STEP2 2000 -- Pure Mathematics

STEP2 2000 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2000  |  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	数论 Number Theory	Standard	因数分解,代数变换,唯一性论证
2	代数 Algebra	Standard	多项式导数,重根条件,方程组求解
3	微积分 Calculus	Standard	偏微分,线性近似,误差传播
4	复数 Complex Numbers	Challenging	De Moivre定理,复数乘法,辐角原理
5	微积分 Calculus	Standard	最小二乘法,积分求极值,极限分析
6	积分 Integration	Challenging	万能代换,三角恒等式,定积分计算
7	向量几何 Vector Geometry	Challenging	向量方程,直线夹角公式,平面方程,轨迹分析
8	微分方程 Differential Equations	Challenging	分离变量法,极限分析,指数函数

---

Question 1

Topic: 数论 Number Theory  |  Difficulty: Standard  |  Marks: 20

Problem

1 A number of the form $1/N$, where $N$ is an integer greater than 1, is called a unit fraction.

Noting that

$\frac{1}{2} = \frac{1}{3} + \frac{1}{6} \quad \text{and} \quad \frac{1}{3} = \frac{1}{4} + \frac{1}{12},$

guess a general result of the form

$\frac{1}{N} = \frac{1}{a} + \frac{1}{b} \qquad \text{(*)}$

and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions.

By writing (*) in the form

$(a - N)(b - N) = N^2$

and by considering the factors of $N^2$, show that if $N$ is prime, then there is only one way of expressing $1/N$ as the sum of two distinct unit fractions.

Prove similarly that any fraction of the form $2/N$, where $N$ is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.

Model Solution

Guessing the general result. From the two examples:

$N = 2: \quad a = 3 = 2+1, \quad b = 6 = 2 \times 3$

$N = 3: \quad a = 4 = 3+1, \quad b = 12 = 3 \times 4$

the pattern suggests:

$\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)} \qquad \text{(**)}$

Proof. Compute the right-hand side with a common denominator:

$\frac{1}{N+1} + \frac{1}{N(N+1)} = \frac{N}{N(N+1)} + \frac{1}{N(N+1)} = \frac{N+1}{N(N+1)} = \frac{1}{N}$

Since $N > 1$, we have $N + 1 \neq N(N+1)$, so the two unit fractions are distinct. This proves that any unit fraction can be expressed as the sum of two distinct unit fractions. $\square$

---

Uniqueness when $N$ is prime. Starting from $\frac{1}{N} = \frac{1}{a} + \frac{1}{b}$ with $a \neq b$, multiply through by $Nab$:

$ab = N(a + b)$

$ab - Na - Nb = 0$

Add $N^2$ to both sides:

$ab - Na - Nb + N^2 = N^2$

$(a - N)(b - N) = N^2$

Since $\frac{1}{a} < \frac{1}{N}$ (because $\frac{1}{a} = \frac{1}{N} - \frac{1}{b}$ and $b > 0$), we have $a > N$, so $a - N > 0$. Similarly $b > N$, so $b - N > 0$.

If $N$ is prime, the positive divisors of $N^2$ are $1, N, N^2$. The factor pairs $(d_1, d_2)$ with $d_1 d_2 = N^2$ and $d_1 \leqslant d_2$ are:

• $(1, N^2)$: gives $a = N + 1$, $b = N^2 + N = N(N+1)$. These are distinct since $N > 1$.
• $(N, N)$: gives $a = b = 2N$. This is excluded since we require distinct fractions.

(The pair $(N^2, 1)$ gives the same solution with $a$ and $b$ swapped.)

Therefore, when $N$ is prime, the only expression of $1/N$ as the sum of two distinct unit fractions (up to order) is:

$\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}$

$\square$

---

The case $2/N$ with $N$ an odd prime. Suppose $\frac{2}{N} = \frac{1}{a} + \frac{1}{b}$ with $a \neq b$. Multiply through by $Nab$:

$2ab = N(a + b)$

$4ab = 2N(a + b)$

$4ab - 2Na - 2Nb = 0$

Add $N^2$ to both sides:

$4ab - 2Na - 2Nb + N^2 = N^2$

$(2a - N)(2b - N) = N^2$

Since $\frac{1}{a} < \frac{2}{N}$, we have $a > N/2$, so $2a - N > 0$. Similarly $2b - N > 0$.

Since $N$ is an odd prime ($N > 2$), the positive factor pairs of $N^2$ are the same as before: $(1, N^2)$, $(N, N)$, $(N^2, 1)$.

• $(N, N)$: $2a - N = N$ and $2b - N = N$, giving $a = b = N$. Not distinct, so excluded.
• $(1, N^2)$: $2a = N + 1$ and $2b = N^2 + N = N(N+1)$. Since $N$ is odd, $N + 1$ is even, so $a = (N+1)/2$ is a positive integer. Since $N + 1$ is even, $N(N+1)$ is also even, so $b = N(N+1)/2$ is a positive integer. And $a \neq b$ since $N > 1$.

(The pair $(N^2, 1)$ gives the same solution with $a$ and $b$ swapped.)

Therefore $2/N$ can be expressed uniquely (up to order) as:

$\frac{2}{N} = \frac{1}{(N+1)/2} + \frac{1}{N(N+1)/2}$

Verification:

$\frac{1}{(N+1)/2} + \frac{1}{N(N+1)/2} = \frac{2}{N+1} + \frac{2}{N(N+1)} = \frac{2N + 2}{N(N+1)} = \frac{2(N+1)}{N(N+1)} = \frac{2}{N} \checkmark$

$\square$

---

Question 2

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

Problem

2 Prove that if $(x - a)^2$ is a factor of the polynomial $p(x)$, then $p'(a) = 0$. Prove a corresponding result if $(x - a)^4$ is a factor of $p(x)$.

