STEP3 2015 -- Pure Mathematics

STEP3 2015 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2015  |  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	积分学 (Integral Calculus)	Challenging	分部积分, 换元积分 $u=x-x^{-1}$, 递推公式推导, 对称性
2	数列与不等式 (Sequences and Inequalities)	Challenging	反例构造, 数学归纳法, 不等式放缩, 序关系的传递性
3	极坐标 (Polar Coordinates)	Hard	极坐标方程分析, 分情况讨论, 三角函数积分, 蚌线的几何性质
4	复数 (Complex Numbers)	Challenging	中间值定理, 韦达定理, 棣莫弗定理, 复数的实部虚部分析
5	数论 (Number Theory)	Challenging	反证法, 最小元原理, 等价命题证明, 集合论方法
6	纯数 (Pure Mathematics)	Challenging	联立方程求解,判别式分析,反例构造,实数与复数条件
7	纯数 (Pure Mathematics)	Challenging	数学归纳法,微分算子运算,二项式展开,x=1代入法
8	纯数 (Pure Mathematics)	Hard	极坐标变换,链式法则,分离变量,部分分式积分,解曲线分类讨论

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Question 1

Topic: 积分学 (Integral Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

1 (i) Let $I_n = \int_0^\infty \frac{1}{(1 + u^2)^n} \, du \, ,$ where $n$ is a positive integer. Show that $I_n - I_{n+1} = \frac{1}{2n} I_n$ and deduce that $I_{n+1} = \frac{(2n)! \, \pi}{2^{2n+1}(n!)^2} \, .$

(ii) Let $J = \int_0^\infty f((x - x^{-1})^2) \, dx \, ,$ where $f$ is any function for which the integral exists. Show that $J = \int_0^\infty x^{-2}f((x - x^{-1})^2) \, dx = \frac{1}{2} \int_0^\infty (1 + x^{-2})f((x - x^{-1})^2) \, dx = \int_0^\infty f(u^2) \, du \, .$

(iii) Hence evaluate $\int_0^\infty \frac{x^{2n-2}}{(x^4 - x^2 + 1)^n} \, dx \, ,$ where $n$ is a positive integer.

Hint

The first result can be obtained by simplifying the LHS and then writing it as $\int_{0}^{\infty} u \frac{u}{(1+u^2)^{n+1}} du$ and integrating this by parts. To obtain the evaluation of $I_{n+1}$, the first result can be re-arranged to make $I_{n+1}$ the subject, and then iterating the result to express it in terms of $I_1$ which is a standard integral. The expression can be tidied by multiplying numerator and denominator by $(2n)(2n-2) \dots (2)$. The first result for (ii) is obtained by means of the substitution $u = x^{-1}$, the second by adding the two versions of $J$, and the third by the substitution $u = x - x^{-1}$, being careful with limits of integration and employing symmetry. Part (iii) is solved by expressing the integrand as $\frac{x^{-2}}{((x-x^{-1})^2+1)^n}$ and then employing first part (ii) then part (i) to obtain $I_n$, which is $\frac{(2n-2)!\pi}{2^{2n-1}((n-1)!)^2}$.

Model Solution

Part (i)

We have

$I_n - I_{n+1} = \int_0^\infty \frac{1}{(1+u^2)^n} \, du - \int_0^\infty \frac{1}{(1+u^2)^{n+1}} \, du = \int_0^\infty \frac{(1+u^2) - 1}{(1+u^2)^{n+1}} \, du = \int_0^\infty \frac{u^2}{(1+u^2)^{n+1}} \, du.$

We rewrite this as $\int_0^\infty u \cdot \frac{u}{(1+u^2)^{n+1}} \, du$ and integrate by parts with

$p = u, \qquad \frac{dq}{du} = \frac{u}{(1+u^2)^{n+1}}.$

Then $\frac{dp}{du} = 1$ and $q = -\frac{1}{2n(1+u^2)^n}$ (found by substituting $w = 1 + u^2$). So

$\int_0^\infty \frac{u^2}{(1+u^2)^{n+1}} \, du = \left[-\frac{u}{2n(1+u^2)^n}\right]_0^\infty + \int_0^\infty \frac{1}{2n(1+u^2)^n} \, du.$

The boundary term vanishes: at $u = 0$ the numerator is $0$, and as $u \to \infty$ the power $(1+u^2)^n$ in the denominator dominates. Therefore

$I_n - I_{n+1} = 0 + \frac{1}{2n} I_n \, ,$

as required.

Rearranging: $I_{n+1} = I_n - \frac{1}{2n} I_n = \frac{2n-1}{2n} I_n$.

We now iterate this recurrence starting from $I_1 = \int_0^\infty \frac{du}{1+u^2} = \left[\arctan u\right]_0^\infty = \frac{\pi}{2}$.

$I_2 = \frac{1}{2} \cdot \frac{\pi}{2}, \qquad I_3 = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}, \qquad \ldots$

In general, for $n \geq 1$:

$I_{n+1} = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} \cdot \frac{\pi}{2} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \cdot \frac{\pi}{2}.$

To simplify the numerator, multiply top and bottom by $2 \cdot 4 \cdot 6 \cdots (2n) = 2^n \cdot n!$:

$1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{(2n)!}{2^n \cdot n!}.$

The denominator is $2^n \cdot n!$. Therefore

$I_{n+1} = \frac{(2n)!}{2^n \cdot n! \cdot 2^n \cdot n!} \cdot \frac{\pi}{2} = \frac{(2n)! \, \pi}{2^{2n+1} (n!)^2}.$

Part (ii)

Let $J = \int_0^\infty f((x - x^{-1})^2) \, dx$.

First, we show $J = \int_0^\infty x^{-2} f((x - x^{-1})^2) \, dx$.

Substitute $x = u^{-1}$, so $dx = -u^{-2} \, du$. When $x = 0$, $u \to \infty$; when $x \to \infty$, $u = 0$. Also $(x - x^{-1})^2 = (u^{-1} - u)^2 = (u - u^{-1})^2$. So

$J = \int_\infty^0 f((u - u^{-1})^2) \left(-u^{-2}\right) du = \int_0^\infty u^{-2} f((u - u^{-1})^2) \, du.$

Renaming the dummy variable $u$ back to $x$ gives $J = \int_0^\infty x^{-2} f((x-x^{-1})^2) \, dx$.

Next, adding this to the original $J$:

$2J = \int_0^\infty (1 + x^{-2}) f((x - x^{-1})^2) \, dx,$

so $J = \frac{1}{2} \int_0^\infty (1 + x^{-2}) f((x - x^{-1})^2) \, dx$.

Finally, substitute $u = x - x^{-1}$. Then $du = (1 + x^{-2}) \, dx$. As $x \to 0^+$, $u \to -\infty$; as $x \to \infty$, $u \to +\infty$. So

$J = \frac{1}{2} \int_{-\infty}^{\infty} f(u^2) \, du.$

Since $f(u^2)$ is an even function of $u$, this equals $\int_0^\infty f(u^2) \, du$.

Part (iii)

We evaluate $\int_0^\infty \frac{x^{2n-2}}{(x^4 - x^2 + 1)^n} \, dx$.

Divide numerator and denominator by $x^{2n}$:

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

Using $(x - x^{-1})^2 = x^2 - 2 + x^{-2}$, we get $x^2 - 1 + x^{-2} = (x - x^{-1})^2 + 1$. So the integral becomes

$\int_0^\infty \frac{x^{-2}}{\left((x - x^{-1})^2 + 1\right)^n} \, dx.$

Set $f(t) = (t + 1)^{-n}$, so the integrand is $x^{-2} f((x - x^{-1})^2)$. By the result of part (ii),

$\int_0^\infty x^{-2} f((x-x^{-1})^2) \, dx = \int_0^\infty f(u^2) \, du = \int_0^\infty \frac{du}{(u^2 + 1)^n} = I_n.$

Using the formula from part (i) (replacing $n+1$ by $n$, i.e., replacing $n$ by $n-1$):

$I_n = \frac{(2n-2)! \, \pi}{2^{2n-1} ((n-1)!)^2}.$

Examiner Notes

This was the most popular question, being attempted by 85% of candidates, it was however only moderately successful although a number achieved full marks. Quite often, candidates ignored the helpful approach suggested by the LHS of the first required result, though, of course, it was possible to start from the first defined integral and achieve the same result. Many needlessly lost marks through omitting fairly straightforward steps such as the final evaluation in the last part of the question and failing to substantiate the simplified form of the result of part (i). Some got very carried away with tan or sinh substitutions in part (i), usually unsuccessfully and leading to monstrous amounts of algebraic working. A few failed to change the limits of integration in part (ii).

