STEP3 2012 -- Pure Mathematics

STEP3 2012 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2012  |  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	微分方程 Differential Equations	Challenging	链式法则, 变量替换降阶, 初值问题求解
2	无穷级数与无穷乘积 Infinite Series and Products	Challenging	裂项消去, 对数微分, 无穷乘积转级数, 部分分式分解
3	坐标几何与曲线分析 Coordinate Geometry and Curve Analysis	Challenging	曲线作图, 联立方程, 因式分解, 参数分类讨论
4	无穷级数求和 Infinite Series Summation	Challenging	泰勒展开, 阶乘级数求和, 部分分式分解, 裂项消去
5	数论与有理点 Number Theory and Rational Points	Challenging	三角代换, 有理点参数化, 二次扩域, 构造性证明
6	复数与复平面 Complex Numbers and Argand Diagram	Standard	复数实部虚部分离,参数消去法,根轨迹分析,阿尔冈图作图
7	微分方程 Differential Equations	Challenging	方程组化高阶ODE,特征方程法,初值与边值求解,无穷级数求和
8	数列与级数 Sequences and Series	Challenging	Fibonacci递推关系,分奇偶讨论,反正切加法公式,telescoping级数

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Question 1

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

Problem

1 Given that $z = y^n \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2$, show that

$\frac{\mathrm{d}z}{\mathrm{d}x} = y^{n-1} \frac{\mathrm{d}y}{\mathrm{d}x} \left( n \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 + 2y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right) .$

(i) Use the above result to show that the solution to the equation

$\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 + 2y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \sqrt{y} \quad \quad (y > 0)$

that satisfies $y = 1$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$ is $y = \left( \frac{3}{8}x^2 + 1 \right)^{\frac{2}{3}}$.

(ii) Find the solution to the equation

$\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 - y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + y^2 = 0$

that satisfies $y = 1$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$.

Hint

1. Considering either $[(F_n F_{n+3} - F_{n+1} F_{n+2}) - (F_{n-2} F_{n+1} - F_{n-1} F_n)]$ or $(F_n F_{n+3} - F_{n+1} F_{n+2}) + (F_{n-1} F_{n+2} - F_n F_{n+1})$, and applying the recurrence relation, both can be found to be zero. \so the given expression is shown to be 1 if $n$ is odd and -1 if $n$ is even. The tan compound angle formula enables the proof in part (iii) to be completed once the initial recurrence relation and the result from part (ii) have been applied to the expression obtained following algebraic simplification. Rearranging the result and substituting into the required sum gives, by the method of differences, $\sum_{r=1}^{\infty} \tan^{-1} \left( \frac{1}{F_{2r+1}} \right) = \frac{\pi}{4}$

Model Solution

Establishing the derivative formula

Given $z = y^n \left(\frac{dy}{dx}\right)^2$, we differentiate using the product rule:

$\frac{dz}{dx} = ny^{n-1}\frac{dy}{dx}\left(\frac{dy}{dx}\right)^2 + y^n \cdot 2\frac{dy}{dx}\frac{d^2y}{dx^2}$

$= y^{n-1}\frac{dy}{dx}\left[n\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2}\right]$

Part (i)

The equation is $\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} = \sqrt{y}$.

Comparing with the bracket in our formula, set $n = 1$ so that $z = y\left(\frac{dy}{dx}\right)^2$ and:

$\frac{dz}{dx} = \frac{dy}{dx}\left[\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2}\right] = \frac{dy}{dx}\sqrt{y}$

With the initial conditions $y = 1$, $\frac{dy}{dx} = 0$ at $x = 0$, we have $z(0) = 1 \cdot 0 = 0$.

Since $z = y(y')^2$, we can write $\frac{dz}{dx} = y'\sqrt{y}$. But $z = y(y')^2$ means $y' = \pm\sqrt{z/y}$, so:

$\frac{dz}{dx} = \pm\frac{\sqrt{z}}{\sqrt{y}} \cdot \sqrt{y} = \pm\sqrt{z}$

Wait — let me reconsider. From $z = y(y')^2$ we get $y' = \pm\sqrt{z/y}$, and the equation gives $\frac{dz}{dx} = y'\sqrt{y}$. Since $y' = \pm\sqrt{z/y}$:

$\frac{dz}{dx} = \pm\sqrt{z/y} \cdot \sqrt{y} = \pm\sqrt{z}$

Taking the positive branch (since we need $z \geq 0$ and the equation is consistent):

$\frac{dz}{dx} = \sqrt{z}$

Separating variables: $\frac{dz}{\sqrt{z}} = dx$, so $2\sqrt{z} = x + C$. At $x = 0$: $z = 0$, so $C = 0$.

$2\sqrt{z} = x \implies z = \frac{x^2}{4}$

Since $z = y(y')^2 = \frac{x^2}{4}$, we have $y(y')^2 = \frac{x^2}{4}$.

Now set $w = y^{3/2}$. Then $\frac{dw}{dx} = \frac{3}{2}y^{1/2}y'$, so $(y')^2 = \frac{4}{9y}\left(\frac{dw}{dx}\right)^2$.

Substituting: $y \cdot \frac{4}{9y}\left(\frac{dw}{dx}\right)^2 = \frac{x^2}{4}$, giving $\left(\frac{dw}{dx}\right)^2 = \frac{9x^2}{16}$, so $\frac{dw}{dx} = \frac{3x}{4}$ (taking positive root).

Integrating: $w = \frac{3x^2}{8} + C$. At $x = 0$: $w = y(0)^{3/2} = 1$, so $C = 1$.

$y^{3/2} = \frac{3x^2}{8} + 1 = \frac{3x^2 + 8}{8}$

$y = \left(\frac{3x^2 + 8}{8}\right)^{2/3} = \left(\frac{3}{8}x^2 + 1\right)^{2/3}$

Part (ii)

The equation is $\left(\frac{dy}{dx}\right)^2 - y\frac{d^2y}{dx^2} + y^2 = 0$, with $y = 1$, $y' = 0$ at $x = 0$.

Using the formula with $n = -1$: $z = y^{-1}(y')^2 = \frac{(y')^2}{y}$.

$\frac{dz}{dx} = y^{-2}y'\left[-(y')^2 + 2y y''\right] = \frac{y'}{y^2}\left[2yy'' - (y')^2\right]$

From the equation: $(y')^2 + y^2 = yy''$, so $yy'' = (y')^2 + y^2$, giving $2yy'' = 2(y')^2 + 2y^2$.

Therefore: $2yy'' - (y')^2 = (y')^2 + 2y^2$.

$\frac{dz}{dx} = \frac{y'}{y^2}\left[(y')^2 + 2y^2\right] = \frac{y'}{y^2}(y')^2 + 2y' = \frac{(y')^3}{y^2} + 2y'$

Since $z = (y')^2/y$, we have $(y')^2 = zy$, so $(y')^3 = y' \cdot zy$.

$\frac{dz}{dx} = \frac{y' \cdot zy}{y^2} + 2y' = \frac{zy'}{y} + 2y'$

From $z = (y')^2/y$: $y' = \pm\sqrt{zy}$. This is getting complicated. Let me try a different substitution.

Alternative approach for part (ii): Let $p = \frac{dy}{dx}$. The equation is $p^2 - yp\frac{dp}{dy} + y^2 = 0$ (using $y'' = p\frac{dp}{dy}$).

Rearranging: $yp\frac{dp}{dy} = p^2 + y^2$.

If $p \neq 0$: $y\frac{dp}{dy} = p + \frac{y^2}{p}$.

Let $p = yv$: $y\left(v + y\frac{dv}{dy}\right) = yv + \frac{y}{v}$, so $y^2\frac{dv}{dy} = \frac{y}{v}$, giving $y\frac{dv}{dy} = \frac{1}{v}$.

Separating: $v\,dv = \frac{dy}{y}$, so $\frac{v^2}{2} = \ln y + C$.

At $x = 0$: $y = 1$, $p = 0$, so $v = p/y = 0$. Thus $C = 0$.

$v^2 = 2\ln y \implies \frac{p^2}{y^2} = 2\ln y \implies p^2 = 2y^2\ln y$

Since $p = dy/dx$: $\frac{dy}{dx} = \pm y\sqrt{2\ln y}$.

