STEP3 2017 -- Pure Mathematics

STEP3 2017 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2017  |  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	组合数学与级数 (Combinatorics and Series)	Standard	telescoping 级数, 二项式系数展开, 不等式放缩, 级数求和
2	复数变换 (Complex Transformations)	Hard	复指数旋转公式, 变换复合, 三角与复数形式转换, 分类讨论
3	代数 (Algebra)	Challenging	预解三次方程, Vieta 公式, 对称函数, 二次方程求根
4	微积分 (Calculus)	Challenging	对数换底, 积分与指数运算, 微分方程分离变量, 反证法
5	微积分 (Calculus)	Challenging	极坐标导数, 正交条件推导, 分离变量法, 心形线作图
6	积分与换元 (Integration and Substitution)	Challenging	换元积分（u=1/v 和 Möbius 变换）,函数方程推导,递推与代入消元
7	解析几何：椭圆与切线 (Analytic Geometry: Ellipse and Tangents)	Challenging	参数方程代入,隐函数求导,切线方程推导,判别式与韦达定理
8	求和与三角恒等式 (Summation and Trigonometric Identities)	Challenging	分部求和（Abel summation）,三角恒等式（积化和差）,裂项求和技巧

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Question 1

Topic: 组合数学与级数 (Combinatorics and Series)  |  Difficulty: Standard  |  Marks: 20

Problem

1 (i) Prove that, for any positive integers $n$ and $r$,

$\frac{1}{{}^{n+r}C_{r+1}} = \frac{r+1}{r} \left( \frac{1}{{}^{n+r-1}C_r} - \frac{1}{{}^{n+r}C_r} \right).$

Hence determine

$\sum_{n=1}^{\infty} \frac{1}{{}^{n+r}C_{r+1}},$

and deduce that $\sum_{n=2}^{\infty} \frac{1}{{}^{n+2}C_3} = \frac{1}{2}$.

(ii) Show that, for $n \geqslant 3$,

$\frac{3!}{n^3} < \frac{1}{{}^{n+1}C_3} \quad \text{and} \quad \frac{20}{{}^{n+1}C_3} - \frac{1}{{}^{n+2}C_5} < \frac{5!}{n^3}.$

By summing these inequalities for $n \geqslant 3$, show that

$\frac{115}{96} < \sum_{n=1}^{\infty} \frac{1}{n^3} < \frac{116}{96}.$

Note: ${}^nC_r$ is another notation for $\begin{pmatrix} n \\ r \end{pmatrix}$.

Hint

(i) Expanding the RHS using binomial coefficient definitions:

$\frac{r+1}{r} \left( \frac{(n-1)!r!}{(n+r-1)!} - \frac{n!r!}{(n+r)!} \right) = \frac{r+1}{r} \cdot \frac{(n-1)!r![(n+r)-n]}{(n+r)!} = \frac{(n-1)!(r+1)!}{(n+r)!} = \frac{1}{{}^{n+r}C_{r+1}}$

Telescoping sum:

$\sum_{n=1}^{\infty} \frac{1}{{}^{n+r}C_{r+1}} = \frac{r+1}{r} \left( \frac{1}{{}^{r}C_r} - \lim_{n\to\infty} \frac{1}{{}^{n+r}C_r} \right) = \frac{r+1}{r}$

Setting $r=2$: $\sum_{n=2}^{\infty} \frac{1}{{}^{n+2}C_3} = \frac{3}{2} - \frac{1}{{}^{3}C_3} = \frac{1}{2}$.

(ii) First inequality: ${}^{n+1}C_3 = \frac{n^3-n}{3!} < \frac{n^3}{3!}$, so $\frac{3!}{n^3} < \frac{1}{{}^{n+1}C_3}$.

Second inequality: $\frac{20}{{}^{n+1}C_3} - \frac{1}{{}^{n+2}C_5} - \frac{5!}{n^3} = \frac{-480}{n^3(n^2-1)(n^2-4)} < 0$ for $n \ge 3$.

Summing the first inequality for $n \ge 3$: $\sum_{n=1}^{\infty} \frac{1}{n^3} < 1 + \frac{1}{8} + \frac{1}{2} \cdot \frac{1}{6} = \frac{116}{96}$.

Summing the second inequality for $n \ge 3$: $\sum_{n=1}^{\infty} \frac{1}{n^3} > \frac{35}{4 \times 5!} + 1 + \frac{1}{8} = \frac{115}{96}$.

