STEP3 2024 -- Pure Mathematics

STEP3 2024 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2024  |  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	纯数	Routine	部分分式分解,伸缩求和(telescoping series),极限运算,级数拆分
2	纯数	Standard	两边平方去根号,渐近分析,图像法,构造法
3	纯数	Standard	求导分析单调性,极限分析,分类讨论(c≥1/2, 0<c<1/2, c=0)
4	纯数	Standard	切线方程推导,两直线夹角公式(m₁-m₂)/(1+m₁m₂)=±1,轨迹方程消参
5	纯数	Challenging	矩阵迹的循环性质,行列式求导(Jacobi公式),矩阵微分方程求解,构造反例
6	纯数	Hard	微分方程组求解,变量消元(d(x-y)/dt等),必要条件分析,控制理论思想
7	纯数	Challenging	几何级数估计,交错级数配对,阶乘与e的级数展开关系,反证法,适当选取n值
8	纯数	Challenging	直线对的代数判据(p²-q²=4r),曲线交点分析,因式分解,蕴含方向论证

---

Question 1

Topic: 纯数  |  Difficulty: Routine  |  Marks: 20

Problem

1 Throughout this question, $N$ is an integer with $N \geqslant 1$ and $S_N = \sum_{r=1}^{N} \frac{1}{r^2}$.

You may assume that $\lim_{N \to \infty} S_N$ exists and is equal to $\frac{1}{6}\pi^2$.

(i) Show that $\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{1}{r^2(r+1)}.$

Hence show that $\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r^2} - 1 + \frac{1}{N+1}.$

Show further that $\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1$.

(ii) Find $\sum_{r=1}^{N} \frac{1}{r^2(r+1)(r+2)}$ in terms of $S_N$, and hence evaluate $\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)}$.

(iii) Show that $\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)^2} = \sum_{r=1}^{\infty} \frac{2}{r^2(r+1)} - 1.$

Hint

(i)

$\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{1}{r^2(r+1)} [r^2 - r(r+1) + (r+1)]$ $= \frac{1}{r^2(r+1)} [r^2 - r^2 - r + r + 1] = \frac{1}{r^2(r+1)}$

*B1

Thus

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r+1} - \sum_{r=1}^{N} \frac{1}{r} + \sum_{r=1}^{N} \frac{1}{r^2}$ $= \sum_{r=2}^{N+1} \frac{1}{r} - \sum_{r=1}^{N} \frac{1}{r} + \sum_{r=1}^{N} \frac{1}{r^2}$

M1

$= \frac{1}{N+1} - 1 + \sum_{r=1}^{N} \frac{1}{r^2}$

as required. *A1

So, as $N \rightarrow \infty$,

$LHS \rightarrow \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)}, \frac{1}{N+1} \rightarrow 0, \text{and } \sum_{r=1}^{N} \frac{1}{r^2} \rightarrow \frac{1}{6}\pi^2$

and hence

$\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1$

*B1 (4)

(ii)

$\frac{1}{r^2(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2}$

M1

$1 = Ar(r+1)(r+2) + B(r+1)(r+2) + Cr^2(r+2) + Dr^2(r+1)$

$r = 0 \quad 1 = 2B \quad B = \frac{1}{2}$

$r = -1 \quad 1 = C$

$r = -2 \quad 1 = -4D \quad D = -\frac{1}{4}$

$r^3 \text{ terms} \quad 0 = A + C + D \quad A = -\frac{3}{4}$

M1 A1 (3)

---

Thus

$\sum_{r=1}^{N} \frac{1}{r^{2}(r+1)(r+2)} = -\frac{3}{4} \sum_{r=1}^{N} \frac{1}{r} + \frac{1}{2} \sum_{r=1}^{N} \frac{1}{r^{2}} + \sum_{r=1}^{N} \frac{1}{r+1} - \frac{1}{4} \sum_{r=1}^{N} \frac{1}{r+2}$

M1

$= -\frac{3}{4} - \frac{3}{8} + \frac{1}{2} S_{N} + \frac{1}{2} + \frac{1}{N+1} - \frac{1}{4} \frac{1}{N+1} - \frac{1}{4} \frac{1}{N+2}$

dM1 A1ft

So

$\sum_{r=1}^{N} \frac{1}{r^{2}(r+1)(r+2)} = \frac{1}{2} S_{N} + \frac{3}{4} \frac{1}{N+1} - \frac{1}{4} \frac{1}{N+2} - \frac{5}{8}$

and taking limits as $N \rightarrow \infty$

M1

$\sum_{r=1}^{\infty} \frac{1}{r^{2}(r+1)(r+2)} = \frac{1}{2} \times \frac{1}{6} \pi^{2} - \frac{5}{8} = \frac{1}{12} \pi^{2} - \frac{5}{8}$

A1 (5)

(iii)

$\frac{1}{r^{2}(r+1)^{2}} = \frac{A}{r} + \frac{B}{r^{2}} + \frac{C}{r+1} + \frac{D}{(r+1)^{2}}$

M1

$1 = Ar(r+1)^{2} + B(r+1)^{2} + Cr^{2}(r+1) + Dr^{2}$ $r = 0 \quad 1 = B$ $r = -1 \quad 1 = D$ $r^{3} \text{ terms} \quad 0 = A + C$ $r^{2} \text{ terms} \quad 0 = 2A + B + C + D$

M1

$-2 = 2A + C$ $A = -2 \quad C = 2$

Thus

$\frac{1}{r^{2}(r+1)^{2}} = \frac{-2}{r} + \frac{1}{r^{2}} + \frac{2}{r+1} + \frac{1}{(r+1)^{2}}$

A1 (3)

---

$\sum_{r=1}^{N} \frac{1}{r^{2}(r+1)^{2}} = \sum_{r=1}^{N} \frac{-2}{r} + \sum_{r=1}^{N} \frac{1}{r^{2}} + \sum_{r=1}^{N} \frac{2}{(r+1)} + \sum_{r=1}^{N} \frac{1}{(r+1)^{2}}$ $= \sum_{r=1}^{N} \frac{-2}{r} + \sum_{r=1}^{N} \frac{1}{r^{2}} + \sum_{r=2}^{N+1} \frac{2}{r} + \sum_{r=2}^{N+1} \frac{1}{r^{2}}$

M1

$= -2 + \frac{2}{N+1} + 2 \sum_{r=1}^{N} \frac{1}{r^{2}} - 1 + \frac{1}{(N+1)^{2}}$

That is

$\sum_{r=1}^{N} \frac{1}{r^{2}(r+1)^{2}} = 2S_{N} - 3 + \frac{2}{N+1} + \frac{1}{(N+1)^{2}}$

A1

From part (i),

$\sum_{r=1}^{N} \frac{1}{r^{2}(r+1)} = \frac{1}{N+1} - 1 + \sum_{r=1}^{N} \frac{1}{r^{2}} = S_{N} - 1 + \frac{1}{N+1}$

Thus

$\sum_{r=1}^{N} \frac{1}{r^{2}(r+1)^{2}} = 2 \left( \sum_{r=1}^{N} \frac{1}{r^{2}(r+1)} + 1 - \frac{1}{N+1} \right) - 3 + \frac{2}{N+1} + \frac{1}{(N+1)^{2}}$ M1 $= 2 \sum_{r=1}^{N} \frac{1}{r^{2}(r+1)} - 1 + \frac{1}{(N+1)^{2}}$ A1

Letting $N \rightarrow \infty$ ,

$\sum_{r=1}^{\infty} \frac{1}{r^{2}(r+1)^{2}} = \sum_{r=1}^{\infty} \frac{2}{r^{2}(r+1)} - 1$ B1* (5)

---

Model Solution

Part (i)

We show the identity by combining the left-hand side over a common denominator $r^2(r+1)$:

$\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{r^2 - r(r+1) + (r+1)}{r^2(r+1)} = \frac{r^2 - r^2 - r + r + 1}{r^2(r+1)} = \frac{1}{r^2(r+1)}$

as required.

Now summing from $r = 1$ to $N$:

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r+1} - \sum_{r=1}^{N} \frac{1}{r} + \sum_{r=1}^{N} \frac{1}{r^2}$

Re-indexing the first sum by setting $s = r + 1$:

$\sum_{r=1}^{N} \frac{1}{r+1} = \sum_{s=2}^{N+1} \frac{1}{s}$

Therefore:

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=2}^{N+1} \frac{1}{r} - \sum_{r=1}^{N} \frac{1}{r} + \sum_{r=1}^{N} \frac{1}{r^2}$

The first two sums telescope:

$\sum_{r=2}^{N+1} \frac{1}{r} - \sum_{r=1}^{N} \frac{1}{r} = \frac{1}{N+1} - 1$

Hence:

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r^2} - 1 + \frac{1}{N+1}$

as required.

Taking the limit as $N \to \infty$: since $\frac{1}{N+1} \to 0$ and $\sum_{r=1}^{N} \frac{1}{r^2} \to \frac{1}{6}\pi^2$:

$\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1$

---

Part (ii)

We perform a partial fraction decomposition:

$\frac{1}{r^2(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2}$

Multiplying both sides by $r^2(r+1)(r+2)$:

$1 = Ar(r+1)(r+2) + B(r+1)(r+2) + Cr^2(r+2) + Dr^2(r+1)$

Setting $r = 0$: $1 = 2B$, so $B = \frac{1}{2}$.

Setting $r = -1$: $1 = C$, so $C = 1$.

Setting $r = -2$: $1 = D(-2)^2(-1) = -4D$, so $D = -\frac{1}{4}$.

Comparing $r^3$ coefficients: $0 = A + C + D = A + 1 - \frac{1}{4}$, so $A = -\frac{3}{4}$.

Therefore:

$\frac{1}{r^2(r+1)(r+2)} = -\frac{3}{4r} + \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)}$

Summing from $r = 1$ to $N$:

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)(r+2)} = -\frac{3}{4}\sum_{r=1}^{N}\frac{1}{r} + \frac{1}{2}\sum_{r=1}^{N}\frac{1}{r^2} + \sum_{r=1}^{N}\frac{1}{r+1} - \frac{1}{4}\sum_{r=1}^{N}\frac{1}{r+2}$

Re-indexing the third and fourth sums:

$\sum_{r=1}^{N} \frac{1}{r+1} = \sum_{r=2}^{N+1} \frac{1}{r} = \sum_{r=1}^{N} \frac{1}{r} - 1 + \frac{1}{N+1}$

$\sum_{r=1}^{N} \frac{1}{r+2} = \sum_{r=3}^{N+2} \frac{1}{r} = \sum_{r=1}^{N} \frac{1}{r} - 1 - \frac{1}{2} + \frac{1}{N+1} + \frac{1}{N+2}$

Substituting and collecting the coefficients of $\sum_{r=1}^{N} \frac{1}{r}$:

$-\frac{3}{4} + 1 - \frac{1}{4} = 0$

so the harmonic sum terms cancel entirely. We are left with:

$\frac{1}{2}S_N + \left(-1 + \frac{1}{N+1}\right) - \frac{1}{4}\left(-\frac{3}{2} + \frac{1}{N+1} + \frac{1}{N+2}\right)$

Collecting constant terms and $N$-dependent terms:

$= \frac{1}{2}S_N - 1 + \frac{1}{N+1} + \frac{3}{8} - \frac{1}{4(N+1)} - \frac{1}{4(N+2)}$

$= \frac{1}{2}S_N + \frac{3}{4(N+1)} - \frac{1}{4(N+2)} - \frac{5}{8}$

Taking the limit as $N \to \infty$, the terms $\frac{3}{4(N+1)}$ and $\frac{1}{4(N+2)}$ vanish:

$\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)} = \frac{1}{2} \cdot \frac{\pi^2}{6} - \frac{5}{8} = \frac{\pi^2}{12} - \frac{5}{8}$

---

Part (iii)

We perform a partial fraction decomposition:

$\frac{1}{r^2(r+1)^2} = \frac{A}{r} + \frac{B}{r^2} + \frac{C}{r+1} + \frac{D}{(r+1)^2}$

Multiplying both sides by $r^2(r+1)^2$:

$1 = Ar(r+1)^2 + B(r+1)^2 + Cr^2(r+1) + Dr^2$

Setting $r = 0$: $1 = B$, so $B = 1$.

