STEP2 2021 -- Pure Mathematics

Exam: STEP2  |  Year: 2021  |  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	三角学 (Trigonometry)	Challenging	积化和差公式,三角方程求解,多角恒等式变换,因式分解
2	代数与函数 (Algebra & Functions)	Challenging	牛顿恒等式,对称多项式,韦达定理,构造多项式方程
3	数论 (Number Theory)	Standard	取整函数性质,小数部分性质,整数与小数分离法,方程组消元
4	微积分 (Calculus)	Challenging	反函数求法,导数与驻点,超越方程求解,对数变换
5	微分方程 (Differential Equations)	Challenging	换元法解ODE,切线方程建模,分离变量法,函数图像分析
6	几何与向量 (Geometry & Vectors)	Challenging	余弦定理,弧长公式,小量近似(sinα≈α),几何分析
7	矩阵与线性代数 (Matrices & Linear Algebra)	Challenging	旋转矩阵性质,矩阵特征方程,Cayley-Hamilton定理,行列式计算
8	微积分 (Calculus)	Hard	分部积分,递推关系建立,积分不等式估计,极限证明

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Question 1

Topic: 三角学 (Trigonometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Prove, from the identities for $\cos(A \pm B)$, that $\cos a \cos 3a \equiv \frac{1}{2}(\cos 4a + \cos 2a)$. Find a similar identity for $\sin a \cos 3a$.

(i) Solve the equation

$4 \cos x \cos 2x \cos 3x = 1$

for $0 \leqslant x \leqslant \pi$.

(ii) Prove that if

$\tan x = \tan 2x \tan 3x \tan 4x \qquad \text{(*)}$

then $\cos 6x = \frac{1}{2}$ or $\sin 4x = 0$.

Hence determine the solutions of equation $(\star)$ with $0 \leqslant x \leqslant \pi$.

Hint

$\cos(3a + a) \equiv \cos 3a \cos a - \sin 3a \sin a$ $\cos(3a - a) \equiv \cos 3a \cos a + \sin 3a \sin a$ M1 $\cos 4a + \cos 2a \equiv 2 \cos 3a \cos a$

$\cos a \cos 3a \equiv \frac{1}{2}(\cos 4a + \cos 2a) \quad \mathbf{AG}$ A1

$\sin(3a + a) \equiv \sin 3a \cos a + \cos 3a \sin a$ $\sin(3a - a) \equiv \sin 3a \cos a - \cos 3a \sin a$ $\sin 4a - \sin 2a \equiv 2 \cos 3a \sin a$ $\sin a \cos 3a \equiv \frac{1}{2}(\sin 4a - \sin 2a)$ B1

(i) $2 \cos 2x (2 \cos x \cos 3x) = 1$ $2 \cos 2x (\cos 4x + \cos 2x) = 1$ M1 $2 \cos 2x (2 \cos^2 2x + \cos 2x - 1) = 1$ M1 $4 \cos^3 2x + 2 \cos^2 2x - 2 \cos 2x - 1 = 0$ $(2 \cos^2 2x - 1)(2 \cos 2x + 1) = 0$ M1 A1

Either $\cos^2 2x = \frac{1}{2}$: $2x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ $x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$ A1

Or $\cos 2x = -\frac{1}{2}$: $2x = \frac{2\pi}{3}, \frac{4\pi}{3}$ $x = \frac{\pi}{3}, \frac{2\pi}{3}$ A1

Therefore: $x = \frac{\pi}{8}, \frac{\pi}{3}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{2\pi}{3}, \frac{7\pi}{8}$

(ii) $2 \cos x \sin 3x \equiv \sin 4x + \sin 2x$ B1

$\tan x = \tan 2x \tan 3x \tan 4x$ M1 $\sin x \cos 2x \cos 3x \cos 4x = \cos x \sin 2x \sin 3x \sin 4x$ $(2 \sin x \cos 3x) \cos 2x \cos 4x = (2 \cos x \sin 3x) \sin 2x \sin 4x$ M1 $(\sin 4x - \sin 2x) \cos 2x \cos 4x = (\sin 4x + \sin 2x) \sin 2x \sin 4x$ $\sin 4x (\cos 2x \cos 4x - \sin 2x \sin 4x) = \sin 2x (\cos 2x \cos 4x + \sin 2x \sin 4x)$ $\sin 4x \cos 6x = \sin 2x \cos 2x$ M1 $\sin 4x \cos 6x = \frac{1}{2} \sin 4x$ M1 $\sin 4x (2 \cos 6x - 1) = 0$ M1 Therefore $\cos 6x = \frac{1}{2}$ or $\sin 4x = 0$. AG A1

$\cos 6x = \frac{1}{2}$: $6x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}$ $x = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}$ A1

$\sin 4x = 0$: $4x = 0, \pi, 2\pi, 3\pi, 4\pi$ $x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$ A1

$\tan x$ is undefined at $x = \frac{\pi}{2}$ B1

$\tan 2x$ is undefined at $x = \frac{\pi}{4}, \frac{3\pi}{4}$ B1

So these are not solutions of the equation. $x = 0, \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \pi$

Model Solution

Proof of the identity

From the addition and subtraction formulas:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$ $\cos(A - B) = \cos A \cos B + \sin A \sin B$

Adding these two identities with $A = 3a$ and $B = a$:

$\cos 4a + \cos 2a = 2 \cos 3a \cos a$

Dividing both sides by 2:

$\cos a \cos 3a \equiv \frac{1}{2}(\cos 4a + \cos 2a) \qquad \blacksquare$

Similar identity for $\sin a \cos 3a$

From the addition and subtraction formulas for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$ $\sin(A - B) = \sin A \cos B - \cos A \sin B$

Subtracting the second from the first with $A = 3a$ and $B = a$:

$\sin 4a - \sin 2a = 2 \cos 3a \sin a$

Therefore:

$\sin a \cos 3a \equiv \frac{1}{2}(\sin 4a - \sin 2a)$

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Part (i)

We solve $4 \cos x \cos 2x \cos 3x = 1$ for $0 \leqslant x \leqslant \pi$.

Rearranging:

$2 \cos 2x \cdot (2 \cos x \cos 3x) = 1$

Using the identity just proved, $2 \cos x \cos 3x = \cos 4x + \cos 2x$:

$2 \cos 2x (\cos 4x + \cos 2x) = 1$

Now use the double-angle identity $\cos 4x = 2 \cos^2 2x - 1$:

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

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

Expanding:

$4 \cos^3 2x + 2 \cos^2 2x - 2 \cos 2x = 1$

$4 \cos^3 2x + 2 \cos^2 2x - 2 \cos 2x - 1 = 0$

We factor this cubic in $\cos 2x$. Trying to factor as a product:

$(2 \cos^2 2x - 1)(2 \cos 2x + 1) = 0$

We can verify: $4\cos^3 2x + 2\cos^2 2x - 2\cos 2x - 1$, which matches.

Case 1: $2 \cos^2 2x - 1 = 0$, so $\cos^2 2x = \frac{1}{2}$, giving $\cos 2x = \pm \frac{1}{\sqrt{2}}$.

Since $0 \leqslant x \leqslant \pi$, we have $0 \leqslant 2x \leqslant 2\pi$.

For $\cos 2x = \frac{1}{\sqrt{2}}$: $2x = \frac{\pi}{4}, \frac{7\pi}{4}$, so $x = \frac{\pi}{8}, \frac{7\pi}{8}$.

For $\cos 2x = -\frac{1}{\sqrt{2}}$: $2x = \frac{3\pi}{4}, \frac{5\pi}{4}$, so $x = \frac{3\pi}{8}, \frac{5\pi}{8}$.

Case 2: $2 \cos 2x + 1 = 0$, so $\cos 2x = -\frac{1}{2}$.

$2x = \frac{2\pi}{3}, \frac{4\pi}{3}$, so $x = \frac{\pi}{3}, \frac{2\pi}{3}$.

Solutions:

$x = \frac{\pi}{8}, \quad \frac{\pi}{3}, \quad \frac{3\pi}{8}, \quad \frac{5\pi}{8}, \quad \frac{2\pi}{3}, \quad \frac{7\pi}{8}$

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Part (ii)

We first prove a useful identity. From the product-to-sum formulas:

$2 \cos x \sin 3x = \sin 4x + \sin 2x$

This follows from $\sin(A+B) + \sin(A-B) = 2\cos A \sin B$ with $A = x, B = 3x$.

Now, given $\tan x = \tan 2x \tan 3x \tan 4x$, rewrite in terms of sine and cosine:

$\frac{\sin x}{\cos x} = \frac{\sin 2x \sin 3x \sin 4x}{\cos 2x \cos 3x \cos 4x}$

Cross-multiplying:

$\sin x \cos 2x \cos 3x \cos 4x = \cos x \sin 2x \sin 3x \sin 4x$

Multiply both sides by 2:

$(2 \sin x \cos 3x) \cos 2x \cos 4x = (2 \cos x \sin 3x) \sin 2x \sin 4x$

Using the identities $2 \sin x \cos 3x = \sin 4x - \sin 2x$ and $2 \cos x \sin 3x = \sin 4x + \sin 2x$:

$(\sin 4x - \sin 2x) \cos 2x \cos 4x = (\sin 4x + \sin 2x) \sin 2x \sin 4x$

Expanding both sides:

$\sin 4x \cos 2x \cos 4x - \sin 2x \cos 2x \cos 4x = \sin 4x \sin 2x \sin 4x + \sin 2x \sin 2x \sin 4x$

Rearranging — collect terms with $\sin 4x$ as a factor on the left and terms with $\sin 2x$ as a factor on the right:

$\sin 4x (\cos 2x \cos 4x - \sin 2x \sin 4x) = \sin 2x (\cos 2x \cos 4x + \sin 2x \sin 4x)$

Recognizing the cosine addition formulas:

$\cos 2x \cos 4x - \sin 2x \sin 4x = \cos(2x + 4x) = \cos 6x$ $\cos 2x \cos 4x + \sin 2x \sin 4x = \cos(4x - 2x) = \cos 2x$

So:

$\sin 4x \cos 6x = \sin 2x \cos 2x$

Using the double-angle identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$:

$\sin 4x \cos 6x = \frac{1}{2} \sin 4x$

$\sin 4x (2 \cos 6x - 1) = 0$

Therefore either $\sin 4x = 0$ or $\cos 6x = \frac{1}{2}$. $\qquad \blacksquare$

Finding the solutions for $0 \leqslant x \leqslant \pi$:

Case 1: $\cos 6x = \frac{1}{2}$

Since $0 \leqslant x \leqslant \pi$, we have $0 \leqslant 6x \leqslant 6\pi$.

$6x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}$

$x = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}$

Case 2: $\sin 4x = 0$

Since $0 \leqslant x \leqslant \pi$, we have $0 \leqslant 4x \leqslant 4\pi$.

$4x = 0, \pi, 2\pi, 3\pi, 4\pi$

$x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$

Checking for undefined tangents:

The original equation involves $\tan x$, $\tan 2x$, $\tan 3x$, and $\tan 4x$, so we must exclude values where any of these is undefined.

• $\tan x$ undefined when $x = \frac{\pi}{2}$
• $\tan 2x$ undefined when $2x = \frac{\pi}{2}, \frac{3\pi}{2}$, i.e., $x = \frac{\pi}{4}, \frac{3\pi}{4}$
• $\tan 4x$ undefined when $4x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$, i.e., $x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$ (none of which are in our solution set)

From Case 2, $x = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$ must be rejected. The values $x = 0$ and $x = \pi$ give $\tan x = \tan 2x = \tan 3x = \tan 4x = 0$, which satisfy the equation.

Final solutions:

$x = 0, \quad \frac{\pi}{18}, \quad \frac{5\pi}{18}, \quad \frac{7\pi}{18}, \quad \frac{11\pi}{18}, \quad \frac{13\pi}{18}, \quad \frac{17\pi}{18}, \quad \pi$

Examiner Notes

无官方评述。易错点：(1) 积化和差方向选择不当导致展开后更复杂；(2) 求解tan等式时忽略分母为零的情况（sin4x=0的解）；(3) 最终求解cos6x=1/2时遗漏区间内的解。

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Question 2

Topic: 代数与函数 (Algebra & Functions)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 In this question, the numbers $a$, $b$ and $c$ may be complex.