Model Solution

Part (i)

We expand the RHS using ${}^nC_r = \frac{n!}{r!(n-r)!}$.

$\frac{1}{{}^{n+r-1}C_r} = \frac{r!(n-1)!}{(n+r-1)!}, \qquad \frac{1}{{}^{n+r}C_r} = \frac{r!\,n!}{(n+r)!}$

$\frac{1}{{}^{n+r-1}C_r} - \frac{1}{{}^{n+r}C_r} = \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!\,n!}{(n+r)!}$

Writing the first fraction with denominator $(n+r)!$:

$= \frac{r!(n-1)!(n+r)}{(n+r)!} - \frac{r!\,n!}{(n+r)!} = \frac{r!(n-1)!\bigl[(n+r) - n\bigr]}{(n+r)!} = \frac{r \cdot r!(n-1)!}{(n+r)!}$

Multiplying by $\frac{r+1}{r}$:

$\frac{r+1}{r}\left(\frac{1}{{}^{n+r-1}C_r} - \frac{1}{{}^{n+r}C_r}\right) = \frac{(r+1)!\,(n-1)!}{(n+r)!} = \frac{1}{{}^{n+r}C_{r+1}} \qquad \blacksquare$

Hence determine $\sum_{n=1}^{\infty} \frac{1}{{}^{n+r}C_{r+1}}$:

The partial sum telescopes:

$\sum_{n=1}^{N} \frac{1}{{}^{n+r}C_{r+1}} = \frac{r+1}{r}\sum_{n=1}^{N}\left(\frac{1}{{}^{n+r-1}C_r} - \frac{1}{{}^{n+r}C_r}\right)$

$= \frac{r+1}{r}\left(\frac{1}{{}^{r}C_r} - \frac{1}{{}^{N+r}C_r}\right) = \frac{r+1}{r}\left(1 - \frac{r!\,N!}{(N+r)!}\right)$

Since ${}^{N+r}C_r \to \infty$ as $N \to \infty$, the second term vanishes:

$\sum_{n=1}^{\infty} \frac{1}{{}^{n+r}C_{r+1}} = \frac{r+1}{r} \qquad \text{(...)}$

Deduce $\sum_{n=2}^{\infty} \frac{1}{{}^{n+2}C_3} = \frac{1}{2}$:

Setting $r = 2$: $\sum_{n=1}^{\infty} \frac{1}{{}^{n+2}C_3} = \frac{3}{2}$.

The $n = 1$ term is $\frac{1}{{}^{3}C_3} = 1$, so

$\sum_{n=2}^{\infty} \frac{1}{{}^{n+2}C_3} = \frac{3}{2} - 1 = \frac{1}{2} \qquad \blacksquare$

Part (ii)

First inequality. For $n \geq 3$:

${}^{n+1}C_3 = \frac{(n+1)n(n-1)}{6} = \frac{n^3 - n}{6}$

Since $n \geq 3 > 0$, we have $n^3 - n < n^3$, so

${}^{n+1}C_3 < \frac{n^3}{6}$

All quantities are positive for $n \geq 3$, so taking reciprocals:

$\frac{6}{n^3} < \frac{1}{{}^{n+1}C_3} \qquad \text{(...)}$

Second inequality. We compute

${}^{n+2}C_5 = \frac{(n+2)(n+1)n(n-1)(n-2)}{120}$

Consider the difference

$\frac{20}{{}^{n+1}C_3} - \frac{1}{{}^{n+2}C_5} - \frac{5!}{n^3} = \frac{120}{n(n^2-1)} - \frac{120}{n(n^2-1)(n^2-4)} - \frac{120}{n^3}$

$= \frac{120\bigl[n^2(n^2-4) - n^2 - (n^2-1)(n^2-4)\bigr]}{n^3(n^2-1)(n^2-4)}$

Expanding the numerator:

$n^4 - 4n^2 - n^2 - (n^4 - 5n^2 + 4) = -4$

So

$\frac{20}{{}^{n+1}C_3} - \frac{1}{{}^{n+2}C_5} - \frac{5!}{n^3} = \frac{-480}{n^3(n^2-1)(n^2-4)}$

For $n \geq 3$, the denominator is positive, so this expression is negative:

$\frac{20}{{}^{n+1}C_3} - \frac{1}{{}^{n+2}C_5} < \frac{5!}{n^3} \qquad \text{(...)}$

Summing the first inequality for $n = 3, 4, 5, \ldots$:

$6\sum_{n=3}^{\infty} \frac{1}{n^3} < \sum_{n=3}^{\infty} \frac{1}{{}^{n+1}C_3}$

From part (i) with $r = 2$: $\sum_{n=1}^{\infty} \frac{1}{{}^{n+2}C_3} = \frac{3}{2}$, so

$\sum_{n=3}^{\infty} \frac{1}{{}^{n+1}C_3} = \sum_{n=1}^{\infty} \frac{1}{{}^{n+2}C_3} - \frac{1}{{}^{3}C_3} = \frac{3}{2} - 1 = \frac{1}{2}$

Also $\sum_{n=3}^{\infty} \frac{1}{n^3} = \sum_{n=1}^{\infty} \frac{1}{n^3} - 1 - \frac{1}{8}$, so

$6\left(\sum_{n=1}^{\infty} \frac{1}{n^3} - \frac{9}{8}\right) < \frac{1}{2}$

$\sum_{n=1}^{\infty} \frac{1}{n^3} < \frac{9}{8} + \frac{1}{12} = \frac{27+2}{24} = \frac{116}{96} \qquad \text{(...)}$

Summing the second inequality for $n = 3, 4, 5, \ldots$:

$20\sum_{n=3}^{\infty} \frac{1}{{}^{n+1}C_3} - \sum_{n=3}^{\infty} \frac{1}{{}^{n+2}C_5} < 120\sum_{n=3}^{\infty} \frac{1}{n^3}$

From part (i) with $r = 4$: $\sum_{n=1}^{\infty} \frac{1}{{}^{n+4}C_5} = \frac{5}{4}$.

Since $\sum_{n=3}^{\infty} \frac{1}{{}^{n+2}C_5}$ has the same terms as $\sum_{n=1}^{\infty} \frac{1}{{}^{n+4}C_5}$ (both start at ${}^{5}C_5$):

$\sum_{n=3}^{\infty} \frac{1}{{}^{n+2}C_5} = \frac{5}{4}$

Substituting:

$20 \cdot \frac{1}{2} - \frac{5}{4} < 120\left(\sum_{n=1}^{\infty} \frac{1}{n^3} - \frac{9}{8}\right)$

$\frac{35}{4} < 120\left(\sum_{n=1}^{\infty} \frac{1}{n^3} - \frac{9}{8}\right)$

$\sum_{n=1}^{\infty} \frac{1}{n^3} > \frac{9}{8} + \frac{35}{480} = \frac{108}{96} + \frac{7}{96} = \frac{115}{96} \qquad \text{(...)}$

Combining the two bounds:

$\frac{115}{96} < \sum_{n=1}^{\infty} \frac{1}{n^3} < \frac{116}{96} \qquad \blacksquare$

Examiner Notes

The most popular question on the paper, attempted by about 84% of the candidates, it was also the most successfully answered with an average mark of about 12/20. The first result was generally well answered with a few candidates attempting to use induction, and then proving the result directly. The summations were usually done well, though often lacked explanation. Usually, the inequalities were not well argued, there was poor layout, and no mention of positivity. Those who spotted the link with part (i) did well in general summing the inequalities, though there were some problems with the indices.

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Question 2

Topic: 复数变换 (Complex Transformations)  |  Difficulty: Hard  |  Marks: 20

Problem

2 The transformation $R$ in the complex plane is a rotation (anticlockwise) by an angle $\theta$ about the point represented by the complex number $a$. The transformation $S$ in the complex plane is a rotation (anticlockwise) by an angle $\phi$ about the point represented by the complex number $b$.

(i) The point $P$ is represented by the complex number $z$. Show that the image of $P$ under $R$ is represented by

$e^{i\theta}z + a(1 - e^{i\theta}).$

(ii) Show that the transformation $SR$ (equivalent to $R$ followed by $S$) is a rotation about the point represented by $c$, where

$c \sin \tfrac{1}{2}(\theta + \phi) = a e^{i\phi/2} \sin \tfrac{1}{2}\theta + b e^{-i\theta/2} \sin \tfrac{1}{2}\phi,$

provided $\theta + \phi \neq 2n\pi$ for any integer $n$.

What is the transformation $SR$ if $\theta + \phi = 2\pi$?

(iii) Under what circumstances is $RS = SR$?

Hint

(i) Rotation $R$ anticlockwise by $\theta$ about $a$: $z' - a = e^{i\theta}(z - a)$, so $z' = e^{i\theta}z + a(1 - e^{i\theta})$.