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Question 2

Topic: 数列与不等式 (Sequences and Inequalities)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 If $s_1, s_2, s_3, \dots$ and $t_1, t_2, t_3, \dots$ are sequences of positive numbers, we write

$(s_n) \leqslant (t_n)$

to mean

“there exists a positive integer $m$ such that $s_n \leqslant t_n$ whenever $n \geqslant m$”.

Determine whether each of the following statements is true or false. In the case of a true statement, you should give a proof which includes an explicit determination of an appropriate $m$; in the case of a false statement, you should give a counterexample.

(i) $(1000n) \leqslant (n^2)$.

(ii) If it is not the case that $(s_n) \leqslant (t_n)$, then it is the case that $(t_n) \leqslant (s_n)$.

(iii) If $(s_n) \leqslant (t_n)$ and $(t_n) \leqslant (u_n)$, then $(s_n) \leqslant (u_n)$.

(iv) $(n^2) \leqslant (2^n)$.

Hint

Part (ii) is the only false statement, and a simple counter-example is $s_n = 1$ and $t_n = 2$ for $n$ odd, and $s_n = 2$ and $t_n = 1$ for $n$ even. Part (i) $m = 1000$ is a suitable value, then $1000 \le n$ and as $n$ is positive, the inequality can be multiplied by it giving the required result. Part (iii) requires the use of the definition twice with values $m_1$ and $m_2$ say, and then using $m = \max(m_1, m_2)$. For part (iv), we can choose $m = 4$, and an inductive argument such as

$(k+1)^2 = \left(1 + \frac{1}{k}\right)^2 k^2 \le \left(1 + \frac{1}{4}\right)^2 k^2 < 2k^2 \le 2 \times 2^k = 2^{k+1}$ works.

Model Solution

Part (i) (TRUE)

We need $1000n \leq n^2$ for all $n \geq m$. Choose $m = 1000$. For $n \geq 1000$, since $n > 0$ we may divide both sides by $n$ to get $1000 \leq n$, which holds by our choice of $m$. Multiplying back by $n > 0$ gives $1000n \leq n^2$. $\blacksquare$

Part (ii) (FALSE)

The statement says: if it is not the case that $(s_n) \leq (t_n)$, then $(t_n) \leq (s_n)$.

Counterexample: Define $s_n = 1$ for all $n$, and $t_n = (-1)^n + 1$ (so $t_n = 2$ for even $n$ and $t_n = 0$ for odd $n$). Both are sequences of positive numbers… Actually, $t_n = 0$ for odd $n$ is not positive. Let me adjust.

Set $s_n = 1$ and $t_n = 2$ when $n$ is odd, and $s_n = 2$ and $t_n = 1$ when $n$ is even. Both sequences consist of positive numbers.

We check: $(s_n) \leq (t_n)$ requires some $m$ such that $s_n \leq t_n$ for all $n \geq m$. But for any $m$, there exists an even $n \geq m$ where $s_n = 2 > 1 = t_n$. So $(s_n) \leq (t_n)$ fails.

Similarly, $(t_n) \leq (s_n)$ requires some $m$ such that $t_n \leq s_n$ for all $n \geq m$. But for any $m$, there exists an odd $n \geq m$ where $t_n = 2 > 1 = s_n$. So $(t_n) \leq (s_n)$ also fails.

Since neither relation holds, the statement is false. $\blacksquare$

Part (iii) (TRUE)

Suppose $(s_n) \leq (t_n)$ and $(t_n) \leq (u_n)$. By definition, there exist positive integers $m_1$ and $m_2$ such that

$s_n \leq t_n \quad \text{for all } n \geq m_1, \qquad t_n \leq u_n \quad \text{for all } n \geq m_2.$

Let $m = \max(m_1, m_2)$. For all $n \geq m$, both inequalities hold, so $s_n \leq t_n \leq u_n$, giving $s_n \leq u_n$. Therefore $(s_n) \leq (u_n)$ with this $m$. $\blacksquare$

Part (iv) (TRUE)

We prove by induction that $n^2 \leq 2^n$ for all $n \geq 4$ (taking $m = 4$).

Base case: $n = 4$. We have $16 = 16$, so $4^2 = 2^4$. The inequality holds.

Inductive step: Suppose $k^2 \leq 2^k$ for some $k \geq 4$. We need $(k+1)^2 \leq 2^{k+1}$.

Write

$(k+1)^2 = k^2 \left(1 + \frac{1}{k}\right)^2.$

Since $k \geq 4$, we have $1 + \frac{1}{k} \leq 1 + \frac{1}{4} = \frac{5}{4}$, so $\left(1 + \frac{1}{k}\right)^2 \leq \frac{25}{16} < 2$.

Therefore $(k+1)^2 < 2k^2$. By the inductive hypothesis $k^2 \leq 2^k$, so

$(k+1)^2 < 2 \cdot 2^k = 2^{k+1}.$

By induction, $n^2 \leq 2^n$ for all $n \geq 4$. We can take $m = 4$. $\blacksquare$

Examiner Notes

Nearly three quarters attempted this, though again with moderate success as the main feature of the question was proof, and this was frequently handled cavalierly. Whilst it was not a crucial aspect of the question, ignoring the fact that the question deals with sequences of positive numbers was careless. Answers to the first part suffered at times from lack of argument or backwards logic. Part (ii) was generally well answered, although there were some silly counter-examples. This part suffered from those who completely missed the point of what the question was all about, forgetting the initial definition. Whilst most appreciated that part (iv) was true, there were many different methods used to attempt to prove it, and often unsuccessfully. Whilst induction using algebra is fairly straightforward, differentiation with or without logarithms and graphical methods frequently came to grief.

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Question 3

Topic: 极坐标 (Polar Coordinates)  |  Difficulty: Hard  |  Marks: 20

Problem

3 In this question, $r$ and $\theta$ are polar coordinates with $r \geqslant 0$ and $-\pi < \theta \leqslant \pi$, and $a$ and $b$ are positive constants.

Let $L$ be a fixed line and let $A$ be a fixed point not lying on $L$. Then the locus of points that are a fixed distance (call it $d$) from $L$ measured along lines through $A$ is called a conchoid of Nicomedes.

(i) Show that if

$|r - a \sec \theta| = b, \qquad \text{(*)}$

where $a > b$, then $\sec \theta > 0$. Show that all points with coordinates satisfying $(\ast)$ lie on a certain conchoid of Nicomedes (you should identify $L$, $d$ and $A$). Sketch the locus of these points.

(ii) In the case $a < b$, sketch the curve (including the loop for which $\sec \theta < 0$) given by

$|r - a \sec \theta| = b.$

Find the area of the loop in the case $a = 1$ and $b = 2$.

[Note: $\int \sec \theta \, \mathrm{d}\theta = \ln |\sec \theta + \tan \theta| + C$.]

Hint

The part (i) inequality for $\sec \theta$ can be obtained by making $r$ the subject of the formula as $r = a \sec \theta \pm b$ and invoking $a > b$ remembering that $r < 0$ is not permitted.

Then the points lie on a conchoid of Nicomedes with A being the pole (origin), d being $b$, and L being the line $r = a \sec \theta$ ("$x = a$").

In part (ii), the extra feature is the loop as specified with end-points at the pole corresponding to $\sec \theta = \frac{-b}{a}$.

So in the given case, the area is given by $2 \times \frac{1}{2} \int_{\frac{2\pi}{3}}^{\pi} (\sec \theta + 2)^2 d\theta$ which is $\frac{4\pi}{3} + \sqrt{3} - 4 \ln |2 + \sqrt{3}|$.

Model Solution

Part (i)

Showing $\sec\theta > 0$:

The equation $|r - a\sec\theta| = b$ gives two cases:

$r = a\sec\theta + b \qquad \text{or} \qquad r = a\sec\theta - b$

Since $r \geq 0$:

For $r = a\sec\theta - b$: we need $a\sec\theta \geq b$, so $\sec\theta \geq \frac{b}{a} > 0$. Thus $\sec\theta > 0$ immediately.

For $r = a\sec\theta + b$: we need $a\sec\theta \geq -b$, so $\sec\theta \geq -\frac{b}{a}$. Suppose for contradiction that $\sec\theta < 0$. Then $|\sec\theta| \geq 1$ (since $|\cos\theta| \leq 1$), giving $\sec\theta \leq -1$. But $a > b > 0$ implies $-\frac{b}{a} > -1$, so $\sec\theta \leq -1 < -\frac{b}{a}$, contradicting $\sec\theta \geq -\frac{b}{a}$. Therefore $\sec\theta > 0$.

So in both cases, $\sec\theta > 0$. $\blacksquare$

Identifying the conchoid:

Since $\sec\theta > 0$, we have $\cos\theta > 0$, and the polar equation becomes $r = a\sec\theta \pm b$.

The line $r = a\sec\theta$ is equivalent to $r\cos\theta = a$, i.e., the vertical line $x = a$ in Cartesian coordinates. For a ray from the origin (the pole) at angle $\theta$, the point where this ray meets $x = a$ is at distance $a\sec\theta$ from the origin. Points satisfying $|r - a\sec\theta| = b$ are at distance $b$ from this intersection point, measured along the ray.

