STEP2 2022 -- Pure Mathematics

STEP2 2022 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2022  |  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	微积分 (Calculus)	Standard	分部积分,函数图像分析,换元技巧,定积分比较,反证法
2	数列与级数 (Sequences & Series)	Challenging	递推关系,代数恒等变形,待定系数法,多项式函数
3	数列与级数 (Sequences & Series)	Challenging	数学归纳法,级数求和,极限估计,Fibonacci恒等式,小数展开
4	代数与函数 (Algebra & Functions)	Standard	绝对值分段讨论,分段函数转换,图像绘制,代数验证
5	几何与向量 (Geometry & Vectors)	Hard	面积法,Cauchy-Schwarz不等式,代数恒等式,约束优化,不等式证明
6	微积分 (Calculus)	Challenging	隐函数微分,微分方程求解,正交轨线,曲线族分析,斜率关系
7	复数 (Complex Numbers)	Hard	复数模的计算,反证法,多项式因式分解,Viete定理,共轭根定理
8	矩阵与线性代数 (Matrices & Linear Algebra)	Challenging	不变直线,特征向量,二次方程根的关系,角度条件,矩阵构造

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Question 1

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

Problem

1 (i) By integrating one of the two terms in the integrand by parts, or otherwise, find $\int \left( 2\sqrt{1 + x^3} + \frac{3x^3}{\sqrt{1 + x^3}} \right) \, dx \, .$

(ii) Find $\int (x^2 + 2) \frac{\sin x}{x^3} \, dx \, .$

(iii) (a) Sketch the graph with equation $y = \frac{e^x}{x}$, giving the coordinates of any stationary points.

(b) Find $a$ if $\int_{a}^{2a} \frac{e^x}{x} \, dx = \int_{a}^{2a} \frac{e^x}{x^2} \, dx \, .$

(c) Show that it is not possible to find distinct integers $m$ and $n$ such that $\int_{m}^{n} \frac{e^x}{x} \, dx = \int_{m}^{n} \frac{e^x}{x^2} \, dx \, .$

Hint

(i) [$\int \frac{3x^3}{\sqrt{1+x^3}} dx = u \cdot v - \int u'v \ dx$] [$\int \frac{3x^3}{\sqrt{1+x^3}} dx = x \cdot k\sqrt{1+x^3} - \int k\sqrt{1+x^3} \ dx$] [$\int \frac{3x^3}{\sqrt{1+x^3}} dx = x \cdot 2\sqrt{1+x^3} - \int 2\sqrt{1+x^3} \ dx$] [so $\int \left( 2\sqrt{1+x^3} + \frac{3x^3}{\sqrt{1+x^3}} \right) dx = 2x\sqrt{1+x^3} + c$] (ii) [$\frac{(x^2 + 2) \sin x}{x^3} = \frac{\sin x}{x} + \frac{2 \sin x}{x^3}$] [$\int \frac{2 \sin x}{x^3} dx = -\frac{p}{x^2} \cdot \sin x + \int \frac{q \cos x}{x^2} dx$] [$= -\frac{p}{x^2} \cdot \sin x - \frac{r}{x} \cdot \cos x - \int \frac{s \sin x}{x} dx$] [$-\frac{1}{x^2} \cdot \sin x - \frac{1}{x} \cdot \cos x - \int \frac{\sin x}{x} dx$] [so $\int \left( (x^2 + 2) \frac{\sin x}{x^3} \right) dx = -\frac{\sin x + x \cos x}{x^2} + c$] (iii) (a) [$\frac{dy}{dx} = \frac{(x-1)e^x}{x^2}$] [Therefore there is a stationary point at (1, e).] [[Graph of the function showing a minimum at (1, e) in the first quadrant and a vertical asymptote at x=0. The curve approaches negative infinity as x approaches 0 from the left and positive infinity as x approaches 0 from the right.]] [Vertical asymptote at $x = 0$] [Minimum in first quadrant and correct behaviour as $x \to \infty$] [Correct behaviour as $x \to -\infty$] (b) $\int_{a}^{2a} \frac{e^x}{x^2} dx = \left[ -\frac{p}{x} \cdot e^x \right]_{a}^{2a} + \int_{a}^{2a} \frac{qe^x}{x} dx$ [M1] $\int_{a}^{2a} \frac{e^x}{x^2} dx = \left[ -\frac{1}{x} \cdot e^x \right]_{a}^{2a} + \int_{a}^{2a} \frac{e^x}{x} dx$ [A1] Therefore for integrals to be equal we need $\left[ -\frac{1}{x} \cdot e^x \right]_{a}^{2a} = 0$ [M1] $-\frac{1}{2a} \cdot e^{2a} + \frac{1}{a} \cdot e^a = 0$ $\frac{1}{2a} \cdot e^a (-e^a + 2) = 0$ [M1] so $a = \ln 2$ [A1] [[5]] (c) As before, this means we would need $\left[ -\frac{1}{x} \cdot e^x \right]_{m}^{n}$ i.e. $\frac{e^n}{n} = \frac{e^m}{m}$ [B1] From the graph in part (iii) (a) this would mean that the smaller of $n, m$ must lie in the range $(0, 1)$. Hence this is not an integer. [E1] [[2]]

Model Solution

Part (i)

We integrate $\frac{3x^3}{\sqrt{1+x^3}}$ by parts. Write

$\int \frac{3x^3}{\sqrt{1+x^3}}\,dx = \int x \cdot \frac{3x^2}{\sqrt{1+x^3}}\,dx.$

Set $u = x$ and $dv = \frac{3x^2}{\sqrt{1+x^3}}\,dx$. Then $du = dx$ and $v = 2\sqrt{1+x^3}$ (since $\frac{d}{dx}\sqrt{1+x^3} = \frac{3x^2}{2\sqrt{1+x^3}}$). By parts:

$\int \frac{3x^3}{\sqrt{1+x^3}}\,dx = 2x\sqrt{1+x^3} - \int 2\sqrt{1+x^3}\,dx.$

Therefore

$\int\!\left(2\sqrt{1+x^3} + \frac{3x^3}{\sqrt{1+x^3}}\right)dx = \int 2\sqrt{1+x^3}\,dx + 2x\sqrt{1+x^3} - \int 2\sqrt{1+x^3}\,dx = 2x\sqrt{1+x^3} + c.$

$\boxed{\int\!\left(2\sqrt{1+x^3} + \frac{3x^3}{\sqrt{1+x^3}}\right)dx = 2x\sqrt{1+x^3} + c}$

Part (ii)

Split the integrand:

$\frac{(x^2+2)\sin x}{x^3} = \frac{\sin x}{x} + \frac{2\sin x}{x^3}.$

For $\int \frac{2\sin x}{x^3}\,dx$, integrate by parts with $u = 2\sin x$, $dv = x^{-3}\,dx$, so $du = 2\cos x\,dx$, $v = -\frac{1}{2x^2}$:

$\int \frac{2\sin x}{x^3}\,dx = -\frac{\sin x}{x^2} + \int \frac{\cos x}{x^2}\,dx.$

For $\int \frac{\cos x}{x^2}\,dx$, integrate by parts with $u = \cos x$, $dv = x^{-2}\,dx$, so $du = -\sin x\,dx$, $v = -\frac{1}{x}$:

$\int \frac{\cos x}{x^2}\,dx = -\frac{\cos x}{x} - \int \frac{\sin x}{x}\,dx.$

Combining:

$\int\!\left(\frac{\sin x}{x} + \frac{2\sin x}{x^3}\right)dx = \int \frac{\sin x}{x}\,dx - \frac{\sin x}{x^2} - \frac{\cos x}{x} - \int \frac{\sin x}{x}\,dx = -\frac{\sin x}{x^2} - \frac{\cos x}{x} + c.$

$\boxed{\int\!\frac{(x^2+2)\sin x}{x^3}\,dx = -\frac{\sin x + x\cos x}{x^2} + c}$

Part (iii)

(a) We have $y = \frac{e^x}{x}$ with domain $x \neq 0$.

Differentiating: $\frac{dy}{dx} = \frac{xe^x - e^x}{x^2} = \frac{(x-1)e^x}{x^2}$.

Setting $\frac{dy}{dx} = 0$: since $e^x > 0$ and $x^2 > 0$, we need $x = 1$, giving $y = e$.

Second derivative test: $\frac{d^2y}{dx^2} = \frac{e^x(x^2 - 2x + 2)}{x^3}$, so at $x = 1$: $\frac{d^2y}{dx^2} = e > 0$, confirming a minimum at $(1, e)$.

Behaviour:

• Vertical asymptote at $x = 0$.
• As $x \to 0^-$: $e^x \to 1$ and $x \to 0^-$, so $y \to -\infty$.
• As $x \to 0^+$: $y \to +\infty$.
• As $x \to +\infty$: $y = \frac{e^x}{x} \to +\infty$.
• As $x \to -\infty$: $e^x \to 0$ and $|x| \to \infty$, so $y \to 0^-$.

The graph has a minimum at $(1, e)$ in the first quadrant, a vertical asymptote at $x = 0$, approaches $-\infty$ from the left of the axis, and approaches $0^-$ as $x \to -\infty$.

(b) Integrate $\int \frac{e^x}{x^2}\,dx$ by parts with $u = e^x$, $dv = x^{-2}\,dx$, giving $du = e^x\,dx$, $v = -\frac{1}{x}$:

$\int \frac{e^x}{x^2}\,dx = -\frac{e^x}{x} + \int \frac{e^x}{x}\,dx.$

Evaluating the definite integrals:

$\int_a^{2a}\frac{e^x}{x^2}\,dx = \left[-\frac{e^x}{x}\right]_a^{2a} + \int_a^{2a}\frac{e^x}{x}\,dx.$

For the two integrals to be equal, we need

$\left[-\frac{e^x}{x}\right]_a^{2a} = 0 \implies -\frac{e^{2a}}{2a} + \frac{e^a}{a} = 0 \implies \frac{e^a}{2a}(-e^a + 2) = 0.$

Since $e^a > 0$ and $a \neq 0$, we need $e^a = 2$, so $a = \ln 2$.

(c) As in part (b), the condition $\int_m^n \frac{e^x}{x}\,dx = \int_m^n \frac{e^x}{x^2}\,dx$ requires

$\left[-\frac{e^x}{x}\right]_m^n = 0 \implies \frac{e^n}{n} = \frac{e^m}{m}.$

From the graph in part (a), the function $y = \frac{e^x}{x}$ has a minimum at $x = 1$ and is strictly decreasing on $(-\infty, 0)$ and on $(0, 1)$, and strictly increasing on $(1, +\infty)$. The only way to have $\frac{e^n}{n} = \frac{e^m}{m}$ with $m \neq n$ is if one value lies in $(0, 1)$ and the other lies in $(1, +\infty)$. But there is no integer in the interval $(0, 1)$, so no distinct integers $m$ and $n$ can satisfy the equation.

Examiner Notes

无官方评述。易错点：(i) 分部积分u和dv的选择，若选错方向会越做越复杂；(iii)(c) 需要结合(a)的图像性质和(b)的积分等式来论证整数m,n不存在，学生容易遗漏对整数条件的充分利用。

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Question 2

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

Problem

2 A sequence $u_n$, where $n = 1, 2, \dots$, is said to have degree d if $u_n$, as a function of $n$, is a polynomial of degree $d$.

(i) Show that, in any sequence $u_n$ ($n = 1, 2, \dots$) that satisfies $u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)$ for all $n \geqslant 1$, there is a constant difference between successive terms.

Deduce that any sequence $u_n$ for which $u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)$, for all $n \geqslant 1$, has degree at most 1.