(i) Let $p$, $q$ and $r$ be real numbers. Given that there are numbers $a$ and $b$ such that $a + b = p, \quad a^2 + b^2 = q \text{ and } a^3 + b^3 = r, \qquad \text{(*)}$ show that $3pq - p^3 = 2r$.

(ii) Conversely, you are given that the real numbers $p$, $q$ and $r$ satisfy $3pq - p^3 = 2r$. By considering the equation $2x^2 - 2px + (p^2 - q) = 0$, show that there exist numbers $a$ and $b$ such that the three equations (*) hold.

(iii) Let $s$, $t$, $u$ and $v$ be real numbers. Given that there are distinct numbers $a$, $b$ and $c$ such that $a + b + c = s, \quad a^2 + b^2 + c^2 = t, \quad a^3 + b^3 + c^3 = u \text{ and } abc = v,$ show, using part (i), that $c$ is a root of the equation $6x^3 - 6sx^2 + 3(s^2 - t)x + 3st - s^3 - 2u = 0$ and write down the other two roots.

Deduce that $s^3 - 3st + 2u = 6v$.

(iv) Find numbers $a$, $b$ and $c$ such that $a + b + c = 3, \quad a^2 + b^2 + c^2 = 1, \quad a^3 + b^3 + c^3 = -3 \text{ and } abc = 2, \qquad \text{(**)}$ and verify that your solution satisfies the four equations (**).

Hint

(i) $3pq - p^3 = 3(a + b)(a^2 + b^2) - (a + b)^3$ M1 $= 2a^3 + 2b^3$ $= 2r \quad \mathbf{AG}$ A1

(ii) $2x^2 - 2px + (p^2 - q) = 0$ M1 The roots of the equation $a$ and $b$ satisfy: $a + b = p$ B1 $2ab = p^2 - q$ B1 $a^2 + b^2 = (a + b)^2 - 2ab$ B1 $= p^2 - (p^2 - q) = q$ $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ $= p^3 - \frac{3}{2}(p^2 - q)p$ $= \frac{1}{2}(3pq - p^3) = r$ B1 So the three equations hold. E1

(iii) $a + b = s - c (= p)$ $a^2 + b^2 = t - c^2 (= q)$ $a^3 + b^3 = u - c^3 (= r)$ M1 By part (i): $3(s - c)(t - c^2) - (s - c)^3 = 2(u - c^3)$ $3st - 3ct - 3c^2s + 3c^3 - s^3 + 3cs^2 - 3c^2s + c^3 = 2u - 2c^3$ M1 $6c^3 - 6sc^2 + 3(s^2 - t)c + 3st - s^3 - 2u = 0$ A1 Therefore $c$ is a root of the equation $6x^3 - 6sx^2 + 3(s^2 - t)x + 3st - s^3 - 2u = 0 \quad \mathbf{AG}$ E1 The other roots are $a$ and $b$. B1

The constant term is $-6 \times$ the product of the roots: M1 $-6abc = 3st - s^3 - 2u$ $s^3 - 3st + 2u = 6v \quad \mathbf{AG}$ A1

(iv) By (iii) $a, b$ and $c$ are the roots of M1 $6x^3 - 18x^2 + 24x - 12 = 0$ A1 $6(x - 1)(x^2 - 2x + 2) = 0$ M1 $1, 1 + i, 1 - i$ A1 $1 + (1 + i) + (1 - i) = 3$ $1^2 + (1 + i)^2 + (1 - i)^2 = 1 + (1 + 2i - 1) + (1 - 2i - 1) = 1$ $1^3 + (1 + i)^3 + (1 - i)^3 = 1 + (-2 + 2i) + (-2 - 2i) = -3$ $1(1 + i)(1 - i) = 2$ B1

Model Solution

Part (i)

Given $a + b = p$, $a^2 + b^2 = q$, $a^3 + b^3 = r$, we show $3pq - p^3 = 2r$.

$3pq - p^3 = 3(a + b)(a^2 + b^2) - (a + b)^3$

Expanding $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$3pq - p^3 = 3(a^3 + ab^2 + a^2b + b^3) - (a^3 + 3a^2b + 3ab^2 + b^3)$

$= 3a^3 + 3ab^2 + 3a^2b + 3b^3 - a^3 - 3a^2b - 3ab^2 - b^3$

$= 2a^3 + 2b^3 = 2r \qquad \blacksquare$

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Part (ii)

Given that $3pq - p^3 = 2r$, we show numbers $a, b$ exist satisfying the three equations.

Consider the quadratic equation $2x^2 - 2px + (p^2 - q) = 0$. By the quadratic formula, the roots are:

$x = \frac{2p \pm \sqrt{4p^2 - 8(p^2 - q)}}{4} = \frac{p \pm \sqrt{2q - p^2}}{2}$

These roots exist (as complex numbers if necessary). Let $a$ and $b$ be the roots.

Checking the first equation: By Vieta’s formulas, $a + b = \frac{2p}{2} = p$. $\checkmark$

Checking the second equation: Also by Vieta’s, $ab = \frac{p^2 - q}{2}$, so:

$a^2 + b^2 = (a + b)^2 - 2ab = p^2 - (p^2 - q) = q \qquad \checkmark$

Checking the third equation:

$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = p^3 - 3 \cdot \frac{p^2 - q}{2} \cdot p = p^3 - \frac{3p(p^2 - q)}{2}$

$= \frac{2p^3 - 3p^3 + 3pq}{2} = \frac{3pq - p^3}{2} = \frac{2r}{2} = r \qquad \checkmark$

So the three equations (*) hold. $\qquad \blacksquare$

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Part (iii)

Given $a + b + c = s$, $a^2 + b^2 + c^2 = t$, $a^3 + b^3 + c^3 = u$, $abc = v$, where $a, b, c$ are distinct.

Define $p = a + b = s - c$, $\quad q = a^2 + b^2 = t - c^2$, $\quad r = a^3 + b^3 = u - c^3$.

By part (i), the relation $3pq - p^3 = 2r$ must hold. Substituting:

$3(s - c)(t - c^2) - (s - c)^3 = 2(u - c^3)$

Expanding $3(s - c)(t - c^2)$:

$3(st - sc^2 - ct + c^3) = 3st - 3sc^2 - 3ct + 3c^3$

Expanding $(s - c)^3$:

$s^3 - 3s^2c + 3sc^2 - c^3$

Substituting:

$3st - 3sc^2 - 3ct + 3c^3 - s^3 + 3s^2c - 3sc^2 + c^3 = 2u - 2c^3$

Collecting terms:

$6c^3 - 6sc^2 + 3(s^2 - t)c + 3st - s^3 - 2u = 0$

So $c$ is a root of:

$6x^3 - 6sx^2 + 3(s^2 - t)x + 3st - s^3 - 2u = 0 \qquad \blacksquare$

Since the relation $3pq - p^3 = 2r$ is symmetric in $a$ and $b$ (it holds whenever we pair up any two of the three numbers), by the same argument with $a$ or $b$ in the role of the “third” variable, the other two roots are $a$ and $b$.

Deducing $s^3 - 3st + 2u = 6v$:

By Vieta’s formulas for the cubic $6x^3 - 6sx^2 + 3(s^2-t)x + (3st - s^3 - 2u) = 0$, the product of the roots equals:

$abc = -\frac{3st - s^3 - 2u}{6}$

Therefore:

$6abc = s^3 - 3st + 2u$

Since $abc = v$:

$s^3 - 3st + 2u = 6v \qquad \blacksquare$

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Part (iv)

We need $a + b + c = 3$, $a^2 + b^2 + c^2 = 1$, $a^3 + b^3 + c^3 = -3$, $abc = 2$.

Here $s = 3$, $t = 1$, $u = -3$, $v = 2$.

First verify the consistency condition from part (iii):

$s^3 - 3st + 2u = 27 - 9 - 6 = 12 = 6(2) = 6v \qquad \checkmark$

By part (iii), $a, b, c$ are roots of:

$6x^3 - 18x^2 + 3(9 - 1)x + (9 - 27 + 6) = 0$

$6x^3 - 18x^2 + 24x - 12 = 0$

$x^3 - 3x^2 + 4x - 2 = 0$

Testing $x = 1$: $1 - 3 + 4 - 2 = 0$. $\checkmark$

Factoring out $(x - 1)$:

$x^3 - 3x^2 + 4x - 2 = (x - 1)(x^2 - 2x + 2) = 0$

The quadratic $x^2 - 2x + 2 = 0$ gives $x = 1 \pm i$.

Therefore $a, b, c$ are $1, 1 + i, 1 - i$ (in any order).

Verification:

$a + b + c = 1 + (1 + i) + (1 - i) = 3 \qquad \checkmark$

$a^2 + b^2 + c^2 = 1 + (1 + i)^2 + (1 - i)^2 = 1 + (2i) + (-2i) = 1 \qquad \checkmark$

$a^3 + b^3 + c^3 = 1 + (1 + i)^3 + (1 - i)^3$

$(1+i)^3 = (1+i)^2(1+i) = 2i(1+i) = -2 + 2i$

$(1-i)^3 = (1-i)^2(1-i) = -2i(1-i) = -2 - 2i$

$a^3 + b^3 + c^3 = 1 + (-2 + 2i) + (-2 - 2i) = -3 \qquad \checkmark$

$abc = 1 \cdot (1+i)(1-i) = 1 \cdot 2 = 2 \qquad \checkmark$

Examiner Notes

无官方评述。易错点：(1) part(ii)中不理解从方程2x²-2px+(p²-q)=0构造a,b的思路；(2) part(iii)推导c满足的三次方程时计算量大容易出错；(3) part(iv)验证解时漏验某个方程。

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Question 3

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

Problem

3 In this question, $x$, $y$ and $z$ are real numbers.

Let $\lfloor x \rfloor$ denote the largest integer that satisfies $\lfloor x \rfloor \leqslant x$ and let $\{x\}$ denote the fractional part of $x$, so that $x = \lfloor x \rfloor + \{x\}$ and $0 \leqslant \{x\} < 1$. For example, if $x = 4.2$, then $\lfloor x \rfloor = 4$ and $\{x\} = 0.2$ and if $x = -4.2$, then $\lfloor x \rfloor = -5$ and $\{x\} = 0.8$.

(i) Solve the simultaneous equations

$\begin{cases} \lfloor x \rfloor + \{y\} = 4.9, \\ \{x\} + \lfloor y \rfloor = -1.4. \end{cases}$

(ii) Given that $x$, $y$ and $z$ satisfy the simultaneous equations

$\begin{cases} x + \lfloor y \rfloor + \{z\} = 3.9, \\ \{x\} + y + \lfloor z \rfloor = 5.3, \\ \lfloor x \rfloor + \{y\} + z = 5, \end{cases}$

show that $\{y\} + \lfloor z \rfloor = 3.2$ and solve the equations.

(iii) Solve the simultaneous equations

$\begin{cases} x + 2\lfloor y \rfloor + \{z\} = 3.9, \\ \{x\} + 2y + \lfloor z \rfloor = 5.3, \\ \lfloor x \rfloor + 2\{y\} + z = 5. \end{cases}$

Hint

(i) From the 1ˢᵗ eqn: $\lfloor x \rfloor = 4$ and $\{y\} = 0.9$ B1 From the 2ⁿᵈ eqn: $\{x\} = 0.6$ and $\lfloor y \rfloor = -2$ B1 Clear use of $x = \lfloor x \rfloor + \{x\}$ etc. M1 Solution is $x = 4.6, y = -1.1$ A1 NB: for candidates scoring none of the above marks, allow a B1 for adding both eqns. to obtain $x + y = 3.5$.

