STEP2 2019 -- Pure Mathematics

STEP2 2019 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2019  |  Questions: Q1—Q8  |  Total marks per question: 20

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

Overview

Q	Topic	Difficulty	Key Techniques
1	代数与函数 (Algebra & Functions)	Challenging	乘积求导法则,切线方程建立,代数化简与因式分解,充要条件证明
2	微积分 (Calculus)	Challenging	反函数的几何意义,面积加法原理,积分换元法,隐函数求值
3	代数与函数 (Algebra & Functions)	Challenging	数学归纳法,三角不等式,多项式估值技巧,有理根定理
4	三角学 (Trigonometry)	Hard	倍角公式连锁消去法,对数求导法,极限计算,无穷乘积与无穷级数
5	数列与级数 (Sequences & Series)	Challenging	不动点方程,递推序列分析,判别式法,充要条件证明
6	微分方程 (Differential Equations)	Challenging	积分因子法,变量代换,驻点与极值分析,定性微分方程理论
7	几何与向量 (Geometry & Vectors)	Challenging	点积运算,向量模长计算,几何条件与向量条件的转化,对称性分析
8	矩阵与线性代数 (Matrices & Linear Algebra)	Hard	矩阵乘法性质,初等矩阵分解,函数方程推理,行列式性质

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Question 1

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

Problem

1 Let $f(x) = (x - p)g(x)$, where $g$ is a polynomial. Show that the tangent to the curve $y = f(x)$ at the point with $x = a$, where $a \neq p$, passes through the point $(p, 0)$ if and only if $g'(a) = 0$.

The curve $C$ has equation

$y = A(x - p)(x - q)(x - r),$

where $p$, $q$ and $r$ are constants with $p < q < r$, and $A$ is a non-zero constant.

(i) The tangent to $C$ at the point with $x = a$, where $a \neq p$, passes through the point $(p, 0)$. Show that $2a = q + r$ and find an expression for the gradient of this tangent in terms of $A$, $q$ and $r$.

(ii) The tangent to $C$ at the point with $x = c$, where $c \neq r$, passes through the point $(r, 0)$. Show that this tangent is parallel to the tangent in part (i) if and only if the tangent to $C$ at the point with $x = q$ does not meet the curve again.

Hint

$f'(x) = g(x) + (x - p)g'(x)$ [M1] Tangent passes through $(a, (a - p)g(a))$ Equation of tangent is $y = (g(a) + (a - p)g'(a))(x - a) + (a - p)g(a)$ [M1] (or equivalent equation) [A1] Substitution of $x = p$ into equation of tangent [E1] $y = -(a - p)^2 g'(a)$ Verification that if $g'(a) = 0$, then $y = 0$ [E1] If $y = 0$ then $g'(a) = 0$ because $a \neq p$ [E1 (AG)] [(6 marks)] (i) $g(x) = A(x - q)(x - r)$ identified [M1] $g'(a) = 0 \Rightarrow 2a = r + q$ (legitimately obtained) [A1 (AG)] Gradient of tangent is $g(a) + (a - p)g'(a)$ $= A(a - q)(a - r)$ [M1] $= -\frac{1}{4}A(r - q)^2$ [A1] [(4 marks)] (ii) By symmetry, the gradient of the second tangent is $-\frac{1}{4}A(p - q)^2$ (can be implied) [B1] Parallel iff $(p - q)^2 = (q - r)^2$ [M1] $\Leftrightarrow q - p = r - q$ [A1] since $p < q < r$. [E1] Tangent at $x = q$, $y = A(q - p)(q - r)(x - q)$, [M1] Meets curve again when $(q - p)(q - r)(x - q) = (x - p)(x - r)(x - q)$ $\Leftrightarrow (q - p)(q - r) = (x - p)(x - r)$ since $x \neq q$ [M1] (cancellation must be justified for M1, can be awarded if used correctly on $(x - q)^2(x - p - r + q)$ later) $\Leftrightarrow (x - q)(x - p - r + q) = 0$ [M1] $\Leftrightarrow x = p + r - q$ or $x = q$ [A1] Therefore there is only one point of intersection between the tangent and the curve if and only if $p + r - q = q$, which is if and only if the [E1] tangents are parallel. [E1 (AG)] One E mark for each direction. [(10 marks)]

Model Solution

General result.

Since $f(x) = (x-p)g(x)$, by the product rule:

$f'(x) = g(x) + (x-p)g'(x)$

The point on the curve at $x = a$ is $(a,\; (a-p)g(a))$, and the gradient is $f'(a) = g(a) + (a-p)g'(a)$. The tangent line at this point is:

$y - (a-p)g(a) = \bigl(g(a) + (a-p)g'(a)\bigr)(x - a)$

Setting $x = p$:

$y = (a-p)g(a) + \bigl(g(a) + (a-p)g'(a)\bigr)(p - a)$

$= (a-p)g(a) - (a-p)g(a) - (a-p)^2 g'(a)$

$= -(a-p)^2 g'(a)$

Since $a \neq p$, we have $(a-p)^2 > 0$, so $y = 0$ if and only if $g'(a) = 0$. $\square$

Part (i)

Here $f(x) = A(x-p)(x-q)(x-r)$, so we identify $g(x) = A(x-q)(x-r)$.

By the general result, the tangent at $x = a$ passes through $(p, 0)$ if and only if $g'(a) = 0$.

Computing $g'(x)$:

$g'(x) = A\bigl[(x-r) + (x-q)\bigr] = A(2x - q - r)$

Setting $g'(a) = 0$:

$A(2a - q - r) = 0$

Since $A \neq 0$, we get $2a = q + r$, i.e. $a = \dfrac{q+r}{2}$. $\square$

The gradient of the tangent is $f'(a) = g(a) + (a-p)g'(a) = g(a) + 0 = g(a)$, so:

$\text{gradient} = A(a-q)(a-r)$

With $a = \dfrac{q+r}{2}$:

$a - q = \frac{r-q}{2}, \qquad a - r = \frac{q-r}{2} = -\frac{r-q}{2}$

$\text{gradient} = A \cdot \frac{r-q}{2} \cdot \left(-\frac{r-q}{2}\right) = -\frac{A(r-q)^2}{4}$

Part (ii)

By the same reasoning with the roles of $p$ and $r$ swapped, the tangent at $x = c$ passes through $(r, 0)$ when $2c = p + q$, and its gradient is:

$A(c-p)(c-q) = -\frac{A(q-p)^2}{4}$

The two tangents are parallel if and only if their gradients are equal:

$-\frac{A(r-q)^2}{4} = -\frac{A(q-p)^2}{4}$

$(r-q)^2 = (q-p)^2$

Since $p < q < r$, both $r - q > 0$ and $q - p > 0$, so taking positive square roots:

$r - q = q - p \qquad \text{(i.e.\ } p + r = 2q\text{)}$

Now consider the tangent to $C$ at $x = q$. The gradient is $f'(q) = A(q-p)(q-r)$, so the tangent line is:

$y = A(q-p)(q-r)(x - q)$

To find where this tangent meets $C$ again, set $A(x-p)(x-q)(x-r) = A(q-p)(q-r)(x-q)$.

If $x = q$ this is trivially satisfied (the point of tangency). For $x \neq q$, we may divide by $A(x-q)$:

$(x-p)(x-r) = (q-p)(q-r)$

Expanding both sides:

$x^2 - (p+r)x + pr = q^2 - (p+r)q + pr$

$x^2 - (p+r)x = q^2 - (p+r)q$

$x^2 - q^2 - (p+r)(x - q) = 0$

$(x-q)\bigl[(x+q) - (p+r)\bigr] = 0$

$(x-q)(x + q - p - r) = 0$

So $x = q$ (the tangency point) or $x = p + r - q$.

The tangent meets the curve again if and only if $p + r - q \neq q$, i.e.\ $p + r \neq 2q$, i.e.\ $q - p \neq r - q$.

Equivalently, the tangent at $x = q$ does not meet the curve again if and only if $p + r = 2q$, which is exactly the condition $q - p = r - q$ for the two tangents to be parallel. $\square$

Examiner Notes

无官方评述。易错点：(1) 乘积求导时遗漏项；(2) part(ii) 中”切线不再与曲线相交”的几何含义理解不清，需联立切线与曲线方程分析重根条件；(3) 题目要求证明”if and only if”，需注意双向推导。

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Question 2

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

Problem

2 The function $f$ satisfies $f(0) = 0$ and $f'(t) > 0$ for $t > 0$. Show by means of a sketch that, for $x > 0$,

$\int_{0}^{x} f(t) \, dt + \int_{0}^{f(x)} f^{-1}(y) \, dy = xf(x).$

(i) The (real) function $g$ is defined, for all $t$, by

$(g(t))^3 + g(t) = t.$

Prove that $g(0) = 0$, and that $g'(t) > 0$ for all $t$.

Evaluate $\int_{0}^{2} g(t) \, dt$.

(ii) The (real) function $h$ is defined, for all $t$, by

$(h(t))^3 + h(t) = t + 2.$

Evaluate $\int_{0}^{8} h(t) \, dt$.

Hint

Sketch with areas $\int_0^x f(t) dt$, $\int_0^{f(x)} f^{-1}(y) dy$ and rectangle correctly identified. (One mark any one) [G1 G1 (2 marks)] (i) $g(0)(g(0)^2 + 1) = 0$ factorised $g(0)$ real so $g(0) = 0$ (must be justified) [M1 A1 (AG)] $1 = (3g(t)^2 + 1)g'(t)$ $(3g(t)^2 + 1) > 0$ so $g'(t) > 0$ [M1 A1 (AG)] $g(2)^3 + g(2) - 2 = 0$ $(g(2) - 1)(g(2)^2 + g(2) + 2) = 0$ $\Delta = -7 < 0$ so $g(2) = 1$ or $g(2)>0$ justified [M1 A1] $g^{-1}(s) = s^3 + s$ $\int_0^2 g(t) dt = 2g(2) - \int_0^{g(2)} g^{-1}(s) ds$ $= \frac{5}{4}$ [B1 M1 A1 (9 marks)] (ii) $h(t) = g(t + 2)$ so $h(0) = g(2) = 1$ and $h'(t) > 0$ [M1 A1] $(h(8) - 2)(h(8)^2 + 2h(8) + 5) = 0$ $h(8) = 2$ correctly justified [M1 A1] $h^{-1}(s) = s^3 + s - 2$ $\int_0^8 h(t) dt + \int_{h(0)}^{h(8)} h^{-1}(s) ds = 16$ (or similar correct equation) $\int_0^8 h(t) dt = 16 - \int_1^2 (s^3 + s - 2) ds$ $= 16 - [\frac{s^4}{4} + \frac{s^2}{2} - 2s]_1^2$ (integration) $= 12 \frac{3}{4}$ [B1 M1 A1 M1 A1 (9 marks)]

Model Solution

Sketch argument for the identity.