Separating: $\frac{dy}{y\sqrt{2\ln y}} = \pm dx$. Let $u = \sqrt{2\ln y}$, so $u^2 = 2\ln y$, $2u\,du = \frac{2}{y}dy$, giving $\frac{dy}{y} = u\,du$.

$\frac{u\,du}{u} = \pm dx \implies du = \pm dx \implies u = \pm x + C$

At $x = 0$: $y = 1$, $u = 0$, so $C = 0$.

$\sqrt{2\ln y} = x \quad (\text{taking positive branch for } x > 0)$

$2\ln y = x^2 \implies \ln y = \frac{x^2}{2} \implies y = e^{x^2/2}$

By symmetry ($x$ appears as $x^2$ in the equation), this works for all $x$:

$y = e^{x^2/2}$

Verification: $y' = xe^{x^2/2}$, $y'' = (1 + x^2)e^{x^2/2}$.

$(y')^2 - yy'' + y^2 = x^2 e^{x^2} - e^{x^2/2}(1+x^2)e^{x^2/2} + e^{x^2} = x^2 e^{x^2} - (1+x^2)e^{x^2} + e^{x^2} = 0$. $\checkmark$

Examiner Notes

In spite of the printing error at the start of the question, two thirds of the candidates attempted this question. Most candidates earned a quarter of the marks by obtaining $z$ in terms of $y$ in part (i), and then went no further. Some candidates realised the significance of the first line of the question that the expression given was an exact differential, and those that did frequently then scored highly. Some candidates found their own way through having obtained $z$ in terms of $y$ in part (i), then making $y$ the subject substituted back to find a second order differential equation for $z$, which they then solved and hence completed the solution to each part.

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Question 2

Topic: 无穷级数与无穷乘积 Infinite Series and Products  |  Difficulty: Challenging  |  Marks: 20

Problem

2 In this question, $|x| < 1$ and you may ignore issues of convergence.

(i) Simplify

$(1 - x)(1 + x)(1 + x^2)(1 + x^4) \cdots (1 + x^{2^n}) ,$

where $n$ is a positive integer, and deduce that

$\frac{1}{1 - x} = (1 + x)(1 + x^2)(1 + x^4) \cdots (1 + x^{2^n}) + \frac{x^{2^{n+1}}}{1 - x} .$

Deduce further that

$\ln(1 - x) = - \sum_{r=0}^{\infty} \ln \left( 1 + x^{2^r} \right) ,$

and hence that

$\frac{1}{1 - x} = \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \cdots .$

(ii) Show that

$\frac{1 + 2x}{1 + x + x^2} = \frac{1 - 2x}{1 - x + x^2} + \frac{2x - 4x^3}{1 - x^2 + x^4} + \frac{4x^3 - 8x^7}{1 - x^4 + x^8} + \cdots .$

Hint

2. The simplification in the opening is $(1 - x^{2^{n+1}})$, obtained by repeated use of the difference of two squares. A simple algebraic rearrangement, followed by taking a limit, the logarithm of both expressions, and differentiation produces the other three results in part (i). Part (ii) can be obtained by replacing $x$ by $x^3$ in $\ln(1 - x) = -\sum_{r=0}^{\infty} \ln(1 + x^{2^r})$ from part (i), factorising the difference and the sums of cubes and subtracting that part (i) result before differentiating. An alternative is to replicate part (i) using instead the product

$(1 + x + x^2)(1 - x + x^2)(1 - x^2 + x^4)(1 - x^4 + x^8) \dots (1 - x^{2^n} + x^{2^{n+1}})$, but then a little extra care is required with the rearrangement, and consideration of the limit.

Model Solution

Part (i)

Simplification: Using the difference of two squares repeatedly:

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

Continuing: $(1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) = 1 - x^{2^{n+1}}$.

First deduction: Dividing by $(1-x)$:

$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) = \frac{1 - x^{2^{n+1}}}{1-x} = \frac{1}{1-x} - \frac{x^{2^{n+1}}}{1-x}$

Therefore:

$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) + \frac{x^{2^{n+1}}}{1-x}$

Second deduction: Taking logarithms of the product identity:

$\ln(1-x) = \ln\left[(1-x)(1+x)(1+x^2)\cdots(1+x^{2^n})\right] = \ln(1 - x^{2^{n+1}})$

Wait — let me be more careful. From $(1-x)\prod_{r=0}^{n}(1+x^{2^r}) = 1 - x^{2^{n+1}}$:

$\ln(1-x) + \sum_{r=0}^{n}\ln(1+x^{2^r}) = \ln(1-x^{2^{n+1}})$

As $n \to \infty$ with $|x| < 1$: $x^{2^{n+1}} \to 0$, so $\ln(1-x^{2^{n+1}}) \to 0$.

$\ln(1-x) = -\sum_{r=0}^{\infty}\ln(1+x^{2^r})$

Final result: Differentiating both sides:

$\frac{-1}{1-x} = -\sum_{r=0}^{\infty}\frac{2^r x^{2^r - 1}}{1+x^{2^r}}$

$\frac{1}{1-x} = \sum_{r=0}^{\infty}\frac{2^r x^{2^r - 1}}{1+x^{2^r}} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots$

Part (ii)

We want to show:

$\frac{1+2x}{1+x+x^2} = \frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} + \cdots$

Strategy: Replace $x$ by $x^3$ in the identity $\ln(1-x) = -\sum_{r=0}^{\infty}\ln(1+x^{2^r})$ from part (i):

$\ln(1-x^3) = -\sum_{r=0}^{\infty}\ln(1+x^{3\cdot 2^r})$

Now use $1 - x^3 = (1-x)(1+x+x^2)$:

$\ln(1-x) + \ln(1+x+x^2) = -\sum_{r=0}^{\infty}\ln(1+x^{3\cdot 2^r})$

Subtracting the part (i) identity $\ln(1-x) = -\sum_{r=0}^{\infty}\ln(1+x^{2^r})$:

$\ln(1+x+x^2) = -\sum_{r=0}^{\infty}\ln(1+x^{3\cdot 2^r}) + \sum_{r=0}^{\infty}\ln(1+x^{2^r})$

$= \sum_{r=0}^{\infty}\left[\ln(1+x^{2^r}) - \ln(1+x^{3\cdot 2^r})\right]$

Using $1 + x^{3\cdot 2^r} = (1+x^{2^r})(1 - x^{2^r} + x^{2^{r+1}})$ (sum of cubes: $a^3 + b^3 = (a+b)(a^2-ab+b^2)$ with $a = x^{2^r}$, $b = 1$):

Wait, $1 + (x^{2^r})^3 = (1 + x^{2^r})(1 - x^{2^r} + x^{2\cdot 2^r})$. So $1 + x^{3\cdot 2^r} = (1+x^{2^r})(1 - x^{2^r} + x^{2^{r+1}})$.

Therefore $\ln(1+x^{3\cdot 2^r}) = \ln(1+x^{2^r}) + \ln(1 - x^{2^r} + x^{2^{r+1}})$.

$\ln(1+x+x^2) = \sum_{r=0}^{\infty}\left[-\ln(1 - x^{2^r} + x^{2^{r+1}})\right]$

$= -\ln(1-x+x^2) - \ln(1-x^2+x^4) - \ln(1-x^4+x^8) - \cdots$

So:

$\ln(1+x+x^2) + \ln(1-x+x^2) = -\ln(1-x^2+x^4) - \ln(1-x^4+x^8) - \cdots$

Note that $(1+x+x^2)(1-x+x^2) = (1+x^2)^2 - x^2 = 1 + x^2 + x^4$. Hmm, that’s not quite right. Let me compute: $(1+x+x^2)(1-x+x^2) = 1 - x + x^2 + x - x^2 + x^3 + x^2 - x^3 + x^4 = 1 + x^2 + x^4$.

So $\ln(1+x^2+x^4) = -\ln(1-x^2+x^4) - \ln(1-x^4+x^8) - \cdots$

Hmm, this doesn’t directly give what we want. Let me try a different approach.

Direct approach: Differentiate $\ln(1+x+x^2) = -\sum_{r=0}^{\infty}\ln(1-x^{2^r}+x^{2^{r+1}})$.

LHS: $\frac{d}{dx}\ln(1+x+x^2) = \frac{1+2x}{1+x+x^2}$.

