STEP3 2009 -- Pure Mathematics

STEP3 2009 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2009  |  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	坐标几何与圆 (Coordinate Geometry and Circles)	Standard	相似三角形,韦达定理,根与系数的关系,对称字母替换
2	微分方程与幂级数 (Differential Equations and Power Series)	Standard	幂级数法,比较系数,递推公式,泰勒级数识别
3	函数分析 (Function Analysis)	Challenging	洛必达法则,泰勒展开,偶函数判定,导数符号分析,渐近线
4	拉普拉斯变换 (Laplace Transforms)	Standard	拉普拉斯变换定义,换元积分,分部积分,逆变换
5	对称多项式与递推关系 (Symmetric Polynomials and Recurrence Relations)	Standard	对称多项式恒等式,韦达定理,特征方程,递推关系
6	复数 (Complex Numbers)	Challenging	欧拉公式,半角公式,因式分解公式,Ptolemy定理
7	微积分 (Calculus)	Challenging	数学归纳法,莱布尼茨公式,乘积求导,递推关系
8	积分 (Integration)	Challenging	极限,分部积分,递推关系,幂级数展开

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Question 1

Topic: 坐标几何与圆 (Coordinate Geometry and Circles)  |  Difficulty: Standard  |  Marks: 20

Problem

1 The points $S, T, U$ and $V$ have coordinates $(s, ms)$, $(t, mt)$, $(u, nu)$ and $(v, nv)$, respectively. The lines $SV$ and $UT$ meet the line $y = 0$ at the points with coordinates $(p, 0)$ and $(q, 0)$, respectively. Show that

$p = \frac{(m - n)sv}{ms - nv} ,$

and write down a similar expression for $q$.

Given that $S$ and $T$ lie on the circle $x^2 + (y - c)^2 = r^2$, find a quadratic equation satisfied by $s$ and by $t$, and hence determine $st$ and $s + t$ in terms of $m, c$ and $r$.

Given that $S, T, U$ and $V$ lie on the above circle, show that $p + q = 0$.

Hint

The result for $p$ can be found via calculating the equation of the line $SV$

$(y - ms = \frac{ms - nv}{s - v}(x - s))$ or similar triangles. The result for $q$ follows from that for $p$

(given in the question) by suitable interchange of letters to give

$q = \frac{(m - n)tu}{mt - nu}$

As $S$ and $T$ lie on the circle, $s$ and $t$ are solutions of the equation

$\lambda^2 + (m\lambda - c)^2 = r^2 \quad \text{i.e. } (1 + m^2)\lambda^2 - 2mc\lambda + (c^2 - r^2) = 0$

and so from considering sum and product of roots, $st = \frac{c^2 - r^2}{1 + m^2}$, and $s + t = \frac{2mc}{1 + m^2}$

Similarly $uv = \frac{c^2 - r^2}{1 + n^2}$, and $u + v = \frac{2nc}{1 + n^2}$ can be deduced by interchanging letters.

Substituting from the earlier results $p + q = \frac{(m - n)sv}{ms - nv} + \frac{(m - n)tu}{mt - nu}$ which can

be simplified to $\frac{(m - n)}{(ms - nv)(mt - nu)}(stm(u + v) - nuv(s + t))$

and then substituting the sum and product results yields the required result.

Model Solution

Finding $p$:

The line $SV$ passes through $S(s, ms)$ and $V(v, nv)$. Its gradient is

$\frac{nv - ms}{v - s} .$

The equation of line $SV$ is

$y - ms = \frac{nv - ms}{v - s}(x - s) .$

Setting $y = 0$ to find where $SV$ meets the $x$-axis:

$-ms = \frac{nv - ms}{v - s}(x - s)$

$x - s = \frac{-ms(v - s)}{nv - ms} = \frac{ms(s - v)}{nv - ms}$

$x = s + \frac{ms(s - v)}{nv - ms} = \frac{s(nv - ms) + ms(s - v)}{nv - ms} = \frac{snv - s \cdot ms + ms \cdot s - ms \cdot v}{nv - ms}$

$= \frac{snv - msv}{nv - ms} = \frac{sv(n - m)}{nv - ms} = \frac{sv(m - n)}{ms - nv} .$

Therefore

$p = \frac{(m - n)sv}{ms - nv} . \qquad \text{(shown)}$

Finding $q$:

By symmetry, $q$ is obtained from $p$ by interchanging $s \leftrightarrow t$ and $v \leftrightarrow u$ (since line $UT$ connects $U(u, nu)$ and $T(t, mt)$). This gives

$q = \frac{(m - n)tu}{mt - nu} .$

Finding the quadratic equation:

Since $S(s, ms)$ lies on $x^2 + (y - c)^2 = r^2$:

$s^2 + (ms - c)^2 = r^2$

$s^2 + m^2 s^2 - 2mcs + c^2 = r^2$

$(1 + m^2)s^2 - 2mcs + (c^2 - r^2) = 0 .$

Since $T(t, mt)$ also lies on the circle, $t$ satisfies the same equation. So $s$ and $t$ are roots of

$(1 + m^2)\lambda^2 - 2mc\lambda + (c^2 - r^2) = 0 .$

By Vieta’s formulas:

$s + t = \frac{2mc}{1 + m^2}, \qquad st = \frac{c^2 - r^2}{1 + m^2} .$

Similarly, since $U(u, nu)$ and $V(v, nv)$ lie on the circle, $u$ and $v$ are roots of

$(1 + n^2)\lambda^2 - 2nc\lambda + (c^2 - r^2) = 0 ,$

so

$u + v = \frac{2nc}{1 + n^2}, \qquad uv = \frac{c^2 - r^2}{1 + n^2} .$

Showing $p + q = 0$:

$p + q = \frac{(m - n)sv}{ms - nv} + \frac{(m - n)tu}{mt - nu}$

$= (m - n)\left[\frac{sv}{ms - nv} + \frac{tu}{mt - nu}\right]$

$= (m - n) \cdot \frac{sv(mt - nu) + tu(ms - nv)}{(ms - nv)(mt - nu)} .$

Expanding the numerator:

$sv(mt - nu) + tu(ms - nv) = stm \cdot v - snv \cdot u + stm \cdot u - tnv \cdot u$

Wait, let me expand carefully:

$sv(mt - nu) + tu(ms - nv) = svmt - svnu + tums - tmsv \cdot \frac{u}{u}$

Let me redo this. We have:

$sv(mt - nu) = svmt - svnu$

$tu(ms - nv) = tums - tunv$

Adding:

$svmt - svnu + tums - tunv = mt(sv + tu) - nu(sv + tv)$

Hmm, let me factor differently:

$= mt \cdot sv + ms \cdot tu - nu \cdot sv - nv \cdot tu$

$= ms \cdot tu + mt \cdot sv - nu \cdot sv - nv \cdot tu$

$= m(tu \cdot s + sv \cdot t) - n(u \cdot sv + v \cdot tu)$

$= mt \cdot us + ms \cdot tu - nu \cdot sv - nv \cdot tu$

Let me try grouping by symmetric sums. The numerator is:

$st \cdot mv + st \cdot mu - nu \cdot sv - nv \cdot tu$

$= st \cdot m(u + v) - uv \cdot n(s + t) .$

Therefore:

$p + q = \frac{(m - n)\left[st \cdot m(u + v) - uv \cdot n(s + t)\right]}{(ms - nv)(mt - nu)} .$

Substituting the expressions for $s + t$, $st$, $u + v$, $uv$:

$st \cdot m(u + v) = \frac{c^2 - r^2}{1 + m^2} \cdot m \cdot \frac{2nc}{1 + n^2} = \frac{2mnc(c^2 - r^2)}{(1 + m^2)(1 + n^2)}$

$uv \cdot n(s + t) = \frac{c^2 - r^2}{1 + n^2} \cdot n \cdot \frac{2mc}{1 + m^2} = \frac{2mnc(c^2 - r^2)}{(1 + m^2)(1 + n^2)}$

These are equal, so

$st \cdot m(u + v) - uv \cdot n(s + t) = 0 ,$

and hence $p + q = 0$. $\qquad \text{(shown)}$

Examiner Notes

A popular question attempted by more than four fifths of the candidates, and scoring as well as any question, and most successfully obtained expressions for $p$ and $q$. Quite a lot also obtained the quadratic equation and from it the sum and product of roots for $s$ and $t$. However, a common error at this stage was to overlook the coefficient of the second degree term not being 1. For this reason, or otherwise, because they didn’t know what to do many “fell at the last hurdle”, although a good number completed the question successfully.

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Question 2

Topic: 微分方程与幂级数 (Differential Equations and Power Series)  |  Difficulty: Standard  |  Marks: 20

Problem

2 (i) Let $y = \sum_{n=0}^{\infty} a_n x^n$, where the coefficients $a_n$ are independent of $x$ and are such that this series and all others in this question converge. Show that

$y' = \sum_{n=1}^{\infty} n a_n x^{n-1} ,$

and write down a similar expression for $y''$.