(ii) ② + ③ - ① M1 $\Rightarrow y + \{y\} - \lfloor y \rfloor + z + \lfloor z \rfloor - \{z\} = 6.4$ $\Rightarrow 2\{y\} + 2\lfloor z \rfloor = 6.4$ M1 $\Rightarrow \{y\} + \lfloor z \rfloor = 3.2$ AG or $\{x\} + \lfloor y \rfloor = 2.1$ or $\lfloor x \rfloor + \{z\} = 1.8$ A1 Similar attempts at ① + ② - ③ $\Rightarrow \{x\} + \lfloor y \rfloor = 2.1$ M1 and ① + ③ - ② $\Rightarrow \lfloor x \rfloor + \{z\} = 1.8$ The remaining two 2-variable eqns. correct A1 $\Rightarrow \{y\} = 0.2$ and $\lfloor z \rfloor = 3$ B1 Also (respectively) $\{x\} = 0.1$ and $\lfloor y \rfloor = 2$ B1 and $\lfloor x \rfloor = 1$ and $\{z\} = 0.8$ Solution is $x = 1.1, y = 2.2, z = 3.8$ A1

(iii) From ② + ③ - ①, we now get $2\{y\} + \lfloor z \rfloor = 3.2$ B1 From ① + ③ - ②, we still get $\lfloor x \rfloor + \{z\} = 1.8$ B1 From ① + ② - ③, we now get $\{x\} + 2\lfloor y \rfloor = 2.1$ B1

First solution follows immediately from (ii): namely, B1 $x = 1.1, y = 1.1, z = 3.8$

For clear evidence that the second possibility exists M1 namely: $2\{y\} + \lfloor z \rfloor = 3.2 \Rightarrow \{y\} = 0.6$ and $\lfloor z \rfloor = 2$ A1 and $\{x\} + 2\lfloor y \rfloor = 2.1 \Rightarrow \{x\} = 0.1$ and $\lfloor y \rfloor = 1$ A1 NB $\lfloor x \rfloor = 1$ and $\{z\} = 0.8$ follows as before

Second solution is $x = 1.1, y = 1.6, z = 2.8$ A1

Model Solution

Part (i)

We solve the simultaneous equations:

$\lfloor x \rfloor + \{y\} = 4.9 \qquad \text{...(1)}$ $\{x\} + \lfloor y \rfloor = -1.4 \qquad \text{...(2)}$

Key property: $\lfloor x \rfloor$ is an integer and $0 \leqslant \{x\} < 1$.

From equation (1): $\lfloor x \rfloor$ is an integer and $\{y\}$ satisfies $0 \leqslant \{y\} < 1$. Since their sum is $4.9$, the integer part of $4.9$ must come from $\lfloor x \rfloor$ and the fractional part from $\{y\}$. Therefore:

$\lfloor x \rfloor = 4, \qquad \{y\} = 0.9$

From equation (2): Similarly, $\lfloor y \rfloor$ is an integer and $\{x\}$ satisfies $0 \leqslant \{x\} < 1$. Since their sum is $-1.4$, we need $\lfloor y \rfloor$ to be the integer part and $\{x\}$ the fractional part. Writing $-1.4 = -2 + 0.6$:

$\lfloor y \rfloor = -2, \qquad \{x\} = 0.6$

Reconstructing $x$ and $y$:

$x = \lfloor x \rfloor + \{x\} = 4 + 0.6 = 4.6$ $y = \lfloor y \rfloor + \{y\} = -2 + 0.9 = -1.1$

Solution: $x = 4.6, \quad y = -1.1$

---

Part (ii)

We have the system:

$x + \lfloor y \rfloor + \{z\} = 3.9 \qquad \text{...(1)}$ $\{x\} + y + \lfloor z \rfloor = 5.3 \qquad \text{...(2)}$ $\lfloor x \rfloor + \{y\} + z = 5 \qquad \text{...(3)}$

Step 1: Compute (2) + (3) - (1):

$(\{x\} + y + \lfloor z \rfloor) + (\lfloor x \rfloor + \{y\} + z) - (x + \lfloor y \rfloor + \{z\}) = 5.3 + 5 - 3.9$

Using $x = \lfloor x \rfloor + \{x\}$, so $\{x\} + \lfloor x \rfloor - x = 0$, and similarly for $y$ and $z$:

$(\{x\} + \lfloor x \rfloor - x) + (y + \{y\} - \lfloor y \rfloor) + (\lfloor z \rfloor + z - \{z\}) = 6.4$

$0 + 2\{y\} + 2\lfloor z \rfloor - (y - y) - \ldots$

Let me be more careful. We have $\{x\} + \lfloor x \rfloor = x$, so $\{x\} + \lfloor x \rfloor - x = 0$.

For the $y$ terms: $y + \{y\} - \lfloor y \rfloor$. Since $y = \lfloor y \rfloor + \{y\}$, this becomes $\lfloor y \rfloor + \{y\} + \{y\} - \lfloor y \rfloor = 2\{y\}$.

For the $z$ terms: $\lfloor z \rfloor + z - \{z\} = \lfloor z \rfloor + \lfloor z \rfloor + \{z\} - \{z\} = 2\lfloor z \rfloor$.

Therefore:

$2\{y\} + 2\lfloor z \rfloor = 6.4$

$\{y\} + \lfloor z \rfloor = 3.2 \qquad \blacksquare$

Step 2: Compute (1) + (3) - (2):

$(x + \lfloor y \rfloor + \{z\}) + (\lfloor x \rfloor + \{y\} + z) - (\{x\} + y + \lfloor z \rfloor) = 3.9 + 5 - 5.3$

By similar reasoning:

$x$ terms: $x + \lfloor x \rfloor - \{x\} = \lfloor x \rfloor + \{x\} + \lfloor x \rfloor - \{x\} = 2\lfloor x \rfloor$

$y$ terms: $\lfloor y \rfloor + \{y\} - y = 0$

$z$ terms: $\{z\} + z - \lfloor z \rfloor = \{z\} + \{z\} = 2\{z\}$

$2\lfloor x \rfloor + 2\{z\} = 3.6$

$\lfloor x \rfloor + \{z\} = 1.8 \qquad \text{...(A)}$

Step 3: Compute (1) + (2) - (3):

$(x + \lfloor y \rfloor + \{z\}) + (\{x\} + y + \lfloor z \rfloor) - (\lfloor x \rfloor + \{y\} + z) = 3.9 + 5.3 - 5$

$x$ terms: $x + \{x\} - \lfloor x \rfloor = 2\{x\}$

$y$ terms: $\lfloor y \rfloor + y - \{y\} = 2\lfloor y \rfloor$

$z$ terms: $\{z\} + \lfloor z \rfloor - z = 0$

$2\{x\} + 2\lfloor y \rfloor = 4.2$

$\{x\} + \lfloor y \rfloor = 2.1 \qquad \text{...(B)}$

Step 4: Solving the system

From $\{y\} + \lfloor z \rfloor = 3.2$: Since $\lfloor z \rfloor$ is an integer and $0 \leqslant \{y\} < 1$, we get $\lfloor z \rfloor = 3$ and $\{y\} = 0.2$.

From (A): $\lfloor x \rfloor + \{z\} = 1.8$. Since $\lfloor x \rfloor$ is an integer and $0 \leqslant \{z\} < 1$, we get $\lfloor x \rfloor = 1$ and $\{z\} = 0.8$.

From (B): $\{x\} + \lfloor y \rfloor = 2.1$. Since $\lfloor y \rfloor$ is an integer and $0 \leqslant \{x\} < 1$, we get $\lfloor y \rfloor = 2$ and $\{x\} = 0.1$.

Reconstructing:

$x = \lfloor x \rfloor + \{x\} = 1 + 0.1 = 1.1$ $y = \lfloor y \rfloor + \{y\} = 2 + 0.2 = 2.2$ $z = \lfloor z \rfloor + \{z\} = 3 + 0.8 = 3.8$

Solution: $x = 1.1, \quad y = 2.2, \quad z = 3.8$

---

Part (iii)

We solve:

$x + 2\lfloor y \rfloor + \{z\} = 3.9 \qquad \text{...(1)}$ $\{x\} + 2y + \lfloor z \rfloor = 5.3 \qquad \text{...(2)}$ $\lfloor x \rfloor + 2\{y\} + z = 5 \qquad \text{...(3)}$

Step 1: Compute (2) + (3) - (1):

$\{x\} + 2y + \lfloor z \rfloor + \lfloor x \rfloor + 2\{y\} + z - x - 2\lfloor y \rfloor - \{z\} = 6.4$

$x$ terms: $\{x\} + \lfloor x \rfloor - x = 0$

$y$ terms: $2y + 2\{y\} - 2\lfloor y \rfloor = 2(\lfloor y \rfloor + \{y\}) + 2\{y\} - 2\lfloor y \rfloor = 4\{y\}$

$z$ terms: $\lfloor z \rfloor + z - \{z\} = 2\lfloor z \rfloor$

$4\{y\} + 2\lfloor z \rfloor = 6.4$

$2\{y\} + \lfloor z \rfloor = 3.2 \qquad \text{...(A)}$

Step 2: Compute (1) + (3) - (2):

$x + 2\lfloor y \rfloor + \{z\} + \lfloor x \rfloor + 2\{y\} + z - \{x\} - 2y - \lfloor z \rfloor = 3.6$

$x$ terms: $x + \lfloor x \rfloor - \{x\} = 2\lfloor x \rfloor$

$y$ terms: $2\lfloor y \rfloor + 2\{y\} - 2y = 0$

$z$ terms: $\{z\} + z - \lfloor z \rfloor = 2\{z\}$

$2\lfloor x \rfloor + 2\{z\} = 3.6$

$\lfloor x \rfloor + \{z\} = 1.8 \qquad \text{...(B)}$

Step 3: Compute (1) + (2) - (3):

$x + 2\lfloor y \rfloor + \{z\} + \{x\} + 2y + \lfloor z \rfloor - \lfloor x \rfloor - 2\{y\} - z = 4.2$

$x$ terms: $x + \{x\} - \lfloor x \rfloor = 2\{x\}$

$y$ terms: $2\lfloor y \rfloor + 2y - 2\{y\} = 2\lfloor y \rfloor + 2(\lfloor y \rfloor + \{y\}) - 2\{y\} = 4\lfloor y \rfloor$

$z$ terms: $\{z\} + \lfloor z \rfloor - z = 0$

$2\{x\} + 4\lfloor y \rfloor = 4.2$

$\{x\} + 2\lfloor y \rfloor = 2.1 \qquad \text{...(C)}$

Step 4: Finding all solutions

From (B): $\lfloor x \rfloor + \{z\} = 1.8$, so $\lfloor x \rfloor = 1$ and $\{z\} = 0.8$. (This is the same as part (ii).)

From (A): $2\{y\} + \lfloor z \rfloor = 3.2$.

Since $\lfloor z \rfloor$ is an integer and $0 \leqslant 2\{y\} < 2$, we need $3.2 - \lfloor z \rfloor = 2\{y\}$ with $0 \leqslant 2\{y\} < 2$.

• If $\lfloor z \rfloor = 2$: $2\{y\} = 1.2$, so $\{y\} = 0.6$. Valid since $0 \leqslant 0.6 < 1$. $\checkmark$
• If $\lfloor z \rfloor = 3$: $2\{y\} = 0.2$, so $\{y\} = 0.1$. Valid since $0 \leqslant 0.1 < 1$. $\checkmark$
• If $\lfloor z \rfloor = 4$: $2\{y\} = -0.8$. Invalid.

Case 1: $\{y\} = 0.1, \quad \lfloor z \rfloor = 3$

From (C): $\{x\} + 2\lfloor y \rfloor = 2.1$, so $\{x\} = 0.1$ and $\lfloor y \rfloor = 1$.

$x = 1 + 0.1 = 1.1$, $\quad y = 1 + 0.1 = 1.1$, $\quad z = 3 + 0.8 = 3.8$.

This matches the solution from part (ii) with $y$ halved (as expected from the coefficient change).

