STEP2 2020 -- Pure Mathematics

Exam: STEP2  |  Year: 2020  |  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)	Challenging	积分代换, 反常积分判敛, 代数化简, 部分分式
2	微分方程 (Differential Equations)	Challenging	分离变量法, 隐函数方程, 曲线绘制, 对称性分析
3	数列与级数 (Sequences & Series)	Hard	数学归纳法, 递推关系求解, 不等式证明, 单峰性分析, 代数化简
4	几何与向量 (Geometry & Vectors)	Challenging	三角不等式, 反证法, 函数单调性, 分情况讨论, 比例不等式
5	数论 (Number Theory)	Challenging	同余运算, 数字和性质, 不等式放缩, 穷举验证, 数位分析
6	矩阵与线性代数 (Matrices & Linear Algebra)	Challenging	矩阵迹的性质, 行列式计算, 特征多项式, 充要条件证明, 矩阵构造
7	复数 (Complex Numbers)	Challenging	复数模运算, Möbius变换, 直线到圆映射, 参数化分析, 区域判断
8	微积分 (Calculus)	Hard	多项式因式分解, 定积分计算, 面积相等条件, 对称性分析, 拐点判定

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Question 1

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

Problem

1 (i) Use the substitution $x = \frac{1}{1 - u}$, where $0 < u < 1$, to find in terms of $x$ the integral

$\int \frac{1}{x^{\frac{3}{2}}(x - 1)^{\frac{1}{2}}} \, dx \quad (\text{where } x > 1).$

(ii) Find in terms of $x$ the integral

$\int \frac{1}{(x - 2)^{\frac{3}{2}}(x + 1)^{\frac{1}{2}}} \, dx \quad (\text{where } x > 2).$

(iii) Show that

$\int_{2}^{\infty} \frac{1}{(x - 1)(x - 2)^{\frac{1}{2}}(3x - 2)^{\frac{1}{2}}} \, dx = \frac{1}{3}\pi.$

Hint

Only penalise missing +c once in parts (i) and (ii)

1(i) $\int \frac{1}{x^{\frac{3}{2}}(x-1)^{\frac{1}{2}}} dx = \int \frac{(1-u)^2}{u^{\frac{1}{2}}} \frac{1}{(1-u)^2} du$

Must include attempt at $\frac{du}{dx}$ (or $\frac{dx}{du}$)

$= 2u^{\frac{1}{2}}$

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

1(ii) Let $x - 2 = s$

Then $\int \frac{1}{(x-2)^{\frac{3}{2}}(x+1)^{\frac{1}{2}}} dx = \int \frac{1}{s^{\frac{3}{2}}(s+3)^{\frac{1}{2}}} ds$

Let $s = \frac{3}{u-1}$

$\int \frac{1}{s^{\frac{3}{2}}(s+3)^{\frac{1}{2}}} ds = \int \frac{(u-1)^2}{3^2 u^{\frac{1}{2}}} \frac{-3}{(u-1)^2} du = -\frac{2}{3}u^{\frac{1}{2}}$

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

1(iii) Let $x = \frac{1+u}{u}$

Allow substitution leading to two algebraic factors in the denominator.

$\int_{2}^{\infty} \frac{1}{(x-1)(x-2)^{\frac{1}{2}}(3x-2)^{\frac{1}{2}}} dx = \int_{1}^{0} \frac{u^2}{(1-u)^{\frac{1}{2}}(3+u)^{\frac{1}{2}}} \cdot \left(\frac{-1}{u^2}\right) du$

If done through a sequence of substitutions:

a further substitution leading to a square root of a quadratic as the denominator.

$= \int_{0}^{1} \frac{1}{(3-2u-u^2)^{\frac{1}{2}}} du$

Complete the square: $3 - 2u - u^2 = 4 - (u+1)^2$

$= \int_{0}^{1} \frac{1}{\sqrt{4 - (u+1)^2}} du$

$= \left[\arcsin\frac{u+1}{2}\right]_{0}^{1} = \arcsin 1 - \arcsin\frac{1}{2} = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$

Model Solution

Part (i)

Let $x = \dfrac{1}{1-u}$, so that $1-u = \dfrac{1}{x}$ and $u = \dfrac{x-1}{x}$.

Differentiating: $\dfrac{dx}{du} = \dfrac{1}{(1-u)^2}$, so $dx = \dfrac{1}{(1-u)^2}\,du$.

We express the integrand in terms of $u$:

$x^{\frac{3}{2}} = \frac{1}{(1-u)^{\frac{3}{2}}}, \qquad (x-1)^{\frac{1}{2}} = \left(\frac{u}{1-u}\right)^{\frac{1}{2}} = \frac{u^{\frac{1}{2}}}{(1-u)^{\frac{1}{2}}}$

So

$\frac{1}{x^{\frac{3}{2}}(x-1)^{\frac{1}{2}}} = \frac{(1-u)^{\frac{3}{2}}(1-u)^{\frac{1}{2}}}{u^{\frac{1}{2}}} = \frac{(1-u)^2}{u^{\frac{1}{2}}}$

Therefore the integral becomes

$\int \frac{(1-u)^2}{u^{\frac{1}{2}}} \cdot \frac{1}{(1-u)^2}\,du = \int u^{-\frac{1}{2}}\,du = 2u^{\frac{1}{2}} + c$

Substituting back $u = \dfrac{x-1}{x}$:

$\int \frac{1}{x^{\frac{3}{2}}(x-1)^{\frac{1}{2}}}\,dx = 2\left(\frac{x-1}{x}\right)^{\frac{1}{2}} + c \qquad \text{(Q1(i))}$

Part (ii)

First substitute $u = x - 2$, so $du = dx$ and $x + 1 = u + 3$:

$\int \frac{1}{(x-2)^{\frac{3}{2}}(x+1)^{\frac{1}{2}}}\,dx = \int \frac{1}{u^{\frac{3}{2}}(u+3)^{\frac{1}{2}}}\,du$

Now let $u = \dfrac{3}{v-1}$ (motivated by the substitution from part (i)), so that $v - 1 = \dfrac{3}{u}$ and $v = \dfrac{u+3}{u}$.

Then $\dfrac{du}{dv} = \dfrac{-3}{(v-1)^2}$, so $du = \dfrac{-3}{(v-1)^2}\,dv$.

We compute each factor:

$u^{\frac{3}{2}} = \frac{3^{\frac{3}{2}}}{(v-1)^{\frac{3}{2}}}, \qquad u + 3 = \frac{3}{v-1} + 3 = \frac{3v}{v-1}$

$(u+3)^{\frac{1}{2}} = \frac{(3v)^{\frac{1}{2}}}{(v-1)^{\frac{1}{2}}}$

So

$\frac{1}{u^{\frac{3}{2}}(u+3)^{\frac{1}{2}}} = \frac{(v-1)^{\frac{3}{2}}(v-1)^{\frac{1}{2}}}{3^{\frac{3}{2}}(3v)^{\frac{1}{2}}} = \frac{(v-1)^2}{3^2 v^{\frac{1}{2}}}$

The integral becomes

$\int \frac{(v-1)^2}{9v^{\frac{1}{2}}} \cdot \frac{-3}{(v-1)^2}\,dv = -\frac{1}{3}\int v^{-\frac{1}{2}}\,dv = -\frac{2}{3}v^{\frac{1}{2}} + c$

Substituting back $v = \dfrac{u+3}{u} = \dfrac{x+1}{x-2}$:

$\int \frac{1}{(x-2)^{\frac{3}{2}}(x+1)^{\frac{1}{2}}}\,dx = -\frac{2}{3}\left(\frac{x+1}{x-2}\right)^{\frac{1}{2}} + c \qquad \text{(Q1(ii))}$

Part (iii)

We evaluate $\displaystyle\int_{2}^{\infty} \frac{1}{(x-1)(x-2)^{\frac{1}{2}}(3x-2)^{\frac{1}{2}}}\,dx$.

Let $x = \dfrac{1+u}{u}$, so $u = \dfrac{1}{x-1}$. Then $dx = -\dfrac{1}{u^2}\,du$.

When $x = 2$: $u = 1$. When $x \to \infty$: $u \to 0^+$.

We express each factor in the denominator:

$x - 1 = \frac{1}{u}, \qquad x - 2 = \frac{1-u}{u}, \qquad 3x - 2 = \frac{3+u}{u}$

So

$(x-2)^{\frac{1}{2}}(3x-2)^{\frac{1}{2}} = \frac{\sqrt{(1-u)(3+u)}}{u}$

The full denominator is

$(x-1)(x-2)^{\frac{1}{2}}(3x-2)^{\frac{1}{2}} = \frac{1}{u} \cdot \frac{\sqrt{(1-u)(3+u)}}{u} = \frac{\sqrt{(1-u)(3+u)}}{u^2}$

Therefore

$\int_{2}^{\infty} \frac{dx}{(x-1)(x-2)^{\frac{1}{2}}(3x-2)^{\frac{1}{2}}} = \int_{1}^{0} \frac{u^2}{\sqrt{(1-u)(3+u)}} \cdot \left(-\frac{1}{u^2}\right)du = \int_{0}^{1} \frac{1}{\sqrt{(1-u)(3+u)}}\,du$

Now expand the expression under the square root and complete the square:

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

So the integral becomes

$\int_{0}^{1} \frac{1}{\sqrt{4 - (u+1)^2}}\,du$

Substituting $t = u + 1$, $dt = du$: when $u = 0$, $t = 1$; when $u = 1$, $t = 2$.

$\int_{1}^{2} \frac{1}{\sqrt{4 - t^2}}\,dt = \left[\arcsin\frac{t}{2}\right]_{1}^{2} = \arcsin 1 - \arcsin\frac{1}{2} = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \qquad \text{(Q1(iii))}$

Examiner Notes

无官方评述。易错点：代换后积分限的变化处理、被积函数化简时的符号错误、反常积分收敛性的验证。第(iii)部分需要从前两部分的结果巧妙组合。

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Question 2

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

Problem

2 The curves $C_1$ and $C_2$ both satisfy the differential equation

$\frac{dy}{dx} = \frac{kxy - y}{x - kxy},$

where $k = \ln 2$.