(ii) The sequence $v_n$ ($n = 1, 2, \dots$) satisfies $v_{n+1} = \frac{1}{2}(v_{n+2} + v_n) - p$ for all $n \geqslant 1$, where $p$ is a non-zero constant. By writing $v_n = t_n + pn^2$, show that the sequence $v_n$ has degree 2.

Given that $v_1 = v_2 = 0$, find $v_n$ in terms of $n$ and $p$.

(iii) The sequence $w_n$ ($n = 1, 2, \dots$) satisfies $w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b$ for all $n \geqslant 1$, where $a$ and $b$ are constants with $a \neq 0$. Show that the sequence $w_n$ has degree 3.

Given that $w_1 = w_2 = 0$, find $w_n$ in terms of $n$, $a$ and $b$.

Hint

(i) $u_{n+2} - u_{n+1} = u_{n+1} - u_n$ [M1] so constant differences. [A1] If $u_n - u_{n-1} = d$, then $u_n = u_1 + (n - 1)d$ which is of degree at most 1 [B1]

(ii) $t_{n+1} + p(n + 1)^2$ $= \frac{1}{2}(t_{n+2} + p(n + 2)^2 + t_n + pn^2) - p$ [M1] so $t_{n+1} = \frac{1}{2}(t_{n+2} + t_n)$ [A1] so $t_n$ has degree at most 1 [A1] Hence since $p \neq 0$, $v_n$ has degree 2. [A1] Taking $v_n = pn^2 + qn + r$, gives: $p + q + r = 0$ $4p + 2q + r = 0$ [M1] so $q = -3p$ [A1] And $r = 2p$ [A1]

(iii) Substitutes $w_n = t_n + kn^3$, so [B1] $t_{n+1} + k(n + 1)^3$ $= \frac{1}{2}(t_{n+2} + k(n + 2)^3 + t_n + kn^3) - an - b$ [M1] LHS and RHS both give $kn^3 + 3kn^2$ terms [A1] $t_{n+1} = \frac{1}{2}(t_{n+2} + t_n) + (3k - a)n - (b - 3k)$ [A1] Choosing $k = \frac{1}{3}a$ [A1] gives case (ii) (with $p = b - a$) so $t_n$ has degree at most 2 and $w_n$ has degree 3, as $a \neq 0$. [A1] unless $b = a$, when case (i) applies so $t_n$ has degree at most 2 and $w_n$ has degree 3, as $a \neq 0$. [A1] Taking $w_n = \frac{1}{3}an^3 + (b - a)n^2 + qn + r$ gives: $b - \frac{2}{3}a + q + r = 0$ $-\frac{4}{3}a + 4b + 2q + r = 0$ [M1] so $q = \frac{2}{3}a - 3b$ [A1] and $r = 2b$ [A1] $w_n = \frac{1}{3}an^3 + (b - a)n^2 + (\frac{2}{3}a - 3b)n + 2b$

Model Solution

Part (i)

From $u_{n+1} = \frac{1}{2}(u_{n+2} + u_n)$, rearranging:

$2u_{n+1} = u_{n+2} + u_n \implies u_{n+2} - u_{n+1} = u_{n+1} - u_n.$

This holds for all $n \geqslant 1$, so the difference between successive terms is the same for every pair. The sequence has constant common difference $d = u_2 - u_1$.

Therefore $u_n = u_1 + (n-1)d$ for all $n \geqslant 1$, which is a polynomial of degree at most 1 in $n$. $\blacksquare$

Part (ii)

Substitute $v_n = t_n + pn^2$ into $v_{n+1} = \frac{1}{2}(v_{n+2} + v_n) - p$:

$t_{n+1} + p(n+1)^2 = \frac{1}{2}\bigl(t_{n+2} + p(n+2)^2 + t_n + pn^2\bigr) - p.$

Expand the $p$-terms on the right:

$p(n+2)^2 + pn^2 = p(n^2 + 4n + 4 + n^2) = p(2n^2 + 4n + 4),$

so $\frac{1}{2}p(2n^2 + 4n + 4) = p(n^2 + 2n + 2)$.

The left-hand side has $p(n+1)^2 = p(n^2 + 2n + 1)$.

Cancelling the $p$-terms from both sides:

$t_{n+1} + p(n^2 + 2n + 1) = t_{n+1} \text{ (from $t$ terms) } + p(n^2 + 2n + 2) + \frac{1}{2}(t_{n+2} + t_n) - p$

Wait, let me redo this carefully. The full equation is:

$t_{n+1} + p(n+1)^2 = \frac{1}{2}(t_{n+2} + t_n) + \frac{1}{2}p\bigl((n+2)^2 + n^2\bigr) - p.$

Compute the $p$-coefficient on the right:

$\frac{1}{2}\bigl((n+2)^2 + n^2\bigr)p - p = \frac{1}{2}(2n^2 + 4n + 4)p - p = (n^2 + 2n + 2)p - p = (n^2 + 2n + 1)p = p(n+1)^2.$

So the equation becomes $t_{n+1} + p(n+1)^2 = \frac{1}{2}(t_{n+2} + t_n) + p(n+1)^2$, which simplifies to

$t_{n+1} = \frac{1}{2}(t_{n+2} + t_n).$

By part (i), $t_n$ has degree at most 1. Since $p \neq 0$, the dominant term in $v_n = t_n + pn^2$ is $pn^2$, so $v_n$ has degree exactly 2.

To find $v_n$ explicitly, write $v_n = pn^2 + qn + r$.

From $v_1 = 0$: $p + q + r = 0$. $\qquad (\star)$

From $v_2 = 0$: $4p + 2q + r = 0$. $\qquad (\star\star)$

Subtracting $(\star)$ from $(\star\star)$: $3p + q = 0$, so $q = -3p$.

From $(\star)$: $r = -p - q = -p + 3p = 2p$.

$\boxed{v_n = pn^2 - 3pn + 2p = p(n-1)(n-2)}$

Part (iii)

Substitute $w_n = t_n + kn^3$ into $w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b$:

$t_{n+1} + k(n+1)^3 = \frac{1}{2}\bigl(t_{n+2} + k(n+2)^3 + t_n + kn^3\bigr) - an - b.$

We need to choose $k$ so that the cubic $p$-terms produce a remainder of the form $-p'n - q'$ (matching part (ii)‘s structure).

Expand the cubic $k$-terms on the right:

$\frac{1}{2}k\bigl((n+2)^3 + n^3\bigr) = \frac{1}{2}k(2n^3 + 6n^2 + 12n + 8) = k(n^3 + 3n^2 + 6n + 4).$

The left side has $k(n+1)^3 = k(n^3 + 3n^2 + 3n + 1)$.

Subtracting the left $k$-terms from the right $k$-terms:

$k(n^3 + 3n^2 + 6n + 4) - k(n^3 + 3n^2 + 3n + 1) = k(3n + 3) = 3k(n+1).$

So the equation becomes:

$t_{n+1} = \frac{1}{2}(t_{n+2} + t_n) + 3k(n+1) - an - b = \frac{1}{2}(t_{n+2} + t_n) + (3k - a)n + (3k - b).$

Choose $k = \frac{a}{3}$ (valid since $a \neq 0$). Then

$t_{n+1} = \frac{1}{2}(t_{n+2} + t_n) + \left(a - b\right).$

This is the form of part (ii) with $p = b - a$ (replacing $-p$ by $a - b$, i.e., $p = b - a$). By part (ii), $t_n$ has degree at most 2, so $w_n = t_n + \frac{a}{3}n^3$ has degree at most 3. Since $a \neq 0$, the $n^3$ term is nonzero, so $w_n$ has degree exactly 3.

To find $w_n$ explicitly, write $w_n = \frac{a}{3}n^3 + \alpha n^2 + \beta n + \gamma$.

From $w_1 = 0$: $\frac{a}{3} + \alpha + \beta + \gamma = 0$. $\qquad (\star)$

From $w_2 = 0$: $\frac{8a}{3} + 4\alpha + 2\beta + \gamma = 0$. $\qquad (\star\star)$

Subtracting $(\star)$ from $(\star\star)$: $\frac{7a}{3} + 3\alpha + \beta = 0$, so $\beta = -\frac{7a}{3} - 3\alpha$.

We need one more condition. Substitute into the recurrence $w_{n+1} = \frac{1}{2}(w_{n+2} + w_n) - an - b$ for a general $n$. Comparing the $n^2$ coefficient on both sides:

Left side: coefficient of $n^2$ in $w_{n+1}$ is $3 \cdot \frac{a}{3} + 2\alpha = a + 2\alpha$ (from expanding $(n+1)^3$ and $(n+1)^2$).

Actually, let me substitute $w_n = \frac{a}{3}n^3 + \alpha n^2 + \beta n + \gamma$ into the recurrence directly and match coefficients.

$w_{n+1} = \frac{a}{3}(n+1)^3 + \alpha(n+1)^2 + \beta(n+1) + \gamma$

$\frac{1}{2}(w_{n+2} + w_n) = \frac{1}{2}\left[\frac{a}{3}((n+2)^3 + n^3) + \alpha((n+2)^2 + n^2) + \beta((n+2) + n) + 2\gamma\right] - an - b$

The $n^2$ coefficients: on the left, from $\frac{a}{3} \cdot 3n^2 + \alpha \cdot n^2$: coefficient is $a + \alpha$. On the right, from $\frac{1}{2} \cdot \frac{a}{3} \cdot 6n^2 + \frac{1}{2} \cdot \alpha \cdot 2n^2$: coefficient is $a + \alpha$. These match automatically.

The $n$ coefficients: on the left, from $\frac{a}{3}(3n) + \alpha(2n) + \beta$: coefficient is $a + 2\alpha + \beta$. On the right: $\frac{1}{2}\left[\frac{a}{3}(12n) + \alpha(4n) + 2\beta n\right] - an$, which gives $2a + 2\alpha + \beta - a = a + 2\alpha + \beta$. These match.

The constant terms: on the left, $\frac{a}{3} + \alpha + \beta + \gamma$. On the right: $\frac{1}{2}\left[\frac{a}{3} \cdot 8 + 4\alpha + 2\beta + 2\gamma\right] - b = \frac{4a}{3} + 2\alpha + \beta + \gamma - b$.

Setting them equal: $\frac{a}{3} + \alpha + \beta + \gamma = \frac{4a}{3} + 2\alpha + \beta + \gamma - b$, which gives $-a - \alpha + b = 0$, so $\alpha = b - a$.

From $\beta = -\frac{7a}{3} - 3\alpha = -\frac{7a}{3} - 3(b-a) = -\frac{7a}{3} + 3a - 3b = \frac{2a}{3} - 3b$.

From $(\star)$: $\gamma = -\frac{a}{3} - \alpha - \beta = -\frac{a}{3} - (b-a) - \left(\frac{2a}{3} - 3b\right) = -\frac{a}{3} - b + a - \frac{2a}{3} + 3b = 2b$.

$\boxed{w_n = \frac{a}{3}n^3 + (b-a)n^2 + \left(\frac{2a}{3} - 3b\right)n + 2b}$

We can verify: $w_1 = \frac{a}{3} + b - a + \frac{2a}{3} - 3b + 2b = 0$ and $w_2 = \frac{8a}{3} + 4(b-a) + \frac{4a}{3} - 6b + 2b = \frac{8a+4a}{3} + 4b - 4a - 6b + 2b = 4a - 4a = 0$. $\checkmark$

Examiner Notes

无官方评述。易错点：(i) 从u_{n+1}=(u_{n+2}+u_n)/2推出等差需要正确移项；(ii) 代换v_n=t_n+pn^2后需验证t_n满足齐次递推；(iii) 三次情形的代换需要更仔细的代数运算。

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Question 3

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

Problem

3 The Fibonacci numbers are defined by $F_0 = 0$, $F_1 = 1$ and, for $n \geqslant 0$, $F_{n+2} = F_{n+1} + F_n$.

(i) Prove that $F_r \leqslant 2^{r-n}F_n$ for all $n \geqslant 1$ and all $r \geqslant n$.

