STEP2 2020 -- Pure Mathematics

STEP2 2020 — Section A (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.

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

	=
	=
	=

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)	$\frac{1 - ky}{y} \frac{dy}{dx} = \frac{kx - 1}{x}$
	$\ln	y	- ky = kx - \ln	x	+ c$
	Hence, $\ln	xy	= k(x + y) + c$

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

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

	In both cases, the equation is invariant under $(x, y) \mapsto (y, x)$, so symmetrical in $y = x$.
2(ii)
	**Graphs**: Correct shapes of curves
	**Graphs**: Intersections at (2,4) and (4,16)
	$(x - y)^2 \geq 0$, so $(x + y)^2 > 2^{x+y}$
	Therefore, $(x + y)$ must lie between 2 and 4
	**Graph**: Symmetry about $y = x$
	**Graph**: Closed curve lying between $x + y = 3 \pm 1$
	**Graph**: 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 $y = x$
	Graph: Passes through (-1,-1)
	Graph: $y \to 0$ 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 of the terms are positive, these imply that $u_k^2 < u_{k-1}u_{k+1}$, so the sequence does not have property L.
	Therefore, if the sequence has property L, once a value $k$ has been reached such that $u_{k-1} \ge u_k$, it must be the case that all subsequent terms also have that property (which is the given definition of 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 that $u_r > 0$ for all $r$ if $u_2 > \alpha u_1 > 0$.
	Since the right hand side of the second identity is always non-negative, the sequence has property L, and is hence unimodal.
3(iii)	$u_1 = (2 - 1)\alpha^{1-1} + 2(1 - 1)\alpha^{1-2} = 1$, which is correct. $u_2 = (2 - 2)\alpha^{2-1} + 2(2 - 1)\alpha^{2-2} = 2$, which is correct.
	Suppose that: $u_{k-2} = (4 - k)\alpha^{k-3} + 2(k - 3)\alpha^{k-4}$, and $u_{k-1} = (3 - k)\alpha^{k-2} + 2(k - 2)\alpha^{k-3}$.
	$u_k = 2\alpha\left((3 - k)\alpha^{k-2} + 2(k - 2)\alpha^{k-3}\right) - \alpha^2((4 - k)\alpha^{k-3} + 2(k - 3)\alpha^{k-4})$ $= \alpha^{k-1}(6 - 2k - 4 + k) + \alpha^{k-2}(4k - 8 - 2k + 6)$ $= \alpha^{k-1}(2 - k) + 2\alpha^{k-2}(k - 1)$ which is the correct expression for $u_k$
	Hence, by induction $u_r = (2 - r)\alpha^{r-1} + 2(r - 1)\alpha^{r-2}$
	$u_r - u_{r+1} = \left((2 - r)\alpha^{r-1} + 2(r - 1)\alpha^{r-2}\right) - \left((1 - r)\alpha^r + 2r\alpha^{r-1}\right)$
	$= \alpha^{r-2}\left(2(r - 1) + (2 - 3r)\alpha + (r - 1)\alpha^2\right)$
	$= \frac{\alpha^{r-2}}{N^2}\left(2N^2(r - 1) + (2 - 3r)N(N - 1) + (r - 1)(N - 1)^2\right)$
	$= \frac{\alpha^{r-2}}{N^2}\left((r - 1) + rN - N^2\right)$
	when $r = N$, $u_N - u_{N+1} = \frac{\alpha^{r-2}(N - 1)}{N^2} > 0$
	when $r = N - 1$, $u_{N-1} - u_N = \frac{-2\alpha^{r-2}}{N^2} < 0$
	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 between the points.
4(ii)
	Diagram showing two circles and straight line joining their centres. Length of line and radii of circles are $a$, $b$ and $c$ in some order.
	Either statement that the straight line is the longest of the lengths, or explanation that one circle cannot be contained inside the other.
	Explanation that the circles must meet.
4(iii)	(A) If $a + b > c$ then $(a + 1) + (b + 1) > c + 2 > c + 1$ et cycl., so $a + 1$, $b + 1$, $c + 1$ can always form the sides of a triangle.
	(B) If $a = b = c = 1$ we have $1, 1, 1$ which can form the sides of a triangle.
	If $a = 1, b = c = 2$ we have $\frac{1}{2}, 1, 2$ which cannot form the sides of a triangle.
	Therefore, $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ can sometimes, but not always form the sides of a triangle.
	(C) If $p \ge q \ge r$ then $|p - q| + |q - r| = p - q + q - r = p - r = |p - r|$
	So two of $|p - q|, |q - r|, |p - r|$ will always sum to the third, so they never form the sides of a triangle.
	(D) If $a + b > c$ then $a^2 + bc + b^2 + ca = a^2 + b^2 - 2ab + c(a + b) + 2ab$
	$= (a - b)^2 + c(a + b) + 2ab > c^2 + ab$ et cycl. so $a^2 + bc, b^2 + ca, c^2 + ab$ can always form the sides of a triangle.
4(iv)	Since $a + b > a$ and $b$, $\frac{f(a)}{a} > \frac{f(a+b)}{a+b}$ and $\frac{f(b)}{b} > \frac{f(a+b)}{a+b}$
	Since $c < a + b, f(c) < f(a + b)$
	Thus $f(a) + f(b) > \frac{af(a)}{a+b} + \frac{bf(b)}{a+b} = f(a + b) > f(c)$ et cycl. So $f(a), f(b)$ and $f(c)$ can form the sides of a triangle.