Verification: Eq (1): $1.1 + 2(1) + 0.8 = 3.9$. $\checkmark$ $\quad$ Eq (2): $0.1 + 2(1.1) + 3 = 5.3$. $\checkmark$ $\quad$ Eq (3): $1 + 2(0.1) + 3.8 = 5$. $\checkmark$

Case 2: $\{y\} = 0.6, \quad \lfloor z \rfloor = 2$

From (C): $\{x\} + 2\lfloor y \rfloor = 2.1$. Since $0 \leqslant \{x\} < 1$, we need $2\lfloor y \rfloor$ to have fractional part $0.1$, which requires $\lfloor y \rfloor$ to have fractional part $0.05$ — impossible since $\lfloor y \rfloor$ is an integer.

Wait, let me reconsider. $\{x\} + 2\lfloor y \rfloor = 2.1$. With $\lfloor y \rfloor$ an integer, $2\lfloor y \rfloor$ is an even integer, so $\{x\} = 2.1 - 2\lfloor y \rfloor$.

For $0 \leqslant \{x\} < 1$: $1.1 < 2\lfloor y \rfloor \leqslant 2.1$, so $\lfloor y \rfloor = 1$.

$\{x\} = 2.1 - 2 = 0.1$.

$x = 1 + 0.1 = 1.1$, $\quad y = 1 + 0.6 = 1.6$, $\quad z = 2 + 0.8 = 2.8$.

Verification: Eq (1): $1.1 + 2(1) + 0.8 = 3.9$. $\checkmark$ $\quad$ Eq (2): $0.1 + 2(1.6) + 2 = 5.3$. $\checkmark$ $\quad$ Eq (3): $1 + 2(0.6) + 2.8 = 5$. $\checkmark$

Solutions:

$x = 1.1, \quad y = 1.1, \quad z = 3.8$ $x = 1.1, \quad y = 1.6, \quad z = 2.8$

Examiner Notes

无官方评述。易错点：(1) 忽略floor值必须为整数、frac值必须在[0,1)的约束；(2) 处理负数的floor时出错（如floor(-4.2)=-5而非-4）；(3) part(iii)系数变化后未重新验证约束条件。

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Question 4

Topic: 微积分 (Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 (i) Sketch the curve $y = xe^x$, giving the coordinates of any stationary points.

(ii) The function $f$ is defined by $f(x) = xe^x$ for $x \geqslant a$, where $a$ is the minimum possible value such that $f$ has an inverse function. What is the value of $a$?

Let $g$ be the inverse of $f$. Sketch the curve $y = g(x)$.

(iii) For each of the following equations, find a real root in terms of a value of the function $g$, or demonstrate that the equation has no real root. If the equation has two real roots, determine whether the root you have found is greater than or less than the other root.

(a) $e^{-x} = 5x$ (b) $2x \ln x + 1 = 0$ (c) $3x \ln x + 1 = 0$ (d) $x = 3 \ln x$

(iv) Given that the equation $x^x = 10$ has a unique positive root, find this root in terms of a value of the function $g$.

Hint

(i) $\frac{dy}{dx} = xe^x + e^x$ M1

Since $e^x > 0$ for all $x$, the only stationary point is when $x = -1$ A1 Coordinates of stationary point are $(-1, -\frac{1}{e})$

Sketch showing: $y \rightarrow \infty$ as $x \rightarrow \infty$ and $y \rightarrow 0^-$ as $x \rightarrow -\infty$ G1 Curve passing through $(0,0)$ with stationary point at $(-1, -\frac{1}{e})$ indicated. G1

(ii) -1 B1

Sketch showing reflection of the correct portion of the graph in the line $y = x$. G1 domain $[-\frac{1}{e}, \infty)$ and range $[-1, \infty)$

(iii) (a) $e^{-x} = 5x$ $xe^x = \frac{1}{5}$

$f(x) = \frac{1}{5}$ M1 Since $f(x) > 0$ there is only one solution A1 $x = g(\frac{1}{5})$

(b) $2x \ln x + 1 = 0$ Let $u = \ln x$:

$ue^u = -\frac{1}{2}$ M1 M1 The minimum value of $f(x)$ is $-\frac{1}{e}$ and $-\frac{1}{2} < -\frac{1}{e}$, so there are no solutions. E1

(c) $3x \ln x + 1 = 0$ Let $u = \ln x$:

$ue^u = -\frac{1}{3}$ M1

$-\frac{1}{e} < -\frac{1}{3} < 0$ so there are two solutions for $u$ and the greater of the two will be when $u = g\left(-\frac{1}{3}\right)$. E1

$x = e^{g\left(-\frac{1}{3}\right)}$ is the larger value. A1

(d) $x = 3 \ln x$ Let $u = \ln x$:

$ue^{-u} = \frac{1}{3}$ M1

$(-u)e^{-u} = -\frac{1}{3}$, so (as in (c)) $g\left(-\frac{1}{3}\right)$ is the greater of the two possible values for $-u$. M1

Therefore $x = e^{-g\left(-\frac{1}{3}\right)}$ is the smaller value. A1

E1

(iv) $x \ln x = \ln 10$ Let $u = \ln x$:

$ue^u = \ln 10$ M1

$u = g(\ln 10)$ $x = e^{g(\ln 10)}$ A1

Model Solution

Part (i)

We sketch $y = xe^x$.

Derivative:

$\frac{dy}{dx} = xe^x + e^x = e^x(x + 1)$

Since $e^x > 0$ for all $x$, the sign of $\frac{dy}{dx}$ is determined by $(x + 1)$:

• $\frac{dy}{dx} < 0$ for $x < -1$ (function decreasing)
• $\frac{dy}{dx} = 0$ at $x = -1$ (stationary point)
• $\frac{dy}{dx} > 0$ for $x > -1$ (function increasing)

The stationary point is at $x = -1$, where $y = -e^{-1} = -\frac{1}{e}$.

Behaviour at extremes:

• As $x \to +\infty$: $y \to +\infty$
• As $x \to -\infty$: $xe^x \to 0^-$ (since $e^x$ decays faster than $x$ grows)

Key points:

• Passes through the origin $(0, 0)$
• Minimum at $\left(-1, -\frac{1}{e}\right) \approx (-1, -0.368)$

The curve decreases from $0^-$ to $-\frac{1}{e}$ as $x$ goes from $-\infty$ to $-1$, then increases through the origin and off to $+\infty$.

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Part (ii)

For $f(x) = xe^x$ to have an inverse, we need $f$ to be one-to-one on its domain. From the sketch, $f$ is decreasing on $(-\infty, -1)$ and increasing on $(-1, \infty)$. The minimum value of $a$ that makes $f$ one-to-one on $[a, \infty)$ is:

$a = -1$

The inverse function $g$ satisfies $g(xe^x) = x$ for $x \geqslant -1$, with domain $\left[-\frac{1}{e}, \infty\right)$ and range $[-1, \infty)$.

The curve $y = g(x)$ is the reflection of $y = f(x)$ (restricted to $x \geqslant -1$) in the line $y = x$. It passes through $(0, 0)$ and $\left(-\frac{1}{e}, -1\right)$, with a vertical tangent at $\left(-\frac{1}{e}, -1\right)$.

---

Part (iii)

(a) $e^{-x} = 5x$

Rewrite: $e^{-x} = 5x$ implies $1 = 5xe^x$, so $xe^x = \frac{1}{5}$.

Since $\frac{1}{5} > 0 > -\frac{1}{e}$ and $f$ is increasing on $[-1, \infty)$ with $f(0) = 0 < \frac{1}{5}$, there is exactly one solution (the other branch of $f$ gives negative values only).

$x = g\!\left(\frac{1}{5}\right)$

(b) $2x \ln x + 1 = 0$

Let $u = \ln x$, so $x = e^u$. The equation becomes:

$2e^u \cdot u + 1 = 0 \implies ue^u = -\frac{1}{2}$

We need $f(u) = ue^u = -\frac{1}{2}$. The minimum value of $f$ is $f(-1) = -\frac{1}{e} \approx -0.368$.

Since $-\frac{1}{2} < -\frac{1}{e}$, the value $-\frac{1}{2}$ is below the range of $f$.

The equation has no real root.

(c) $3x \ln x + 1 = 0$

Let $u = \ln x$, so $x = e^u$. The equation becomes:

$3e^u \cdot u + 1 = 0 \implies ue^u = -\frac{1}{3}$

We need $f(u) = -\frac{1}{3}$. Since $-\frac{1}{e} < -\frac{1}{3} < 0$, there are two solutions for $u$: one with $u < -1$ and one with $-1 < u < 0$.

The value $u = g\!\left(-\frac{1}{3}\right)$ gives the larger solution for $u$ (since $g$ is the inverse on $[-1, \infty)$ and $-1 < g\!\left(-\frac{1}{3}\right) < 0$).

Therefore $x = e^u = e^{g(-1/3)}$ is the larger real root.

(d) $x = 3 \ln x$

Let $u = \ln x$, so $x = e^u$. The equation becomes:

$e^u = 3u \implies ue^{-u} = \frac{1}{3}$

Multiply both sides by $-1$: $(-u)e^{-u} = -\frac{1}{3}$.

This is of the form $f(v) = ve^v = -\frac{1}{3}$ where $v = -u$.

Since $-\frac{1}{e} < -\frac{1}{3} < 0$, there are two solutions for $v$. The larger one is $v = g\!\left(-\frac{1}{3}\right)$, so $u = -g\!\left(-\frac{1}{3}\right)$ and $x = e^{-g(-1/3)}$.

Since $v = g\!\left(-\frac{1}{3}\right)$ is the larger $v$-value (closer to 0), $u = -v$ is the smaller $u$-value, so $x = e^{-g(-1/3)}$ is the smaller real root.

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Part (iv)

Given $x^x = 10$ has a unique positive root, find it in terms of $g$.

Taking logarithms: $x \ln x = \ln 10$.

Let $u = \ln x$, so $x = e^u$:

$e^u \cdot u = \ln 10 \implies ue^u = \ln 10$

Since $\ln 10 > 0 > -\frac{1}{e}$, there is exactly one solution (on the increasing branch of $f$):

$u = g(\ln 10)$

Therefore:

$x = e^{g(\ln 10)}$

Examiner Notes

无官方评述。易错点：(1) 未正确识别a=-1为f可逆的最小值；(2) 求解方程时未能正确转化为xe^x的形式；(3) part(iii)(a)中e^{-x}=5x需要巧妙变形为(-x)e^{-x}=-1/5再用g；(4) part(iv)中x^x=10的变形步骤容易出错。

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Question 5

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

Problem

5 (i) Use the substitution $y = (x - a)u$, where $u$ is a function of $x$, to solve the differential equation

$(x - a) \frac{dy}{dx} = y - x ,$

where $a$ is a constant.

(ii) The curve $C$ with equation $y = f(x)$ has the property that, for all values of $t$ except $t = 1$, the tangent at the point $(t, f(t))$ passes through the point $(1, t)$.

(a) Given that $f(0) = 0$, find $f(x)$ for $x < 1$.

Sketch $C$ for $x < 1$. You should find the co-ordinates of any stationary points and consider the gradient of $C$ as $x \to 1$. You may assume that $z \ln |z| \to 0$ as $z \to 0$.

(b) Given that $f(2) = 2$, sketch $C$ for $x > 1$, giving the co-ordinates of any stationary points.