Given that the polynomial

$x^6 + 4x^5 - 5x^4 - 40x^3 - 40x^2 + 32x + k$

has a factor of the form $(x - a)^4$, find $k$.

Model Solution

Proof for $(x - a)^2$. If $(x - a)^2$ is a factor of $p(x)$, then $p(x) = (x - a)^2 q(x)$ for some polynomial $q(x)$. Differentiating by the product rule:

$p'(x) = 2(x - a) \, q(x) + (x - a)^2 \, q'(x)$

Evaluating at $x = a$:

$p'(a) = 2(0) \cdot q(a) + 0^2 \cdot q'(a) = 0 \qquad \square$

---

Corresponding result for $(x - a)^4$. If $(x - a)^4$ is a factor of $p(x)$, then $p(x) = (x - a)^4 q(x)$ for some polynomial $q(x)$. Differentiate:

$p'(x) = 4(x - a)^3 \, q(x) + (x - a)^4 \, q'(x) = (x - a)^3 \underbrace{\left[ 4q(x) + (x - a) q'(x) \right]}_{s(x)}$

So $p'(x) = (x - a)^3 s(x)$, which means $(x - a)^2$ is a factor of $p'(x)$. By the result just proved (applied to $p'$ instead of $p$):

$p''(a) = 0$

Now differentiate $p'(x) = (x - a)^3 s(x)$:

$p''(x) = 3(x - a)^2 \, s(x) + (x - a)^3 \, s'(x) = (x - a)^2 \underbrace{\left[ 3s(x) + (x - a) s'(x) \right]}_{t(x)}$

So $p''(x) = (x - a)^2 t(x)$, meaning $(x - a)^2$ is a factor of $p''(x)$. Applying the same result to $p''$:

$p'''(a) = 0$

Conclusion: if $(x - a)^4$ is a factor of $p(x)$, then $p'(a) = p''(a) = p'''(a) = 0$. $\square$

---

Finding $k$. Let $p(x) = x^6 + 4x^5 - 5x^4 - 40x^3 - 40x^2 + 32x + k$. Compute the first three derivatives:

$p'(x) = 6x^5 + 20x^4 - 20x^3 - 120x^2 - 80x + 32$

$p''(x) = 30x^4 + 80x^3 - 60x^2 - 240x - 80$

$p'''(x) = 120x^3 + 240x^2 - 120x - 240$

Step 1. Solve $p'''(a) = 0$:

$120(a^3 + 2a^2 - a - 2) = 0$

Factor by grouping:

$a^2(a + 2) - 1(a + 2) = 0 \implies (a + 2)(a^2 - 1) = 0 \implies (a + 2)(a - 1)(a + 1) = 0$

So the candidates are $a = -2, \, -1, \, 1$.

Step 2. Check which candidates satisfy $p''(a) = 0$:

$p''(-2) = 30(16) + 80(-8) - 60(4) - 240(-2) - 80 = 480 - 640 - 240 + 480 - 80 = 0 \checkmark$

$p''(-1) = 30 - 80 - 60 + 240 - 80 = 50 \neq 0$

$p''(1) = 30 + 80 - 60 - 240 - 80 = -270 \neq 0$

So $a = -2$.

Step 3. Verify $p'(-2) = 0$:

$p'(-2) = 6(-32) + 20(16) - 20(-8) - 120(4) - 80(-2) + 32$ $= -192 + 320 + 160 - 480 + 160 + 32 = 0 \checkmark$

Step 4. Find $k$ from $p(-2) = 0$:

$p(-2) = 64 - 128 - 80 + 320 - 160 - 64 + k = -48 + k = 0$

$\boxed{k = 48}$

Verification. We check that $(x + 2)^4 (x^2 - 4x + 3) = x^6 + 4x^5 - 5x^4 - 40x^3 - 40x^2 + 32x + 48$.

$(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16$ and $x^2 - 4x + 3$. Multiplying term by term:

Term from $(x+2)^4$	$\times \, x^2$	$\times \, (-4x)$	$\times \, 3$
$x^4$	$x^6$	$-4x^5$	$3x^4$
$8x^3$	$8x^5$	$-32x^4$	$24x^3$
$24x^2$	$24x^4$	$-96x^3$	$72x^2$
$32x$	$32x^3$	$-128x^2$	$96x$
$16$	$16x^2$	$-64x$	$48$

Collecting:

• $x^6$: $1$
• $x^5$: $-4 + 8 = 4$
• $x^4$: $3 - 32 + 24 = -5$
• $x^3$: $24 - 96 + 32 = -40$
• $x^2$: $72 - 128 + 16 = -40$
• $x$: $96 - 64 = 32$
• constant: $48$

This matches the original polynomial with $k = 48$. Note also that $x^2 - 4x + 3 = (x-1)(x-3)$, so the full factorisation is:

$p(x) = (x + 2)^4 (x - 1)(x - 3)$

---

Question 3

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

Problem

3 The lengths of the sides $BC, CA, AB$ of the triangle $ABC$ are denoted by $a, b, c$, respectively. Given that

$b = 8 + \epsilon_1, c = 3 + \epsilon_2, A = \pi/3 + \epsilon_3,$

where $\epsilon_1, \epsilon_2$, and $\epsilon_3$ are small, show that $a \approx 7 + \eta$, where $\eta = (13 \, \epsilon_1 - 2 \, \epsilon_2 + 24\sqrt{3} \, \epsilon_3)/14$.