Since $f(0) = 0$ and $f'(t) > 0$ for $t > 0$, the function $f$ is strictly increasing for $t > 0$, so $f^{-1}$ exists. Consider the rectangle with vertices $(0,0)$, $(x,0)$, $(x, f(x))$, $(0, f(x))$. This rectangle has area $x \cdot f(x)$.

The region under $y = f(t)$ from $t = 0$ to $t = x$ has area $\int_0^x f(t)\,dt$.

The region to the left of $t = f^{-1}(y)$ from $y = 0$ to $y = f(x)$ has area $\int_0^{f(x)} f^{-1}(y)\,dy$.

These two regions tile the rectangle exactly (they share only the boundary curve $y = f(t)$), so:

$\int_0^x f(t)\,dt + \int_0^{f(x)} f^{-1}(y)\,dy = x\,f(x) \qquad \square$

Part (i)

Showing $g(0) = 0$: Setting $t = 0$ in $(g(t))^3 + g(t) = t$:

$(g(0))^3 + g(0) = 0$

$g(0)\bigl((g(0))^2 + 1\bigr) = 0$

Since $(g(0))^2 + 1 \geq 1 > 0$ for real $g(0)$, we must have $g(0) = 0$. $\square$

Showing $g'(t) > 0$: Differentiating $(g(t))^3 + g(t) = t$ implicitly with respect to $t$:

$3(g(t))^2 g'(t) + g'(t) = 1$

$g'(t)\bigl(3(g(t))^2 + 1\bigr) = 1$

Since $3(g(t))^2 + 1 \geq 1 > 0$, we have $g'(t) = \dfrac{1}{3(g(t))^2 + 1} > 0$ for all $t$. $\square$

Evaluating $\int_0^2 g(t)\,dt$: Since $g$ is strictly increasing (as $g'(t) > 0$) with $g(0) = 0$, we can apply the identity with $f = g$ and $x = 2$:

$\int_0^2 g(t)\,dt + \int_0^{g(2)} g^{-1}(s)\,ds = 2\,g(2) \qquad (\ast)$

We first find $g(2)$. Setting $t = 2$: $(g(2))^3 + g(2) = 2$, so $(g(2))^3 + g(2) - 2 = 0$. Testing $g(2) = 1$: $1 + 1 - 2 = 0$. Factoring:

$(g(2) - 1)\bigl((g(2))^2 + g(2) + 2\bigr) = 0$

The discriminant of $s^2 + s + 2$ is $1 - 8 = -7 < 0$, so there are no other real roots. Thus $g(2) = 1$.

The inverse function satisfies $s = (g^{-1}(s))^3 + g^{-1}(s)$… equivalently, from $(g(t))^3 + g(t) = t$, if $s = g(t)$ then $t = s^3 + s$, so $g^{-1}(s) = s^3 + s$.

Substituting into $(\ast)$:

$\int_0^2 g(t)\,dt + \int_0^1 (s^3 + s)\,ds = 2 \cdot 1 = 2$

$\int_0^1 (s^3 + s)\,ds = \left[\frac{s^4}{4} + \frac{s^2}{2}\right]_0^1 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

$\int_0^2 g(t)\,dt = 2 - \frac{3}{4} = \frac{5}{4}$

Part (ii)

The function $h$ satisfies $(h(t))^3 + h(t) = t + 2$, while $g$ satisfies $(g(t))^3 + g(t) = t$.

Comparing: if we set $u = t + 2$, then $(g(u))^3 + g(u) = u = t + 2$, so $h(t) = g(t + 2)$.

Therefore $h(0) = g(2) = 1$ and $h(8) = g(10)$.

We find $h(8) = g(10)$: setting $s = g(10)$, we need $s^3 + s = 10$. Testing $s = 2$: $8 + 2 = 10$. Factoring:

$(s - 2)(s^2 + 2s + 5) = 0$

The discriminant of $s^2 + 2s + 5$ is $4 - 20 = -16 < 0$, so $h(8) = g(10) = 2$.

Since $h(0) = 1 \neq 0$, the identity does not apply directly to $h$. Instead, use the substitution $u = t + 2$:

$\int_0^8 h(t)\,dt = \int_0^8 g(t+2)\,dt = \int_2^{10} g(u)\,du$

Now apply the identity to $g$ (which satisfies $g(0) = 0$) at $x = 10$ and at $x = 2$:

$\int_0^{10} g(u)\,du + \int_0^{g(10)} g^{-1}(s)\,ds = 10 \cdot g(10) = 20$

$\int_0^2 g(u)\,du + \int_0^{g(2)} g^{-1}(s)\,ds = 2 \cdot g(2) = 2$

Subtracting:

$\int_2^{10} g(u)\,du = 18 - \int_{g(2)}^{g(10)} g^{-1}(s)\,ds = 18 - \int_1^2 (s^3 + s)\,ds$

$\int_1^2 (s^3 + s)\,ds = \left[\frac{s^4}{4} + \frac{s^2}{2}\right]_1^2 = (4 + 2) - \left(\frac{1}{4} + \frac{1}{2}\right) = 6 - \frac{3}{4} = \frac{21}{4}$

$\int_0^8 h(t)\,dt = 18 - \frac{21}{4} = \frac{72 - 21}{4} = \frac{51}{4} = 12\frac{3}{4}$

Examiner Notes

无官方评述。易错点：(1) 面积图示法的严格性——需说明 f 单调递增确保反函数存在；(2) part(ii) 中 h(t) 与 g(t) 的关系容易忽略平移变换 h(t) = g(t-2)+? 实际上 (h(t))³+h(t)=t+2 意味着 h(t)=g(t+2) 的某种变形，需仔细处理；(3) 积分限的变换要准确。

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Question 3

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

Problem

3 For any two real numbers $x_1$ and $x_2$, show that

$|x_1 + x_2| \leqslant |x_1| + |x_2|.$

Show further that, for any real numbers $x_1, x_2, \dots, x_n$,

$|x_1 + x_2 + \dots + x_n| \leqslant |x_1| + |x_2| + \dots + |x_n|.$

(i) The polynomial f is defined by

$f(x) = 1 + a_1x + a_2x^2 + \dots + a_{n-1}x^{n-1} + x^n$

where the coefficients are real and satisfy $|a_i| \leqslant A$ for $i = 1, 2, \dots, n-1$, where $A \geqslant 1$.

(a) If $|x| < 1$, show that

$|f(x) - 1| \leqslant \frac{A|x|}{1 - |x|}.$

(b) Let $\omega$ be a real root of f, so that $f(\omega) = 0$. In the case $|\omega| < 1$, show that

$\frac{1}{1 + A} \leqslant |\omega| \leqslant 1 + A. \qquad \text{(*)}$

(c) Show further that the inequalities (*) also hold if $|\omega| \geqslant 1$.

(ii) Find the integer root or roots of the quintic equation

$135x^5 - 135x^4 - 100x^3 - 91x^2 - 126x + 135 = 0.$

Hint

$|x_1 + x_2|$ is maximised when both have the same sign, [E1] In which case $|x_1 + x_2| = |x_1| + |x_2|$. Thus, $|x_1 + x_2| \le |x_1 + x_2|$ (or by consideration of all four combinations of signs separately) $|x_1 + \dots + x_{n-1} + x_n| \le |x_1 + \dots + x_{n-1}| + |x_n|$ $\le \dots$ $\le |x_1| + \dots + |x_{n-1}| + |x_n|$ by induction [E1] [(2 marks)] (i) (a) $|f(x) - 1| = |a_1x + \dots + a_{n-1}x^{n-1} + x^n|$ $\le |a_1x| + \dots + |a_{n-1}x^{n-1}| + |x^n|$ $= |a_1||x| + \dots + |a_{n-1}||x|^{n-1} + |x|^n$ $\le A(|x| + \dots + |x|^{n-1}) + |x|^n$ $\le A(|x| + \dots + |x|^{n-1} + |x|^n)$ (justified) $= A \frac{|x|(1-|x|^n)}{1-|x|}$ $\le A \frac{|x|}{1-|x|}$ (justified) [M1 M1 M1 M1 M1 A1 (AG)] (6 marks) (b) $1 \le \frac{A|\omega|}{1-|\omega|}$ using $f(\omega) = 0$ $1 \le (A + 1)|\omega|$ (with sign of $1 - |\omega|$ justified) $A + 1 \ge 1 \ge |\omega|$ [M1 A1 (AG) B1 (AG)] (3 marks) (c) If $|\omega| > 1$, $0 = \omega^n f\left(\frac{1}{\omega}\right)$ $= 1 + a_{n-1}\omega + \dots + a_1\omega^{n-1} + \omega^n$ Inequalities continue to hold since $|a_i| \le A$ If $|\omega| = 1$, then $1 + A \ge 1 \ge \frac{1}{1+A}$ since $A > 0$ [M1 E1 E1] (3 marks) (ii) $f(x) = x^5 - x^4 - \frac{100}{135}x^3 - \frac{91}{135}x^2 - \frac{126}{135}x + 1$ Use $A = 1$. Integer roots with $\frac{1}{2} \le |\omega| \le 2$ could only be $\pm 1$ or $\pm 2$ $f(\pm 2) \ne 0$ because numerator is odd (or any valid justification) $f(1) = -\frac{182}{135} \ne 0$ $f(1) = 0$ $x = 1$ is the only integer root. [B1 M1 M1 E1 A1 A1] (6 marks)

Model Solution

Preliminary: Triangle Inequality

For real numbers $x_1, x_2$, both sides of $|x_1 + x_2| \leqslant |x_1| + |x_2|$ are non-negative, so squaring preserves the inequality. It suffices to show

$(x_1 + x_2)^2 \leqslant (|x_1| + |x_2|)^2$

$x_1^2 + 2x_1 x_2 + x_2^2 \leqslant x_1^2 + 2|x_1||x_2| + x_2^2$

$x_1 x_2 \leqslant |x_1||x_2|.$

This holds since $x_1 x_2 \leqslant |x_1 x_2| = |x_1||x_2|$ for all real $x_1, x_2$. Hence $|x_1 + x_2| \leqslant |x_1| + |x_2|$.