Hint

(i) $\frac{dy}{dx} = (x - a) \frac{du}{dx} + u$ M1 $(x - a) \left( (x - a) \frac{du}{dx} + u \right) = (x - a)u - x$ A1 $(x - a)^2 \frac{du}{dx} = -x$ $u = \int \frac{-x}{(x - a)^2} dx = \int \frac{-(x - a) - a}{(x - a)^2} dx$ M1 $u = -\ln|x - a| + \frac{a}{x - a} + c$ A1 $y = -(x - a) \ln|x - a| + a + c(x - a)$ A1 (ft)

(ii) (a) The gradient of the line through $(1, t)$ and $(t, f(t))$ is $\frac{f(t) - t}{t - 1} = f'(t)$ M1 Applying the result from (i), with $a=1$ or solving the d.e. directly: $f(x) = -(x - 1) \ln|x - 1| + 1 + c(x - 1)$ B1 (ft) $f(0) = 0$, so $c = 1$ B1 (ft) $y = -(x - 1) \ln|x - 1| + x$ $\frac{dy}{dx} = -\ln|x - 1|$ M1

$-\ln|x - 1| = 0$ when $x = 0$ only (since $x < 1$) and $y = 0$ A1 (ft)

As $x \to 1^-$, $y \to 1^-$ and $\frac{dy}{dx} \to \infty$. B1 (ft)

Sketch showing: Curve approaching $(1,1)$ with a vertical tangent at that point. G1 (ft) Minimum point at $(0,0)$. G1 (ft) $y \to \infty$ as $x \to -\infty$ G1 (ft)

x	y
-2.5	1.5
-2.0	1.1
-1.5	0.75
-1.0	0.4
-0.5	0.15
0.0	0.0
0.5	0.15
1.0	1.0

(b) $f(2) = 2$, so $c = 1$ B1 (ft)

$y = -(x - 1) \ln|x - 1| + x$ $\frac{dy}{dx} = -\ln|x - 1|$

$-\ln|x - 1| = 0$ when $x = 2$ only (since $x > 1$) and $y = 2$. B1 (ft)

As $x \to 1^+$, $y \to 1^+$ and $\frac{dy}{dx} \to \infty$. B1 (ft)

Sketch showing: Curve approaching (1,1) with a vertical tangent at that point. G1 (ft) Maximum point at (2,2). G1 (ft) The curve crossing the $x$-axis for some $x > 2$ and $y \to -\infty$ as $x \to \infty$ G1 (ft)

x	y
1	1
1.5	1.7
2	2
2.5	1.9
3	1.6
3.5	1.1
4	0.5
4.5	-0.2
5	-1

Model Solution

Part (i)

We solve $(x - a)\dfrac{dy}{dx} = y - x$ using the substitution $y = (x - a)u$.

Differentiating by the product rule:

$\frac{dy}{dx} = (x - a)\frac{du}{dx} + u$

Substituting into the ODE:

$(x - a)\left[(x - a)\frac{du}{dx} + u\right] = (x - a)u - x$

$(x - a)^2 \frac{du}{dx} + (x - a)u = (x - a)u - x$

$(x - a)^2 \frac{du}{dx} = -x$

Separating variables:

$du = \frac{-x}{(x - a)^2}\, dx$

To integrate the right side, write $-x = -(x - a) - a$:

$\int du = \int \frac{-(x - a) - a}{(x - a)^2}\, dx = \int \left[\frac{-1}{x - a} - \frac{a}{(x - a)^2}\right] dx$

$u = -\ln|x - a| + \frac{a}{x - a} + c$

Substituting back $y = (x - a)u$:

$y = (x - a)\left[-\ln|x - a| + \frac{a}{x - a} + c\right]$

$y = -(x - a)\ln|x - a| + a + c(x - a)$

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Part (ii)(a)

The tangent to $C$ at $(t, f(t))$ passes through $(1, t)$ for all $t \neq 1$. The gradient of this tangent is $f'(t)$, so:

$f'(t) = \frac{f(t) - t}{t - 1}$

Rewriting (replacing $t$ by $x$, and noting $y = f(x)$):

$(x - 1)\frac{dy}{dx} = y - x$

This is exactly the ODE from part (i) with $a = 1$. The general solution is:

$y = -(x - 1)\ln|x - 1| + 1 + c(x - 1)$

For $x < 1$, we have $|x - 1| = 1 - x$, so:

$y = -(x - 1)\ln(1 - x) + 1 + c(x - 1)$

Applying the condition $f(0) = 0$:

$0 = -(0 - 1)\ln(1 - 0) + 1 + c(0 - 1) = (1)(0) + 1 - c = 1 - c$

$c = 1$

Therefore for $x < 1$:

$f(x) = -(x - 1)\ln(1 - x) + x$

Stationary points: Differentiating:

$f'(x) = -\ln(1 - x) - (x - 1) \cdot \frac{-1}{1 - x} = -\ln(1 - x) + 1 - 1 = -\ln(1 - x)$

Setting $f'(x) = 0$: $\ln(1 - x) = 0$, so $1 - x = 1$, giving $x = 0$.

At $x = 0$: $f(0) = 0$. The stationary point is $(0, 0)$.

Since $f'(x) = -\ln(1 - x)$: for $x < 0$ we have $1 - x > 1$ so $\ln(1-x) > 0$ and $f'(x) < 0$; for $0 < x < 1$ we have $0 < 1 - x < 1$ so $\ln(1-x) < 0$ and $f'(x) > 0$. Thus $(0, 0)$ is a minimum.

Behaviour as $x \to 1^-$: Let $z = 1 - x \to 0^+$. Then $y = z\ln z + 1 - z \to 0 + 1 - 0 = 1$, using the given result $z\ln|z| \to 0$. So $y \to 1^-$.

The gradient $f'(x) = -\ln(1 - x) \to +\infty$ as $x \to 1^-$, so the curve has a vertical tangent at $(1, 1)$.

Behaviour as $x \to -\infty$: The term $-(x-1)\ln(1-x)$ dominates and grows to $+\infty$ (since $-(x-1) > 0$ and $\ln(1-x) > 0$ both grow), so $y \to +\infty$.

Sketch for $x < 1$: The curve decreases from $+\infty$ (as $x \to -\infty$) to the minimum at $(0, 0)$, then increases toward the point $(1, 1)$ with a vertical tangent.

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Part (ii)(b)

For $x > 1$, we have $|x - 1| = x - 1$, so the general solution becomes:

$y = -(x - 1)\ln(x - 1) + 1 + c(x - 1)$

Applying the condition $f(2) = 2$:

$2 = -(2 - 1)\ln(2 - 1) + 1 + c(2 - 1) = -(1)(0) + 1 + c = 1 + c$

$c = 1$

Therefore for $x > 1$:

$f(x) = -(x - 1)\ln(x - 1) + x$

Stationary points: Differentiating:

$f'(x) = -\ln(x - 1) - 1 + 1 = -\ln(x - 1)$

Setting $f'(x) = 0$: $\ln(x - 1) = 0$, so $x - 1 = 1$, giving $x = 2$.

At $x = 2$: $f(2) = -(1)\ln 1 + 2 = 2$. The stationary point is $(2, 2)$.

Since $f'(x) = -\ln(x - 1)$: for $1 < x < 2$ we have $0 < x - 1 < 1$ so $\ln(x-1) < 0$ and $f'(x) > 0$; for $x > 2$ we have $x - 1 > 1$ so $\ln(x-1) > 0$ and $f'(x) < 0$. Thus $(2, 2)$ is a maximum.

Behaviour as $x \to 1^+$: Let $z = x - 1 \to 0^+$. Then $y = -z\ln z + 1 + z \to 0 + 1 + 0 = 1$, and $f'(x) = -\ln(x - 1) \to +\infty$. The curve approaches $(1, 1)$ with a vertical tangent.

Behaviour as $x \to +\infty$: The term $-(x-1)\ln(x-1)$ dominates and drives $y \to -\infty$.

Sketch for $x > 1$: The curve increases from $(1, 1)$ (with a vertical tangent) to the maximum at $(2, 2)$, then decreases through the $x$-axis and down to $-\infty$.

Examiner Notes

无官方评述。易错点：(1) part(ii)中从切线过(1,t)的条件建立微分方程时符号易错；(2) 分段讨论x<1和x>1时积分常数的确定；(3) 忽略x=1处的奇点行为分析。

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Question 6

Topic: 几何与向量 (Geometry & Vectors)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 A plane circular road is bounded by two concentric circles with centres at point $O$. The inner circle has radius $R$ and the outer circle has radius $R + w$. The points $A$ and $B$ lie on the outer circle, as shown in the diagram, with $\angle AOB = 2\alpha$, $\frac{1}{3}\pi \leqslant \alpha \leqslant \frac{1}{2}\pi$ and $0 < w < R$.

(i) Show that I cannot cycle from $A$ to $B$ in a straight line, while remaining on the road.

(ii) I take a path from $A$ to $B$ that is an arc of a circle. This circle is tangent to the inner edge of the road, and has radius $R + d$ (where $d > w$) and centre $O'$.

My path is represented by the dashed arc in the above diagram.

Let $\angle AO'B = 2\theta$.

(a) Use the cosine rule to find $d$ in terms of $w$, $R$ and $\cos \alpha$.

(b) Find also an expression for $\sin(\alpha - \theta)$ in terms of $R$, $d$ and $\sin \alpha$.

You are now given that $\frac{w}{R}$ is much less than 1.

(iii) Show that $\frac{d}{R}$ and $\alpha - \theta$ are also both much less than 1.

(iv) My friend cycles from $A$ to $B$ along the outer edge of the road.

Let my path be shorter than my friend’s path by distance $S$. Show that

$S = 2(R + d)(\alpha - \theta) + 2\alpha(w - d).$

Hence show that $S$ is approximately a fraction

$\left( \frac{\sin \alpha - \alpha \cos \alpha}{\alpha(1 - \cos \alpha)} \right) \frac{w}{R}$

of the length of my friend’s path.

Hint

(i) The shortest distance from $O$ to the line $AB$ is $(R + w) \cos \alpha$ B1 Since $\frac{1}{3}\pi \le \alpha \le \frac{1}{2}\pi$, $0 \le \cos \alpha \le \frac{1}{2}$. M1 Since $w < R$, $(R + w) \cos \alpha < \frac{1}{2}(R + R) = R$, so the midpoint of the line $AB$ lies inside the smaller circle. E1

(ii) (a) $(R + d)^2 = (R + w)^2 + d^2 - 2d(R + w) \cos(\pi - \alpha)$ M1 A1 $R^2 + 2Rd + d^2 = R^2 + 2Rw + w^2 + d^2 + 2d(R + w) \cos \alpha$ $d = \frac{w(2R + w)}{2(R - (R + w) \cos \alpha)}$ M1 A1

(b) $\angle O'AO = \alpha - \theta$ $\frac{\sin(\alpha - \theta)}{d} = \frac{\sin(\pi - \alpha)}{R + d}$ M1 $\sin(\alpha - \theta) = \frac{d \sin \alpha}{R + d}$ A1

(iii) $\frac{d}{R} = \frac{\left(\frac{w}{R}\right)\left(2 + \frac{w}{R}\right)}{2\left(1 - \left(1 + \frac{w}{R}\right) \cos \alpha\right)} \approx \frac{1}{1 - \cos \alpha} \times \frac{w}{R}$ M1 A1 $1 - \cos \alpha > \frac{1}{2}$ and $\frac{w}{R}$ is much less than 1, so $\frac{d}{R}$ is much less than 1. E1

$\sin(\alpha - \theta) = \frac{\left(\frac{d}{R}\right) \sin \alpha}{1 + \left(\frac{d}{R}\right)} < \frac{d}{R}$ M1

$\sin(\alpha - \theta)$ is much less than 1 and so $(\alpha - \theta)$ is a small angle. M1 Therefore $\sin(\alpha - \theta) \approx \alpha - \theta$, so $\alpha - \theta$ is much less than 1. E1

(iv) The longer length is $(R + w) \times 2\alpha$ The shorter length is $(R + d) \times 2\theta$ $S = 2\alpha(R + w) - 2\theta(R + d)$ $S = 2(R + d + w - d)\alpha - 2(R + d)\theta$ $S = 2(R + d)(\alpha - \theta) + 2(w - d)\alpha$ B1

$\alpha - \theta \approx \frac{w \sin \alpha}{R(1 - \cos \alpha)}$ M1 $d - w \approx \frac{\cos \alpha}{(1 - \cos \alpha)} \times \frac{w}{R}$ M1 So $S \approx 2(R + d) \frac{w \sin \alpha}{R(1 - \cos \alpha)} - 2\left(\frac{\cos \alpha}{(1 - \cos \alpha)} \times \frac{w}{R}\right)\alpha$

As a fraction of the longer path length: $\frac{S}{2\alpha(R + w)} = \frac{R + d}{R + w} \times \frac{\alpha - \theta}{\alpha} + \frac{w - d}{R + w} \approx \frac{\sin \alpha}{\alpha(1 - \cos \alpha)} \frac{w}{R} - \frac{\cos \alpha}{(1 - \cos \alpha)} \frac{w}{R}$ M1 $S \approx \left(\frac{\sin \alpha - \alpha \cos \alpha}{\alpha(1 - \cos \alpha)}\right) \frac{w}{R} \quad \mathbf{AG}$ A1

Model Solution

Part (i)

Let $M$ be the midpoint of $AB$. Since $O$ is the centre of the outer circle, the line $OM$ is the perpendicular bisector of $AB$.