Given now that

$|\epsilon_1| \leqslant 2 \times 10^{-3}, \quad |\epsilon_2| \leqslant 4 \cdot 9 \times 10^{-2}, \quad |\epsilon_3| \leqslant \sqrt{3} \times 10^{-3},$

find the range of possible values of $\eta$.

Model Solution

Part (a): Showing $a \approx 7 + \eta$

By the cosine rule,

$a^2 = b^2 + c^2 - 2bc\cos A. \qquad \text{(*)}$

When $\epsilon_1 = \epsilon_2 = \epsilon_3 = 0$, we have $b = 8$, $c = 3$, $A = \pi/3$, so

$a^2 = 64 + 9 - 48\cos(\pi/3) = 73 - 24 = 49,$

giving $a = 7$.

To find the first-order perturbation, we differentiate (*). Since $a$, $b$, $c$, and $A$ all vary:

$2a\,da = 2b\,db + 2c\,dc - 2\cos A\,(c\,db + b\,dc) - 2bc(-\sin A)\,dA.$

Note the last term: $\frac{\partial}{\partial A}(-2bc\cos A) = 2bc\sin A$. Dividing by 2:

$a\,da = b\,db + c\,dc - \cos A\,(c\,db + b\,dc) + bc\sin A\,dA.$

Collecting terms in $db$, $dc$, $dA$:

$a\,da = (b - c\cos A)\,db + (c - b\cos A)\,dc + bc\sin A\,dA.$

Now substitute $db = d\epsilon_1$, $dc = d\epsilon_2$, $dA = d\epsilon_3$, and evaluate at $b = 8$, $c = 3$, $A = \pi/3$, $a = 7$:

• $b - c\cos A = 8 - 3 \cdot \tfrac{1}{2} = \tfrac{13}{2}$
• $c - b\cos A = 3 - 8 \cdot \tfrac{1}{2} = -1$
• $bc\sin A = 24 \cdot \tfrac{\sqrt{3}}{2} = 12\sqrt{3}$

Therefore

$7\,d\eta = \tfrac{13}{2}\,d\epsilon_1 - d\epsilon_2 + 12\sqrt{3}\,d\epsilon_3,$

which integrates (to first order) to

$\eta = \frac{13\epsilon_1 - 2\epsilon_2 + 24\sqrt{3}\,\epsilon_3}{14}.$

Hence $a \approx 7 + \eta$ where $\eta = (13\epsilon_1 - 2\epsilon_2 + 24\sqrt{3}\,\epsilon_3)/14$.

Part (b): Finding the range of $\eta$

Since $\eta = (13\epsilon_1 - 2\epsilon_2 + 24\sqrt{3}\,\epsilon_3)/14$ is linear in the $\epsilon_i$, its maximum and minimum occur at the extreme values of each $\epsilon_i$.

The maximum of $|\eta|$ is achieved when each term $13\epsilon_1$, $-2\epsilon_2$, and $24\sqrt{3}\,\epsilon_3$ has the same sign. By the triangle inequality:

$|\eta| \leqslant \frac{13|\epsilon_1| + 2|\epsilon_2| + 24\sqrt{3}\,|\epsilon_3|}{14}.$

Substituting the given bounds:

$|\eta| \leqslant \frac{13 \times 2 \times 10^{-3} + 2 \times 4.9 \times 10^{-2} + 24\sqrt{3} \times \sqrt{3} \times 10^{-3}}{14}.$

Evaluating each term:

• $13 \times 2 \times 10^{-3} = 26 \times 10^{-3}$
• $2 \times 4.9 \times 10^{-2} = 98 \times 10^{-3}$
• $24\sqrt{3} \times \sqrt{3} \times 10^{-3} = 24 \times 3 \times 10^{-3} = 72 \times 10^{-3}$

Therefore

$|\eta| \leqslant \frac{(26 + 98 + 72) \times 10^{-3}}{14} = \frac{196 \times 10^{-3}}{14} = 14 \times 10^{-3} = 0.014.$

This bound is attained (for example, when $\epsilon_1 = 2 \times 10^{-3}$, $\epsilon_2 = -4.9 \times 10^{-2}$, $\epsilon_3 = \sqrt{3} \times 10^{-3}$), so

$-0.014 \leqslant \eta \leqslant 0.014.$

---

Question 4

Topic: 复数 Complex Numbers  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Prove that $(\cos \theta + i \sin \theta)(\cos \phi + i \sin \phi) = \cos(\theta + \phi) + i \sin(\theta + \phi)$ and that, for every positive integer $n$, $(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta.$ By considering $(5 - i)^2(1 + i)$, or otherwise, prove that $\arctan(7/17) + 2 \arctan(1/5) = \pi/4.$ Prove also that $3 \arctan(1/4) + \arctan(1/20) + \arctan(1/1985) = \pi/4.$ [Note that $\arctan \theta$ is another notation for $\tan^{-1} \theta$.]

Model Solution

Part (a): Proving the multiplication formula

$(\cos\theta + i\sin\theta)(\cos\phi + i\sin\phi)$ $= \cos\theta\cos\phi + i\cos\theta\sin\phi + i\sin\theta\cos\phi + i^2\sin\theta\sin\phi$ $= (\cos\theta\cos\phi - \sin\theta\sin\phi) + i(\sin\theta\cos\phi + \cos\theta\sin\phi).$

By the angle addition formulae $\cos(\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi$ and $\sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi$, this equals

$\cos(\theta + \phi) + i\sin(\theta + \phi). \qquad \text{(*)}$

Part (b): Proving de Moivre’s theorem by induction

We prove $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ for all positive integers $n$.