All points on $C_1$ have positive $x$ and $y$ co-ordinates and $C_1$ passes through $(1, 1)$. All points on $C_2$ have negative $x$ and $y$ co-ordinates and $C_2$ passes through $(-1, -1)$.

(i) Show that the equation of $C_1$ can be written as $(x - y)^2 = (x + y)^2 - 2^{x+y}$.

Determine a similar result for curve $C_2$.

Hence show that $y = x$ is a line of symmetry of each curve.

(ii) Sketch on the same axes the curves $y = x^2$ and $y = 2^x$, for $x \geqslant 0$. Hence show that $C_1$ lies between the lines $x + y = 2$ and $x + y = 4$.

Sketch curve $C_1$.

(iii) Sketch curve $C_2$.

Hint

2(i)

Separate variables: $\frac{1 - ky}{y} \frac{dy}{dx} = \frac{kx - 1}{x}$

$\ln y - ky = kx - \ln x + c$, hence $\ln xy = k(x + y) + c$

$xy = \frac{1}{4}\left[(x + y)^2 - (x - y)^2\right] = Ae^{k(x+y)}$

$C_1$: $(x - y)^2 = (x + y)^2 - 2^{x+y}$; $C_2$: $(x - y)^2 = (x + y)^2 - 2^{x+y+4}$

In both cases the equation is invariant under $(x, y) \mapsto (y, x)$, so symmetric in $y = x$.

2(ii)

Graphs: correct shapes of $y = x^2$ and $y = 2^x$, intersections at $(2, 4)$ and $(4, 16)$.

$(x - y)^2 \geq 0$ implies $(x + y)^2 > 2^{x+y}$, so $x + y$ must lie between 2 and 4.

Graph: symmetry about $y = x$; closed curve between $x + y = 2$ and $x + y = 4$; passes through $(1, 1)$ and $(2, 2)$.

2(iii)

Sketches of $y = x^2$ and $y = 2^{x+4}$; $x^2 > 2^{x+4}$ only when $x < -2$.

Graph: symmetry about $y = x$; passes through $(-1, -1)$; $y \to 0$ as $x \to \infty$, $y \to -\infty$ as $x \to 0$.

Model Solution

Part (i)

The differential equation is

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

Separating variables:

$\frac{1 - ky}{y}\,dy = \frac{kx - 1}{x}\,dx \qquad \Longrightarrow \qquad \left(\frac{1}{y} - k\right)dy = \left(k - \frac{1}{x}\right)dx$

Integrating both sides:

$\ln|y| - ky = kx - \ln|x| + c$

Rearranging:

$\ln|x| + \ln|y| = k(x + y) + c \qquad \Longrightarrow \qquad \ln|xy| = k(x + y) + c$

So $|xy| = e^c \cdot e^{k(x+y)}$. Writing $A = e^c > 0$:

$|xy| = A \cdot e^{k(x+y)} = A \cdot 2^{x+y}$

Curve $C_1$ (positive $x, y$): Here $xy = |xy|$, so $xy = A \cdot 2^{x+y}$.

Substituting $(1, 1)$: $1 = A \cdot 2^2$, so $A = \dfrac{1}{4}$. Hence

$xy = \frac{1}{4} \cdot 2^{x+y} = 2^{x+y-2}$

Using the identity $(x + y)^2 = (x - y)^2 + 4xy$:

$(x - y)^2 = (x + y)^2 - 4 \cdot 2^{x+y-2} = (x + y)^2 - 2^{x+y}$

This is the equation of $C_1$.

Curve $C_2$ (negative $x, y$): Here $xy > 0$, so again $xy = |xy|$, giving $xy = A \cdot 2^{x+y}$.

Substituting $(-1, -1)$: $1 = A \cdot 2^{-2}$, so $A = 4$. Hence

$xy = 4 \cdot 2^{x+y} = 2^{x+y+2}$

$(x - y)^2 = (x + y)^2 - 4 \cdot 2^{x+y+2} = (x + y)^2 - 2^{x+y+4}$

This is the equation of $C_2$.

Symmetry: In both cases the equation involves only $(x-y)^2$ and $(x+y)^2$, both of which are invariant under the swap $(x, y) \mapsto (y, x)$. Therefore $y = x$ is a line of symmetry for each curve.

Part (ii)

The curves $y = x^2$ and $y = 2^x$ (for $x \geqslant 0$) intersect at $x = 2$ (both equal 4) and $x = 4$ (both equal 16). For $2 < x < 4$, we have $x^2 > 2^x$, and outside this interval $2^x > x^2$.

For any point on $C_1$, we have $(x - y)^2 \geqslant 0$, so from the equation of $C_1$:

$(x + y)^2 - 2^{x+y} \geqslant 0 \qquad \Longrightarrow \qquad (x + y)^2 \geqslant 2^{x+y}$

Setting $t = x + y > 0$, this requires $t^2 \geqslant 2^t$, which holds if and only if $2 \leqslant t \leqslant 4$. Therefore

$2 \leqslant x + y \leqslant 4$

so $C_1$ lies between the lines $x + y = 2$ and $x + y = 4$.

Sketch of $C_1$: The curve is a closed oval, symmetric about $y = x$, lying entirely in the first quadrant between the two parallel lines $x + y = 2$ and $x + y = 4$. It passes through $(1, 1)$ (where $x + y = 2$) and $(2, 2)$ (where $x + y = 4$). At $(1,1)$ the curve is tangent to $x + y = 2$, and at $(2,2)$ it is tangent to $x + y = 4$.

Part (iii)

For $C_2$ with $(x - y)^2 = (x + y)^2 - 2^{x+y+4}$, the condition $(x - y)^2 \geqslant 0$ gives

$(x + y)^2 \geqslant 2^{x+y+4}$

Setting $u = x + y < 0$, we need $u^2 \geqslant 2^{u+4}$. At $u = -2$: $4 = 4$. For $u < -2$: $u^2$ grows while $2^{u+4}$ shrinks, so the inequality holds. For $-2 < u < 0$: $2^{u+4} > u^2$, so the inequality fails. Therefore $x + y \leqslant -2$.

From $xy = 2^{x+y+2}$: as $x \to -\infty$, $2^{x+y+2} \to 0$, so $y \to 0^-$. By symmetry, as $y \to -\infty$, $x \to 0^-$.

Sketch of $C_2$: The curve is an open curve in the third quadrant, symmetric about $y = x$, with $x + y \leqslant -2$. It passes through $(-1, -1)$ where it touches the boundary line $x + y = -2$. As $x \to 0^-$, $y \to -\infty$ (and vice versa by symmetry). The curve has both axes as asymptotes.

Examiner Notes

无官方评述。易错点：分离变量时的代数处理、C1(正坐标)与C2(负坐标)的区别处理、图像绘制时边界线x+y=2和x+y=4的推导。

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Question 3

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

Problem

3 A sequence $u_1, u_2, \dots, u_n$ of positive real numbers is said to be unimodal if there is a value $k$ such that

$u_1 \leqslant u_2 \leqslant \dots \leqslant u_k$

and

$u_k \geqslant u_{k+1} \geqslant \dots \geqslant u_n.$

So the sequences 1, 2, 3, 2, 1; 1, 2, 3, 4, 5; 1, 1, 3, 3, 2 and 2, 2, 2, 2, 2 are all unimodal, but 1, 2, 1, 3, 1 is not.

A sequence $u_1, u_2, \dots, u_n$ of positive real numbers is said to have property $L$ if $u_{r-1}u_{r+1} \leqslant u_r^2$ for all $r$ with $2 \leqslant r \leqslant n - 1$.

(i) Show that, in any sequence of positive real numbers with property $L$,

$u_{r-1} \geqslant u_r \implies u_r \geqslant u_{r+1}.$

Prove that any sequence of positive real numbers with property $L$ is unimodal.

(ii) A sequence $u_1, u_2, \dots, u_n$ of real numbers satisfies $u_r = 2\alpha u_{r-1} - \alpha^2 u_{r-2}$ for $3 \leqslant r \leqslant n$, where $\alpha$ is a positive real constant. Prove that, for $2 \leqslant r \leqslant n$,

$u_r - \alpha u_{r-1} = \alpha^{r-2}(u_2 - \alpha u_1)$

and, for $2 \leqslant r \leqslant n - 1$,

$u_r^2 - u_{r-1}u_{r+1} = (u_r - \alpha u_{r-1})^2.$

Hence show that the sequence consists of positive terms and is unimodal, provided $u_2 > \alpha u_1 > 0$.

In the case $u_1 = 1$ and $u_2 = 2$, prove by induction that $u_r = (2 - r)\alpha^{r-1} + 2(r - 1)\alpha^{r-2}$.

Let $\alpha = 1 - \frac{1}{N}$, where $N$ is an integer with $2 \leqslant N \leqslant n$.

In the case $u_1 = 1$ and $u_2 = 2$, prove that $u_r$ is largest when $r = N$.

Hint

3(i) Suppose $\exists k: 2 \le k \le n - 1$ such that $u_{k-1} \ge u_k$ but $u_k < u_{k+1}$. Since all terms are positive, $u_k^2 < u_{k-1}u_{k+1}$, so the sequence does not have property L. Therefore, once $u_{k-1} \ge u_k$, all subsequent terms must also decrease (unimodality).

3(ii) $u_r - \alpha u_{r-1} = \alpha(u_{r-1} - \alpha u_{r-2})$, so $u_r - \alpha u_{r-1} = \alpha^{r-2}(u_2 - \alpha u_1)$.

$u_r^2 - u_{r-1}u_{r+1} = u_r^2 - u_{r-1}(2\alpha u_r - \alpha^2 u_{r-1}) = (u_r - \alpha u_{r-1})^2$ for $r \ge 2$.

The first identity shows $u_r > 0$ for all $r$ if $u_2 > \alpha u_1 > 0$. The second identity shows property L holds, hence unimodal.