(ii) Let $S_n = \sum_{r=1}^{n} \frac{F_r}{10^r}$.

Show that

$\sum_{r=1}^{n} \frac{F_{r+1}}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_r}{10^{r-1}} - \sum_{r=1}^{n} \frac{F_{r-1}}{10^{r-1}} = 89S_n - 10F_1 - F_0 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}} .$

(iii) Show that $\sum_{r=1}^{\infty} \frac{F_r}{10^r} = \frac{10}{89}$ and that $\sum_{r=7}^{\infty} \frac{F_r}{10^r} < 2 \times 10^{-6}$. Hence find, with justification, the first six digits after the decimal point in the decimal expansion of $\frac{1}{89}$.

(iv) Find, with justification, a number of the form $\frac{r}{s}$ with $r$ and $s$ both positive integers less than 10000 whose decimal expansion starts

0.0001010203050813213455 …

Hint

(i) Base case: $F_n \le 2^{n-n}F_n$ [B1] (i) For $n \ge 1$, $F_{n-1} \le F_n$, so if $r \ge n$ and $F_r \le 2^{r-n}F_n$ [M1] (i) $F_{r+1} \le 2F_r \le 2^{(r+1)-n}F_n$ [A1] (i) Logical structure correct, with conclusion. [A1] (ii) $\sum_{r=1}^n \frac{F_{r+1}}{10^{r-1}} - \sum_{r=1}^n \frac{F_r}{10^{r-1}} - \sum_{r=1}^n \frac{F_{r-1}}{10^{r-1}}$ $= 100 \sum_{r=1}^n \frac{F_{r+1}}{10^{r+1}} - 10 \sum_{r=1}^n \frac{F_r}{10^r} - \sum_{r=1}^n \frac{F_{r-1}}{10^{r-1}}$ [M1] (ii) $100 \sum_{r=2}^{n+1} \frac{F_r}{10^r} - 10 \sum_{r=1}^n \frac{F_r}{10^r} - \sum_{r=0}^{n-1} \frac{F_r}{10^r}$ [M1] (ii) $= 100(S_n + \dots) - 10S_n - (S_n + \dots)$ [A1] (ii) $= 100 \left( \dots + \frac{F_{n+1}}{10^{n+1}} - \frac{F_1}{10} \right) - \left( \dots + F_0 - \frac{F_n}{10^n} \right)$ [A1] (iii) In (ii), the left hand side is equal to zero, so $89S_n = 10F_1 + F_0 - \frac{F_n}{10^n} - \frac{F_{n+1}}{10^{n-1}}$ [M1] (iii) but $\frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}} \to 0$ as $n \to \infty$, from (i) [B1] (iii) and $F_0 = 0$, so $S_\infty = \frac{10}{89}$ [A1] (iii) $\sum_{r=7}^\infty \frac{F_r}{10^r} \le \frac{F_7}{10^7} \sum_{r=0}^\infty \frac{2^r}{10^r} = \frac{13}{10^7(1 - \frac{2}{10})} < 2 \times 10^{-6}$ [M1] (iii) $\frac{1}{89} = \frac{1}{10} \left( \frac{1}{10} + \frac{1}{10^2} + \frac{2}{10^3} + \frac{3}{10^4} + \frac{5}{10^5} + \frac{8}{10^6} + \sum_{r=7}^\infty \frac{F_r}{10^r} \right)$ $= 0.0112358 + \varepsilon$ [A1] (iii) with $\varepsilon < 2 \times 10^{-7}$, so the first six digits of the decimal expansion of $\frac{1}{89}$ are $0.011235$ [A1] (iv) Let $T_n = \sum_{r=1}^n \frac{F_r}{100^r}$ [M1] (iv) then $0 = \sum_{r=1}^n \frac{F_{r+1}}{100^{r-1}} - \sum_{r=1}^n \frac{F_r}{100^{r-1}} - \sum_{r=1}^n \frac{F_{r-1}}{100^{r-1}}$ $= 10000 \left( T_n + \frac{F_{n+1}}{100^{n+1}} - \frac{1}{100} \right) - 100T_n - \left( T_n - \frac{F_n}{100^n} \right)$ [M1] (iv) $9899T_n = 100 - \frac{F_n}{100^n} - \frac{F_{n+1}}{100^{n-1}}$ [A1] (iv) so $T_\infty = \frac{100}{9899}$ and $\frac{1}{9899}$ is the required fraction [A1] (iv) as $\frac{1}{9899} = 0.0001010203050813213455 + \varepsilon$ [M1] (iv) where $\varepsilon \le \frac{89}{1 - \frac{2}{100}} \times 10^{-24} < 10^{-22}$ [A1]

Model Solution

Part (i)

We prove by strong induction on $r$ that $F_r \leqslant 2^{r-n}F_n$ for all $r \geqslant n$, where $n \geqslant 1$ is fixed.

Base cases. For $r = n$: $F_n = 2^{n-n}F_n = F_n$. $\checkmark$

For $r = n+1$: $F_{n+1} = F_n + F_{n-1} \leqslant F_n + F_n = 2F_n = 2^{(n+1)-n}F_n$, using $F_{n-1} \leqslant F_n$ (which holds since the Fibonacci sequence is non-decreasing for $n \geqslant 1$). $\checkmark$

Inductive step. Suppose $F_k \leqslant 2^{k-n}F_n$ for all $k$ with $n \leqslant k \leqslant r$ (where $r \geqslant n+1$). Then

$F_{r+1} = F_r + F_{r-1} \leqslant 2^{r-n}F_n + 2^{(r-1)-n}F_n = 2^{r-n}F_n\left(1 + \frac{1}{2}\right) = \frac{3}{2}\cdot 2^{r-n}F_n < 2^{r+1-n}F_n.$

So $F_{r+1} \leqslant 2^{(r+1)-n}F_n$. By induction, the result holds for all $r \geqslant n$. $\blacksquare$

Part (ii)

We evaluate the left-hand side by re-indexing each sum. Set $S_n = \sum_{r=1}^{n}\frac{F_r}{10^r}$.

First sum: $\sum_{r=1}^{n}\frac{F_{r+1}}{10^{r-1}} = 100\sum_{r=1}^{n}\frac{F_{r+1}}{10^{r+1}} = 100\sum_{s=2}^{n+1}\frac{F_s}{10^s} = 100\left(S_n + \frac{F_{n+1}}{10^{n+1}} - \frac{F_1}{10}\right)$.

Second sum: $\sum_{r=1}^{n}\frac{F_r}{10^{r-1}} = 10\sum_{r=1}^{n}\frac{F_r}{10^r} = 10S_n$.

Third sum: $\sum_{r=1}^{n}\frac{F_{r-1}}{10^{r-1}} = \sum_{s=0}^{n-1}\frac{F_s}{10^s} = S_n - \frac{F_n}{10^n} + F_0 = S_n - \frac{F_n}{10^n}$.

(Here $F_0 = 0$.)

Subtracting:

$\text{LHS} = 100\left(S_n + \frac{F_{n+1}}{10^{n+1}} - \frac{F_1}{10}\right) - 10S_n - \left(S_n - \frac{F_n}{10^n}\right)$

$= 100S_n + \frac{100F_{n+1}}{10^{n+1}} - 10F_1 - 10S_n - S_n + \frac{F_n}{10^n}$

$= 89S_n - 10F_1 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}}.$

Since $F_0 = 0$, we can write $-10F_1 = -10F_1 - F_0$, confirming:

$\text{LHS} = 89S_n - 10F_1 - F_0 + \frac{F_n}{10^n} + \frac{F_{n+1}}{10^{n-1}}. \qquad \blacksquare$

Part (iii)

The left-hand side of part (ii) telescopes. Writing out the terms using $F_{r+1} - F_r - F_{r-1} = 0$ (the Fibonacci recurrence), each group of three consecutive terms cancels:

$\sum_{r=1}^{n}\left(\frac{F_{r+1}}{10^{r-1}} - \frac{F_r}{10^{r-1}} - \frac{F_{r-1}}{10^{r-1}}\right) = \sum_{r=1}^{n}\frac{F_{r+1} - F_r - F_{r-1}}{10^{r-1}} = 0.$

So from part (ii): $89S_n = 10F_1 + F_0 - \frac{F_n}{10^n} - \frac{F_{n+1}}{10^{n-1}}$.

As $n \to \infty$, from part (i) with $n = 7$ (say), $F_r \leqslant 2^{r-7}F_7 = 13 \cdot 2^{r-7}$, so $\frac{F_r}{10^r} \leqslant \frac{13}{2^7}\left(\frac{2}{10}\right)^r \to 0$. Therefore $\frac{F_n}{10^n} \to 0$ and $\frac{F_{n+1}}{10^{n-1}} = \frac{10F_{n+1}}{10^n} \to 0$.

Taking the limit: $89S_\infty = 10$, so $\boxed{\sum_{r=1}^{\infty}\frac{F_r}{10^r} = \frac{10}{89}}$.

Bounding the tail. Using $F_r \leqslant 2^{r-7}F_7 = 13 \cdot 2^{r-7}$ for $r \geqslant 7$:

$\sum_{r=7}^{\infty}\frac{F_r}{10^r} \leqslant \frac{13}{10^7}\sum_{j=0}^{\infty}\left(\frac{2}{10}\right)^j = \frac{13}{10^7} \cdot \frac{1}{1 - \frac{1}{5}} = \frac{13}{10^7} \cdot \frac{5}{4} = \frac{65}{4 \times 10^7} = 1.625 \times 10^{-6} < 2 \times 10^{-6}.$

Finding $1/89$. Since $\frac{10}{89} = \sum_{r=1}^{\infty}\frac{F_r}{10^r}$, we have $\frac{1}{89} = \frac{1}{10}\sum_{r=1}^{\infty}\frac{F_r}{10^r}$.

Computing the partial sum up to $r = 6$:

$\frac{1}{89} = \frac{1}{10}\left(\frac{1}{10} + \frac{1}{10^2} + \frac{2}{10^3} + \frac{3}{10^4} + \frac{5}{10^5} + \frac{8}{10^6} + \sum_{r=7}^{\infty}\frac{F_r}{10^r}\right)$

$= 0.011235 + \frac{1}{10}\sum_{r=7}^{\infty}\frac{F_r}{10^r}.$

The error term is $\frac{\varepsilon}{10}$ where $\varepsilon < 2 \times 10^{-6}$, so the error is less than $2 \times 10^{-7} < 5 \times 10^{-7}$.

Therefore $\frac{1}{89} = 0.011235\ldots$, and the first six digits after the decimal point are $\boxed{011235}$.

Part (iv)

Define $T_n = \sum_{r=1}^{n}\frac{F_r}{100^r}$. By the same telescoping argument as in parts (ii)—(iii), but with $10$ replaced by $100$:

$0 = \sum_{r=1}^{n}\frac{F_{r+1} - F_r - F_{r-1}}{100^{r-1}} = 10000\left(T_n + \frac{F_{n+1}}{100^{n+1}} - \frac{F_1}{100}\right) - 100T_n - \left(T_n - \frac{F_n}{100^n}\right).$

Simplifying: $(10000 - 100 - 1)T_n = 100F_1 - \frac{F_n}{100^n} - \frac{100F_{n+1}}{100^{n+1}} \cdot 100$, so

$9899T_n = 100 - \frac{F_n}{100^n} - \frac{F_{n+1}}{100^{n-1}}.$

As $n \to \infty$, the last two terms vanish (by the same argument as in part (iii)), so

$T_\infty = \frac{100}{9899}.$

Now $\sum_{r=1}^{\infty}\frac{F_r}{100^r} = \frac{F_1}{100} + \frac{F_2}{100^2} + \frac{F_3}{100^3} + \cdots = 0.01010203050813213455\ldots$

And $\frac{1}{9899} = \frac{1}{100} \cdot \frac{100}{9899} = \frac{1}{100}\sum_{r=1}^{\infty}\frac{F_r}{100^r} = 0.0001010203050813213455\ldots$

The required number is $\boxed{\dfrac{1}{9899}}$.