Write out explicitly each of the three series as far as the term containing $a_3$.

(ii) It is given that $y$ satisfies the differential equation

$xy'' - y' + 4x^3y = 0 .$

By substituting the series of part (i) into the differential equation and comparing coefficients, show that $a_1 = 0$.

Show that, for $n \geqslant 4$,

$a_n = -\frac{4}{n(n - 2)} a_{n-4} ,$

and that, if $a_0 = 1$ and $a_2 = 0$, then $y = \cos(x^2)$.

Find the corresponding result when $a_0 = 0$ and $a_2 = 1$.

Hint

(i) The five required results are straightforward to write down, merely observing that initial terms in the summations are zero.

(ii) Substituting the series from (i) in the differential equation yields that

$-a_1 + 3a_3x^2 + (8a_4 + 4a_0)x^3 + \dots = 0$, after having collected like terms.

Thus, comparing constants and $x^2$ coefficients $a_1 = 0$ and $a_3 = 0$

Comparing coefficients of $x^{n-1}$, for $n \ge 4$, $n(n - 1)a_n - na_n + 4a_{n-4} = 0$ which gives the required result upon rearrangement.

With $a_0 = 1$, $a_2 = 0$, and as $a_1 = 0$, and $a_3 = 0$, we find $a_4 = \frac{-1}{2!}$, $a_5 = 0$, $a_6 = 0$,

$a_7 = 0$, $a_8 = \frac{1}{4!}$, etc.

Thus $y = 1 - \frac{1}{2!}(x^2)^2 + \frac{1}{4!}(x^2)^4 - \frac{1}{6!}(x^2)^6 + \dots = \cos(x^2)$

With $a_0 = 0$, $a_2 = 1$, $y = (x^2) - \frac{1}{3!}(x^2)^3 + \frac{1}{5!}(x^2)^5 - \frac{1}{7!}(x^2)^7 + \dots = \sin(x^2)$

Model Solution

Part (i)

Given $y = \sum_{n=0}^{\infty} a_n x^n$, we differentiate term by term:

$y' = \sum_{n=1}^{\infty} n a_n x^{n-1} . \qquad \text{(shown)}$

The sum starts at $n = 1$ because the $n = 0$ term is a constant, which vanishes on differentiation. Similarly:

$y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} .$

Writing out explicitly:

$y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$

$y' = a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3 + \cdots$

$y'' = 2a_2 + 6a_3 x + 12a_4 x^2 + 20a_5 x^3 + \cdots$

Part (ii)

We substitute into $xy'' - y' + 4x^3 y = 0$.

First, $xy'' = x(2a_2 + 6a_3 x + 12a_4 x^2 + \cdots) = 2a_2 x + 6a_3 x^2 + 12a_4 x^3 + \cdots$

In general, $xy'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1}$.

So $xy'' - y' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} - \sum_{n=1}^{\infty} n a_n x^{n-1}$.

For $n \geqslant 2$, the coefficient of $x^{n-1}$ in $xy'' - y'$ is $n(n-1)a_n - na_n = n(n-2)a_n$.

For $n = 1$: the coefficient of $x^0$ is $-a_1$ (only the $y'$ term contributes).

Therefore $xy'' - y' = -a_1 + \sum_{n=2}^{\infty} n(n-2) a_n x^{n-1}$.

Next, $4x^3 y = 4x^3 \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} 4a_n x^{n+3}$.

Setting $k = n + 3$, this is $\sum_{k=3}^{\infty} 4a_{k-3} x^k$.

Re-indexing $xy'' - y'$ with $k = n - 1$ (so $n = k + 1$): the sum becomes $\sum_{k=1}^{\infty} (k+1)(k-1) a_{k+1} x^k$.

So the differential equation $xy'' - y' + 4x^3 y = 0$ gives:

$-a_1 + \sum_{k=1}^{\infty} (k+1)(k-1) a_{k+1} x^k + \sum_{k=3}^{\infty} 4a_{k-3} x^k = 0 .$

Comparing coefficients:

• Constant term ($x^0$): $-a_1 = 0$, so $a_1 = 0$. $\qquad \text{(shown)}$

• Coefficient of $x^1$: $(2)(0) a_2 = 0$, which is automatically satisfied.

• Coefficient of $x^2$: $(3)(1) a_3 = 0$, so $a_3 = 0$.

• Coefficient of $x^k$ for $k \geqslant 3$: $(k+1)(k-1) a_{k+1} + 4a_{k-3} = 0$.

Setting $n = k + 1$ (so $n \geqslant 4$ and $k = n - 1$):

$n(n-2) a_n + 4a_{n-4} = 0$

$a_n = -\frac{4}{n(n-2)} a_{n-4} . \qquad \text{(shown)}$

Case $a_0 = 1$, $a_2 = 0$:

We have $a_1 = 0$, $a_3 = 0$. Using the recurrence:

$a_4 = -\frac{4}{4 \cdot 2} a_0 = -\frac{1}{2} \cdot 1 = -\frac{1}{2}$

$a_5 = -\frac{4}{5 \cdot 3} a_1 = 0$

$a_6 = -\frac{4}{6 \cdot 4} a_2 = 0$

$a_7 = -\frac{4}{7 \cdot 5} a_3 = 0$

$a_8 = -\frac{4}{8 \cdot 6} a_4 = -\frac{1}{12} \cdot \left(-\frac{1}{2}\right) = \frac{1}{24}$

$a_{12} = -\frac{4}{12 \cdot 10} a_8 = -\frac{1}{30} \cdot \frac{1}{24} = -\frac{1}{720}$

So only $a_0, a_4, a_8, a_{12}, \ldots$ are nonzero. In general, $a_{4k} = \frac{(-1)^k}{(2k)!}$ (verified: $a_0 = 1 = \frac{1}{0!}$, $a_4 = -\frac{1}{2} = \frac{-1}{2!}$, $a_8 = \frac{1}{24} = \frac{1}{4!}$).

Therefore:

$y = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{4k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} (x^2)^{2k} = \cos(x^2) . \qquad \text{(shown)}$

Case $a_0 = 0$, $a_2 = 1$:

We have $a_1 = 0$, $a_3 = 0$. Using the recurrence:

$a_6 = -\frac{4}{6 \cdot 4} a_2 = -\frac{1}{6}$

$a_{10} = -\frac{4}{10 \cdot 8} a_6 = -\frac{1}{20} \cdot \left(-\frac{1}{6}\right) = \frac{1}{120}$

So only $a_2, a_6, a_{10}, \ldots$ are nonzero. In general, $a_{4k+2} = \frac{(-1)^k}{(2k+1)!}$ (verified: $a_2 = 1 = \frac{1}{1!}$, $a_6 = -\frac{1}{6} = \frac{-1}{3!}$, $a_{10} = \frac{1}{120} = \frac{1}{5!}$).

Therefore:

$y = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{4k+2} = x^2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} (x^2)^{2k} = \sin(x^2) .$

Examiner Notes

This was similar to question 1 in popularity and success. Virtually all got part (i) correct, and many used the series correctly to obtain the value for $a_1$. Quite a few completed the question, although frequently candidates dropped 2 marks through not looking at terms properly.

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Question 3

Topic: 函数分析 (Function Analysis)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 The function $f(t)$ is defined, for $t \neq 0$, by

$f(t) = \frac{t}{e^t - 1} .$

(i) By expanding $e^t$, show that $\lim_{t \to 0} f(t) = 1$. Find $f'(t)$ and evaluate $\lim_{t \to 0} f'(t)$.

(ii) Show that $f(t) + \frac{1}{2}t$ is an even function. [Note: A function $g(t)$ is said to be even if $g(t) \equiv g(-t)$.]

(iii) Show with the aid of a sketch that $e^t(1 - t) \leqslant 1$ and deduce that $f'(t) \neq 0$ for $t \neq 0$.

Sketch the graph of $f(t)$.

Hint

(i) Substituting the power series and tidying up the algebra yields

$f(t) = \frac{1}{\left( 1 + \frac{t}{2!} + \dots \right)} \text{ and so } \lim_{t \to 0} f(t) = 1.$

Similarly, $f'(t) = \frac{(e^t - 1) - te^t}{(e^t - 1)^2} = \frac{-\frac{1}{2} - t \left( \frac{1}{2!} - \frac{1}{3!} \right) - \dots}{\left( 1 + \frac{t}{2!} + \dots \right)^2}$ and so $\lim_{t \to 0} f'(t) = \frac{-1}{2}$

(Alternatively, this can be obtained by de l’Hopital.)

(ii) If we let $g(t) = f(t) + \frac{1}{2}t$, then simplifying the algebra gives

$g(t) = \frac{t(e^t + 1)}{2(e^t - 1)}$

after which it is can be shown by substituting $-t$ for $t$ that $g(-t)$ is the same expression.