For the general case, we proceed by induction. The base case $n = 2$ is established above. Suppose the inequality holds for $n - 1$ terms. Then

$|x_1 + \dots + x_{n-1} + x_n| \leqslant |x_1 + \dots + x_{n-1}| + |x_n| \leqslant (|x_1| + \dots + |x_{n-1}|) + |x_n|$

by the two-term inequality (applied to $x_1 + \dots + x_{n-1}$ and $x_n$) and the inductive hypothesis. Hence $|x_1 + \dots + x_n| \leqslant |x_1| + \dots + |x_n|$ for all $n \geqslant 2$. $\qquad \text{(shown)}$

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Part (i)(a)

We have $f(x) - 1 = a_1 x + a_2 x^2 + \dots + a_{n-1}x^{n-1} + x^n$. Applying the triangle inequality:

$|f(x) - 1| \leqslant |a_1||x| + |a_2||x|^2 + \dots + |a_{n-1}||x|^{n-1} + |x|^n$

Since $|a_i| \leqslant A$ for each $i$:

$|f(x) - 1| \leqslant A|x| + A|x|^2 + \dots + A|x|^{n-1} + |x|^n$

Since $A \geqslant 1$, we have $|x|^n \leqslant A|x|^n$, so

$|f(x) - 1| \leqslant A\bigl(|x| + |x|^2 + \dots + |x|^n\bigr) = A \cdot \frac{|x|(1 - |x|^n)}{1 - |x|}$

using the geometric series formula (valid for $|x| < 1$). Since $|x|^n > 0$, we have $1 - |x|^n < 1$, giving

$|f(x) - 1| \leqslant \frac{A|x|(1 - |x|^n)}{1 - |x|} \leqslant \frac{A|x|}{1 - |x|}. \qquad \text{(shown)}$

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Part (i)(b)

If $\omega$ is a real root with $|\omega| < 1$, then $f(\omega) = 0$, so $|f(\omega) - 1| = 1$. By part (a):

$1 \leqslant \frac{A|\omega|}{1 - |\omega|}$

Since $|\omega| < 1$, we have $1 - |\omega| > 0$. Multiplying both sides by $(1 - |\omega|)$:

$1 - |\omega| \leqslant A|\omega|$

$1 \leqslant (1 + A)|\omega|$

$|\omega| \geqslant \frac{1}{1 + A}.$

Also, since $A \geqslant 1$ and $|\omega| < 1 < 1 + A$, we have $|\omega| \leqslant 1 + A$. Combining:

$\frac{1}{1 + A} \leqslant |\omega| \leqslant 1 + A. \qquad \text{(shown)}$

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Part (i)(c)

Case $|\omega| > 1$: Since $f(\omega) = 0$, we have $\omega^n + a_{n-1}\omega^{n-1} + \dots + a_1 \omega + 1 = 0$. Dividing by $\omega^n$:

$1 + \frac{a_{n-1}}{\omega} + \frac{a_{n-2}}{\omega^2} + \dots + \frac{a_1}{\omega^{n-1}} + \frac{1}{\omega^n} = 0$

So $\omega' = 1/\omega$ is a root of $g(t) = 1 + a_{n-1}t + a_{n-2}t^2 + \dots + a_1 t^{n-1} + t^n$, which is a polynomial of the same form as $f$ with $|a_i| \leqslant A$.

Since $|\omega'| = 1/|\omega| < 1$, part (b) gives $\frac{1}{1+A} \leqslant |\omega'| \leqslant 1+A$, i.e.

$\frac{1}{1+A} \leqslant \frac{1}{|\omega|} \leqslant 1 + A.$

Taking reciprocals (reversing the first inequality):

$\frac{1}{1+A} \leqslant |\omega| \leqslant 1 + A.$

Case $|\omega| = 1$: We need $\frac{1}{1+A} \leqslant 1 \leqslant 1 + A$. Since $A \geqslant 1$, we have $1 + A \geqslant 2 > 1$ and $\frac{1}{1+A} \leqslant \frac{1}{2} < 1$. Both inequalities hold.

Hence the bounds $(*)$ hold for all real roots $\omega$. $\qquad \text{(shown)}$

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Part (ii)

The equation is $135x^5 - 135x^4 - 100x^3 - 91x^2 - 126x + 135 = 0$. Dividing through by 135:

$x^5 - x^4 - \frac{20}{27}x^3 - \frac{91}{135}x^2 - \frac{14}{15}x + 1 = 0$

This has the form $1 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + x^5 = 0$ with $a_1 = -\frac{14}{15}$, $a_2 = -\frac{91}{135}$, $a_3 = -\frac{20}{27}$, $a_4 = -1$.

We check $|a_i| \leqslant 1$: $\frac{14}{15} < 1$, $\frac{91}{135} < 1$, $\frac{20}{27} < 1$, $1 = 1$. So $A = 1$.

By the bounds $(*)$, any root $\omega$ satisfies $\frac{1}{2} \leqslant |\omega| \leqslant 2$. The only integers in this range are $\pm 1$ and $\pm 2$.

Checking $x = 1$:

$135 - 135 - 100 - 91 - 126 + 135 = -182 \neq 0.$

Checking $x = -1$:

$-135 - 135 + 100 - 91 + 126 + 135 = 0.$

So $x = -1$ is a root. For completeness, we verify the remaining candidates.

Checking $x = 2$: $\;135(32) - 135(16) - 100(8) - 91(4) - 126(2) + 135 = 4320 - 2160 - 800 - 364 - 252 + 135 = 767 \neq 0$.

Checking $x = -2$: $\;135(-32) - 135(16) - 100(-8) - 91(4) - 126(-2) + 135 = -4320 - 2160 + 800 - 364 + 252 + 135 = -5657 \neq 0$.

The only integer root is $x = -1$.

Examiner Notes

无官方评述。易错点：(1) part(i)(a) 中 |f(x)-1| 的估计需逐项拆分后用等比级数求和；(2) part(i)(b) 和 (c) 中 |ω|<1 与 |ω|≥1 两种情况的分界处理，特别是 (*) 中 1/(1+A)≤|ω| 的下界推导；(3) part(ii) 的整数根需先用有理根定理缩小候选范围再逐一验证。

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Question 4

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

Problem

4 You are not required to consider issues of convergence in this question.

For any sequence of numbers $a_1, a_2, \dots, a_m, \dots, a_n$, the notation $\displaystyle \prod_{i=m}^{n} a_i$ denotes the product $a_m a_{m+1} \cdots a_n$.

(i) Use the identity $2 \cos x \sin x = \sin(2x)$ to evaluate the product $\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})$.

(ii) Simplify the expression

$\prod_{k=0}^{n} \cos \left( \frac{x}{2^k} \right) \qquad (0 < x < \frac{1}{2}\pi).$

Using differentiation, or otherwise, show that, for $0 < x < \frac{1}{2}\pi$,

$\sum_{k=0}^{n} \frac{1}{2^k} \tan \left( \frac{x}{2^k} \right) = \frac{1}{2^n} \cot \left( \frac{x}{2^n} \right) - 2 \cot(2x).$

(iii) Using the results $\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\displaystyle \lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$, show that

$\prod_{k=1}^{\infty} \cos \left( \frac{x}{2^k} \right) = \frac{\sin x}{x}$

and evaluate

$\sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan \left( \frac{\pi}{2^j} \right).$

Hint

(i) $\sin \frac{\pi}{9} \cos \frac{\pi}{9} \cos \frac{2\pi}{9} \cos \frac{4\pi}{9}$ [B1] $= \frac{1}{2} \sin \frac{2\pi}{9} \cos \frac{2\pi}{9} \cos \frac{4\pi}{9}$ [M1] $= \frac{1}{8} \sin \frac{8\pi}{9}$ $= \frac{1}{8} \sin \frac{\pi}{9}$ (use of $\sin(\pi - x) = \sin(x)$) [M1] $\cos \frac{\pi}{9} \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} = \frac{1}{8}$ [A1] (4 marks) (ii) $\sin \left( \frac{x}{2^n} \right) \prod_{k=0}^{n} \cos \left( \frac{x}{2^k} \right)$ [B1] $= \frac{1}{2} \sin \left( \frac{x}{2^{n-1}} \right) \prod_{k=0}^{n-1} \cos \left( \frac{x}{2^k} \right)$ [M1] $= \dots$ (convincing use of induction or repeated application) [E1] $= \frac{\sin(2x)}{2^{n+1}}$ (induction end point correct) $\prod_{k=0}^{n} \cos \left( \frac{x}{2^k} \right) = \frac{\sin(2x)}{2^{n+1} \sin \left( \frac{x}{2^n} \right)}$ [A1] $\sum_{k=0}^{n} \log \left( \cos \left( \frac{x}{2^k} \right) \right) = \log(\sin(2x)) - \log \left( \sin \left( \frac{x}{2^n} \right) \right) - \log(2^{n+1})$ [M1 (diff)] $\sum_{k=0}^{n} \frac{1}{2^k} \tan \left( \frac{x}{2^k} \right) = -2 \cot(2x) + \frac{1}{2^n} \cot \left( \frac{x}{2^n} \right)$ [M1 (division)] (justified with differentiation) [A1 (AG)] (7 marks) (iii) B1 – switch to product starting at 0 M1 – set up as limiting case of product to n M1 – apply small angle for sin A1 – correct answer $\prod_{k=1}^{n} \cos \left( \frac{x}{2^k} \right) = \frac{\sin(2x)}{2^{n+1} \sin \left( \frac{x}{2^n} \right) \cos(x)}$ [M1] $= \frac{2 \sin(x)}{2^{n+1} \sin \left( \frac{x}{2^n} \right)}$ [M1] $\sim \frac{\sin(x)}{2^n \times \left( \frac{x}{2^n} \right)}$ [M1] $= \frac{\sin(x)}{x}$ [A1 (AG)]