Base case ($n = 1$): $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta$. True.

Inductive step: Assume the result holds for $n = k$, i.e., $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$. Then

$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta)$ $= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta).$

Applying (*) with $\phi = \theta$:

$= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos(k+1)\theta + i\sin(k+1)\theta.$

By induction, the result holds for all positive integers $n$.

Part (c): Proving $\arctan(7/17) + 2\arctan(1/5) = \pi/4$

Consider the complex number $(5 - i)^2(1 + i)$.

First, compute $(5 - i)^2$:

$(5 - i)^2 = 25 - 10i + i^2 = 24 - 10i.$

Then multiply by $(1 + i)$:

$(24 - 10i)(1 + i) = 24 + 24i - 10i - 10i^2 = 24 + 14i + 10 = 34 + 14i.$

Now we find the argument of this product in two ways.

Method 1 (direct): Since $34 + 14i$ has positive real and imaginary parts,

$\arg(34 + 14i) = \arctan\!\left(\frac{14}{34}\right) = \arctan\!\left(\frac{7}{17}\right).$

Method 2 (via arguments of factors): We have $\arg(1 + i) = \pi/4$. Also $5 - i$ has positive real part and negative imaginary part, so $\arg(5 - i) = -\arctan(1/5)$. By de Moivre’s theorem:

$\arg\bigl((5-i)^2(1+i)\bigr) = 2\arg(5-i) + \arg(1+i) = -2\arctan(1/5) + \pi/4.$

Equating the two expressions:

$\arctan(7/17) = -2\arctan(1/5) + \pi/4,$

hence

$\arctan(7/17) + 2\arctan(1/5) = \pi/4. \qquad \blacksquare$

Part (d): Proving $3\arctan(1/4) + \arctan(1/20) + \arctan(1/1985) = \pi/4$

Consider the product $(4 + i)^3(20 + i)(1985 + i)$.

Step 1: Compute $(4 + i)^3$.

$(4 + i)^2 = 16 + 8i + i^2 = 15 + 8i.$

$(4 + i)^3 = (15 + 8i)(4 + i) = 60 + 15i + 32i + 8i^2 = 52 + 47i.$

Step 2: Multiply by $(20 + i)$.

$(52 + 47i)(20 + i) = 1040 + 52i + 940i + 47i^2 = 993 + 992i.$

Step 3: Multiply by $(1985 + i)$.

$(993 + 992i)(1985 + i)$ $= 993 \times 1985 + 993i + 992 \times 1985\,i + 992i^2$ $= (993 \times 1985 - 992) + (993 + 992 \times 1985)i.$

Evaluating:

• Real part: $993 \times 1985 - 992 = 1971105 - 992 = 1970113$.
• Imaginary part: $993 + 992 \times 1985 = 993 + 1969120 = 1970113$.

So

$(4 + i)^3(20 + i)(1985 + i) = 1970113(1 + i).$

Since $\arg(1 + i) = \pi/4$, we have

$\arg\bigl((4+i)^3(20+i)(1985+i)\bigr) = \pi/4.$

On the other hand, since each factor $4 + i$, $20 + i$, $1985 + i$ has positive real and imaginary parts, their arguments lie in $(0, \pi/2)$, and

$\arg(4+i) = \arctan(1/4), \quad \arg(20+i) = \arctan(1/20), \quad \arg(1985+i) = \arctan(1/1985).$

The sum $3\arctan(1/4) + \arctan(1/20) + \arctan(1/1985) < 3 \cdot \pi/4 + \pi/2 < 2\pi$, so there is no ambiguity in adding the arguments. Therefore

$3\arctan(1/4) + \arctan(1/20) + \arctan(1/1985) = \pi/4. \qquad \blacksquare$

---

Question 5

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

Problem

5 It is required to approximate a given function $f(x)$, over the interval $0 \leqslant x \leqslant 1$, by the linear function $\lambda x$, where $\lambda$ is chosen to minimise $\int_{0}^{1} \left( f(x) - \lambda x \right)^2 dx.$ Show that $\lambda = 3 \int_{0}^{1} xf(x) dx.$ The residual error, $R$, of this approximation process is such that $R^2 = \int_{0}^{1} \left( f(x) - \lambda x \right)^2 dx.$ Show that $R^2 = \int_{0}^{1} \left( f(x) \right)^2 dx - \frac{1}{3} \lambda^2.$ Given now that $f(x) = \sin(\pi x/n)$, show that (i) for large $n$, $\lambda \approx \pi/n$ and (ii) $\lim_{n \to \infty} R = 0$. Explain why, prior to any calculation, these results are to be expected. [You may assume that, when $\theta$ is small, $\sin \theta \approx \theta - \theta^3/6$ and $\cos \theta \approx 1 - \theta^2/2$.]

Model Solution

Step 1: Find $\lambda$.