Setting $r = -1$: $1 = D$, so $D = 1$.

Comparing $r^3$ coefficients: $0 = A + C$.

Comparing $r^2$ coefficients: $0 = 2A + B + C + D = 2A + C + 2$.

Substituting $C = -A$ into the second equation: $0 = 2A - A + 2 = A + 2$, so $A = -2$ and $C = 2$.

Therefore:

$\frac{1}{r^2(r+1)^2} = -\frac{2}{r} + \frac{1}{r^2} + \frac{2}{r+1} + \frac{1}{(r+1)^2}$

Summing from $r = 1$ to $N$:

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)^2} = -2\sum_{r=1}^{N}\frac{1}{r} + \sum_{r=1}^{N}\frac{1}{r^2} + 2\sum_{r=1}^{N}\frac{1}{r+1} + \sum_{r=1}^{N}\frac{1}{(r+1)^2}$

The first and third sums telescope:

$-2\sum_{r=1}^{N}\frac{1}{r} + 2\sum_{r=2}^{N+1}\frac{1}{r} = -2 \cdot 1 + \frac{2}{N+1} = -2 + \frac{2}{N+1}$

The second and fourth sums combine:

$\sum_{r=1}^{N}\frac{1}{r^2} + \sum_{r=2}^{N+1}\frac{1}{r^2} = 2S_N - 1 + \frac{1}{(N+1)^2}$

Combining all parts:

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)^2} = 2S_N - 3 + \frac{2}{N+1} + \frac{1}{(N+1)^2} \qquad \text{(...)}$

From part (i), we know $\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = S_N - 1 + \frac{1}{N+1}$, so $S_N = \sum_{r=1}^{N} \frac{1}{r^2(r+1)} + 1 - \frac{1}{N+1}$. Substituting into (…):

$\sum_{r=1}^{N} \frac{1}{r^2(r+1)^2} = 2\left(\sum_{r=1}^{N} \frac{1}{r^2(r+1)} + 1 - \frac{1}{N+1}\right) - 3 + \frac{2}{N+1} + \frac{1}{(N+1)^2}$

$= 2\sum_{r=1}^{N} \frac{1}{r^2(r+1)} + 2 - \frac{2}{N+1} - 3 + \frac{2}{N+1} + \frac{1}{(N+1)^2}$

$= 2\sum_{r=1}^{N} \frac{1}{r^2(r+1)} - 1 + \frac{1}{(N+1)^2}$

Taking the limit as $N \to \infty$:

$\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)^2} = 2\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} - 1$

Examiner Notes

全卷最受欢迎且得分最高的题目（均分约15/20，98%+考生尝试）。考官报告指出：关键易错点包括将调和级数误当作收敛级数处理、将∞当作数字代入表达式。部分分式分解有多种正确方法，但解释需要清晰完整。

---

Question 2

Topic: 纯数  |  Difficulty: Standard  |  Marks: 20

Problem

2 (i) Solve the inequalities

         **(a)**    $\sqrt{4x^2 - 8x + 64} \leqslant |x + 8|$,

         **(b)**    $\sqrt{4x^2 - 8x + 64} \leqslant |3x - 8|$.

   **(ii)**    **(a)**    Let $f(x) = \sqrt{4x^2 - 8x + 64} - 2(x - 1)$.

                 Show, by considering $(\sqrt{4x^2 - 8x + 64} + 2(x - 1))f(x)$ or otherwise, that $f(x) \to 0$ as $x \to \infty$.

         **(b)**    Sketch $y = \sqrt{4x^2 - 8x + 64}$ and $y = 2(x - 1)$ on the same axes.

   **(iii)** Find a value of $m$ and the corresponding value of $c$ such that the solution set of the inequality

$\sqrt{4x^2 - 5x + 4} \leqslant |mx + c|$

         is $\{x : x \geqslant 3\}$.

   **(iv)** Find values of $p, q, m$ and $c$ such that the solution set of the inequality

$|x^2 + px + q| \leqslant mx + c$

         is $\{x : -5 \leqslant x \leqslant 1\} \cup \{x : 5 \leqslant x \leqslant 7\}$.

Hint

(i) (a)**

$\sqrt{4x^2 - 8x + 64} \leq |x + 8|$

$4x^2 - 8x + 64 \leq (x + 8)^2 = x^2 + 16x + 64$

Thus

$3x^2 - 24x = 3x(x - 8) \leq 0$

M1

G1 or consideration of intervals M1

Alternative method Solve for critical values M1

Sketch graph

![Graph showing a parabola and an absolute value function intersecting on a grid

G1 ]

So $0 \leq x \leq 8$ A1 (3)

(b)

$\sqrt{4x^2 - 8x + 64} \leq |3x - 8|$

$4x^2 - 8x + 64 \leq (3x - 8)^2 = 9x^2 - 48x + 64$

Thus

$5x^2 - 40x = 5x(x - 8) \geq 0$

M1

---

GRAPH G1 or consideration of intervals M1

Alternative method Solve for critical values M1

Sketch graph

![A blue curve and a red V-shaped curve intersecting on a coordinate grid G1 ]

So $x \le 0$ or $x \ge 8$ A1 (3)

(ii) (a)

$(\sqrt{4x^2 - 8x + 64} + 2(x - 1)) f(x)$ $= (\sqrt{4x^2 - 8x + 64} + 2(x - 1))(\sqrt{4x^2 - 8x + 64} - 2(x - 1))$ $= 4x^2 - 8x + 64 - 4x^2 + 8x - 4 = 60$

Thus $f(x) = \frac{60}{(\sqrt{4x^2 - 8x + 64} + 2(x - 1))}$

and so $f(x) \to 0$ as $x \to \infty$

E1 (1)

---

(b)

G2 (2)

(iii) Require one critical value 3, so $3m + c = \pm 5$

M1

and as only one critical value choose $m = 2$ and $c = -1$, or $m = -2$ and $c = 1$

dM1 A1

$\sqrt{4x^2 - 5x + 4} \le |2x - 1|$

$4x^2 - 5x + 4 \le (2x - 1)^2 = 4x^2 - 4x + 1$

Giving $x \ge 3$

M1

G1 (5)

---

(iv) To obtain 4 critical values require quadratic to cross x axis and so

E1

$x^2 + px + q = mx + c$ has roots -5 and 7 giving $p - m = -2$ and $q - c = -35$

M1 A1

and

$-(x^2 + px + q) = mx + c$

has roots 1 and 5 giving $p + m = -6$ and $q + c = 5$

B1

Thus, $p = -4$, $q = -15, m = -2$, $c = 20$

A1

x	y (quadratic)	y (linear)
-5	0	30
-2	11	24
0	15	20
1	12	18
3	6	14
5	0	10
7	0	6

G1 (6)

---

Model Solution

Part (i)(a)

We need to solve $\sqrt{4x^2 - 8x + 64} \leqslant |x + 8|$.

The left-hand side is a square root, so it is non-negative for all real $x$. The right-hand side is an absolute value, also non-negative. Since both sides are non-negative, we may square both sides without changing the direction of the inequality:

$4x^2 - 8x + 64 \leqslant (x + 8)^2 = x^2 + 16x + 64$

Rearranging:

$3x^2 - 24x \leqslant 0$

$3x(x - 8) \leqslant 0$

The product $3x(x-8)$ is zero at $x = 0$ and $x = 8$, and is negative when $0 < x < 8$ (one factor positive, one negative). Therefore the solution set is $0 \leqslant x \leqslant 8$.

---

Part (i)(b)

We need to solve $\sqrt{4x^2 - 8x + 64} \leqslant |3x - 8|$.

Again, both sides are non-negative, so squaring preserves the inequality:

$4x^2 - 8x + 64 \leqslant (3x - 8)^2 = 9x^2 - 48x + 64$

Rearranging:

$0 \leqslant 5x^2 - 40x$

$5x(x - 8) \geqslant 0$

The product $5x(x-8)$ is zero at $x = 0$ and $x = 8$, and is non-negative when $x \leqslant 0$ or $x \geqslant 8$ (both factors same sign). Therefore the solution set is $x \leqslant 0$ or $x \geqslant 8$.

---

Part (ii)(a)

Let $f(x) = \sqrt{4x^2 - 8x + 64} - 2(x - 1)$. Consider the product:

$\left(\sqrt{4x^2 - 8x + 64} + 2(x - 1)\right) f(x)$

$= \left(\sqrt{4x^2 - 8x + 64}\right)^2 - \left(2(x - 1)\right)^2$

$= (4x^2 - 8x + 64) - 4(x^2 - 2x + 1)$

$= 4x^2 - 8x + 64 - 4x^2 + 8x - 4 = 60$

Therefore:

$f(x) = \frac{60}{\sqrt{4x^2 - 8x + 64} + 2(x - 1)}$

As $x \to \infty$, the denominator $\sqrt{4x^2 - 8x + 64} + 2(x - 1) \to \infty$ (both terms grow without bound), so $f(x) \to 0$.

---

Part (ii)(b)

Complete the square: $4x^2 - 8x + 64 = 4(x - 1)^2 + 60$, so $y = \sqrt{4(x-1)^2 + 60}$.

Key features:

• Symmetric about $x = 1$.
• At $x = 1$: $y = \sqrt{60} = 2\sqrt{15} \approx 7.75$.
• As $x \to \pm\infty$: $y \to \infty$.

The line $y = 2(x - 1)$ passes through $(1, 0)$ with gradient 2. From part (ii)(a), this is an asymptote of the curve for large $x$. Since $f(x) > 0$ for all $x$, the curve lies strictly above this asymptote.

The sketch shows the curve starting at $(0, 2\sqrt{15})$, decreasing to its minimum $2\sqrt{15}$ at $x = 1$, then increasing and approaching the dashed line $y = 2(x-1)$ from above as $x \to \infty$. The left branch rises steeply as $x \to -\infty$.

---

Part (iii)

We need $m$ and $c$ such that $\sqrt{4x^2 - 5x + 4} \leqslant |mx + c|$ has solution set $\{x : x \geqslant 3\}$.

Squaring both sides (valid since both are non-negative):

$4x^2 - 5x + 4 \leqslant m^2 x^2 + 2mcx + c^2$

$(m^2 - 4)x^2 + (2mc + 5)x + (c^2 - 4) \geqslant 0$

For the solution set to be $x \geqslant 3$ (a single boundary point), the quadratic must reduce to a linear expression with positive coefficient and root at $x = 3$. This requires $m^2 = 4$, i.e. $m = \pm 2$.

The inequality becomes linear: $(2mc + 5)x + (c^2 - 4) \geqslant 0$, giving $x \geqslant \frac{4 - c^2}{2mc + 5}$ (when $2mc + 5 > 0$).

Setting $\frac{4 - c^2}{2mc + 5} = 3$: $4 - c^2 = 6mc + 15$, i.e. $c^2 + 6mc + 11 = 0$.

With $m = 2$: $c^2 + 12c + 11 = (c+1)(c+11) = 0$, so $c = -1$ or $c = -11$.

• $c = -1$: $2(2)(-1) + 5 = 1 > 0$. Solution: $x \geqslant 3$. Check at $x = 3$: $\sqrt{36 - 15 + 4} = 5$ and $|6 - 1| = 5$. Equality holds. Valid.
• $c = -11$: $2(2)(-11) + 5 = -39 < 0$. Direction flips. Not valid.

With $m = -2$: $c^2 - 12c + 11 = (c-1)(c-11) = 0$, giving $c = 1$ or $c = 11$.

• $c = 1$: $2(-2)(1) + 5 = 1 > 0$. But $|mx + c| = |-2x + 1| = |2x - 1|$, same as $m = 2, c = -1$.

So $m = 2$, $c = -1$.

Verification: $\sqrt{4x^2 - 5x + 4} \leqslant |2x - 1|$. Squaring: $4x^2 - 5x + 4 \leqslant 4x^2 - 4x + 1$, giving $x \geqslant 3$.

---

Part (iv)

We need $p, q, m, c$ such that $|x^2 + px + q| \leqslant mx + c$ has solution set $\{-5 \leqslant x \leqslant 1\} \cup \{5 \leqslant x \leqslant 7\}$.