Model Solution

Part (i)

We use the double-angle identity $2\cos\theta\sin\theta = \sin(2\theta)$ repeatedly. The key trick is to multiply the product by $\sin(\pi/9)$:

$\sin\frac{\pi}{9} \cdot \cos\frac{\pi}{9} \cdot \cos\frac{2\pi}{9} \cdot \cos\frac{4\pi}{9}$

Applying $2\cos\theta\sin\theta = \sin(2\theta)$ with $\theta = \pi/9$:

$= \frac{1}{2}\sin\frac{2\pi}{9} \cdot \cos\frac{2\pi}{9} \cdot \cos\frac{4\pi}{9}$

Applying the identity again with $\theta = 2\pi/9$:

$= \frac{1}{4}\sin\frac{4\pi}{9} \cdot \cos\frac{4\pi}{9}$

Applying the identity once more with $\theta = 4\pi/9$:

$= \frac{1}{8}\sin\frac{8\pi}{9}$

Since $\sin(\pi - x) = \sin x$, we have $\sin\frac{8\pi}{9} = \sin\frac{\pi}{9}$. Therefore:

$\sin\frac{\pi}{9} \cdot \cos\frac{\pi}{9} \cdot \cos\frac{2\pi}{9} \cdot \cos\frac{4\pi}{9} = \frac{1}{8}\sin\frac{\pi}{9}$

Dividing both sides by $\sin(\pi/9) \neq 0$:

$\cos\frac{\pi}{9} \cdot \cos\frac{2\pi}{9} \cdot \cos\frac{4\pi}{9} = \frac{1}{8}$

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Part (ii) — Simplification

We claim that

$\prod_{k=0}^{n} \cos\frac{x}{2^k} = \frac{\sin(2x)}{2^{n+1}\sin\!\left(\dfrac{x}{2^n}\right)}. \qquad \text{(*)}$

Proof by induction. For $n = 0$: the left side is $\cos x$ and the right side is $\frac{\sin(2x)}{2\sin x} = \frac{2\sin x\cos x}{2\sin x} = \cos x$. So the base case holds.

Suppose the formula holds for $n - 1$. Then

$\prod_{k=0}^{n} \cos\frac{x}{2^k} = \cos\frac{x}{2^n} \cdot \prod_{k=0}^{n-1} \cos\frac{x}{2^k} = \cos\frac{x}{2^n} \cdot \frac{\sin(2x)}{2^n \sin\!\left(\dfrac{x}{2^{n-1}}\right)}$

by the inductive hypothesis. Using $2\sin\frac{x}{2^{n-1}}\cos\frac{x}{2^n} = \sin\frac{x}{2^{n-1}} \cdot 2\cos\frac{x}{2^n}$… let us instead write $2\cos\frac{x}{2^n}\sin\frac{x}{2^n} = \sin\frac{x}{2^{n-1}}$, so $\cos\frac{x}{2^n} = \frac{\sin\frac{x}{2^{n-1}}}{2\sin\frac{x}{2^n}}$. Substituting:

$= \frac{\sin\frac{x}{2^{n-1}}}{2\sin\frac{x}{2^n}} \cdot \frac{\sin(2x)}{2^n \sin\frac{x}{2^{n-1}}} = \frac{\sin(2x)}{2^{n+1}\sin\frac{x}{2^n}}.$

This completes the induction. $\qquad \text{(shown)}$

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Part (ii) — Tan sum formula

Taking logarithms of both sides of $(*)$:

$\sum_{k=0}^{n} \ln\!\left(\cos\frac{x}{2^k}\right) = \ln(\sin 2x) - \ln\!\left(\sin\frac{x}{2^n}\right) - (n+1)\ln 2$

Differentiating both sides with respect to $x$:

$\sum_{k=0}^{n} \frac{1}{2^k} \cdot \frac{-\sin\frac{x}{2^k}}{\cos\frac{x}{2^k}} = \frac{2\cos 2x}{\sin 2x} - \frac{1}{2^n} \cdot \frac{\cos\frac{x}{2^n}}{\sin\frac{x}{2^n}}$

$-\sum_{k=0}^{n} \frac{1}{2^k}\tan\frac{x}{2^k} = 2\cot 2x - \frac{1}{2^n}\cot\frac{x}{2^n}$

Multiplying both sides by $-1$:

$\sum_{k=0}^{n} \frac{1}{2^k}\tan\frac{x}{2^k} = \frac{1}{2^n}\cot\frac{x}{2^n} - 2\cot 2x. \qquad \text{(shown)}$

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Part (iii) — Infinite product

From part (ii), the product $\prod_{k=0}^{n} \cos\frac{x}{2^k}$ includes the $k = 0$ factor $\cos x$. So

$\prod_{k=1}^{n} \cos\frac{x}{2^k} = \frac{\prod_{k=0}^{n} \cos\frac{x}{2^k}}{\cos x} = \frac{\sin 2x}{2^{n+1}\sin\frac{x}{2^n}\cos x}.$

Using $\sin 2x = 2\sin x\cos x$:

$\prod_{k=1}^{n} \cos\frac{x}{2^k} = \frac{2\sin x\cos x}{2^{n+1}\sin\frac{x}{2^n}\cos x} = \frac{\sin x}{2^n \sin\frac{x}{2^n}}.$

As $n \to \infty$, $x/2^n \to 0$, so $\sin\frac{x}{2^n} \sim \frac{x}{2^n}$ (using $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$). Therefore

$\lim_{n\to\infty} \prod_{k=1}^{n} \cos\frac{x}{2^k} = \lim_{n\to\infty} \frac{\sin x}{2^n \cdot \frac{x}{2^n}} = \frac{\sin x}{x}.$

$\prod_{k=1}^{\infty} \cos\frac{x}{2^k} = \frac{\sin x}{x}. \qquad \text{(shown)}$

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Part (iii) — Evaluating the sum

We evaluate $\displaystyle S = \sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\frac{\pi}{2^j}$.

Setting $k = j - 2$ (so $j = k + 2$), the sum becomes

$S = \sum_{k=0}^{\infty} \frac{1}{2^k} \tan\frac{\pi}{2^{k+2}} = \sum_{k=0}^{\infty} \frac{1}{2^k} \tan\frac{\pi}{4 \cdot 2^k}$

This matches the formula from part (ii) with $x = \pi/4$. Taking the limit $n \to \infty$:

$S = \lim_{n\to\infty}\left(\frac{1}{2^n}\cot\frac{\pi/4}{2^n} - 2\cot\frac{\pi}{2}\right)$

Since $\cot(\pi/2) = 0$, the second term vanishes. For the first term, let $\theta = \frac{\pi/4}{2^n} \to 0$:

$\frac{1}{2^n}\cot\theta = \frac{\cos\theta}{2^n \sin\theta} = \frac{\cos\theta}{2^n \theta} \cdot \frac{\theta}{\sin\theta} \cdot \frac{1}{1} = \frac{\cos\theta}{\pi/4} \cdot \frac{\theta}{\sin\theta}$

since $2^n\theta = \pi/4$. As $\theta \to 0$: $\cos\theta \to 1$ and $\frac{\theta}{\sin\theta} \to 1$ (using $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$). Therefore

$\lim_{n\to\infty} \frac{1}{2^n}\cot\frac{\pi/4}{2^n} = \frac{1}{\pi/4} = \frac{4}{\pi}$

$S = \frac{4}{\pi}$

Examiner Notes

无官方评述。易错点：(1) part(i) 的技巧——乘以 sin(π/9) 构造连锁消去——不易想到；(2) part(ii) 中对乘积取对数后求导得到 tan 求和公式，需注意乘积与求和的对应关系；(3) part(iii) 取极限时需用 lim(sinθ/θ)=1 的条件，且求和公式的指标变换（j 从 2 开始）容易出错。

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Question 5

Topic: 数列与级数 (Sequences & Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The sequence $u_0, u_1, \dots$ is said to be a constant sequence if $u_n = u_{n+1}$ for $n = 0, 1, 2, \dots$. The sequence is said to be a sequence of period 2 if $u_n = u_{n+2}$ for $n = 0, 1, 2, \dots$ and the sequence is not constant.

(i) A sequence of real numbers is defined by $u_0 = a$ and $u_{n+1} = f(u_n)$ for $n = 0, 1, 2, \dots$, where

$f(x) = p + (x - p)x,$

and $p$ is a given real number.

Find the values of $a$ for which the sequence is constant.

Show that the sequence has period 2 for some value of $a$ if and only if $p > 3$ or $p < -1$.

(ii) A sequence of real numbers is defined by $u_0 = a$ and $u_{n+1} = f(u_n)$ for $n = 0, 1, 2, \dots$, where

$f(x) = q + (x - p)x,$

and $p$ and $q$ are given real numbers.

Show that there is no value of $a$ for which the sequence is constant if and only if $f(x) > x$ for all $x$.

Deduce that, if there is no value of $a$ for which the sequence is constant, then there is no value of $a$ for which the sequence has period 2.

Is it true that, if there is no value of $a$ for which the sequence has period 2, then there is no value of $a$ for which the sequence is constant?