RHS: $-\sum_{r=0}^{\infty}\frac{d}{dx}\ln(1-x^{2^r}+x^{2^{r+1}}) = -\sum_{r=0}^{\infty}\frac{-2^r x^{2^r-1} + 2^{r+1}x^{2^{r+1}-1}}{1-x^{2^r}+x^{2^{r+1}}}$

$= \sum_{r=0}^{\infty}\frac{2^r x^{2^r-1} - 2^{r+1}x^{2^{r+1}-1}}{1-x^{2^r}+x^{2^{r+1}}}$

For $r = 0$: $\frac{1 - 2x}{1-x+x^2}$

For $r = 1$: $\frac{2x - 4x^3}{1-x^2+x^4}$

For $r = 2$: $\frac{4x^3 - 8x^7}{1-x^4+x^8}$

Therefore:

$\frac{1+2x}{1+x+x^2} = \frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} + \cdots$ $\blacksquare$

Examiner Notes

Three quarter s of the candidates attempted this question, making it the second most popular, and one of the two most successful. Generally, part (i) was successful attempted, though at times marks were dropped through insufficient explanation and quite a few struggled to deal with the “remainder term”. Some candidates expanding the brackets worked with the second and third, the fourth brackets etc., only including the first bracket last. About half the candidates considered the product of all of the denominators in part (ii) and replicated the method for the first part, whilst others used the results from part (i), replacing $x$ by $x^3$ and employing the factorisations of sums and differences of cubes. Full marks were not uncommon on this question, nor were half marks.

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Question 3

Topic: 坐标几何与曲线分析 Coordinate Geometry and Curve Analysis  |  Difficulty: Challenging  |  Marks: 20

Problem

3 It is given that the two curves $y = 4 - x^2 \quad \text{and} \quad mx = k - y^2,$ where $m > 0$, touch exactly once.

(i) In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes: (a) $k < 0$; (b) $0 < k < 16, k/m < 2$; (c) $k > 16, k/m > 2$; (d) $k > 16, k/m < 2$.

(ii) Now set $m = 12$. Show that the $x$-coordinate of any point at which the two curves meet satisfies $x^4 - 8x^2 + 12x + 16 - k = 0.$ Let $a$ be the value of $x$ at the point where the curves touch. Show that $a$ satisfies $a^3 - 4a + 3 = 0$ and hence find the three possible values of $a$. Derive also the equation $k = -4a^2 + 9a + 16.$ Which of the four sketches in part (i) arise?

Hint

3. The two parabolas, with vertices oriented in the direction of the positive axes, touch in the third quadrant in case (a), and in the first quadrant in the other three cases. In case (b), there are intersections in the second and third quadrants, in (c) in the third and fourth, and in (d), the trickiest case, they are in the third quadrant and in the first, between the touching point and the vertex of the parabola on the $x$ axis. The first result of part (ii)is obtained by eliminating $y$ between the two equations, the second by differentiating and equating gradients, and the third, by eliminating $x^4$, $(a^4 = aa^3)$, from the first result using the second.

The cases that arise are $a = 1, m = 21, \frac{k}{m} = \frac{21}{12} < 2$ (d), $a = \frac{-1+\sqrt{13}}{2}, k = \frac{13}{2}\sqrt{13} - \frac{5}{2}, \frac{k}{m} < 2$ (d), and $a = \frac{-1-\sqrt{13}}{2}, k = \frac{-13}{2}\sqrt{13} - \frac{5}{2}, k < 0$ (a).

Model Solution

Part (i)

The first curve $y = 4 - x^2$ is a downward-opening parabola with vertex $(0, 4)$, crossing the $x$-axis at $x = \pm 2$.

The second curve $mx = k - y^2$ can be written as $x = \frac{k - y^2}{m}$, which is a leftward-opening parabola with vertex $\left(\frac{k}{m}, 0\right)$. It crosses the $y$-axis ($x = 0$) at $y = \pm\sqrt{k}$ when $k > 0$.

(a) $k < 0$: The second parabola has vertex $\left(\frac{k}{m}, 0\right)$ with $\frac{k}{m} < 0$ and no $y$-intercepts. The two curves touch once in the third quadrant.

(b) $0 < k < 16$, $k/m < 2$: The second parabola has vertex $\left(\frac{k}{m}, 0\right)$ with $0 < \frac{k}{m} < 2$, between the origin and the right $x$-intercept $(2, 0)$ of the first parabola. It crosses the $y$-axis at $y = \pm\sqrt{k}$ with $0 < \sqrt{k} < 4$. The curves touch once in the first quadrant and also intersect in the second and third quadrants.

(c) $k > 16$, $k/m > 2$: The second parabola has vertex $\left(\frac{k}{m}, 0\right)$ with $\frac{k}{m} > 2$, to the right of $(2, 0)$. It crosses the $y$-axis at $y = \pm\sqrt{k}$ with $\sqrt{k} > 4$. The curves touch once in the first quadrant and also intersect in the third and fourth quadrants.

(d) $k > 16$, $k/m < 2$: The second parabola has vertex $\left(\frac{k}{m}, 0\right)$ with $\frac{k}{m} < 2$ but $\sqrt{k} > 4$. The curves touch once in the first quadrant. They also intersect in the third quadrant and in the first quadrant between the touching point and the vertex of the second parabola.

Part (ii)

Showing the quartic. Substituting $y = 4 - x^2$ into $12x = k - y^2$:

$12x = k - (4 - x^2)^2 = k - (16 - 8x^2 + x^4) = k - 16 + 8x^2 - x^4.$

Rearranging:

$x^4 - 8x^2 + 12x + 16 - k = 0. \qquad \text{(*)}$

Showing the cubic. At the touching point $(a, b)$, the gradients are equal. From $y = 4 - x^2$:

$\frac{\mathrm{d}y}{\mathrm{d}x} = -2x.$

From $12x = k - y^2$, differentiating implicitly:

$12 = -2y\frac{\mathrm{d}y}{\mathrm{d}x} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{6}{y}.$

Equating gradients at $x = a$, $y = b = 4 - a^2$:

$-2a = -\frac{6}{4 - a^2}.$

Since $a \neq 0$ (otherwise $0 \neq -6/4$), rearranging gives $2a(4 - a^2) = 6$, i.e., $4a - a^3 = 3$, so:

$a^3 - 4a + 3 = 0.$

Finding the three values of $a$. Testing $a = 1$: $1 - 4 + 3 = 0$, so $(a - 1)$ is a factor:

$a^3 - 4a + 3 = (a - 1)(a^2 + a - 3) = 0.$

The quadratic gives $a = \frac{-1 \pm \sqrt{13}}{2}$, so the three values are $a = 1$, $a = \frac{-1 + \sqrt{13}}{2}$, $a = \frac{-1 - \sqrt{13}}{2}$.

Deriving $k = -4a^2 + 9a + 16$. From (*), $k = a^4 - 8a^2 + 12a + 16$. Using $a^3 = 4a - 3$:

$a^4 = a \cdot a^3 = a(4a - 3) = 4a^2 - 3a.$

Substituting:

$k = (4a^2 - 3a) - 8a^2 + 12a + 16 = -4a^2 + 9a + 16.$

Which sketches arise?

• $a = 1$: $k = -4 + 9 + 16 = 21 > 16$ and $\frac{k}{m} = \frac{21}{12} = \frac{7}{4} < 2$. This is case (d).

• $a = \frac{-1 + \sqrt{13}}{2}$: $a^2 = \frac{7 - \sqrt{13}}{2}$, so $k = -2(7 - \sqrt{13}) + \frac{9(-1 + \sqrt{13})}{2} + 16 = \frac{-5 + 13\sqrt{13}}{2} \approx 20.9$. Since $k > 16$ and $\frac{k}{m} \approx 1.74 < 2$, this is case (d).

• $a = \frac{-1 - \sqrt{13}}{2}$: $a^2 = \frac{7 + \sqrt{13}}{2}$, so $k = -2(7 + \sqrt{13}) + \frac{9(-1 - \sqrt{13})}{2} + 16 = \frac{-5 - 13\sqrt{13}}{2} \approx -25.9$. Since $k < 0$, this is case (a).

The sketches that arise are case (d) (twice) and case (a) (once).

