ in terms of $V_0$.

What is the solution for $c = 0$? Show that this is the limit (for fixed $N$) as $c \to 0$ of your solution for $c > 0$.

Model Solution

We seek a solution of the form $V_T = Ak^T + B$ for constants $A$, $B$, and $k$.

Part (i): Finding $V_N$ for $c > 0$

Substituting $V_T = Ak^T + B$ into the recurrence $V_{T+1} = (1+c)V_T - d$:

$Ak^{T+1} + B = (1+c)(Ak^T + B) - d = (1+c)Ak^T + (1+c)B - d.$

Comparing coefficients of $k^T$:

$Ak = (1+c)A \implies k = 1 + c \qquad (A \neq 0).$

Comparing constant terms:

$B = (1+c)B - d \implies cB = d \implies B = \frac{d}{c}.$

So the general solution is $V_N = A(1+c)^N + \dfrac{d}{c}$ for arbitrary constant $A$.

Setting $N = 0$: $V_0 = A + \dfrac{d}{c}$, so $A = V_0 - \dfrac{d}{c}$.

Therefore:

$V_N = \left(V_0 - \frac{d}{c}\right)(1+c)^N + \frac{d}{c} \qquad (c > 0).$

Part (ii): Solution for $c = 0$

When $c = 0$, the recurrence becomes $V_{N+1} = V_N - d$, which is a simple arithmetic progression:

$V_N = V_0 - Nd.$

Part (iii): Showing the limit as $c \to 0$

We need to show that for fixed $N$:

$\lim_{c \to 0} \left[\left(V_0 - \frac{d}{c}\right)(1+c)^N + \frac{d}{c}\right] = V_0 - Nd.$

Using the binomial expansion (valid for fixed $N$ and small $c$):

$(1+c)^N = 1 + Nc + \binom{N}{2}c^2 + \cdots$

Substituting:

$\left(V_0 - \frac{d}{c}\right)\!\left(1 + Nc + O(c^2)\right) + \frac{d}{c}$

$= V_0 + V_0 Nc + V_0 \cdot O(c^2) - \frac{d}{c} - dN - d \cdot O(c) + \frac{d}{c}$

The $\dfrac{d}{c}$ terms cancel:

$= V_0 + V_0 Nc + O(c^2) - dN - d \cdot O(c)$

$= V_0 - dN + c\bigl(V_0 N - d \cdot O(1)\bigr) + O(c^2).$

As $c \to 0$:

$\lim_{c \to 0} V_N = V_0 - Nd,$

which is exactly the solution for $c = 0$. $\blacksquare$

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Question 4

Topic: 极坐标与面积计算 Polar Coordinates and Area  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Show that the equation (in plane polar coordinates) $r = \cos \theta$, for $-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}$, represents a circle.

Sketch the curve $r = \cos 2\theta$ for $0 \leqslant \theta \leqslant 2\pi$, and describe the curves $r = \cos 2n\theta$, where $n$ is an integer. Show that the area enclosed by such a curve is independent of $n$.

Sketch also the curve $r = \cos 3\theta$ for $0 \leqslant \theta \leqslant 2\pi$.

Model Solution

Part (i): $r = \cos\theta$ is a circle

Multiply both sides by $r$:

$r^2 = r\cos\theta.$

Using $r^2 = x^2 + y^2$ and $r\cos\theta = x$:

$x^2 + y^2 = x.$

Completing the square:

$\left(x - \frac{1}{2}\right)^2 + y^2 = \frac{1}{4}.$

This is a circle with centre $\left(\dfrac{1}{2}, 0\right)$ and radius $\dfrac{1}{2}$.

For $-\dfrac{\pi}{2} \leqslant \theta \leqslant \dfrac{\pi}{2}$, we have $r = \cos\theta \geqslant 0$, so no extra points arise from negative $r$. The curve traces out the full circle exactly once. $\blacksquare$

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Part (ii): $r = \cos 2\theta$ and the general case $r = \cos 2n\theta$

Sketch of $r = \cos 2\theta$:

The function $\cos 2\theta$ has period $\pi$. Over $[0, 2\pi]$ it completes two full periods. The curve has $r > 0$ in the intervals $\left[0, \frac{\pi}{4}\right]$, $\left[\frac{3\pi}{4}, \frac{5\pi}{4}\right]$, $\left[\frac{7\pi}{4}, 2\pi\right]$ and $r < 0$ in $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$, $\left[\frac{5\pi}{4}, \frac{7\pi}{4}\right]$. When $r < 0$, the point is plotted in the opposite direction, which traces the same petals again. The result is a four-petalled rose with petals along the $x$- and $y$-axes.