Hint

(i) Constant iff $a = f(a)$ [M1] $\Leftrightarrow a = p + (a - p)a$ $\Leftrightarrow 0 = (a - p)(a - 1)$ [M1] $\Leftrightarrow a = p$ or $a = 1$. [A1] Period 2 $\Leftrightarrow a = f(f(a))$ [M1] $\Leftrightarrow 0 = (a - p)(-1 + 2ap - pa^2 + a^3)$ (factorisation) [M1] $\Leftrightarrow 0 = (a - p)(a - 1)(a^2 + (1 - p)a + 1)$ [A1] If $a = p$ or $a = 1$, then sequence is constant. [B1] The quadratic has solutions when $(p - 1)^2 \ge 4$. [M1] If $(p - 1)^2 > 4$, i.e. $p > 3$ or $p < -1$, the solutions are distinct. They are not both $1, p$ since the sum of the roots is $p - 1 \ne p + 1$ [E1] So for $p > 3$ or $p < -1$, one of the roots of the quadratic gives a sequence of period 2. [E1 (AG)] If $p = 3, a = 1$ so not period 2. [B1] If $p = -1, a = -1 = p$ so not period 2. [B1] (12 marks) (ii) No value of $a$ for which the sequence is constant $\Leftrightarrow f(a) = a$ has no solution [E1 ($\rightarrow$)] $\Leftrightarrow f(x) > x$ or $f(x) < x$ for all $x$ [E1 ($\leftarrow$)] But $f(x) > x$ for large $x$. So cannot have $f(x) < x$ for all $x$. [E1] If no value of $a$ for which sequence constant, then $f(x) > x$ for all $x$ [E1] So $f(f(x)) > f(x) > x$ for all $x$ [E1] And hence no solution to $f(f(a)) = a$. [E1] Setting $p = q$, gives (i). [E1] Then if $-1 \le p \le 3$, there is no period 2 sequence but a constant sequence exists. [E1] (8 marks)

Model Solution

Part (i): Constant Sequences

The sequence is constant if and only if $a = f(a)$, i.e.

$a = p + (a - p)a = p + a^2 - pa$

$0 = a^2 - (p + 1)a + p = (a - p)(a - 1)$

So the sequence is constant if and only if $a = p$ or $a = 1$. $\qquad \text{(shown)}$

Part (i): Period 2

The sequence has period 2 if and only if $a = f(f(a))$ but $a \neq f(a)$ (i.e. the sequence is not constant).

First, compute $f(f(a))$. Let $b = f(a) = a^2 - pa + p$. Then

$f(b) = b^2 - pb + p = (a^2 - pa + p)^2 - p(a^2 - pa + p) + p$

Setting $f(f(a)) = a$ and rearranging:

$f(f(a)) - a = 0$

Since $a = p$ and $a = 1$ are solutions of $f(a) = a$ (hence also of $f(f(a)) = a$), we can factor out $(a - p)(a - 1)$. Working through the algebra:

$f(f(a)) - a = (a - p)(a - 1)\bigl(a^2 + (1 - p)a + 1\bigr)$

So $f(f(a)) = a$ if and only if $a = p$, $a = 1$, or $a$ is a root of $a^2 + (1 - p)a + 1 = 0$.

If $a = p$ or $a = 1$, the sequence is constant, not period 2. For a genuine period-2 sequence, we need a root of the quadratic that is distinct from both $p$ and $1$.

The quadratic $a^2 + (1 - p)a + 1 = 0$ has real roots if and only if its discriminant is non-negative:

$(1 - p)^2 - 4 \geqslant 0 \qquad \Longleftrightarrow \qquad (p - 1)^2 \geqslant 4$

This gives $|p - 1| \geqslant 2$, i.e. $p \geqslant 3$ or $p \leqslant -1$.

Checking $p = 3$: The quadratic becomes $a^2 - 2a + 1 = (a - 1)^2 = 0$, so $a = 1$, which gives a constant sequence. No period-2 sequence.

Checking $p = -1$: The quadratic becomes $a^2 + 2a + 1 = (a + 1)^2 = 0$, so $a = -1 = p$, which gives a constant sequence. No period-2 sequence.

For $p > 3$ or $p < -1$: The quadratic has two distinct real roots. We verify neither root equals $p$ or $1$:

• If $a = p$ were a root: $p^2 + (1 - p)p + 1 = p^2 + p - p^2 + 1 = p + 1 = 0$, so $p = -1$. But we are in the case $p < -1$ or $p > 3$, so $p \neq -1$. (At $p = -1$, the root $a = -1 = p$ gives a constant sequence, as checked above.)

• If $a = 1$ were a root: $1 + (1 - p) + 1 = 3 - p = 0$, so $p = 3$. But we are in the case $p < -1$ or $p > 3$, so $p \neq 3$. (At $p = 3$, the root $a = 1$ gives a constant sequence, as checked above.)

Therefore, for $p > 3$ or $p < -1$, the quadratic has two distinct real roots, neither equal to $p$ or $1$. Either root gives a starting value $a$ for which the sequence has period 2.

The sequence has period 2 for some value of $a$ if and only if $p > 3$ or $p < -1$. $\qquad \text{(shown)}$

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Part (ii): No Constant Sequence iff $f(x) > x$ for all $x$

We have $f(x) = q + x^2 - px$, so $f(x) - x = x^2 - (p + 1)x + q$. This is a quadratic in $x$ with positive leading coefficient.

$(\Rightarrow)$ Suppose there is no value of $a$ for which the sequence is constant. Then $f(a) = a$ has no real solution, so $x^2 - (p+1)x + q = 0$ has no real roots. Since the leading coefficient is positive, the quadratic is strictly positive for all $x$:

$f(x) - x = x^2 - (p+1)x + q > 0 \qquad \text{for all } x$

i.e. $f(x) > x$ for all $x$.

$(\Leftarrow)$ Suppose $f(x) > x$ for all $x$. Then for any $a$, $f(a) > a$, so $f(a) \neq a$. Hence there is no constant sequence.

Therefore, there is no value of $a$ for which the sequence is constant if and only if $f(x) > x$ for all $x$. $\qquad \text{(shown)}$

Deduction: No Period-2 Sequence

Suppose there is no value of $a$ for which the sequence is constant. By the result above, $f(x) > x$ for all $x$. In particular, for any $x$:

$f(f(x)) > f(x) > x$

So $f(f(x)) > x$ for all $x$, which means $f(f(a)) = a$ has no solution. Hence there is no value of $a$ for which the sequence has period 2. $\qquad \text{(shown)}$

Final Question: Converse

Is it true that if there is no period-2 sequence, then there is no constant sequence?

No. Take $p = q = 0$, so $f(x) = x^2$. Then $f(x) = x$ gives $x^2 - x = 0$, so $x = 0$ and $x = 1$ are both constant sequences.

For period 2: $f(f(x)) = x$ gives $x^4 = x$, i.e. $x(x^3 - 1) = 0$. The real solutions are $x = 0$ and $x = 1$, both of which give $f(x) = x$ (constant sequences). There is no period-2 sequence.

So we have a case with no period-2 sequence but constant sequences do exist. $\qquad \text{(shown)}$

Examiner Notes

无官方评述。易错点：(1) “常数序列”和”周期2序列”的定义需严格区分——周期2要求非常数；(2) part(i) 中周期2的条件 f(f(x))=x 且 f(x)≠x 的分析需仔细因式分解；(3) part(ii) 最后一问是反问句，答案为”否”，需构造反例。

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Question 6

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

Problem

6 Note: You may assume that if the functions $y_1(x)$ and $y_2(x)$ both satisfy one of the differential equations in this question, then the curves $y = y_1(x)$ and $y = y_2(x)$ do not intersect.

(i) Find the solution of the differential equation

$\frac{dy}{dx} = y + x + 1$

that has the form $y = mx + c$, where $m$ and $c$ are constants.

Let $y_3(x)$ be the solution of this differential equation with $y_3(0) = k$. Show that any stationary point on the curve $y = y_3(x)$ lies on the line $y = -x - 1$. Deduce that solution curves with $k < -2$ cannot have any stationary points.

Show further that any stationary point on the solution curve is a local minimum.

Use the substitution $Y = y + x$ to solve the differential equation, and sketch, on the same axes, the solutions with $k = 0$, $k = -2$ and $k = -3$.

(ii) Find the two solutions of the differential equation

$\frac{dy}{dx} = x^2 + y^2 - 2xy - 4x + 4y + 3$

that have the form $y = mx + c$.

Let $y_4(x)$ be the solution of this differential equation with $y_4(0) = -2$. (Do not attempt to find this solution.)

Show that any stationary point on the curve $y = y_4(x)$ lies on one of two lines that you should identify. What can be said about the gradient of the curve at points between these lines?

Sketch the curve $y = y_4(x)$. You should include on your sketch the two straight line solutions and the two lines of stationary points.

Hint

(i) If $y = mx + c$, Then the differential equation becomes $m = mx + c + x + 1$ [M1] $m = -1, c = -2$ $y = -x - 2$ [A1] $\frac{dy}{dx} = 0 \Rightarrow y + x + 1 = 0 \Rightarrow y = -x - 1$ [E1 (AG)] $y = y_3(x)$ cannot cross the line $y = -x - 2$. So if $y_3(0) < -2$, it cannot reach the line $y = -x - 1$ and hence has no stationary points. [E1] At a stationary point, $\frac{d^2y}{dx^2} = \frac{dy}{dx} + 1 = y + x + 2 = 1 > 0$ so minimum [E1] $\frac{dY}{dx} = Y + 2$ [M1] $\log(Y + 2) = x + c$ [M1] $Y = -2 + Ae^x$ $y = -x - 2 + Ae^x$ [A1] $y(0) = 0 \Rightarrow A = 2$ $y(0) = -3 \Rightarrow A = -1$ (attempt at both) [M1] So $y = -x - 2 + 2e^x$ So $y = -x - 2 - e^x$ (both) Curves tending to asymptote to the left [G1] Curve above line through origin tending to $\infty$ [G1] Curve below line tending to $-\infty$ [G1] [(12 marks)] (ii) If $y = mx + c$, Then the differential equation becomes $m = (mx + c)^2 + 4(mx + c) + x^2 - 4x - 2x(mx + c) + 3$ $0 = (m^2 - 2x + 1)x^2 + (2mc + 4m - 4 - 2c) + c^2 + 4c + 3 - m$ From $x^2$: $m = 1$ From $x$: $2mc + 4m - 4 - 2c = 2c + 4 - 4 - 2c = 0$

Model Solution

Part (i): Linear Solution

Try $y = mx + c$. Then $\frac{dy}{dx} = m$, and the equation $m = (mx + c) + x + 1$ must hold for all $x$. Comparing coefficients:

• Coefficient of $x$: $0 = m + 1$, so $m = -1$.
• Constant: $m = c + 1$, so $c = m - 1 = -2$.