3(iii) Base cases: $u_1 = 1$, $u_2 = 2$. Inductive step: assume for $u_{k-2}, u_{k-1}$, then $u_k = 2\alpha u_{k-1} - \alpha^2 u_{k-2} = \alpha^{k-1}(2-k) + 2\alpha^{k-2}(k-1)$.

$u_r - u_{r+1} = \alpha^{r-2}\left(2(r-1) + (2-3r)\alpha + (r-1)\alpha^2\right) = \frac{\alpha^{r-2}}{N^2}\left((r-1) + rN - N^2\right)$.

When $r = N$: $u_N - u_{N+1} = \frac{\alpha^{N-2}(N-1)}{N^2} > 0$. When $r = N-1$: $u_{N-1} - u_N = \frac{-2\alpha^{N-3}}{N^2} < 0$. So $u_r$ is largest when $r = N$.

Model Solution

Part (i)

We are given that a sequence of positive real numbers has property $L$: $u_{r-1}u_{r+1} \leqslant u_r^2$ for all $2 \leqslant r \leqslant n - 1$.

We want to show $u_{r-1} \geqslant u_r \implies u_r \geqslant u_{r+1}$.

Suppose $u_{r-1} \geqslant u_r > 0$. From property $L$:

$u_{r-1}u_{r+1} \leqslant u_r^2$

Since $u_{r-1} > 0$, we can divide both sides:

$u_{r+1} \leqslant \frac{u_r^2}{u_{r-1}}$

Since $u_{r-1} \geqslant u_r > 0$, we have $\frac{1}{u_{r-1}} \leqslant \frac{1}{u_r}$, so:

$u_{r+1} \leqslant \frac{u_r^2}{u_{r-1}} \leqslant \frac{u_r^2}{u_r} = u_r$

Therefore $u_r \geqslant u_{r+1}$, as required.

Now we prove any sequence with property $L$ is unimodal. Consider two cases:

Case 1: $u_1 \leqslant u_2 \leqslant \dots \leqslant u_n$ (the sequence is non-decreasing). Then $k = n$ satisfies the definition.

Case 2: The sequence is not non-decreasing. Then there exists a smallest index $k$ such that $u_k < u_{k+1}$ is false, i.e., $u_k \geqslant u_{k+1}$. By the minimality of $k$, we have $u_1 \leqslant u_2 \leqslant \dots \leqslant u_k$.

By the result above, $u_k \geqslant u_{k+1} \implies u_{k+1} \geqslant u_{k+2}$, and applying this repeatedly gives $u_k \geqslant u_{k+1} \geqslant \dots \geqslant u_n$.

Therefore the sequence is unimodal.

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Part (ii)

First identity. Define $v_r = u_r - \alpha u_{r-1}$ for $r \geqslant 2$. From the recurrence $u_r = 2\alpha u_{r-1} - \alpha^2 u_{r-2}$:

$v_r = u_r - \alpha u_{r-1} = (2\alpha u_{r-1} - \alpha^2 u_{r-2}) - \alpha u_{r-1} = \alpha u_{r-1} - \alpha^2 u_{r-2} = \alpha(u_{r-1} - \alpha u_{r-2}) = \alpha \, v_{r-1}$

So $\{v_r\}$ is a geometric sequence with common ratio $\alpha$ and first term $v_2 = u_2 - \alpha u_1$. Therefore:

$u_r - \alpha u_{r-1} = v_r = \alpha^{r-2}(u_2 - \alpha u_1) \qquad \text{(for } 2 \leqslant r \leqslant n\text{)}$

Second identity. For $2 \leqslant r \leqslant n - 1$, using $u_{r+1} = 2\alpha u_r - \alpha^2 u_{r-1}$:

$u_r^2 - u_{r-1}u_{r+1} = u_r^2 - u_{r-1}(2\alpha u_r - \alpha^2 u_{r-1}) = u_r^2 - 2\alpha u_{r-1}u_r + \alpha^2 u_{r-1}^2 = (u_r - \alpha u_{r-1})^2$

Positivity and unimodality. Since $u_2 > \alpha u_1 > 0$, we have $u_2 - \alpha u_1 > 0$, so from the first identity:

$u_r = \alpha u_{r-1} + \alpha^{r-2}(u_2 - \alpha u_1)$

Since $\alpha > 0$ and $u_1 > 0$, by induction every $u_r > 0$.

The second identity gives $u_r^2 - u_{r-1}u_{r+1} = (u_r - \alpha u_{r-1})^2 \geqslant 0$, so property $L$ holds. By part (i), the sequence is unimodal.

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Induction proof. We prove $u_r = (2 - r)\alpha^{r-1} + 2(r - 1)\alpha^{r-2}$ for $r \geqslant 1$.

Base cases:

• $r = 1$: $(2 - 1)\alpha^0 + 2(0)\alpha^{-1} = 1 = u_1$. Check.
• $r = 2$: $(2 - 2)\alpha^1 + 2(1)\alpha^0 = 2 = u_2$. Check.

Inductive step: Assume the formula holds for $u_{k-2}$ and $u_{k-1}$ (where $k \geqslant 3$). Then:

$u_k = 2\alpha u_{k-1} - \alpha^2 u_{k-2}$

$= 2\alpha\Big[(3 - k)\alpha^{k-2} + 2(k - 2)\alpha^{k-3}\Big] - \alpha^2\Big[(4 - k)\alpha^{k-3} + 2(k - 3)\alpha^{k-4}\Big]$

$= (6 - 2k)\alpha^{k-1} + 4(k - 2)\alpha^{k-2} - (4 - k)\alpha^{k-1} - 2(k - 3)\alpha^{k-2}$

$= \alpha^{k-1}\big[(6 - 2k) - (4 - k)\big] + \alpha^{k-2}\big[4(k - 2) - 2(k - 3)\big]$

$= \alpha^{k-1}(2 - k) + \alpha^{k-2}(2k - 2) = (2 - k)\alpha^{k-1} + 2(k - 1)\alpha^{k-2}$

This is exactly the formula for $u_k$. By induction, the result holds for all $r \geqslant 1$.

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Maximum at $r = N$. Let $\alpha = 1 - \frac{1}{N}$. We compute $u_r - u_{r+1}$:

$u_r - u_{r+1} = \Big[(2 - r)\alpha^{r-1} + 2(r - 1)\alpha^{r-2}\Big] - \Big[(1 - r)\alpha^r + 2r\alpha^{r-1}\Big]$

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

$= \alpha^{r-2}\Big[2(r - 1) + (2 - 3r)\alpha + (r - 1)\alpha^2\Big]$

Substituting $\alpha = 1 - \frac{1}{N}$, i.e., $\alpha = \frac{N - 1}{N}$:

$= \frac{\alpha^{r-2}}{N^2}\Big[2N^2(r - 1) + (2 - 3r)N(N - 1) + (r - 1)(N - 1)^2\Big]$

Expanding the bracket:

$2N^2(r-1) + N(N-1)(2 - 3r) + (r-1)(N-1)^2$

$= 2N^2 r - 2N^2 + 2N^2 - 2N - 3N^2 r + 3Nr + (r-1)(N^2 - 2N + 1)$

$= -N^2 r - 2N + 3Nr + rN^2 - 2Nr + r - N^2 + 2N - 1$

$= rN - N^2 + r - 1 = (r - 1)(N + 1) - N^2 + rN - N + r - 1$

Let me redo this more carefully. Grouping by powers of $r$ and $N$:

Coefficient of $r$: $2N^2 - 3N^2 + 3N + N^2 - 2N + 1 = N + 1$

Constant terms: $-2N^2 + 2N^2 - 2N - N^2 + 2N - 1 = -N^2 - 1$

Wait, let me recompute. Constant (no $r$): $-2N^2 + 2N^2 - 2N - N^2 + 2N - 1 = -N^2 - 1$. Hmm, that gives $(N+1)r - N^2 - 1$.

Checking: $(N+1)r - (N^2 + 1) = (N+1)(r - N + 1) - 2$? Let me just evaluate directly.

When $r = N$:

$(N+1)N - N^2 - 1 = N^2 + N - N^2 - 1 = N - 1 > 0 \qquad \text{(since } N \geqslant 2\text{)}$

So $u_N - u_{N+1} > 0$, meaning $u_N > u_{N+1}$.

When $r = N - 1$:

$(N+1)(N-1) - N^2 - 1 = N^2 - 1 - N^2 - 1 = -2 < 0$

So $u_{N-1} - u_N < 0$, meaning $u_{N-1} < u_N$.

Together: $u_{N-1} < u_N > u_{N+1}$, and since the sequence is unimodal (proved above), $u_r$ is largest when $r = N$.

Examiner Notes

无官方评述。易错点：归纳法基础步骤的正确设置、递推关系的代数化简（u_r-αu_{r-1}的公式）、极值位置r=N的精确判断。此题逻辑链长、综合性极强。

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Question 4

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

Problem

4 (i) Given that $a, b$ and $c$ are the lengths of the sides of a triangle, explain why $c < a + b$, $a < b + c$ and $b < a + c$.

(ii) Use a diagram to show that the converse of the result in part (i) also holds: if $a, b$ and $c$ are positive numbers such that $c < a + b, a < b + c$ and $b < c + a$ then it is possible to construct a triangle with sides of length $a, b$ and $c$.

(iii) When $a, b$ and $c$ are the lengths of the sides of a triangle, determine in each case whether the following sets of three lengths can

• always
• sometimes but not always
• never

form the sides of a triangle. Prove your claims.

(A) $a + 1, b + 1, c + 1$.

(B) $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$.

(C) $|a - b|, |b - c|, |c - a|$.

(D) $a^2 + bc, b^2 + ca, c^2 + ab$.

(iv) Let f be a function defined on the positive real numbers and such that, whenever $x > y > 0$,

$f(x) > f(y) > 0 \text{ but } \frac{f(x)}{x} < \frac{f(y)}{y}.$

Show that, whenever $a, b$ and $c$ are the lengths of the sides of a triangle, then $f(a), f(b)$ and $f(c)$ can also be the lengths of the sides of a triangle.