(iii) If we let $h(t) = e^t(1 - t)$, and find its stationary point, sketching the graph gives

Hence $e^t(1 - t) \leq 1$ and so $e^t(1 - t) - 1 \leq 0$. (Alternatively, a sketch with $e^t$ and $\frac{1}{1-t}$ will yield the result.)

Thus $f'(t) = \frac{(1-t)e^t - 1}{(e^t - 1)^2} \leq 0$, with equality only possible for $t = 0$, but we know

$\lim_{t \to 0} f'(t) = \frac{-1}{2}$ and so, in fact, $f(t)$ is always decreasing i.e. has no turning points.

Considering the graph of $g(t) = f(t) + \frac{1}{2}t$. It passes through $(0,1)$, is symmetrical and approaches $y = \frac{1}{2}t$ as $t \to \infty$ and thus is

Therefore the graph of $f(t) = g(t) - \frac{1}{2}t$ also passes through $(0,1)$, and has asymptotes $y = 0$ and $y = -t$ and thus is

Model Solution

Part (i)

We expand $e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots$, so

$e^t - 1 = t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots = t\left(1 + \frac{t}{2!} + \frac{t^2}{3!} + \cdots\right) .$

Therefore

$f(t) = \frac{t}{e^t - 1} = \frac{t}{t\left(1 + \frac{t}{2!} + \frac{t^2}{3!} + \cdots\right)} = \frac{1}{1 + \frac{t}{2!} + \frac{t^2}{3!} + \cdots} .$

As $t \to 0$, the denominator tends to $1$, so

$\lim_{t \to 0} f(t) = 1 . \qquad \text{(shown)}$

Now we find $f'(t)$ using the quotient rule:

$f'(t) = \frac{(e^t - 1) \cdot 1 - t \cdot e^t}{(e^t - 1)^2} = \frac{e^t - 1 - te^t}{(e^t - 1)^2} = \frac{(1 - t)e^t - 1}{(e^t - 1)^2} .$

To evaluate $\lim_{t \to 0} f'(t)$, we substitute the power series. The numerator is

$(1 - t)e^t - 1 = (1 - t)\left(1 + t + \frac{t^2}{2!} + \cdots\right) - 1$

$= 1 + t + \frac{t^2}{2!} + \cdots - t - t^2 - \frac{t^3}{2!} - \cdots - 1$

$= -\frac{t^2}{2} - \frac{t^3}{3} - \cdots$

and the denominator is

$(e^t - 1)^2 = \left(t + \frac{t^2}{2!} + \cdots\right)^2 = t^2 + t^3 + \cdots$

Dividing numerator and denominator by $t^2$:

$f'(t) = \frac{-\frac{1}{2} - \frac{t}{3} - \cdots}{1 + t + \cdots} .$

As $t \to 0$, this tends to $-\frac{1}{2}$, so

$\lim_{t \to 0} f'(t) = -\frac{1}{2} .$

Part (ii)

Let $g(t) = f(t) + \frac{1}{2}t$. We compute:

$g(t) = \frac{t}{e^t - 1} + \frac{t}{2} = \frac{2t + t(e^t - 1)}{2(e^t - 1)} = \frac{2t + te^t - t}{2(e^t - 1)} = \frac{t(e^t + 1)}{2(e^t - 1)} .$

Now we evaluate $g(-t)$:

$g(-t) = \frac{(-t)(e^{-t} + 1)}{2(e^{-t} - 1)} .$

Multiplying numerator and denominator by $e^t$:

$g(-t) = \frac{(-t)(1 + e^t)}{2(1 - e^t)} = \frac{(-t)(e^t + 1)}{-(2(e^t - 1))} = \frac{t(e^t + 1)}{2(e^t - 1)} = g(t) .$

Since $g(-t) = g(t)$ for all $t \neq 0$, the function $g(t) = f(t) + \frac{1}{2}t$ is even. $\qquad \text{(shown)}$

Part (iii)

Let $h(t) = e^t(1 - t)$. Then

$h'(t) = e^t(1 - t) + e^t(-1) = -te^t .$

Since $e^t > 0$ for all $t$:

• For $t > 0$: $h'(t) < 0$, so $h$ is decreasing.
• For $t < 0$: $h'(t) > 0$, so $h$ is increasing.
• At $t = 0$: $h'(0) = 0$, so $t = 0$ is a stationary point (in fact a global maximum).

Therefore $h(t) \leqslant h(0) = e^0(1 - 0) = 1$ for all $t$, with equality only at $t = 0$:

$e^t(1 - t) \leqslant 1 \qquad \text{for all } t . \qquad \text{(shown)}$

From part (i), we have

$f'(t) = \frac{(1 - t)e^t - 1}{(e^t - 1)^2} .$

Since $(1 - t)e^t \leqslant 1$, the numerator satisfies $(1 - t)e^t - 1 \leqslant 0$ for all $t$. The denominator $(e^t - 1)^2 > 0$ for all $t \neq 0$. Therefore $f'(t) \leqslant 0$ for $t \neq 0$.

Equality $f'(t) = 0$ would require $(1 - t)e^t = 1$, which holds only at $t = 0$. Since $t \neq 0$, we conclude

$f'(t) \neq 0 \text{ for } t \neq 0 . \qquad \text{(shown)}$

In fact, $f'(t) < 0$ for all $t \neq 0$, so $f$ is strictly decreasing on $(-\infty, 0)$ and on $(0, \infty)$.

Sketch of $f(t)$:

From part (i): $f(0) = 1$ and $f'(0) = -\frac{1}{2}$ (the function has a removable singularity at $t = 0$).

From part (ii): $f(t) = g(t) - \frac{1}{2}t$ where $g$ is even.

As $t \to +\infty$: $f(t) = \frac{t}{e^t - 1} \to 0^+$ (the exponential dominates).

As $t \to -\infty$: $f(t) = \frac{t}{e^t - 1} \approx \frac{t}{-1} = -t \to +\infty$, so $f(t) \sim -t$ is an asymptote.

The graph passes through $(0, 1)$, is strictly decreasing, approaches $0$ from above as $t \to +\infty$, and approaches the line $y = -t$ as $t \to -\infty$.

Examiner Notes

Though slightly more popular than the first 2 questions, the attempts scored marginally less well. Candidates began well, though the limit of $f'(t)$ was not well done. The even function in (ii) was usually correctly justified. Part (iii) was frequently not quite correctly justified, though some did so by sketching $y = e^{-t^2}$ and $y = 1 - \frac{1}{2}t^2$. The sketch of $f(t)$ often had the wrong gradient as it approached the y axis, and asymptotes were frequently not identified.

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Question 4

Topic: 拉普拉斯变换 (Laplace Transforms)  |  Difficulty: Standard  |  Marks: 20

Problem

4 For any given (suitable) function $f$, the Laplace transform of $f$ is the function $F$ defined by

$F(s) = \int_0^\infty e^{-st} f(t) dt \quad (s > 0) .$

(i) Show that the Laplace transform of $e^{-bt} f(t)$, where $b > 0$, is $F(s + b)$.

(ii) Show that the Laplace transform of $f(at)$, where $a > 0$, is $a^{-1} F(\frac{s}{a})$.

(iii) Show that the Laplace transform of $f'(t)$ is $sF(s) - f(0)$.

(iv) In the case $f(t) = \sin t$, show that $F(s) = \frac{1}{s^2 + 1}$.

Using only these four results, find the Laplace transform of $e^{-pt} \cos qt$, where $p > 0$ and $q > 0$.

Hint

(i) Substituting into the definition yields the Laplace transform as $\int_{0}^{\infty} e^{-st} e^{-bt} f(t) dt = \int_{0}^{\infty} e^{-t(s+b)} f(t) dt = F(s+b)$

(ii) Similarly, a change of variable in the integral using $u = at$ yields the result.

(iii) Integrating by parts yields this answer.

(iv) A repeated integration by parts obtains $F(s) = 1 - s^2 F(s)$ which leads to the stated result.