The linear solution is $y = -x - 2$. $\qquad \text{(shown)}$

Part (i): Stationary Points on $y = -x - 1$

At a stationary point, $\frac{dy}{dx} = 0$, so $y + x + 1 = 0$, i.e.

$y = -x - 1. \qquad \text{(shown)}$

Part (i): No Stationary Points for $k < -2$

The linear solution $y = -x - 2$ is itself a solution curve. By the given assumption, solution curves do not intersect. So any other solution curve lies entirely above or entirely below $y = -x - 2$.

If $k = y_3(0) < -2$, then at $x = 0$ the curve $y = y_3(x)$ lies below the line $y = -x - 2$. Since the curves do not intersect, $y_3(x) < -x - 2$ for all $x$. But stationary points require $y = -x - 1$, and

$-x - 2 < y_3(x) < -x - 2 \quad \text{is impossible when} \quad y_3(x) = -x - 1$

since $-x - 1 > -x - 2$. More precisely: $y_3(x) < -x - 2 < -x - 1$ for all $x$, so $y_3(x) \neq -x - 1$ for any $x$. Hence the curve has no stationary points. $\qquad \text{(shown)}$

Part (i): Stationary Points are Local Minima

At a stationary point, $\frac{dy}{dx} = 0$, so $y + x + 1 = 0$. Differentiating the ODE:

$\frac{d^2 y}{dx^2} = \frac{dy}{dx} + 1 = 0 + 1 = 1 > 0$

So every stationary point is a local minimum. $\qquad \text{(shown)}$

Part (i): Solving via Substitution $Y = y + x$

Let $Y = y + x$, so $y = Y - x$ and $\frac{dy}{dx} = \frac{dY}{dx} - 1$. Substituting:

$\frac{dY}{dx} - 1 = (Y - x) + x + 1 = Y + 1$

$\frac{dY}{dx} = Y + 2$

This is a separable ODE:

$\int \frac{dY}{Y + 2} = \int dx$

$\ln|Y + 2| = x + C$

$Y + 2 = Ae^x \qquad (A \neq 0)$

$y + x + 2 = Ae^x$

$y = -x - 2 + Ae^x$

With $y(0) = k$: $k = 0 - 2 + A$, so $A = k + 2$.

$y_3(x) = -x - 2 + (k + 2)e^x$

Sketch for $k = 0$, $k = -2$, $k = -3$:

All three curves share the asymptote $y = -x - 2$ (the linear solution) as $x \to -\infty$, since $e^x \to 0$.

• $k = 0$: $y = -x - 2 + 2e^x$. At $x = 0$: $y = -2$. For large positive $x$, the exponential dominates and $y \to +\infty$. Stationary point at $2e^x = 1$, i.e. $x = -\ln 2$, with $y = \ln 2 - 2 + \ln 2 = 2\ln 2 - 2$. This is a local minimum. The curve lies above the asymptote.

• $k = -2$: $y = -x - 2$. This is the straight line solution itself. No stationary points (since $-x - 2 \neq -x - 1$).

• $k = -3$: $y = -x - 2 - e^x$. At $x = 0$: $y = -3$. For large positive $x$, $y \to -\infty$. The curve lies below the asymptote. Since $k < -2$, there are no stationary points (as shown above). $\frac{dy}{dx} = -1 - e^x < 0$ for all $x$, so the curve is strictly decreasing.

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Part (ii): Straight-Line Solutions

First, factor the right-hand side. Let $u = x - y$:

$x^2 + y^2 - 2xy - 4x + 4y + 3 = (x - y)^2 - 4(x - y) + 3 = (u - 1)(u - 3)$

So the ODE is $\frac{dy}{dx} = (x - y - 1)(x - y - 3)$.

Try $y = x + c$ (i.e. $m = 1$). Then $\frac{dy}{dx} = 1$ and $x - y = -c$, so:

$1 = (-c - 1)(-c - 3) = (c + 1)(c + 3) = c^2 + 4c + 3$

$c^2 + 4c + 2 = 0$

$c = \frac{-4 \pm \sqrt{16 - 8}}{2} = -2 \pm \sqrt{2}$

The two straight-line solutions are

$y = x - 2 + \sqrt{2} \qquad \text{and} \qquad y = x - 2 - \sqrt{2}$

Verification: For $y = x + c$ with $x - y = -c$:

$(x-y-1)(x-y-3) = (-c-1)(-c-3) = c^2 + 4c + 3 = (c^2 + 4c + 2) + 1 = 1 = \frac{dy}{dx} \quad \checkmark$

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Part (ii): Stationary Points

Stationary points occur when $\frac{dy}{dx} = 0$:

$(x - y - 1)(x - y - 3) = 0$

$x - y = 1 \quad \text{or} \quad x - y = 3$

i.e. on the lines $y = x - 1$ and $y = x - 3$.

Gradient between the lines: When $1 < x - y < 3$:

• $(x - y - 1) > 0$ and $(x - y - 3) < 0$
• So $\frac{dy}{dx} = (x-y-1)(x-y-3) < 0$

The gradient is strictly negative between the two lines.

Gradient outside the lines:

• If $x - y < 1$: both factors are negative, so $\frac{dy}{dx} > 0$.
• If $x - y > 3$: both factors are positive, so $\frac{dy}{dx} > 0$.

Behaviour of $y_4$: At $x = 0$, $y_4(0) = -2$, so $x - y = 2$, which lies between the two lines ($1 < 2 < 3$). The gradient there is $(2 - 1)(2 - 3) = -1 < 0$.

Since solution curves do not cross, and the gradient is positive outside the strip while zero on the boundaries, the curve $y = y_4(x)$ is trapped in the region $1 < x - y < 3$ for all $x$. Within this region, $\frac{dy}{dx} < 0$ always, so the curve is strictly decreasing and has no stationary points.

As $x \to -\infty$: $x - y$ decreases toward 1, so the curve approaches $y = x - 1$ from below (the gradient tends to 0).

As $x \to +\infty$: $x - y$ increases toward 3, so the curve approaches $y = x - 3$ from above (the gradient tends to 0).

The curve passes through $(0, -2)$ with gradient $-1$, decreasing from the neighbourhood of $y = x - 1$ (upper left) toward $y = x - 3$ (lower right).

Sketch: The sketch should show:

1. The two straight-line solutions $y = x - 2 + \sqrt{2}$ and $y = x - 2 - \sqrt{2}$ (slope 1, parallel to each other, outside the strip).
2. The two lines of stationary points $y = x - 1$ and $y = x - 3$ (slope 1, also parallel, bounding the strip).
3. The curve $y = y_4(x)$ trapped between $y = x - 1$ and $y = x - 3$, passing through $(0, -2)$, approaching $y = x - 1$ as $x \to -\infty$ and $y = x - 3$ as $x \to +\infty$, always decreasing.

The ordering from top to bottom is: $y = x - 2 + \sqrt{2}$, then $y = x - 1$, then $y = x - 3$, then $y = x - 2 - \sqrt{2}$.

Examiner Notes

无官方评述。易错点：(1) part(i) 中 k<-2 时无驻点的证明需结合驻点在 y=-x-1 上且该直线与解曲线不相交的条件；(2) part(ii) 中 dy/dx 可因式分解为 (x-y-1)(x-y-3)，识别此结构是关键；(3) 定性画图时需标注直线解和驻点轨迹线。

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Question 7

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

Problem

7 (i) The points $A$, $B$ and $C$ have position vectors a, b and c, respectively. Each of these vectors is a unit vector (so $\mathbf{a} \cdot \mathbf{a} = 1$, for example) and

$\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}.$

Show that $\mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$. What can be said about the triangle $ABC$? You should justify your answer.

(ii) The four distinct points $A_i$ ($i = 1, 2, 3, 4$) have unit position vectors $\mathbf{a}_i$ and

$\sum_{i=1}^{4} \mathbf{a}_i = \mathbf{0}.$

Show that $\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4$.

(a) Given that the four points lie in a plane, determine the shape of the quadrilateral with vertices $A_1, A_2, A_3$ and $A_4$.

(b) Given instead that the four points are the vertices of a regular tetrahedron, find the length of the sides of this tetrahedron.