Examiner Notes

Two thirds of candidates attempted this question, but generally, with only moderate success earning just less than half marks. The vast majority of candidates (more than 85%) did not observe that, regardless of the case, the two parabolas “touch exactly once”, dropping 4 or 5 marks immediately. However, most managed to obtain the three results in part (ii), though a few seemed to forget to derive that for $k$. Unaccountably, many threw away the final marks, only considering the case $a = 1$.

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Question 4

Topic: 无穷级数求和 Infinite Series Summation  |  Difficulty: Challenging  |  Marks: 20

Problem

4 (i) Show that $\sum_{n=1}^{\infty} \frac{n+1}{n!} = 2e - 1$ and $\sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} = 5e - 1.$ Sum the series $\displaystyle \sum_{n=1}^{\infty} \frac{(2n-1)^3}{n!}$.

(ii) Sum the series $\displaystyle \sum_{n=0}^{\infty} \frac{(n^2+1)2^{-n}}{(n+1)(n+2)}$, giving your answer in terms of natural logarithms.

Hint

4. Writing $\frac{n+1}{n!}$ as $\frac{n}{n!} + \frac{1}{n!}$ and then cancelling the first fraction, give exponential series. Similarly, $(n + 1)^2$ can be written as $n(n - 1) + 3n + 1$, and $(2n - 1)^3$ as $8n(n - 1)(n - 2) + 12n(n - 1) + 2n - 1$, the latter giving the result $21e + 1$. Using partial fractions, $\frac{n^2+1}{(n+1)(n+2)}$ can be written as $1 + \frac{2}{n+1} - \frac{5}{n+2}$, the first term giving a GP, and the other two, log series. The result for part (ii) is thus $12 - 16 \ln 2$.

Model Solution

Part (i)

First sum. We split the fraction:

$\sum_{n=1}^{\infty} \frac{n+1}{n!} = \sum_{n=1}^{\infty} \frac{n}{n!} + \sum_{n=1}^{\infty} \frac{1}{n!}.$

For the first sum, $\frac{n}{n!} = \frac{1}{(n-1)!}$, so re-indexing with $m = n - 1$:

$\sum_{n=1}^{\infty} \frac{1}{(n-1)!} = \sum_{m=0}^{\infty} \frac{1}{m!} = e.$

For the second sum:

$\sum_{n=1}^{\infty} \frac{1}{n!} = e - 1.$

Therefore $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n!} = e + (e - 1) = 2e - 1$.

Second sum. We expand $(n+1)^2 = n^2 + 2n + 1$ and write $n^2 = n(n-1) + n$:

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

So:

$\sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} = \sum_{n=1}^{\infty} \frac{n(n-1)}{n!} + 3\sum_{n=1}^{\infty} \frac{n}{n!} + \sum_{n=1}^{\infty} \frac{1}{n!}.$

Since $\frac{n(n-1)}{n!} = \frac{1}{(n-2)!}$ (zero for $n = 1$, valid from $n = 2$):

$\sum_{n=2}^{\infty} \frac{1}{(n-2)!} = \sum_{m=0}^{\infty} \frac{1}{m!} = e.$

Using the results from the first sum:

$\sum_{n=1}^{\infty} \frac{(n+1)^2}{n!} = e + 3e + (e - 1) = 5e - 1.$

Third sum. We expand $(2n-1)^3 = 8n^3 - 12n^2 + 6n - 1$ and write $n^3$ in terms of falling factorials:

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

So:

$(2n-1)^3 = 8[n(n-1)(n-2) + 3n(n-1) + n] - 12[n(n-1) + n] + 6n - 1$ $= 8n(n-1)(n-2) + 24n(n-1) + 8n - 12n(n-1) - 12n + 6n - 1$ $= 8n(n-1)(n-2) + 12n(n-1) + 2n - 1.$

Therefore:

$\sum_{n=1}^{\infty} \frac{(2n-1)^3}{n!} = 8\sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 12\sum_{n=1}^{\infty} \frac{1}{(n-1)!} + 2\sum_{n=1}^{\infty} \frac{1}{(n-1)!} - \sum_{n=1}^{\infty} \frac{1}{n!}$ $= 8e + 12e + 2e - (e - 1) = 21e + 1.$

Part (ii)

Partial fractions. We write:

$\frac{n^2+1}{(n+1)(n+2)} = A + \frac{B}{n+1} + \frac{C}{n+2}.$

Multiplying through by $(n+1)(n+2)$: $n^2 + 1 = A(n+1)(n+2) + B(n+2) + C(n+1)$.

Setting $n = -1$: $2 = B$. Setting $n = -2$: $5 = -C$, so $C = -5$. Comparing coefficients of $n^2$: $A = 1$.

So:

$\frac{n^2+1}{(n+1)(n+2)} = 1 + \frac{2}{n+1} - \frac{5}{n+2}.$

Splitting the series. With $x = \frac{1}{2}$:

$S = \sum_{n=0}^{\infty} \frac{n^2+1}{(n+1)(n+2)} \cdot x^n = \underbrace{\sum_{n=0}^{\infty} x^n}_{S_1} + 2\underbrace{\sum_{n=0}^{\infty} \frac{x^n}{n+1}}_{S_2} - 5\underbrace{\sum_{n=0}^{\infty} \frac{x^n}{n+2}}_{S_3}.$

Evaluating $S_1$: This is a geometric series:

$S_1 = \frac{1}{1 - x} = \frac{1}{1 - 1/2} = 2.$

Evaluating $S_2$: From the Maclaurin series $-\ln(1 - x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$, we have $\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = -\ln(1-x)$, so:

$S_2 = \sum_{n=0}^{\infty} \frac{x^n}{n+1} = \frac{1}{x} \cdot (-\ln(1-x)) = \frac{-\ln(1-x)}{x}.$

At $x = \frac{1}{2}$: $S_2 = \frac{-\ln(1/2)}{1/2} = 2\ln 2$.

Evaluating $S_3$: Since $\sum_{n=0}^{\infty} \frac{x^{n+2}}{n+2} = -\ln(1-x) - x$ (the log series missing its first term):

$S_3 = \sum_{n=0}^{\infty} \frac{x^n}{n+2} = \frac{-\ln(1-x) - x}{x^2}.$

At $x = \frac{1}{2}$: $S_3 = \frac{-\ln(1/2) - 1/2}{1/4} = 4\left(\ln 2 - \frac{1}{2}\right) = 4\ln 2 - 2$.

Combining:

$S = 2 + 2(2\ln 2) - 5(4\ln 2 - 2) = 2 + 4\ln 2 - 20\ln 2 + 10 = 12 - 16\ln 2.$

Examiner Notes

Just over 70% of the candidates attempted this, with marginally less success than question 3. Lots of attempts relied on manipulating series for $e$, and would have struggled had the first two results not been given, and even so, there were varying levels of success and conviction. This approach fell apart in the this part with the cubic term. Some candidates used a generating function method successfully with $an(x) = \sum_{n=1}^{\infty} \frac{n+1}{n!} x^n$. However, whilst this worked well for part (i), it got very nasty for part (ii). There were lots of sign errors with the log series in part (ii), having begun well with partial fractions.

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Question 5

Topic: 数论与有理点 Number Theory and Rational Points  |  Difficulty: Challenging  |  Marks: 20

Problem

5 (i) The point with coordinates $(a, b)$, where $a$ and $b$ are rational numbers, is called: an integer rational point if both $a$ and $b$ are integers; a non-integer rational point if neither $a$ nor $b$ is an integer.

(a) Write down an integer rational point and a non-integer rational point on the circle $x^2 + y^2 = 1$. (b) Write down an integer rational point on the circle $x^2 + y^2 = 2$. Simplify $(\cos \theta + \sqrt{m} \sin \theta)^2 + (\sin \theta - \sqrt{m} \cos \theta)^2$ and hence obtain a non-integer rational point on the circle $x^2 + y^2 = 2$.

(ii) The point with coordinates $(p + \sqrt{2}q, r + \sqrt{2}s)$, where $p, q, r$ and $s$ are rational numbers, is called: an integer 2-rational point if all of $p, q, r$ and $s$ are integers; a non-integer 2-rational point if none of $p, q, r$ and $s$ is an integer.

(a) Write down an integer 2-rational point, and obtain a non-integer 2-rational point, on the circle $x^2 + y^2 = 3$. (b) Obtain a non-integer 2-rational point on the circle $x^2 + y^2 = 11$. (c) Obtain a non-integer 2-rational point on the hyperbola $x^2 - y^2 = 7$.