Using the results obtained in the question, the transform of $\cos qt$ is

$q^{-1}\left(\frac{s/q}{s^2/q^2+1}\right) = \frac{s}{s^2+q^2}, \text{and so the transform of } e^{-pt} \cos qt \text{ is } \frac{(s+p)}{(s+p)^2+q^2}$

Model Solution

Part (i)

The Laplace transform of $e^{-bt}f(t)$ is

$\int_0^\infty e^{-st} \cdot e^{-bt} f(t) \, dt = \int_0^\infty e^{-(s+b)t} f(t) \, dt = F(s + b) . \qquad \text{(shown)}$

Part (ii)

The Laplace transform of $f(at)$ is

$\int_0^\infty e^{-st} f(at) \, dt .$

Let $u = at$, so $t = \frac{u}{a}$ and $dt = \frac{du}{a}$. When $t = 0$, $u = 0$; when $t \to \infty$, $u \to \infty$ (since $a > 0$). Substituting:

$\int_0^\infty e^{-s \cdot u/a} f(u) \cdot \frac{du}{a} = \frac{1}{a} \int_0^\infty e^{-(s/a)u} f(u) \, du = \frac{1}{a} F\!\left(\frac{s}{a}\right) . \qquad \text{(shown)}$

Part (iii)

The Laplace transform of $f'(t)$ is

$\int_0^\infty e^{-st} f'(t) \, dt .$

We integrate by parts with $u = e^{-st}$ and $dv = f'(t)\,dt$, so $du = -se^{-st}\,dt$ and $v = f(t)$:

$\int_0^\infty e^{-st} f'(t) \, dt = \left[ e^{-st} f(t) \right]_0^\infty - \int_0^\infty (-s)e^{-st} f(t) \, dt .$

For the boundary term: as $t \to \infty$, $e^{-st} f(t) \to 0$ (since the Laplace transform is assumed to converge, meaning $f(t)$ grows no faster than exponentially, and $e^{-st}$ decays for $s > 0$). At $t = 0$: $e^0 f(0) = f(0)$. Therefore

$\left[ e^{-st} f(t) \right]_0^\infty = 0 - f(0) = -f(0) .$

Hence

$\int_0^\infty e^{-st} f'(t) \, dt = -f(0) + s \int_0^\infty e^{-st} f(t) \, dt = sF(s) - f(0) . \qquad \text{(shown)}$

Part (iv)

Let $f(t) = \sin t$, so $f'(t) = \cos t$ and $f(0) = \sin 0 = 0$. Let $F(s)$ denote the Laplace transform of $\sin t$, and let $G(s)$ denote the Laplace transform of $\cos t$.

From part (iii), the Laplace transform of $f'(t) = \cos t$ is $sF(s) - f(0) = sF(s)$. Therefore

$G(s) = sF(s) . \qquad (*)$

Now apply part (iii) again to $f(t) = \cos t$ (so $f'(t) = -\sin t$ and $f(0) = \cos 0 = 1$): the Laplace transform of $-\sin t$ is $sG(s) - 1$, i.e.

$-F(s) = sG(s) - 1 . \qquad (**)$

Substituting $(*)$ into $(**)$:

$-F(s) = s \cdot sF(s) - 1 = s^2 F(s) - 1 .$

Rearranging:

$F(s) + s^2 F(s) = 1$

$F(s)(s^2 + 1) = 1$

$F(s) = \frac{1}{s^2 + 1} . \qquad \text{(shown)}$

Finding the Laplace transform of $e^{-pt}\cos qt$:

Step 1. From $(*)$, $G(s) = sF(s) = \frac{s}{s^2 + 1}$, so the Laplace transform of $\cos t$ is $\frac{s}{s^2 + 1}$.

Step 2. Apply part (ii) with $a = q$ to find the Laplace transform of $\cos qt$: this is

$\frac{1}{q} G\!\left(\frac{s}{q}\right) = \frac{1}{q} \cdot \frac{s/q}{(s/q)^2 + 1} = \frac{1}{q} \cdot \frac{s/q}{s^2/q^2 + 1} = \frac{1}{q} \cdot \frac{s/q \cdot q^2}{s^2 + q^2} = \frac{s}{s^2 + q^2} .$

Step 3. Apply part (i) with $b = p$ to find the Laplace transform of $e^{-pt}\cos qt$: replace $s$ by $s + p$:

$\mathcal{L}\{e^{-pt}\cos qt\} = \frac{s + p}{(s + p)^2 + q^2} .$

Examiner Notes

About half the candidates attempted this, with similar levels of success to question 3. Parts (i) and (iii) caused few problems though part (ii) did. There were some errors in part (iv), but it was the last part using the four results that usually went wrong.

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Question 5

Topic: 对称多项式与递推关系 (Symmetric Polynomials and Recurrence Relations)  |  Difficulty: Standard  |  Marks: 20

Problem

5 The numbers $x, y$ and $z$ satisfy

$$\begin{aligned} x + y + z &= 1 \\ x^2 + y^2 + z^2 &= 2 \\ x^3 + y^3 + z^3 &= 3 \, . \end{aligned}$$

Show that

$yz + zx + xy = -\frac{1}{2} \, .$

Show also that $x^2y + x^2z + y^2z + y^2x + z^2x + z^2y = -1$ , and hence that

$xyz = \frac{1}{6} \, .$

Let $S_n = x^n + y^n + z^n$ . Use the above results to find numbers $a, b$ and $c$ such that the relation

$S_{n+1} = aS_n + bS_{n-1} + cS_{n-2} \, ,$

holds for all $n$.

Hint

The first result may be obtained by considering

$(x+y+z)^2 - (x^2+y^2+z^2) = 2(yz+zx+xy)$ the second by $(x^2+y^2+z^2)(x+y+z) = x^3+y^3+z^3 + (x^2y+x^2z+y^2z+y^2z+z^2x+z^2y)$ and the third by $(x+y+z)^3 = (x^3+y^3+z^3) + 3(x^2y+x^2z+y^2z+y^2z+z^2x+z^2y) + 6xyz$

Considering sums and products of roots, we can deduce that $x$ satisfies the cubic equation $x^3 - x^2 - \frac{1}{2}x - \frac{1}{6} = 0$, as do $y$ and $z$ by symmetry. Multiplying by $x^{n-2}$, $x^{n+1} = x^n + \frac{1}{2}x^{n-1} + \frac{1}{6}x^{n-2}$, with similar results for $y$ and $z$. Summing these yields

$S_{n+1} = S_n + \frac{1}{2}S_{n-1} + \frac{1}{6}S_{n-2}$

Alternatively, $x^{n+1} + y^{n+1} + z^{n+1} = (x+y+z)(x^n+y^n+z^n) - (xy^n+xz^n+yx^n+yz^n+zx^n+zy^n)$ $= 1.S_n - (xy+yz+zx)(x^{n-1}+y^{n-1}+z^{n-1}) + xyz(x^{n-2}+y^{n-2}+z^{n-2})$ to give the result.

Model Solution

Showing $yz + zx + xy = -\frac{1}{2}$:

We use the identity

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) .$

Substituting the given values:

$1^2 = 2 + 2(xy + yz + zx)$

$1 = 2 + 2(xy + yz + zx)$

$xy + yz + zx = -\frac{1}{2} . \qquad \text{(shown)}$

Showing $x^2y + x^2z + y^2z + y^2x + z^2x + z^2y = -1$:

We use the identity

$(x^2 + y^2 + z^2)(x + y + z) = x^3 + y^3 + z^3 + (x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) .$

Substituting the given values:

$2 \cdot 1 = 3 + (x^2y + x^2z + y^2x + y^2z + z^2x + z^2y)$

$x^2y + x^2z + y^2x + y^2z + z^2x + z^2y = -1 . \qquad \text{(shown)}$

Showing $xyz = \frac{1}{6}$:

We use the identity

$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) + 6xyz .$

Substituting the known values:

$1^3 = 3 + 3(-1) + 6xyz$

$1 = 3 - 3 + 6xyz$

$xyz = \frac{1}{6} . \qquad \text{(shown)}$

Finding the recurrence relation for $S_n$:

By Vieta’s formulas, $x$, $y$ and $z$ are roots of the cubic

$\lambda^3 - (x+y+z)\lambda^2 + (xy+yz+zx)\lambda - xyz = 0$

$\lambda^3 - \lambda^2 - \frac{1}{2}\lambda - \frac{1}{6} = 0 .$

Since $x$ satisfies this equation:

$x^3 = x^2 + \frac{1}{2}x + \frac{1}{6} .$

Multiplying both sides by $x^{n-2}$ (for $n \geqslant 2$):

$x^{n+1} = x^n + \frac{1}{2}x^{n-1} + \frac{1}{6}x^{n-2} .$

Similarly for $y$ and $z$:

$y^{n+1} = y^n + \frac{1}{2}y^{n-1} + \frac{1}{6}y^{n-2}$

$z^{n+1} = z^n + \frac{1}{2}z^{n-1} + \frac{1}{6}z^{n-2}$

Adding the three equations:

$S_{n+1} = S_n + \frac{1}{2}S_{n-1} + \frac{1}{6}S_{n-2} .$

Therefore $a = 1$, $b = \frac{1}{2}$, $c = \frac{1}{6}$.

Examiner Notes

This was the most popular question, with a few more attempts than question 3, but with a level of success matching the first two questions. Many showed the first two results correctly, and quite a few the third one. The last part tripped up many candidates, though the most successful used the first approach in the mark scheme. A number of candidates understood “independent of $n$” in the question to be given information, and attempted to find $a$, $b$, and $c$ by solving three simultaneous equations for specific values of $n$. However, there were commonly errors in the values of the $S_n$ used. An efficient alternative solution is given in the mark scheme.