Hint

(i) $\boldsymbol{a} \cdot (\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c}) = 0$ [M1] $\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c} = -1$ and cyclic permutations [M1] $\boldsymbol{a} \cdot \boldsymbol{b} = -\frac{1}{2}$ legitimately obtained [A1] $\cos \theta = -\frac{1}{2}$ where $\theta$ is the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ [M1] $\theta = 120^{\circ}$ [A1] Similarly, the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$ is $120^{\circ}$. [M1] Justification of equilateral triangle by sketch or otherwise [M1] ABC is equilateral [A1] (8 marks) (ii) $\boldsymbol{a}_1 \cdot \boldsymbol{a}_2 + \boldsymbol{a}_1 \cdot \boldsymbol{a}_3 + \boldsymbol{a}_1 \cdot \boldsymbol{a}_4 = -1$ and cyclic permutations [M1] Linear combination of these equations [M1] $\boldsymbol{a}_1 \cdot \boldsymbol{a}_2 = \boldsymbol{a}_3 \cdot \boldsymbol{a}_4$ (legitimately obtained) [A1 (AG)] (3 marks) (a) [Angles $\angle A_1OA_2 = \angle A_3OA_4$] By symmetry, $\angle A_2OA_3 = \angle A_4OA_1$ The $\boldsymbol{a}_i$ are distinct and unit length so no angles are zero (accept justification by sketch) [M1] $A_1A_2A_3A_4$ is a rectangle [A1] (3 marks) (b) [$(A_1A_2)^2 = (\boldsymbol{a}_1 - \boldsymbol{a}_2)^2$] $= \boldsymbol{a}_1^2 + \boldsymbol{a}_2^2 - 2\boldsymbol{a}_1 \cdot \boldsymbol{a}_2$ [M1] $= 2 - 2\boldsymbol{a}_1 \cdot \boldsymbol{a}_2$ [M1] By symmetry, $\boldsymbol{a}_1 \cdot \boldsymbol{a}_2 = \boldsymbol{a}_1 \cdot \boldsymbol{a}_3 = \boldsymbol{a}_1 \cdot \boldsymbol{a}_4$ [M1] So $\boldsymbol{a}_1 \cdot \boldsymbol{a}_2 = -\frac{1}{3}$ [A1] So $(A_1A_2)^2 = \frac{8}{3}$ [M1] $A_1A_2 = \frac{2\sqrt{2}}{\sqrt{3}}$ [A1] (6 marks)

Model Solution

Part (i)

Since $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$, we take the dot product of both sides with $\mathbf{a}$:

$\mathbf{a} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{0} = 0$

$\mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} = 0$

Since $\mathbf{a}$ is a unit vector, $\mathbf{a} \cdot \mathbf{a} = 1$, so:

$1 + \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} = 0 \qquad \text{(... 1)}$

Similarly, taking the dot product with $\mathbf{b}$:

$\mathbf{a} \cdot \mathbf{b} + 1 + \mathbf{b} \cdot \mathbf{c} = 0 \qquad \text{(... 2)}$

And with $\mathbf{c}$:

$\mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} + 1 = 0 \qquad \text{(... 3)}$

Adding equations (1), (2), and (3):

$2(\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}) + 3 = 0$

$\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} = -\frac{3}{2} \qquad \text{(... 4)}$

Subtracting equation (1) from equation (4):

$\mathbf{b} \cdot \mathbf{c} = -\frac{3}{2} - (-1) = -\frac{1}{2}$

By symmetry of equations (1), (2), (3) (each has the same structure), all pairwise dot products are equal:

$\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{c} = -\frac{1}{2}$

Therefore $\mathbf{a} \cdot \mathbf{b} = -\dfrac{1}{2}$.

The side length $|AB|$ is:

$|AB|^2 = |\mathbf{b} - \mathbf{a}|^2 = (\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a}) = \mathbf{b} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{a} = 1 - 2\left(-\frac{1}{2}\right) + 1 = 3$

So $|AB| = \sqrt{3}$. By identical calculations:

$|BC|^2 = 1 - 2\left(-\frac{1}{2}\right) + 1 = 3 \qquad \Rightarrow \quad |BC| = \sqrt{3}$

$|CA|^2 = 1 - 2\left(-\frac{1}{2}\right) + 1 = 3 \qquad \Rightarrow \quad |CA| = \sqrt{3}$

Since $|AB| = |BC| = |CA| = \sqrt{3}$, triangle $ABC$ is equilateral.

Alternatively, $\cos(\angle AOB) = \mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$ gives $\angle AOB = 120^\circ$. Similarly $\angle BOC = \angle COA = 120^\circ$. The three points are equally spaced at equal distance from the origin, confirming the triangle is equilateral.

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Part (ii)

From $\mathbf{a}_1 + \mathbf{a}_2 + \mathbf{a}_3 + \mathbf{a}_4 = \mathbf{0}$, we take the dot product with each $\mathbf{a}_i$:

$1 + \mathbf{a}_1 \cdot \mathbf{a}_2 + \mathbf{a}_1 \cdot \mathbf{a}_3 + \mathbf{a}_1 \cdot \mathbf{a}_4 = 0 \qquad \text{(... 1)}$

$\mathbf{a}_2 \cdot \mathbf{a}_1 + 1 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_2 \cdot \mathbf{a}_4 = 0 \qquad \text{(... 2)}$

$\mathbf{a}_3 \cdot \mathbf{a}_1 + \mathbf{a}_3 \cdot \mathbf{a}_2 + 1 + \mathbf{a}_3 \cdot \mathbf{a}_4 = 0 \qquad \text{(... 3)}$

$\mathbf{a}_4 \cdot \mathbf{a}_1 + \mathbf{a}_4 \cdot \mathbf{a}_2 + \mathbf{a}_4 \cdot \mathbf{a}_3 + 1 = 0 \qquad \text{(... 4)}$

Adding (1) and (3):

$2 + \mathbf{a}_1 \cdot \mathbf{a}_2 + 2(\mathbf{a}_1 \cdot \mathbf{a}_3) + \mathbf{a}_1 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = 0 \qquad \text{(... 5)}$

Adding (2) and (4):

$2 + \mathbf{a}_1 \cdot \mathbf{a}_2 + 2(\mathbf{a}_2 \cdot \mathbf{a}_4) + \mathbf{a}_1 \cdot \mathbf{a}_4 + \mathbf{a}_2 \cdot \mathbf{a}_3 + \mathbf{a}_3 \cdot \mathbf{a}_4 = 0 \qquad \text{(... 6)}$

Subtracting (6) from (5):

$2(\mathbf{a}_1 \cdot \mathbf{a}_3) - 2(\mathbf{a}_2 \cdot \mathbf{a}_4) = 0$

$\mathbf{a}_1 \cdot \mathbf{a}_3 = \mathbf{a}_2 \cdot \mathbf{a}_4 \qquad \text{(... 7)}$

Subtracting (2) from (1):

$(\mathbf{a}_1 \cdot \mathbf{a}_3 - \mathbf{a}_2 \cdot \mathbf{a}_4) + (\mathbf{a}_1 \cdot \mathbf{a}_4 - \mathbf{a}_2 \cdot \mathbf{a}_3) = 0$

Using (7), we get $\mathbf{a}_1 \cdot \mathbf{a}_4 = \mathbf{a}_2 \cdot \mathbf{a}_3 \qquad \text{(... 8)}$.

From (1): $\mathbf{a}_1 \cdot \mathbf{a}_2 + (\mathbf{a}_1 \cdot \mathbf{a}_3) + (\mathbf{a}_1 \cdot \mathbf{a}_4) = -1$

From (3): $(\mathbf{a}_1 \cdot \mathbf{a}_3) + (\mathbf{a}_2 \cdot \mathbf{a}_3) + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1$

Using (8) to replace $\mathbf{a}_2 \cdot \mathbf{a}_3$ with $\mathbf{a}_1 \cdot \mathbf{a}_4$ in the second equation:

$(\mathbf{a}_1 \cdot \mathbf{a}_3) + (\mathbf{a}_1 \cdot \mathbf{a}_4) + \mathbf{a}_3 \cdot \mathbf{a}_4 = -1$

Comparing with the first equation, we see that $\mathbf{a}_1 \cdot \mathbf{a}_2$ and $\mathbf{a}_3 \cdot \mathbf{a}_4$ satisfy the same equation, so:

$\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4 \qquad \text{(shown)}$

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(a) Suppose the four points lie in a plane. From the result above and equation (8), let:

$\mathbf{a}_1 \cdot \mathbf{a}_2 = \mathbf{a}_3 \cdot \mathbf{a}_4 = p, \qquad \mathbf{a}_2 \cdot \mathbf{a}_3 = \mathbf{a}_1 \cdot \mathbf{a}_4 = q$

The side lengths of quadrilateral $A_1A_2A_3A_4$:

$|A_1A_2|^2 = |\mathbf{a}_2 - \mathbf{a}_1|^2 = 2 - 2p, \qquad |A_3A_4|^2 = |\mathbf{a}_4 - \mathbf{a}_3|^2 = 2 - 2p$

$|A_2A_3|^2 = |\mathbf{a}_3 - \mathbf{a}_2|^2 = 2 - 2q, \qquad |A_4A_1|^2 = |\mathbf{a}_1 - \mathbf{a}_4|^2 = 2 - 2q$

So opposite sides are equal: $|A_1A_2| = |A_3A_4|$ and $|A_2A_3| = |A_4A_1|$.

Since the four points are coplanar with the origin as centroid (their position vectors sum to zero), we can set $\mathbf{s} = \mathbf{a}_1 + \mathbf{a}_3$, so $\mathbf{a}_2 + \mathbf{a}_4 = -\mathbf{s}$. For four points on the unit circle in a plane with centroid at the origin, the configuration forces $\mathbf{a}_3 = -\mathbf{a}_1$ and $\mathbf{a}_4 = -\mathbf{a}_2$ (opposite vertices are antipodal). This means the diagonals $A_1A_3$ and $A_2A_4$ both pass through the origin and have length 2 (each is a diameter of the unit circle).

A quadrilateral whose diagonals bisect each other is a parallelogram. Since the diagonals are equal (both of length 2), the parallelogram is a rectangle.

(b) If the four points are vertices of a regular tetrahedron, by symmetry every pair of distinct vertices has the same dot product: $\mathbf{a}_i \cdot \mathbf{a}_j = \alpha$ for all $i \neq j$. There are $\binom{4}{2} = 6$ such pairs. Adding equations (1) through (4):

$4 + 2 \cdot 6\alpha = 0 \qquad \Rightarrow \qquad \alpha = -\frac{1}{3}$

The side length satisfies:

$|A_iA_j|^2 = |\mathbf{a}_j - \mathbf{a}_i|^2 = 1 - 2\left(-\frac{1}{3}\right) + 1 = \frac{8}{3}$

$|A_iA_j| = \sqrt{\frac{8}{3}} = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$

Examiner Notes

无官方评述。易错点：(1) part(i) 中 a·b=-1/2 后需进一步说明 |a-b|=√3 从而三角形等边；(2) part(ii)(a) 中平行四面体的证明需利用 a₁·a₂=a₃·a₄ 和共面条件；(3) part(ii)(b) 中正四面体边长 √(8/3) 的计算需注意 a₁·a₂=-1/3。

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Question 8

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

Problem

8 The domain of the function $f$ is the set of all $2 \times 2$ matrices and its range is the set of real numbers. Thus, if $\mathbf{M}$ is a $2 \times 2$ matrix, then $f(\mathbf{M}) \in \mathbb{R}$.