Hint

5. For non-integer rational points, it makes sense to use values of $\cos \theta$ and $\sin \theta$ based on Pythagorean triples such as 3, 4, 5 or 5, 12,

Model Solution

Part (i)(a)

An integer rational point on $x^2 + y^2 = 1$: $(1, 0)$.

A non-integer rational point: $\left(\dfrac{3}{5}, \dfrac{4}{5}\right)$, since $\dfrac{9}{25} + \dfrac{16}{25} = 1$.

Part (i)(b)

An integer rational point on $x^2 + y^2 = 2$: $(1, 1)$, since $1 + 1 = 2$.

Simplifying the expression:

$(\cos\theta + \sqrt{m}\sin\theta)^2 + (\sin\theta - \sqrt{m}\cos\theta)^2$

$= \cos^2\theta + 2\sqrt{m}\sin\theta\cos\theta + m\sin^2\theta + \sin^2\theta - 2\sqrt{m}\sin\theta\cos\theta + m\cos^2\theta$

$= (\cos^2\theta + \sin^2\theta) + m(\sin^2\theta + \cos^2\theta) = 1 + m.$

For this to equal $2$, we need $m = 1$.

To find a non-integer rational point, we choose $\cos\theta = \dfrac{3}{5}$, $\sin\theta = \dfrac{4}{5}$ (from the 3-4-5 Pythagorean triple). With $m = 1$:

$x = \cos\theta + \sin\theta = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}, \qquad y = \sin\theta - \cos\theta = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}.$

Check: $\left(\dfrac{7}{5}\right)^2 + \left(\dfrac{1}{5}\right)^2 = \dfrac{49}{25} + \dfrac{1}{25} = 2$. $\checkmark$

A non-integer rational point is $\left(\dfrac{7}{5}, \dfrac{1}{5}\right)$.

Part (ii)(a)

An integer 2-rational point on $x^2 + y^2 = 3$: $(1 + \sqrt{2}, 1 - \sqrt{2})$ (here $p = 1, q = 1, r = 1, s = -1$, all integers).

Check: $(1 + \sqrt{2})^2 + (1 - \sqrt{2})^2 = (1 + 2\sqrt{2} + 2) + (1 - 2\sqrt{2} + 2) = 6$. This gives $6$, not $3$.

Let us try $(1, 1 + \sqrt{2})$: $1 + (1 + \sqrt{2})^2 = 1 + 3 + 2\sqrt{2} = 4 + 2\sqrt{2}$. No.

Try $(1 + \sqrt{2}, 1)$: $(1 + \sqrt{2})^2 + 1 = 3 + 2\sqrt{2} + 1 = 4 + 2\sqrt{2}$. No.

We need $(p + \sqrt{2}q)^2 + (r + \sqrt{2}s)^2 = 3$, i.e., $p^2 + 2q^2 + r^2 + 2s^2 + 2\sqrt{2}(pq + rs) = 3$.

So $pq + rs = 0$ and $p^2 + 2q^2 + r^2 + 2s^2 = 3$.

Take $p = 1, q = 1, r = 1, s = -1$: $pq + rs = 1 - 1 = 0$, $p^2 + 2q^2 + r^2 + 2s^2 = 1 + 2 + 1 + 2 = 6 \neq 3$.

Take $p = 1, q = 0, r = 0, s = 0$: gives $(1, 0)$ with $1 + 0 = 1 \neq 3$.

Take $p = 1, q = 0, r = \sqrt{2}$ — but $r$ must be rational/integer.

We need integer solutions to $p^2 + 2q^2 + r^2 + 2s^2 = 3$ and $pq + rs = 0$.

Try $q = 0, s = 0$: $p^2 + r^2 = 3$. No integer solution.

Try $p = 1, q = 1$: then $rs = -1$, so $(r,s) = (1,-1)$ or $(-1,1)$. Then $1 + 2 + 1 + 2 = 6 \neq 3$.

Try $p = 1, q = -1$: $rs = 1$, $(r,s) = (1,1)$. Same sum: $6$.

Try smaller: $p = 1, q = 0, r = 1, s = 0$: $1 + 0 + 1 + 0 = 2 \neq 3$.

$p = 0, q = 0, r = 1, s = 0$: $0 + 0 + 1 + 0 = 1$.

$p = 0, q = 1, r = 1, s = 0$: $0 + 2 + 1 + 0 = 3$. And $pq + rs = 0 + 0 = 0$. $\checkmark$

So $(0 + \sqrt{2} \cdot 1, 1 + \sqrt{2} \cdot 0) = (\sqrt{2}, 1)$.

Check: $2 + 1 = 3$. $\checkmark$

But this is $(\sqrt{2}, 1)$, which has $p = 0, q = 1, r = 1, s = 0$ — all integers, so this is an integer 2-rational point.

For a non-integer 2-rational point, we use the rotation trick. Take $(\sqrt{2}, 1)$ and rotate by choosing rational values for $\cos\theta$ and $\sin\theta$.

Let $\cos\theta = \dfrac{3}{5}$, $\sin\theta = \dfrac{4}{5}$. The transformation:

$(x', y') = (x\cos\theta + y\sin\theta,\; -x\sin\theta + y\cos\theta)$

preserves $x'^2 + y'^2 = x^2 + y^2 = 3$.

$x' = \sqrt{2}\cdot\frac{3}{5} + 1\cdot\frac{4}{5} = \frac{3\sqrt{2} + 4}{5}, \qquad y' = -\sqrt{2}\cdot\frac{4}{5} + 1\cdot\frac{3}{5} = \frac{3 - 4\sqrt{2}}{5}.$

So $x' = \dfrac{4}{5} + \sqrt{2}\cdot\dfrac{3}{5}$ and $y' = \dfrac{3}{5} + \sqrt{2}\cdot\left(-\dfrac{4}{5}\right)$.

Here $p = \dfrac{4}{5}, q = \dfrac{3}{5}, r = \dfrac{3}{5}, s = -\dfrac{4}{5}$ — all non-integers. $\checkmark$

A non-integer 2-rational point is $\left(\dfrac{4 + 3\sqrt{2}}{5}, \dfrac{3 - 4\sqrt{2}}{5}\right)$.

Part (ii)(b)

We need $(p + \sqrt{2}q)^2 + (r + \sqrt{2}s)^2 = 11$ with $p, q, r, s$ all non-integer rationals.

First find an integer 2-rational point. We need $p^2 + 2q^2 + r^2 + 2s^2 = 11$ and $pq + rs = 0$.

Try $p = 1, q = 2, r = 0, s = 0$: $1 + 8 + 0 + 0 = 9 \neq 11$.

Try $p = 3, q = 1, r = 0, s = 0$: $9 + 2 = 11$, $pq + rs = 3 \neq 0$.

Try $p = 1, q = 0, r = 1, s = 2$: $1 + 0 + 1 + 8 = 10 \neq 11$.

Try $p = 1, q = 0, r = 3, s = 0$: $1 + 9 = 10$.

Try $p = 1, q = 1, r = 2, s = 1$: $1 + 2 + 4 + 2 = 9$, $pq + rs = 1 + 2 = 3 \neq 0$.

Try $p = 1, q = 2, r = 2, s = -1$: $1 + 8 + 4 + 2 = 15$, too big.

Try $p = 1, q = 1, r = 0, s = 2$: $1 + 2 + 0 + 8 = 11$, $pq + rs = 1 + 0 = 1 \neq 0$.

Try $p = 3, q = 1, r = 1, s = -3$: $9 + 2 + 1 + 18 = 30$.

Try $p = 0, q = 1, r = 3, s = 0$: $0 + 2 + 9 + 0 = 11$, $pq + rs = 0 + 0 = 0$. $\checkmark$

So $(\sqrt{2}, 3)$ is an integer 2-rational point on $x^2 + y^2 = 11$.

Now rotate with $\cos\theta = \dfrac{3}{5}$, $\sin\theta = \dfrac{4}{5}$:

$x' = \sqrt{2}\cdot\frac{3}{5} + 3\cdot\frac{4}{5} = \frac{12 + 3\sqrt{2}}{5}, \qquad y' = -\sqrt{2}\cdot\frac{4}{5} + 3\cdot\frac{3}{5} = \frac{9 - 4\sqrt{2}}{5}.$

Here $p = \dfrac{12}{5}, q = \dfrac{3}{5}, r = \dfrac{9}{5}, s = -\dfrac{4}{5}$ — all non-integers.