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Question 6

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

Problem

6 Show that $|e^{i\beta} - e^{i\alpha}| = 2 \sin \frac{1}{2}(\beta - \alpha)$ for $0 < \alpha < \beta < 2\pi$ . Hence show that

$|e^{i\alpha} - e^{i\beta}| \ |e^{i\gamma} - e^{i\delta}| + |e^{i\beta} - e^{i\gamma}| \ |e^{i\alpha} - e^{i\delta}| = |e^{i\alpha} - e^{i\gamma}| \ |e^{i\beta} - e^{i\delta}| \, ,$

where $0 < \alpha < \beta < \gamma < \delta < 2\pi$.

Interpret this result as a theorem about cyclic quadrilaterals.

Hint

6. Using Euler, $e^{i\beta} - e^{i\alpha} = (\cos\beta - \cos\alpha) + i(\sin\beta - \sin\alpha)$ and so $|e^{i\beta} - e^{i\alpha}|^2 = (\cos\beta - \cos\alpha)^2 + (\sin\beta - \sin\alpha)^2$

which can be expanded, and then using Pythagoras, compound and half angle formulae this becomes

$4\sin^2 \frac{1}{2}(\beta - \alpha)$

$|e^{i\beta} - e^{i\alpha}| = 2\sin \frac{1}{2}(\beta - \alpha) \text{ as both expressions are positive.}$

Alternative methods employ the factor formulae.

$|e^{i\alpha} - e^{i\beta}||e^{i\gamma} - e^{i\delta}| + |e^{i\beta} - e^{i\gamma}||e^{i\alpha} - e^{i\delta}|$ $= 2\sin\left(\frac{1}{2}(\alpha - \beta)\right)2\sin\left(\frac{1}{2}(\gamma - \delta)\right) + 2\sin\left(\frac{1}{2}(\beta - \gamma)\right)2\sin\left(\frac{1}{2}(\alpha - \delta)\right)$ which by use of the factor formulae and cancelling terms may be written $2\left(\cos\left(\frac{1}{2}(\alpha - \beta - \gamma + \delta)\right) - \cos\left(\frac{1}{2}(\beta - \gamma + \alpha - \delta)\right)\right)$

and then again by factor formulae,

$2 \sin \left( \frac{1}{2} (\alpha - \gamma) \right) 2 \sin \left( \frac{1}{2} (\beta - \delta) \right)$

which is

$|e^{i\alpha} - e^{i\gamma}| |e^{i\beta} - e^{i\delta}| \text{ as required.}$

Thus, the product of the diagonals of a cyclic quadrilateral is equal to the sum of the products of the opposite pairs of sides (Ptolemy’s Theorem).

Model Solution

Showing $|e^{i\beta} - e^{i\alpha}| = 2\sin\frac{1}{2}(\beta - \alpha)$:

Using Euler’s formula:

$|e^{i\beta} - e^{i\alpha}|^2 = (\cos\beta - \cos\alpha)^2 + (\sin\beta - \sin\alpha)^2$

$= \cos^2\beta - 2\cos\alpha\cos\beta + \cos^2\alpha + \sin^2\beta - 2\sin\alpha\sin\beta + \sin^2\alpha$

$= 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta)$

$= 2 - 2\cos(\beta - \alpha) .$

Using the identity $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$:

$|e^{i\beta} - e^{i\alpha}|^2 = 2 \cdot 2\sin^2\frac{\beta - \alpha}{2} = 4\sin^2\frac{\beta - \alpha}{2} .$

Since $0 < \alpha < \beta < 2\pi$, we have $0 < \frac{\beta - \alpha}{2} < \pi$, so $\sin\frac{\beta - \alpha}{2} > 0$. Therefore

$|e^{i\beta} - e^{i\alpha}| = 2\sin\frac{\beta - \alpha}{2} . \qquad \text{(shown)}$

Proving the identity:

We need to show

$|e^{i\alpha} - e^{i\beta}| \cdot |e^{i\gamma} - e^{i\delta}| + |e^{i\beta} - e^{i\gamma}| \cdot |e^{i\alpha} - e^{i\delta}| = |e^{i\alpha} - e^{i\gamma}| \cdot |e^{i\beta} - e^{i\delta}| .$

Using the result above, each modulus can be written as a sine:

$|e^{i\alpha} - e^{i\beta}| = 2\sin\frac{\beta - \alpha}{2}, \quad |e^{i\gamma} - e^{i\delta}| = 2\sin\frac{\delta - \gamma}{2}$

$|e^{i\beta} - e^{i\gamma}| = 2\sin\frac{\gamma - \beta}{2}, \quad |e^{i\alpha} - e^{i\delta}| = 2\sin\frac{\delta - \alpha}{2}$

$|e^{i\alpha} - e^{i\gamma}| = 2\sin\frac{\gamma - \alpha}{2}, \quad |e^{i\beta} - e^{i\delta}| = 2\sin\frac{\delta - \beta}{2}$

So we need to prove:

$4\sin\frac{\beta - \alpha}{2}\sin\frac{\delta - \gamma}{2} + 4\sin\frac{\gamma - \beta}{2}\sin\frac{\delta - \alpha}{2} = 4\sin\frac{\gamma - \alpha}{2}\sin\frac{\delta - \beta}{2} .$

Dividing by 4, we need:

$\sin\frac{\beta - \alpha}{2}\sin\frac{\delta - \gamma}{2} + \sin\frac{\gamma - \beta}{2}\sin\frac{\delta - \alpha}{2} = \sin\frac{\gamma - \alpha}{2}\sin\frac{\delta - \beta}{2} . \qquad (\dagger)$

We use the product-to-sum formula $\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]$.

Let $A = \frac{\beta - \alpha}{2}$ and $B = \frac{\delta - \gamma}{2}$. Then:

$\sin A \sin B = \frac{1}{2}\left[\cos\left(\frac{\beta - \alpha - \delta + \gamma}{2}\right) - \cos\left(\frac{\beta - \alpha + \delta - \gamma}{2}\right)\right] .$

Let $C = \frac{\gamma - \beta}{2}$ and $D = \frac{\delta - \alpha}{2}$. Then:

$\sin C \sin D = \frac{1}{2}\left[\cos\left(\frac{\gamma - \beta - \delta + \alpha}{2}\right) - \cos\left(\frac{\gamma - \beta + \delta - \alpha}{2}\right)\right] .$

Notice that $A - B = \frac{\beta - \alpha - \delta + \gamma}{2}$ and $C - D = \frac{\gamma - \beta - \delta + \alpha}{2}$, so $\cos(A - B) = \cos(D - C)$ and $\cos(C - D) = \cos(D - C)$. These terms are equal.

Also $A + B = \frac{\beta - \alpha + \delta - \gamma}{2}$ and $C + D = \frac{\gamma - \beta + \delta - \alpha}{2}$, so $A + B + C + D = \delta - \alpha$.

Adding the two products:

$\sin A \sin B + \sin C \sin D = \frac{1}{2}\left[2\cos(A - B) - \cos(A + B) - \cos(C + D)\right]$

$= \cos(A - B) - \frac{1}{2}[\cos(A + B) + \cos(C + D)] .$

Using $\cos P + \cos Q = 2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2}$ with $P = A + B$ and $Q = C + D$:

$A + B + C + D = \frac{\beta - \alpha + \delta - \gamma + \gamma - \beta + \delta - \alpha}{2} = \delta - \alpha$

$A + B - C - D = \frac{\beta - \alpha + \delta - \gamma - \gamma + \beta - \delta + \alpha}{2} = \beta - \gamma$

So:

$\cos(A+B) + \cos(C+D) = 2\cos\frac{\delta - \alpha}{2}\cos\frac{\beta - \gamma}{2} .$

Also:

$A - B = \frac{\beta - \alpha - \delta + \gamma}{2} = \frac{(\gamma - \alpha) - (\delta - \beta)}{2} .$

Let $P = \frac{\gamma - \alpha}{2}$ and $Q = \frac{\delta - \beta}{2}$. Then $A - B = P - Q$, and $A + B + C + D = \delta - \alpha = 2P + 2Q - (\gamma - \beta) = \ldots$ Let me just directly verify by computing the RHS.

The RHS of $(\dagger)$ is:

$\sin\frac{\gamma - \alpha}{2}\sin\frac{\delta - \beta}{2} = \frac{1}{2}\left[\cos\frac{(\gamma - \alpha) - (\delta - \beta)}{2} - \cos\frac{(\gamma - \alpha) + (\delta - \beta)}{2}\right]$

$= \frac{1}{2}\left[\cos\frac{\gamma - \alpha - \delta + \beta}{2} - \cos\frac{\gamma - \alpha + \delta - \beta}{2}\right] .$

Note $\frac{\gamma - \alpha - \delta + \beta}{2} = A - B$ and $\frac{\gamma - \alpha + \delta - \beta}{2} = \frac{\delta - \alpha}{2} + \frac{\gamma - \beta}{2} = D + C$.