Since $mx + c \geqslant 0$ on the solution set, and the solution has two separate intervals, the quadratic $x^2 + px + q$ must be negative in the gap $(1, 5)$. At the boundary points, equality holds.

At $x = -5$ and $x = 7$ (where the quadratic is non-negative): $x^2 + px + q = mx + c$.

Substituting: $25 - 5p + q = -5m + c$ and $49 + 7p + q = 7m + c$.

Subtracting: $24 + 12p = 12m$, so $p - m = -2$.

At $x = 1$ and $x = 5$ (where the quadratic is negative): $-(x^2 + px + q) = mx + c$.

Substituting: $-(1 + p + q) = m + c$ and $-(25 + 5p + q) = 5m + c$.

Subtracting: $-24 - 4p = 4m$, so $p + m = -6$.

From $p - m = -2$ and $p + m = -6$: $p = -4$, $m = -2$.

From $-(1 + p + q) = m + c$: $3 - q = -2 + c$, so $q + c = 5$.

From $25 - 5p + q = -5m + c$: $45 + q = 10 + c$, so $q - c = -35$.

Adding: $2q = -30$, so $q = -15$ and $c = 20$.

Therefore $p = -4$, $q = -15$, $m = -2$, $c = 20$.

Verification: The quadratic is $x^2 - 4x - 15 = (x + 5)(x - 7)$, with roots at $x = -5$ and $x = 7$, negative on $(-5, 7)$. The linear function is $-2x + 20$.

• At $x = -5$: $|0| = 0 \leqslant 30$. OK.
• At $x = 1$: $|1 - 4 - 15| = 18 \leqslant 18$. OK (boundary).
• At $x = 5$: $|25 - 20 - 15| = 10 \leqslant 10$. OK (boundary).
• At $x = 7$: $|0| = 0 \leqslant 6$. OK.
• At $x = 3$ (in gap): $|9 - 12 - 15| = 18$ and $-6 + 20 = 14$. So $18 \leqslant 14$ is false. Correctly excluded.
• At $x = 8$ (outside): $|64 - 32 - 15| = 17$ and $-16 + 20 = 4$. So $17 \leqslant 4$ is false. Correctly excluded.

Examiner Notes

约75%考生尝试，均分约10/20。常见问题：平方前未说明LHS为实数且两边为正；(ii)(a)中仅凭首项抵消就断言f(x)→0不得满分；作图需体现渐近线和关于x=1的对称性；(iv)部分使用图像法的考生表现明显更好，常见错误是四个根未能正确对应二次函数的正负号。

---

Question 3

Topic: 纯数  |  Difficulty: Standard  |  Marks: 20

Problem

3 Throughout this question, consider only $x > 0$.

(i) Let $g(x) = \ln \left( 1 + \frac{1}{x} \right) - \frac{x + c}{x(x + 1)}$ where $c \geqslant 0$.

(a) Show that $y = g(x)$ has positive gradient for all $x > 0$ when $c \geqslant \frac{1}{2}$.

(b) Find the values of $x$ for which $y = g(x)$ has negative gradient when $0 \leqslant c < \frac{1}{2}$.

(ii) It is given that, for all $c > 0$, $g(x) \to -\infty$ as $x \to 0$.

Sketch, for $x > 0$, the graphs of $y = g(x)$ in the cases

(a) $c = \frac{3}{4}$,

(b) $c = \frac{1}{4}$.

(iii) The function f is defined as $f(x) = \left( 1 + \frac{1}{x} \right)^{x+c}.$ Show that, for $x > 0$,

(a) f is a decreasing function when $c \geqslant \frac{1}{2}$;

(b) f has a turning point when $0 < c < \frac{1}{2}$;

(c) f is an increasing function when $c = 0$.

Hint

(i) (a)

$y = g(x) = \ln \left( 1 + \frac{1}{x} \right) - \frac{x + c}{x(x + 1)} = \ln(x + 1) - \ln x - \frac{x + c}{x(x + 1)}$

$\frac{dy}{dx} = \frac{1}{x + 1} - \frac{1}{x} - \frac{x(x + 1) - (x + c)(2x + 1)}{x^2(x + 1)^2}$

M1

$= \frac{x^2(x + 1) - x(x + 1)^2 - x(x + 1) + (x + c)(2x + 1)}{x^2(x + 1)^2}$

$= \frac{(2c - 1)x + c}{x^2(x + 1)^2}$

A1

When $c \geq \frac{1}{2}$, as $x > 0$, $(2c - 1)x \geq 0$, and $c > 0$, so the numerator is positive and the denominator is a non-zero square so also positive, so $y = g(x)$ has positive gradient.

E1 (3)

(b) $y = g(x)$ has negative gradient for $0 \leq c < \frac{1}{2}$, if

$(2c - 1)x + c < 0$

That is

$(2c - 1)x < -c$

So

$x > \frac{-c}{2c - 1} = \frac{c}{1 - 2c}$

B1 (1)

(ii) (a)

If $c = \frac{3}{4}$, then from (i) (a) the gradient is positive

E1

and we are given that $g(x) \rightarrow -\infty$ as $x \rightarrow 0$.

The gradient tends to zero as $x \rightarrow \infty$, and to $\infty$ as $x \rightarrow 0$. E1

Also, $g(x) \rightarrow 0$ as $x \rightarrow \infty$. E1

---

G1 (4)

(b)

If $c = \frac{1}{4}$, then from (i)(b) the gradient is negative for $x > \frac{1}{2}$.

B1

The gradient is zero when $x = \frac{1}{2}$, and positive when $x < \frac{1}{2}$, and tending to zero as $x \rightarrow \infty$, and to $\infty$ as $x \rightarrow 0$. B1

Again, we are given that $g(x) \rightarrow -\infty$ as $x \rightarrow 0$, and $g(x) \rightarrow 0$ as $x \rightarrow \infty$.

There is a turning point (maximum) at $(\frac{1}{2}, \ln 3 - 1) = (\frac{1}{2}, \ln \frac{3}{e})$ which is above the x-axis.

M1 A1

G1 (5)

---

(iii)

$f(x) = \left( 1 + \frac{1}{x} \right)^{x+c}$

$\ln(f(x)) = (x + c) \ln \left( 1 + \frac{1}{x} \right)$

Thus

$\frac{f'(x)}{f(x)} = \ln \left( 1 + \frac{1}{x} \right) + (x + c) \left( \frac{1}{x + 1} - \frac{1}{x} \right) = g(x)$

$f'(x) = f(x)g(x)$

M1 A1 (2)

Also, $f(x)$ is positive for $x > 0$.

(a) As has been demonstrated in (i) (a) and (ii) (a), $g(x) < 0$ for $x > 0$ when $c \geq \frac{1}{2}$, so

$f'(x) < 0$ and f is a decreasing function. [E1]

(b) As has been demonstrated in (i) (b) and (ii) (b), $g(x) = 0$ for some $x > 0$ when $0 < c < \frac{1}{2}$, so $f'(x) = 0$ for some x and f has a turning point. [E1]

(c) When $c = 0$,

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

Is always negative and $\rightarrow -\infty$ as $x \rightarrow 0$, and $\rightarrow 0$ as $x \rightarrow \infty$

[E1]

whilst $g(x) \rightarrow \infty$ as $x \rightarrow 0$, and $g(x) \rightarrow 0$ as $x \rightarrow \infty$

so $g(x)$ is positive for all $x > 0$, [E1]

thus $f'(x)$ is too and thus f is an increasing function

for all $x > 0$ [E1 (5)]

---

Model Solution

Part (i)(a)

We write $g(x) = \ln(x+1) - \ln x - \frac{x+c}{x(x+1)}$ and compute $g'(x)$.

The first two terms give: $\frac{d}{dx}[\ln(x+1) - \ln x] = \frac{1}{x+1} - \frac{1}{x} = \frac{-1}{x(x+1)}$

For the third term, using the quotient rule with numerator $x + c$ and denominator $x^2 + x$: $\frac{d}{dx}\left[\frac{x+c}{x^2+x}\right] = \frac{1 \cdot (x^2+x) - (x+c)(2x+1)}{(x^2+x)^2}$

Expanding the numerator: $x^2 + x - (2x^2 + x + 2cx + c) = -x^2 - 2cx - c$

So the third term contributes $\frac{-x^2 - 2cx - c}{x^2(x+1)^2}$ to $g'(x)$ (with a minus sign from the definition of $g$).

Combining over the common denominator $x^2(x+1)^2$: $g'(x) = \frac{-x(x+1)}{x^2(x+1)^2} - \frac{-x^2 - 2cx - c}{x^2(x+1)^2} = \frac{-x^2 - x + x^2 + 2cx + c}{x^2(x+1)^2} = \frac{(2c-1)x + c}{x^2(x+1)^2}$

When $c \geqslant \frac{1}{2}$ and $x > 0$:

• $2c - 1 \geqslant 0$, so $(2c-1)x \geqslant 0$
• $c \geqslant \frac{1}{2} > 0$

Hence the numerator $(2c-1)x + c > 0$. The denominator $x^2(x+1)^2 > 0$, so $g'(x) > 0$ for all $x > 0$. $\qquad \blacksquare$

---

Part (i)(b)

When $0 \leqslant c < \frac{1}{2}$, the gradient is negative when: $(2c-1)x + c < 0$

Since $2c - 1 < 0$, dividing by this negative quantity reverses the inequality: $x > \frac{-c}{2c-1} = \frac{c}{1-2c}$

When $c = 0$: this gives $x > 0$, so $g'(x) < 0$ for all $x > 0$.

When $0 < c < \frac{1}{2}$: $g(x)$ has negative gradient for $x > \frac{c}{1-2c}$.

---

Part (ii)(a): $c = \frac{3}{4}$

Since $c = \frac{3}{4} \geqslant \frac{1}{2}$, part (i)(a) tells us $g$ has positive gradient for all $x > 0$, so $g$ is strictly increasing.

We are given $g(x) \to -\infty$ as $x \to 0$. As $x \to \infty$: $\ln(1+1/x) \to 0$ and $\frac{x+c}{x(x+1)} \to 0$, so $g(x) \to 0$.

Since $g$ is increasing and approaches $0$ from below, we have $g(x) < 0$ for all $x > 0$.

The graph rises monotonically from $-\infty$, asymptotic to $y = 0$ from below.

Part (ii)(b): $c = \frac{1}{4}$

Since $c = \frac{1}{4} < \frac{1}{2}$, part (i)(b) gives $\frac{c}{1-2c} = \frac{1/4}{1/2} = \frac{1}{2}$.

• $g'(x) > 0$ for $0 < x < \frac{1}{2}$ (increasing)
• $g'(x) = 0$ at $x = \frac{1}{2}$ (turning point)
• $g'(x) < 0$ for $x > \frac{1}{2}$ (decreasing)

At the turning point: $g\!\left(\tfrac{1}{2}\right) = \ln\!\left(1 + 2\right) - \frac{\frac{1}{2} + \frac{1}{4}}{\frac{1}{2} \cdot \frac{3}{2}} = \ln 3 - \frac{3/4}{3/4} = \ln 3 - 1$

Since $\ln 3 \approx 1.099 > 1$, this maximum is positive. The graph rises from $-\infty$ to a maximum at $\left(\frac{1}{2},\, \ln 3 - 1\right)$ above the $x$-axis, then decreases towards $y = 0$ from above.

---

Part (iii)

Taking logarithms: $\ln f(x) = (x+c)\ln\!\left(1 + \frac{1}{x}\right)$

Differentiating using the product rule: $\frac{f'(x)}{f(x)} = \ln\!\left(1 + \frac{1}{x}\right) + (x+c) \cdot \frac{d}{dx}\ln\!\left(1 + \frac{1}{x}\right)$

Now $\frac{d}{dx}\ln\!\left(1 + \frac{1}{x}\right) = \frac{1}{1+1/x} \cdot \left(-\frac{1}{x^2}\right) = \frac{x}{x+1} \cdot \left(-\frac{1}{x^2}\right) = \frac{-1}{x(x+1)}$.

Therefore: $\frac{f'(x)}{f(x)} = \ln\!\left(1 + \frac{1}{x}\right) - \frac{x+c}{x(x+1)} = g(x)$

So $f'(x) = f(x)\,g(x)$.