A non-integer 2-rational point is $\left(\dfrac{12 + 3\sqrt{2}}{5}, \dfrac{9 - 4\sqrt{2}}{5}\right)$.

Part (ii)(c)

We need $(p + \sqrt{2}q)^2 - (r + \sqrt{2}s)^2 = 7$ with all non-integer.

Expanding: $p^2 + 2q^2 - r^2 - 2s^2 + 2\sqrt{2}(pq - rs) = 7$.

So $pq - rs = 0$ and $p^2 + 2q^2 - r^2 - 2s^2 = 7$.

First find an integer 2-rational point. Try $p = 3, q = 0, r = \sqrt{2}, s = 0$ — no, $r$ must be integer.

Try $p = 3, q = 0, r = 1, s = 0$: $9 - 1 = 8 \neq 7$.

Try $p = 3, q = 0, r = 0, s = 1$: $9 - 2 = 7$, $pq - rs = 0 - 0 = 0$. $\checkmark$

So $(3, \sqrt{2})$ is an integer 2-rational point on $x^2 - y^2 = 7$.

Now rotate. For hyperbolas, we use a hyperbolic rotation. Alternatively, use the parametrization approach. Consider the point $(3, \sqrt{2})$. Let us parametrize using $\cosh t$ and $\sinh t$:

$(x, y) = (3\cosh t + \sqrt{2}\sinh t,\; 3\sinh t + \sqrt{2}\cosh t).$

Check: $x^2 - y^2 = (3\cosh t + \sqrt{2}\sinh t)^2 - (3\sinh t + \sqrt{2}\cosh t)^2$.

Expanding: $9\cosh^2 t + 6\sqrt{2}\cosh t\sinh t + 2\sinh^2 t - 9\sinh^2 t - 6\sqrt{2}\sinh t\cosh t - 2\cosh^2 t$

$= 7\cosh^2 t - 7\sinh^2 t = 7$. $\checkmark$

For a non-integer 2-rational point, choose $t$ such that $\cosh t$ and $\sinh t$ are rational. Take $\cosh t = \dfrac{5}{4}$, $\sinh t = \dfrac{3}{4}$ (since $\cosh^2 t - \sinh^2 t = \dfrac{25}{16} - \dfrac{9}{16} = 1$).

$x = 3\cdot\frac{5}{4} + \sqrt{2}\cdot\frac{3}{4} = \frac{15 + 3\sqrt{2}}{4}, \qquad y = 3\cdot\frac{3}{4} + \sqrt{2}\cdot\frac{5}{4} = \frac{9 + 5\sqrt{2}}{4}.$

Here $p = \dfrac{15}{4}, q = \dfrac{3}{4}, r = \dfrac{9}{4}, s = \dfrac{5}{4}$ — all non-integers.

A non-integer 2-rational point is $\left(\dfrac{15 + 3\sqrt{2}}{4}, \dfrac{9 + 5\sqrt{2}}{4}\right)$.

Examiner Notes

This was only very slightly more popular than question 4, though with the same level of success. A lot of candidates scored just the first 5 marks, getting as far as completing the simplification in part (i) (b), but then, being unable to apply it for the final result, and then making no progress with part (ii). The biggest problem was that candidates ignored the definitions given at the start of the question, most notably that “$a$ and $b$ are rational numbers”. The other common problem was that candidates chose a simple value for $\theta$ such as $\frac{\pi}{4}$ or $\frac{\pi}{3}$ rather than for $\cos \theta$ such as $\frac{4}{5}$. In part (ii), quite frequently, candidates substituted $x = p + \sqrt{2}q$, and $y = r + \sqrt{2}s$ and some then successfully found solutions. For part (ii) (c), a method using $\cosh \theta$ and $\sinh \theta$ was not unexpected, although the comparable one with $\sec \theta$ and $\tan \theta$ was quite commonly used too.

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Question 6

Topic: 复数与复平面 Complex Numbers and Argand Diagram  |  Difficulty: Standard  |  Marks: 20

Problem

6 Let $x + iy$ be a root of the quadratic equation $z^2 + pz + 1 = 0$, where $p$ is a real number. Show that $x^2 - y^2 + px + 1 = 0$ and $(2x + p)y = 0$. Show further that $\text{either } p = -2x \quad \text{or} \quad p = -(x^2 + 1)/x \text{ with } x \neq 0.$

Hence show that the set of points in the Argand diagram that can (as $p$ varies) represent roots of the quadratic equation consists of the real axis with one point missing and a circle. This set of points is called the root locus of the quadratic equation.

Obtain and sketch in the Argand diagram the root locus of the equation $pz^2 + z + 1 = 0$ and the root locus of the equation $pz^2 + p^2z + 2 = 0.$

Hint

6. Substituting $x + iy$ for $z$ in the quadratic equation and equating real and imaginary parts yields the first two results, the imaginary gives two situations, one as required and the other substituted into the real gives the second. The first of these two results substituted into the real gives a circle radius 1, centre the origin, whilst the second gives the real axis without the origin. The second quadratic equation succumbs to the same approach giving the real axis without the origin (again), and a circle centre $(-1,0)$ radius 1 also omitting the origin. The same approach in the third case yields the real axis with $p = \frac{-x^2 \pm \sqrt{x^4 - 8x}}{2x}$ and considering the discriminant, $x < 0$ and $x \geq 2$.

On the other hand, $p = -2x$ produces $y^2 = -\frac{x^3 + 1}{x}$

Model Solution

First equation: $z^2 + pz + 1 = 0$

Let $z = x + iy$ be a root. Substituting:

$(x+iy)^2 + p(x+iy) + 1 = 0$

$(x^2 - y^2 + px + 1) + i(2xy + py) = 0$

Equating real and imaginary parts:

$x^2 - y^2 + px + 1 = 0 \qquad \text{and} \qquad (2x + p)y = 0$

From the imaginary part: either $y = 0$ or $p = -2x$.

Case 1: $p = -2x$. Substituting into the real part:

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

This is the unit circle (excluding the origin, since $x^2 + y^2 = 1 \neq 0$).

Case 2: $y = 0$. Then $z = x$ is real, and the real part gives $x^2 + px + 1 = 0$, so $p = -(x^2+1)/x$ with $x \neq 0$.

This is the real axis with the origin removed (since $x = 0$ would require $p$ to be undefined, and indeed $z = 0$ is never a root since the constant term is 1).

Root locus: The unit circle $\cup$ the real axis minus $\{0\}$.

Second equation: $pz^2 + z + 1 = 0$

Substituting $z = x + iy$:

$p(x^2 - y^2) + x + 1 + i(2pxy + y) = 0$

Real: $p(x^2 - y^2) + x + 1 = 0$ Imaginary: $y(2px + 1) = 0$

Case 1: $2px + 1 = 0$, i.e., $p = -\frac{1}{2x}$ (with $x \neq 0$).

Substituting: $-\frac{x^2 - y^2}{2x} + x + 1 = 0$

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

$-x^2 + y^2 + 2x^2 + 2x = 0$

$x^2 + 2x + y^2 = 0 \implies (x+1)^2 + y^2 = 1$

This is the circle centre $(-1, 0)$, radius 1. The origin $(0,0)$ is on this circle but requires $p = -\frac{1}{0}$ which is undefined, so the origin is excluded.

Case 2: $y = 0$. Then $px^2 + x + 1 = 0$, so $p = -(x+1)/x^2$ with $x \neq 0$.

This gives the real axis minus the origin.

Root locus: Circle $(x+1)^2 + y^2 = 1$ minus $\{0\}$ $\cup$ real axis minus $\{0\}$.

Third equation: $pz^2 + p^2z + 2 = 0$

Substituting $z = x + iy$:

$p(x^2 - y^2) + p^2x + 2 + i(2pxy + p^2y) = 0$

Real: $p(x^2 - y^2) + p^2x + 2 = 0$ Imaginary: $py(2x + p) = 0$

Case 1: $p = 0$. Then $2 = 0$, contradiction. So $p \neq 0$.

Case 2: $y = 0$. Then $px^2 + p^2x + 2 = 0$. This is a quadratic in $p$: $xp^2 + x^2p + 2 = 0$.