So the RHS is $\frac{1}{2}[\cos(A - B) - \cos(C + D)]$.

For the LHS, from the computation above:

$\text{LHS} = \cos(A - B) - \frac{1}{2}[\cos(A + B) + \cos(C + D)]$

Now we need to check $\cos(A+B)$. We have $A + B = \frac{\beta - \alpha + \delta - \gamma}{2}$.

Note that $A - B = \frac{\beta - \alpha - \delta + \gamma}{2}$, so $\cos(A-B) = \cos\frac{\gamma - \alpha - (\delta - \beta)}{2}$.

And $C + D = \frac{\gamma - \beta + \delta - \alpha}{2}$. So $A + B = \delta - \alpha - (C + D) = (A + B + C + D) - (C + D)$.

Actually, let me just directly verify $A + B = \frac{\delta - \alpha + \beta - \gamma}{2}$ and $C + D = \frac{\delta - \alpha + \gamma - \beta}{2}$, so $A + B$ and $C + D$ sum to $\delta - \alpha$ and differ by $\beta - \gamma$.

Using $\cos(A+B) = \cos(\delta - \alpha - (C+D))$, this makes the algebra messy. Let me try a cleaner approach.

We use the identity: $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$.

LHS:

$\sin A \sin B + \sin C \sin D = \frac{1}{2}[\cos(A-B) - \cos(A+B) + \cos(C-D) - \cos(C+D)]$

Now $A - B = \frac{\beta - \alpha - \delta + \gamma}{2}$ and $C - D = \frac{\gamma - \beta - \delta + \alpha}{2}$.

Note $A - B = -(C - D) + (\gamma - \alpha) - (\delta - \beta) + \ldots$ Actually, let’s just note $(A - B) + (C - D) = \frac{\beta - \alpha - \delta + \gamma + \gamma - \beta - \delta + \alpha}{2} = \gamma - \delta$ and $(A - B) - (C - D) = \frac{\beta - \alpha - \delta + \gamma - \gamma + \beta + \delta - \alpha}{2} = \beta - \alpha$.

So $\cos(A - B) + \cos(C - D) = 2\cos\frac{(A-B)+(C-D)}{2}\cos\frac{(A-B)-(C-D)}{2} = 2\cos\frac{\gamma - \delta}{2}\cos\frac{\beta - \alpha}{2}$.

And $-(A+B) = -\frac{\beta - \alpha + \delta - \gamma}{2}$, $-(C+D) = -\frac{\gamma - \beta + \delta - \alpha}{2}$, so $-(A+B) + (-(C+D)) = -(\delta - \alpha)$ and $-(A+B) - (-(C+D)) = -(\beta - \gamma)$.

So $-\cos(A+B) - \cos(C+D) = -[\cos(A+B) + \cos(C+D)] = -2\cos\frac{(A+B)+(C+D)}{2}\cos\frac{(A+B)-(C+D)}{2} = -2\cos\frac{\delta - \alpha}{2}\cos\frac{\beta - \gamma}{2}$.

Hmm, this is getting complicated. Let me try a more direct approach.

Using the factor formula approach:

Let me set $\alpha = a, \beta = b, \gamma = c, \delta = d$ for brevity and use half-angle substitution. Let $A = \frac{a}{2}, B = \frac{b}{2}, C = \frac{c}{2}, D = \frac{d}{2}$.

Then $\sin\frac{b-a}{2} = \sin(B - A)$, etc.

We need: $\sin(B-A)\sin(D-C) + \sin(C-B)\sin(D-A) = \sin(C-A)\sin(D-B)$.

Expand each using $\sin P \sin Q = \frac{1}{2}[\cos(P-Q) - \cos(P+Q)]$:

LHS $= \frac{1}{2}[\cos(B-A-D+C) - \cos(B-A+D-C) + \sin(C-B)\sin(D-A)]$

This is still messy. Let me just use the addition formulas directly.

$\sin(B-A)\sin(D-C) = \frac{1}{2}[\cos(B-A-D+C) - \cos(B-A+D-C)]$

$\sin(C-B)\sin(D-A) = \frac{1}{2}[\cos(C-B-D+A) - \cos(C-B+D-A)]$

Note: $B-A-D+C = -(D-C) + (B-A) = (B-A) - (D-C)$, and $C-B-D+A = (C-B) - (D-A)$.

Also: $B-A+D-C = (B-C) + (D-A)$ and $C-B+D-A = (D-A) - (B-C)$.

Hmm, let me note:

• $B - A - D + C = (C - A) - (D - B)$
• $B - A + D - C = (D - A) + (B - C)$
• $C - B - D + A = (C - A) - (D - B)$ — Wait: $C - B - D + A = (C + A) - (B + D) = (C - A) - (D - B) + 2A - 2B + 2A$… no.

Let me just carefully compute: $C - B - D + A = A - B + C - D = -(B - A) - (D - C) = -[(B-A) + (D-C)]$.

And $B - A - D + C = (B - A) - (D - C)$.

So $\cos(B-A-D+C) = \cos((B-A)-(D-C))$ and $\cos(C-B-D+A) = \cos((B-A)+(D-C))$.

Therefore: $\sin(B-A)\sin(D-C) = \frac{1}{2}[\cos((B-A)-(D-C)) - \cos((B-A)+(D-C))]$

Wait, that’s just the product formula restated. But the second product:

$C - B + D - A = (D - A) + (C - B) = (D - A) - (B - C)$.

So $\cos(C-B+D-A) = \cos((D-A)-(B-C))$ and $\cos(C-B-D+A) = \cos((D-A)+(B-C))$.

Wait: $C - B - D + A = (A - B) + (C - D) = -[(B-A) + (D-C)]$, so $\cos(C-B-D+A) = \cos((B-A)+(D-C))$.

And $C - B + D - A = (D - A) + (C - B) = (D - A) - (B - C)$. Note $D - A = (D - B) + (B - A)$ and $B - C = -(C - B)$.

OK let me try yet another approach. I’ll use the sine subtraction formula.

The LHS of $(\dagger)$ is: $\sin\frac{b-a}{2}\sin\frac{d-c}{2} + \sin\frac{c-b}{2}\sin\frac{d-a}{2}$

Let $p = \frac{b-a}{2}, q = \frac{c-b}{2}, r = \frac{d-c}{2}$. Then $0 < p, q, r$ and $p + q + r = \frac{d-a}{2}$.

So $\frac{d-a}{2} = p + q + r$ and $\frac{c-a}{2} = p + q$ and $\frac{d-b}{2} = q + r$.

LHS $= \sin p \sin r + \sin q \sin(p + q + r)$

RHS $= \sin(p+q)\sin(q+r)$

Expanding the RHS:

$\sin(p+q)\sin(q+r) = [\sin p \cos q + \cos p \sin q][\sin q \cos r + \cos q \sin r]$

$= \sin p \sin q \cos q \cos r + \sin p \cos^2 q \sin r + \cos p \sin^2 q \cos r + \cos p \sin q \cos q \sin r$

Expanding the LHS:

$\sin p \sin r + \sin q \sin(p + q + r)$

$= \sin p \sin r + \sin q [\sin(p+q)\cos r + \cos(p+q)\sin r]$

$= \sin p \sin r + \sin q [(\sin p \cos q + \cos p \sin q)\cos r + (\cos p \cos q - \sin p \sin q)\sin r]$

$= \sin p \sin r + \sin p \sin q \cos q \cos r + \cos p \sin^2 q \cos r + \cos p \sin q \cos q \sin r - \sin p \sin^2 q \sin r$

$= \sin p \sin r (1 - \sin^2 q) + \sin p \sin q \cos q \cos r + \cos p \sin^2 q \cos r + \cos p \sin q \cos q \sin r$

$= \sin p \sin r \cos^2 q + \sin p \sin q \cos q \cos r + \cos p \sin^2 q \cos r + \cos p \sin q \cos q \sin r$

This matches the RHS exactly! Therefore $(\dagger)$ holds, and we have proved

$|e^{i\alpha} - e^{i\beta}| \cdot |e^{i\gamma} - e^{i\delta}| + |e^{i\beta} - e^{i\gamma}| \cdot |e^{i\alpha} - e^{i\delta}| = |e^{i\alpha} - e^{i\gamma}| \cdot |e^{i\beta} - e^{i\delta}| . \qquad \text{(shown)}$

Geometric interpretation:

The points $e^{i\alpha}, e^{i\beta}, e^{i\gamma}, e^{i\delta}$ lie on the unit circle in the complex plane, with $0 < \alpha < \beta < \gamma < \delta < 2\pi$. The distance between two points on the unit circle is $|e^{i\theta_1} - e^{i\theta_2}|$.

Label the four points $A = e^{i\alpha}$, $B = e^{i\beta}$, $C = e^{i\gamma}$, $D = e^{i\delta}$. Since $0 < \alpha < \beta < \gamma < \delta < 2\pi$, these points are in order around the circle, forming a cyclic quadrilateral $ABCD$.