This is the conchoid of Nicomedes with:

• $A$ = the pole (origin)
• $L$ = the line $x = a$
• $d = b$

Sketch:

The curve has two branches, with $x = a$ as a common vertical asymptote:

• Outer branch ($r = a\sec\theta + b$): lies to the right of $x = a$, since $x = r\cos\theta = a + b\cos\theta > a$. Passes through $(a+b, 0)$ at $\theta = 0$.
• Inner branch ($r = a\sec\theta - b$): lies to the left of $x = a$, since $x = r\cos\theta = a - b\cos\theta < a$ (but $x > a - b > 0$). Passes through $(a-b, 0)$ at $\theta = 0$.

Both branches extend to infinity as $\theta \to \pm\frac{\pi}{2}$, approaching the asymptote $x = a$. Both are symmetric about the $x$-axis.

Part (ii)

When $a < b$, the equation $|r - a\sec\theta| = b$ again gives $r = a\sec\theta + b$ or $r = a\sec\theta - b$.

For $r = a\sec\theta - b \geq 0$: $\sec\theta \geq \frac{b}{a} > 1$, so $\theta \in \left(-\frac{\pi}{2}, -\arccos\frac{a}{b}\right] \cup \left[\arccos\frac{a}{b}, \frac{\pi}{2}\right)$. This branch starts at the origin (where $r = 0$) and extends to infinity as $|\theta| \to \frac{\pi}{2}$.

For $r = a\sec\theta + b$:

• When $\sec\theta > 0$ (i.e., $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$): $r > 0$ always, giving a branch on the right similar to part (i).
• When $\sec\theta < 0$: $r \geq 0$ requires $\sec\theta \geq -\frac{b}{a}$. Since $a < b$, we have $-\frac{b}{a} < -1$. With $\sec\theta < 0$ and $\sec\theta \geq -\frac{b}{a}$, we get $\cos\theta \leq -\frac{a}{b}$ (and $\cos\theta < 0$). So

$\theta \in \left[\pi - \arccos\frac{a}{b},\, \pi\right] \cup \left[-\pi,\, -\pi + \arccos\frac{a}{b}\right].$

This last piece forms a loop. At the endpoints $\theta = \pm\left(\pi - \arccos\frac{a}{b}\right)$:

$r = a\left(-\frac{b}{a}\right) + b = 0$

so the loop passes through the origin. At $\theta = \pm\pi$: $r = -a + b = b - a > 0$, giving the Cartesian point $(a-b, 0)$ on the negative $x$-axis (since $a < b$).

Area of the loop for $a = 1$, $b = 2$:

Here $\arccos\frac{a}{b} = \arccos\frac{1}{2} = \frac{\pi}{3}$, so the loop is traced by $r = \sec\theta + 2$ for $\theta \in \left[\frac{2\pi}{3},\, \pi\right] \cup \left[-\pi,\, -\frac{2\pi}{3}\right]$.

By symmetry of the integrand (since $\sec\theta$ is even, $(\sec\theta + 2)^2$ is even):

$A = 2 \times \frac{1}{2}\int_{2\pi/3}^{\pi} (\sec\theta + 2)^2 \, d\theta = \int_{2\pi/3}^{\pi} (\sec\theta + 2)^2 \, d\theta$

Expanding:

$A = \int_{2\pi/3}^{\pi} \sec^2\theta \, d\theta + 4\int_{2\pi/3}^{\pi} \sec\theta \, d\theta + 4\int_{2\pi/3}^{\pi} d\theta$

First integral:

$\int_{2\pi/3}^{\pi} \sec^2\theta \, d\theta = \left[\tan\theta\right]_{2\pi/3}^{\pi} = \tan\pi - \tan\frac{2\pi}{3} = 0 - (-\sqrt{3}) = \sqrt{3}$

Second integral (using the given antiderivative):

$\int_{2\pi/3}^{\pi} \sec\theta \, d\theta = \left[\ln|\sec\theta + \tan\theta|\right]_{2\pi/3}^{\pi}$

At $\theta = \pi$: $\sec\pi = -1$, $\tan\pi = 0$, so $\ln|-1 + 0| = \ln 1 = 0$.

At $\theta = \frac{2\pi}{3}$: $\sec\frac{2\pi}{3} = -2$, $\tan\frac{2\pi}{3} = -\sqrt{3}$, so $\ln|-2 + (-\sqrt{3})| = \ln(2 + \sqrt{3})$.

Therefore the integral is $0 - \ln(2 + \sqrt{3}) = -\ln(2 + \sqrt{3})$.

Third integral:

$\int_{2\pi/3}^{\pi} d\theta = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$

Combining:

$A = \sqrt{3} + 4\bigl(-\ln(2 + \sqrt{3})\bigr) + 4 \cdot \frac{\pi}{3} = \frac{4\pi}{3} + \sqrt{3} - 4\ln(2 + \sqrt{3})$

$\blacksquare$

Examiner Notes

Under 20% attempted this, making it the least popular Pure question on the paper, and it was the least successfully attempted of all questions on the paper. Candidates seemed to find it intimidating, and many gave up before part (ii). They often got confused when dealing with separate cases and did not seem to understand what was required to show $\sec \theta > 0$ in part (i). Those that did make a stab at (ii) usually omitted a factor of two and most failed to find the correct limit to use.

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Question 4

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

Problem

4 (i) If $a$, $b$ and $c$ are all real, show that the equation

$z^3 + az^2 + bz + c = 0 \qquad \text{(*)}$

has at least one real root.

(ii) Let

$S_1 = z_1 + z_2 + z_3, \quad S_2 = z_1^2 + z_2^2 + z_3^2, \quad S_3 = z_1^3 + z_2^3 + z_3^3,$

where $z_1$, $z_2$ and $z_3$ are the roots of the equation ($*$). Express $a$ and $b$ in terms of $S_1$ and $S_2$, and show that

$6c = -S_1^3 + 3S_1S_2 - 2S_3.$

(iii) The six real numbers $r_k$ and $\theta_k$ ($k = 1, 2, 3$), where $r_k > 0$ and $-\pi < \theta_k < \pi$, satisfy

$\sum_{k=1}^3 r_k \sin(\theta_k) = 0, \quad \sum_{k=1}^3 r_k^2 \sin(2\theta_k) = 0, \quad \sum_{k=1}^3 r_k^3 \sin(3\theta_k) = 0.$

Show that $\theta_k = 0$ for at least one value of $k$.

Show further that if $\theta_1 = 0$ then $\theta_2 = -\theta_3$.

Hint

Part (i) is simply shown by considering the image of the function $f(z) = z^3 + az^2 + bz + c$ as $z \rightarrow \pm\infty$ and then observing that the function is continuous and exhibits a sign-change. Part (ii) can be approached by writing $z^3 + az^2 + bz + c = (z - z_1)(z - z_2)(z - z_3)$ giving $a = -S_1$, $b = \frac{S_1^2 - S_2}{2}$, which can be obtained by considering $(z_1 + z_2 + z_3)^2$ and the required result for $6c$ which can be neatly obtained by considering $f(z_1) + f(z_2) + f(z_3) = 0$.

Writing $z_k = r_k(\cos \theta_k + i \sin \theta_k)$ for $k = 1, 2, 3$, employing de Moivre’s theorem, the three sums imply the reality of $S_1$, $S_2$, and $S_3$, and hence $a$, $b$, and $c$ which by virtue of the result of part (i) yields the reality of $z_1$, $z_2$, or $z_3$ and hence the required result. The final result can be considered as two cases, the trivial one of all three roots being real, and the one where the other two are complex. The latter can be shown to give the required result by considering the real and imaginary parts of the roots of a real quadratic.

Model Solution

Part (i)

Let $f(z) = z^3 + az^2 + bz + c$. Since $a, b, c \in \mathbb{R}$, $f$ is a continuous real-valued function on $\mathbb{R}$.

As $z \to +\infty$: the leading term $z^3$ dominates, so $f(z) \to +\infty$. As $z \to -\infty$: the leading term $z^3$ dominates, so $f(z) \to -\infty$.