Verification. The error is bounded by $\frac{1}{100}\sum_{r=7}^{\infty}\frac{F_r}{100^r}$. Using $F_r \leqslant 89 \cdot 2^{r-7}$ (with $F_{14} = 377$ and noting $F_r < 89 \cdot 2^{r-7}$ for $r \geqslant 7$; actually $F_7 = 13 < 89$, so we can use $F_r \leqslant 89 \cdot 2^{r-7}$ as a loose bound):

$\sum_{r=7}^{\infty}\frac{F_r}{100^r} < \frac{89}{100^7}\cdot\frac{1}{1-\frac{2}{100}} = \frac{89}{100^7}\cdot\frac{100}{98} < 10^{-12}.$

So the error in $1/9899$ is less than $10^{-14}$, far smaller than what is needed to confirm the first 20+ digits. Hence $1/9899 = 0.0001010203050813213455\ldots$ with both $r = 1$ and $s = 9899$ being positive integers less than $10000$.

Examiner Notes

无官方评述。易错点：(ii) 级数恒等式的推导中指标偏移容易出错，需仔细处理求和上下限；(iii) 从无穷级数到小数展开的推理需要严格的误差估计；(iv) 需要逆向思考，从Fibonacci小数构造有理数。

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Question 4

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

Problem

4 (i) Show that the function f, given by the single formula $f(x) = |x| - |x - 5| + 1$, can be written without using modulus signs as

$f(x) = \begin{cases} -4 & x \le 0, \\ 2x - 4 & 0 \le x \le 5, \\ 6 & 5 \le x. \end{cases}$

Sketch the graph with equation $y = f(x)$.

(ii) The function g is given by:

$g(x) = \begin{cases} -x & x \le 0, \\ 3x & 0 \le x \le 5, \\ x + 10 & 5 \le x. \end{cases}$

Use modulus signs to write $g(x)$ as a single formula.

(iii) Sketch the graph with equation $y = h(x)$, where $h(x) = x^2 - x - 4|x| + |x(x - 5)|$.

(iv) The function k is given by:

$k(x) = \begin{cases} 10x & x \le 0, \\ 2x^2 & 0 \le x \le 5, \\ 50 & 5 \le x. \end{cases}$

Use modulus signs to write $k(x)$ as a single formula, explicitly verifying that your formula is correct.

Hint

(i) For $x \le 0$, $|x| = -x$ $|x - 5| = -(x - 5)$ For $0 \le x \le 5$ $|x| = x$ $|x - 5| = -(x - 5)$ For $5 \le x$ $|x| = x$ $|x - 5| = x - 5$ [M1] (i) For $x \le 0$, $f(x) = -x - (-(x - 5)) + 1 = -4$ For $0 \le x \le 5$ $f(x) = x - (-(x - 5)) + 1 = 2x - 4$ For $5 \le x$ $f(x) = x - (x - 5) + 1 = 6$ [A1] (i) [Graph of f(x) showing a horizontal line at y=-4 for x<=0, a line with positive gradient from (2,0) to (5,6), and a horizontal line at y=6 for x>=5.] [G1 G1] (i) [[4]] (ii) Writing $g(x) = a|x| + b|x - 5| + c$ [M1] (ii) For $x \le 0$, $g(x) = -ax + b(-(x - 5)) + c$ For $0 \le x \le 5$ $g(x) = ax + b(-(x - 5)) + c$ For $5 \le x$ $g(x) = ax + b(x - 5) + c$ [M1] (ii) Coefficients of $x$: $-a - b = -1$ $a - b = 3$ $a + b = 1$ (ii) $a = 2, b = -1$ So $c = 5$ (ii) $g(x) = 2|x| - |x - 5| + 5$ [A1] (ii) [[3]] (ii) [Graph showing two convex quadratic curves for x<=0 and x>=5, joined by a horizontal line segment for 0<=x<=5.] (ii) Convex quadratic shapes of appropriate gradient and without vertex in $(-\infty, 0], [5, \infty)$ [G1] (ii) Horizontal section in $[0, 5]$, with discontinuous gradient at endpoints. [G1] (ii) Appropriate asymmetry of quadratic parts [G1] (ii) [[5]] (iv) $k(x) = x^2 - |x(x - 5)| + \text{linear, constant terms}$ [M1] (iv) $k(x) - x^2 + |x(x - 5)|$ is: [M1] (iv) $x \le 0$: $10x - x^2 + x(x - 5) = 5x$ $0 \le x \le 5$: $2x^2 - x^2 - x(x - 5) = 5x$ $5 \le x$: $50 - x^2 + x(x - 5) = 50 - 5x$ [A1] (iv) Set equal to $a + b|x - 5|$ [M1] Determine by substitution necessary values of $a$ and $b$ [M1] $a = 25$ and $b = -5$ [A1] Verification that these are sufficient [A1] Thus $k(x) = x^2 - |x(x - 5)| + 25 - 5|x - 5|$ [A1] [[8]]

Model Solution

Part (i)

We analyse $f(x) = |x| - |x-5| + 1$ by considering three regions determined by the points where the expressions inside the modulus signs change sign: $x = 0$ and $x = 5$.

Case 1: $x \leqslant 0$. Here $|x| = -x$ and $|x-5| = -(x-5) = 5-x$, so

$f(x) = -x - (5-x) + 1 = -x - 5 + x + 1 = -4.$

Case 2: $0 \leqslant x \leqslant 5$. Here $|x| = x$ and $|x-5| = 5-x$, so

$f(x) = x - (5-x) + 1 = x - 5 + x + 1 = 2x - 4.$

Case 3: $x \geqslant 5$. Here $|x| = x$ and $|x-5| = x-5$, so

$f(x) = x - (x-5) + 1 = x - x + 5 + 1 = 6.$

Therefore

$f(x) = \begin{cases} -4 & x \leqslant 0, \\ 2x - 4 & 0 \leqslant x \leqslant 5, \\ 6 & x \geqslant 5. \end{cases}$

Sketch: The graph consists of three pieces:

• A horizontal line at $y = -4$ for $x \leqslant 0$.
• A line with gradient 2 from $(0, -4)$ to $(5, 6)$ for $0 \leqslant x \leqslant 5$.
• A horizontal line at $y = 6$ for $x \geqslant 5$.

The $x$-intercept in the middle section occurs at $2x - 4 = 0$, i.e., $x = 2$.

Part (ii)

We seek $a$, $b$, $c$ such that $g(x) = a|x| + b|x-5| + c$ matches the given piecewise definition.

For each region, expanding $a|x| + b|x-5| + c$:

• $x \leqslant 0$: $\; -ax + b(5-x) + c = -(a+b)x + (5b+c)$.
• $0 \leqslant x \leqslant 5$: $\; ax + b(5-x) + c = (a-b)x + (5b+c)$.
• $x \geqslant 5$: $\; ax + b(x-5) + c = (a+b)x + (-5b+c)$.

Matching with $g(x)$:

$\begin{cases} -(a+b) = -1 \\ a-b = 3 \\ a+b = 1 \end{cases} \qquad \text{and} \qquad \begin{cases} 5b + c = 0 \\ -5b + c = 10 \end{cases}$

From the first system: $a+b = 1$ and $a-b = 3$ give $a = 2$, $b = -1$. From the second system: adding gives $2c = 10$, so $c = 5$; check: $5(-1) + 5 = 0$. $\checkmark$

$\boxed{g(x) = 2|x| - |x-5| + 5}$

Verification:

• $x \leqslant 0$: $-2x + (5-x) + 5 = -3x + 10$… Hmm, that gives $-3x + 10$, not $-x$. Let me re-check.

Wait: with $a=2, b=-1, c=5$: for $x \leqslant 0$: $-(a+b)x + (5b+c) = -(2-1)x + (-5+5) = -x$. $\checkmark$

For $0 \leqslant x \leqslant 5$: $(a-b)x + (5b+c) = 3x + 0 = 3x$. $\checkmark$

For $x \geqslant 5$: $(a+b)x + (-5b+c) = x + 10$. $\checkmark$

Part (iii)

We sketch $h(x) = x^2 - x - 4|x| + |x(x-5)|$. Consider three regions:

Case 1: $x \leqslant 0$. Here $|x| = -x$ and $x(x-5) \geqslant 0$, so $|x(x-5)| = x^2 - 5x$.

$h(x) = x^2 - x + 4x + x^2 - 5x = 2x^2 - 2x = 2x(x-1).$

Case 2: $0 \leqslant x \leqslant 5$. Here $|x| = x$ and $x(x-5) \leqslant 0$, so $|x(x-5)| = -(x^2 - 5x) = 5x - x^2$.

$h(x) = x^2 - x - 4x + 5x - x^2 = 0.$

So $h(x) = 0$ on $[0, 5]$.

Case 3: $x \geqslant 5$. Here $|x| = x$ and $x(x-5) \geqslant 0$, so $|x(x-5)| = x^2 - 5x$.

$h(x) = x^2 - x - 4x + x^2 - 5x = 2x^2 - 10x = 2x(x-5).$

Sketch: The graph is:

• For $x \leqslant 0$: the upward parabola $2x^2 - 2x$ with vertex at $(1/2, -1/2)$; only the portion for $x \leqslant 0$ is shown, passing through $(0, 0)$.
• For $0 \leqslant x \leqslant 5$: $h(x) = 0$ (the $x$-axis).
• For $x \geqslant 5$: the upward parabola $2x^2 - 10x$ with vertex at $(5/2, -25/2)$; only the portion for $x \geqslant 5$ is shown, starting at $(5, 0)$.

The graph is continuous: $h(0) = 0$ and $h(5) = 0$ from all regions.

Part (iv)

We seek a single formula for $k(x)$ using modulus signs. Inspired by the structure, try

$k(x) = x^2 - |x(x-5)| + \alpha + \beta|x-5|$

and determine $\alpha$, $\beta$ by substitution.

Compute $x^2 - |x(x-5)|$ in each region:

• $x \leqslant 0$: $x^2 - (x^2 - 5x) = 5x$.
• $0 \leqslant x \leqslant 5$: $x^2 - (5x - x^2) = 2x^2 - 5x$.
• $x \geqslant 5$: $x^2 - (x^2 - 5x) = 5x$.

So $k(x) - x^2 + |x(x-5)| = \alpha + \beta|x-5|$ should equal:

• $x \leqslant 0$: $10x - 5x = 5x$.
• $0 \leqslant x \leqslant 5$: $2x^2 - (2x^2 - 5x) = 5x$.
• $x \geqslant 5$: $50 - 5x$.

For the first two regions: $\alpha + \beta(5-x) = 5x$ for all $x \leqslant 5$ (since $|x-5| = 5-x$).

Setting $x = 0$: $\alpha + 5\beta = 0$. Setting $x = 5$: $\alpha = 25$.

From $\alpha = 25$: $25 + 5\beta = 0$, so $\beta = -5$.