The identity becomes:

$AB \cdot CD + BC \cdot AD = AC \cdot BD$

This is Ptolemy’s Theorem: for a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides. Here $AC$ and $BD$ are the diagonals, while $AB, CD$ and $BC, AD$ are the two pairs of opposite sides.

Examiner Notes

About a third of the candidates attempted this, though with less success than any of its predecessors. Attempts were mostly “all or nothing”. Some candidates thought that the cyclic quadrilateral property had to be that opposite angles are supplementary, as the only property that they knew.

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Question 7

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

Problem

7 (i) The functions $f_n(x)$ are defined for $n = 0, 1, 2, \dots$, by

$f_0(x) = \frac{1}{1 + x^2} \qquad \text{and} \qquad f_{n+1}(x) = \frac{\mathrm{d}f_n(x)}{\mathrm{d}x} .$

Prove, for $n \geqslant 1$, that

$(1 + x^2)f_{n+1}(x) + 2(n + 1)xf_n(x) + n(n + 1)f_{n-1}(x) = 0 .$

(ii) The functions $P_n(x)$ are defined for $n = 0, 1, 2, \dots$, by

$P_n(x) = (1 + x^2)^{n+1}f_n(x) .$

Find expressions for $P_0(x)$, $P_1(x)$ and $P_2(x)$.

Prove, for $n \geqslant 0$, that

$P_{n+1}(x) - (1 + x^2)\frac{\mathrm{d}P_n(x)}{\mathrm{d}x} + 2(n + 1)xP_n(x) = 0 ,$

and that $P_n(x)$ is a polynomial of degree $n$.

Hint

7. (i) This result is simply obtained using the principle of mathematical induction. The $n = 1$ case can be established merely by obtaining $f_1$ and $f_2$ from the definition, and then substituting these along with $f_0$.

(ii)

$P_0(x) = (1 + x^2) \frac{1}{1 + x^2} = 1$

$P_1(x) = (1 + x^2)^2 \frac{-2x}{(1 + x^2)^2} = -2x$

$P_2(x) = (1 + x^2)^3 \frac{6x^2 - 2}{(1 + x^2)^3} = 6x^2 - 2$

$P_{n+1}(x) - (1 + x^2) \frac{dP_n(x)}{dx} + 2(n + 1)xP_n(x)$

which differentiating $P_n$ by the product rule and substituting

$= (1 + x^2)^{n+2} f_{n+1}(x) - (1 + x^2) \left( (1 + x^2)^{n+1} f_{n+1}(x) + (n + 1)2x(1 + x^2)^n f_n(x) \right) + 2(n + 1)x(1 + x^2)^{n+1} f_n(x)$ which is zero.

Again using the principle of mathematical induction and the result just obtained, it can be found that $P_{k+1}(x)$ is a polynomial of degree not greater than $k + 1$.

Further, assuming that $P_k(x)$ has term of highest degree, $(-1)^k (k + 1)! x^k$, as

$P_{n+1}(x) - (1 + x^2) \frac{dP_n(x)}{dx} + 2(n + 1)xP_n(x) = 0 \text{, the term of highest degree of } P_{k+1}(x) \text{ is}$

$(-1)^k (k + 1)! kx^{k-1} x^2 - 2(k + 1)x(-1)^k (k + 1)! x^k$

$= (-1)^{k+1} (k + 2)! x^{k+1} \text{ as required.}$

(The form of the term need not be determined, but it must be shown to be non-zero.)

Model Solution

Part (i): Proving the recurrence for $f_n$

We prove by induction that for $n \geqslant 1$:

$(1 + x^2)f_{n+1}(x) + 2(n+1)x f_n(x) + n(n+1)f_{n-1}(x) = 0 .$

Base case $n = 1$: We have $f_0(x) = \frac{1}{1+x^2}$. Differentiating:

$f_1(x) = f_0'(x) = \frac{-2x}{(1+x^2)^2}$

$f_2(x) = f_1'(x) = \frac{-2(1+x^2)^2 + 2x \cdot 2(1+x^2) \cdot 2x}{(1+x^2)^4} = \frac{-2(1+x^2) + 8x^2}{(1+x^2)^3} = \frac{6x^2 - 2}{(1+x^2)^3}$

Substituting into the recurrence:

$(1+x^2) \cdot \frac{6x^2 - 2}{(1+x^2)^3} + 4x \cdot \frac{-2x}{(1+x^2)^2} + 2 \cdot \frac{1}{1+x^2}$

$= \frac{6x^2 - 2}{(1+x^2)^2} + \frac{-8x^2}{(1+x^2)^2} + \frac{2(1+x^2)}{(1+x^2)^2}$

$= \frac{6x^2 - 2 - 8x^2 + 2 + 2x^2}{(1+x^2)^2} = \frac{0}{(1+x^2)^2} = 0 . \qquad \checkmark$

Inductive step: Assume the result holds for $n = k$ (where $k \geqslant 1$), i.e.

$(1 + x^2)f_{k+1} + 2(k+1)x f_k + k(k+1)f_{k-1} = 0 . \qquad (*)$

We differentiate $(*)$ with respect to $x$:

$2x f_{k+1} + (1+x^2)f_{k+2} + 2(k+1)f_k + 2(k+1)x f_{k+1} + k(k+1)f_k = 0$

$(1+x^2)f_{k+2} + [2x + 2(k+1)x]f_{k+1} + [2(k+1) + k(k+1)]f_k = 0$

$(1+x^2)f_{k+2} + 2(k+2)x f_{k+1} + (k+1)(k+2)f_k = 0 .$

This is exactly the result for $n = k + 1$. By induction, the recurrence holds for all $n \geqslant 1$. $\qquad \text{(shown)}$

Part (ii)

Finding $P_0$, $P_1$, $P_2$:

$P_0(x) = (1+x^2)^1 f_0(x) = (1+x^2) \cdot \frac{1}{1+x^2} = 1$

$P_1(x) = (1+x^2)^2 f_1(x) = (1+x^2)^2 \cdot \frac{-2x}{(1+x^2)^2} = -2x$

$P_2(x) = (1+x^2)^3 f_2(x) = (1+x^2)^3 \cdot \frac{6x^2 - 2}{(1+x^2)^3} = 6x^2 - 2$

Proving the recurrence for $P_n$:

We want to show that for $n \geqslant 0$:

$P_{n+1}(x) - (1+x^2)\frac{dP_n}{dx} + 2(n+1)xP_n(x) = 0 .$

By definition $P_n(x) = (1+x^2)^{n+1} f_n(x)$, so $f_n(x) = \frac{P_n(x)}{(1+x^2)^{n+1}}$ and $P_{n+1}(x) = (1+x^2)^{n+2} f_{n+1}(x)$.

Differentiating $P_n(x)$ by the product rule:

$\frac{dP_n}{dx} = (n+1) \cdot 2x \cdot (1+x^2)^n f_n(x) + (1+x^2)^{n+1} f_{n+1}(x)$

$= 2(n+1)x(1+x^2)^n f_n(x) + (1+x^2)^{n+1} f_{n+1}(x) .$

Therefore:

$(1+x^2)\frac{dP_n}{dx} = 2(n+1)x(1+x^2)^{n+1} f_n(x) + (1+x^2)^{n+2} f_{n+1}(x)$

$= 2(n+1)x P_n(x) + P_{n+1}(x) .$

Rearranging:

$P_{n+1}(x) - (1+x^2)\frac{dP_n}{dx} + 2(n+1)xP_n(x) = P_{n+1} - [2(n+1)xP_n + P_{n+1}] + 2(n+1)xP_n = 0 . \qquad \text{(shown)}$

Proving $P_n(x)$ is a polynomial of degree $n$:

We prove by induction that $P_n(x)$ is a polynomial of degree exactly $n$.

Base cases: $P_0 = 1$ (degree 0), $P_1 = -2x$ (degree 1), $P_2 = 6x^2 - 2$ (degree 2). $\checkmark$

Inductive step: Assume $P_k(x)$ is a polynomial of degree $k$ with leading term $c_k x^k$ (where $c_k \neq 0$). From the recurrence:

$P_{k+1}(x) = (1+x^2)\frac{dP_k}{dx} - 2(k+1)xP_k(x) .$

Since $P_k$ has degree $k$, $\frac{dP_k}{dx}$ has degree $k - 1$, so $(1+x^2)\frac{dP_k}{dx}$ has degree $k + 1$ with leading term $k \cdot c_k \cdot x^{k+1}$ (from $x^2 \cdot k c_k x^{k-1}$).

The term $2(k+1)xP_k(x)$ has degree $k + 1$ with leading term $2(k+1)c_k x^{k+1}$.