The function $f$ has the property that $f(\mathbf{MN}) = f(\mathbf{M})f(\mathbf{N})$ for any $2 \times 2$ matrices $\mathbf{M}$ and $\mathbf{N}$.

(i) You are given that there is a matrix $\mathbf{M}$ such that $f(\mathbf{M}) \neq 0$. Let $\mathbf{I}$ be the $2 \times 2$ identity matrix. By considering $f(\mathbf{MI})$, show that $f(\mathbf{I}) = 1$.

(ii) Let $\mathbf{J} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. You are given that $f(\mathbf{J}) \neq 1$. By considering $\mathbf{J}^2$, evaluate $f(\mathbf{J})$.

Using $\mathbf{J}$, show that, for any real numbers $a, b, c$ and $d$,

$f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right).$

(iii) Let $\mathbf{K} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$ where $k \in \mathbb{R}$. Use $\mathbf{K}$ to show that, if the second row of the matrix $\mathbf{A}$ is a multiple of the first row, then $f(\mathbf{A}) = 0$.

(iv) Let $\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. By considering the matrices $\mathbf{P}^2$, $\mathbf{P}^{-1}$, and $\mathbf{K}^{-1}\mathbf{PK}$ for suitable values of $k$, evaluate $f(\mathbf{P})$.

Hint

(i) $f(\mathbf{M}) = f(\mathbf{MI}) = f(\mathbf{M})f(\mathbf{I})$ [M1] $\Rightarrow f(\mathbf{I}) = 1$ since $f(\mathbf{M}) \neq 0$ [A1 (AG)] (2 marks) (ii) $f(\mathbf{J})^2 = f(\mathbf{J}^2)$ [M1] $= f(\mathbf{I}) = 1$ [M1] $\Rightarrow f(\mathbf{J}) = -1$ since $f(\mathbf{J}) \neq 1$ [A1] $f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$ [M1] $= -f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$ legitimately obtained [A1 (AG)] $f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) = f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right)$ [M1] $= -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right)$ legitimately obtained [A1 (AG)] (7 marks) (iii) Using first equality in previous part (or otherwise correctly justified) $f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right)$ [M1] $f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = 0$ [M1] $f\left(\begin{pmatrix} a & b \\ ka & kb \end{pmatrix}\right) = f\left(\mathbf{K} \begin{pmatrix} a & b \\ a & b \end{pmatrix}\right)$ [M1] $= 0$ [A1 (AG)] (4 marks) (iv) $\mathbf{K}^{-1}\mathbf{PK} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ [B1] $f(\mathbf{K})f(\mathbf{K}^{-1}) = f(\mathbf{I}) = 1 \Rightarrow f(\mathbf{K}^{-1}) = f(\mathbf{K})^{-1}$ [M1] $f(\mathbf{K}^{-1}\mathbf{PK}) = f(\mathbf{K}^{-1})f(\mathbf{PK})$ (must use two stages) [M1] $= f(\mathbf{K}^{-1})f(\mathbf{P})f(\mathbf{K})$ $= f(\mathbf{P})$ [A1 (AG)?] $\mathbf{P}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ [B1] $f(\mathbf{P}^2) = f(\mathbf{P}) \Rightarrow f(\mathbf{P}) = 0$ or $1$ [M1] $\mathbf{P}^{-1}$ exists so $f(\mathbf{P})f(\mathbf{P}^{-1}) = 1 \Rightarrow f(\mathbf{P}) \neq 0$ [A1] (7 marks)

Model Solution

Part (i)

Since there exists a matrix $\mathbf{M}$ with $f(\mathbf{M}) \neq 0$, we use the multiplicative property with $\mathbf{N} = \mathbf{I}$:

$f(\mathbf{M}) = f(\mathbf{MI}) = f(\mathbf{M})f(\mathbf{I})$

Since $f(\mathbf{M}) \neq 0$, we can divide both sides by $f(\mathbf{M})$:

$f(\mathbf{I}) = 1 \qquad \text{(shown)}$

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Part (ii)

We first evaluate $f(\mathbf{J})$. Note that $\mathbf{J}^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I}$.

Using the multiplicative property:

$f(\mathbf{J}^2) = f(\mathbf{J})f(\mathbf{J}) = f(\mathbf{J})^2$

But also $f(\mathbf{J}^2) = f(\mathbf{I}) = 1$. So $f(\mathbf{J})^2 = 1$, giving $f(\mathbf{J}) = 1$ or $f(\mathbf{J}) = -1$. Since $f(\mathbf{J}) \neq 1$, we have:

$f(\mathbf{J}) = -1$

Now we show the row-swap identities. For any matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$\mathbf{J}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} c & d \\ a & b \end{pmatrix}$

Therefore:

$f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\mathbf{J}\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = f(\mathbf{J}) \cdot f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$

This gives the first equality: $f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right)$.

For the second equality, we multiply on the right by $\mathbf{J}$:

$\begin{pmatrix} c & d \\ a & b \end{pmatrix}\mathbf{J} = \begin{pmatrix} c & d \\ a & b \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} d & c \\ b & a \end{pmatrix}$

Therefore:

$f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) = f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\mathbf{J}\right) = f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) \cdot f(\mathbf{J}) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right)$

Combining: $-f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right)$, so:

$f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) \qquad \text{(shown)}$

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Part (iii)

Suppose the second row of $\mathbf{A}$ is $k$ times the first row, so $\mathbf{A} = \begin{pmatrix} a & b \\ ka & kb \end{pmatrix}$. We can write this as:

$\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\begin{pmatrix} a & b \\ a & b \end{pmatrix} = \mathbf{K}\begin{pmatrix} a & b \\ a & b \end{pmatrix}$

We first show that $f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = 0$. From part (ii), we have:

$f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right)$

(since swapping the two identical rows of $\begin{pmatrix} a & b \\ a & b \end{pmatrix}$ gives the same matrix). Therefore:

$2f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = 0 \qquad \Rightarrow \qquad f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = 0$

Now, by the multiplicative property:

$f(\mathbf{A}) = f\left(\mathbf{K}\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = f(\mathbf{K}) \cdot f\left(\begin{pmatrix} a & b \\ a & b \end{pmatrix}\right) = f(\mathbf{K}) \cdot 0 = 0$

Therefore $f(\mathbf{A}) = 0$. $\qquad \text{(shown)}$

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Part (iv)

We evaluate $f(\mathbf{P})$ where $\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$.

Step 1: Conjugation by $\mathbf{K}$. Let $\mathbf{K} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$ with $k \neq 0$. Then $\mathbf{K}^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1/k \end{pmatrix}$ and:

$\mathbf{K}^{-1}\mathbf{P}\mathbf{K} = \begin{pmatrix} 1 & 0 \\ 0 & 1/k \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1/k \end{pmatrix}\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$

We show $f(\mathbf{K}^{-1}\mathbf{P}\mathbf{K}) = f(\mathbf{P})$. Using the multiplicative property in two stages:

$f(\mathbf{K}^{-1}\mathbf{P}\mathbf{K}) = f(\mathbf{K}^{-1}) \cdot f(\mathbf{P}\mathbf{K}) = f(\mathbf{K}^{-1}) \cdot f(\mathbf{P}) \cdot f(\mathbf{K})$

Since $\mathbf{K}\mathbf{K}^{-1} = \mathbf{I}$: $f(\mathbf{K})f(\mathbf{K}^{-1}) = f(\mathbf{I}) = 1$, so $f(\mathbf{K}^{-1}) = f(\mathbf{K})^{-1}$. Therefore:

$f(\mathbf{K}^{-1}\mathbf{P}\mathbf{K}) = f(\mathbf{K})^{-1} \cdot f(\mathbf{P}) \cdot f(\mathbf{K}) = f(\mathbf{P})$

This means $f\left(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\right) = f(\mathbf{P})$ for all $k \neq 0$. (It also holds for $k = 0$ since $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I}$ and $f(\mathbf{I}) = 1$; we verify this is consistent below.)

Step 2: Using $\mathbf{P}^2$. Setting $k = 2$:

$\mathbf{K}^{-1}\mathbf{P}\mathbf{K} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \mathbf{P}^2$

So $f(\mathbf{P}^2) = f(\mathbf{P})$. But $f(\mathbf{P}^2) = f(\mathbf{P})^2$ by the multiplicative property. Therefore:

$f(\mathbf{P})^2 = f(\mathbf{P})$

$f(\mathbf{P})\bigl(f(\mathbf{P}) - 1\bigr) = 0$

So $f(\mathbf{P}) = 0$ or $f(\mathbf{P}) = 1$.

Step 3: Eliminating $f(\mathbf{P}) = 0$. Since $\mathbf{P}$ is invertible ($\det \mathbf{P} = 1 \neq 0$), we have $\mathbf{P}^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$, and:

$f(\mathbf{P})f(\mathbf{P}^{-1}) = f(\mathbf{P}\mathbf{P}^{-1}) = f(\mathbf{I}) = 1$

This requires $f(\mathbf{P}) \neq 0$. Therefore:

$f(\mathbf{P}) = 1 \qquad \text{(shown)}$

Examiner Notes

无官方评述。易错点：(1) part(ii) 中 J²=I 推出 f(J)²=1 再由 f(J)≠1 得 f(J)=-1，此步推理需严谨；(2) part(iii) 中利用 K=diag(1,k) 让 k→0 证明行成比例时 f=0 的技巧不易想到；(3) part(iv) 中需综合使用 P²、P⁻¹ 和 K⁻¹PK 的性质来锁定 f(P)=1。