Verification for all regions:

• $x \leqslant 0$: $x^2 - |x(x-5)| + 25 - 5|x-5| = 5x + 25 - 5(5-x) = 5x + 25 - 25 + 5x = 10x$. $\checkmark$
• $0 \leqslant x \leqslant 5$: $(2x^2 - 5x) + 25 - 5(5-x) = 2x^2 - 5x + 25 - 25 + 5x = 2x^2$. $\checkmark$
• $x \geqslant 5$: $5x + 25 - 5(x-5) = 5x + 25 - 5x + 25 = 50$. $\checkmark$

$\boxed{k(x) = x^2 - |x(x-5)| + 25 - 5|x-5|}$

Examiner Notes

无官方评述。易错点：(ii) 从分段函数反推绝对值表达式时需注意分段点和系数的对应关系；(iii) h(x)中|x(x-5)|的处理需要同时考虑x=0和x=5两个分段点；(iv) 验证时需对所有区间逐一检查。

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Question 5

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

Problem

5 (i) Given that $a > b > c > 0$ are constants, and that $x, y, z$ are non-negative variables, show that $ax + by + cz \leqslant a(x + y + z).$

In the acute-angled triangle $ABC$, $a, b$ and $c$ are the lengths of sides $BC, CA$ and $AB$, respectively, with $a > b > c$. $P$ is a point inside, or on the sides of, the triangle, and $x, y$ and $z$ are the perpendicular distances from $P$ to $BC, CA$ and $AB$, respectively. The area of the triangle is $\Delta$.

(ii) (a) Find $\Delta$ in terms of $a, b, c, x, y$ and $z$.

(b) Find both the minimum value of the sum of the perpendicular distances from $P$ to the three sides of the triangle and the values of $x, y$ and $z$ which give this minimum sum, expressing your answers in terms of some or all of $a, b, c$ and $\Delta$.

(iii) (a) Show that, for all real $a, b, c, x, y$ and $z$, $(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (bx - ay)^2 + (cy - bz)^2 + (az - cx)^2 + (ax + by + cz)^2.$

(b) Find both the minimum value of the sum of the squares of the perpendicular distances from $P$ to the three sides of the triangle and the values of $x, y$ and $z$ which give this minimum sum, expressing your answers in terms of some or all of $a, b, c$ and $\Delta$.

(iv) Find both the maximum value of the sum of the squares of the perpendicular distances from $P$ to the three sides of the triangle and the values of $x, y$ and $z$ which give this maximum sum, expressing your answers in terms of some or all of $a, b, c$ and $\Delta$.

Hint

(i) As $z, y$ non-negative and $a > b, c$: $ay \ge by$ and $az \ge cz$ [B1] (i) [[1]] (ii) (a) $\Delta = \frac{1}{2}ax + \frac{1}{2}by + \frac{1}{2}cz$ [B1] (ii) (b) By (i), $(x + y + z) \ge \frac{2\Delta}{a}$ [M1] (ii) (b) $\frac{2\Delta}{a}$ is the minimum value [A1] (ii) (b) [as this lower bound is attained at] $(\frac{2\Delta}{a}, 0, 0)$. [A1] (ii) (b) [[4]] (iii) (a) Correct number of terms for expansions of any two of: $(a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$ $(bx - ay)^2 + (cy - bz)^2 + (az - cx)^2$ $(ax + by + cz)^2$ [M1] (iii) (a) Fully correct expansions. [A1] (iii) (a) Given result fully shown. [A1] (iii) (a) [[3]] (iii) (b) By (iii), $(a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$ $= (bx - ay)^2 + (cy - bz)^2 + (az - cx)^2 + (2\Delta)^2$ [M1] (iii) (b) so the minimum value of $x^2 + y^2 + z^2$ is $\frac{4\Delta^2}{a^2 + b^2 + c^2}$ [A1] (iii) (b) This occurs when $bx = ay, cy = bz$ and $az = cx$ [M1] (iii) (b) so when $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \lambda$, say, where $\lambda > 0$. [M1] (iii) (b) Then $\Delta = \frac{1}{2}a(a\lambda) + \frac{1}{2}b(b\lambda) + \frac{1}{2}c(c\lambda)$ [M1] (iii) (b) so $\lambda = \frac{2\Delta}{a^2 + b^2 + c^2}$ [A1] (iii) (b) minimum at $(a\lambda, b\lambda, c\lambda)$ with this value of $\lambda$. [A1] (iii) (b) [[7]] (iv) $(ax + by + cz)^2 \ge (cx + cy + cz)^2$ [M1] (iv) $= c^2(x + y + z)^2 \ge c^2(x^2 + y^2 + z^2)$ [M1] (iv) so $x^2 + y^2 + z^2 \le \frac{4\Delta^2}{c^2}$ [M1] (iv) Maximum of $\frac{4\Delta^2}{c^2}$ [A1] (iv) at $(0, 0, \frac{2\Delta}{c})$. [A1] (iv) [[5]]

Model Solution

Part (i)

Since $a > b$ and $y \geqslant 0$: $ay \geqslant by$.

Since $a > c$ and $z \geqslant 0$: $az \geqslant cz$.

Adding: $ax + ay + az \geqslant ax + by + cz$, i.e., $a(x+y+z) \geqslant ax + by + cz$. $\blacksquare$

Part (ii)

(a) The point $P$ with perpendicular distances $x$, $y$, $z$ to sides $BC$, $CA$, $AB$ respectively divides $\triangle ABC$ into three sub-triangles $PBC$, $PCA$, $PAB$ with areas $\frac{1}{2}ax$, $\frac{1}{2}by$, $\frac{1}{2}cz$. Therefore

$\Delta = \tfrac{1}{2}ax + \tfrac{1}{2}by + \tfrac{1}{2}cz. \qquad (\star)$

(b) From $(\star)$: $ax + by + cz = 2\Delta$. By part (i), $ax + by + cz \leqslant a(x+y+z)$, so

$x + y + z \geqslant \frac{2\Delta}{a}.$

This lower bound is achieved when the inequalities in part (i) are equalities, i.e., when $y = 0$ and $z = 0$ (since $a > b$ and $a > c$). Then $x = \frac{2\Delta}{a}$, which corresponds to the point $P$ being on side $BC$ (specifically at the foot of the altitude from $A$, since $y = z = 0$ means $P$ lies on $BC$, and $x = \frac{2\Delta}{a}$ is the altitude from $A$).

$\boxed{\text{Minimum of } x+y+z = \frac{2\Delta}{a}, \text{ achieved at } (x,y,z) = \left(\frac{2\Delta}{a}, 0, 0\right).}$

Part (iii)

(a) Expand both sides.

Left side:

$(a^2+b^2+c^2)(x^2+y^2+z^2) = a^2x^2 + a^2y^2 + a^2z^2 + b^2x^2 + b^2y^2 + b^2z^2 + c^2x^2 + c^2y^2 + c^2z^2.$

Right side: Expand each square:

$(bx-ay)^2 = b^2x^2 - 2abxy + a^2y^2$

$(cy-bz)^2 = c^2y^2 - 2bcyz + b^2z^2$

$(az-cx)^2 = a^2z^2 - 2acxz + c^2x^2$

$(ax+by+cz)^2 = a^2x^2 + b^2y^2 + c^2z^2 + 2abxy + 2bcyz + 2acxz$

Adding all four:

• Squares: $a^2x^2 + a^2y^2 + a^2z^2 + b^2x^2 + b^2y^2 + b^2z^2 + c^2x^2 + c^2y^2 + c^2z^2$
• Cross terms: $(-2abxy + 2abxy) + (-2bcyz + 2bcyz) + (-2acxz + 2acxz) = 0$

So the right side equals the left side. $\blacksquare$

(b) Using $(\star)$: $ax + by + cz = 2\Delta$. Substituting into the identity from (a):

$(a^2+b^2+c^2)(x^2+y^2+z^2) = (bx-ay)^2 + (cy-bz)^2 + (az-cx)^2 + 4\Delta^2.$

Since the three squared terms are non-negative:

$(a^2+b^2+c^2)(x^2+y^2+z^2) \geqslant 4\Delta^2,$

with equality if and only if $bx = ay$, $cy = bz$, and $az = cx$ simultaneously.

From $bx = ay$: $\frac{x}{a} = \frac{y}{b}$. From $cy = bz$: $\frac{y}{b} = \frac{z}{c}$. So $\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \lambda$ (say), giving $x = a\lambda$, $y = b\lambda$, $z = c\lambda$ with $\lambda > 0$.

Substituting into $(\star)$: $\Delta = \frac{1}{2}a(a\lambda) + \frac{1}{2}b(b\lambda) + \frac{1}{2}c(c\lambda) = \frac{\lambda}{2}(a^2+b^2+c^2)$, so $\lambda = \frac{2\Delta}{a^2+b^2+c^2}$.

$\boxed{\text{Minimum of } x^2+y^2+z^2 = \frac{4\Delta^2}{a^2+b^2+c^2}, \text{ at } (x,y,z) = \frac{2\Delta}{a^2+b^2+c^2}(a,b,c).}$

Part (iv)

We maximize $x^2 + y^2 + z^2$ subject to $ax + by + cz = 2\Delta$ with $x, y, z \geqslant 0$.

Since $c < b < a$, for any $x, y, z \geqslant 0$:

$ax + by + cz \geqslant cx + cy + cz = c(x+y+z).$

So $x + y + z \leqslant \frac{2\Delta}{c}$.

By Cauchy—Schwarz (or the AM-QM inequality): $(x+y+z)^2 \leqslant 3(x^2+y^2+z^2)$, but we can get a sharper bound. Note that for non-negative reals, $x + y + z \geqslant 0$, and since $x, y, z \geqslant 0$:

$x^2 + y^2 + z^2 \leqslant (x+y+z)^2 \leqslant \frac{4\Delta^2}{c^2}.$

The first inequality uses $2(xy+yz+zx) \geqslant 0$; the second uses $x+y+z \leqslant \frac{2\Delta}{c}$.

Equality requires $xy + yz + zx = 0$ and $x + y + z = \frac{2\Delta}{c}$. Since $x, y, z \geqslant 0$, the condition $xy + yz + zx = 0$ forces two of the three to be zero.

• If $x = y = 0$: $z = \frac{2\Delta}{c}$. Check: $ax + by + cz = c \cdot \frac{2\Delta}{c} = 2\Delta$. $\checkmark$
• If $x = z = 0$: $y = \frac{2\Delta}{c}$, but $by = \frac{2b\Delta}{c} > 2\Delta$ since $b > c$. So $ax + by + cz > 2\Delta$. Contradiction.
• If $y = z = 0$: $x = \frac{2\Delta}{c}$, but $ax = \frac{2a\Delta}{c} > 2\Delta$ since $a > c$. Contradiction.

So the unique maximum is at $(x,y,z) = (0, 0, \frac{2\Delta}{c})$, which is the vertex $C$ (opposite the shortest side $AB = c$).

$\boxed{\text{Maximum of } x^2+y^2+z^2 = \frac{4\Delta^2}{c^2}, \text{ at } (x,y,z) = \left(0, 0, \frac{2\Delta}{c}\right).}$

Examiner Notes

无官方评述。易错点：(ii)(b) 求最小距离和时需要正确利用面积约束；(iii)(a) 的恒等式展开项数多，容易遗漏交叉项；(iv) 求最大值时需考虑P在边界上的情况，约束条件的处理容易出错。

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Question 6

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

Problem

6 In this question, you should consider only points lying in the first quadrant, that is with $x > 0$ and $y > 0$.

(i) The equation $x^2 + y^2 = 2ax$ defines a family of curves in the first quadrant, one curve for each positive value of $a$. A second family of curves in the first quadrant is defined by the equation $x^2 + y^2 = 2by$, where $b > 0$.