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)	$x - q(x) = \sum_{r=0}^{n-1} a_r \times 10^r - \sum_{r=0}^{n-1} a_r = \sum_{r=0}^{n-1} a_r \times (10^r - 1)$
	$10^r \ge 1 \quad \forall r$, so $x - q(x)$ is non-negative
	$9 | (10^r - 1) \quad \forall r$
5(ii)	$x - 44q(x) = 44(x - q(x)) - 43x$
	So it is a multiple of 9 iff $43x$ is.
	$(43, 9) = 1$, so $x - 44q(x)$ is a multiple of 9 iff $x$ is
	If $x$ has $n$ digits, $q(x) \le 9n$
	Since $x = 44q(x)$, $x \le 396n$. Any $n$ digit number must be at least $10^{n-1}$.
	These inequalities cannot be simultaneously true for $n \ge 5$ ($396 \times 5 < 10^4$). Therefore $n \le 4$.
	Since $x - 44q(x) = 0$, which is a multiple of 9, $x$ is a multiple of 9.
	$q(x)$ is an integer and $x = 44q(x)$, so $x$ is a multiple of 44. Since $(9, 44) = 1$, $x$ must be a multiple of $44 \times 9 = 396$.
	So $x = 396k$ and therefore (by the result above) $k \le 4$.
	Checking: Only $k = 2$ works.
5(iii)	$x - 107q(q(x)) = 0 = 107(x - q(x)) + 107(q(x) - q(q(x))) - 106x$
	$(x - q(x))$ and $(q(x) - q(q(x)))$ are both divisible by 9 (by part (i)) and so $x$ is divisible by 9
	$x = 107q(q(x))$ and so is divisible by 107, and so is divisible by 963. So $x = 963k$ for some $k$.
	If $x$ has $n$ digits, then $q(x) \le 9n$. By (i), $q(q(x)) \le q(x) \le 9n$. So $x \le 963n$ and $x \ge 10^{n-1}$ which implies that $n \le 4$ and so $k \le 4$
	Checking: Only $k = 1$ works.

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) Let $\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$; then $\mathbf{M}^2 = \begin{pmatrix} a\tau - \delta & b\tau \\ c\tau & d\tau - \delta \end{pmatrix}$, where $\tau = \text{Tr}(\mathbf{M})$ and $\delta = \text{Det}(\mathbf{M})$. 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 $\delta = -1$ Thus $\mathbf{M}^2 = +\mathbf{I} \Leftrightarrow \tau = 0$ and $\delta = -1$. Thus $\mathbf{M}^2 = -\mathbf{I} \Leftrightarrow \tau = 0$ and $\delta = +1$. 6(iii) Part (ii) implies $\text{Det}(\mathbf{M}^2) = -1$, if $\mathbf{M}^4 = \mathbf{I}$, but $\mathbf{M}^2 \neq \pm\mathbf{I}$. However, $\text{Det}(\mathbf{M}^2) = \text{Det}(\mathbf{M})^2$, so this is impossible. Clearly $\mathbf{M}^2 = \pm\mathbf{I} \Rightarrow \mathbf{M}^4 = \mathbf{I}$ Part (ii) implies that $\mathbf{M}^4 = -\mathbf{I} \Leftrightarrow \text{Tr}(\mathbf{M}^2) = 0$ and $\text{Det}(\mathbf{M}^2) = 1$ so from (i) $\Leftrightarrow \text{Tr}(\mathbf{M})^2 = 2\text{Det}(\mathbf{M})$ and $\text{Det}(\mathbf{M}) = \pm 1$ so $\Leftrightarrow \text{Tr}(\mathbf{M}) = \pm\sqrt{2}$
and $\text{Det}(\mathbf{M}) = 1$. Any example, for instance a matrix satisfying the conditions for any of $\mathbf{M}^2 = \mathbf{I}$, $\mathbf{M}^2 = -\mathbf{I}$, $\mathbf{M}^4 = -\mathbf{I}$, 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.

---

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.

9 If the particles collide at time $t$: $Vt + Ut \cos \theta = d$, and $h - \frac{1}{2}gt^2 = Ut \sin \theta - \frac{1}{2}gt^2$ (or $h = Ut \sin \theta$)

Therefore, $d \sin \theta - h \cos \theta = Vt \sin \theta + Ut \sin \theta \cos \theta - Ut \sin \theta \cos \theta$

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。