By the Intermediate Value Theorem, there exists at least one $z_0 \in \mathbb{R}$ such that $f(z_0) = 0$. Therefore the equation has at least one real root. $\blacksquare$

Part (ii)

Since $z_1, z_2, z_3$ are roots of $z^3 + az^2 + bz + c = 0$:

$z^3 + az^2 + bz + c = (z - z_1)(z - z_2)(z - z_3)$

Expanding the right side:

$(z - z_1)(z - z_2)(z - z_3) = z^3 - (z_1 + z_2 + z_3)z^2 + (z_1z_2 + z_1z_3 + z_2z_3)z - z_1z_2z_3$

Comparing coefficients of $z^2$:

$a = -(z_1 + z_2 + z_3) = -S_1$

For $b$, expand $S_1^2$:

$S_1^2 = (z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_1z_3 + z_2z_3) = S_2 + 2b$

Solving for $b$:

$b = \frac{S_1^2 - S_2}{2}$

For $c$: since $z_k$ satisfies $z_k^3 + az_k^2 + bz_k + c = 0$ for each $k = 1, 2, 3$, summing these three equations:

$(z_1^3 + z_2^3 + z_3^3) + a(z_1^2 + z_2^2 + z_3^2) + b(z_1 + z_2 + z_3) + 3c = 0$

$S_3 + aS_2 + bS_1 + 3c = 0$

Substituting $a = -S_1$ and $b = \frac{S_1^2 - S_2}{2}$:

$S_3 - S_1 S_2 + \frac{S_1^2 - S_2}{2} \cdot S_1 + 3c = 0$

$S_3 - S_1 S_2 + \frac{S_1^3 - S_1 S_2}{2} + 3c = 0$

Combining the $S_1 S_2$ terms (writing $-S_1 S_2 = -\frac{2S_1 S_2}{2}$):

$S_3 + \frac{S_1^3 - 2S_1 S_2 - S_1 S_2}{2} + 3c = 0$

$S_3 + \frac{S_1^3 - 3S_1 S_2}{2} + 3c = 0$

$3c = -S_3 - \frac{S_1^3 - 3S_1 S_2}{2} = \frac{-2S_3 - S_1^3 + 3S_1 S_2}{2}$

Multiplying both sides by 2:

$6c = -S_1^3 + 3S_1 S_2 - 2S_3$

$\blacksquare$

Part (iii)

Define $z_k = r_k e^{i\theta_k} = r_k(\cos\theta_k + i\sin\theta_k)$ for $k = 1, 2, 3$.

By de Moivre’s theorem, $z_k^n = r_k^n(\cos(n\theta_k) + i\sin(n\theta_k))$.

Step 1: Show $S_1, S_2, S_3$ are real.

$S_1 = \sum_{k=1}^3 z_k = \sum_{k=1}^3 r_k\cos\theta_k + i\sum_{k=1}^3 r_k\sin\theta_k = \sum_{k=1}^3 r_k\cos\theta_k$

since $\sum r_k\sin\theta_k = 0$. So $S_1 \in \mathbb{R}$.

$S_2 = \sum_{k=1}^3 z_k^2 = \sum_{k=1}^3 r_k^2\cos(2\theta_k) + i\sum_{k=1}^3 r_k^2\sin(2\theta_k) = \sum_{k=1}^3 r_k^2\cos(2\theta_k)$

since $\sum r_k^2\sin(2\theta_k) = 0$. So $S_2 \in \mathbb{R}$.

$S_3 = \sum_{k=1}^3 z_k^3 = \sum_{k=1}^3 r_k^3\cos(3\theta_k) + i\sum_{k=1}^3 r_k^3\sin(3\theta_k) = \sum_{k=1}^3 r_k^3\cos(3\theta_k)$

since $\sum r_k^3\sin(3\theta_k) = 0$. So $S_3 \in \mathbb{R}$.

Step 2: Construct a cubic with real coefficients having $z_1, z_2, z_3$ as roots.

Since $S_1, S_2, S_3$ are real, define:

$a = -S_1, \qquad b = \frac{S_1^2 - S_2}{2}, \qquad 6c = -S_1^3 + 3S_1 S_2 - 2S_3$

These are all real. We verify that $z_1, z_2, z_3$ are roots of $z^3 + az^2 + bz + c = 0$ by checking the Vieta relations:

• $z_1 + z_2 + z_3 = S_1 = -a$ $\checkmark$
• $z_1z_2 + z_1z_3 + z_2z_3 = \frac{S_1^2 - S_2}{2} = b$ $\checkmark$ (since $S_1^2 = S_2 + 2(z_1z_2 + z_1z_3 + z_2z_3)$)
• $z_1z_2z_3 = -c$ $\checkmark$ (from $6c = -S_1^3 + 3S_1S_2 - 2S_3 = -6z_1z_2z_3$, verified by expanding $S_1^3$ and $S_1S_2$)

Step 3: Apply part (i).

By part (i), this cubic with real coefficients has at least one real root, which must be one of $z_1, z_2, z_3$. If $z_k$ is real and $r_k > 0$, then $z_k = r_k(\cos\theta_k + i\sin\theta_k) \in \mathbb{R}$ requires $\sin\theta_k = 0$. Since $-\pi < \theta_k < \pi$, this gives $\theta_k = 0$.

Therefore $\theta_k = 0$ for at least one value of $k$. $\blacksquare$

Showing $\theta_2 = -\theta_3$ when $\theta_1 = 0$:

If $\theta_1 = 0$, then $z_1 = r_1 > 0$ is real. Factor the cubic:

$z^3 + az^2 + bz + c = (z - z_1)(z^2 + \alpha z + \beta)$

where $\alpha = a + z_1$ and $\beta = b + z_1(a + z_1)$ are real (since $a, b, z_1$ are all real). The remaining roots $z_2, z_3$ satisfy $z^2 + \alpha z + \beta = 0$, a quadratic with real coefficients.

Case 1: The discriminant $\alpha^2 - 4\beta \geq 0$. Then $z_2, z_3$ are both real, so $\theta_2 = \theta_3 = 0$, giving $\theta_2 = -\theta_3 = 0$.

Case 2: The discriminant $\alpha^2 - 4\beta < 0$. Then $z_2, z_3$ are complex conjugates: $z_2 = \overline{z_3}$. Writing $z_k = r_k e^{i\theta_k}$:

$r_2 e^{i\theta_2} = r_3 e^{-i\theta_3}$

Taking moduli: $r_2 = r_3$. Then $e^{i\theta_2} = e^{-i\theta_3}$, so $\theta_2 = -\theta_3 + 2n\pi$ for some integer $n$. Since $-\pi < \theta_2, \theta_3 < \pi$, we must have $n = 0$, giving $\theta_2 = -\theta_3$.

In both cases, $\theta_2 = -\theta_3$. $\blacksquare$

Examiner Notes

Along with questions 5 and 7, attempted by just over three quarters, this was the third most popular question, though a little less successful than the most popular question 1. The first part was frequently not well attempted, but the second part was usually mastered. Attempts at the third part suffered from arguments with poor logical structure, though many did not get a start on this part.

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Question 5

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

Problem

5 (i) In the following argument to show that $\sqrt{2}$ is irrational, give proofs appropriate for steps 3, 5 and 6.

1. Assume that $\sqrt{2}$ is rational.

2. Define the set $S$ to be the set of positive integers with the following property:

$n$ is in $S$ if and only if $n\sqrt{2}$ is an integer.

3. Show that the set $S$ contains at least one positive integer.

4. Define the integer $k$ to be the smallest positive integer in $S$.

5. Show that $(\sqrt{2} - 1)k$ is in $S$.

6. Show that steps 4 and 5 are contradictory and hence that $\sqrt{2}$ is irrational.

(ii) Prove that $2^{\frac{1}{3}}$ is rational if and only if $2^{\frac{2}{3}}$ is rational.

Use an argument similar to that of part (i) to prove that $2^{\frac{1}{3}}$ and $2^{\frac{2}{3}}$ are irrational.

Hint

(i) Step 3 is straightforward on the basis of steps 1 and 2, noting that no lowest terms restriction need be made in part 1. Step 5 requires that the given expression is a positive integer as well as well as being integer when multiplied by root two. Step 6 requires justification that $\sqrt{2} - 1 < 1$.

(ii) The rationality of $2^{2/3}$ on the basis of $2^{1/3}$ being rational is simply obtained by squaring the latter, and the opposite implication can be made by squaring the former or dividing 2 by the former. To construct the similar argument, let the set $T$ be the set of positive integers with the following property: $n$ is in $T$ if and only if $n2^{1/3}$ and $n2^{2/3}$ are integers, and taking $t$ to be the smallest positive integer in that set, consider $t(2^{2/3} - 1)$ to produce the argument.

Model Solution

Part (i)

Step 3: Show that $S$ contains at least one positive integer.

By Step 1, $\sqrt{2}$ is rational, so we can write $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are positive integers. Then

$q\sqrt{2} = q \cdot \frac{p}{q} = p,$

which is a positive integer. Therefore $q$ is a positive integer with $q\sqrt{2} \in \mathbb{Z}$, so $q \in S$. In particular, $S$ is non-empty. $\blacksquare$

Step 5: Show that $(\sqrt{2} - 1)k$ is in $S$.

We need to show two things: (a) $(\sqrt{2} - 1)k$ is a positive integer, and (b) $(\sqrt{2} - 1)k \cdot \sqrt{2}$ is an integer.