General curves $r = \cos 2n\theta$:

For integer $n \geqslant 1$, $\cos 2n\theta$ has period $\dfrac{\pi}{n}$. Over $[0, 2\pi]$ the cosine completes $2n$ full periods, producing $4n$ petals (each period gives two intervals where $r > 0$ and two where $r < 0$, but the negative-$r$ intervals trace the same petals). The petals are equally spaced, with angular spacing $\dfrac{\pi}{2n}$ between adjacent petals.

Area enclosed by $r = \cos 2n\theta$:

Using the polar area formula and the $4n$-fold symmetry (each petal is identical):

$A = 4n \cdot \frac{1}{2}\int_0^{\pi/(4n)} r^2 \, d\theta = 2n\int_0^{\pi/(4n)} \cos^2 2n\theta \, d\theta.$

Using $\cos^2 2n\theta = \dfrac{1 + \cos 4n\theta}{2}$:

$A = 2n \int_0^{\pi/(4n)} \frac{1 + \cos 4n\theta}{2} \, d\theta = n\left[\theta + \frac{\sin 4n\theta}{4n}\right]_0^{\pi/(4n)}$

$= n\left[\frac{\pi}{4n} + \frac{\sin\pi}{4n} - 0\right] = n \cdot \frac{\pi}{4n} = \frac{\pi}{4}.$

Since the $n$ cancels, the area $\dfrac{\pi}{4}$ is independent of $n$. $\blacksquare$

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Part (iii): Sketch of $r = \cos 3\theta$

The function $\cos 3\theta$ has period $\dfrac{2\pi}{3}$. Over $[0, 2\pi]$ it completes three full periods. In each period, $\cos 3\theta > 0$ on an interval of length $\dfrac{\pi}{3}$ and $\cos 3\theta < 0$ on another interval of length $\dfrac{\pi}{3}$. The positive-$r$ intervals produce petals in one set of directions, and the negative-$r$ intervals produce petals in between. The result is a three-petalled rose (since $3$ is odd, $r = \cos 3\theta$ produces exactly $3$ petals, not $6$).

The petals are centred at $\theta = 0$, $\dfrac{2\pi}{3}$, and $\dfrac{4\pi}{3}$, each with maximum radius $r = 1$. The petals are equally spaced at $120°$ intervals.

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Question 5

Topic: 矩阵指数与几何变换 Matrix Exponential and Geometric Transformations  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The exponential of a square matrix $\mathbf{A}$ is defined to be

$\exp(\mathbf{A}) = \sum_{r=0}^{\infty} \frac{1}{r!} \mathbf{A}^r ,$

where $\mathbf{A}^0 = \mathbf{I}$ and $\mathbf{I}$ is the identity matrix.

Let

$\mathbf{M} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} .$

Show that $\mathbf{M}^2 = -\mathbf{I}$ and hence express $\exp(\theta\mathbf{M})$ as a single $2 \times 2$ matrix, where $\theta$ is a real number. Explain the geometrical significance of $\exp(\theta\mathbf{M})$.

Let

$\mathbf{N} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} .$

Express similarly $\exp(s\mathbf{N})$, where $s$ is a real number, and explain the geometrical significance of $\exp(s\mathbf{N})$.

For which values of $\theta$ does

$\exp(s\mathbf{N}) \exp(\theta\mathbf{M}) = \exp(\theta\mathbf{M}) \exp(s\mathbf{N})$

for all $s$? Interpret this fact geometrically.