(a) Differentiate the equation $x^2 + y^2 = 2ax$ implicitly with respect to $x$, and hence show that every curve in the first family satisfies the differential equation

$2xy \frac{dy}{dx} = y^2 - x^2.$

Find similarly a differential equation, independent of $b$, for the second family of curves.

(b) Hence, or otherwise, show that, at every point with $y \neq x$ where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.

A curve in the first family meets a curve in the second family at $(c, c)$, where $c > 0$. Find the equations of the tangents to the two curves at this point. Is it true that where a curve in the first family meets a curve in the second family on the line $y = x$, the tangents to the two curves are perpendicular?

(ii) Given the family of curves in the first quadrant $y = c \ln x$, where $c$ takes any non-zero value, find, by solving an appropriate differential equation, a second family of curves with the property that at every point where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.

(iii) A family of curves in the first quadrant is defined by the equation $y^2 = 4k(x + k)$, where $k$ takes any non-zero value.

Show that, at every point where one curve in this family meets a second curve in the family, the tangents to the two curves are perpendicular.

Hint

(i) (a) Differentiating implicitly with respect to $x$ gives $2x + 2y\frac{dy}{dx} = 2a$ so, by substitution, $x^2 + y^2 = x\left(2x + 2y\frac{dy}{dx}\right)$ [B1]

For second family: $2x + 2y\frac{dy}{dx} = 2b\frac{dy}{dx}$ [M1]

so $y\left(2x + 2y\frac{dy}{dx}\right) = (x^2 + y^2)\frac{dy}{dx}$ [A1]

$(x^2 - y^2)\frac{dy}{dx} = 2xy$ [A1]

[4]

(b) The product of the gradients at points $(x, y)$ where the curves meet is $\frac{y^2-x^2}{2xy} \times \frac{2xy}{x^2-y^2} = -1$, provided $x \neq y$, So the tangents to the curves at these points are perpendicular [B1]

At $(c, c)$, for the first family of curves: $2c^2 \frac{dy}{dx} = 0$ and so $\frac{dy}{dx} = 0$. [M1]

For the second family of curves: $2c^2 \frac{dx}{dy} = 0$ and so $\frac{dy}{dx} = \infty$. [A1]

The tangents to the circles $(x - c)^2 + y^2 = c^2$ and $(y - c)^2 + x^2 = c^2$ at this point are $y = c$ and $x = c$, which are indeed perpendicular. [A1]

[4]

(ii) First family $\frac{dy}{dx} = \frac{y}{x}$ [M1]

so $x \ln x \frac{dy}{dx} = y$ [A1]

so orthogonal family has $y \frac{dy}{dx} = -x \ln x$ [A1]

Solving differential equation by separating variables: [M1]

$\int -x \ln x \, dx = -\frac{1}{2}x^2 \ln x - \int -\frac{1}{2}x^2 \cdot \frac{1}{x} \, dx$ $= -\frac{1}{2}x^2 \ln x + \frac{1}{4}x^2$ [M1, A1]

so $\frac{1}{2}y^2 = -\frac{1}{2}x^2 \ln x + \frac{1}{4}x^2 + c$ [A1]

[7]

(iii) If two curves, with parameters $k_1, k_2$ meet, require $4k_1(x + k_1) = 4k_2(x + k_2)$ so $x = -(k_1 + k_2)$ [M1]

$y^2 = -4k_1k_2$ [A1]

for any curve, $2y\frac{dy}{dx} = 4k$ [A1]

so the gradients of the two curves satisfy $\frac{dy}{dx}|_1 \cdot \frac{dy}{dx}|_2 = \frac{2k_1}{y} \cdot \frac{2k_2}{y} = -1$ [M1, A1]

CSO

[5]

Model Solution

Part (i)

(a) For the first family $x^2 + y^2 = 2ax$: differentiating implicitly,

$2x + 2y\frac{dy}{dx} = 2a.$

From the original equation, $a = \frac{x^2+y^2}{2x}$. Substituting:

$2x + 2y\frac{dy}{dx} = \frac{x^2+y^2}{x},$

$2x^2 + 2xy\frac{dy}{dx} = x^2 + y^2,$

$\boxed{2xy\frac{dy}{dx} = y^2 - x^2.} \qquad (\star)$

For the second family $x^2 + y^2 = 2by$: differentiating implicitly,

$2x + 2y\frac{dy}{dx} = 2b\frac{dy}{dx}.$

From the original equation, $b = \frac{x^2+y^2}{2y}$. Substituting:

$2x + 2y\frac{dy}{dx} = \frac{x^2+y^2}{y}\frac{dy}{dx},$

$2xy + 2y^2\frac{dy}{dx} = (x^2+y^2)\frac{dy}{dx},$

$2xy = (x^2 - y^2)\frac{dy}{dx},$

$\boxed{(x^2-y^2)\frac{dy}{dx} = 2xy.} \qquad (\star\star)$

(b) At any point $(x,y)$ where curves from the two families meet (with $y \neq x$, so neither $\frac{dy}{dx}$ is undefined):

From $(\star)$: $\frac{dy}{dx}\bigg|_{\text{family 1}} = \frac{y^2-x^2}{2xy}$.

From $(\star\star)$: $\frac{dy}{dx}\bigg|_{\text{family 2}} = \frac{2xy}{x^2-y^2}$.

The product of the gradients:

$\frac{y^2-x^2}{2xy} \cdot \frac{2xy}{x^2-y^2} = \frac{y^2-x^2}{x^2-y^2} = -1.$

Since the product of slopes is $-1$, the tangents are perpendicular at every such intersection point. $\blacksquare$

At $(c,c)$ on the line $y = x$: For the first family, $(c,c)$ lies on $x^2+y^2 = 2ax$ when $2c^2 = 2ac$, i.e., $a = c$. So the circle is $(x-c)^2 + y^2 = c^2$. Differentiating: $2(x-c) + 2y\frac{dy}{dx} = 0$, so at $(c,c)$: $2c\frac{dy}{dx} = 0$, giving $\frac{dy}{dx} = 0$. The tangent is $y = c$.

For the second family, $(c,c)$ lies on $x^2+y^2 = 2by$ when $2c^2 = 2bc$, i.e., $b = c$. So the circle is $x^2 + (y-c)^2 = c^2$. Differentiating: $2x + 2(y-c)\frac{dy}{dx} = 0$, so at $(c,c)$: $2c = 0$. This gives $\frac{dy}{dx} = \infty$, meaning the tangent is vertical: $x = c$.

The tangents $y = c$ (horizontal) and $x = c$ (vertical) are indeed perpendicular. So the result holds even on the line $y = x$, despite the differential equation $(\star)$ giving $0/0$ at such points. The perpendicularity at points with $y = x$ must be verified directly from the geometry of the circles.

Part (ii)

For the family $y = c\ln x$: differentiating, $\frac{dy}{dx} = \frac{c}{x}$.

Eliminate $c$ using $c = \frac{y}{\ln x}$:

$\frac{dy}{dx} = \frac{y}{x\ln x}.$

The orthogonal family satisfies $\frac{dy}{dx} = -\frac{x\ln x}{y}$ (negative reciprocal), i.e.,

$y\,dy = -x\ln x\,dx.$

Integrate both sides:

$\int y\,dy = -\int x\ln x\,dx.$

Left side: $\frac{y^2}{2}$.

Right side: integrate by parts with $u = \ln x$, $dv = x\,dx$, so $du = \frac{1}{x}dx$, $v = \frac{x^2}{2}$:

$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}.$

Therefore

$\frac{y^2}{2} = -\frac{x^2}{2}\ln x + \frac{x^2}{4} + C,$

$\boxed{y^2 = -x^2\ln x + \frac{x^2}{2} + C'}$

where $C' = 2C$ is an arbitrary constant. This is the orthogonal family.

Part (iii)

Let two curves from the family $y^2 = 4k(x+k)$ with parameters $k_1$ and $k_2$ (both nonzero) meet at a point $(x,y)$.

From $y^2 = 4k_1(x+k_1)$ and $y^2 = 4k_2(x+k_2)$:

$4k_1(x+k_1) = 4k_2(x+k_2),$

$k_1 x + k_1^2 = k_2 x + k_2^2,$

$(k_1 - k_2)x = k_2^2 - k_1^2 = -(k_1-k_2)(k_1+k_2).$

Since $k_1 \neq k_2$: $x = -(k_1+k_2)$.

Then $y^2 = 4k_1(-(k_1+k_2) + k_1) = -4k_1 k_2$.

For the intersection to exist in the first quadrant ($x > 0$, $y > 0$), we need $k_1 + k_2 < 0$ and $k_1 k_2 < 0$, i.e., one parameter is positive and one negative.

Now differentiate $y^2 = 4k(x+k)$ implicitly: $2y\frac{dy}{dx} = 4k$, so $\frac{dy}{dx} = \frac{2k}{y}$.

The product of gradients at the intersection:

$\frac{dy}{dx}\bigg|_{k_1} \cdot \frac{dy}{dx}\bigg|_{k_2} = \frac{2k_1}{y} \cdot \frac{2k_2}{y} = \frac{4k_1 k_2}{y^2} = \frac{4k_1 k_2}{-4k_1 k_2} = -1.$

Since the product of slopes is $-1$, the tangents to the two curves are perpendicular at every intersection point. $\blacksquare$

Remark. This family consists of parabolas. The parameter $k$ is the distance from the vertex to the focus. When $k > 0$, the parabola opens to the left with vertex at $(-k, 0)$; when $k < 0$, it opens to the right. Two parabolas from this family always intersect at right angles.

Examiner Notes

无官方评述。易错点：(i)(a) 隐函数微分后需正确消去参数a得到微分方程；(i)(b) 在y=x上的交点处两切线斜率相同(均为1)，不垂直，需特别说明；(ii) 求解正交轨线微分方程时需注意分离变量的正确性。

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Question 7

Topic: 复数 (Complex Numbers)  |  Difficulty: Hard  |  Marks: 20

Problem

7 Let $h(z) = nz^6 + z^5 + z + n$, where $z$ is a complex number and $n \geqslant 2$ is an integer.

(i) Let $w$ be a root of the equation $h(z) = 0$.

(a) Show that $|w^5| = \sqrt{\frac{f(w)}{g(w)}}$, where

$f(z) = n^2 + 2n\text{Re}(z) + |z|^2 \text{ and } g(z) = n^2|z|^2 + 2n\text{Re}(z) + 1.$

(b) By considering $f(w) - g(w)$, prove by contradiction that $|w| \geqslant 1$.

(c) Show that $|w| = 1$.

(ii) It is given that the equation $h(z) = 0$ has six distinct roots, none of which is purely real.

(a) Show that $h(z)$ can be written in the form

$h(z) = n(z^2 - a_1z + 1)(z^2 - a_2z + 1)(z^2 - a_3z + 1),$

where $a_1$, $a_2$ and $a_3$ are real constants.

(b) Find $a_1 + a_2 + a_3$ in terms of $n$.

(c) By considering the coefficient of $z^3$ in $h(z)$, find $a_1a_2a_3$ in terms of $n$.

(d) How many of the six roots of the equation $h(z) = 0$ have a negative real part? Justify your answer.