Since $x > 0$, we have $1 + \frac{1}{x} > 1 > 0$, so $f(x) > 0$. The sign of $f'(x)$ is determined entirely by the sign of $g(x)$.

(a) When $c \geqslant \frac{1}{2}$: By part (i)(a), $g$ is increasing. By part (ii)(a), $g(x) < 0$ for all $x > 0$ (since $g$ increases towards $0$ from below). Hence $f'(x) < 0$ for all $x > 0$, and $f$ is a decreasing function.

(b) When $0 < c < \frac{1}{2}$: From the analysis in part (ii)(b), $g$ increases from $-\infty$ (as $x \to 0$) to a positive maximum, then decreases towards $0$ (as $x \to \infty$). By the intermediate value theorem, $g$ equals $0$ at exactly one value of $x$. At this point $f'(x) = 0$, so $f$ has a turning point.

(c) When $c = 0$:

$g'(x) = \frac{(2 \cdot 0 - 1)x + 0}{x^2(x+1)^2} = \frac{-x}{x^2(x+1)^2} = \frac{-1}{x(x+1)^2} < 0 \quad \text{for all } x > 0$

So $g$ is strictly decreasing. As $x \to \infty$: $g(x) \to 0$. As $x \to 0^+$: $\ln(1+1/x) \to +\infty$ while $\frac{1}{x+1} \to 1$, so $g(x) \to +\infty$.

Since $g$ decreases from $+\infty$ to $0$, we have $g(x) > 0$ for all $x > 0$. Hence $f'(x) > 0$ for all $x > 0$, and $f$ is an increasing function.

Examiner Notes

第二受欢迎题目，均分约9/20。常见问题：(i)(a)中c≥1/2时正梯度的充分论证常缺失；(i)(b)中忘记不等式在除以负数时需要反向；(ii)作图中多数能正确画出，但(a)中需论证正梯度和渐近线，(b)中需论证拐点前的正梯度；(iii)中忘记提及f>0是丢分常见原因。

---

Question 4

Topic: 纯数  |  Difficulty: Standard  |  Marks: 20

Problem

4 (i) Show that if the acute angle between straight lines with gradients $m_1$ and $m_2$ is $45^\circ$, then

$\frac{m_1 - m_2}{1 + m_1 m_2} = \pm 1.$

The curve $C$ has equation $4ay = x^2$ (where $a \neq 0$).

(ii) If $p \neq q$, show that the tangents to the curve $C$ at the points with $x$-coordinates $p$ and $q$ meet at a point with $x$-coordinate $\frac{1}{2}(p + q)$. Find the $y$-coordinate of this point in terms of $p$ and $q$.

Show further that any two tangents to the curve $C$ which are at $45^\circ$ to each other meet on the curve $(y + 3a)^2 = 8a^2 + x^2$.

(iii) Show that the acute angle between any two tangents to the curve $C$ which meet on the curve $(y + 7a)^2 = 48a^2 + 3x^2$ is constant. Find this acute angle.

Hint

(i)**

Suppose $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$, where $-\frac{1}{2}\pi < \theta_1, \theta_2 \le \frac{1}{2}\pi$, then as the angle between the lines is $45^\circ$, $\theta_1 - \theta_2 = \pm \frac{1}{4}\pi$, or $\pm \frac{3}{4}\pi$. [M1]

Therefore

$\tan(\theta_1 - \theta_2) = \pm 1$

and so

$\tan(\theta_1 - \theta_2) = \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2} = \pm 1$ [M1]

i.e.

$\frac{m_1 - m_2}{1 + m_1 m_2} = \pm 1$ [*A1 (3)]

(ii)

$4ay = x^2$

$4a \frac{dy}{dx} = 2x$

So the tangent at the point with x-coordinate $p$ is

$y - \frac{p^2}{4a} = \frac{p}{2a}(x - p)$

$4ay + p^2 = 2px$ [M1]

The tangents $4ay + p^2 = 2px$, $4ay + q^2 = 2qx$ meet when

$2(p - q)x = p^2 - q^2 = (p - q)(p + q)$ and as $(p - q) \neq 0$,

[M1]

$x = \frac{1}{2}(p + q)$ [*A1]

So

$y = \frac{2p \times \frac{1}{2}(p + q) - p^2}{4a} = \frac{pq}{4a}$ [A1 (4)]

---

and if the tangents meet at $45^\circ$, then

$\frac{\frac{p}{2a} - \frac{q}{2a}}{1 + \frac{p}{2a} \frac{q}{2a}} = \pm 1$

M1 M1

$2a(p - q) = \pm(4a^2 + pq)$ $(4a^2 + pq)^2 = 4a^2(p - q)^2 = 4a^2((p + q)^2 - 4pq)$

M1

Thus the point of intersection satisfies

$(4a^2 + 4ay)^2 = 4a^2((2x)^2 - 16ay)$

M1

That simplifies to

$(a + y)^2 = x^2 - 4ay$ $y^2 + 6ay + a^2 = x^2$ $y^2 + 6ay + 9a^2 = x^2 + 8a^2$ $(y + 3a)^2 = x^2 + 8a^2$

*M1 A1 (6)

(iii)

If

$(y + 7a)^2 = 48a^2 + 3x^2$ $\left(\frac{pq}{4a} + 7a\right)^2 = 48a^2 + 3\left(\frac{1}{2}(p + q)\right)^2$

M1

$(pq + 28a^2)^2 = 768a^4 + 12a^2(p + q)^2$ $p^2q^2 + 56a^2pq + 784a^4 = 768a^4 + 12a^2(p - q)^2 + 48a^2pq$

M1

$p^2q^2 + 8a^2pq + 16a^4 = 12a^2(p - q)^2$

M1 A1

$(pq + 4a^2)^2 = 3(2a(p - q))^2$

$\frac{\frac{p}{2a} - \frac{q}{2a}}{1 + \frac{p}{2a} \frac{q}{2a}} = \pm \frac{1}{\sqrt{3}}$

M1 A1

Thus the tangents are at a constant angle to each other which is $30^\circ$. A1 (7)

---

Model Solution

Part (i)

Let the two lines make angles $\theta_1$ and $\theta_2$ with the positive $x$-axis, where $m_1 = \tan\theta_1$ and $m_2 = \tan\theta_2$. The angle between the two lines is $|\theta_1 - \theta_2|$.

If the acute angle between the lines is $45°$, then: $\theta_1 - \theta_2 = \pm 45° \quad \text{or} \quad \theta_1 - \theta_2 = \pm 135°$

(since $\theta_1 - \theta_2$ and $\theta_1 - \theta_2 + 180°$ give the same pair of lines).

In all cases, $\tan(\theta_1 - \theta_2) = \pm 1$ (as $\tan 45° = 1$ and $\tan 135° = -1$).

Using the tangent subtraction formula: $\tan(\theta_1 - \theta_2) = \frac{\tan\theta_1 - \tan\theta_2}{1 + \tan\theta_1\tan\theta_2} = \frac{m_1 - m_2}{1 + m_1 m_2}$

Therefore: $\frac{m_1 - m_2}{1 + m_1 m_2} = \pm 1 \qquad \blacksquare$

---

Part (ii)

The curve $C$ has equation $4ay = x^2$, so $y = \frac{x^2}{4a}$ and $\frac{dy}{dx} = \frac{x}{2a}$.

The tangent at $x = p$ passes through $\left(p, \frac{p^2}{4a}\right)$ with gradient $\frac{p}{2a}$: $y - \frac{p^2}{4a} = \frac{p}{2a}(x - p)$

Multiply through by $4a$: $4ay - p^2 = 2p(x - p) = 2px - 2p^2$ $4ay = 2px - p^2 \qquad (\star)$

Similarly, the tangent at $x = q$: $4ay = 2qx - q^2 \qquad (\star\star)$

Setting $(\star) = (\star\star)$: $2px - p^2 = 2qx - q^2$ $2(p - q)x = p^2 - q^2 = (p - q)(p + q)$

Since $p \neq q$, divide by $p - q$: $x = \frac{p + q}{2}$

Substituting back into $(\star)$: $4ay = 2p \cdot \frac{p+q}{2} - p^2 = p^2 + pq - p^2 = pq$ $y = \frac{pq}{4a}$

So the tangents meet at $\left(\frac{p+q}{2},\, \frac{pq}{4a}\right)$.

---

Now suppose the two tangents are at $45°$ to each other. Their gradients are $\frac{p}{2a}$ and $\frac{q}{2a}$, so by part (i): $\frac{\frac{p}{2a} - \frac{q}{2a}}{1 + \frac{p}{2a} \cdot \frac{q}{2a}} = \pm 1$

$\frac{\frac{p-q}{2a}}{\frac{4a^2 + pq}{4a^2}} = \pm 1$

$\frac{2a(p-q)}{4a^2 + pq} = \pm 1$

Squaring both sides: $4a^2(p-q)^2 = (4a^2 + pq)^2 \qquad (\dagger)$

Now express $(\dagger)$ in terms of the meeting point. Let $X = \frac{p+q}{2}$ and $Y = \frac{pq}{4a}$, so $p + q = 2X$ and $pq = 4aY$.

Then: $(p-q)^2 = (p+q)^2 - 4pq = 4X^2 - 16aY$

Substituting into $(\dagger)$: $4a^2(4X^2 - 16aY) = (4a^2 + 4aY)^2$

Divide by $16a^2$: $X^2 - 4aY = (a + Y)^2$

$X^2 - 4aY = a^2 + 2aY + Y^2$

$X^2 = Y^2 + 6aY + a^2$

Adding $8a^2$ to both sides: $X^2 + 8a^2 = Y^2 + 6aY + 9a^2 = (Y + 3a)^2$

Therefore the meeting point $(X, Y)$ satisfies $(Y + 3a)^2 = X^2 + 8a^2$, i.e.:

$(y + 3a)^2 = x^2 + 8a^2 \qquad \blacksquare$

---

Part (iii)

The meeting point of tangents at $x = p$ and $x = q$ is $\left(\frac{p+q}{2},\, \frac{pq}{4a}\right)$.

If this point lies on $(y + 7a)^2 = 48a^2 + 3x^2$, then: $\left(\frac{pq}{4a} + 7a\right)^2 = 48a^2 + 3\left(\frac{p+q}{2}\right)^2$

Multiply through by $16a^2$: $(pq + 28a^2)^2 = 768a^4 + 12a^2(p+q)^2$

Expand the left side: $p^2q^2 + 56a^2 pq + 784a^4 = 768a^4 + 12a^2(p+q)^2$

Use $(p+q)^2 = (p-q)^2 + 4pq$: $p^2q^2 + 56a^2 pq + 784a^4 = 768a^4 + 12a^2(p-q)^2 + 48a^2 pq$

$p^2q^2 + 8a^2 pq + 16a^4 = 12a^2(p-q)^2$

The left side factors: $(pq + 4a^2)^2 = 12a^2(p-q)^2 = 3 \cdot [2a(p-q)]^2$

Taking square roots: $pq + 4a^2 = \pm 2a(p-q)\sqrt{3}$

Now recall the tangent angle formula from part (i): $\frac{m_1 - m_2}{1 + m_1 m_2} = \frac{\frac{p}{2a} - \frac{q}{2a}}{1 + \frac{pq}{4a^2}} = \frac{2a(p-q)}{4a^2 + pq}$

From $(pq + 4a^2)^2 = 3 \cdot [2a(p-q)]^2$: $\left(\frac{2a(p-q)}{4a^2 + pq}\right)^2 = \frac{4a^2(p-q)^2}{(4a^2 + pq)^2} = \frac{4a^2(p-q)^2}{12a^2(p-q)^2} = \frac{1}{3}$

Therefore: $\left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \frac{1}{\sqrt{3}} = \tan 30°$

Since $\tan\theta = \frac{1}{\sqrt{3}}$ gives $\theta = 30°$, the acute angle between any two such tangents is $30°$. $\qquad \blacksquare$

Examiner Notes

第四受欢迎题目，均分约10/20，第三成功。Part(i)需要完整论证：定义角度、考虑正切函数的周期性、讨论哪种情况对应锐角。Part(ii)中y坐标错误（代入抛物线方程而非切线方程）是常见失误。Part(iii)对多数考生有挑战性，不少人靠猜测30°/45°/60°作答。

---

Question 5

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

5 In this question, M and N are non-singular $2 \times 2$ matrices.

The trace of the matrix $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is defined as $\text{tr}(\mathbf{M}) = a + d$.