Model Solution

Showing $\mathbf{M}^2 = -\mathbf{I}$:

$\mathbf{M}^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (0)(0)+(-1)(1) & (0)(-1)+(-1)(0) \\ (1)(0)+(0)(1) & (1)(-1)+(0)(0) \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\mathbf{I}$

Computing $\exp(\theta\mathbf{M})$:

From $\mathbf{M}^2 = -\mathbf{I}$, we can compute all powers of $\mathbf{M}$:

$\mathbf{M}^0 = \mathbf{I}, \quad \mathbf{M}^1 = \mathbf{M}, \quad \mathbf{M}^2 = -\mathbf{I}, \quad \mathbf{M}^3 = -\mathbf{M}, \quad \mathbf{M}^4 = \mathbf{I}, \quad \ldots$

In general, for even powers $\mathbf{M}^{2k} = (-1)^k \mathbf{I}$ and for odd powers $\mathbf{M}^{2k+1} = (-1)^k \mathbf{M}$.

Substituting into the definition:

$\exp(\theta\mathbf{M}) = \sum_{r=0}^{\infty} \frac{\theta^r}{r!} \mathbf{M}^r = \sum_{k=0}^{\infty} \frac{\theta^{2k}}{(2k)!} \mathbf{M}^{2k} + \sum_{k=0}^{\infty} \frac{\theta^{2k+1}}{(2k+1)!} \mathbf{M}^{2k+1}$

$= \sum_{k=0}^{\infty} \frac{\theta^{2k}}{(2k)!} (-1)^k \mathbf{I} + \sum_{k=0}^{\infty} \frac{\theta^{2k+1}}{(2k+1)!} (-1)^k \mathbf{M}$

$= \left(\sum_{k=0}^{\infty} \frac{(-1)^k \theta^{2k}}{(2k)!}\right) \mathbf{I} + \left(\sum_{k=0}^{\infty} \frac{(-1)^k \theta^{2k+1}}{(2k+1)!}\right) \mathbf{M}$

Recognising the Taylor series for cosine and sine:

$\exp(\theta\mathbf{M}) = \cos\theta \cdot \mathbf{I} + \sin\theta \cdot \mathbf{M} = \cos\theta \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sin\theta \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

$\exp(\theta\mathbf{M}) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

Geometrical significance: This is the matrix for anticlockwise rotation by angle $\theta$ about the origin. The matrix $\mathbf{M}$ is the generator of rotations in two dimensions.

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Computing $\exp(s\mathbf{N})$:

First we compute the powers of $\mathbf{N}$:

$\mathbf{N}^2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{0}$

Since $\mathbf{N}^2 = \mathbf{0}$, all higher powers are also zero: $\mathbf{N}^r = \mathbf{0}$ for $r \geqslant 2$. The exponential series therefore terminates after two terms:

$\exp(s\mathbf{N}) = \mathbf{I} + s\mathbf{N} + \frac{s^2}{2!}\mathbf{N}^2 + \cdots = \mathbf{I} + s\mathbf{N}$

$\exp(s\mathbf{N}) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & s \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}$

Geometrical significance: The matrix $\begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}$ represents a horizontal shear. It maps $(x, y) \mapsto (x + sy, y)$: each point is shifted horizontally by $s$ times its $y$-coordinate, while the $y$-coordinate is unchanged. Horizontal lines remain fixed, and vertical lines are tilted.

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Finding $\theta$ for commutativity:

We compute both products.

$\exp(s\mathbf{N})\exp(\theta\mathbf{M}) = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = \begin{pmatrix} \cos\theta + s\sin\theta & -\sin\theta + s\cos\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

$\exp(\theta\mathbf{M})\exp(s\mathbf{N}) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos\theta & s\cos\theta - \sin\theta \\ \sin\theta & s\sin\theta + \cos\theta \end{pmatrix}$

Comparing entries:

• Entry $(1,1)$: $\cos\theta + s\sin\theta = \cos\theta \implies s\sin\theta = 0$
• Entry $(2,2)$: $\cos\theta = s\sin\theta + \cos\theta \implies s\sin\theta = 0$

For the equation to hold for all $s$, we need $\sin\theta = 0$, i.e. $\theta = n\pi$ for integer $n$.

(We can verify the remaining entries are automatically satisfied: when $\sin\theta = 0$ and $\cos\theta = (-1)^n$, entry $(1,2)$ gives $s(-1)^n = s(-1)^n$ on both sides, and entry $(2,1)$ gives $0 = 0$.)

Geometrical interpretation: $\theta = n\pi$ corresponds to rotations by multiples of $180°$ (including the identity at $\theta = 0$). A horizontal shear followed by a $180°$ rotation is the same as the rotation followed by the shear: the rotation flips both the shear direction and the $y$-axis, and these two effects cancel. For any other rotation angle, the shear direction changes non-trivially, so the operations do not commute.