In the right-angled triangle $OMA$:

$OM = OA \cos \alpha = (R + w) \cos \alpha$

Since $\frac{1}{3}\pi \leqslant \alpha \leqslant \frac{1}{2}\pi$, we have $0 \leqslant \cos \alpha \leqslant \frac{1}{2}$, so:

$OM = (R + w)\cos \alpha \leqslant \frac{R + w}{2}$

Since $w < R$:

$\frac{R + w}{2} < \frac{R + R}{2} = R$

Therefore $OM < R$, which means the midpoint $M$ of the chord $AB$ lies inside the inner circle. The straight line from $A$ to $B$ must pass through the interior of the inner circle, so I cannot cycle in a straight line while remaining on the road.

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Part (ii)(a)

By symmetry, both $O$ and $O'$ lie on the perpendicular bisector of $AB$. The path circle (centre $O'$, radius $R + d$) is tangent to the inner circle (centre $O$, radius $R$). Since the arc from $A$ to $B$ bulges towards $O$ to achieve tangency, $O'$ lies on the opposite side of $O$ from $M$, with the order $O' - O - M$ along the perpendicular bisector.

The two circles are internally tangent (the path circle encloses the inner circle), so:

$OO' = (R + d) - R = d$

In triangle $OAO'$, the angle at $O$ between $OA$ and $OO'$ is:

$\angle AOO' = \pi - \alpha$

since $OA$ makes angle $\alpha$ with $OM$, and $OO'$ points in the opposite direction from $OM$.

Applying the cosine rule in triangle $OAO'$ with sides $OA = R + w$, $O'A = R + d$, $OO' = d$:

$(R + d)^2 = (R + w)^2 + d^2 - 2d(R + w)\cos(\pi - \alpha)$

Since $\cos(\pi - \alpha) = -\cos \alpha$:

$(R + d)^2 = (R + w)^2 + d^2 + 2d(R + w)\cos \alpha$

Expanding the left side:

$R^2 + 2Rd + d^2 = R^2 + 2Rw + w^2 + d^2 + 2d(R + w)\cos \alpha$

Cancelling $R^2 + d^2$ from both sides:

$2Rd = 2Rw + w^2 + 2d(R + w)\cos \alpha$

Collecting terms in $d$ on the left:

$2Rd - 2d(R + w)\cos \alpha = 2Rw + w^2$

$2d\bigl[R - (R + w)\cos \alpha\bigr] = w(2R + w)$

$d = \frac{w(2R + w)}{2\bigl[R - (R + w)\cos \alpha\bigr]}$

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Part (ii)(b)

In triangle $OAO'$, we identify the three angles.

The angle at $O$ is $\angle AOO' = \pi - \alpha$ (established above).

To find the angle at $A$, note that in right triangle $OMA$:

$\angle OAM = \frac{\pi}{2} - \alpha$

In right triangle $O'MA$ (right-angled at $M$ since $O'M$ is part of the perpendicular bisector):

$\angle O'AM = \frac{\pi}{2} - \theta$

Since $O'$ is further from $M$ than $O$ is (along the same line), we have $\angle O'AM > \angle OAM$, so:

$\angle OAO' = \angle O'AM - \angle OAM = \left(\frac{\pi}{2} - \theta\right) - \left(\frac{\pi}{2} - \alpha\right) = \alpha - \theta$

The angle at $O'$ is then:

$\angle AO'O = \pi - (\pi - \alpha) - (\alpha - \theta) = \theta$

By the sine rule in triangle $OAO'$:

$\frac{OO'}{\sin(\angle OAO')} = \frac{O'A}{\sin(\angle AOO')}$

$\frac{d}{\sin(\alpha - \theta)} = \frac{R + d}{\sin(\pi - \alpha)}$

Since $\sin(\pi - \alpha) = \sin \alpha$:

$\sin(\alpha - \theta) = \frac{d \sin \alpha}{R + d}$

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Part (iii)

Dividing numerator and denominator of the expression for $d$ by $R$:

$\frac{d}{R} = \frac{\left(\frac{w}{R}\right)\left(2 + \frac{w}{R}\right)}{2\left[1 - \left(1 + \frac{w}{R}\right)\cos \alpha\right]}$

Since $w/R \ll 1$, the numerator is approximately $2(w/R)$ and the denominator is approximately $2(1 - \cos \alpha)$:

$\frac{d}{R} \approx \frac{w/R}{1 - \cos \alpha}$

Since $\alpha \geqslant \frac{1}{3}\pi$, we have $\cos \alpha \leqslant \frac{1}{2}$, so $1 - \cos \alpha \geqslant \frac{1}{2}$. Therefore:

$\frac{d}{R} \lesssim 2 \cdot \frac{w}{R} \ll 1$

For $\alpha - \theta$, we use the result from (ii)(b):

$\sin(\alpha - \theta) = \frac{(d/R)\sin \alpha}{1 + d/R}$

Since $d/R \ll 1$, the denominator $1 + d/R \approx 1$, and:

$\sin(\alpha - \theta) \approx \frac{d}{R}\sin \alpha \ll 1$

Since $\sin(\alpha - \theta) \ll 1$, the angle $\alpha - \theta$ is small, so $\sin(\alpha - \theta) \approx \alpha - \theta$. Therefore $\alpha - \theta \ll 1$.

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Part (iv)

My friend’s path is the arc from $A$ to $B$ along the outer circle (radius $R + w$), with length:

$(R + w) \cdot 2\alpha$

My path is the arc from $A$ to $B$ on the circle centred at $O'$ (radius $R + d$), with length:

$(R + d) \cdot 2\theta$

The saving $S$ is:

$S = 2\alpha(R + w) - 2\theta(R + d)$

Writing $R + w = (R + d) + (w - d)$:

$S = 2\alpha\bigl[(R + d) + (w - d)\bigr] - 2\theta(R + d)$

$S = 2(R + d)(\alpha - \theta) + 2\alpha(w - d)$

To find $S$ as a fraction of my friend’s path length $2\alpha(R + w)$:

$\frac{S}{2\alpha(R + w)} = \frac{(R + d)(\alpha - \theta)}{\alpha(R + w)} + \frac{w - d}{R + w}$

Since $w/R \ll 1$ and $d/R \ll 1$, we have $R + w \approx R$ and $R + d \approx R$ to leading order:

$\frac{S}{2\alpha(R + w)} \approx \frac{\alpha - \theta}{\alpha} + \frac{w - d}{R}$

Computing $\alpha - \theta$: From the small-angle approximation and part (ii)(b):

$\alpha - \theta \approx \frac{d \sin \alpha}{R + d} \approx \frac{d \sin \alpha}{R}$

From part (ii)(a), with $w/R \ll 1$:

$d \approx \frac{2Rw}{2R(1 - \cos \alpha)} = \frac{w}{1 - \cos \alpha}$

Therefore:

$\alpha - \theta \approx \frac{w \sin \alpha}{R(1 - \cos \alpha)}$

Computing $w - d$:

$w - d = w - \frac{w}{1 - \cos \alpha} = w\left(\frac{1 - \cos \alpha - 1}{1 - \cos \alpha}\right) = \frac{-w \cos \alpha}{1 - \cos \alpha}$

Substituting into the fraction:

$\frac{S}{2\alpha(R + w)} \approx \frac{1}{\alpha} \cdot \frac{w \sin \alpha}{R(1 - \cos \alpha)} + \frac{1}{R} \cdot \frac{-w \cos \alpha}{1 - \cos \alpha}$

$= \frac{w}{R(1 - \cos \alpha)}\left(\frac{\sin \alpha}{\alpha} - \cos \alpha\right)$

$= \left(\frac{\sin \alpha - \alpha \cos \alpha}{\alpha(1 - \cos \alpha)}\right) \frac{w}{R}$

as required.

Examiner Notes

无官方评述。易错点：(1) part(ii)(a)中用余弦定理建立d与w,R,cosα的关系时几何关系搞混；(2) part(iii)小量近似的应用条件和精度控制；(3) 最终分式化简时分子分母搞反。

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Question 7

Topic: 矩阵与线性代数 (Matrices & Linear Algebra)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 (i) The matrix R represents an anticlockwise rotation through angle $\phi$ ($0^\circ \le \phi < 360^\circ$) in two dimensions, and the matrix $\mathbf{R} + \mathbf{I}$ also represents a rotation in two dimensions. Determine the possible values of $\phi$ and deduce that $\mathbf{R}^3 = \mathbf{I}$.

(ii) Let S be a real matrix with $\mathbf{S}^3 = \mathbf{I}$, but $\mathbf{S} \neq \mathbf{I}$.

Show that $\det(\mathbf{S}) = 1$.

Given that

$\mathbf{S} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

show that $\mathbf{S}^2 = (a + d)\mathbf{S} - \mathbf{I}$.

Hence prove that $a + d = -1$.

(iii) Let S be a real $2 \times 2$ matrix.

Show that if $\mathbf{S}^3 = \mathbf{I}$ and $\mathbf{S} + \mathbf{I}$ represents a rotation, then S also represents a rotation. What are the possible angles of the rotation represented by S?

Hint

(i) $\boldsymbol{R} = \begin{pmatrix} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}$ M1

$\boldsymbol{R} + \boldsymbol{I} = \begin{pmatrix} 1 + \cos \phi & -\sin \phi \\ \sin \phi & 1 + \cos \phi \end{pmatrix}$ A1

This must also be of the form $\begin{pmatrix} \cos & -\sin \\ \sin & \cos \end{pmatrix}$, so $(1 + \cos \phi)^2 + \sin^2 \phi = 1$ M1

$1 + 2 \cos \phi = 0$ $\phi = 120^\circ$ or $240^\circ$ A1

In either case, three consecutive rotations is equivalent to a rotation through $0^\circ$, so $\boldsymbol{R}^3 = \boldsymbol{I} \quad \boldsymbol{AG}$ E1

(ii) $\det(\boldsymbol{S}^3) = 1$ $\det(\boldsymbol{S}^3) = \det(\boldsymbol{S})^3$ Therefore $\det(\boldsymbol{S}) = 1 \quad \boldsymbol{AG}$ B1

$\boldsymbol{S}^2 = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix} = \begin{pmatrix} a^2 + bc & (a + d)b \\ (a + d)c & bc + d^2 \end{pmatrix}$

Since $\det(\boldsymbol{S}) = 1, ad - bc = 1$

$a^2 + bc = a^2 + ad - 1 = a(a + d) - 1$ M1

$bc + d^2 = ad + d^2 - 1 = d(a + d) - 1$ Therefore, $\boldsymbol{S}^2 = (a + d)\boldsymbol{S} - \boldsymbol{I} \quad \boldsymbol{AG}$ A1

$\boldsymbol{S}^3 = \boldsymbol{S}^2\boldsymbol{S} = (a + d)\boldsymbol{S}^2 - \boldsymbol{S}$ M1

$\boldsymbol{I} = (a + d)((a + d)\boldsymbol{S} - \boldsymbol{I}) - \boldsymbol{S}$ M1

$((a + d)^2 - 1)\boldsymbol{S} = (a + d + 1)\boldsymbol{I}$ A1

If $((a + d)^2 - 1)$ and $(a + d + 1)$ are non-zero, then $b = c = 0$ B1

In which case $ad = 1$ M1

since $\det(\boldsymbol{S}) = 1$ and since $\boldsymbol{S}^3 = \boldsymbol{I}, a = d = 1$ A1 But $\boldsymbol{S} \neq \boldsymbol{I}$, so this is not possible. E1

Therefore $a + d = -1$ A1

(iii) If $\boldsymbol{S} = \boldsymbol{I}$, then $\boldsymbol{S} + \boldsymbol{I} = 2\boldsymbol{I}$, which does not represent a rotation. Therefore, the conditions of part (ii) are met and so $a + d = -1$. Suppose that $\boldsymbol{S} + \boldsymbol{I}$ represents an anticlockwise rotation through angle $\theta$: M1 $a + 1 = d + 1 = \cos \theta$ $(a + 1) + (d + 1) = 1$, so $a = d = -\frac{1}{2}$.