Hint

4(i) The straight line distance between two points must be less than the length of any other rectilinear path.

4(ii) Diagram: two circles with radii $a$, $b$ and line of length $c$. Either the straight line is the longest, or one circle cannot be contained inside the other. The circles must meet.

4(iii) (A) If $a + b > c$ then $(a+1)+(b+1) > c+2 > c+1$ et cycl., so always. (B) $a=b=c=1$ gives $1,1,1$ (works); $a=1,b=c=2$ gives $\frac{1}{2},1,2$ (fails). Sometimes but not always. (C) If $p \ge q \ge r$ then $|p-q|+|q-r| = p-r = |p-r|$, so never. (D) $(a-b)^2 + c(a+b) + 2ab > c^2 + ab$ et cycl., so always.

4(iv) Since $a+b > a,b$: $\frac{f(a)}{a} > \frac{f(a+b)}{a+b}$ and $\frac{f(b)}{b} > \frac{f(a+b)}{a+b}$. Since $c < a+b$, $f(c) < f(a+b)$. Thus $f(a)+f(b) > f(a+b) > f(c)$ et cycl.

Model Solution

Part (i)

If $a, b, c$ are the side lengths of a triangle, then the triangle inequality states that the length of any one side is strictly less than the sum of the other two. Geometrically, the straight-line distance between two vertices is shorter than any path that goes through the third vertex (since a straight line is the shortest path between two points). Therefore $c < a + b$, $a < b + c$, and $b < a + c$.

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Part (ii)

Suppose $a, b, c > 0$ satisfy $c < a + b$, $a < b + c$, and $b < c + a$.

Without loss of generality, assume $c$ is the largest of the three values. Then $c < a + b$ ensures that the longest length is strictly less than the sum of the other two.

Draw a line segment $AB$ of length $c$. With centre $A$, draw a circle of radius $b$. With centre $B$, draw a circle of radius $a$.

Since $c < a + b$, neither circle is contained inside the other: circle $A$ extends to distance $b$ from $A$, and circle $B$ extends to distance $a$ from $B$; since $b + a > c$, each circle reaches past the centre of the other.

Since $c > |a - b|$ (which follows from $c + \min(a,b) > \max(a,b)$, true because $c > 0$ and the triangle inequalities hold), the circles are not disjoint.

Therefore the two circles must intersect. Let $C$ be a point of intersection. Then $AC = b$ and $BC = a$, so triangle $ABC$ has side lengths $a$, $b$, $c$.

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Part (iii)

(A) $a + 1, b + 1, c + 1$: Always form a triangle.

We need to show $(a + 1) + (b + 1) > c + 1$ (and cyclically). Since $a + b > c$:

$(a + 1) + (b + 1) = a + b + 2 > c + 2 > c + 1$

The cyclic inequalities follow identically. Therefore $a + 1$, $b + 1$, $c + 1$ always form a triangle.

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(B) $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$: Sometimes but not always form a triangle.

Example where they do: $a = b = c = 1$. Then $\frac{a}{b} = \frac{b}{c} = \frac{c}{a} = 1$, and $1 + 1 > 1$. They form an equilateral triangle.

Example where they do not: $a = 1, b = 2, c = 2$. The three values are $\frac{1}{2}, 1, 2$. Then $\frac{1}{2} + 1 = \frac{3}{2} < 2$, so the triangle inequality fails.

Therefore $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ can sometimes, but not always, form the sides of a triangle.

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(C) $|a - b|, |b - c|, |c - a|$: Never form a triangle.

Let $p = \max(a, b, c)$, $r = \min(a, b, c)$, and let $q$ be the remaining value, so $p \geqslant q \geqslant r$.

Then the three lengths are:

$|p - q| = p - q, \qquad |q - r| = q - r, \qquad |p - r| = p - r$

Now:

$|p - q| + |q - r| = (p - q) + (q - r) = p - r = |p - r|$

So two of the three lengths sum to exactly the third. The triangle inequality requires strict inequality, so these three lengths can never form a triangle.

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(D) $a^2 + bc, \; b^2 + ca, \; c^2 + ab$: Always form a triangle.

We must show $(a^2 + bc) + (b^2 + ca) > c^2 + ab$ (and cyclically).

$(a^2 + bc) + (b^2 + ca) - (c^2 + ab) = a^2 + b^2 - c^2 + bc + ca - ab$

$= a^2 + b^2 - 2ab + c(a + b) - c^2 = (a - b)^2 + c\big[(a + b) - c\big]$

Since $a + b > c$ and $(a - b)^2 \geqslant 0$:

$(a - b)^2 + c(a + b - c) > 0$

By the same argument applied cyclically (with $a, b, c$ permuted), all three triangle inequalities hold. Therefore $a^2 + bc, b^2 + ca, c^2 + ab$ always form a triangle.

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Part (iv)

We are given that $f$ is defined on positive reals, and whenever $x > y > 0$:

• $f(x) > f(y) > 0$ (so $f$ is strictly increasing and positive), and
• $\frac{f(x)}{x} < \frac{f(y)}{y}$ (so $\frac{f(x)}{x}$ is strictly decreasing).

Let $a, b, c$ be the side lengths of a triangle, so $a + b > c$ (and cyclically). We want to show $f(a) + f(b) > f(c)$.

Since $a + b > a$ and $a + b > b$, the decreasing property of $\frac{f(x)}{x}$ gives:

$\frac{f(a)}{a} > \frac{f(a + b)}{a + b} \qquad \text{and} \qquad \frac{f(b)}{b} > \frac{f(a + b)}{a + b}$

Multiplying:

$f(a) > \frac{a \cdot f(a + b)}{a + b} \qquad \text{and} \qquad f(b) > \frac{b \cdot f(a + b)}{a + b}$

Adding:

$f(a) + f(b) > \frac{a \cdot f(a + b)}{a + b} + \frac{b \cdot f(a + b)}{a + b} = \frac{(a + b) \cdot f(a + b)}{a + b} = f(a + b)$

Since $a + b > c$ and $f$ is strictly increasing:

$f(a + b) > f(c)$

Therefore $f(a) + f(b) > f(a + b) > f(c)$.

By the same argument applied cyclically to the other two triangle inequalities, we obtain $f(b) + f(c) > f(a)$ and $f(c) + f(a) > f(b)$. Therefore $f(a), f(b), f(c)$ can form the sides of a triangle.

Examiner Notes

无官方评述。易错点：(B)部分a/b, b/c, c/a的判定需要巧妙构造、(C)部分|a-b|,|b-c|,|c-a|通常不能构成三角形的证明、(iv)中函数性质的正确运用。

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Question 5

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

Problem

5 If $x$ is a positive integer, the value of the function $d(x)$ is the sum of the digits of $x$ in base 10. For example, $d(249) = 2 + 4 + 9 = 15$.

An $n$-digit positive integer $x$ is written in the form $\sum_{r=0}^{n-1} a_r \times 10^r$, where $0 \leqslant a_r \leqslant 9$ for all $0 \leqslant r \leqslant n - 1$ and $a_{n-1} > 0$.

(i) Prove that $x - d(x)$ is non-negative and divisible by 9.

(ii) Prove that $x - 44d(x)$ is a multiple of 9 if and only if $x$ is a multiple of 9.

Suppose that $x = 44d(x)$. Show that if $x$ has $n$ digits, then $x \leqslant 396n$ and $x \geqslant 10^{n-1}$, and hence that $n \leqslant 4$.

Find a value of $x$ for which $x = 44d(x)$. Show that there are no further values of $x$ satisfying this equation.

(iii) Find a value of $x$ for which $x = 107d(d(x))$. Show that there are no further values of $x$ satisfying this equation.

Hint

5(i) $x - d(x) = \sum_{r=0}^{n-1} a_r(10^r - 1)$. Since $10^r \ge 1$ for all $r$, this is non-negative. Since $9 \mid (10^r - 1)$ for all $r$, it is divisible by 9.

5(ii) $x - 44d(x) = 44(x - d(x)) - 43x$. It is a multiple of 9 iff $43x$ is. Since $\gcd(43, 9) = 1$, iff $x$ is a multiple of 9.

If $x$ has $n$ digits, $d(x) \le 9n$, so $x = 44d(x) \le 396n$. Also $x \ge 10^{n-1}$. These cannot hold for $n \ge 5$ ($396 \times 5 < 10^4$), so $n \le 4$.

Since $x = 44d(x) = 0 \pmod{9}$, $x$ is a multiple of 9. Also $44 \mid x$. Since $\gcd(9, 44) = 1$, $396 \mid x$, so $x = 396k$ with $k \le 4$. Checking: only $k = 2$ works ($x = 792$).

5(iii) $x - 107d(d(x)) = 107(x - d(x)) + 107(d(x) - d(d(x))) - 106x$. Both $(x - d(x))$ and $(d(x) - d(d(x)))$ are divisible by 9, so $x$ is divisible by 9. Also $107 \mid x$, so $963 \mid x$. Then $x \le 963n$ and $x \ge 10^{n-1}$ gives $n \le 4$, $k \le 4$. Checking: only $k = 1$ works ($x = 963$).

Model Solution

Part (i)

Write $x = \displaystyle\sum_{r=0}^{n-1} a_r \times 10^r$ where $0 \leqslant a_r \leqslant 9$ for all $r$ and $a_{n-1} > 0$. Then $d(x) = \displaystyle\sum_{r=0}^{n-1} a_r$.

$x - d(x) = \sum_{r=0}^{n-1} a_r \times 10^r - \sum_{r=0}^{n-1} a_r = \sum_{r=0}^{n-1} a_r(10^r - 1)$

Since $10^r \geqslant 1$ for all $r \geqslant 0$, and $a_r \geqslant 0$, every term in the sum is non-negative, so $x - d(x) \geqslant 0$.

Also, $10^r - 1 = \underbrace{99\ldots9}_{r \text{ digits}}$ is divisible by 9 for every $r \geqslant 0$. Therefore $9 \mid a_r(10^r - 1)$ for each $r$, and so $9 \mid (x - d(x))$.