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Question 6

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

Problem

6 (i) Show that four vertices of a cube, no two of which are adjacent, form the vertices of a regular tetrahedron. Hence, or otherwise, find the volume of a regular tetrahedron whose edges are of unit length.

(ii) Find the volume of a regular octahedron whose edges are of unit length.

(iii) Show that the centres of the faces of a cube form the vertices of a regular octahedron. Show that its volume is half that of the tetrahedron whose vertices are the vertices of the cube.

[A regular tetrahedron (octahedron) has four (eight) faces, all equilateral triangles.]

Model Solution

Part (i): Four non-adjacent vertices of a cube form a regular tetrahedron

Take a unit cube with vertices at all points $(x,y,z)$ where each coordinate is $0$ or $1$. A vertex is adjacent to another if they differ in exactly one coordinate. Four non-adjacent vertices (differing in all three coordinates pairwise) are:

$A = (0,0,0), \quad B = (1,1,0), \quad C = (1,0,1), \quad D = (0,1,1).$

We compute all six pairwise distances:

$|AB| = \sqrt{1+1+0} = \sqrt{2}, \quad |AC| = \sqrt{1+0+1} = \sqrt{2}, \quad |AD| = \sqrt{0+1+1} = \sqrt{2},$

$|BC| = \sqrt{0+1+1} = \sqrt{2}, \quad |BD| = \sqrt{1+0+1} = \sqrt{2}, \quad |CD| = \sqrt{1+1+0} = \sqrt{2}.$

All six edges have equal length $\sqrt{2}$, so these four vertices form a regular tetrahedron. $\blacksquare$

Volume of a unit regular tetrahedron:

Scale the tetrahedron so that each edge has length $1$. The scaling factor from edge-length $\sqrt{2}$ to edge-length $1$ is $\frac{1}{\sqrt{2}}$, so we multiply all coordinates by $\frac{1}{\sqrt{2}}$:

$A' = (0,0,0), \quad B' = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right), \quad C' = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right), \quad D' = \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right).$

The volume is:

$V = \frac{1}{6}\left|\overrightarrow{A'B'} \cdot (\overrightarrow{A'C'} \times \overrightarrow{A'D'})\right|.$

$\overrightarrow{A'B'} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right), \quad \overrightarrow{A'C'} = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right), \quad \overrightarrow{A'D'} = \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right).$

Computing the cross product:

$\overrightarrow{A'C'} \times \overrightarrow{A'D'} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{vmatrix} = \left(0 \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right)\mathbf{i} - \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot 0\right)\mathbf{j} + \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - 0 \cdot 0\right)\mathbf{k}$

$= \left(-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}\right).$

Dot product with $\overrightarrow{A'B'}$:

$\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \cdot \left(-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}\right) = -\frac{1}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} + 0 = -\frac{1}{\sqrt{2}}.$

$V = \frac{1}{6} \cdot \frac{1}{\sqrt{2}} = \frac{1}{6\sqrt{2}} = \frac{\sqrt{2}}{12}.$

So the volume of a regular tetrahedron with unit edges is $\dfrac{\sqrt{2}}{12}$. $\blacksquare$

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Part (ii): Volume of a regular octahedron with unit edges

Place the octahedron with vertices at $(\pm a, 0, 0)$, $(0, \pm a, 0)$, $(0, 0, \pm a)$. Adjacent vertices (e.g. $(a,0,0)$ and $(0,a,0)$) have distance $a\sqrt{2}$. Setting $a\sqrt{2} = 1$ gives $a = \dfrac{1}{\sqrt{2}}$.