(i) Prove that, for any two matrices M and N, $\text{tr}(\mathbf{MN}) = \text{tr}(\mathbf{NM})$ and derive an expression for $\text{tr}(\mathbf{M} + \mathbf{N})$ in terms of $\text{tr}(\mathbf{M})$ and $\text{tr}(\mathbf{N})$.

The entries in matrix M are functions of $t$ and $\frac{d\mathbf{M}}{dt}$ denotes the matrix whose entries are the derivatives of the corresponding entries in M.

(ii) Show that

$\frac{1}{\det \mathbf{M}} \frac{d}{dt} (\det \mathbf{M}) = \text{tr} \left( \mathbf{M}^{-1} \frac{d\mathbf{M}}{dt} \right).$

(iii) In this part, matrix M satisfies the differential equation

$\frac{d\mathbf{M}}{dt} = \mathbf{MN} - \mathbf{NM},$

where the entries in matrix N are also functions of $t$.

Show that $\det \mathbf{M}$, $\text{tr}(\mathbf{M})$ and $\text{tr}(\mathbf{M}^2)$ are independent of $t$.

In the case $\mathbf{N} = \begin{pmatrix} t & t \\ 0 & t \end{pmatrix}$, and given that $\mathbf{M} = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ when $t = 0$, find M as a function of $t$.

(iv) In this part, matrix M satisfies the differential equation

$\frac{d\mathbf{M}}{dt} = \mathbf{MN},$

where the entries in matrix N are again functions of $t$.

The trace of M is non-zero and independent of $t$. Is it necessarily true that $\text{tr}(\mathbf{N}) = 0$?

Hint

(i)

Let

$N = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$ $MN = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}$ $NM = \begin{pmatrix} e & f \\ g & h \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ea + fc & eb + fd \\ ga + hc & gb + hd \end{pmatrix}$ $tr(MN) = ae + bg + cf + dh = ea + gb + fc + hd$ $= ea + fc + gb + hd = tr(NM)$

M1A1

$tr(M + N) = tr \left( \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} e & f \\ g & h \end{pmatrix} \right) = tr \begin{pmatrix} a + e & b + f \\ c + g & d + h \end{pmatrix} = a + e + d + h$ $= a + d + e + h = tr(M) + tr(N)$ B1 (3)

(ii) $\frac{1}{\det M} \frac{d}{dt} (\det M) = \frac{1}{ad - bc} \times \frac{d}{dt} (ad - bc) = \frac{1}{ad - bc} \times (\dot{a}d + a\dot{d} - \dot{b}c - b\dot{c})$

M1

$tr \left( M^{-1} \frac{dM}{dt} \right) = tr \left( \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} \dot{a} & \dot{b} \\ \dot{c} & \dot{d} \end{pmatrix} \right) = tr \left( \frac{1}{ad - bc} \begin{pmatrix} d\dot{a} - b\dot{c} & d\dot{b} - b\dot{d} \\ -c\dot{a} + a\dot{c} & -c\dot{b} + a\dot{d} \end{pmatrix} \right)$ $= \frac{1}{ad - bc} \times (d\dot{a} - b\dot{c} - c\dot{b} + a\dot{d}) = \frac{1}{\det M} \frac{d}{dt} (\det M)$ M1 *A1 (3)

(iii)

$tr \left( M^{-1} \frac{dM}{dt} \right) = tr(M^{-1}(MN - NM)) = tr(M^{-1}MN - M^{-1}NM) = tr(N) - tr(NMM^{-1}) = 0$

Thus

$\frac{1}{\det M} \frac{d}{dt} (\det M) = 0$

and so $\det M$ is independent of $t$

E1

$\frac{d}{dt} (tr(M)) = tr \left( \frac{dM}{dt} \right) = tr(MN - NM) = tr(MN) - tr(NM) = 0$

so $tr(M)$ is independent of $t$

E1

---

$tr(M^2) = tr \left( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \right) = tr \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix} = a^2 + bc + bc + d^2$ $= a^2 + 2ad + d^2 + 2(bc - ad) = (a + d)^2 - 2 \det M = (tr M)^2 - 2 \det M$ M1 A1

Therefore

$\frac{d}{dt}(tr(M^2)) = \frac{d}{dt}((tr M)^2) - \frac{d}{dt}(2 \det M) = 0$

and so $tr(M^2)$ is independent of t

E1 (5) $\frac{dM}{dt} = MN - NM$

So

$\begin{pmatrix} \dot{A} & \dot{B} \\ \dot{C} & \dot{D} \end{pmatrix} = \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} t & t \\ 0 & t \end{pmatrix} - \begin{pmatrix} t & t \\ 0 & t \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \begin{pmatrix} -Ct & At - Dt \\ 0 & Ct \end{pmatrix}$

M1 A1

Thus $C$ is a constant,

A1

$A = a - \frac{1}{2}Ct^2, D = d + \frac{1}{2}Ct^2$

A1

and as

$\dot{B} = At - Dt = (a - d)t - Ct^3$

M1 $B = b + \frac{1}{2}(a - d)t^2 - \frac{1}{4}Ct^4$ A1 (6)

(iv) If

$\frac{dM}{dt} = MN$

Then for example,

$M = \begin{pmatrix} e^t & 1 + e^t \\ e^t & 1 - e^t \end{pmatrix}, N = \begin{pmatrix} 1 & -e^{-t} \\ 0 & 1 \end{pmatrix}$

$\frac{dM}{dt} = \begin{pmatrix} e^t & e^t \\ e^t & -e^t \end{pmatrix}$

And

$N = \begin{pmatrix} e^t & 1 + e^t \\ e^t & 1 - e^t \end{pmatrix} \begin{pmatrix} 1 & -e^{-t} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^t & e^t \\ e^t & -e^t \end{pmatrix}$

M1 A1

and

$tr(N) = 2$

so no.

A1 (3)

---

Model Solution

Part (i)

Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\mathbf{N} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$.

Computing the products:

$\mathbf{MN} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}, \qquad \mathbf{NM} = \begin{pmatrix} ea + fc & eb + fd \\ ga + hc & gb + hd \end{pmatrix}$

Taking traces:

$\text{tr}(\mathbf{MN}) = ae + bg + cf + dh, \qquad \text{tr}(\mathbf{NM}) = ea + fc + gb + hd$

Since scalar multiplication is commutative ($ae = ea$, $bg = gb$, $cf = fc$, $dh = hd$), the two expressions are term-by-term equal, so

$\text{tr}(\mathbf{MN}) = \text{tr}(\mathbf{NM}).$

For the trace of the sum:

$\mathbf{M} + \mathbf{N} = \begin{pmatrix} a + e & b + f \\ c + g & d + h \end{pmatrix}$

$\text{tr}(\mathbf{M} + \mathbf{N}) = (a + e) + (d + h) = (a + d) + (e + h) = \text{tr}(\mathbf{M}) + \text{tr}(\mathbf{N}).$

Part (ii)

Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a, b, c, d$ are functions of $t$. Then $\det \mathbf{M} = ad - bc$, and by the product rule:

$\frac{d}{dt}(\det \mathbf{M}) = \frac{d}{dt}(ad - bc) = \dot{a}d + a\dot{d} - \dot{b}c - b\dot{c}. \qquad \text{(...)}$

We have $\mathbf{M}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ and $\frac{d\mathbf{M}}{dt} = \begin{pmatrix} \dot{a} & \dot{b} \\ \dot{c} & \dot{d} \end{pmatrix}$, so

$\mathbf{M}^{-1}\frac{d\mathbf{M}}{dt} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\begin{pmatrix} \dot{a} & \dot{b} \\ \dot{c} & \dot{d} \end{pmatrix} = \frac{1}{ad - bc}\begin{pmatrix} d\dot{a} - b\dot{c} & d\dot{b} - b\dot{d} \\ -c\dot{a} + a\dot{c} & -c\dot{b} + a\dot{d} \end{pmatrix}$

Taking the trace (sum of diagonal entries):

$\text{tr}\left(\mathbf{M}^{-1}\frac{d\mathbf{M}}{dt}\right) = \frac{d\dot{a} - b\dot{c} - c\dot{b} + a\dot{d}}{ad - bc} = \frac{\dot{a}d + a\dot{d} - \dot{b}c - b\dot{c}}{ad - bc}$

The numerator is identical to (…), so

$\frac{1}{\det \mathbf{M}}\frac{d}{dt}(\det \mathbf{M}) = \text{tr}\left(\mathbf{M}^{-1}\frac{d\mathbf{M}}{dt}\right).$

Part (iii)

Given $\frac{d\mathbf{M}}{dt} = \mathbf{MN} - \mathbf{NM}$.

det M: Using part (ii),

$\frac{1}{\det \mathbf{M}}\frac{d}{dt}(\det \mathbf{M}) = \text{tr}\left(\mathbf{M}^{-1}(\mathbf{MN} - \mathbf{NM})\right) = \text{tr}(\mathbf{N}) - \text{tr}(\mathbf{M}^{-1}\mathbf{NM}).$

Setting $\mathbf{X} = \mathbf{M}$ and $\mathbf{Y} = \mathbf{M}^{-1}\mathbf{N}$ in $\text{tr}(\mathbf{XY}) = \text{tr}(\mathbf{YX})$:

$\text{tr}(\mathbf{M} \cdot \mathbf{M}^{-1}\mathbf{N}) = \text{tr}(\mathbf{M}^{-1}\mathbf{N} \cdot \mathbf{M})$

$\text{tr}(\mathbf{N}) = \text{tr}(\mathbf{M}^{-1}\mathbf{NM}).$

So $\frac{1}{\det \mathbf{M}}\frac{d}{dt}(\det \mathbf{M}) = \text{tr}(\mathbf{N}) - \text{tr}(\mathbf{N}) = 0$, hence $\det \mathbf{M}$ is independent of $t$.

tr(M): The derivative of the trace is the trace of the derivative (since differentiation acts entry-wise):

$\frac{d}{dt}\text{tr}(\mathbf{M}) = \text{tr}\left(\frac{d\mathbf{M}}{dt}\right) = \text{tr}(\mathbf{MN} - \mathbf{NM}) = \text{tr}(\mathbf{MN}) - \text{tr}(\mathbf{NM}) = 0$

by part (i). Hence $\text{tr}(\mathbf{M})$ is independent of $t$.

tr(M²): For $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$\mathbf{M}^2 = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix}, \qquad \text{tr}(\mathbf{M}^2) = a^2 + 2bc + d^2$

We also compute:

$(\text{tr}\,\mathbf{M})^2 - 2\det \mathbf{M} = (a + d)^2 - 2(ad - bc) = a^2 + 2ad + d^2 - 2ad + 2bc = a^2 + 2bc + d^2 = \text{tr}(\mathbf{M}^2)$

So $\text{tr}(\mathbf{M}^2) = (\text{tr}\,\mathbf{M})^2 - 2\det \mathbf{M}$. Since both $\text{tr}(\mathbf{M})$ and $\det \mathbf{M}$ are independent of $t$, so is $\text{tr}(\mathbf{M}^2)$.

Finding M(t): With $\mathbf{N} = \begin{pmatrix} t & t \\ 0 & t \end{pmatrix}$ and $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$\mathbf{MN} = \begin{pmatrix} at & (a + b)t \\ ct & (c + d)t \end{pmatrix}, \qquad \mathbf{NM} = \begin{pmatrix} (a + c)t & (b + d)t \\ ct & dt \end{pmatrix}$

$\frac{d\mathbf{M}}{dt} = \mathbf{MN} - \mathbf{NM} = \begin{pmatrix} -ct & (a - d)t \\ 0 & ct \end{pmatrix}$

Reading off each entry of $\frac{d\mathbf{M}}{dt}$:

$\dot{c} = 0$: so $c = C$ (constant, equal to its initial value at $t = 0$).

$\dot{a} = -Ct$: integrating, $a = A - \frac{1}{2}Ct^2$.