For (a): Since $1 < \sqrt{2} < 2$, we have $0 < \sqrt{2} - 1 < 1$. As $k$ is a positive integer, $(\sqrt{2} - 1)k > 0$. Also, $(\sqrt{2} - 1)k < k$, so $(\sqrt{2} - 1)k$ is a positive real number less than $k$.

Now, since $k \in S$, the number $k\sqrt{2}$ is an integer. Therefore

$(\sqrt{2} - 1)k = k\sqrt{2} - k$

is the difference of two integers, hence an integer. Combined with $(\sqrt{2} - 1)k > 0$, this is a positive integer.

For (b): We compute

$(\sqrt{2} - 1)k \cdot \sqrt{2} = k(2 - \sqrt{2}) = 2k - k\sqrt{2}.$

Since $k$ and $k\sqrt{2}$ are both integers, this is an integer.

Therefore $(\sqrt{2} - 1)k$ is a positive integer whose product with $\sqrt{2}$ is an integer, so $(\sqrt{2} - 1)k \in S$. $\blacksquare$

Step 6: Show that Steps 4 and 5 are contradictory.

From Step 5, $(\sqrt{2} - 1)k \in S$, and from the analysis above, $0 < (\sqrt{2} - 1)k < k$. But Step 4 defines $k$ as the smallest positive integer in $S$. This is a contradiction: $(\sqrt{2} - 1)k$ is a positive integer in $S$ that is strictly smaller than $k$.

The contradiction arose from the assumption in Step 1 that $\sqrt{2}$ is rational. Therefore $\sqrt{2}$ is irrational. $\blacksquare$

Part (ii)

First, prove the equivalence: $2^{1/3}$ is rational if and only if $2^{2/3}$ is rational.

$(\Rightarrow)$: If $2^{1/3}$ is rational, then $2^{2/3} = (2^{1/3})^2$ is the square of a rational number, hence rational.

$(\Leftarrow)$: If $2^{2/3}$ is rational, then $2^{1/3} = \frac{2}{2^{2/3}}$ is the quotient of two rational numbers (with non-zero denominator), hence rational. $\blacksquare$

Now prove that $2^{1/3}$ and $2^{2/3}$ are irrational, using an argument similar to part (i).

Suppose for contradiction that $2^{1/3}$ is rational (equivalently, $2^{2/3}$ is rational, by the result above). Define the set

$T = \{n \in \mathbb{Z}^+ : n \cdot 2^{1/3} \in \mathbb{Z} \text{ and } n \cdot 2^{2/3} \in \mathbb{Z}\}.$

Since $2^{1/3}$ is rational, say $2^{1/3} = p/q$ for positive integers $p, q$, then $2^{2/3} = p^2/q^2$. Choose $n = q^2$. Then $n \cdot 2^{1/3} = q^2 \cdot p/q = qp \in \mathbb{Z}$ and $n \cdot 2^{2/3} = q^2 \cdot p^2/q^2 = p^2 \in \mathbb{Z}$. So $q^2 \in T$, and $T$ is non-empty.

Since $T$ is a non-empty set of positive integers, by the well-ordering principle it has a smallest element; call it $t$.

Now consider $t(2^{2/3} - 1)$.

(a) We show $t(2^{2/3} - 1)$ is a positive integer. Since $1^3 = 1 < 4 = 2^2$ and $2^3 = 8 > 4 = 2^2$, we have $1 < 2^{2/3} < 2$, so $0 < 2^{2/3} - 1 < 1$. Now

$t(2^{2/3} - 1) = t \cdot 2^{2/3} - t.$

Since $t \in T$, the number $t \cdot 2^{2/3}$ is an integer, so $t \cdot 2^{2/3} - t$ is an integer. Since $t \cdot 2^{2/3} > t$ (because $2^{2/3} > 1$), this integer is positive. Since $2^{2/3} - 1 < 1$, we also have $t(2^{2/3} - 1) < t$.

(b) We show $t(2^{2/3} - 1) \cdot 2^{1/3}$ is an integer:

$t(2^{2/3} - 1) \cdot 2^{1/3} = t(2 - 2^{1/3}) = 2t - t \cdot 2^{1/3}.$

Since $t \in T$, the number $t \cdot 2^{1/3}$ is an integer, so $2t - t \cdot 2^{1/3}$ is an integer.

(c) We show $t(2^{2/3} - 1) \cdot 2^{2/3}$ is an integer:

$t(2^{2/3} - 1) \cdot 2^{2/3} = t(2^{4/3} - 2^{2/3}) = t(2 \cdot 2^{1/3} - 2^{2/3}) = 2(t \cdot 2^{1/3}) - t \cdot 2^{2/3}.$

Since $t \in T$, both $t \cdot 2^{1/3}$ and $t \cdot 2^{2/3}$ are integers, so this is an integer.

Therefore $t(2^{2/3} - 1)$ is a positive integer in $T$ with $t(2^{2/3} - 1) < t$, contradicting the minimality of $t$.

The contradiction shows that $2^{1/3}$ (and hence $2^{2/3}$) is irrational. $\blacksquare$

Examiner Notes

Marginally less successful than question 2, a lot of candidates earned about half of the marks. Unfortunately, many candidates approached this on the basis of their knowledge of the standard irrationality proof for root two employing rational numbers expressed in lowest terms rather than observing the specified argument. In part (i), proving step 5 was frequently beset with omissions, and simple steps like $0 < \sqrt{2} - 1 < 1$ were not acknowledged let alone justified. The first result of part (ii) caused few problems except to those that did not appreciate ‘if and only if’, but defining a suitable set in order to construct a similar argument to prove the irrationality of the cube roots of 2 and 2 squared was beyond most leading to mostly spurious logic.

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Question 6

Topic: 纯数 (Pure Mathematics)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 (i) Let $w$ and $z$ be complex numbers, and let $u = w + z$ and $v = w^2 + z^2$. Prove that $w$ and $z$ are real if and only if $u$ and $v$ are real and $u^2 \leqslant 2v$.

      **(ii)**   The complex numbers $u$, $w$ and $z$ satisfy the equations

$$\begin{aligned} w + z - u &= 0 \\ w^2 + z^2 - u^2 &= -\frac{2}{3} \\ w^3 + z^3 - \lambda u &= -\lambda \end{aligned}$$

              where $\lambda$ is a positive real number. Show that for all values of $\lambda$ except one (which you should find) there are three possible values of $u$, all real.

              Are $w$ and $z$ necessarily real? Give a proof or counterexample.

Hint

Treating the equations for $u$ and $v$ as simultaneous equations for $w$ and $z$, one finds that $w = \frac{u \pm \sqrt{2v - u^2}}{2}$ and $z = \frac{u \mp \sqrt{2v - u^2}}{2}$ which demonstrates that if $u \in \mathbb{R}$ and $u^2 \le 2v$, i.e. $v \in \mathbb{R}$, $w$ and $z$ are real. If $w$ and $z$ are real, then $u$ and $v$ are (trivially) and $2v - u^2 = (w - z)^2 \ge 0$.

In (ii), the first two equations yield $3wz = 1$, making it possible to write the third equation as $u(u^2 - 1) = \lambda(u - 1)$ which has an obvious factor of $(u - 1)$ leading to $u = 1$ or $u = \frac{-1 \pm \sqrt{1 + 4\lambda}}{2}$ from the quadratic equation. If one of the solutions of the quadratic equation gives the same root $u = 1$, then there are not three possible values, i.e. if $\lambda = 2$. From the first part of the question, for $w$ and $z$ to be real, we would want $u$ to be real, $u^2 - \frac{2}{3}$ to be real, and $u^2 \le 2(u^2 - \frac{2}{3})$, in other words $u^2 \geq \frac{4}{3}$. So a counter-example could be $u = 1$ giving $2w^2 - 2w + \frac{2}{3} = 0$ which has a negative discriminant.

Model Solution

Part (i)

We prove both directions.

$(\Rightarrow)$ Suppose $w$ and $z$ are real. Then $u = w + z$ and $v = w^2 + z^2$ are clearly real. Moreover,

$2v - u^2 = 2(w^2 + z^2) - (w + z)^2 = 2w^2 + 2z^2 - w^2 - 2wz - z^2 = w^2 - 2wz + z^2 = (w - z)^2 \geqslant 0,$

so $u^2 \leqslant 2v$. $\blacksquare$

$(\Leftarrow)$ Suppose $u$ and $v$ are real with $u^2 \leqslant 2v$. Since $w + z = u$ and $w^2 + z^2 = v$, we can find $wz$:

$(w + z)^2 = w^2 + 2wz + z^2 = v + 2wz,$

so $u^2 = v + 2wz$, giving $wz = \frac{u^2 - v}{2}$, which is real.