The octahedron consists of $8$ congruent tetrahedra, each with one vertex at the origin and three vertices on the positive or negative axes. Consider the tetrahedron with vertices $O = (0,0,0)$, $P_1 = (a,0,0)$, $P_2 = (0,a,0)$, $P_3 = (0,0,a)$:

$V_{\text{tet}} = \frac{1}{6} \left|\det\begin{pmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{pmatrix}\right| = \frac{a^3}{6}.$

Total volume:

$V_{\text{oct}} = 8 \cdot \frac{a^3}{6} = \frac{4a^3}{3} = \frac{4}{3} \cdot \frac{1}{2\sqrt{2}} = \frac{4}{6\sqrt{2}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}.$

So the volume of a regular octahedron with unit edges is $\dfrac{\sqrt{2}}{3}$. $\blacksquare$

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Part (iii): Face centres of a cube form a regular octahedron

Take a cube of side $a$ with vertices at all $(x,y,z)$ where each coordinate is $0$ or $a$. The six face centres are:

$\left(\frac{a}{2}, \frac{a}{2}, 0\right), \quad \left(\frac{a}{2}, \frac{a}{2}, a\right), \quad \left(\frac{a}{2}, 0, \frac{a}{2}\right), \quad \left(\frac{a}{2}, a, \frac{a}{2}\right), \quad \left(0, \frac{a}{2}, \frac{a}{2}\right), \quad \left(a, \frac{a}{2}, \frac{a}{2}\right).$

Shifting the origin to the centre of the cube $\left(\frac{a}{2}, \frac{a}{2}, \frac{a}{2}\right)$, these become:

$\left(0, 0, -\frac{a}{2}\right), \quad \left(0, 0, \frac{a}{2}\right), \quad \left(0, -\frac{a}{2}, 0\right), \quad \left(0, \frac{a}{2}, 0\right), \quad \left(-\frac{a}{2}, 0, 0\right), \quad \left(\frac{a}{2}, 0, 0\right).$

These are the six vertices $(\pm \frac{a}{2}, 0, 0)$, $(0, \pm \frac{a}{2}, 0)$, $(0, 0, \pm \frac{a}{2})$, which is a regular octahedron (as shown in Part (ii) with edge length $\frac{a}{\sqrt{2}}$). $\blacksquare$

Comparing volumes:

The tetrahedron from Part (i) inscribed in this cube (side $a$) has edge length $a\sqrt{2}$ and volume:

$V_{\text{tet}} = \left(\frac{a}{\sqrt{2}}\right)^3 \cdot \frac{6\sqrt{2}}{1} \cdot \frac{\sqrt{2}}{12} = a^3 \cdot \frac{\sqrt{2}}{12} \cdot 2\sqrt{2} = \frac{a^3}{3}.$

(Using the unit-edge formula scaled by $(a\sqrt{2})^3$: $V_{\text{tet}} = \frac{\sqrt{2}}{12}(a\sqrt{2})^3 = \frac{\sqrt{2}}{12} \cdot 2\sqrt{2}\,a^3 = \frac{a^3}{3}$.)

The octahedron formed by face centres has edge length $\frac{a}{\sqrt{2}}$ and volume:

$V_{\text{oct}} = \frac{\sqrt{2}}{3} \cdot \left(\frac{a}{\sqrt{2}}\right)^3 = \frac{\sqrt{2}}{3} \cdot \frac{a^3}{2\sqrt{2}} = \frac{a^3}{6}.$

Therefore $V_{\text{oct}} = \dfrac{1}{2}\,V_{\text{tet}}$. The volume of the octahedron is half that of the tetrahedron. $\blacksquare$

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Question 7

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

Problem

7 Sketch the graph of $f(s) = e^s(s - 3) + 3$ for $0 \leqslant s < \infty$. Taking $e \approx 2.7$, find the smallest positive integer, $m$, such that $f(m) > 0$.

Now let

$b(x) = \frac{x^3}{e^{x/T} - 1}$

where $T$ is a positive constant. Show that $b(x)$ has a single turning point in $0 < x < \infty$. By considering the behaviour for small $x$ and for large $x$, sketch $b(x)$ for $0 \leqslant x < \infty$.

Let

$\int_0^\infty b(x) \, dx = B,$

which may be assumed to be finite. Show that $B = KT^n$ where $K$ is a constant, and $n$ is an integer which you should determine.

Given that $B \approx 2 \int\limits_0^{Tm} b(x) dx$, use your graph of $b(x)$ to find a rough estimate for $K$.

Model Solution

Part (i): Sketch of $f(s) = e^s(s-3) + 3$ and finding $m$

Derivative:

$f'(s) = e^s(s-3) + e^s = e^s(s-2).$

Since $e^s > 0$ always: $f'(s) < 0$ for $s < 2$ and $f'(s) > 0$ for $s > 2$. So $f$ has a minimum at $s = 2$.