$\qquad \text{(Q5(i))}$

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Part (ii)

We write:

$x - 44\,d(x) = x - d(x) - 43\,d(x) = \big[x - d(x)\big] - 43\,d(x)$

Hmm, a cleaner decomposition. Alternatively:

$x - 44\,d(x) = \big[x - d(x)\big] - 43\,d(x)$

Wait, let us use the hint’s approach:

$x - 44\,d(x) = 44\big[x - d(x)\big] - 43x$

From part (i), $9 \mid (x - d(x))$, so $9 \mid 44(x - d(x))$. Therefore:

$9 \mid \big[x - 44\,d(x)\big] \iff 9 \mid 43x$

Since $\gcd(43, 9) = 1$, we have $9 \mid 43x \iff 9 \mid x$. Therefore $x - 44\,d(x)$ is a multiple of 9 if and only if $x$ is a multiple of 9.

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Now suppose $x = 44\,d(x)$.

Upper bound. If $x$ has $n$ digits, then $d(x) \leqslant 9n$ (since each digit is at most 9), so:

$x = 44\,d(x) \leqslant 44 \times 9n = 396n$

Lower bound. Any $n$-digit positive integer satisfies $x \geqslant 10^{n-1}$.

Combining: $10^{n-1} \leqslant 396n$. Checking:

• $n = 4$: $10^3 = 1000 \leqslant 396 \times 4 = 1584$. Holds.
• $n = 5$: $10^4 = 10000 > 396 \times 5 = 1980$. Fails.

For $n \geqslant 5$, $10^{n-1}$ grows much faster than $396n$, so the inequality fails for all $n \geqslant 5$. Hence $n \leqslant 4$.

Finding $x$. Since $x = 44\,d(x) = 0$, and $9 \mid 0$, we know $9 \mid x$ from the result above. Also $x = 44\,d(x)$ means $44 \mid x$. Since $\gcd(9, 44) = 1$, we get $396 \mid x$. So $x = 396k$ for some positive integer $k$, and since $x \leqslant 396 \times 4 = 1584$, we have $k \leqslant 4$.

Checking each:

$k$	$x = 396k$	$d(x)$	$44\,d(x)$	Match?
1	396	18	792	No
2	792	18	792	Yes
3	1188	18	792	No
4	1584	18	792	No

The only solution is $x = 792$.

We have shown $n \leqslant 4$, and among the multiples of 396 in the range $[1, 1584]$, only $x = 792$ satisfies the equation. Therefore there are no further solutions.

$\qquad \text{(Q5(ii))}$

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Part (iii)

Claim: $x = 963$ satisfies $x = 107\,d(d(x))$.

Check: $d(963) = 9 + 6 + 3 = 18$, then $d(18) = 1 + 8 = 9$, and $107 \times 9 = 963$.

Uniqueness. Suppose $x = 107\,d(d(x))$. We show $x$ must be 963.

First, $x - 107\,d(d(x)) = 0$, so:

$x - 107\,d(d(x)) = 107\big[x - d(x)\big] + 107\big[d(x) - d(d(x))\big] - 106x$

From part (i), $9 \mid (x - d(x))$ and $9 \mid (d(x) - d(d(x)))$ (applying part (i) with $d(x)$ in place of $x$). Therefore $9 \mid 106x$. Since $\gcd(106, 9) = \gcd(106, 9) = 1$ (as $106 = 11 \times 9 + 7$), we get $9 \mid x$.

Since $x = 107\,d(d(x))$, we also have $107 \mid x$. Since $\gcd(9, 107) = 1$ (107 is prime and not 3), we get $963 \mid x$.

Bounding the number of digits. If $x$ has $n$ digits, then $d(x) \leqslant 9n$, and applying part (i) to $d(x)$: $d(d(x)) \leqslant d(x) \leqslant 9n$. So:

$x = 107\,d(d(x)) \leqslant 107 \times 9n = 963n$

Also $x \geqslant 10^{n-1}$. Since $10^4 = 10000 > 963 \times 4 = 3852$, the inequality $10^{n-1} \leqslant 963n$ fails for $n \geqslant 5$, so $n \leqslant 4$. This gives $x \leqslant 963 \times 4 = 3852$, and since $963 \mid x$, the candidates are $x = 963k$ with $k = 1, 2, 3, 4$.

$k$	$x$	$d(x)$	$d(d(x))$	$107\,d(d(x))$	Match?
1	963	18	9	963	Yes
2	1926	18	9	963	No
3	2889	27	9	963	No
4	3852	18	9	963	No

The only solution is $x = 963$.

$\qquad \text{(Q5(iii))}$

Examiner Notes

无官方评述。易错点：数字和上界的估计（x≤396n的推导）、n的范围缩小到n≤4、穷举时遗漏解或重复计算。

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Question 6

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

Problem

6 A $2 \times 2$ matrix $\mathbf{M}$ is real if it can be written as $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a, b, c$ and $d$ are real. In this case, the trace of matrix $\mathbf{M}$ is defined to be $\text{tr}(\mathbf{M}) = a + d$ and $\text{det}(\mathbf{M})$ is the determinant of matrix $\mathbf{M}$. In this question, $\mathbf{M}$ is a real $2 \times 2$ matrix.

(i) Prove that $\text{tr}(\mathbf{M}^2) = \text{tr}(\mathbf{M})^2 - 2\text{det}(\mathbf{M}).$

(ii) Prove that $\mathbf{M}^2 = \mathbf{I} \text{ but } \mathbf{M} \neq \pm\mathbf{I} \iff \text{tr}(\mathbf{M}) = 0 \text{ and } \text{det}(\mathbf{M}) = -1,$ and that $\mathbf{M}^2 = -\mathbf{I} \iff \text{tr}(\mathbf{M}) = 0 \text{ and } \text{det}(\mathbf{M}) = 1.$

(iii) Use part (ii) to prove that $\mathbf{M}^4 = \mathbf{I} \iff \mathbf{M}^2 = \pm\mathbf{I}.$

Find a necessary and sufficient condition on $\text{det}(\mathbf{M})$ and $\text{tr}(\mathbf{M})$ so that $\mathbf{M}^4 = -\mathbf{I}$.

(iv) Give an example of a matrix $\mathbf{M}$ for which $\mathbf{M}^8 = \mathbf{I}$, but which does not represent a rotation or reflection. [Note that the matrices $\pm\mathbf{I}$ are both rotations.]

Hint

6(i) Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$; then $\mathbf{M}^2 = \begin{pmatrix} a^2 + bc & b(a+d) \\ c(a+d) & d^2 + bc \end{pmatrix}$, so $\text{tr}(\mathbf{M}^2) = a^2 + d^2 + 2bc = (a+d)^2 - 2(ad - bc)$.

6(ii) $\mathbf{M}^2 = \begin{pmatrix} a\tau - \delta & b\tau \\ c\tau & d\tau - \delta \end{pmatrix}$ where $\tau = \text{tr}(\mathbf{M})$, $\delta = \det(\mathbf{M})$. Thus $\mathbf{M}^2 = \pm\mathbf{I} \Leftrightarrow \tau = 0$ and $\delta = \mp 1$, or $b = c = 0$ and $a^2 = d^2 = \pm 1$. If $b = c = 0$ and $a = d = \pm 1$, then $\mathbf{M} = \pm\mathbf{I}$. If $b = c = 0$ and $a = -d = \pm 1$, then $\tau = 0$ and $\delta = -1$. So $\mathbf{M}^2 = \mathbf{I} \Leftrightarrow \tau = 0, \delta = -1$; $\mathbf{M}^2 = -\mathbf{I} \Leftrightarrow \tau = 0, \delta = +1$.

6(iii) Part (ii) implies $\det(\mathbf{M}^2) = -1$ if $\mathbf{M}^4 = \mathbf{I}$ but $\mathbf{M}^2 \neq \pm\mathbf{I}$. But $\det(\mathbf{M}^2) = \det(\mathbf{M})^2 \ge 0$, so impossible. Clearly $\mathbf{M}^2 = \pm\mathbf{I} \Rightarrow \mathbf{M}^4 = \mathbf{I}$. For $\mathbf{M}^4 = -\mathbf{I}$: need $\text{tr}(\mathbf{M}^2) = 0$ and $\det(\mathbf{M}^2) = 1$, so $\text{tr}(\mathbf{M})^2 = 2\det(\mathbf{M})$ and $\det(\mathbf{M}) = \pm 1$, giving $\text{tr}(\mathbf{M}) = \pm\sqrt{2}$ and $\det(\mathbf{M}) = 1$.

6(iv) Any example satisfying conditions for $\mathbf{M}^2 = \mathbf{I}$, $\mathbf{M}^2 = -\mathbf{I}$, or $\mathbf{M}^4 = -\mathbf{I}$, which is not a rotation or reflection.

Model Solution

Part (i)

Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then:

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

So:

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

Now:

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

Therefore $\text{tr}(\mathbf{M}^2) = \text{tr}(\mathbf{M})^2 - 2\det(\mathbf{M})$.

$\qquad \text{(Q6(i))}$

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Part (ii)

Every $2 \times 2$ matrix satisfies its own characteristic equation (Cayley—Hamilton):

$\mathbf{M}^2 - \text{tr}(\mathbf{M})\,\mathbf{M} + \det(\mathbf{M})\,\mathbf{I} = \mathbf{0} \qquad \Longrightarrow \qquad \mathbf{M}^2 = \text{tr}(\mathbf{M})\,\mathbf{M} - \det(\mathbf{M})\,\mathbf{I}$

First statement: $\mathbf{M}^2 = \mathbf{I}$ but $\mathbf{M} \neq \pm\mathbf{I}$ $\iff$ $\text{tr}(\mathbf{M}) = 0$ and $\det(\mathbf{M}) = -1$.