Hint

(i) (a) $w^5 = -\frac{w+n}{nw+1}$ [M1] (i) (a) $|w^5| = \left| \frac{w+n}{nw+1} \right| = \sqrt{\left( \frac{w+n}{nw+1} \right) \left( \frac{\bar{w}+n}{n\bar{w}+1} \right)}$ [M1] (i) (a) $= \sqrt{\frac{w\bar{w} + n(w+\bar{w}) + n^2}{n^2w\bar{w} + n(w+\bar{w}) + 1}}$ [A1] (i) (a) which gives the required result, as $w + \bar{w} = 2\ Re(w)$ [A1] (i) (a) [[4]] (i) (b) $f(w) - g(w) = (n^2 - 1)(1 - |w|^2)$ [M1] (i) (b) and $n > 1$, so if $|w| < 1$, $f(w) - g(w) > 0$ [A1] (i) (b) but since $f(w)$ and $g(w)$ are both positive (each is the square of the magnitude of a complex number) $f(w) > g(w) > 0$ [A1] (i) (b) so $\frac{f(w)}{g(w)} > 1$ and so $|w| = \sqrt[10]{\frac{f(w)}{g(w)}} > 1$ # [A1] (i) (b) Hence $|w| \ge 1$ (i) (b) [[4]] (i) (c) if $|w| > 1$, $f(w) - g(w) < 0$ [M1] (i) (c) so $\frac{f(w)}{g(w)} < 1$ [A1] (i) (c) so $|w| = \sqrt[10]{\frac{f(w)}{g(w)}} < 1$ #. Hence $|w| \le 1$ [A1] (i) (c) and, in combination with (b), this gives $|w| = 1$ [A1] (i) (c) [[4]] (ii) (a) Since the coefficients of $h(z)$ are real, but none of the roots is purely real, the six roots occur in conjugate pairs [B1] (ii) (a) Suppose $p \pm iq$ are roots; then quadratic factor of $(z - p - iq)(z - p + iq) = (z^2 - 2pz + p^2 + q^2)$ with $2p$ real and $p^2 + q^2 = |z|^2 = 1$ by (i)(c) [M1] (ii) (a) Hence the algebraic factors are as stated, and the only remaining possibility is a numerical factor, which must be $n$ by comparison of the $z^6$ term [A1] (ii) (a) [[3]] (ii) (b) $a_1 + a_2 + a_3$ is the sum of all six roots, so equal to $-\frac{1}{n}$ [B1] (ii) (b) [[1]] (ii) (c) The coefficient of $z^3$ in $h$ is $-a_1a_2a_3 - 2a_1 - 2a_2 - 2a_3$ [M1] (ii) (c) which must be zero so $a_1a_2a_3 = \frac{2}{n}$ [A1] (ii) (c) [[2]] (ii) (d) The sum of $a_1, a_2, a_3$ is negative, so they cannot all be positive, but their product is positive, so exactly two of them are negative [B1] (ii) (d) hence exactly four roots of the equation have negative real part [B1] (ii) (d) [[2]]

Model Solution

Part (i)

(a) If $w$ is a root of $h(z) = 0$, then $nw^6 + w^5 + w + n = 0$, so

$w^5(nw + 1) = -(w + n),$

$w^5 = -\frac{w+n}{nw+1}.$

Taking moduli:

$|w^5| = \left|\frac{w+n}{nw+1}\right| = \sqrt{\frac{w+n}{nw+1} \cdot \frac{\bar{w}+n}{n\bar{w}+1}}.$

Numerator:

$(w+n)(\bar{w}+n) = w\bar{w} + n(w+\bar{w}) + n^2 = |w|^2 + 2n\operatorname{Re}(w) + n^2 = f(w).$

Denominator:

$(nw+1)(n\bar{w}+1) = n^2 w\bar{w} + n(w+\bar{w}) + 1 = n^2|w|^2 + 2n\operatorname{Re}(w) + 1 = g(w).$

Therefore $|w^5| = \sqrt{\frac{f(w)}{g(w)}}$, so $|w|^{10} = \frac{f(w)}{g(w)}$, i.e.,

$|w| = \left(\frac{f(w)}{g(w)}\right)^{1/10}. \qquad \blacksquare$

(b) Compute:

$f(w) - g(w) = (|w|^2 + 2n\operatorname{Re}(w) + n^2) - (n^2|w|^2 + 2n\operatorname{Re}(w) + 1)$

$= |w|^2(1-n^2) + (n^2-1) = (1-n^2)(|w|^2 - 1) = -(n^2-1)(|w|^2-1).$

Since $n \geqslant 2$, $n^2 - 1 > 0$.

Suppose for contradiction that $|w| < 1$. Then $|w|^2 - 1 < 0$, so

$f(w) - g(w) = -(n^2-1)(|w|^2-1) > 0.$

Note that $f(w) = |w+n|^2 > 0$ and $g(w) = |nw+1|^2 > 0$ (since $w+n \neq 0$ and $nw+1 \neq 0$ for any root of $h$).

From $f(w) > g(w) > 0$: $\frac{f(w)}{g(w)} > 1$, so $|w|^{10} > 1$, giving $|w| > 1$.

This contradicts $|w| < 1$. Therefore $|w| \geqslant 1$. $\blacksquare$

(c) Suppose $|w| > 1$. Then $|w|^2 - 1 > 0$, so

$f(w) - g(w) = -(n^2-1)(|w|^2-1) < 0,$

i.e., $f(w) < g(w)$. Since $g(w) > 0$:

$\frac{f(w)}{g(w)} < 1 \implies |w|^{10} < 1 \implies |w| < 1.$

This contradicts $|w| > 1$. Therefore $|w| \leqslant 1$.

Combining with part (b): $|w| \geqslant 1$ and $|w| \leqslant 1$, so $|w| = 1$. $\blacksquare$

Part (ii)

(a) Since $h(z)$ has real coefficients and none of its six roots is purely real, the roots come in conjugate pairs: $w_1, \bar{w}_1, w_2, \bar{w}_2, w_3, \bar{w}_3$.

From part (i)(c), each root has $|w_i| = 1$. Each conjugate pair contributes a quadratic factor:

$(z - w_i)(z - \bar{w}_i) = z^2 - (w_i + \bar{w}_i)z + |w_i|^2 = z^2 - 2\operatorname{Re}(w_i)z + 1.$

Set $a_i = 2\operatorname{Re}(w_i)$ (real). Then

$h(z) = n(z^2 - a_1 z + 1)(z^2 - a_2 z + 1)(z^2 - a_3 z + 1).$

The leading coefficient must be $n$ since the $z^6$ coefficient of $h(z)$ is $n$. $\blacksquare$

(b) Expanding the product, the coefficient of $z^5$ comes from choosing the $(-a_i z)$ term from exactly one factor and the $z^2$ term from the other two:

$\text{coeff of } z^5 = n(-a_1 - a_2 - a_3) = -n(a_1+a_2+a_3).$

From $h(z) = nz^6 + z^5 + z + n$, the coefficient of $z^5$ is $1$. So

$-n(a_1+a_2+a_3) = 1 \implies \boxed{a_1+a_2+a_3 = -\frac{1}{n}}.$

(c) The coefficient of $z^3$ in the expansion:

• Choose $(-a_1 z)(-a_2 z)(-a_3 z) = -a_1 a_2 a_3 z^3$ from the linear terms of all three factors.
• Choose $(-a_i z)(1)(1) = -a_i z$ from one linear term and two constant terms: this contributes $z$ not $z^3$, so does not contribute.
• Choose $(-a_i z)(-a_j z)(1) = a_i a_j z^2$ from two linear terms and one constant: this contributes $z^2$ not $z^3$.

Wait, I need to be more careful. Expanding $(z^2 - a_1 z + 1)(z^2 - a_2 z + 1)(z^2 - a_3 z + 1)$:

The $z^3$ coefficient comes from: (i) choosing $z^2$ from one factor, $(-a_j z)$ from another, $(-a_k z)$ from the third; and (ii) choosing $(-a_1 z)(-a_2 z)(-a_3 z)$ — but that gives $z^3$ only if each contributes $z^1$, but the factors are degree 2, so I need to think more carefully.

Each factor is $z^2 - a_i z + 1$. To get $z^3$ from the product of three degree-2 polynomials, the powers from the three factors must sum to 3. The possible triples are $(2,1,0)$ and its permutations, and $(1,1,1)$.

For $(2,1,0)$: choose $z^2$ from factor $i$, $(-a_j z)$ from factor $j$, and $1$ from factor $k$. This gives $-a_j z^3$. There are $3! / 1 = 6$ such terms (3 choices for which factor contributes $z^2$, 2 for which contributes $(-a_j z)$, 1 for the rest), but actually it’s: choose which factor gives $z^2$ (3 choices), which gives $(-a_j z)$ (2 remaining choices). The contribution is $-a_j$ for each such choice. Total: $-(a_1+a_2+a_3) \cdot 2$… Let me just enumerate.

$(z^2)(-a_2 z)(1) = -a_2 z^3$ $(z^2)(1)(-a_3 z) = -a_3 z^3$ $(-a_1 z)(z^2)(1) = -a_1 z^3$ $(-a_1 z)(1)(z^2) = -a_1 z^3$ $(1)(z^2)(-a_3 z) = -a_3 z^3$ $(1)(-a_2 z)(z^2) = -a_2 z^3$

Hmm, wait. Factor 1 is $z^2 - a_1 z + 1$, factor 2 is $z^2 - a_2 z + 1$, factor 3 is $z^2 - a_3 z + 1$.

From factor $i$, I can pick $z^2$, $-a_i z$, or $1$. To get total degree 3: need to pick terms whose degrees sum to 3.

Possibilities for (degree from factor 1, degree from factor 2, degree from factor 3):

• $(2, 1, 0)$: $(z^2)(-a_2 z)(1) = -a_2 z^3$
• $(2, 0, 1)$: $(z^2)(1)(-a_3 z) = -a_3 z^3$
• $(1, 2, 0)$: $(-a_1 z)(z^2)(1) = -a_1 z^3$
• $(0, 2, 1)$: $(1)(z^2)(-a_3 z) = -a_3 z^3$
• $(1, 0, 2)$: $(-a_1 z)(1)(z^2) = -a_1 z^3$
• $(0, 1, 2)$: $(1)(-a_2 z)(z^2) = -a_2 z^3$
• $(1, 1, 1)$: $(-a_1 z)(-a_2 z)(-a_3 z) = -a_1 a_2 a_3 z^3$

Total coefficient of $z^3$ (before multiplying by $n$):

$-2a_1 - 2a_2 - 2a_3 - a_1 a_2 a_3.$

From $h(z) = nz^6 + z^5 + z + n$, the coefficient of $z^3$ is $0$. So

$n(-2(a_1+a_2+a_3) - a_1 a_2 a_3) = 0.$

Since $n \geqslant 2$:

$-2(a_1+a_2+a_3) - a_1 a_2 a_3 = 0.$

Using $a_1+a_2+a_3 = -\frac{1}{n}$:

$\frac{2}{n} = a_1 a_2 a_3 \implies \boxed{a_1 a_2 a_3 = \frac{2}{n}}.$

(d) We have $a_1 + a_2 + a_3 = -\frac{1}{n} < 0$ and $a_1 a_2 a_3 = \frac{2}{n} > 0$.

Since the product is positive, either all three $a_i$ are positive, or exactly one is positive and two are negative.

But the sum is negative, so they cannot all be positive (that would give a positive sum). Therefore exactly one $a_i$ is positive and two are negative.

Each root $w$ of $z^2 - a_i z + 1 = 0$ has $\operatorname{Re}(w) = \frac{a_i}{2}$. So:

• If $a_i > 0$: both roots have positive real part.
• If $a_i < 0$: both roots have negative real part.

Since exactly two $a_i$ are negative, exactly $2 \times 2 = 4$ roots have negative real part.

$\boxed{4 \text{ roots have a negative real part.}}$

Examiner Notes

无官方评述。易错点：(i)(a) 将h(w)=0转化为|w^5|的表达式需要巧妙的代数操作；(i)(b) 用f(w)-g(w)进行反证时需注意n>=2的条件；(ii)(a) 因式分解需利用|w|=1和共轭的性质；(ii)(d) 判断负实部根的个数需结合a_i的正负分析。

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Question 8

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

Problem

8 Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be a real matrix with \neq d$. The transformation represented by$\mathbf{M}$ has exactly two distinct invariant lines through the origin.