$\dot{d} = Ct$: integrating, $d = D + \frac{1}{2}Ct^2$.

$\dot{b} = (a - d)t$: substituting the expressions for $a$ and $d$:

$\dot{b} = \left(\left(A - \tfrac{1}{2}Ct^2\right) - \left(D + \tfrac{1}{2}Ct^2\right)\right)t = (A - D)t - Ct^3$

Integrating: $b = B + \frac{1}{2}(A - D)t^2 - \frac{1}{4}Ct^4$.

Therefore

$\mathbf{M}(t) = \begin{pmatrix} A - \frac{1}{2}Ct^2 & B + \frac{1}{2}(A - D)t^2 - \frac{1}{4}Ct^4 \\[6pt] C & D + \frac{1}{2}Ct^2 \end{pmatrix}.$

Part (iv)

No. We construct a counterexample.

Let $\mathbf{N} = \begin{pmatrix} 1 & -e^{-t} \\ 0 & 1 \end{pmatrix}$ and $\mathbf{M} = \begin{pmatrix} e^t & 1 + e^t \\ e^t & 1 - e^t \end{pmatrix}$.

First, both matrices are non-singular: $\det \mathbf{M} = e^t(1 - e^t) - (1 + e^t)e^t = -2e^{2t} \neq 0$ and $\det \mathbf{N} = 1 \neq 0$.

We have $\text{tr}(\mathbf{M}) = e^t + 1 - e^t = 1$, which is non-zero and independent of $t$.

We verify $\frac{d\mathbf{M}}{dt} = \mathbf{MN}$:

$\frac{d\mathbf{M}}{dt} = \begin{pmatrix} e^t & e^t \\ e^t & -e^t \end{pmatrix}$

$\mathbf{MN} = \begin{pmatrix} e^t & 1 + e^t \\ e^t & 1 - e^t \end{pmatrix}\begin{pmatrix} 1 & -e^{-t} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^t & -1 + 1 + e^t \\ e^t & -1 + 1 - e^t \end{pmatrix} = \begin{pmatrix} e^t & e^t \\ e^t & -e^t \end{pmatrix}$

Since $\frac{d\mathbf{M}}{dt} = \mathbf{MN}$, all conditions of the problem are satisfied. But $\text{tr}(\mathbf{N}) = 1 + 1 = 2 \neq 0$, so it is not necessarily true that $\text{tr}(\mathbf{N}) = 0$.

Examiner Notes

均分约8/20。Part(i)很直接但部分考生遗漏tr(M+N)的结果。Part(iii)常见错误：忽略矩阵乘法的非交换性、将A/B/C/D视为常数、对标量链规则直接应用于M²。Part(iv)需要构造反例，仅6人满分——多数考生要么未能给出满足所有条件的反例，要么错误地试图证明命题成立。

---

Question 6

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

6 (i) A particle moves in two-dimensional space. Its position is given by coordinates $(x, y)$ which satisfy

$$\begin{aligned} \frac{dx}{dt} &= -x + 3y + u \\ \frac{dy}{dt} &= x + y + u \end{aligned}$$

where $t$ is the time and $u$ is a function of time. At time $t = 0$, the particle has position $(x_0, y_0)$.

(a) By considering $\frac{dx}{dt} - \frac{dy}{dt}$, show that if the particle is at the origin $(0, 0)$ at some time $t > 0$, then it is necessary that $x_0 = y_0$.

(b) Given that $x_0 = y_0$, find a constant value of $u$ that ensures that the particle is at the origin at a time $t = T$, where $T > 0$.

(ii) A particle whose position in three-dimensional space is given by co-ordinates $(x, y, z)$ moves with time $t$ such that

$$\begin{aligned} \frac{dx}{dt} &= 4y - 5z + u \\ \frac{dy}{dt} &= x - 2z + u \\ \frac{dz}{dt} &= x - 2y + u \end{aligned}$$

where $u$ is a function of time. At time $t = 0$, the particle has position $(x_0, y_0, z_0)$.

(a) Show that, if the particle is at the origin $(0, 0, 0)$ at some time $t > 0$, it is necessary that $y_0$ is the mean of $x_0$ and $z_0$.

(b) Show further that, if the particle is at the origin $(0, 0, 0)$ at some time $t > 0$, it is necessary that $x_0 = y_0 = z_0$.

(c) Given that $x_0 = y_0 = z_0$, find a constant value of $u$ that ensures that the particle is at the origin at a time $t = T$, where $T > 0$.

Hint

(i) (a)**

$\frac{dx}{dt} = -x + 3y + u$

$\frac{dy}{dt} = x + y + u$ $\frac{dx}{dt} - \frac{dy}{dt} = \frac{d(x - y)}{dt} = -2(x - y)$

M1

$x - y = Ae^{-2t}$

A1

If $x = y = 0$ at some time $t > 0$, then $A = 0$, [A1]

so considering $t = 0$, $x_0 - y_0 = 0$ which gives the required result. [E1 (4)]

(b) If $x_0 = y_0$, then at $t = 0$, $x - y = 0$ so $A = 0$ and hence $x = y$ for all $t$

E1

Thus

$\frac{dx}{dt} = 2x + u$

$\frac{dx}{dt} - 2x = u$

$e^{-2t} \frac{dx}{dt} - 2e^{-2t}x = e^{-2t}u$

M1 A1

$t = 0$, $x = x_0$ so $x_0 = -\frac{1}{2}u + c$ and we want $x = 0$ when $t = T$

so

$0 = -\frac{1}{2}u + ce^{2T}$

Thus $c = \frac{1}{2}ue^{-2T}$, $x_0 = -\frac{1}{2}u + \frac{1}{2}ue^{-2T}$

and hence,

$u = \frac{2x_0 e^{2T}}{1 - e^{2T}}$

dM1 A1 (5)

---

(ii) (a)

$\frac{dx}{dt} - 2\frac{dy}{dt} + \frac{dz}{dt} = \frac{d(x - 2y + z)}{dt} = -(x - 2y + z)$

Thus

$x - 2y + z = Ae^{-t}$

M1 A1

If $x = y = z = 0$ at some time $t > 0$, then $A = 0$, so considering $t = 0$, $x_0 - 2y_0 + z_0 = 0$ which gives the required result. E1 (3)

(b) we know from (a) that if $x = y = z = 0$ at some time $t > 0$, then $A = 0$, and so

$x - 2y + z = 0$ or $2y = x + z$ E1

Thus

$\frac{dx}{dt} = 2x - 3z + u$

and

$\frac{dz}{dt} = -z + u$

So

$\frac{dx}{dt} - \frac{dz}{dt} = \frac{d}{dt}(x - z) = 2(x - z)$

and so

$x - z = Be^{2t}$

M1 A1

But as $x = z = 0$ at some time $t > 0$, $B = 0$ and so $x = z$ for all $t$

and thus $x = y = z$ for all $t$

Hence

$x_0 = y_0 = z_0$

E1 (4)

---

(c)

Given $x_0 = y_0 = z_0$ we know that (a) and (b) apply (as similarly in (i)), so $\frac{dz}{dt} = -z + u$

M1

Thus $z = u + ce^{-t}$

A1

$t = 0$ , $z = z_0$ so $z_0 = u + c$ and $0 = u + ce^{-T}$

dM1

$c = -ue^T$

$u = \frac{z_0}{1 - e^T}$

A1 (4)

---

Model Solution

Part (i)(a)

Subtracting the two equations:

$\frac{dx}{dt} - \frac{dy}{dt} = (-x + 3y + u) - (x + y + u) = -2x + 2y = -2(x - y)$

So $\frac{d}{dt}(x - y) = -2(x - y)$, which is a separable ODE with solution

$x - y = Ae^{-2t}$

where $A$ is a constant. At $t = 0$: $A = x_0 - y_0$.

If the particle is at the origin at some time $t > 0$, then $x = y = 0$ at that time, so $x - y = 0$, giving $Ae^{-2t} = 0$. Since $e^{-2t} \neq 0$ for all $t$, we must have $A = 0$, hence $x_0 = y_0$.

Part (i)(b)

Given $x_0 = y_0$, we have $A = 0$, so $x = y$ for all $t \geq 0$.

Substituting $y = x$ into $\frac{dx}{dt} = -x + 3y + u$:

$\frac{dx}{dt} = -x + 3x + u = 2x + u$

Rearranging: $\frac{dx}{dt} - 2x = u$. Multiplying both sides by the integrating factor $e^{-2t}$:

$\frac{d}{dt}\left(xe^{-2t}\right) = ue^{-2t}$

Integrating:

$xe^{-2t} = -\frac{u}{2}e^{-2t} + K$

where $K$ is a constant of integration. At $t = 0$: $x_0 = -\frac{u}{2} + K$, so $K = x_0 + \frac{u}{2}$.

Thus

$x = -\frac{u}{2} + \left(x_0 + \frac{u}{2}\right)e^{2t}$

Setting $x = 0$ at $t = T$:

$0 = -\frac{u}{2} + \left(x_0 + \frac{u}{2}\right)e^{2T}$

$\frac{u}{2} = \left(x_0 + \frac{u}{2}\right)e^{2T}$

$\frac{u}{2}\left(1 - e^{2T}\right) = x_0 e^{2T}$

$u = \frac{2x_0 e^{2T}}{1 - e^{2T}}$

Part (ii)(a)

Consider $\frac{dx}{dt} - 2\frac{dy}{dt} + \frac{dz}{dt}$:

$= (4y - 5z + u) - 2(x - 2z + u) + (x - 2y + u)$

$= 4y - 5z + u - 2x + 4z - 2u + x - 2y + u$

$= -x + 2y - z = -(x - 2y + z)$

So $\frac{d}{dt}(x - 2y + z) = -(x - 2y + z)$, giving $x - 2y + z = Ae^{-t}$.

At $t = 0$: $A = x_0 - 2y_0 + z_0$.

If the particle is at the origin at some $t > 0$, then $x = y = z = 0$, so $Ae^{-t} = 0$, hence $A = 0$, giving $x_0 - 2y_0 + z_0 = 0$, i.e., $y_0 = \frac{x_0 + z_0}{2}$.

Part (ii)(b)

From part (a), $x - 2y + z = 0$ for all $t$, so $2y = x + z$.

Substituting $y = \frac{x + z}{2}$ into the first and third equations:

$\frac{dx}{dt} = 4 \cdot \frac{x + z}{2} - 5z + u = 2x + 2z - 5z + u = 2x - 3z + u$

$\frac{dz}{dt} = x - 2 \cdot \frac{x + z}{2} + u = x - x - z + u = -z + u$

Subtracting:

$\frac{dx}{dt} - \frac{dz}{dt} = (2x - 3z + u) - (-z + u) = 2x - 2z = 2(x - z)$

So $\frac{d}{dt}(x - z) = 2(x - z)$, giving $x - z = Be^{2t}$.

At $t = 0$: $B = x_0 - z_0$.

If $x = z = 0$ at some $t > 0$, then $Be^{2t} = 0$. Since $e^{2t} \neq 0$, we need $B = 0$, hence $x_0 = z_0$.

Combined with $y_0 = \frac{x_0 + z_0}{2} = \frac{x_0 + x_0}{2} = x_0$, we conclude $x_0 = y_0 = z_0$.

Part (ii)(c)

Given $x_0 = y_0 = z_0$, from parts (a) and (b) we have $x - 2y + z = 0$ and $x - z = 0$ for all $t$. So $x = z$ and $2y = x + z = 2x$, giving $y = x = z$ for all $t$.

From $\frac{dz}{dt} = -z + u$:

$\frac{dz}{dt} + z = u$

Multiplying by the integrating factor $e^t$:

$\frac{d}{dt}\left(ze^t\right) = ue^t$

Integrating: $ze^t = ue^t + K$.

At $t = 0$: $z_0 = u + K$, so $K = z_0 - u$.

Thus $z = u + (z_0 - u)e^{-t}$.