Define $I(\lambda) = \displaystyle\int_{0}^{1}(f(x) - \lambda x)^2\,dx$. Expand the integrand:

$I(\lambda) = \int_{0}^{1}f(x)^2\,dx - 2\lambda\int_{0}^{1}xf(x)\,dx + \lambda^2\int_{0}^{1}x^2\,dx$

Evaluating the last integral:

$\int_{0}^{1}x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$

So

$I(\lambda) = \int_{0}^{1}f(x)^2\,dx - 2\lambda\int_{0}^{1}xf(x)\,dx + \frac{\lambda^2}{3}$

This is a quadratic in $\lambda$ with positive leading coefficient, so it has a unique minimum. Differentiate and set to zero:

$\frac{dI}{d\lambda} = -2\int_{0}^{1}xf(x)\,dx + \frac{2\lambda}{3} = 0$

$\therefore\quad \lambda = 3\int_{0}^{1}xf(x)\,dx \qquad \text{(as required)}$

Step 2: Show $R^2 = \displaystyle\int_0^1 f(x)^2\,dx - \tfrac{1}{3}\lambda^2$.

From Step 1, we have $\lambda = 3\displaystyle\int_0^1 xf(x)\,dx$, so $\displaystyle\int_0^1 xf(x)\,dx = \frac{\lambda}{3}$.

Substituting back:

$R^2 = I(\lambda) = \int_{0}^{1}f(x)^2\,dx - 2\lambda\cdot\frac{\lambda}{3} + \frac{\lambda^2}{3}$

$= \int_{0}^{1}f(x)^2\,dx - \frac{2\lambda^2}{3} + \frac{\lambda^2}{3} = \int_{0}^{1}f(x)^2\,dx - \frac{\lambda^2}{3} \qquad \text{(as required)}$

Step 3: Now let $f(x) = \sin(\pi x/n)$. Part (i) — show $\lambda \approx \pi/n$ for large $n$.

$\lambda = 3\int_{0}^{1}x\sin\!\left(\frac{\pi x}{n}\right)dx$

For large $n$, use $\sin\theta \approx \theta - \theta^3/6$ with $\theta = \pi x/n$:

$\lambda \approx 3\int_{0}^{1}x\left(\frac{\pi x}{n} - \frac{\pi^3 x^3}{6n^3}\right)dx = 3\int_{0}^{1}\left(\frac{\pi x^2}{n} - \frac{\pi^3 x^4}{6n^3}\right)dx$

$= 3\left[\frac{\pi x^3}{3n} - \frac{\pi^3 x^5}{30n^3}\right]_0^1 = 3\left(\frac{\pi}{3n} - \frac{\pi^3}{30n^3}\right) = \frac{\pi}{n} - \frac{\pi^3}{10n^3}$

For large $n$, the second term is negligible, so $\lambda \approx \dfrac{\pi}{n}$.

Step 4: Part (ii) — show $\lim_{n\to\infty}R = 0$.

Compute $\displaystyle\int_0^1 f(x)^2\,dx = \int_0^1 \sin^2\!\left(\frac{\pi x}{n}\right)dx$.

Using $\sin^2 u = \tfrac{1}{2}(1 - \cos 2u)$ and substituting $u = \pi x/n$:

$\int_0^1 \sin^2\!\left(\frac{\pi x}{n}\right)dx = \frac{n}{2\pi}\int_0^{\pi/n}(1 - \cos 2u)\,du = \frac{n}{2\pi}\left[u - \frac{\sin 2u}{2}\right]_0^{\pi/n}$

$= \frac{n}{2\pi}\left(\frac{\pi}{n} - \frac{\sin(2\pi/n)}{2}\right) = \frac{1}{2} - \frac{n\sin(2\pi/n)}{4\pi}$

Expand $\sin(2\pi/n) = \dfrac{2\pi}{n} - \dfrac{(2\pi)^3}{6n^3} + \cdots$:

$\frac{n\sin(2\pi/n)}{4\pi} = \frac{n}{4\pi}\left(\frac{2\pi}{n} - \frac{8\pi^3}{6n^3} + \cdots\right) = \frac{1}{2} - \frac{\pi^2}{3n^2} + \cdots$

So

$\int_0^1 f(x)^2\,dx = \frac{1}{2} - \left(\frac{1}{2} - \frac{\pi^2}{3n^2} + \cdots\right) = \frac{\pi^2}{3n^2} + O\!\left(\frac{1}{n^4}\right)$

And $\frac{1}{3}\lambda^2 = \frac{1}{3}\left(\frac{\pi}{n} + O\!\left(\frac{1}{n^3}\right)\right)^2 = \frac{\pi^2}{3n^2} + O\!\left(\frac{1}{n^4}\right)$.

Therefore

$R^2 = \frac{\pi^2}{3n^2} - \frac{\pi^2}{3n^2} + O\!\left(\frac{1}{n^4}\right) = O\!\left(\frac{1}{n^4}\right) \to 0$

as $n \to \infty$, so $\lim_{n\to\infty}R = 0$.

Explanation. For large $n$, $\sin(\pi x/n) \approx \pi x/n$ over $[0,1]$, since $\pi x/n$ is small. But $\pi x/n$ is already a linear function of $x$, so the best linear approximation $\lambda x$ fits $f(x)$ almost perfectly and the residual error tends to zero.

---

Question 6

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

Problem

6 Show that $\sin \theta = \frac{2t}{1 + t^2}, \quad \cos \theta = \frac{1 - t^2}{1 + t^2}, \quad \frac{1 + \cos \theta}{\sin \theta} = \tan(\pi/2 - \theta/2),$ where $t = \tan(\theta/2)$.

Use the substitution $t = \tan(\theta/2)$ to show that, for $0 < \alpha < \pi/2$, $\int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \cos \alpha \sin \theta} \, d\theta = \frac{\alpha}{\sin \alpha},$ and deduce a similar result for $\int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \sin \alpha \cos \theta} \, d\theta.$

Model Solution

Part (a): Proving the three identities with $t = \tan(\theta/2)$.