$p = \frac{-x^2 \pm \sqrt{x^4 - 8x}}{2x}$

For real $p$: $x^4 - 8x \geq 0$, i.e., $x(x^3 - 8) \geq 0$. This gives $x \leq 0$ or $x \geq 2$.

At $x = 2$: $p = \frac{-4 \pm 0}{4} = -1$. At $x = 0$: no solution (denominator zero, and $2 = 0$ contradiction).

So the real axis part is $\{x : x < 0 \text{ or } x \geq 2\}$.

Case 3: $p = -2x$ (with $x \neq 0$). Substituting into the real part:

$-2x(x^2 - y^2) + 4x^2 \cdot x + 2 = 0$

$-2x^3 + 2xy^2 + 4x^3 + 2 = 0$

$2x^3 + 2xy^2 + 2 = 0$

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

For this to have solutions, need $x < 0$ (since $x^2 + y^2 \geq 0$). When $x < 0$: $x^2 + y^2 = |1/x|$, which is a curve in the left half-plane. At $x = -1$: $y^2 = 1 - 1 = 0$, so $(-1, 0)$ is on this curve. As $x \to 0^-$: $x^2 + y^2 \to \infty$. As $x \to -\infty$: $x^2 + y^2 \approx x^2$, so $y^2 \approx -1/x \to 0$.

This curve $x^2 + y^2 = -1/x$ combined with the real axis part $\{x < 0 \text{ or } x \geq 2\}$ gives the complete root locus.

Examiner Notes

Two thirds attempted this, with less success than its three predecessors. Very few indeed scored full marks, for even those that mastered the question rarely sketched the last locus correctly, putting in a non-existent cusp. Most candidates managed the first part, good ones the second part too, and only the very best the third part. Quite a few assumed the roots were complex and then used complex conjugates, with varying success. Many candidates lost marks through careless arithmetic and algebraic errors. Given that most could do the first part, it was possible for candidates to score reasonably if they took care and took real parts and imaginary parts correctly.

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Question 7

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

Problem

7 A pain-killing drug is injected into the bloodstream. It then diffuses into the brain, where it is absorbed. The quantities at time $t$ of the drug in the blood and the brain respectively are $y(t)$ and $z(t)$. These satisfy

$\dot{y} = -2(y - z) , \qquad \dot{z} = -\dot{y} - 3z ,$

where the dot denotes differentiation with respect to $t$.

Obtain a second order differential equation for $y$ and hence derive the solution

$y = Ae^{-t} + Be^{-6t} , \qquad z = \frac{1}{2}Ae^{-t} - 2Be^{-6t} ,$

where $A$ and $B$ are arbitrary constants.

(i) Obtain the solution that satisfies $z(0) = 0$ and $y(0) = 5$. The quantity of the drug in the brain for this solution is denoted by $z_1(t)$.

(ii) Obtain the solution that satisfies $z(0) = z(1) = c$, where $c$ is a given constant. The quantity of the drug in the brain for this solution is denoted by $z_2(t)$.

(iii) Show that for $0 \leqslant t \leqslant 1$,

$z_2(t) = \sum_{n=-\infty}^{0} z_1(t - n) ,$

provided $c$ takes a particular value that you should find.

Hint

7. The second order differential equation $\ddot{y} + 7\dot{y} + 6y = 0$ may be obtained by differentiating the first equation, then substituting for $\dot{z}$ using the other equation, and then doing likewise for $z$ using the first again. The solutions for $y$ and $z$ follow in the usual manner. Parts (i) and (ii) yield $z_1(t) = 2e^{-t} - 2e^{-6t}$ and $z_2(t) = \frac{c(e^6 - 1)}{(e^5 - 1)}e^{-t} - \frac{ce^5(e - 1)}{(e^5 - 1)}e^{-6t}$ respectively. Part (iii) merely requires the sum to be expanded and two geometric progressions emerge so that $c = \frac{2e(e^5 - 1)}{e - 1(e^6 - 1)}$.

Model Solution

Deriving the second-order ODE and general solution.

From the first equation, $\dot{y} = -2y + 2z$, so

$z = \frac{\dot{y} + 2y}{2} . \qquad \text{(*)}$

Differentiating the first equation with respect to $t$:

$\ddot{y} = -2(\dot{y} - \dot{z}) .$

Substituting $\dot{z} = -\dot{y} - 3z$ from the second given equation:

$\ddot{y} = -2\bigl(\dot{y} - (-\dot{y} - 3z)\bigr) = -2(2\dot{y} + 3z) = -4\dot{y} - 6z .$

Now substituting (*) for $z$:

$\ddot{y} = -4\dot{y} - 6 \cdot \frac{\dot{y} + 2y}{2} = -4\dot{y} - 3\dot{y} - 6y = -7\dot{y} - 6y .$

So the second-order ODE is

$\ddot{y} + 7\dot{y} + 6y = 0 .$

The characteristic equation is $\lambda^2 + 7\lambda + 6 = 0$, i.e. $(\lambda + 1)(\lambda + 6) = 0$, giving $\lambda = -1$ and $\lambda = -6$. Thus

$y = Ae^{-t} + Be^{-6t} .$

Then $\dot{y} = -Ae^{-t} - 6Be^{-6t}$, and from (*):

$z = \frac{-Ae^{-t} - 6Be^{-6t} + 2Ae^{-t} + 2Be^{-6t}}{2} = \frac{Ae^{-t} - 4Be^{-6t}}{2} = \frac{1}{2}Ae^{-t} - 2Be^{-6t} .$

Part (i)

We need $y(0) = 5$ and $z(0) = 0$.

$y(0) = A + B = 5$ and $z(0) = \tfrac{A}{2} - 2B = 0$, so $A = 4B$.

Substituting: $4B + B = 5$, giving $B = 1$ and $A = 4$.

$\boxed{z_1(t) = 2e^{-t} - 2e^{-6t}} .$

Part (ii)

We need $z(0) = c$ and $z(1) = c$.

$z(0) = \tfrac{A}{2} - 2B = c \qquad \Longrightarrow \qquad A = 2c + 4B . \qquad \text{(**)}$

$z(1) = \tfrac{A}{2}e^{-1} - 2Be^{-6} = c .$

Substituting (**) into the second equation:

$\frac{2c + 4B}{2}e^{-1} - 2Be^{-6} = c$

$(c + 2B)e^{-1} - 2Be^{-6} = c$

$ce^{-1} + 2Be^{-1} - 2Be^{-6} = c$

$2B(e^{-1} - e^{-6}) = c(1 - e^{-1})$

$B = \frac{c(1 - e^{-1})}{2(e^{-1} - e^{-6})} = \frac{c \cdot \frac{e - 1}{e}}{2 \cdot \frac{e^5 - 1}{e^6}} = \frac{c(e - 1)e^5}{2(e^5 - 1)} .$

From (**):

$A = 2c + \frac{2c(e - 1)e^5}{e^5 - 1} = \frac{2c(e^5 - 1) + 2c(e - 1)e^5}{e^5 - 1} = \frac{2c(e^6 - 1)}{e^5 - 1} .$

Therefore

$\boxed{z_2(t) = \frac{c(e^6 - 1)}{e^5 - 1}e^{-t} - \frac{c(e - 1)e^5}{e^5 - 1}e^{-6t}} .$

Part (iii)

Expanding the sum $\displaystyle\sum_{n=-\infty}^{0} z_1(t - n)$ for $0 \leqslant t \leqslant 1$:

$\sum_{n=-\infty}^{0} z_1(t - n) = \sum_{n=-\infty}^{0} \bigl(2e^{-(t-n)} - 2e^{-6(t-n)}\bigr) = 2e^{-t}\sum_{n=-\infty}^{0} e^{n} - 2e^{-6t}\sum_{n=-\infty}^{0} e^{6n} .$

Each sum is a geometric series (with $n = 0, -1, -2, \ldots$):

$\sum_{n=-\infty}^{0} e^{n} = 1 + e^{-1} + e^{-2} + \cdots = \frac{1}{1 - e^{-1}} = \frac{e}{e - 1} ,$

$\sum_{n=-\infty}^{0} e^{6n} = 1 + e^{-6} + e^{-12} + \cdots = \frac{1}{1 - e^{-6}} = \frac{e^6}{e^6 - 1} .$

So

$\sum_{n=-\infty}^{0} z_1(t - n) = \frac{2e}{e - 1}\,e^{-t} - \frac{2e^6}{e^6 - 1}\,e^{-6t} .$

Comparing with $z_2(t) = \dfrac{c(e^6 - 1)}{e^5 - 1}\,e^{-t} - \dfrac{c(e - 1)e^5}{e^5 - 1}\,e^{-6t}$, we equate coefficients.