Setting $z = 0$ at $t = T$:

$0 = u + (z_0 - u)e^{-T}$

$u - ue^{-T} = -z_0 e^{-T}$

$u(1 - e^{-T}) = -z_0 e^{-T}$

$u = \frac{-z_0 e^{-T}}{1 - e^{-T}}$

Multiplying numerator and denominator by $e^T$:

$u = \frac{z_0}{1 - e^T}$

Examiner Notes

纯数部分最不受欢迎且得分最低。考官报告：成功考生通常将方程改写为d(x-y)/dt=-2(x-y)；也有考生用积分因子法。常见错误：将x=y=0代入微分方程后积分（混淆了条件）；正确得出x=y但未说明这蕴含x₀=y₀。Part(ii)中巧妙组合后两个方程得出y=z再推x=z的方法值得学习。

---

Question 7

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

7 In this question, you need not consider issues of convergence.

For positive integer $n$ let

$f(n) = \frac{1}{n + 1} + \frac{1}{(n + 1)(n + 2)} + \frac{1}{(n + 1)(n + 2)(n + 3)} + \dots$

and

$g(n) = \frac{1}{n + 1} - \frac{1}{(n + 1)(n + 2)} + \frac{1}{(n + 1)(n + 2)(n + 3)} - \dots$

(i) Show, by considering a geometric series, that $0 < f(n) < \frac{1}{n}$.

(ii) Show, by comparing consecutive terms, that $0 < g(n) < \frac{1}{n + 1}$.

(iii) Show, for positive integer $n$, that $(2n)! e - f(2n)$ and $\frac{(2n)!}{e} + g(2n)$ are both integers.

(iv) Show that if $qe = \frac{p}{e}$ for some positive integers $p$ and $q$, then $qf(2n) + pg(2n)$ is an integer for all positive integers $n$.

(v) Hence show that the number $e^2$ is irrational.

Hint

(i)** Each term of $f(n) > 0$ so their sum is too.

[E1]

$\frac{1}{(n+1)(n+2)...(n+r)} < \frac{1}{(n+1)^r}$ so $f(n) = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \dots < \frac{1}{n+1} + \frac{1}{(n+1)^2} + \dots = \frac{1/n+1}{1-1/n+1} = \frac{1}{n}$

[M1 A1(3)]

Thus $0 < f(n) < \frac{1}{n}$

(ii) $\frac{1}{n+1} - \frac{1}{(n+1)(n+2)} > 0$, $\frac{1}{(n+1)(n+2)(n+3)} - \frac{1}{(n+1)(n+2)(n+3)(n+4)} > 0$, etc so $g(n) > 0$

[M1 A1]

Also, $\frac{1}{(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)} > 0$, $\frac{1}{(n+1)(n+2)(n+3)(n+4)} - \frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)} > 0$, etc

so $g(n) = \frac{1}{n+1} - \text{a sum of positive terms} < \frac{1}{n+1}$

[M1 A1 (4)]

Thus $0 < g(n) < \frac{1}{n+1}$

(iii)

$(2n)! e - f(2n)$ $= (2n)! \left( 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots \right) - \frac{1}{2n+1} - \frac{1}{(2n+1)(2n+2)} - \frac{1}{(2n+1)(2n+2)(2n+3)} - \dots$

[M1 A1]

$= (2n)! \left( 1 + 1 + \frac{1}{2!} + \dots + \frac{1}{(2n)!} \right)$ $+ \frac{(2n)!}{(2n+1)!} + \frac{(2n)!}{(2n+2)!} + \dots - \frac{1}{2n+1} - \frac{1}{(2n+1)(2n+2)} - \frac{1}{(2n+1)(2n+2)(2n+3)} - \dots$ $= (2n)! \left( 1 + 1 + \frac{1}{2!} + \dots + \frac{1}{(2n)!} \right)$

which is an integer. [M1 A1(4)]

---

$\frac{(2n)!}{e} + g(2n) = (2n)! e^{-1} + g(2n)$ $= (2n)! \left( 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \cdots \right) + \frac{1}{2n+1} - \frac{1}{(2n+1)(2n+2)} + \frac{1}{(2n+1)(2n+2)(2n+3)} - \cdots$

M1 A1

$= (2n)! \left( 1 - 1 + \frac{1}{2!} + \cdots + \frac{1}{(2n)!} \right)$ $- \frac{(2n)!}{(2n+1)!} + \frac{(2n)!}{(2n+2)!} - \cdots + \frac{1}{2n+1} - \frac{1}{(2n+1)(2n+2)} + \frac{1}{(2n+1)(2n+2)(2n+3)} - \cdots$ $= (2n)! \left( 1 - 1 + \frac{1}{2!} + \cdots + \frac{1}{(2n)!} \right)$

M1 A1 (4)

which is an integer.

(iv) $q((2n)! e - f(2n))$ is an integer as is $p \left( \frac{(2n)!}{e} + g(2n) \right)$

Thus $\left( p \left( \frac{(2n)!}{e} + g(2n) \right) \right) - (q((2n)! e - f(2n)))$ is an integer.

$\left( p \left( \frac{(2n)!}{e} + g(2n) \right) \right) - (q((2n)! e - f(2n))) = (2n)! \left( \frac{p}{e} - qe \right) + pg(2n) + qf(2n)$

M1

$= pg(2n) + qf(2n)$

so $pg(2n) + qf(2n)$ is an integer as required. A1(2)

(v) As (iv) is true for all positive integers $n$, it must be true for $n = \max(p, q)$

By (ii) $pg(2n) < \frac{p}{2n+1} \leq \frac{n}{2n+1} < \frac{1}{2}$

By (i) $qf(2n) < \frac{q}{2n} \leq \frac{n}{2n} = \frac{1}{2}$

M1

Therefore, $pg(2n) + qf(2n) < \frac{1}{2} + \frac{1}{2} = 1$

and trivially by (i) and (ii)

$pg(2n) + qf(2n) > 0$

A1

This means that $pg(2n) + qf(2n)$ cannot be an integer which contradicts the result of (iv)

and hence there are no integers such that $\frac{p}{e} = qe$, that is such that $\frac{p}{q} = e^2$ and so $e^2$ is irrational. E1 (3)

---

Model Solution

Part (i)

Each term of $f(n)$ is positive (since $n$ is a positive integer), so $f(n) > 0$.

For the upper bound, observe that for $r \geq 1$:

$$(n + 1)(n + 2) \cdots (n + r) \geq (n + 1)^r$$

since each factor $n + k \geq n + 1$ for $k = 1, 2, \ldots, r$, with strict inequality for $r \geq 2$. Therefore:

$$\frac{1}{(n + 1)(n + 2) \cdots (n + r)} \leq \frac{1}{(n + 1)^r}$$

with strict inequality for $r \geq 2$. Summing over all terms:

$$f(n) = \sum_{r=1}^{\infty} \frac{1}{(n+1)(n+2)\cdots(n+r)} < \sum_{r=1}^{\infty} \frac{1}{(n+1)^r}$$

The right-hand side is a geometric series with first term $\frac{1}{n+1}$ and common ratio $\frac{1}{n+1}$:

$$\sum_{r=1}^{\infty} \frac{1}{(n+1)^r} = \frac{\frac{1}{n+1}}{1 - \frac{1}{n+1}} = \frac{\frac{1}{n+1}}{\frac{n}{n+1}} = \frac{1}{n}$$

Therefore $0 < f(n) < \frac{1}{n}$. $\qquad \blacksquare$

---

Part (ii)

Write $g(n) = T_1 - T_2 + T_3 - T_4 + \cdots$ where

$$T_r = \frac{1}{(n+1)(n+2)\cdots(n+r)}$$

Showing $g(n) > 0$: Group consecutive pairs from the start:

$$g(n) = (T_1 - T_2) + (T_3 - T_4) + (T_5 - T_6) + \cdots$$

Each pair is positive since $T_r > T_{r+1}$ (each $T_{r+1}$ is $T_r$ multiplied by $\frac{1}{n+r+1} < 1$). A sum of positive terms is positive, so $g(n) > 0$.

Showing $g(n) < \frac{1}{n+1}$: Group differently by taking the first term and pairing the rest:

$$g(n) = T_1 - (T_2 - T_3) - (T_4 - T_5) - \cdots$$

Each bracketed pair is positive (again since $T_r > T_{r+1}$). Therefore $g(n) = T_1 - (\text{positive terms}) < T_1 = \frac{1}{n+1}$.

Therefore $0 < g(n) < \frac{1}{n+1}$. $\qquad \blacksquare$

---

Part (iii)

Using $e = \sum_{k=0}^{\infty} \frac{1}{k!}$:

$$(2n)!\, e = (2n)! \left(1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \right)$$

Split this into two parts: terms up to $k = 2n$ and terms for $k > 2n$:

$$(2n)!\, e = \underbrace{\sum_{k=0}^{2n} \frac{(2n)!}{k!}}_{\text{integer}} + \sum_{k=2n+1}^{\infty} \frac{(2n)!}{k!}$$

The first sum is an integer since $\frac{(2n)!}{k!}$ is an integer for $0 \leq k \leq 2n$. For the second sum, substitute $k = 2n + r$ with $r \geq 1$:

$$\sum_{k=2n+1}^{\infty} \frac{(2n)!}{k!} = \sum_{r=1}^{\infty} \frac{(2n)!}{(2n+r)!} = \sum_{r=1}^{\infty} \frac{1}{(2n+1)(2n+2)\cdots(2n+r)} = f(2n)$$

Therefore $(2n)!\, e = \text{integer} + f(2n)$, which gives:

$$(2n)!\, e - f(2n) = \text{integer} \qquad \blacksquare$$

---

For the second expression, use $e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}$:

$$\frac{(2n)!}{e} = (2n)! \left(1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \cdots \right)$$

Split into terms up to $k = 2n$ and terms for $k > 2n$:

$$\frac{(2n)!}{e} = \underbrace{\sum_{k=0}^{2n} \frac{(-1)^k (2n)!}{k!}}_{\text{integer}} + \sum_{k=2n+1}^{\infty} \frac{(-1)^k (2n)!}{k!}$$

The first sum is again an integer. For the second sum, with $k = 2n + r$, $r \geq 1$:

$$\sum_{k=2n+1}^{\infty} \frac{(-1)^k (2n)!}{k!} = \sum_{r=1}^{\infty} \frac{(-1)^{2n+r}}{(2n+1)(2n+2)\cdots(2n+r)} = \sum_{r=1}^{\infty} \frac{(-1)^r}{(2n+1)(2n+2)\cdots(2n+r)} = -g(2n)$$

since $(-1)^{2n} = 1$. Therefore:

$$\frac{(2n)!}{e} = \text{integer} - g(2n)$$

which gives:

$$\frac{(2n)!}{e} + g(2n) = \text{integer} \qquad \blacksquare$$

---

Part (iv)

From part (iii), both $(2n)!\, e - f(2n)$ and $\frac{(2n)!}{e} + g(2n)$ are integers. Therefore:

$$q\left((2n)!\, e - f(2n)\right) \quad \text{and} \quad p\left(\frac{(2n)!}{e} + g(2n)\right)$$

are both integers (products of integers). Their difference is also an integer:

$$p\left(\frac{(2n)!}{e} + g(2n)\right) - q\left((2n)!\, e - f(2n)\right) = (2n)! \left(\frac{p}{e} - qe\right) + pg(2n) + qf(2n)$$

Since $\frac{p}{e} = qe$, i.e. $\frac{p}{e} - qe = 0$, the first term vanishes. Therefore:

$$pg(2n) + qf(2n) \text{ is an integer} \qquad \blacksquare$$

---

Part (v)

Suppose for contradiction that $e^2$ is rational, so $e^2 = \frac{p}{q}$ for some positive integers $p, q$. Then $qe = \frac{p}{e}$, and by part (iv), $qf(2n) + pg(2n)$ is an integer for all positive integers $n$.

Choose $n = \max(p, q)$, which ensures $n \geq p$ and $n \geq q$. Then:

• From part (i): $qf(2n) < \frac{q}{2n} \leq \frac{n}{2n} = \frac{1}{2}$
• From part (ii): $pg(2n) < \frac{p}{2n+1} < \frac{p}{2n} \leq \frac{n}{2n} = \frac{1}{2}$

Also, since both $f(2n)$ and $g(2n)$ are positive and $p, q$ are positive:

$$0 < qf(2n) + pg(2n) < \frac{1}{2} + \frac{1}{2} = 1$$

But $qf(2n) + pg(2n)$ is supposed to be an integer, and no integer lies strictly between 0 and 1. This is a contradiction.