Also, $b = -c$ and $b^2 = c^2 = \frac{3}{4}$ M1

Therefore $\boldsymbol{S} = \begin{pmatrix} -\frac{1}{2} & \pm \frac{1}{2}\sqrt{3} \\ \mp \frac{1}{2}\sqrt{3} & -\frac{1}{2} \end{pmatrix}$ A1

Which represents a rotation through $120^\circ$ or $240^\circ$ A1

Model Solution

Part (i)

The rotation matrix through angle $\phi$ is:

$\mathbf{R} = \begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix}$

So:

$\mathbf{R} + \mathbf{I} = \begin{pmatrix} 1 + \cos\phi & -\sin\phi \\ \sin\phi & 1 + \cos\phi \end{pmatrix}$

For $\mathbf{R} + \mathbf{I}$ to represent a rotation, it must have the standard form $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ for some angle $\theta$. This requires the diagonal entries to be equal (already satisfied) and:

$(1 + \cos\phi)^2 + \sin^2\phi = 1$

Expanding the left side:

$1 + 2\cos\phi + \cos^2\phi + \sin^2\phi = 1$

$1 + 2\cos\phi + 1 = 1$

$2\cos\phi = -1$

$\cos\phi = -\frac{1}{2}$

Since $0^\circ \le \phi < 360^\circ$:

$\phi = 120^\circ \quad \text{or} \quad \phi = 240^\circ$

In either case, $\mathbf{R}$ is a rotation through one-third of a full turn. Applying $\mathbf{R}$ three times gives a total rotation of $3 \times 120^\circ = 360^\circ$ or $3 \times 240^\circ = 720^\circ$, both equivalent to the identity rotation. Therefore:

$\mathbf{R}^3 = \mathbf{I} \qquad \blacksquare$

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Part (ii)

Showing $\det(\mathbf{S}) = 1$:

Since $\mathbf{S}^3 = \mathbf{I}$:

$\det(\mathbf{S}^3) = \det(\mathbf{I}) = 1$

Using the multiplicative property of determinants, $\det(\mathbf{S}^3) = \det(\mathbf{S})^3$:

$\det(\mathbf{S})^3 = 1$

Since $\mathbf{S}$ is a real matrix, $\det(\mathbf{S})$ is real. The only real cube root of $1$ is $1$, so:

$\det(\mathbf{S}) = 1 \qquad \blacksquare$

Showing $\mathbf{S}^2 = (a + d)\mathbf{S} - \mathbf{I}$:

Computing $\mathbf{S}^2$ directly:

$\mathbf{S}^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix}$

Since $\det(\mathbf{S}) = ad - bc = 1$, we have $bc = ad - 1$. Substituting into each entry:

Top-left: $a^2 + bc = a^2 + ad - 1 = a(a + d) - 1$

Top-right: $ab + bd = b(a + d)$

Bottom-left: $ac + cd = c(a + d)$

Bottom-right: $bc + d^2 = ad - 1 + d^2 = d(a + d) - 1$

Therefore:

$\mathbf{S}^2 = \begin{pmatrix} a(a+d) - 1 & (a+d)b \\ (a+d)c & d(a+d) - 1 \end{pmatrix} = (a + d)\begin{pmatrix} a & b \\ c & d \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$\mathbf{S}^2 = (a + d)\mathbf{S} - \mathbf{I} \qquad \blacksquare$

Proving $a + d = -1$:

Multiply both sides of $\mathbf{S}^2 = (a+d)\mathbf{S} - \mathbf{I}$ on the right by $\mathbf{S}$:

$\mathbf{S}^3 = (a + d)\mathbf{S}^2 - \mathbf{S}$

Substituting $\mathbf{S}^3 = \mathbf{I}$ and replacing $\mathbf{S}^2$ using the relation again:

$\mathbf{I} = (a + d)\big[(a + d)\mathbf{S} - \mathbf{I}\big] - \mathbf{S}$

$\mathbf{I} = (a + d)^2\,\mathbf{S} - (a + d)\mathbf{I} - \mathbf{S}$

$\mathbf{I} = \big[(a + d)^2 - 1\big]\mathbf{S} - (a + d)\mathbf{I}$

$\big[(a + d)^2 - 1\big]\mathbf{S} = \big[(a + d) + 1\big]\mathbf{I} \qquad (\star)$

Suppose for contradiction that $a + d + 1 \ne 0$. Then we can divide both sides by $(a + d + 1)$:

$\big[(a + d) - 1\big]\mathbf{S} = \mathbf{I}$

This means $\mathbf{S} = \frac{1}{(a+d) - 1}\,\mathbf{I}$, i.e., $\mathbf{S}$ is a scalar matrix. Then:

$\mathbf{S}^3 = \frac{1}{\big[(a+d)-1\big]^3}\,\mathbf{I} = \mathbf{I}$

So $(a + d - 1)^3 = 1$. Since $a + d$ is real, $a + d - 1 = 1$, giving $a + d = 2$. Then $\mathbf{S} = \mathbf{I}$, which contradicts $\mathbf{S} \ne \mathbf{I}$.

Therefore $a + d + 1 = 0$, i.e.:

$a + d = -1 \qquad \blacksquare$

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Part (iii)

First, note that if $\mathbf{S} = \mathbf{I}$, then $\mathbf{S} + \mathbf{I} = 2\mathbf{I}$. For $2\mathbf{I}$ to be a rotation matrix, we would need $2^2 + 0^2 = 1$, which is false. So $\mathbf{S} \ne \mathbf{I}$, and the conditions of part (ii) are met: $a + d = -1$ and $\det(\mathbf{S}) = 1$.

Since $\mathbf{S} + \mathbf{I}$ represents a rotation through some angle $\theta$:

$\mathbf{S} + \mathbf{I} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

So:

$\mathbf{S} = \begin{pmatrix} \cos\theta - 1 & -\sin\theta \\ \sin\theta & \cos\theta - 1 \end{pmatrix}$

From $a + d = -1$:

$(\cos\theta - 1) + (\cos\theta - 1) = -1$

$2\cos\theta - 2 = -1$

$\cos\theta = \frac{1}{2}$

$\theta = 60^\circ \quad \text{or} \quad \theta = 300^\circ$

Then $\sin^2\theta = 1 - \frac{1}{4} = \frac{3}{4}$, so $\sin\theta = \pm\frac{\sqrt{3}}{2}$.

Case 1: $\theta = 60^\circ$ (so $\sin\theta = \frac{\sqrt{3}}{2}$)

$\mathbf{S} = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\[4pt] \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} \cos 120^\circ & -\sin 120^\circ \\ \sin 120^\circ & \cos 120^\circ \end{pmatrix}$

This is a rotation through $120^\circ$.

Case 2: $\theta = 300^\circ$ (so $\sin\theta = -\frac{\sqrt{3}}{2}$)

$\mathbf{S} = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\[4pt] -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} \cos 240^\circ & -\sin 240^\circ \\ \sin 240^\circ & \cos 240^\circ \end{pmatrix}$

This is a rotation through $240^\circ$.

In both cases, $\mathbf{S}$ has the form of a rotation matrix with $\det(\mathbf{S}) = \frac{1}{4} + \frac{3}{4} = 1$, confirming it represents a rotation.

The possible angles of rotation represented by $\mathbf{S}$ are $120^\circ$ or $240^\circ$.

Examiner Notes

无官方评述。易错点：(1) part(i)中R+I也是旋转矩阵这一条件的几何含义理解困难；(2) part(ii)中证明S²=(a+d)S-I时需要利用S³=I和det(S)=1的组合；(3) 忽略S≠I的条件排除平凡解。

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Question 8

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

Problem

8 (i) Show that, for $n = 2, 3, 4, \dots$,

$\frac{\mathrm{d}^2}{\mathrm{d}t^2} \left( t^n(1 - t)^n \right) = nt^{n-2}(1 - t)^{n-2} \Big[ (n - 1) - 2(2n - 1)t(1 - t) \Big].$

(ii) The sequence $T_0, T_1, \dots$ is defined by

$T_n = \int_0^1 \frac{t^n(1 - t)^n}{n!} \mathrm{e}^t \, \mathrm{d}t \, .$

Show that, for $n \ge 2$,

$T_n = T_{n-2} - 2(2n - 1)T_{n-1} \, .$

(iii) Evaluate $T_0$ and $T_1$ and deduce that, for $n \ge 0$, $T_n$ can be written in the form

$T_n = a_n + b_n \mathrm{e} \, ,$

where $a_n$ and $b_n$ are integers (which you should not attempt to evaluate).

(iv) Show that $0 < T_n < \frac{\mathrm{e}}{n!}$ for $n \ge 0$. Given that $b_n$ is non-zero for all $n$, deduce that $\frac{-a_n}{b_n}$ tends to $\mathrm{e}$ as $n$ tends to infinity.

Hint

(i) $\frac{d}{dt}(t^n(1-t)^n) = nt^{n-1}(1-t)^n - nt^n(1-t)^{n-1}$ M1 $\frac{d^2}{dt^2}(t^n(1-t)^n) = n(n-1)t^{n-2}(1-t)^n - n^2t^{n-1}(1-t)^{n-1}$ $- n^2t^{n-1}(1-t)^{n-1} + n(n-1)t^n(1-t)^{n-2}$ A1 $= nt^{n-2}(1-t)^{n-2}[(n-1)(1-t)^2 - 2nt(1-t) + (n-1)t^2]$ M1 $= nt^{n-2}(1-t)^{n-2}[(4n-2)t^2 - (4n-2)t + (n-1)]$ $= nt^{n-2}(1-t)^{n-2}[(n-1) - 2(2n-1)t(1-t)] \quad \mathbf{AG}$ A1

(ii) Integrating by parts: $u = t^n(1-t)^n, \frac{dv}{dx} = \frac{e^t}{n!}$ M1 $\frac{du}{dx} = nt^{n-1}(1-t)^{n-1}(1-2t), v = \frac{e^t}{n!}$

$T_n = \left[ t^n(1-t)^n \frac{e^t}{n!} \right]_0^1 - \int_0^1 nt^{n-1}(1-t)^{n-1}(1-2t) \frac{e^t}{n!} dt$ $= - \int_0^1 nt^{n-1}(1-t)^{n-1}(1-2t) \frac{e^t}{n!} dt$ M1

Integrating by parts:

$u = nt^{n-1}(1-t)^{n-1}(1-2t), \frac{dv}{dx} = \frac{e^t}{n!}$ $\frac{du}{dx} = nt^{n-2}(1-t)^{n-2}[(n-1) - 2(2n-1)t(1-t)], v = \frac{e^t}{n!}$ $T_n = - \left[ nt^{n-1}(1-t)^{n-1} \frac{e^t}{n!} \right]_0^1$ $+ \int_0^1 nt^{n-2}(1-t)^{n-2}[(n-1) - 2(2n-1)t(1-t)] \frac{e^t}{n!} dt$ M1 $= \int_0^1 nt^{n-2}(1-t)^{n-2}[(n-1) - 2(2n-1)t(1-t)] \frac{e^t}{n!} dt$ $= \int_0^1 t^{n-2}(1-t)^{n-2} \frac{e^t}{(n-2)!} - 2(2n-1)t^{n-1}(1-t)^{n-1} \frac{e^t}{(n-1)!} dt$ M1 M1 $T_n = T_{n-2} - 2(2n-1)T_{n-1} \quad for \ n \geq 2 \quad \mathbf{AG}$ A1

(iii) $T_0 = \int_0^1 e^t \, dt = e - 1$ B1

$T_1 = \int_0^1 t(1 - t)e^t \, dt$ $= \int_0^1 te^t - t^2e^t \, dt$ M1

$\int_0^1 te^t dt = [te^t]_0^1 - \int_0^1 e^t \, dt = 1$ M1

$\int_0^1 t^2e^t dt = [t^2e^t]_0^1 - 2 \int_0^1 te^t \, dt = e - 2$ M1

$T_1 = 1 - (e - 2) = 3 - e$ A1

$T_0$ and $T_1$ are both of the given form. B1

If $T_{n-2}$ and $T_{n-1}$ are both of the given form, then by part (ii):

$a_n = a_{n-2} - 2(2n - 1)a_{n-1}$ $b_n = b_{n-2} - 2(2n - 1)b_{n-1}$

If $a_{n-2}, a_{n-1}, b_{n-2}$ and $b_{n-1}$ are all integers, so $a_n$ and $b_n$ will also be integers. E1

(iv) For $0 \le t \le 1$: $0 \le t^n(1 - t)^n \le 1$ M1

$0 \le e^t \le e$ M1

$0 \le \frac{t^n(1-t)^n}{n!} e^t \le \frac{e}{n!}$ and equality can only occur at t=0 or t=1, so $T_n > 0$ and is less

than the area of a rectangle with width 1 and height $\frac{e}{n!}$.