From the $e^{-t}$ coefficient:

$\frac{c(e^6 - 1)}{e^5 - 1} = \frac{2e}{e - 1} \qquad \Longrightarrow \qquad c = \frac{2e(e^5 - 1)}{(e - 1)(e^6 - 1)} .$

We verify with the $e^{-6t}$ coefficient:

$\frac{c(e - 1)e^5}{e^5 - 1} = \frac{2e(e^5 - 1)}{(e - 1)(e^6 - 1)} \cdot \frac{(e - 1)e^5}{e^5 - 1} = \frac{2e \cdot e^5}{e^6 - 1} = \frac{2e^6}{e^6 - 1} \checkmark$

Both coefficients match, so $z_2(t) = \displaystyle\sum_{n=-\infty}^{0} z_1(t - n)$ for $0 \leqslant t \leqslant 1$ provided

$\boxed{c = \frac{2e(e^5 - 1)}{(e - 1)(e^6 - 1)}} .$

Examiner Notes

Two thirds attempted this too, with marginally greater success than question 2. Most did very well with the stem, though a few were unable to obtain a proper second order equation. Those that attempted part (i) were usually successful. The non-trivial exponential calculations in part (ii) caused problems for some making computation mistakes whilst others were totally on top of this. Part (iii) tested the candidates on two levels, interpreting the sigma notation correctly, and recognising and using the geometric series. Some managed this excellently.

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Question 8

Topic: 数列与级数 Sequences and Series  |  Difficulty: Challenging  |  Marks: 20

Problem

8 The sequence $F_0, F_1, F_2, \dots$ is defined by $F_0 = 0, F_1 = 1$ and, for $n \geqslant 0$,

$F_{n+2} = F_{n+1} + F_n .$

(i) Show that $F_0F_3 - F_1F_2 = F_2F_5 - F_3F_4$.

(ii) Find the values of $F_nF_{n+3} - F_{n+1}F_{n+2}$ in the two cases that arise.

(iii) Prove that, for $r = 1, 2, 3, \dots$,

$\arctan \left( \frac{1}{F_{2r}} \right) = \arctan \left( \frac{1}{F_{2r+1}} \right) + \arctan \left( \frac{1}{F_{2r+2}} \right)$

and hence evaluate the following sum (which you may assume converges):

$\sum_{r=1}^{\infty} \arctan \left( \frac{1}{F_{2r+1}} \right) .$

Hint

8. $F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5$ so both expressions in part (i) equal -

Model Solution

The Fibonacci sequence begins $F_0 = 0,\; F_1 = 1,\; F_2 = 1,\; F_3 = 2,\; F_4 = 3,\; F_5 = 5,\; F_6 = 8,\; F_7 = 13, \ldots$

Part (i)

$F_0F_3 - F_1F_2 = 0 \cdot 2 - 1 \cdot 1 = -1$.

$F_2F_5 - F_3F_4 = 1 \cdot 5 - 2 \cdot 3 = 5 - 6 = -1$.

Both sides equal $-1$, so $F_0F_3 - F_1F_2 = F_2F_5 - F_3F_4$.

Part (ii)

We first simplify the expression $F_nF_{n+3} - F_{n+1}F_{n+2}$. Using $F_{n+3} = F_{n+2} + F_{n+1}$:

$F_nF_{n+3} - F_{n+1}F_{n+2} = F_n(F_{n+2} + F_{n+1}) - F_{n+1}F_{n+2} = F_nF_{n+2} + F_nF_{n+1} - F_{n+1}F_{n+2} .$

Rearranging:

$= F_nF_{n+2} - F_{n+1}(F_{n+2} - F_n) = F_nF_{n+2} - F_{n+1}^2 ,$

since $F_{n+2} - F_n = F_{n+1}$.

We now prove by induction that $F_nF_{n+2} - F_{n+1}^2 = (-1)^{n+1}$.

Base case ($n = 0$): $F_0F_2 - F_1^2 = 0 - 1 = -1 = (-1)^1$. ✓

Inductive step: Assume $F_nF_{n+2} - F_{n+1}^2 = (-1)^{n+1}$. Then

$F_{n+1}F_{n+3} - F_{n+2}^2 = F_{n+1}(F_{n+1} + F_{n+2}) - F_{n+2}(F_n + F_{n+1})$

$= F_{n+1}^2 + F_{n+1}F_{n+2} - F_nF_{n+2} - F_{n+1}F_{n+2} = F_{n+1}^2 - F_nF_{n+2} = -(-1)^{n+1} = (-1)^{n+2} .$

This completes the induction.

Therefore:

$\boxed{F_nF_{n+3} - F_{n+1}F_{n+2} = \begin{cases} -1 & \text{if } n \text{ is even,} \\ 1 & \text{if } n \text{ is odd.} \end{cases}}$

Part (iii)

We wish to prove that $\arctan\!\left(\dfrac{1}{F_{2r}}\right) = \arctan\!\left(\dfrac{1}{F_{2r+1}}\right) + \arctan\!\left(\dfrac{1}{F_{2r+2}}\right)$.

Using the addition formula $\arctan a + \arctan b = \arctan\!\left(\dfrac{a + b}{1 - ab}\right)$ (valid when $ab < 1$):

$\arctan\!\left(\frac{1}{F_{2r+1}}\right) + \arctan\!\left(\frac{1}{F_{2r+2}}\right) = \arctan\!\left(\frac{\frac{1}{F_{2r+1}} + \frac{1}{F_{2r+2}}}{1 - \frac{1}{F_{2r+1}F_{2r+2}}}\right) = \arctan\!\left(\frac{F_{2r+2} + F_{2r+1}}{F_{2r+1}F_{2r+2} - 1}\right) .$

The numerator is $F_{2r+2} + F_{2r+1} = F_{2r+3}$. For the denominator, from part (ii) with $n = 2r$ (even):

$F_{2r}F_{2r+3} - F_{2r+1}F_{2r+2} = -1 \qquad \Longrightarrow \qquad F_{2r+1}F_{2r+2} - 1 = F_{2r}F_{2r+3} .$

Therefore

$\arctan\!\left(\frac{1}{F_{2r+1}}\right) + \arctan\!\left(\frac{1}{F_{2r+2}}\right) = \arctan\!\left(\frac{F_{2r+3}}{F_{2r}F_{2r+3}}\right) = \arctan\!\left(\frac{1}{F_{2r}}\right) .$

This proves the identity.

Evaluating the sum. Rearranging the identity:

$\arctan\!\left(\frac{1}{F_{2r+1}}\right) = \arctan\!\left(\frac{1}{F_{2r}}\right) - \arctan\!\left(\frac{1}{F_{2r+2}}\right) .$

The sum telescopes:

$\sum_{r=1}^{N} \arctan\!\left(\frac{1}{F_{2r+1}}\right) = \sum_{r=1}^{N} \left[\arctan\!\left(\frac{1}{F_{2r}}\right) - \arctan\!\left(\frac{1}{F_{2r+2}}\right)\right]$

$= \arctan\!\left(\frac{1}{F_2}\right) - \arctan\!\left(\frac{1}{F_{2N+2}}\right) = \arctan(1) - \arctan\!\left(\frac{1}{F_{2N+2}}\right) .$

As $N \to \infty$, $F_{2N+2} \to \infty$, so $\arctan\!\left(\dfrac{1}{F_{2N+2}}\right) \to 0$. Therefore

$\boxed{\sum_{r=1}^{\infty} \arctan\!\left(\frac{1}{F_{2r+1}}\right) = \frac{\pi}{4}} .$

Examiner Notes

This was the most popular question attempted by over 83% of candidates, and the third most successful with, on average, half marks being scored. Part (i) caused no problems, though some chose to obtain the result algebraically. Part (ii) was not well attempted, with a number stating the two values the expression can take but failing to do anything else or failing with the algebra. Part (iii) was generally fairly well done although frequently the details were not quite tied up fully.