Now $w$ and $z$ are roots of the quadratic

$t^2 - (w+z)t + wz = t^2 - ut + \frac{u^2 - v}{2} = 0.$

The discriminant is

$\Delta = u^2 - 4 \cdot \frac{u^2 - v}{2} = u^2 - 2(u^2 - v) = 2v - u^2.$

Since $u^2 \leqslant 2v$, we have $\Delta = 2v - u^2 \geqslant 0$, so the quadratic has real roots. Therefore $w$ and $z$ are real. $\blacksquare$

Part (ii)

From the first equation: $w + z = u$.

From the second equation: $w^2 + z^2 = u^2 - \frac{2}{3}$.

Using $(w + z)^2 = w^2 + 2wz + z^2$:

$u^2 = \left(u^2 - \frac{2}{3}\right) + 2wz \implies wz = \frac{1}{3}.$

For the third equation, we use the identity

$w^3 + z^3 = (w + z)(w^2 - wz + z^2) = (w + z)\left((w^2 + z^2) - wz\right).$

Substituting:

$w^3 + z^3 = u\left(\left(u^2 - \frac{2}{3}\right) - \frac{1}{3}\right) = u(u^2 - 1).$

The third equation $w^3 + z^3 - \lambda u = -\lambda$ becomes

$u(u^2 - 1) - \lambda u = -\lambda.$

Rearranging:

$u^3 - u - \lambda u + \lambda = 0 \implies u^3 - (1 + \lambda)u + \lambda = 0.$

We observe that $u = 1$ is a root: $1 - (1 + \lambda) + \lambda = 0$. Factoring out $(u - 1)$:

$u^3 - (1 + \lambda)u + \lambda = (u - 1)(u^2 + u - \lambda) = 0.$

The quadratic $u^2 + u - \lambda = 0$ has solutions

$u = \frac{-1 \pm \sqrt{1 + 4\lambda}}{2}.$

Since $\lambda > 0$, we have $1 + 4\lambda > 1 > 0$, so these are real and distinct. They are also distinct from $u = 1$ unless $u = 1$ satisfies $u^2 + u - \lambda = 0$, i.e., $1 + 1 - \lambda = 0$, giving $\lambda = 2$.

Therefore: for all positive real $\lambda$ except $\lambda = 2$, there are three distinct real values of $u$. When $\lambda = 2$, the values $u = 1$ coincide and there are only two distinct values: $u = 1$ and $u = -2$.

Are $w$ and $z$ necessarily real?

No. We use part (i): $w$ and $z$ are real if and only if $u$ and $v = u^2 - \frac{2}{3}$ are real and $u^2 \leqslant 2v$, i.e.,

$u^2 \leqslant 2\left(u^2 - \frac{2}{3}\right) = 2u^2 - \frac{4}{3} \implies u^2 \geqslant \frac{4}{3}.$

For the root $u = 1$ (which exists for every positive $\lambda$): $u^2 = 1 < \frac{4}{3}$, so the condition fails, and $w, z$ need not be real.

Counterexample: Take $\lambda = 1$ and $u = 1$. Then $w + z = 1$ and $wz = \frac{1}{3}$, so $w$ and $z$ are roots of

$t^2 - t + \frac{1}{3} = 0 \implies t = \frac{1 \pm \sqrt{1 - \frac{4}{3}}}{2} = \frac{1 \pm \sqrt{-\frac{1}{3}}}{2} = \frac{1}{2} \pm \frac{i}{2\sqrt{3}}.$

These are not real. We verify all three equations:

• $w + z = 1 = u$. $\checkmark$
• $w^2 + z^2 = (w + z)^2 - 2wz = 1 - \frac{2}{3} = \frac{1}{3} = u^2 - \frac{2}{3}$. $\checkmark$
• $w^3 + z^3 = (w + z)^3 - 3wz(w + z) = 1 - 3 \cdot \frac{1}{3} \cdot 1 = 0$, and $\lambda u - \lambda = 1 \cdot 1 - 1 = 0$. $\checkmark$

Therefore $w$ and $z$ are not necessarily real. $\blacksquare$

Examiner Notes

约 3/5 的考生尝试此题但成功率不高。(i) 部分需要证明双向蕴含（if and only if），难倒许多考生；(ii) 第一小题得分机会较好，但当 lambda=2 时无法产生三个不同 u 值的情况需注意；反例部分需明确展示其确实为反例。

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Question 7

Topic: 纯数 (Pure Mathematics)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 An operator D is defined, for any function $f$, by

$Df(x) = x \frac{df(x)}{dx} .$

      The notation $D^n$ means that D is applied $n$ times; for example

$D^2f(x) = x \frac{d}{dx} \left( x \frac{df(x)}{dx} \right) .$

      Show that, for any constant $a$, $D^2x^a = a^2x^a$.

      **(i)**     Show that if $P(x)$ is a polynomial of degree $r$ (where $r \geqslant 1$) then, for any positive integer $n$, $D^nP(x)$ is also a polynomial of degree $r$.

      **(ii)**    Show that if $n$ and $m$ are positive integers with $n < m$, then $D^n(1 - x)^m$ is divisible by $(1 - x)^{m-n}$.

      **(iii)**   Deduce that, if $m$ and $n$ are positive integers with $n < m$, then

$\sum_{r=0}^{m} (-1)^r \binom{m}{r} r^n = 0 .$

Hint

The opening result is simply achieved by following the given explanation for $D^2 f(x)$ with $f(x) = x^a$. Parts (i) and (ii) can both be shown using the principle of mathematical induction with initial statements

“Suppose $D^k P(x)$ is a polynomial of degree $r$ i.e. $D^k P(x) = a_r x^r + a_{r-1} x^{r-1} + \dots + a_0$ for some integer $k$.” and “Suppose $D^k (1 - x)^m$ is divisible by $(1 - x)^{m-k}$ i.e. $D^k (1 - x)^m = f(x)(1 - x)^{m-k}$ for some integer $k$, with $k < m - 1$.” Part (iii) is obtained by expressing $(1 - x)^m$ in sigma notation (by the binomial theorem), then carrying out $D^n (1 - x)^m$ using the idea in the stem, and finally invoking the result of part (ii) and then substituting $x = 1$.

Model Solution

We first show $D^2 x^a = a^2 x^a$.

$D(x^a) = x \cdot ax^{a-1} = ax^a.$

$D^2(x^a) = D(ax^a) = x \cdot \frac{d}{dx}(ax^a) = x \cdot a^2 x^{a-1} = a^2 x^a.$

More generally, by induction one can show $D^n x^a = a^n x^a$ for any positive integer $n$.

Part (i)

We prove by induction on $n$ that if $P(x)$ is a polynomial of degree $r \geq 1$, then $D^n P(x)$ is a polynomial of degree $r$.

Base case ($n = 1$): Let $P(x) = a_r x^r + a_{r-1} x^{r-1} + \cdots + a_0$ with $a_r \neq 0$. Then

$DP(x) = x \cdot P'(x) = x(ra_r x^{r-1} + (r-1)a_{r-1} x^{r-2} + \cdots + a_1) = ra_r x^r + (r-1)a_{r-1} x^{r-1} + \cdots + a_1 x.$

Since $r \geq 1$ and $a_r \neq 0$, the leading coefficient $ra_r \neq 0$, so $DP(x)$ is a polynomial of degree $r$.

Inductive step: Suppose $D^k P(x)$ is a polynomial of degree $r$ for some positive integer $k$. Write $D^k P(x) = b_r x^r + b_{r-1} x^{r-1} + \cdots + b_0$ with $b_r \neq 0$. Then by the same argument as the base case, $D^{k+1} P(x) = D(D^k P(x))$ is a polynomial of degree $r$ with leading coefficient $rb_r \neq 0$.

By induction, $D^n P(x)$ is a polynomial of degree $r$ for all positive integers $n$. $\blacksquare$

Part (ii)

We prove by induction on $n$ that if $n$ and $m$ are positive integers with $n < m$, then $D^n(1-x)^m$ is divisible by $(1-x)^{m-n}$.

Base case ($n = 0$): $D^0(1-x)^m = (1-x)^m$, which is divisible by $(1-x)^{m-0} = (1-x)^m$. $\checkmark$

Inductive step: Suppose $D^k(1-x)^m = g(x)(1-x)^{m-k}$ for some polynomial $g(x)$ and some non-negative integer $k < m$. Then

$D^{k+1}(1-x)^m = D\bigl(g(x)(1-x)^{m-k}\bigr) = x \frac{d}{dx}\bigl(g(x)(1-x)^{m-k}\bigr).$

Using the product rule:

$\frac{d}{dx}\bigl(g(x)(1-x)^{m-k}\bigr) = g'(x)(1-x)^{m-k} + g(x)(m-k)(1-x)^{m-k-1}(-1).$

So

$D^{k+1}(1-x)^m = x\bigl[g'(x)(1-x)^{m-k} - (m-k)g(x)(1-x)^{m-k-1}\bigr]$ $= x(1-x)^{m-k-1}\bigl[g'(x)(1-x) - (m-k)g(x)\bigr].$

The bracketed expression is a polynomial in $x$, and the power of $(1-x)$ is $m - k - 1 = m - (k+1)$. Since $k + 1 \leq m$ (as $k < m$), this is non-negative, so $D^{k+1}(1-x)^m$ is divisible by $(1-x)^{m-(k+1)}$.