($\Rightarrow$) Suppose $\mathbf{M}^2 = \mathbf{I}$ and $\mathbf{M} \neq \pm\mathbf{I}$. From Cayley—Hamilton:

$\mathbf{I} = \text{tr}(\mathbf{M})\,\mathbf{M} - \det(\mathbf{M})\,\mathbf{I} \qquad \Longrightarrow \qquad \text{tr}(\mathbf{M})\,\mathbf{M} = \big(1 + \det(\mathbf{M})\big)\,\mathbf{I}$

If $\text{tr}(\mathbf{M}) \neq 0$, then $\mathbf{M} = \frac{1 + \det(\mathbf{M})}{\text{tr}(\mathbf{M})}\,\mathbf{I} = \lambda\mathbf{I}$ for some scalar $\lambda$. Then $\mathbf{M}^2 = \lambda^2\mathbf{I} = \mathbf{I}$ forces $\lambda = \pm 1$, giving $\mathbf{M} = \pm\mathbf{I}$, contradicting our assumption.

Therefore $\text{tr}(\mathbf{M}) = 0$. Substituting into the trace identity from part (i):

$\text{tr}(\mathbf{M}^2) = \text{tr}(\mathbf{I}) = 2 = 0 - 2\det(\mathbf{M}) \qquad \Longrightarrow \qquad \det(\mathbf{M}) = -1$

($\Leftarrow$) Suppose $\text{tr}(\mathbf{M}) = 0$ and $\det(\mathbf{M}) = -1$. From Cayley—Hamilton:

$\mathbf{M}^2 = 0 \cdot \mathbf{M} - (-1)\,\mathbf{I} = \mathbf{I}$

We verify $\mathbf{M} \neq \pm\mathbf{I}$: if $\mathbf{M} = \pm\mathbf{I}$, then $\text{tr}(\mathbf{M}) = \pm 2 \neq 0$, a contradiction.

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Second statement: $\mathbf{M}^2 = -\mathbf{I}$ $\iff$ $\text{tr}(\mathbf{M}) = 0$ and $\det(\mathbf{M}) = 1$.

($\Rightarrow$) Suppose $\mathbf{M}^2 = -\mathbf{I}$. From Cayley—Hamilton:

$-\mathbf{I} = \text{tr}(\mathbf{M})\,\mathbf{M} - \det(\mathbf{M})\,\mathbf{I} \qquad \Longrightarrow \qquad \text{tr}(\mathbf{M})\,\mathbf{M} = \big(\det(\mathbf{M}) - 1\big)\,\mathbf{I}$

If $\text{tr}(\mathbf{M}) \neq 0$, then $\mathbf{M} = \lambda\mathbf{I}$ and $\mathbf{M}^2 = \lambda^2\mathbf{I} = -\mathbf{I}$ requires $\lambda^2 = -1$, which has no real solution. Contradiction.

Therefore $\text{tr}(\mathbf{M}) = 0$. Then:

$\text{tr}(\mathbf{M}^2) = \text{tr}(-\mathbf{I}) = -2 = 0 - 2\det(\mathbf{M}) \qquad \Longrightarrow \qquad \det(\mathbf{M}) = 1$

($\Leftarrow$) Suppose $\text{tr}(\mathbf{M}) = 0$ and $\det(\mathbf{M}) = 1$. From Cayley—Hamilton:

$\mathbf{M}^2 = 0 \cdot \mathbf{M} - 1 \cdot \mathbf{I} = -\mathbf{I}$

$\qquad \text{(Q6(ii))}$

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Part (iii)

Claim: $\mathbf{M}^4 = \mathbf{I} \iff \mathbf{M}^2 = \pm\mathbf{I}$.

($\Leftarrow$) If $\mathbf{M}^2 = \mathbf{I}$, then $\mathbf{M}^4 = \mathbf{I}^2 = \mathbf{I}$. If $\mathbf{M}^2 = -\mathbf{I}$, then $\mathbf{M}^4 = (-\mathbf{I})^2 = \mathbf{I}$.

($\Rightarrow$) Suppose $\mathbf{M}^4 = \mathbf{I}$, so $(\mathbf{M}^2)^2 = \mathbf{I}$. Applying part (ii) to the matrix $\mathbf{N} = \mathbf{M}^2$: either $\mathbf{N} = \pm\mathbf{I}$ (which means $\mathbf{M}^2 = \pm\mathbf{I}$, and we are done), or $\text{tr}(\mathbf{N}) = 0$ and $\det(\mathbf{N}) = -1$.

But $\det(\mathbf{M}^2) = \det(\mathbf{M})^2 \geqslant 0$ for any real matrix $\mathbf{M}$. So $\det(\mathbf{M}^2) = -1$ is impossible.

Therefore $\mathbf{M}^2 = \pm\mathbf{I}$.

---

Condition for $\mathbf{M}^4 = -\mathbf{I}$.

By part (ii) applied to $\mathbf{N} = \mathbf{M}^2$:

$\mathbf{M}^4 = -\mathbf{I} \iff \text{tr}(\mathbf{M}^2) = 0 \text{ and } \det(\mathbf{M}^2) = 1$

From part (i): $\text{tr}(\mathbf{M}^2) = \text{tr}(\mathbf{M})^2 - 2\det(\mathbf{M})$.

Also: $\det(\mathbf{M}^2) = \det(\mathbf{M})^2$.

So the conditions are:

$\text{tr}(\mathbf{M})^2 - 2\det(\mathbf{M}) = 0 \qquad \text{and} \qquad \det(\mathbf{M})^2 = 1$

From $\det(\mathbf{M})^2 = 1$: $\det(\mathbf{M}) = \pm 1$.

If $\det(\mathbf{M}) = -1$: $\text{tr}(\mathbf{M})^2 = -2$, which has no real solution.

If $\det(\mathbf{M}) = 1$: $\text{tr}(\mathbf{M})^2 = 2$, so $\text{tr}(\mathbf{M}) = \pm\sqrt{2}$.

Therefore:

$\mathbf{M}^4 = -\mathbf{I} \iff \det(\mathbf{M}) = 1 \text{ and } \text{tr}(\mathbf{M}) = \pm\sqrt{2}$

$\qquad \text{(Q6(iii))}$

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Part (iv)

We need a real $2 \times 2$ matrix $\mathbf{M}$ with $\mathbf{M}^8 = \mathbf{I}$ that is neither a rotation nor a reflection.

Take:

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

Verification that $\mathbf{M}^8 = \mathbf{I}$: We have $\text{tr}(\mathbf{M}) = 0$ and $\det(\mathbf{M}) = -1$. By part (ii), $\mathbf{M}^2 = \mathbf{I}$, so $\mathbf{M}^8 = (\mathbf{M}^2)^4 = \mathbf{I}^4 = \mathbf{I}$.

Verification that $\mathbf{M}$ is not a rotation: A rotation matrix has determinant $+1$. Since $\det(\mathbf{M}) = -1$, $\mathbf{M}$ is not a rotation.

Verification that $\mathbf{M}$ is not a reflection: A reflection about a line through the origin is a symmetric matrix (it has the form $\mathbf{I} - 2\mathbf{n}\mathbf{n}^T$). Since $\mathbf{M}$ is not symmetric ($b = 2 \neq 0 = c$), $\mathbf{M}$ is not a reflection.

Therefore $\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix}$ is a valid example.

$\qquad \text{(Q6(iv))}$

Examiner Notes

无官方评述。易错点：M²=-I的充分性证明需考虑复特征值、M⁴=-I的条件推导涉及二次方程判别式、构造反例时矩阵元素的选取需验证M⁸=I。

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Question 7

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

Problem

7 In this question, $w = \frac{2}{z - 2}$.

(i) Let $z$ be the complex number $3 + ti$, where $t \in \mathbb{R}$. Show that $|w - 1|$ is independent of $t$. Hence show that, if $z$ is a complex number on the line $\text{Re}(z) = 3$ in the Argand diagram, then $w$ lies on a circle in the Argand diagram with centre 1.

Let $V$ be the line $\text{Re}(z) = p$, where $p$ is a real constant not equal to 2. Show that, if $z$ lies on $V$, then $w$ lies on a circle whose centre and radius you should give in terms of $p$. For which $z$ on $V$ is $\text{Im}(w) > 0$?

(ii) Let $H$ be the line $\text{Im}(z) = q$, where $q$ is a non-zero real constant. Show that, if $z$ lies on $H$, then $w$ lies on a circle whose centre and radius you should give in terms of $q$. For which $z$ on $H$ is $\text{Re}(w) > 0$?

Hint

7(i) $|w - 1|^2 = \left| \frac{1 - ti}{1 + ti} \right|^2 = \frac{(1 - ti)(1 + ti)}{(1 + ti)(1 - ti)} = 1$, which is independent of $t$.

Points on the line $Re(z) = 3$ have the form $z = 3 + ti$ and the points satisfying $|w - 1| = 1$ lie on a circle with centre 1.

If $z = p + ti$, then

$|w - c|^2 = \left| \frac{2 - (p - 2)c - cti}{(p - 2) + ti} \right|^2 = \frac{(2 - (p - 2)c)^2 + c^2t^2}{(p - 2)^2 + t^2}$

which is independent of $t$ when $(2 - (p - 2)c)^2 = c^2(p - 2)^2$

which is when $c = \frac{1}{p - 2}$.

Thus the circle has centre at $\frac{1}{p - 2}$ and radius $\frac{1}{|p - 2|}$.

$w = \frac{2}{(p - 2) + ti} = \frac{2(p - 2) - 2ti}{(p - 2)^2 + t^2}$,

so $Im(w) > 0$ when $t < 0$; that is, for those $z$ on $V$ with negative imaginary part.

7(ii) If $z = t + qi$ then

$|w - ci|^2 = \left| \frac{2 + cq - (t - 2)ci}{(t - 2) + qi} \right|^2 = \frac{c^2(t - 2)^2 + (cq + 2)^2}{(t - 2)^2 + q^2}$

which is independent of $t$ when $(cq + 2)^2 = c^2q^2$

which is when $c = -\frac{1}{q}$

so the circle has centre $-\frac{1}{q}i$ A1 and radius $\sqrt{c^2} = \frac{1}{|q|}$ A1.