For the first two identities, start from $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$. Divide numerator and denominator by $\cos^2(\theta/2)$:

$\sin\theta = \frac{2\sin(\theta/2)\cos(\theta/2)}{1} = \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos^2(\theta/2) + \sin^2(\theta/2)}$

Dividing top and bottom by $\cos^2(\theta/2)$:

$= \frac{2\tan(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{2t}{1 + t^2}. \qquad \text{(i)}$

Similarly for cosine, start from $\cos\theta = \cos^2(\theta/2) - \sin^2(\theta/2)$:

$\cos\theta = \frac{\cos^2(\theta/2) - \sin^2(\theta/2)}{\cos^2(\theta/2) + \sin^2(\theta/2)}$

Dividing top and bottom by $\cos^2(\theta/2)$:

$= \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - t^2}{1 + t^2}. \qquad \text{(ii)}$

For the third identity, use $\cos\theta = 2\cos^2(\theta/2) - 1$, so $1 + \cos\theta = 2\cos^2(\theta/2)$. Also $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$.

$\frac{1 + \cos\theta}{\sin\theta} = \frac{2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \frac{\cos(\theta/2)}{\sin(\theta/2)} = \cot(\theta/2)$

Since $\cot(\theta/2) = \tan(\pi/2 - \theta/2)$, we have

$\frac{1 + \cos\theta}{\sin\theta} = \tan(\pi/2 - \theta/2). \qquad \text{(iii)}$

Part (b): Evaluating $\displaystyle\int_0^{\pi/2}\frac{d\theta}{1 + \cos\alpha\,\sin\theta}$.

Use the substitution $t = \tan(\theta/2)$, so $d\theta = \dfrac{2}{1+t^2}\,dt$. The limits transform as: $\theta = 0 \Rightarrow t = 0$; $\theta = \pi/2 \Rightarrow t = \tan(\pi/4) = 1$.

Substituting $\sin\theta = \dfrac{2t}{1+t^2}$ from (i):

$\int_0^{\pi/2}\frac{d\theta}{1 + \cos\alpha\,\sin\theta} = \int_0^1 \frac{1}{1 + \cos\alpha\cdot\dfrac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt$

$= \int_0^1 \frac{2}{(1+t^2) + 2t\cos\alpha}\,dt = \int_0^1 \frac{2}{t^2 + 2t\cos\alpha + 1}\,dt$

Complete the square in the denominator:

$t^2 + 2t\cos\alpha + 1 = (t + \cos\alpha)^2 + 1 - \cos^2\alpha = (t + \cos\alpha)^2 + \sin^2\alpha$

So the integral becomes

$\int_0^1 \frac{2}{(t+\cos\alpha)^2 + \sin^2\alpha}\,dt$

Substitute $u = \dfrac{t + \cos\alpha}{\sin\alpha}$, so $du = \dfrac{dt}{\sin\alpha}$, i.e., $dt = \sin\alpha\,du$.

Limits: $t = 0 \Rightarrow u = \cot\alpha$; $t = 1 \Rightarrow u = \dfrac{1 + \cos\alpha}{\sin\alpha} = \cot(\alpha/2)$ by identity (iii).

$= \int_{\cot\alpha}^{\cot(\alpha/2)} \frac{2\sin\alpha}{\sin^2\alpha\,(u^2 + 1)}\,du = \frac{2}{\sin\alpha}\int_{\cot\alpha}^{\cot(\alpha/2)}\frac{du}{u^2 + 1}$

$= \frac{2}{\sin\alpha}\Big[\arctan u\Big]_{\cot\alpha}^{\cot(\alpha/2)} = \frac{2}{\sin\alpha}\left(\arctan(\cot(\alpha/2)) - \arctan(\cot\alpha)\right)$

Now use the identity: for $0 < x < \pi/2$, $\arctan(\cot x) = \arctan(\tan(\pi/2 - x)) = \pi/2 - x$.

$= \frac{2}{\sin\alpha}\left(\frac{\pi}{2} - \frac{\alpha}{2} - \frac{\pi}{2} + \alpha\right) = \frac{2}{\sin\alpha}\cdot\frac{\alpha}{2} = \frac{\alpha}{\sin\alpha}. \qquad \blacksquare$

Part (c): Deducing $\displaystyle\int_0^{\pi/2}\frac{d\theta}{1 + \sin\alpha\,\cos\theta}$.

Substitute $\phi = \pi/2 - \theta$ in the integral. Then $\cos\theta = \sin\phi$, $d\theta = -d\phi$, and the limits reverse: $\theta = 0 \to \phi = \pi/2$, $\theta = \pi/2 \to \phi = 0$.

$\int_0^{\pi/2}\frac{d\theta}{1 + \sin\alpha\,\cos\theta} = \int_0^{\pi/2}\frac{d\phi}{1 + \sin\alpha\,\sin\phi}$

This has exactly the form $\displaystyle\int_0^{\pi/2}\frac{d\phi}{1 + \cos\beta\,\sin\phi}$ with $\cos\beta = \sin\alpha$.

Since $0 < \alpha < \pi/2$, we have $\cos\beta = \sin\alpha = \cos(\pi/2 - \alpha)$, so $\beta = \pi/2 - \alpha$.