Therefore $e^2$ is irrational. $\qquad \blacksquare$

Examiner Notes

第三受欢迎题目，均分约9/20。Part(i)(ii)容易猜出几何级数形式但缺乏严格论证无法得满分。Part(ii)中配对连续项的论证常缺乏必要细节。Part(iii)中将1/e展开为倒数而非e⁻¹是常见错误，容易迷失方向。Part(v)反证法中选取合适的n值是难点。

---

Question 8

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

8 (i) Explain why the equation $(y - x + 3)(y + x - 5) = 0$ represents a pair of straight lines with gradients 1 and $-1$. Show further that the equation

$y^2 - x^2 + py + qx + r = 0$

represents a pair of straight lines with gradients 1 and $-1$ if and only if $p^2 - q^2 = 4r$.

In the remainder of this question, $C_1$ is the curve with equation $x = y^2 + 2sy + s(s + 1)$ and $C_2$ is the curve with equation $y = x^2$.

(ii) Explain why the coordinates of any point which lies on both of the curves $C_1$ and $C_2$ also satisfy the equation

$y^2 + 2sy + s(s + 1) - x + k(y - x^2) = 0$

for any real number $k$.

Given that $s$ is such that $C_1$ and $C_2$ intersect at four distinct points, show that choosing $k = 1$ gives an equation representing a pair of straight lines, with gradients 1 and $-1$, on which all four points of intersection lie.

(iii) Show that if $C_1$ and $C_2$ intersect at four distinct points, then $s < -\frac{3}{4}$.

(iv) Show that if $s < -\frac{3}{4}$, then $C_1$ and $C_2$ intersect at four distinct points.

Hint

(i)** $(y - x + 3)(y + x - 5) = 0$ if and only if either $y - x + 3 = 0$ or $y + x - 5 = 0$. These are the equations of two straight lines with gradients 1 and -1.

E1

A pair of straight lines with gradients 1 and -1 can be expressed as $y - x + a = 0$ and

$y + x + b = 0$. Thus $(y - x + a)(y + x + b) = 0$ can be expressed

$y^2 - x^2 + py + qx + r = 0$ if and only if $a + b = p$, $a - b = q$, and $ab = r$. M1 A1

Hence, $a = \frac{1}{2}(p + q)$, $b = \frac{1}{2}(p - q)$ and so $\frac{1}{2}(p + q)\frac{1}{2}(p - q) = r$ which can be written

$p^2 - q^2 = 4r$. *M1 A1 (5)

(ii) If a point $(x, y)$ lies on $C_1$, then $x = y^2 + 2sy + s(s + 1) = 0$ which can be rearranged as $y^2 + 2sy + s(s + 1) - x = 0$. If a point $(x, y)$ lies on $C_2$, then $y = x^2$ which can be expressed as $k(y - x^2) = 0$ for any real number $k$. Thus, if it lies on both

$y^2 + 2sy + s(s + 1) - x + k(y - x^2) = 0$ for any real number $k$

E1 E1

If $k = 1$,

$y^2 - x^2 + (2s + 1)y - x + s(s + 1) = 0$

and

$(2s + 1)^2 - (-1)^2 = 4s^2 + 4s = 4s(s + 1)$

satisfying the condition as derived in (i). M1 A1 (4)

(iii) If $C_1$ and $C_2$ intersect at four distinct points, then they do so on the pair of straight lines,

$y^2 - x^2 + (2s + 1)y - x + s(s + 1) = 0$.

E1

$y^2 - x^2 + (2s + 1)y - x + s(s + 1) = (y + x + s + 1)(y - x + s)$ B1

Therefore, $y + x + s + 1 = 0$ must meet $C_2$ at two distinct points and $y - x + s = 0$ must meet $C_2$ at two different distinct points. E1

Thus, solving $x^2 + x + s + 1 = 0$ having two distinct roots, the discriminant $1 - 4(s + 1) > 0$

That is $s < -\frac{3}{4}$ M1 A1

and solving $x^2 - x + s = 0$ having two distinct roots, the discriminant $1 - 4s > 0$ i.e. $s < \frac{1}{4}$

So it is necessary that $s < -\frac{3}{4}$. *A1 (6)

---

(iv) If $s < -\frac{3}{4}$ for $C_1$ and $C_2$ to intersect at four points, they do so on the pair of straight lines, two distinct on each of the lines in (iii) as shown by the non-zero discriminants in (iii) and that will be four distinct points provided that the point of intersection of the two lines is not one of them. E1 E1

The intersection of

$y + x + s + 1 = 0$ and $y - x + s = 0$ is at $(-\frac{1}{2}, -\frac{2s+1}{2})$ which will only lie on $C_2$ if $-\frac{2s+1}{2} = (-\frac{1}{2})^2$ that is if $s = -\frac{3}{4}$ which is prohibited. M1 A1 E1 (5)

---

Model Solution

Part (i)

The equation $(y - x + 3)(y + x - 5) = 0$ holds if and only if either $y - x + 3 = 0$ or $y + x - 5 = 0$. These are equations of straight lines. The line $y = x - 3$ has gradient 1, and the line $y = -x + 5$ has gradient $-1$.

---

For the general case, suppose we have a pair of straight lines with gradients 1 and $-1$. These can be written as:

$$y - x + a = 0 \quad \text{and} \quad y + x + b = 0$$

for some constants $a, b$. Their combined equation is:

$$(y - x + a)(y + x + b) = 0$$

Expanding:

$$y^2 + xy + by - xy - x^2 - bx + ay + ax + ab = 0$$ $$y^2 - x^2 + (a + b)y + (a - b)x + ab = 0$$

Comparing with $y^2 - x^2 + py + qx + r = 0$, we need:

$$p = a + b, \quad q = a - b, \quad r = ab$$

From the first two equations: $a = \frac{p + q}{2}$ and $b = \frac{p - q}{2}$.

Therefore:

$$r = ab = \frac{p + q}{2} \cdot \frac{p - q}{2} = \frac{p^2 - q^2}{4}$$

which gives $p^2 - q^2 = 4r$.

Conversely, if $p^2 - q^2 = 4r$, set $a = \frac{p+q}{2}$ and $b = \frac{p-q}{2}$. Then $ab = \frac{p^2 - q^2}{4} = r$, and the equation factors as $(y - x + a)(y + x + b) = 0$, representing a pair of lines with gradients 1 and $-1$.

Therefore $y^2 - x^2 + py + qx + r = 0$ represents a pair of straight lines with gradients 1 and $-1$ if and only if $p^2 - q^2 = 4r$. $\qquad \blacksquare$

---

Part (ii)

If a point $(x, y)$ lies on $C_1$, then $x = y^2 + 2sy + s(s+1)$, which rearranges to:

$$y^2 + 2sy + s(s+1) - x = 0$$

If a point $(x, y)$ lies on $C_2$, then $y = x^2$, i.e. $y - x^2 = 0$, so $k(y - x^2) = 0$ for any real number $k$.

Therefore, if a point lies on both curves, both equations hold simultaneously, and adding them gives:

$$y^2 + 2sy + s(s+1) - x + k(y - x^2) = 0 \qquad \text{for any } k \in \mathbb{R} \qquad \blacksquare$$

---

Setting $k = 1$:

$$y^2 - x^2 + (2s + 1)y - x + s(s + 1) = 0$$

This is of the form $y^2 - x^2 + py + qx + r = 0$ with $p = 2s + 1$, $q = -1$, $r = s(s+1)$.

Checking the condition from part (i):

$$p^2 - q^2 = (2s+1)^2 - 1 = 4s^2 + 4s + 1 - 1 = 4s^2 + 4s = 4s(s+1) = 4r$$

So the condition $p^2 - q^2 = 4r$ is satisfied, and the equation represents a pair of straight lines with gradients 1 and $-1$. All four intersection points of $C_1$ and $C_2$ lie on these lines. $\qquad \blacksquare$

---

Part (iii)

We show: if $C_1$ and $C_2$ intersect at four distinct points, then $s < -\frac{3}{4}$.

From part (ii), with $k = 1$, the equation becomes:

$$y^2 - x^2 + (2s+1)y - x + s(s+1) = 0$$

The pair of lines with gradients 1 and $-1$ takes the form $y - x + a = 0$ and $y + x + b = 0$. Expanding their product:

$$(y - x + a)(y + x + b) = y^2 - x^2 + (a + b)y + (a - b)x + ab$$

Comparing with $y^2 - x^2 + (2s+1)y - x + s(s+1) = 0$:

$$a + b = 2s + 1, \quad a - b = -1, \quad ab = s(s+1)$$

From the first two: $a = s$, $b = s + 1$. Checking: $ab = s(s+1)$. Correct.

The equation factors as:

$$(y - x + s)(y + x + s + 1) = 0$$

giving the two lines $y - x + s = 0$ (i.e. $y = x - s$) and $y + x + s + 1 = 0$ (i.e. $y = -x - s - 1$).

For four distinct intersection points, each line must meet $C_2$ ($y = x^2$) at two distinct points, and no intersection point is shared between the two lines.

Line 1: $y = x - s$. Substituting into $y = x^2$:

$$x^2 = x - s \implies x^2 - x + s = 0$$

Discriminant: $\Delta_1 = 1 - 4s$.

For two distinct real roots: $1 - 4s > 0$, i.e. $s < \frac{1}{4}$.

Line 2: $y = -x - s - 1$. Substituting into $y = x^2$:

$$x^2 = -x - s - 1 \implies x^2 + x + s + 1 = 0$$

Discriminant: $\Delta_2 = 1 - 4(s + 1) = -3 - 4s$.

For two distinct real roots: $-3 - 4s > 0$, i.e. $s < -\frac{3}{4}$.

Since $s < -\frac{3}{4}$ is the more restrictive condition, and both are necessary for four intersection points, we conclude that if $C_1$ and $C_2$ intersect at four distinct points, then $s < -\frac{3}{4}$. $\qquad \blacksquare$

---

Part (iv)

We show: if $s < -\frac{3}{4}$, then $C_1$ and $C_2$ intersect at four distinct points.

From part (ii), when $k = 1$, the equation factors as:

$$(y - x + s)(y + x + s + 1) = 0$$

and any intersection point of $C_1$ and $C_2$ lies on these two lines.

Step 1: Each line meets $C_2$ at two distinct points.

For $y - x + s = 0$ with $y = x^2$: the equation $x^2 - x + s = 0$ has discriminant $1 - 4s$.

Since $s < -\frac{3}{4}$, we have $1 - 4s > 1 + 3 = 4 > 0$. Two distinct real roots.

For $y + x + s + 1 = 0$ with $y = x^2$: the equation $x^2 + x + s + 1 = 0$ has discriminant $-3 - 4s$.

Since $s < -\frac{3}{4}$, we have $-3 - 4s > -3 + 3 = 0$. Two distinct real roots.

So each line meets $C_2$ at two points, giving at most four intersection points.

Step 2: No intersection point is shared (the four points are distinct).

The two lines $y - x + s = 0$ and $y + x + s + 1 = 0$ intersect at the point obtained by solving simultaneously:

Adding: $2y + 2s + 1 = 0$, so $y = -\frac{2s+1}{2}$. Subtracting: $2x + 1 = 0$, so $x = -\frac{1}{2}$.

This intersection point lies on $C_2$ ($y = x^2$) if and only if:

$$-\frac{2s + 1}{2} = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$$ $$-(2s + 1) = \frac{1}{2} \implies 2s + 1 = -\frac{1}{2} \implies s = -\frac{3}{4}$$

Since $s < -\frac{3}{4}$, the intersection point of the two lines does not lie on $C_2$. Therefore none of the four points is shared, and $C_1$ and $C_2$ intersect at exactly four distinct points. $\qquad \blacksquare$

Examiner Notes

纯数中较不受欢迎的题目。Part(iii)(iv)中考官特别指出蕴含方向(implication direction)让很多考生困惑：Part(iii)需要’四交点→s<-3/4’，Part(iv)需要’s<-3/4→四交点’。因式分解时的符号错误是常见问题（建议展开验证）。少数考生通过考虑直线与y=x²相切的情况同时解决(iii)(iv)两部分。