$w = \frac{2}{(t - 2) + qi} = \frac{2(t - 2) - 2qi}{(t - 2)^2 + q^2}$,

so $Re(w) > 0$ when $t > 2$; that is, for those $z$ on $H$ with real part greater than 2.

Model Solution

Part (i)

Let $z = 3 + ti$ where $t \in \mathbb{R}$. Then

$w = \frac{2}{z - 2} = \frac{2}{(3 + ti) - 2} = \frac{2}{1 + ti}$

We compute $w - 1$:

$w - 1 = \frac{2}{1 + ti} - 1 = \frac{2 - (1 + ti)}{1 + ti} = \frac{1 - ti}{1 + ti}$

Taking the modulus:

$|w - 1| = \frac{|1 - ti|}{|1 + ti|} = \frac{\sqrt{1 + t^2}}{\sqrt{1 + t^2}} = 1$

This is independent of $t$, as required.

Since $z = 3 + ti$ parametrises all points on the line $\operatorname{Re}(z) = 3$ as $t$ ranges over $\mathbb{R}$, and $|w - 1| = 1$ for every such $z$, the image of this line under $w = \frac{2}{z-2}$ is the circle with centre $1$ and radius $1$ in the Argand diagram.

General line $V$: $\operatorname{Re}(z) = p$, $p \neq 2$.

Let $z = p + ti$ with $t \in \mathbb{R}$. Then

$w = \frac{2}{(p - 2) + ti} = \frac{2\big[(p - 2) - ti\big]}{(p - 2)^2 + t^2}$

So

$\operatorname{Re}(w) = \frac{2(p - 2)}{(p - 2)^2 + t^2}, \qquad \operatorname{Im}(w) = \frac{-2t}{(p - 2)^2 + t^2}$

We seek a real constant $c$ such that $|w - c|^2$ is independent of $t$. We have

$|w - c|^2 = \left(\frac{2(p-2)}{(p-2)^2 + t^2} - c\right)^2 + \frac{4t^2}{\big[(p-2)^2 + t^2\big]^2}$

Writing this over a common denominator:

$|w - c|^2 = \frac{\big[2(p-2) - c(p-2)^2 - ct^2\big]^2 + 4t^2}{\big[(p-2)^2 + t^2\big]^2}$

Expanding the numerator, the coefficient of $t^2$ is $-2c\big[2(p-2) - c(p-2)^2\big] + 4$ and the coefficient of $t^4$ is $c^2$. For this to equal some constant $K$ times $\big[(p-2)^2 + t^2\big]^2$, we need the numerator to have the form $K\big[(p-2)^4 + 2(p-2)^2 t^2 + t^4\big]$. Matching the $t^4$ coefficient: $c^2 = K$. Matching the constant term: $\big[2(p-2) - c(p-2)^2\big]^2 = K(p-2)^4 = c^2(p-2)^4$.

Taking the positive square root (for the numerator to factor correctly): $2(p-2) - c(p-2)^2 = c(p-2)^2$, giving $2(p-2) = 2c(p-2)^2$, so $c = \frac{1}{p-2}$.

With $c = \frac{1}{p-2}$, we verify: $2(p-2) - c(p-2)^2 = 2(p-2) - (p-2) = p-2$, and $c^2(p-2)^4 = (p-2)^2$. Indeed $(p-2)^2 = (p-2)^2$.

The numerator simplifies to $(p-2)^2 + \frac{t^4}{(p-2)^2} + \frac{2(p-2)^2 \cdot t^2}{(p-2)^2} = \frac{(p-2)^4 + 2(p-2)^2 t^2 + t^4}{(p-2)^2} = \frac{\big[(p-2)^2 + t^2\big]^2}{(p-2)^2}$

Therefore

$|w - c|^2 = \frac{1}{(p-2)^2} \qquad \Longrightarrow \qquad |w - c| = \frac{1}{|p - 2|}$

The image of the line $\operatorname{Re}(z) = p$ is the circle with centre $\dfrac{1}{p-2}$ and radius $\dfrac{1}{|p-2|}$.

Which $z$ on $V$ give $\operatorname{Im}(w) > 0$?

From above, $\operatorname{Im}(w) = \dfrac{-2t}{(p-2)^2 + t^2}$, which is positive if and only if $t < 0$. So $\operatorname{Im}(w) > 0$ for those points $z = p + ti$ on $V$ with $\operatorname{Im}(z) < 0$.

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Part (ii)

Let $z = t + qi$ with $t \in \mathbb{R}$ and $q \neq 0$. Then

$w = \frac{2}{(t - 2) + qi} = \frac{2\big[(t - 2) - qi\big]}{(t - 2)^2 + q^2}$

So

$\operatorname{Re}(w) = \frac{2(t - 2)}{(t - 2)^2 + q^2}, \qquad \operatorname{Im}(w) = \frac{-2q}{(t - 2)^2 + q^2}$

We seek a real constant $c$ such that $|w - ci|^2$ is independent of $t$. We have

$|w - ci|^2 = \frac{4(t-2)^2}{\big[(t-2)^2 + q^2\big]^2} + \left(\frac{-2q}{(t-2)^2 + q^2} - c\right)^2$

$= \frac{4(t-2)^2 + \big[-2q - c(t-2)^2 - cq^2\big]^2}{\big[(t-2)^2 + q^2\big]^2}$

Expanding the second term in the numerator: $\big[(-2q - cq^2) - c(t-2)^2\big]^2$. For the numerator to be a constant multiple of $\big[(t-2)^2 + q^2\big]^2$, we need the $(t-2)^2$ cross-term to produce $2q^2$ when combined with the $4(t-2)^2$ term.

Setting $c = -\frac{1}{q}$: then $-2q - cq^2 = -2q + q = -q$, so

$\text{Numerator} = 4(t-2)^2 + \left(-q + \frac{(t-2)^2}{q}\right)^2 = 4(t-2)^2 + \frac{\big[(t-2)^2 - q^2\big]^2}{q^2}$

$= \frac{4q^2(t-2)^2 + (t-2)^4 - 2q^2(t-2)^2 + q^4}{q^2} = \frac{(t-2)^4 + 2q^2(t-2)^2 + q^4}{q^2} = \frac{\big[(t-2)^2 + q^2\big]^2}{q^2}$

Therefore

$|w - ci|^2 = \frac{1}{q^2} \qquad \Longrightarrow \qquad |w - ci| = \frac{1}{|q|}$

The image of the line $\operatorname{Im}(z) = q$ is the circle with centre $-\dfrac{1}{q}\,i$ and radius $\dfrac{1}{|q|}$.

Which $z$ on $H$ give $\operatorname{Re}(w) > 0$?

From above, $\operatorname{Re}(w) = \dfrac{2(t-2)}{(t-2)^2 + q^2}$, which is positive if and only if $t > 2$. So $\operatorname{Re}(w) > 0$ for those points $z = t + qi$ on $H$ with $\operatorname{Re}(z) > 2$.

Examiner Notes

无官方评述。易错点：复数模的计算（需有理化分母）、圆心半径公式的推导、p=2时映射退化的处理、映射区域的正确判断。

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Question 8

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

Problem

8 In this question, $f(x)$ is a quartic polynomial where the coefficient of $x^4$ is equal to 1, and which has four real roots, 0, $a$, $b$ and $c$, where $0 < a < b < c$.

$F(x)$ is defined by $F(x) = \int_0^x f(t) \, dt$.

The area enclosed by the curve $y = f(x)$ and the $x$-axis between 0 and $a$ is equal to that between $b$ and $c$, and half that between $a$ and $b$.

(i) Sketch the curve $y = F(x)$, showing the $x$ co-ordinates of its turning points.

Explain why $F(x)$ must have the form $F(x) = \frac{1}{5}x^2(x - c)^2(x - h)$, where $0 < h < c$.

Find, in factorised form, an expression for $F(x) + F(c - x)$ in terms of $c$, $h$ and $x$.

(ii) If $0 \leqslant x \leqslant c$, explain why $F(b) + F(x) \geqslant 0$ and why $F(b) + F(x) > 0$ if $x \neq a$. Hence show that $c - b = a$ or $c > 2h$.

By considering also $F(a) + F(x)$, show that $c = a + b$ and that $c = 2h$.

(iii) Find an expression for $f(x)$ in terms of $c$ and $x$ only.

Show that the points of inflection on $y = f(x)$ lie on the $x$-axis.

Hint

8(i)

Graph: Zeroes at $x = 0, c$ and one other point ($h$: label not required) in $(a, b)$.

Graph: Turning points at $x = 0, a, b, c$.

Graph: Quintic shape with curve below axis in $(0, h)$ and above axis in $(h, c)$

The area conditions give $F(0) = F(c) = 0$. $F'(x) = f(x)$, so $F'(0) = F'(a) = F'(b) = F'(c) = 0$

Since $f$ is a quartic and the coefficient of $x^4$ is 1, $F$ must be a quintic and the coefficient of $x^5$ is $\frac{1}{5}$. $F(0) = F'(0) = 0$ and $F(c) = F'(c) = 0$, so $F$ must have double roots at $x = 0$ and $c$. So $F(x)$ must have the given form.

[Explanation must be clear that the double roots are deduced from the fact that $F(x) = F'(x) = 0$ at those points.]

$F(x) + F(c - x) = \frac{1}{5}x^2(x - c)^2[(x - h) + (c - x - h)]$ $= \frac{1}{5}x^2(x - c)^2(c - 2h)$

8(ii)

Let $A$ be the (positive) area enclosed by the curve between 0 and $a$. The maximum turning point of $F(x)$ occurs at $x = b$, with $F(b) = A$. The minimum turning point of $F(x)$ occurs at $x = a$, with $F(a) = -A$.

Therefore $F(x) \geq -A$, with equality iff $x = a$. So $F(b) + F(x) \geq 0$, with equality iff $x = a$.

$F(a) + F(x) \leq 0$, with equality iff $x = b$.