By induction, $D^n(1-x)^m$ is divisible by $(1-x)^{m-n}$ for all positive integers $n < m$. $\blacksquare$

Part (iii)

By the binomial theorem:

$(1-x)^m = \sum_{r=0}^{m} \binom{m}{r}(-x)^r = \sum_{r=0}^{m} (-1)^r \binom{m}{r} x^r.$

We compute $D^n(1-x)^m$ by applying $D^n$ to this sum. Since $D(x^r) = r \cdot x^r$ (as shown in the stem), repeated application gives $D^n(x^r) = r^n x^r$. Therefore

$D^n(1-x)^m = D^n\left(\sum_{r=0}^{m} (-1)^r \binom{m}{r} x^r\right) = \sum_{r=0}^{m} (-1)^r \binom{m}{r} r^n x^r.$

By part (ii), $D^n(1-x)^m$ is divisible by $(1-x)^{m-n}$. Since $n < m$, we have $m - n \geq 1$, which means $D^n(1-x)^m$ has a factor of $(1-x)$ and therefore vanishes at $x = 1$:

$D^n(1-x)^m\Big|_{x=1} = 0.$

Substituting $x = 1$ into the sum expression:

$\sum_{r=0}^{m} (-1)^r \binom{m}{r} r^n \cdot 1^r = \sum_{r=0}^{m} (-1)^r \binom{m}{r} r^n = 0.$

$\blacksquare$

Examiner Notes

与 Q2 并列第三受欢迎和第二成功。(i) 部分存在许多缺陷，很多考生无法进行正确的形式归纳；(ii) 部分一些考生认为 x 与 d/dx 可交换，经常从前几项推断出错误的 D^n(1-x)^m 公式。不令人意外地，朝给定结果推导时很多人通过错误的步骤（如在使用微分算子之前先将 x=1 代入 (1-x)^m）得到正确答案。

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Question 8

Topic: 纯数 (Pure Mathematics)  |  Difficulty: Hard  |  Marks: 20

Problem

8 (i) Show that under the changes of variable $x = r \cos \theta$ and $y = r \sin \theta$, where $r$ is a function of $\theta$ with $r > 0$, the differential equation

$(y + x) \frac{dy}{dx} = y - x$

becomes

$\frac{dr}{d\theta} + r = 0 \, .$

Sketch a solution in the $x$-$y$ plane.

(ii) Show that the solutions of

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

can be written in the form

$r^2 = \frac{1}{1 + Ae^{2\theta}}$

and sketch the different forms of solution that arise according to the value of $A$.

Hint

Transforming the differential equation in part (i) is made by substituting for $x$ and $y$ as given, for $\frac{dy}{dx}$ using $\frac{dy}{d\theta} = r \cos \theta + \frac{dr}{d\theta} \sin \theta$ and a similar result for $\frac{dx}{d\theta}$, and then simplifying the algebra by multiplying out and collecting like terms bearing in mind that a factor $r$ can be cancelled as $r \neq 0$. The transformed equation can be solved by separating variables or using an integrating factor, to give $r = k e^{-\theta}$, the sketch of which is a logarithmic spiral.

The same techniques for part (ii) yields a differential equation $r - r^3 + \frac{dr}{d\theta} = 0$ which is solved by separating the variables and then employing partial fractions giving a variety of possible solution sketches depending on the value of $A$.

Model Solution

Part (i)

We substitute $x = r\cos\theta$ and $y = r\sin\theta$ into $(y+x)\frac{dy}{dx} = y - x$, where $r = r(\theta)$.

First, compute the derivatives using the chain rule. Since $r$ is a function of $\theta$:

$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta.$

Therefore

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}.$

Substituting into the ODE $(y+x)\frac{dy}{dx} = y - x$:

$(r\sin\theta + r\cos\theta) \cdot \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} = r\sin\theta - r\cos\theta.$

Since $r > 0$, we cancel $r$ from both sides:

$(\sin\theta + \cos\theta) \cdot \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} = \sin\theta - \cos\theta.$

Cross-multiplying (assuming the denominator $\frac{dr}{d\theta}\cos\theta - r\sin\theta \neq 0$):

(\sin\theta + \cos\theta)\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta) = (\sin\theta - \cos\theta)\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right).

Expand the left side:

$\frac{dr}{d\theta}\sin^2\theta + r\sin\theta\cos\theta + \frac{dr}{d\theta}\sin\theta\cos\theta + r\cos^2\theta$ $= \frac{dr}{d\theta}(\sin^2\theta + \sin\theta\cos\theta) + r(\sin\theta\cos\theta + \cos^2\theta).$

Expand the right side:

$\frac{dr}{d\theta}\sin\theta\cos\theta - r\sin^2\theta - \frac{dr}{d\theta}\cos^2\theta + r\sin\theta\cos\theta$ $= \frac{dr}{d\theta}(\sin\theta\cos\theta - \cos^2\theta) + r(-\sin^2\theta + \sin\theta\cos\theta).$

Setting left = right and collecting terms with $\frac{dr}{d\theta}$ on one side:

$\frac{dr}{d\theta}\bigl[(\sin^2\theta + \sin\theta\cos\theta) - (\sin\theta\cos\theta - \cos^2\theta)\bigr] = r\bigl[(-\sin^2\theta + \sin\theta\cos\theta) - (\sin\theta\cos\theta + \cos^2\theta)\bigr].$

Simplifying the brackets:

$\frac{dr}{d\theta}(\sin^2\theta + \cos^2\theta) = r(-\sin^2\theta - \cos^2\theta).$

Since $\sin^2\theta + \cos^2\theta = 1$:

$\frac{dr}{d\theta} = -r,$

that is,

$\frac{dr}{d\theta} + r = 0.$

$\blacksquare$

Sketch: This ODE has solution $r = Ce^{-\theta}$ for a constant $C > 0$. This is a logarithmic spiral: as $\theta$ increases, $r$ decreases exponentially, and the curve spirals inwards toward the origin. The curve winds around the origin infinitely many times, getting closer and closer without ever reaching it.

Part (ii)

We substitute $x = r\cos\theta$, $y = r\sin\theta$ into

$(y + x - x(x^2 + y^2))\frac{dy}{dx} = y - x - y(x^2 + y^2).$

Note that $x^2 + y^2 = r^2$, so $x(x^2+y^2) = r^3\cos\theta$ and $y(x^2+y^2) = r^3\sin\theta$. The ODE becomes:

$(r\sin\theta + r\cos\theta - r^3\cos\theta)\frac{dy}{dx} = r\sin\theta - r\cos\theta - r^3\sin\theta.$

Factor out $r$ (using $r > 0$):

$(\sin\theta + \cos\theta - r^2\cos\theta)\frac{dy}{dx} = \sin\theta - \cos\theta - r^2\sin\theta.$

Using the expression for $\frac{dy}{dx}$ from part (i):

$(\sin\theta + \cos\theta - r^2\cos\theta) \cdot \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} = \sin\theta - \cos\theta - r^2\sin\theta.$

Cross-multiplying:

$(\sin\theta + \cos\theta - r^2\cos\theta)\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right)$ $= (\sin\theta - \cos\theta - r^2\sin\theta)\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right).$

We can shortcut this algebra by noticing that the part (i) calculation is embedded here. Write the equation as

$(\sin\theta + \cos\theta - r^2\cos\theta)\frac{dy}{dx} = (\sin\theta - \cos\theta - r^2\sin\theta).$

This equals the part (i) equation with an extra "$-r^2\cos\theta$" on the left and "$-r^2\sin\theta$" on the right. So we can write

$\bigl[(\sin\theta + \cos\theta) - r^2\cos\theta\bigr]\frac{dy}{dx} = (\sin\theta - \cos\theta) - r^2\sin\theta.$

Following the same algebraic steps as in part (i), the $(\sin\theta + \cos\theta)\frac{dy}{dx} = \sin\theta - \cos\theta$ part yields $\frac{dr}{d\theta} = -r$ as before. The extra terms contribute additional pieces. Let us work through this systematically.