$0 < T_n < \frac{e}{n!}$ E1

Therefore $a_n + b_n e \to 0$ as $n \to \infty$ Therefore $-\frac{a_n}{b_n} \to e$ as $n \to \infty$ E1

Model Solution

Part (i)

Let $f(t) = t^n(1-t)^n$. We compute the first and second derivatives.

First derivative using the product rule on $t^n \cdot (1-t)^n$:

$f'(t) = nt^{n-1}(1-t)^n + t^n \cdot n(1-t)^{n-1}(-1)$

$= nt^{n-1}(1-t)^n - nt^n(1-t)^{n-1}$

$= nt^{n-1}(1-t)^{n-1}\big[(1-t) - t\big]$

$= nt^{n-1}(1-t)^{n-1}(1 - 2t)$

Second derivative using the product rule on $nt^{n-1}(1-t)^{n-1} \cdot (1-2t)$:

$f''(t) = n \cdot \frac{\mathrm{d}}{\mathrm{d}t}\!\Big[t^{n-1}(1-t)^{n-1}\Big] \cdot (1-2t) \;+\; nt^{n-1}(1-t)^{n-1} \cdot (-2)$

Computing $\frac{\mathrm{d}}{\mathrm{d}t}\!\big[t^{n-1}(1-t)^{n-1}\big]$:

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

$= (n-1)t^{n-2}(1-t)^{n-2}\big[(1-t) - t\big]$

$= (n-1)t^{n-2}(1-t)^{n-2}(1-2t)$

Substituting back:

$f''(t) = n(n-1)t^{n-2}(1-t)^{n-2}(1-2t)^2 - 2nt^{n-1}(1-t)^{n-1}$

$= nt^{n-2}(1-t)^{n-2}\Big[(n-1)(1-2t)^2 - 2t(1-t)\Big]$

Now expand the bracket. Using $(1-2t)^2 = 1 - 4t + 4t^2$:

$(n-1)(1 - 4t + 4t^2) - 2t(1-t)$

$= (n-1) - 4(n-1)t + 4(n-1)t^2 - 2t + 2t^2$

$= (n-1) - \big[4(n-1) + 2\big]t + \big[4(n-1) + 2\big]t^2$

$= (n-1) - (4n-2)t + (4n-2)t^2$

$= (n-1) - (4n-2)t(1-t)$

$= (n-1) - 2(2n-1)t(1-t)$

Therefore:

$f''(t) = nt^{n-2}(1-t)^{n-2}\Big[(n-1) - 2(2n-1)t(1-t)\Big] \qquad \blacksquare$

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Part (ii)

We use integration by parts twice. Let:

$T_n = \int_0^1 \frac{t^n(1-t)^n}{n!}\,e^t\,\mathrm{d}t$

First integration by parts. Let $u = t^n(1-t)^n$ and $\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{e^t}{n!}$.

Then $\frac{\mathrm{d}u}{\mathrm{d}t} = nt^{n-1}(1-t)^{n-1}(1-2t)$ and $v = \frac{e^t}{n!}$.

$T_n = \left[\frac{t^n(1-t)^n\,e^t}{n!}\right]_0^1 - \int_0^1 \frac{nt^{n-1}(1-t)^{n-1}(1-2t)}{n!}\,e^t\,\mathrm{d}t$

The boundary term vanishes: at $t = 0$, $t^n = 0$; at $t = 1$, $(1-t)^n = 0$ (for $n \ge 1$).

$T_n = -\int_0^1 \frac{nt^{n-1}(1-t)^{n-1}(1-2t)}{n!}\,e^t\,\mathrm{d}t \qquad (\star)$

Second integration by parts. Let $u = nt^{n-1}(1-t)^{n-1}(1-2t)$ and $\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{e^t}{n!}$.

From part (i), $\frac{\mathrm{d}u}{\mathrm{d}t} = f''(t) = nt^{n-2}(1-t)^{n-2}\big[(n-1) - 2(2n-1)t(1-t)\big]$ and $v = \frac{e^t}{n!}$.

Applying integration by parts to $(\star)$:

$T_n = -\left[\frac{nt^{n-1}(1-t)^{n-1}(1-2t)\,e^t}{n!}\right]_0^1 + \int_0^1 \frac{nt^{n-2}(1-t)^{n-2}\big[(n-1) - 2(2n-1)t(1-t)\big]}{n!}\,e^t\,\mathrm{d}t$

The boundary term vanishes for $n \ge 2$: at $t = 0$, $t^{n-1} = 0$; at $t = 1$, $(1-t)^{n-1} = 0$.

Splitting the remaining integral:

$T_n = \int_0^1 \frac{n(n-1)\,t^{n-2}(1-t)^{n-2}}{n!}\,e^t\,\mathrm{d}t - \int_0^1 \frac{2n(2n-1)\,t^{n-1}(1-t)^{n-1}}{n!}\,e^t\,\mathrm{d}t$

Simplifying the factorials. For the first integral: $\frac{n(n-1)}{n!} = \frac{1}{(n-2)!}$. For the second: $\frac{2n(2n-1)}{n!}$ — wait, let me be more careful. The second integral has the factor $2(2n-1)$ from the bracket and $n$ from the original $f''$ expression:

$\frac{n \cdot 2(2n-1)}{n!} = \frac{2(2n-1)}{(n-1)!}$

Hmm, that’s not quite right. Let me re-check.

From the integral split:

$T_n = \int_0^1 \frac{n \cdot (n-1) \cdot t^{n-2}(1-t)^{n-2}}{n!}\,e^t\,\mathrm{d}t - \int_0^1 \frac{n \cdot 2(2n-1) \cdot t^{n-1}(1-t)^{n-1}}{n!}\,e^t\,\mathrm{d}t$

For the first integral: $\frac{n(n-1)}{n!} = \frac{1}{(n-2)!}$, so:

$\int_0^1 \frac{t^{n-2}(1-t)^{n-2}}{(n-2)!}\,e^t\,\mathrm{d}t = T_{n-2}$

For the second integral: $\frac{2n(2n-1)}{n!} = \frac{2(2n-1)}{(n-1)!}$, so:

$\int_0^1 \frac{2(2n-1)\,t^{n-1}(1-t)^{n-1}}{(n-1)!}\,e^t\,\mathrm{d}t = 2(2n-1)\,T_{n-1}$

Therefore:

$T_n = T_{n-2} - 2(2n-1)\,T_{n-1} \qquad \text{for } n \ge 2 \qquad \blacksquare$

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Part (iii)

Evaluating $T_0$:

$T_0 = \int_0^1 e^t\,\mathrm{d}t = \Big[e^t\Big]_0^1 = e - 1$

So $a_0 = -1$, $b_0 = 1$.

Evaluating $T_1$:

$T_1 = \int_0^1 t(1-t)\,e^t\,\mathrm{d}t = \int_0^1 (t - t^2)\,e^t\,\mathrm{d}t$

For $\int_0^1 t\,e^t\,\mathrm{d}t$: integration by parts with $u = t$, $v' = e^t$:

$\Big[t\,e^t\Big]_0^1 - \int_0^1 e^t\,\mathrm{d}t = e - (e - 1) = 1$

For $\int_0^1 t^2\,e^t\,\mathrm{d}t$: integration by parts with $u = t^2$, $v' = e^t$:

$\Big[t^2\,e^t\Big]_0^1 - 2\int_0^1 t\,e^t\,\mathrm{d}t = e - 2(1) = e - 2$

Therefore:

$T_1 = 1 - (e - 2) = 3 - e$

So $a_1 = 3$, $b_1 = -1$.

Deducing the integer form by strong induction:

Base cases: $T_0 = -1 + 1 \cdot e$ and $T_1 = 3 + (-1) \cdot e$, with $a_0, b_0, a_1, b_1$ all integers.

Inductive step: Suppose $T_{n-2} = a_{n-2} + b_{n-2}\,e$ and $T_{n-1} = a_{n-1} + b_{n-1}\,e$ where all four coefficients are integers. By the recurrence from part (ii):

$T_n = T_{n-2} - 2(2n-1)\,T_{n-1}$

$= \big[a_{n-2} + b_{n-2}\,e\big] - 2(2n-1)\big[a_{n-1} + b_{n-1}\,e\big]$

$= \underbrace{\big[a_{n-2} - 2(2n-1)\,a_{n-1}\big]}_{a_n} + \underbrace{\big[b_{n-2} - 2(2n-1)\,b_{n-1}\big]}_{b_n}\,e$

Since $2(2n-1)$ is an integer and the inductive hypothesis gives integer coefficients, $a_n$ and $b_n$ are integers.

By induction, for all $n \ge 0$:

$T_n = a_n + b_n\,e \qquad \text{with } a_n, b_n \in \mathbb{Z} \qquad \blacksquare$

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Part (iv)

Showing $0 < T_n < \frac{e}{n!}$:

For $0 < t < 1$: both $t$ and $1-t$ are positive and less than $1$, so $t^n(1-t)^n > 0$. Also $e^t > 0$. Therefore the integrand is strictly positive on $(0,1)$, and:

$T_n = \int_0^1 \frac{t^n(1-t)^n}{n!}\,e^t\,\mathrm{d}t > 0$

For the upper bound: on $0 \le t \le 1$, we have $0 \le t \le 1$ and $0 \le 1-t \le 1$, so $t^n(1-t)^n \le 1$. Also $e^t \le e$. Therefore:

$\frac{t^n(1-t)^n}{n!}\,e^t \le \frac{e}{n!}$

for all $t \in [0,1]$. Since the inequality is strict on $(0,1)$ (where $t^n(1-t)^n < 1$):

$T_n < \int_0^1 \frac{e}{n!}\,\mathrm{d}t = \frac{e}{n!}$

Therefore $0 < T_n < \frac{e}{n!}$ for all $n \ge 0$. $\qquad \blacksquare$

Deducing $\frac{-a_n}{b_n} \to e$:

Since $0 < T_n < \frac{e}{n!}$ and $\frac{e}{n!} \to 0$ as $n \to \infty$, by the squeeze theorem:

$T_n = a_n + b_n\,e \to 0 \quad \text{as } n \to \infty$

Since $b_n \ne 0$ for all $n$, we can divide by $b_n$:

$\frac{a_n}{b_n} + e \to 0 \quad \text{as } n \to \infty$

$\frac{-a_n}{b_n} \to e \quad \text{as } n \to \infty \qquad \blacksquare$

Examiner Notes

无官方评述。易错点：(1) part(i)的二阶导数计算繁琐，符号易错；(2) part(ii)中两次分部积分的边界项处理；(3) part(iv)中估计Tn的上下界时积分放缩方向；(4) 从Tn的界推出aₙ/bₙ收敛性的逻辑链条。