Therefore the leading term of $P_{k+1}(x)$ is:

$k \cdot c_k \cdot x^{k+1} - 2(k+1) \cdot c_k \cdot x^{k+1} = c_k[k - 2(k+1)] x^{k+1} = -c_k(k+2) x^{k+1} .$

Since $c_k \neq 0$ and $k + 2 \neq 0$, the leading coefficient $-c_k(k+2) \neq 0$, so $P_{k+1}$ has degree exactly $k + 1$.

By induction, $P_n(x)$ is a polynomial of degree $n$ for all $n \geqslant 0$. $\qquad \text{(shown)}$

(As a check: the leading coefficients follow $c_0 = 1$, $c_1 = -2$, $c_2 = 6$, and in general $c_{k+1} = -(k+2)c_k$, giving $c_n = (-1)^n(n+1)!$, consistent with the computed values.)

Examiner Notes

Approximately two thirds of the candidates attempted this, earning roughly half marks in doing so. Part (i) and finding the three expressions for $P_0$, $P_1$ & $P_2$ from part (ii) largely went well. The result involving $P_{n+1}$ saw most falling by the wayside, especially those who attempted it by induction. Quite a few candidates did score all but two marks in proving that $P_n$ was a polynomial of degree $n$ or less, but not appreciating that there was still something to do regarding the leading term.

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Question 8

Topic: 积分 (Integration)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Let $m$ be a positive integer and let $n$ be a non-negative integer.

(i) Use the result $\lim_{t \to \infty} e^{-mt}t^n = 0$ to show that

$\lim_{x \to 0} x^m(\ln x)^n = 0 .$

By writing $x^x$ as $e^{x \ln x}$ show that

$\lim_{x \to 0} x^x = 1 .$

(ii) Let $I_n = \int_0^1 x^m(\ln x)^n \mathrm{d}x$. Show that

$I_{n+1} = -\frac{n + 1}{m + 1}I_n$

and hence evaluate $I_n$.

(iii) Show that

$\int_0^1 x^x \mathrm{d}x = 1 - \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^3 - \left(\frac{1}{4}\right)^4 + \dots .$

Hint

8. (i) Letting $x = e^{-t}$, $\lim_{x \to 0} [x^m (\ln x)^n] = \lim_{t \to \infty} [(e^{-t})^m (-t)^n] = (-1)^n \lim_{x \to 0} [e^{-mt} t^n] = 0$

and so letting $m = n = 1$, $\lim_{x \to 0} [x \ln x] = 0$.

Thus, $\lim_{x \to 0} x^x = \lim_{x \to 0} e^{x \ln x} = e^{\lim_{x \to 0} x \ln x} = e^0 = 1$

(ii) Integrating by parts,

$I_{n+1} = \int_{0}^{1} x^{m} (\ln x)^{n+1} dx = \left[ \frac{x^{m+1} (\ln x)^{n+1}}{m+1} \right]_{0}^{1} - \int_{0}^{1} \frac{x^{m+1}}{m+1} \frac{(n+1) (\ln x)^{n}}{x} dx$

$= 0 - 0 \text{ (using the first result)} - \int_{0}^{1} \frac{n+1}{m+1} x^{m} (\ln x)^{n} dx = -\frac{n+1}{m+1} I_{n}$

So $I_{n} = \frac{-n}{m+1} \times \frac{-(n-1)}{m+1} \times \frac{-(n-2)}{m+1} \times \dots \times \frac{-1}{m+1} I_{0} = \frac{(-1)^{n} n!}{(m+1)^{n}} \int_{0}^{1} x^{m} dx$

$= \frac{(-1)^{n} n!}{(m+1)^{n+1}}$

(iii) $\int_{0}^{1} x^{x} dx = \int_{0}^{1} e^{x \ln x} dx = \int_{0}^{1} 1 + x \ln x + \frac{x^{2} (\ln x)^{2}}{2!} + \dots dx$$

$= 1 + I_{1} + \frac{1}{2!} I_{2} + \dots = 1 - \left( \frac{1}{2} \right)^{2} + \left( \frac{1}{3} \right)^{3} - \left( \frac{1}{4} \right)^{4} + \dots \text{ as required.}$

Model Solution

Part (i): Showing $\lim_{x \to 0} x^m(\ln x)^n = 0$

Let $x = e^{-t}$, so as $x \to 0^+$, $t \to \infty$, and $\ln x = -t$. Then:

$x^m(\ln x)^n = (e^{-t})^m(-t)^n = (-1)^n e^{-mt} t^n .$

Therefore:

$\lim_{x \to 0^+} x^m(\ln x)^n = (-1)^n \lim_{t \to \infty} e^{-mt} t^n = 0 ,$

using the given result $\lim_{t \to \infty} e^{-mt}t^n = 0$. $\qquad \text{(shown)}$

Showing $\lim_{x \to 0} x^x = 1$:

We write $x^x = e^{x \ln x}$. Since $e^u$ is continuous, it suffices to show $\lim_{x \to 0^+} x\ln x = 0$.

Setting $m = 1$, $n = 1$ in the result above:

$\lim_{x \to 0^+} x^1 (\ln x)^1 = \lim_{x \to 0^+} x \ln x = 0 .$

Therefore:

$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x\ln x} = e^0 = 1 . \qquad \text{(shown)}$

Part (ii): Evaluating $I_n$

We integrate $I_{n+1} = \int_0^1 x^m (\ln x)^{n+1}\, dx$ by parts. Let $u = (\ln x)^{n+1}$ and $dv = x^m\,dx$, so $du = \frac{(n+1)(\ln x)^n}{x}\,dx$ and $v = \frac{x^{m+1}}{m+1}$.

$I_{n+1} = \left[\frac{x^{m+1}(\ln x)^{n+1}}{m+1}\right]_0^1 - \int_0^1 \frac{x^{m+1}}{m+1} \cdot \frac{(n+1)(\ln x)^n}{x}\,dx$

$= \left[\frac{x^{m+1}(\ln x)^{n+1}}{m+1}\right]_0^1 - \frac{n+1}{m+1}\int_0^1 x^m(\ln x)^n\,dx .$

For the boundary term at $x = 1$: $\ln 1 = 0$, so the term is $0$.

At $x = 0$: $\frac{x^{m+1}(\ln x)^{n+1}}{m+1}$. This is $\frac{1}{m+1} \cdot x^{m+1}(\ln x)^{n+1}$. Setting the result from part (i) with $m' = m + 1$ (which is a positive integer since $m$ is) and $n' = n + 1$ (a non-negative integer), we get $\lim_{x \to 0} x^{m+1}(\ln x)^{n+1} = 0$.

Therefore the boundary term vanishes at both ends, giving:

$I_{n+1} = -\frac{n+1}{m+1} I_n . \qquad \text{(shown)}$

Evaluating $I_n$:

Applying the recurrence repeatedly:

$I_n = -\frac{n}{m+1} I_{n-1} = -\frac{n}{m+1} \cdot \left(-\frac{n-1}{m+1}\right) I_{n-2} = \cdots = \frac{(-1)^n \cdot n!}{(m+1)^n} I_0$

where

$I_0 = \int_0^1 x^m\,dx = \left[\frac{x^{m+1}}{m+1}\right]_0^1 = \frac{1}{m+1} .$

Therefore:

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

Part (iii): Evaluating $\int_0^1 x^x\,dx$

We write $x^x = e^{x\ln x}$ and expand as a power series:

$x^x = e^{x\ln x} = 1 + x\ln x + \frac{(x\ln x)^2}{2!} + \frac{(x\ln x)^3}{3!} + \cdots = \sum_{k=0}^{\infty} \frac{x^k(\ln x)^k}{k!} .$

Integrating term by term (justified since the series converges uniformly on $[0, 1]$):

$\int_0^1 x^x\,dx = \sum_{k=0}^{\infty} \frac{1}{k!}\int_0^1 x^k(\ln x)^k\,dx = \sum_{k=0}^{\infty} \frac{1}{k!} I_k \Big|_{m=k}$

where $I_k\big|_{m=k} = \int_0^1 x^k(\ln x)^k\,dx$. Using the result from part (ii) with $m = k$ and $n = k$:

$\int_0^1 x^k(\ln x)^k\,dx = \frac{(-1)^k \, k!}{(k+1)^{k+1}} .$

Therefore:

$\int_0^1 x^x\,dx = \sum_{k=0}^{\infty} \frac{1}{k!} \cdot \frac{(-1)^k \, k!}{(k+1)^{k+1}} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^{k+1}} .$

Writing out the first few terms (with $k = 0, 1, 2, 3, \ldots$):

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

$= 1 - \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^3 - \left(\frac{1}{4}\right)^4 + \cdots . \qquad \text{(shown)}$

Examiner Notes

Roughly the same number attempted this as question 7, with slightly less success. Usually, a candidate did not properly obtain the first three results, and so would end up having apparently finished the whole question but in fact scoring only two thirds marks. The problem was often that the limiting process was not fully understood. In part (ii), there was often odd splitting going on to attempt the integration by parts and this part often went wrong.