By the result of Part (b):

$\int_0^{\pi/2}\frac{d\phi}{1 + \sin\alpha\,\sin\phi} = \frac{\beta}{\sin\beta} = \frac{\pi/2 - \alpha}{\sin(\pi/2 - \alpha)} = \frac{\pi/2 - \alpha}{\cos\alpha}$

Therefore

$\int_0^{\pi/2}\frac{d\theta}{1 + \sin\alpha\,\cos\theta} = \frac{\pi/2 - \alpha}{\cos\alpha}. \qquad \blacksquare$

---

Question 7

Topic: 向量几何 Vector Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

7 The line $l$ has vector equation $\mathbf{r} = \lambda\mathbf{s}$, where $\mathbf{s} = (\cos \theta + \sqrt{3} \, ) \, \mathbf{i} + (\sqrt{2} \, \sin \theta) \, \mathbf{j} + (\cos \theta - \sqrt{3} \, ) \, \mathbf{k}$ and $\lambda$ is a scalar parameter. Find an expression for the angle between $l$ and the line $\mathbf{r} = \mu(a \, \mathbf{i} + b \, \mathbf{j} + c \, \mathbf{k})$. Show that there is a line $m$ through the origin such that, whatever the value of $\theta$, the acute angle between $l$ and $m$ is $\pi/6$.

A plane has equation $x - z = 4\sqrt{3}$. The line $l$ meets this plane at $P$. Show that, as $\theta$ varies, $P$ describes a circle, with its centre on $m$. Find the radius of this circle.

Model Solution

Part 1: Angle between $l$ and a general line through the origin.

The direction vector of $l$ is $\mathbf{s} = (\cos\theta + \sqrt{3})\,\mathbf{i} + \sqrt{2}\sin\theta\,\mathbf{j} + (\cos\theta - \sqrt{3})\,\mathbf{k}$. Let the second line have direction $\mathbf{t} = a\,\mathbf{i} + b\,\mathbf{j} + c\,\mathbf{k}$.

First, compute $|\mathbf{s}|$:

$|\mathbf{s}|^2 = (\cos\theta + \sqrt{3})^2 + 2\sin^2\theta + (\cos\theta - \sqrt{3})^2$

$= \cos^2\theta + 2\sqrt{3}\cos\theta + 3 + 2\sin^2\theta + \cos^2\theta - 2\sqrt{3}\cos\theta + 3$

$= 2\cos^2\theta + 2\sin^2\theta + 6 = 2 + 6 = 8$

So $|\mathbf{s}| = 2\sqrt{2}$.

The acute angle $\phi$ between the two lines is given by

$\cos\phi = \frac{|\mathbf{s} \cdot \mathbf{t}|}{|\mathbf{s}||\mathbf{t}|} = \frac{|a(\cos\theta + \sqrt{3}) + b\sqrt{2}\sin\theta + c(\cos\theta - \sqrt{3})|}{2\sqrt{2}\sqrt{a^2 + b^2 + c^2}}$

$= \frac{|(a + c)\cos\theta + \sqrt{2}\,b\sin\theta + \sqrt{3}(a - c)|}{2\sqrt{2}\sqrt{a^2 + b^2 + c^2}} \qquad \text{(*)}$

Part 2: Showing line $m$ exists with constant angle $\pi/6$.

We need a fixed direction $\mathbf{m} = p\,\mathbf{i} + q\,\mathbf{j} + r\,\mathbf{k}$ such that the angle between $l$ and $m$ is $\pi/6$ for all $\theta$. From (*), setting $a = p, b = q, c = r$:

$\mathbf{s} \cdot \mathbf{m} = (p + r)\cos\theta + \sqrt{2}\,q\sin\theta + \sqrt{3}(p - r)$

For this dot product to be independent of $\theta$, the coefficients of $\cos\theta$ and $\sin\theta$ must vanish:

$p + r = 0 \quad \text{and} \quad q = 0$

So $r = -p$ and $q = 0$, giving $\mathbf{m} = p\,\mathbf{i} - p\,\mathbf{k} = p(\mathbf{i} - \mathbf{k})$.

With this choice, $\mathbf{s} \cdot \mathbf{m} = \sqrt{3}(p - (-p)) = 2\sqrt{3}\,p$, and $|\mathbf{m}| = |p|\sqrt{2}$. The angle satisfies:

$\cos\alpha = \frac{|2\sqrt{3}\,p|}{2\sqrt{2} \cdot |p|\sqrt{2}} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$

Since $\cos\alpha = \sqrt{3}/2$, we get $\alpha = \pi/6$ for every value of $\theta$. The line $m$ through the origin has direction $\mathbf{i} - \mathbf{k}$.

Part 3: Locus of $P$ and the circle.

Points on $l$ have coordinates $x = \lambda(\cos\theta + \sqrt{3})$, $y = \lambda\sqrt{2}\sin\theta$, $z = \lambda(\cos\theta - \sqrt{3})$. Substituting into the plane equation $x - z = 4\sqrt{3}$:

$\lambda(\cos\theta + \sqrt{3}) - \lambda(\cos\theta - \sqrt{3}) = 4\sqrt{3}$

$2\sqrt{3}\,\lambda = 4\sqrt{3} \implies \lambda = 2$

So $P$ has coordinates:

$P = \big(2\cos\theta + 2\sqrt{3},\; 2\sqrt{2}\sin\theta,\; 2\cos\theta - 2\sqrt{3}\big)$

The centre of the circle lies on both $m$ and the plane. A general point on $m$ is $(t, 0, -t)$; substituting into $x - z = 4\sqrt{3}$ gives $t - (-t) = 4\sqrt{3}$, so $t = 2\sqrt{3}$. The centre is $C = (2\sqrt{3}, 0, -2\sqrt{3})$.