Key values:

$f(0) = e^0(0-3) + 3 = -3 + 3 = 0.$

$f(2) = e^2(2-3) + 3 = 3 - e^2 \approx 3 - 7.29 = -4.29.$

$f(3) = e^3(3-3) + 3 = 3.$

Behaviour: $f$ starts at $0$, decreases to a minimum of $3 - e^2 \approx -4.29$ at $s = 2$, then increases. At $s = 3$ we have $f(3) = 3 > 0$. For large $s$, $f(s) \approx se^s \to \infty$.

Sketch: The curve starts at the origin, dips down to a minimum near $s = 2$, then rises steeply through $f(3) = 3$ and continues to infinity.

Finding $m$: We need the smallest positive integer with $f(m) > 0$:

• $f(1) = e^1(1-3) + 3 = -2e + 3 \approx -5.4 + 3 = -2.4 < 0$
• $f(2) = 3 - e^2 \approx -4.3 < 0$
• $f(3) = 3 > 0$

So $m = 3$.

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Part (ii): $b(x)$ has a single turning point

$b(x) = \frac{x^3}{e^{x/T} - 1}.$

Using the quotient rule:

$b'(x) = \frac{3x^2(e^{x/T} - 1) - x^3 \cdot \frac{1}{T}e^{x/T}}{(e^{x/T} - 1)^2} = \frac{x^2\left[3(e^{x/T} - 1) - \frac{x}{T}e^{x/T}\right]}{(e^{x/T} - 1)^2}.$

Setting the numerator to zero (for $x > 0$):

$3(e^{x/T} - 1) - \frac{x}{T}e^{x/T} = 0 \qquad \Longleftrightarrow \qquad e^{x/T}\left(3 - \frac{x}{T}\right) = 3. \qquad (\star)$

Let $u = x/T > 0$ and define $h(u) = e^u(3 - u) - 3$. Then:

$h'(u) = e^u(3-u) - e^u = e^u(2 - u).$

So $h'(u) > 0$ for $u < 2$ and $h'(u) < 0$ for $u > 2$: $h$ increases on $(0, 2)$ and decreases on $(2, \infty)$.

At the boundaries: $h(0) = 3 - 3 = 0$ and $h(3) = e^3 \cdot 0 - 3 = -3 < 0$. Also $h(2) = e^2 \cdot 1 - 3 = e^2 - 3 > 0$.

Since $h(0) = 0$, $h$ increases from $0$ to a positive maximum at $u = 2$, then decreases to negative values. It crosses zero exactly once for $u > 0$ (at some $u_0 \in (2, 3)$). This gives a unique turning point of $b(x)$ at $x_0 = Tu_0$. $\blacksquare$

Behaviour of $b(x)$:

• Near $x = 0$: Using $e^{x/T} - 1 \approx \frac{x}{T}$: $\quad b(x) \approx \frac{x^3}{x/T} = Tx^2 \to 0$.
• At $x = 0$: By L’Hopital (or the above), $b(0) = 0$.
• For large $x$: $e^{x/T}$ grows exponentially, so $b(x) \approx x^3 e^{-x/T} \to 0$.
• At the turning point $x_0$: $b$ attains its maximum.

The curve starts at the origin, rises to a single maximum, then decays back toward zero.

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Part (iii): Show $B = KT^4$

Substituting $u = x/T$ (so $x = Tu$, $dx = T\,du$):

$B = \int_0^\infty \frac{x^3}{e^{x/T} - 1}\,dx = \int_0^\infty \frac{T^3 u^3}{e^u - 1}\,T\,du = T^4 \int_0^\infty \frac{u^3}{e^u - 1}\,du.$

The integral $\displaystyle K = \int_0^\infty \frac{u^3}{e^u - 1}\,du$ is a dimensionless constant (it equals $\dfrac{\pi^4}{15}$, though this is not required here). Therefore $B = KT^4$, so $n = 4$. $\blacksquare$

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Part (iv): Rough estimate for $K$

Using $m = 3$ and $B \approx 2\int_0^{Tm} b(x)\,dx = 2\int_0^{3T} b(x)\,dx$, substitute $u = x/T$:

$KT^4 \approx 2T^4 \int_0^3 \frac{u^3}{e^u - 1}\,du \qquad \Longrightarrow \qquad K \approx 2\int_0^3 \frac{u^3}{e^u - 1}\,du.$

We estimate the integral using Simpson’s rule with $h = 0.5$ and values at $u = 0, 0.5, 1, 1.5, 2, 2.5, 3$:

$u$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$	$3$
$f(u) = \frac{u^3}{e^u - 1}$	$1$	$0.238$	$0.582$	$0.753$	$0.767$	$0.692$	$0.588$

(Near $u = 0$: $\frac{u^3}{e^u-1} \approx u^2$, so $f(0) = 0$; at $u = 0.5$: $\frac{0.125}{e^{0.5}-1} = \frac{0.125}{0.6487} = 0.193$.)

Let me recompute more carefully. At $u = 0$: $f(0) = 0$ (by L’Hopital, $\lim_{u\to 0} \frac{u^3}{e^u-1} = 0$).

$u$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$	$3$
$f(u)$	$0$	$0.193$	$0.582$	$0.753$	$0.767$	$0.692$	$0.588$

Simpson’s rule with $h = 0.5$:

$\int_0^3 f(u)\,du \approx \frac{0.5}{3}\left[f(0) + 4f(0.5) + 2f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + f(3)\right]$

$= \frac{1}{6}\left[0 + 4(0.193) + 2(0.582) + 4(0.753) + 2(0.767) + 4(0.692) + 0.588\right]$

$= \frac{1}{6}\left[0 + 0.772 + 1.164 + 3.012 + 1.534 + 2.768 + 0.588\right] = \frac{9.838}{6} \approx 1.64.$

Therefore $K \approx 2 \times 1.64 = 3.3$.

(For comparison, the exact value is $K = \frac{\pi^4}{15} \approx 6.49$; our approximation underestimates because we are missing the tail integral from $3$ to $\infty$, which contributes about half the total.)

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Question 8

Topic: 向量与几何 Vectors and Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

8 (i) Show that the line $\mathbf{r} = \mathbf{b} + \lambda\mathbf{m}$, where $\mathbf{m}$ is a unit vector, intersects the sphere $\mathbf{r}.\mathbf{r} = a^2$ at two points if

$a^2 > \mathbf{b}.\mathbf{b} - (\mathbf{b}.\mathbf{m})^2.$

Write down the corresponding condition for there to be precisely one point of intersection. If this point has position vector $\mathbf{p}$, show that $\mathbf{m}.\mathbf{p} = 0$.

(ii) Now consider a second sphere of radius $a$ and a plane perpendicular to a unit vector $\mathbf{n}$. The centre of the sphere has position vector $\mathbf{d}$ and the minimum distance from the origin to the plane is $l$. What is the condition for the plane to be tangential to this second sphere?

(iii) Show that the first and second spheres intersect at right angles (i.e. the two radii to each point of intersection are perpendicular) if

$\mathbf{d}.\mathbf{d} = 2a^2.$

Model Solution

Part (i): Line-sphere intersection

The line is $\mathbf{r} = \mathbf{b} + \lambda\mathbf{m}$ where $|\mathbf{m}| = 1$. The sphere is $\mathbf{r} \cdot \mathbf{r} = a^2$. Substituting:

$(\mathbf{b} + \lambda\mathbf{m}) \cdot (\mathbf{b} + \lambda\mathbf{m}) = a^2$

$\mathbf{b} \cdot \mathbf{b} + 2\lambda(\mathbf{b} \cdot \mathbf{m}) + \lambda^2(\mathbf{m} \cdot \mathbf{m}) = a^2.$

Since $|\mathbf{m}| = 1$:

$\lambda^2 + 2(\mathbf{b} \cdot \mathbf{m})\lambda + (\mathbf{b} \cdot \mathbf{b} - a^2) = 0. \qquad (\star)$

This is a quadratic in $\lambda$ with discriminant:

$\Delta = 4(\mathbf{b} \cdot \mathbf{m})^2 - 4(\mathbf{b} \cdot \mathbf{b} - a^2) = 4\left[a^2 - \mathbf{b} \cdot \mathbf{b} + (\mathbf{b} \cdot \mathbf{m})^2\right].$