From the cross-multiplied equation, expand both sides. The left side is:

$\left[(\sin\theta + \cos\theta) - r^2\cos\theta\right]\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right)$

$= (\sin\theta + \cos\theta)\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right) - r^2\cos\theta\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right).$

The right side is:

$\left[(\sin\theta - \cos\theta) - r^2\sin\theta\right]\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right)$

$= (\sin\theta - \cos\theta)\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right) - r^2\sin\theta\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right).$

From part (i), the first terms on each side are equal (both equal $\frac{dr}{d\theta}$ after simplification). So we are left with:

$-r^2\cos\theta\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right) = -r^2\sin\theta\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right).$

Dividing by $-r^2$ (valid since $r > 0$):

$\cos\theta\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right) = \sin\theta\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right).$

Expanding:

$\frac{dr}{d\theta}\sin\theta\cos\theta + r\cos^2\theta = \frac{dr}{d\theta}\sin\theta\cos\theta - r\sin^2\theta.$

The $\frac{dr}{d\theta}\sin\theta\cos\theta$ terms cancel, leaving:

$r\cos^2\theta = -r\sin^2\theta \qquad \Longrightarrow \qquad r(\cos^2\theta + \sin^2\theta) = 0 \qquad \Longrightarrow \qquad r = 0.$

But $r > 0$, so this seems contradictory. We need to be more careful: the part (i) simplification and the extra terms do not decouple in this way because $\frac{dr}{d\theta}$ appears in the extra terms too.

Let us redo the full expansion properly. The cross-multiplied equation is:

$LHS = (\sin\theta + \cos\theta - r^2\cos\theta)\left(\frac{dr}{d\theta}\sin\theta + r\cos\theta\right),$

$RHS = (\sin\theta - \cos\theta - r^2\sin\theta)\left(\frac{dr}{d\theta}\cos\theta - r\sin\theta\right).$

Expanding $LHS$:

$= \frac{dr}{d\theta}\sin^2\theta + r\sin\theta\cos\theta + \frac{dr}{d\theta}\sin\theta\cos\theta + r\cos^2\theta - r^2\frac{dr}{d\theta}\sin\theta\cos\theta - r^3\cos^2\theta.$

Expanding $RHS$:

$= \frac{dr}{d\theta}\sin\theta\cos\theta - r\sin^2\theta - \frac{dr}{d\theta}\cos^2\theta + r\sin\theta\cos\theta - r^2\frac{dr}{d\theta}\sin\theta\cos\theta + r^3\sin^2\theta.$

Setting $LHS = RHS$ and moving everything to the left:

$\frac{dr}{d\theta}\sin^2\theta + r\sin\theta\cos\theta + \frac{dr}{d\theta}\sin\theta\cos\theta + r\cos^2\theta - r^2\frac{dr}{d\theta}\sin\theta\cos\theta - r^3\cos^2\theta$ $- \frac{dr}{d\theta}\sin\theta\cos\theta + r\sin^2\theta + \frac{dr}{d\theta}\cos^2\theta - r\sin\theta\cos\theta + r^2\frac{dr}{d\theta}\sin\theta\cos\theta - r^3\sin^2\theta = 0.$

Collect $\frac{dr}{d\theta}$ terms:

$\frac{dr}{d\theta}\bigl[\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta - \sin\theta\cos\theta - \sin\theta\cos\theta + \cos^2\theta\bigr]$

Wait, let me be more careful. The $\frac{dr}{d\theta}$ terms from $LHS$ are: $\sin^2\theta + \sin\theta\cos\theta - r^2\sin\theta\cos\theta$. From $-RHS$: $-\sin\theta\cos\theta + \cos^2\theta + r^2\sin\theta\cos\theta$. So the coefficient of $\frac{dr}{d\theta}$ is:

$\sin^2\theta + \sin\theta\cos\theta - r^2\sin\theta\cos\theta - \sin\theta\cos\theta + \cos^2\theta + r^2\sin\theta\cos\theta = \sin^2\theta + \cos^2\theta = 1.$

Collect $r$ terms (without $\frac{dr}{d\theta}$):

From $LHS$: $\sin\theta\cos\theta + \cos^2\theta - r^3\cos^2\theta$. From $-RHS$: $\sin^2\theta - \sin\theta\cos\theta - r^3\sin^2\theta$.

The coefficient of $r$ is $\sin\theta\cos\theta + \cos^2\theta + \sin^2\theta - \sin\theta\cos\theta = 1$.

The coefficient of $r^3$ is $-\cos^2\theta - \sin^2\theta = -1$.

So the equation simplifies to:

$\frac{dr}{d\theta} + r - r^3 = 0,$

that is,

$\frac{dr}{d\theta} = r^3 - r = r(r^2 - 1).$

$\blacksquare$

Solving by separation of variables:

$\frac{dr}{r(r^2 - 1)} = d\theta.$

Using partial fractions:

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

Multiplying through: $1 = A(r-1)(r+1) + Br(r+1) + Cr(r-1)$.

Setting $r = 0$: $1 = A(-1)(1)$, so $A = -1$.

Setting $r = 1$: $1 = B(1)(2)$, so $B = \frac{1}{2}$.

Setting $r = -1$: $1 = C(-1)(-2)$, so $C = \frac{1}{2}$.

Therefore

$\int\left(\frac{-1}{r} + \frac{1/2}{r-1} + \frac{1/2}{r+1}\right) dr = \int d\theta,$

$-\ln r + \frac{1}{2}\ln|r-1| + \frac{1}{2}\ln|r+1| = \theta + \text{const},$

$\frac{1}{2}\ln|r^2 - 1| - \ln r = \theta + \text{const},$

$\ln\frac{|r^2 - 1|}{r^2} = 2\theta + \text{const},$

$\frac{r^2 - 1}{r^2} = \pm e^{2\theta + \text{const}} = Be^{2\theta}$

where $B$ is a non-zero constant (absorbing the $\pm$ into the constant, and noting $B$ can be positive or negative; also $B = 0$ corresponds to $r^2 = 1$, i.e., the unit circle $r = 1$).

Solving for $r^2$:

$1 - \frac{1}{r^2} = Be^{2\theta} \qquad \Longrightarrow \qquad \frac{1}{r^2} = 1 - Be^{2\theta} \qquad \Longrightarrow \qquad r^2 = \frac{1}{1 - Be^{2\theta}}.$

Writing $A = -B$ (so $A$ is an arbitrary real constant):

$r^2 = \frac{1}{1 + Ae^{2\theta}}.$

$\blacksquare$

Sketch of solutions according to the value of $A$:

The solutions are $r^2 = \frac{1}{1 + Ae^{2\theta}}$. Since $r^2 > 0$, we need $1 + Ae^{2\theta} > 0$.

$A = 0$: $r^2 = 1$, i.e., $r = 1$. The solution is the unit circle.

$A > 0$: As $\theta \to -\infty$, $Ae^{2\theta} \to 0$, so $r^2 \to 1$. As $\theta$ increases, $Ae^{2\theta}$ grows, so $r^2$ decreases toward $0$. The curve starts near the unit circle for large negative $\theta$ and spirals inward, reaching $r = 0$ when $Ae^{2\theta} \to \infty$, i.e., as $\theta \to \infty$. These are spirals that start near the unit circle and spiral in to the origin.

$A < 0$: Write $A = -|A|$. Then $r^2 = \frac{1}{1 - |A|e^{2\theta}}$. We need $1 - |A|e^{2\theta} > 0$, i.e., $e^{2\theta} < \frac{1}{|A|}$, i.e., $\theta < -\frac{1}{2}\ln|A|$.

• If $|A| < 1$: The curve exists for $\theta < -\frac{1}{2}\ln|A| > 0$. As $\theta \to -\infty$, $r \to 1$ again. As $\theta$ increases toward $-\frac{1}{2}\ln|A|$, the denominator $1 - |A|e^{2\theta} \to 0^+$, so $r^2 \to \infty$. These are spirals that start near the unit circle and spiral outward to infinity.

• If $|A| = 1$: The curve exists for $\theta < 0$. As $\theta \to 0^-$, $r \to \infty$. This is a spiral from the unit circle outward to infinity, defined for negative $\theta$.

• If $|A| > 1$: The curve exists for $\theta < -\frac{1}{2}\ln|A| < 0$. Same qualitative behaviour as above but defined on a shorter interval.

In all cases with $A < 0$, the solutions spiral outward from the unit circle to infinity (over a finite range of $\theta$).

For $A > 0$, the solutions spiral inward from the unit circle to the origin. For $A = 0$, the solution is the unit circle itself. The unit circle acts as a separatrix between the inward-spiralling and outward-spiralling families.

Examiner Notes

超过 80% 的考生尝试此题，是得分最高的题目。(i) 部分许多考生未意识到 r 是 theta 的函数；(ii) 部分变换后的微分方程通常正确，但分离变量后正确使用部分分式积分的频率低于预期。令人惊讶的是，尽管其余部分做得很好或完美，很多人没有尝试画任何解的图。没有人意识到常数 A 在 (ii) 中是真正任意的（因为积分中的对数项有绝对值）。解曲线的绘制考查了最优秀的学生。