The radius is $|CP|$:

$|CP|^2 = (2\cos\theta + 2\sqrt{3} - 2\sqrt{3})^2 + (2\sqrt{2}\sin\theta - 0)^2 + (2\cos\theta - 2\sqrt{3} - (-2\sqrt{3}))^2$

$= (2\cos\theta)^2 + (2\sqrt{2}\sin\theta)^2 + (2\cos\theta)^2$

$= 4\cos^2\theta + 8\sin^2\theta + 4\cos^2\theta = 8\cos^2\theta + 8\sin^2\theta = 8$

So $|CP| = 2\sqrt{2}$, which is independent of $\theta$. As $\theta$ varies, $P$ traces out a circle of radius $2\sqrt{2}$ with centre on $m$.

---

Question 8

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

Problem

8 (i) Let $y$ be the solution of the differential equation $\frac{dy}{dx} + 4x \, e^{-x^2}(y + 3)^{\frac{1}{2}} = 0 \quad (x \geqslant 0),$ that satisfies the condition $y = 6$ when $x = 0$. Find $y$ in terms of $x$ and show that $y \to 1$ as $x \to \infty$.

(ii) Let $y$ be any solution of the differential equation $\frac{dy}{dx} - x \, e^{6x^2}(y + 3)^{1-k} = 0 \quad (x \geqslant 0).$ Find a value of $k$ such that, as $x \to \infty$, $e^{-3x^2}y$ tends to a finite non-zero limit, which you should determine.

Model Solution

Part (i): Solving the ODE and finding the limit.

The equation is:

$\frac{dy}{dx} = -4x\,e^{-x^2}(y + 3)^{1/2}$

This is separable. Divide both sides by $(y + 3)^{1/2}$:

$(y + 3)^{-1/2}\,\frac{dy}{dx} = -4x\,e^{-x^2}$

Integrate both sides with respect to $x$:

$\int (y + 3)^{-1/2}\,dy = \int -4x\,e^{-x^2}\,dx$

The left side: let $u = y + 3$, so $du = dy$:

$\int u^{-1/2}\,du = 2u^{1/2} = 2(y + 3)^{1/2}$

The right side: let $v = -x^2$, so $dv = -2x\,dx$, hence $-4x\,dx = 2\,dv$:

$\int 2e^v\,dv = 2e^v = 2e^{-x^2}$

Combining:

$2(y + 3)^{1/2} = 2e^{-x^2} + C$

Apply the initial condition $y = 6$ when $x = 0$:

$2(6 + 3)^{1/2} = 2e^0 + C \implies 2 \cdot 3 = 2 + C \implies C = 4$

So:

$2(y + 3)^{1/2} = 2e^{-x^2} + 4$

$(y + 3)^{1/2} = e^{-x^2} + 2$

Squaring both sides:

$y + 3 = (e^{-x^2} + 2)^2 = e^{-2x^2} + 4e^{-x^2} + 4$

$y = e^{-2x^2} + 4e^{-x^2} + 1$

As $x \to \infty$, both $e^{-2x^2} \to 0$ and $e^{-x^2} \to 0$, so $y \to 0 + 0 + 1 = 1$.

Part (ii): Finding $k$ for a finite non-zero limit of $e^{-3x^2}y$.

The equation is:

$\frac{dy}{dx} = x\,e^{6x^2}(y + 3)^{1-k}$

Separate variables:

$(y + 3)^{k-1}\,dy = x\,e^{6x^2}\,dx$

Integrate both sides. The right side: let $w = 6x^2$, so $dw = 12x\,dx$:

$\int x\,e^{6x^2}\,dx = \frac{1}{12}\int e^w\,dw = \frac{1}{12}e^{6x^2}$

Case $k \neq 0$: The left side integrates to $\frac{(y+3)^k}{k}$, giving:

$\frac{(y + 3)^k}{k} = \frac{1}{12}e^{6x^2} + C_1$

$\implies (y + 3)^k = \frac{k}{12}e^{6x^2} + C \qquad \text{where } C = kC_1$

For large $x$, if $C \geq 0$:

$(y + 3)^k \approx \frac{k}{12}e^{6x^2}$

$y + 3 \approx \left(\frac{k}{12}\right)^{1/k} e^{6x^2/k}$

$y \approx \left(\frac{k}{12}\right)^{1/k} e^{6x^2/k}$

Now examine $e^{-3x^2}y$:

$e^{-3x^2}y \approx \left(\frac{k}{12}\right)^{1/k} e^{(6/k - 3)x^2}$

For this to tend to a finite non-zero limit as $x \to \infty$, the exponent of $e$ must be exactly zero:

$\frac{6}{k} - 3 = 0 \implies k = 2$

Verification and determination of the limit. With $k = 2$:

$(y + 3)^2 = \frac{1}{6}e^{6x^2} + C$

$y + 3 = \sqrt{\frac{1}{6}e^{6x^2} + C} \qquad \text{(taking positive root since } y + 3 > 0\text{)}$

As $x \to \infty$:

$y + 3 = \frac{e^{3x^2}}{\sqrt{6}}\sqrt{1 + 6Ce^{-6x^2}} \to \frac{e^{3x^2}}{\sqrt{6}}$

Therefore:

$e^{-3x^2}(y + 3) \to \frac{1}{\sqrt{6}} \implies e^{-3x^2}y = e^{-3x^2}(y+3) - 3e^{-3x^2} \to \frac{1}{\sqrt{6}} - 0 = \frac{1}{\sqrt{6}}$

The value of $k$ is $2$, and the limit is $\dfrac{1}{\sqrt{6}}$.