Two distinct real roots (two intersection points) exist if and only if $\Delta > 0$:

$a^2 > \mathbf{b} \cdot \mathbf{b} - (\mathbf{b} \cdot \mathbf{m})^2. \qquad \blacksquare$

Condition for exactly one intersection (tangency):

$a^2 = \mathbf{b} \cdot \mathbf{b} - (\mathbf{b} \cdot \mathbf{m})^2.$

When $\Delta = 0$, the unique solution of $(\star)$ is $\lambda = -\mathbf{b} \cdot \mathbf{m}$, giving:

$\mathbf{p} = \mathbf{b} - (\mathbf{b} \cdot \mathbf{m})\mathbf{m}.$

We verify $\mathbf{m} \cdot \mathbf{p} = 0$:

$\mathbf{m} \cdot \mathbf{p} = \mathbf{m} \cdot \mathbf{b} - (\mathbf{b} \cdot \mathbf{m})(\mathbf{m} \cdot \mathbf{m}) = \mathbf{b} \cdot \mathbf{m} - \mathbf{b} \cdot \mathbf{m} = 0. \qquad \blacksquare$

(Geometrically, $\mathbf{p}$ is the foot of the perpendicular from the centre of the sphere to the line, and tangency occurs when this perpendicular has length $a$.)

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Part (ii): Tangency condition for a plane and a sphere

The plane has equation $\mathbf{r} \cdot \mathbf{n} = l$ (where $|\mathbf{n}| = 1$ and $l > 0$ is the minimum distance from the origin to the plane). The second sphere has centre $\mathbf{d}$ and radius $a$, so its equation is:

$(\mathbf{r} - \mathbf{d}) \cdot (\mathbf{r} - \mathbf{d}) = a^2.$

The distance from the centre $\mathbf{d}$ to the plane $\mathbf{r} \cdot \mathbf{n} = l$ is $|\mathbf{d} \cdot \mathbf{n} - l|$. The plane is tangential to the sphere if and only if this distance equals the radius:

$(\mathbf{d} \cdot \mathbf{n} - l)^2 = a^2 \qquad \Longleftrightarrow \qquad |\mathbf{d} \cdot \mathbf{n} - l| = a. \qquad \blacksquare$

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Part (iii): Two spheres intersect at right angles

The first sphere (centre $O$, radius $a$): $\mathbf{r} \cdot \mathbf{r} = a^2$.

The second sphere (centre $\mathbf{d}$, radius $a$): $(\mathbf{r} - \mathbf{d}) \cdot (\mathbf{r} - \mathbf{d}) = a^2$.

Finding the intersection: Expanding the second equation:

$\mathbf{r} \cdot \mathbf{r} - 2\mathbf{r} \cdot \mathbf{d} + \mathbf{d} \cdot \mathbf{d} = a^2.$

Substituting $\mathbf{r} \cdot \mathbf{r} = a^2$ from the first sphere:

$a^2 - 2\mathbf{r} \cdot \mathbf{d} + \mathbf{d} \cdot \mathbf{d} = a^2 \qquad \Longrightarrow \qquad \mathbf{r} \cdot \mathbf{d} = \frac{\mathbf{d} \cdot \mathbf{d}}{2}. \qquad (\star)$

So the intersection lies in the plane $\mathbf{r} \cdot \mathbf{d} = \frac{|\mathbf{d}|^2}{2}$.

Orthogonality condition: Two spheres intersect at right angles if, at every point of intersection $\mathbf{r}$, the radius vectors from the two centres are perpendicular. The radius vectors are $\mathbf{r}$ (from $O$) and $\mathbf{r} - \mathbf{d}$ (from $\mathbf{d}$). We need:

$\mathbf{r} \cdot (\mathbf{r} - \mathbf{d}) = 0 \qquad \Longleftrightarrow \qquad \mathbf{r} \cdot \mathbf{r} - \mathbf{r} \cdot \mathbf{d} = 0.$

Using $\mathbf{r} \cdot \mathbf{r} = a^2$ and $(\star)$:

$a^2 - \frac{\mathbf{d} \cdot \mathbf{d}}{2} = 0 \qquad \Longleftrightarrow \qquad \mathbf{d} \cdot \mathbf{d} = 2a^2. \qquad \blacksquare$