(i) Show that, if neither invariant line is the hBcaxis, then the gradients of the invariant lines are the roots of the equation

1435583bm^2 + (a - d)m - c = 0.1435583

If one invariant line is the hBcaxis, what is the gradient of the other?

(ii) Show that, if the angle between the two invariant lines is 5^\circ$, then

1435583(a - d)^2 = (b - c)^2 - 4bc.1435583

(iii) Find a necessary and sufficient condition, on some or all of , b, c$and$, for the two invariant lines to make equal angles with the line = x$.

(iv) Give an example of a matrix which satisfies both the conditions in parts (ii) and (iii).

Hint

(i) If neither parallel to the $y$-axis, their gradients satisfy $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ m \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ m \end{pmatrix}$ [M1]

with $\lambda \neq 0$ [M1]

Eliminating $\lambda$ from $c + dm = \lambda m, a + bm = \lambda$: $c + dm = m(a + bm)$ [A1]

If $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ is invariant, then $b = 0$ [M1]

and the gradient of the other line satisfies $(a - d)m = c$ [A1]

[5]

(ii) If $b \neq 0$, and the angle $\theta$ between the lines is $45°$, then $\cos^2 \theta = \frac{1}{2}$, so using the scalar product: [B1]

$\left( \begin{pmatrix} 1 \\ m_1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ m_2 \end{pmatrix} \right)^2 = \frac{1}{2} (1 + m_1^2)(1 + m_2^2)$ [B1]

so $(1 + m_1 m_2)^2 + 4m_1 m_2 = (m_1 + m_2)^2$ [M1]

so $\left( 1 - \frac{c}{b} \right)^2 - 4\frac{c}{b} = \frac{(a-d)^2}{b^2}$ [A1]

If $b = 0$, the condition is: $\left( \begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} a-d \\ c \end{pmatrix} \right)^2 = \frac{1}{2} ((a-d)^2 + c^2)$ [M1]

so $c^2 = (a - d)^2$ as required [A1]

[6]

(iii) If $b \neq 0$, the angles with $y = x$ are equal iff $\begin{pmatrix} 1 \\ m_1 \end{pmatrix}, \begin{pmatrix} 1 \\ m_2 \end{pmatrix}$ make equal angles with $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ [B1]

so $\frac{\left( \begin{pmatrix} 1 \\ m_1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right)^2}{2(1+m_1^2)} = \frac{\left( \begin{pmatrix} 1 \\ m_2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right)^2}{2(1+m_2^2)}$ [M1]

$(1 + m_2^2)(1 + m_1)^2 = (1 + m_1^2)(1 + m_2)^2$ so $(m_1 - m_2)(1 - m_1 m_2) = 0$ [A1]

but $m_1 \neq m_2$ so requirement is $m_1 m_2 = 1$ [B1]

which is $b + c = 0$ [A1]

If $b = 0$, require $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ also invariant [M1]

so $c = 0$, which is the same condition [A1]

[7]

(iv) Require $c = -b$ and $(a - d)^2 = 8b^2$ [M1]

so e.g. $\begin{pmatrix} 2\sqrt{2} & 1 \\ -1 & 0 \end{pmatrix}$ or $\begin{pmatrix} \sqrt{2} & 1 \\ -1 & -\sqrt{2} \end{pmatrix}$ [A1]

[2]

Model Solution

Part (i)

Case: neither invariant line is the $y$-axis. A line through the origin with gradient $m$ consists of points $\lambda\binom{1}{m}$. For this line to be invariant under $\mathbf{M}$, the image $\mathbf{M}\binom{1}{m}$ must lie on the same line, i.e.,

$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ m \end{pmatrix} = \lambda\begin{pmatrix} 1 \\ m \end{pmatrix}$

for some scalar $\lambda$. This gives

$\lambda = a + bm \qquad \text{and} \qquad \lambda m = c + dm.$

Eliminating $\lambda$: $(a+bm)m = c + dm$, so $am + bm^2 = c + dm$, i.e.,

$\boxed{bm^2 + (a-d)m - c = 0.} \qquad (\star)$

Since $\mathbf{M}$ has exactly two distinct invariant lines and neither is the $y$-axis, both gradients satisfy $(\star)$, and the quadratic has two distinct real roots (requiring $b \neq 0$, since $b = 0$ would give at most one root).

Case: one invariant line is the $y$-axis. The $y$-axis has direction $\binom{0}{1}$, so invariance requires $\mathbf{M}\binom{0}{1} = \lambda\binom{0}{1}$, i.e., $\binom{b}{d} = \binom{0}{\lambda}$, so $b = 0$.

With $b = 0$, the other invariant line has gradient $m$ satisfying the linear equation $(a-d)m = c$, i.e., $m = \frac{c}{a-d}$ (well-defined since $a \neq d$; if $a = d$ with $b = 0$ and $c = 0$ we would have $\mathbf{M} = a\mathbf{I}$ with every line invariant, contradicting “exactly two”).

$\boxed{m = \frac{c}{a-d}}$

Part (ii)

We need to show: if the angle between the two invariant lines is $45^\circ$, then $(a-d)^2 = (b-c)^2 - 4bc$.

Case $b \neq 0$. Let $m_1, m_2$ be the two roots of $(\star)$. By Vieta’s formulas:

$m_1 + m_2 = -\frac{a-d}{b}, \qquad m_1 m_2 = -\frac{c}{b}.$

The angle $\theta$ between lines with gradients $m_1$ and $m_2$ satisfies

$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|.$

With $\theta = 45^\circ$: $\tan^2\theta = 1$, so

$(m_1 - m_2)^2 = (1 + m_1 m_2)^2.$

Expanding: $(m_1+m_2)^2 - 4m_1 m_2 = 1 + 2m_1 m_2 + m_1^2 m_2^2$.

Substituting Vieta’s:

$\frac{(a-d)^2}{b^2} + \frac{4c}{b} = 1 - \frac{2c}{b} + \frac{c^2}{b^2}.$

Multiply by $b^2$:

$(a-d)^2 + 4bc = b^2 - 2bc + c^2 = (b-c)^2.$

Therefore

$(a-d)^2 = (b-c)^2 - 4bc. \qquad \blacksquare$

Case $b = 0$. One invariant line is the $y$-axis (direction $\binom{0}{1}$) and the other has direction $\binom{a-d}{c}$. The angle condition using the dot product:

$\cos^2 45^\circ = \frac{\left(\binom{0}{1}\cdot\binom{a-d}{c}\right)^2}{|\binom{0}{1}|^2 \cdot |\binom{a-d}{c}|^2} = \frac{c^2}{(a-d)^2 + c^2}.$

Setting $\cos^2 45^\circ = \frac{1}{2}$:

$2c^2 = (a-d)^2 + c^2 \implies c^2 = (a-d)^2.$

Check: $(b-c)^2 - 4bc = c^2 - 0 = c^2 = (a-d)^2$. $\checkmark$

Part (iii)

We find the condition for the two invariant lines to make equal angles with $y = x$.

Case $b \neq 0$. The line $y = x$ has gradient $1$. Two lines with gradients $m_1$ and $m_2$ make equal angles with the line of gradient $1$ if and only if

$\frac{|1 - m_1|}{|1 + m_1|} = \frac{|1 - m_2|}{|1 + m_2|},$

i.e., $(1-m_1)^2(1+m_2)^2 = (1-m_2)^2(1+m_1)^2$.

Expanding: $(1 - m_1 - m_2 + m_1 m_2)^2 \cdot (\text{cross terms cancel})$. Alternatively, rearranging:

$(1+m_1^2)(1+m_2)^2 = (1+m_2^2)(1+m_1)^2$

(this comes from the squared cosine formula). Expanding and simplifying:

$(1+m_1^2)(1+2m_2+m_2^2) - (1+m_2^2)(1+2m_1+m_1^2) = 0.$

The constant and $m_1^2 m_2^2$ terms cancel. Collecting:

$2(m_2 - m_1) + (m_1^2 - m_2^2) + 2(m_1^2 m_2 - m_1 m_2^2) = 0,$

$(m_2 - m_1)\bigl[-2 + (m_1 + m_2) - 2m_1 m_2\bigr] = 0.$

Hmm, let me redo this more carefully. Alternatively, the condition that two lines make equal angles with $y = x$ is equivalent to saying $y = x$ bisects the angle between them. A standard result is that this holds iff $m_1 m_2 = 1$ (the lines are “symmetric about $y = x$”).

Proof: The angle that a line of gradient $m$ makes with $y = x$ (gradient 1) satisfies $\tan\alpha = \frac{m-1}{1+m}$. Equal angles means $\frac{m_1-1}{1+m_1} = -\frac{m_2-1}{1+m_2}$ (opposite signs), i.e., $(m_1-1)(1+m_2) = -(m_2-1)(1+m_1)$. Expanding: $m_1 + m_1 m_2 - 1 - m_2 = -(m_2 + m_1 m_2 - 1 - m_1) = -m_2 - m_1 m_2 + 1 + m_1$. So $m_1 m_2 = 1$. (We used $m_1 \neq m_2$ to get the non-trivial case.)

By Vieta’s: $m_1 m_2 = -\frac{c}{b} = 1$, so $\boxed{b + c = 0}$.

Case $b = 0$. One invariant line is the $y$-axis (gradient $\infty$), which makes angle $45^\circ$ with $y = x$. The other line has gradient $m = \frac{c}{a-d}$. For equal angles, this line must also make $45^\circ$ with $y = x$:

$\left|\frac{m-1}{1+m}\right| = 1 \implies |m-1| = |1+m| \implies m = 0.$

So $c = 0$, which again gives $b + c = 0$.

Summary: The necessary and sufficient condition is $\boxed{b + c = 0}$.

Part (iv)

We need both conditions:

1. $(a-d)^2 = (b-c)^2 - 4bc$ (from part (ii))
2. $b + c = 0$ (from part (iii))

Substituting $c = -b$ into condition 1:

$(a-d)^2 = (b-(-b))^2 - 4b(-b) = 4b^2 + 4b^2 = 8b^2.$

Choose $b = 1$ (so $c = -1$) and $a - d = 2\sqrt{2}$. Setting $d = 0$ gives $a = 2\sqrt{2}$:

$\boxed{\mathbf{M} = \begin{pmatrix} 2\sqrt{2} & 1 \\ -1 & 0 \end{pmatrix}}$

Verification. The invariant lines satisfy $m^2 + 2\sqrt{2}\,m + 1 = 0$, giving $m = -\sqrt{2} \pm 1$.

• Product: $m_1 m_2 = (-\sqrt{2}+1)(-\sqrt{2}-1) = 2-1 = 1$. $\checkmark$ (equal angles with $y = x$)
• $m_1 - m_2 = 2$, $1 + m_1 m_2 = 2$, so $\left|\frac{m_1-m_2}{1+m_1 m_2}\right| = 1$. $\checkmark$ (angle between them is $45^\circ$)

Alternative example: $\mathbf{M} = \begin{pmatrix} \sqrt{2} & 1 \\ -1 & -\sqrt{2} \end{pmatrix}$ also works, with $a - d = 2\sqrt{2}$ and the same verification.

Examiner Notes

无官方评述。易错点：(i) 需要正确推导bm^2+(a-d)m-c=0，同时处理y轴作为特殊情况；(ii) 夹角公式的使用需注意正负号；(iii) 关于y=x对称的条件是m1*m2=1，结合根与系数关系可得充要条件。