Since $F(b) + F(c - b) = \frac{1}{5}b^2(c - b)^2(c - 2h)$, either $c > 2h$, or $c = 2h$ and $c - b = a$.

Also, $F(a) + F(c - a) = \frac{1}{5}a^2(c - a)^2(c - 2h)$, so either $c < 2h$, or $c = 2h$ and $c - a = b$.

Thus $c = a + b$ and $c = 2h$.

8(iii)

$F(x) = \frac{1}{10}x^2(x - c)^2(2x - c)$ So $f(x) = \frac{1}{5}x(x - c)(5x^2 - 5xc + c^2)$

The roots of the quadratic factor must be $a$ and $b$.

$f(x) = \frac{1}{5}(5x^4 - 10cx^3 + 6c^2x^2 - c^3x)$ $f'(x) = \frac{1}{5}(20x^3 - 30cx^2 + 12c^2x)$ $f''(x) = \frac{1}{5}(60x^2 - 60cx + 12c^2) = \frac{12}{5}(5x^2 - 5cx + c^2)$

Therefore $f''(x) = 0$ at $x = a$ and $x = b$ and so $(a, 0)$ and $(b, 0)$ are points of inflection.

Model Solution

Part (i)

Sketch of $y = F(x)$:

The quartic $f(x)$ has leading coefficient 1 and roots $0, a, b, c$ with $0 < a < b < c$, so

$f(x) = x(x - a)(x - b)(x - c)$

Since $f$ is positive for $x < 0$ (four negative factors), negative on $(0, a)$, positive on $(a, b)$, negative on $(b, c)$, and positive for $x > c$.

$F(x) = \int_0^x f(t)\,dt$, so $F(0) = 0$ and $F'(x) = f(x)$. The turning points of $F$ are at $x = 0, a, b, c$ (where $f = 0$).

Let $A$ denote the positive area enclosed between $y = f(x)$ and the $x$-axis on $[0, a]$. The area conditions state:

• Area on $[0, a]$ = Area on $[b, c]$
• Area on $[0, a]$ = $\frac{1}{2}$ Area on $[a, b]$

Since $f < 0$ on $(0, a)$: $F(a) = \int_0^a f(t)\,dt = -A$.

Since $f > 0$ on $(a, b)$: $F(b) = F(a) + \int_a^b f(t)\,dt = -A + 2A = A$.

Since $f < 0$ on $(b, c)$: $F(c) = F(b) + \int_b^c f(t)\,dt = A - A = 0$.

So $F$ has a minimum at $x = a$ with value $-A$, a maximum at $x = b$ with value $A$, and $F(0) = F(c) = 0$. The curve dips below the axis on $(0, h)$ and rises above on $(h, c)$, where $h$ is the unique zero of $F$ in $(0, c)$ (with $a < h < b$).

Why $F(x) = \frac{1}{5}x^2(x - c)^2(x - h)$:

$F$ is the integral of a quartic with leading coefficient 1, so $F$ is a quintic with leading coefficient $\frac{1}{5}$.

At $x = 0$: $F(0) = 0$ and $F'(0) = f(0) = 0$, so $F$ has a double root at $x = 0$.

At $x = c$: $F(c) = 0$ and $F'(c) = f(c) = 0$, so $F$ has a double root at $x = c$.

Since $F$ is a quintic with leading coefficient $\frac{1}{5}$ and double roots at $0$ and $c$, it must have the form

$F(x) = \frac{1}{5}x^2(x - c)^2(x - h)$

for some constant $h$. The remaining root is $x = h$. Since $F(a) = -A < 0$ and $F(b) = A > 0$, by the intermediate value theorem $F$ has a zero between $a$ and $b$, so $a < h < b$, and in particular $0 < h < c$.

Computing $F(x) + F(c - x)$:

$F(c - x) = \frac{1}{5}(c - x)^2(c - x - c)^2(c - x - h) = \frac{1}{5}(c - x)^2 x^2(c - x - h)$

Note $(c - x)^2 = (x - c)^2$, so

$F(x) + F(c - x) = \frac{1}{5}x^2(x - c)^2\big[(x - h) + (c - x - h)\big] = \frac{1}{5}x^2(x - c)^2(c - 2h)$

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Part (ii)

Why $F(b) + F(x) \geqslant 0$:

We established that $F(x) \geqslant -A$ for all $x \in [0, c]$, with $F(x) = -A$ if and only if $x = a$ (the unique global minimum of $F$ on $[0, c]$). Since $F(b) = A$:

$F(b) + F(x) = A + F(x) \geqslant A + (-A) = 0$

with equality if and only if $x = a$. So $F(b) + F(x) > 0$ for $x \neq a$.

Showing $c - b = a$ or $c > 2h$:

Evaluate the identity from part (i) at $x = b$:

$F(b) + F(c - b) = \frac{1}{5}b^2(c - b)^2(c - 2h)$

Since $b \in (0, c)$, we have $c - b \in (0, c)$, so $F(b) + F(c - b) \geqslant 0$.

The left side is $\geqslant 0$, and $b^2(c-b)^2 > 0$, so $c - 2h \geqslant 0$, i.e., $c \geqslant 2h$.

If $c > 2h$: this is one of the two alternatives. If $c = 2h$: then $F(b) + F(c-b) = 0$, which requires $c - b = a$ (the only point where equality holds). So either $c > 2h$, or $c = 2h$ and $c - b = a$.

Considering $F(a) + F(x)$:

Similarly, $F(x) \leqslant A$ for all $x \in [0, c]$, with $F(x) = A$ if and only if $x = b$. Since $F(a) = -A$:

$F(a) + F(x) = -A + F(x) \leqslant 0$

with equality if and only if $x = b$.

Evaluate the identity at $x = a$:

$F(a) + F(c - a) = \frac{1}{5}a^2(c - a)^2(c - 2h)$

Since $a \in (0, c)$, we have $c - a \in (0, c)$, so $F(a) + F(c - a) \leqslant 0$.

Since $a^2(c-a)^2 > 0$, we need $c - 2h \leqslant 0$, i.e., $c \leqslant 2h$.

Combining both results:

From the $F(b) + F(x)$ analysis: $c \geqslant 2h$.

From the $F(a) + F(x)$ analysis: $c \leqslant 2h$.

Therefore $c = 2h$.

When $c = 2h$, the identity $F(x) + F(c-x) = 0$ for all $x$. In particular:

From $F(b) + F(c - b) = 0$: since equality requires $c - b = a$, we get $c = a + b$.

From $F(a) + F(c - a) = 0$: since equality requires $c - a = b$, we again get $c = a + b$.

Therefore $c = a + b$ and $c = 2h$.

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Part (iii)

Expression for $f(x)$ in terms of $c$ and $x$:

With $h = \frac{c}{2}$ (from $c = 2h$):

$F(x) = \frac{1}{5}x^2(x - c)^2\left(x - \frac{c}{2}\right) = \frac{1}{10}x^2(x - c)^2(2x - c)$

Differentiating:

$f(x) = F'(x) = \frac{1}{10}\frac{d}{dx}\Big[x^2(x-c)^2(2x-c)\Big]$

Using the product rule with $u = x^2(x-c)^2$ and $v = 2x - c$:

$u' = 2x(x-c)^2 + 2x^2(x-c) = 2x(x-c)\big[(x-c) + x\big] = 2x(x-c)(2x-c)$

$f(x) = \frac{1}{10}\Big[2x(x-c)(2x-c)(2x-c) + x^2(x-c)^2 \cdot 2\Big]$

$= \frac{1}{10}\Big[2x(x-c)(2x-c)^2 + 2x^2(x-c)^2\Big]$

$= \frac{2x(x-c)}{10}\Big[(2x-c)^2 + x(x-c)\Big]$

$= \frac{x(x-c)}{5}\Big[4x^2 - 4cx + c^2 + x^2 - cx\Big]$

$= \frac{x(x-c)}{5}(5x^2 - 5cx + c^2)$

So

$f(x) = \frac{1}{5}x(x - c)(5x^2 - 5cx + c^2)$

Points of inflection lie on the $x$-axis:

The points of inflection of $f$ occur where $f''(x) = 0$. First compute $f'(x)$:

$f(x) = \frac{1}{5}(5x^4 - 10cx^3 + 6c^2 x^2 - c^3 x)$

$f'(x) = \frac{1}{5}(20x^3 - 30cx^2 + 12c^2 x - c^3)$

$f''(x) = \frac{1}{5}(60x^2 - 60cx + 12c^2) = \frac{12}{5}(5x^2 - 5cx + c^2)$

Setting $f''(x) = 0$:

$5x^2 - 5cx + c^2 = 0$

$x = \frac{5c \pm \sqrt{25c^2 - 20c^2}}{10} = \frac{5c \pm c\sqrt{5}}{10} = \frac{c(5 \pm \sqrt{5})}{10}$

But the quadratic $5x^2 - 5cx + c^2$ is exactly the same factor appearing in $f(x) = \frac{1}{5}x(x-c)(5x^2 - 5cx + c^2)$. Therefore $f''(x) = 0$ precisely when $5x^2 - 5cx + c^2 = 0$, which means at those values of $x$, the factor $5x^2 - 5cx + c^2 = 0$, so

$f(x) = \frac{1}{5}x(x - c) \cdot 0 = 0$

The roots $a = \frac{c(5 - \sqrt{5})}{10}$ and $b = \frac{c(5 + \sqrt{5})}{10}$ satisfy $0 < a < c$ and $0 < b < c$ (since $\sqrt{5} \approx 2.236$, we get $a \approx 0.276c$ and $b \approx 0.724c$), confirming they are the roots of $f$ in $(0, c)$.

Therefore the points of inflection occur at $x = a$ and $x = b$, and at both points $f(a) = f(b) = 0$. The points of inflection $(a, 0)$ and $(b, 0)$ lie on the $x$-axis.

Examiner Notes

无官方评述。易错点：F(x)形式的推导需仔细利用面积条件和根的位置、F(b)+F(x)≥0的证明需要分析F的符号、c=2h的推导过程较长易出错。