(ii) Applying $R$ then $S$: $z'' = e^{i\phi}z' + b(1 - e^{i\phi}) = e^{i(\phi+\theta)}z + ae^{i\phi}(1-e^{i\theta}) + b(1-e^{i\phi})$.

For this to be a rotation about $c$: $c(1-e^{i(\phi+\theta)}) = ae^{i\phi}(1-e^{i\theta}) + b(1-e^{i\phi})$.

Multiplying by $-e^{-i(\phi+\theta)/2}$ and using $e^{ix} - e^{-ix} = 2i\sin x$:

$c \sin \tfrac{1}{2}(\theta+\phi) = ae^{i\phi/2}\sin\tfrac{1}{2}\theta + be^{-i\theta/2}\sin\tfrac{1}{2}\phi$

If $\theta+\phi = 2n\pi$: $z'' = z + (b-a)(1-e^{i\phi})$, a translation by $(b-a)(1-e^{i\phi})$.

(iii) $RS = SR$ requires:

• If $\theta+\phi = 2n\pi$: $(b-a)(1-e^{i\phi}) = (a-b)(1-e^{i\theta})$, giving $a=b$ or ($a\neq b$ and $\theta = 2m\pi$).
• If $\theta+\phi \neq 2n\pi$: $2i(a-b)\sin\tfrac{1}{2}\phi\sin\tfrac{1}{2}\theta = 0$, so $a=b$, $\theta = 2n\pi$, or $\phi = 2n\pi$.

Model Solution

Part (i)

The rotation $R$ anticlockwise by $\theta$ about $a$ maps $z$ to $z'$ satisfying $z' - a = e^{i\theta}(z - a)$:

$z' = e^{i\theta}z - ae^{i\theta} + a = e^{i\theta}z + a(1 - e^{i\theta}) \qquad \blacksquare$

Part (ii)

Let $z' = R(z) = e^{i\theta}z + a(1 - e^{i\theta})$. Then

$SR(z) = S(z') = e^{i\phi}z' + b(1 - e^{i\phi})$

$= e^{i\phi}\bigl[e^{i\theta}z + a(1 - e^{i\theta})\bigr] + b(1 - e^{i\phi})$

$= e^{i(\theta+\phi)}z + ae^{i\phi}(1 - e^{i\theta}) + b(1 - e^{i\phi}) \qquad \text{(...)}$

A rotation by $\theta + \phi$ about $c$ would map $z$ to $e^{i(\theta+\phi)}z + c(1 - e^{i(\theta+\phi)})$. So $SR$ is such a rotation if and only if

$c(1 - e^{i(\theta+\phi)}) = ae^{i\phi}(1 - e^{i\theta}) + b(1 - e^{i\phi}) \qquad \text{(...)}$

Since $\theta + \phi \neq 2n\pi$, we have $1 - e^{i(\theta+\phi)} \neq 0$, so $c$ exists and is unique.

To convert to the required form, multiply both sides by $-e^{-i(\theta+\phi)/2}$.

LHS:

$-e^{-i(\theta+\phi)/2} \cdot c(1 - e^{i(\theta+\phi)}) = c\bigl(e^{i(\theta+\phi)/2} - e^{-i(\theta+\phi)/2}\bigr) = 2ic\sin\tfrac{1}{2}(\theta+\phi)$

First term on RHS: using $1 - e^{i\theta} = -e^{i\theta/2}(e^{i\theta/2} - e^{-i\theta/2}) = -2ie^{i\theta/2}\sin\tfrac{1}{2}\theta$:

$-e^{-i(\theta+\phi)/2} \cdot ae^{i\phi}(1 - e^{i\theta}) = ae^{i\phi - i(\theta+\phi)/2} \cdot 2ie^{i\theta/2}\sin\tfrac{1}{2}\theta = 2iae^{i\phi/2}\sin\tfrac{1}{2}\theta$

Second term on RHS: using $1 - e^{i\phi} = -2ie^{i\phi/2}\sin\tfrac{1}{2}\phi$:

$-e^{-i(\theta+\phi)/2} \cdot b(1 - e^{i\phi}) = be^{-i(\theta+\phi)/2} \cdot 2ie^{i\phi/2}\sin\tfrac{1}{2}\phi = 2ibe^{-i\theta/2}\sin\tfrac{1}{2}\phi$

Combining and dividing by $2i$:

$c\sin\tfrac{1}{2}(\theta+\phi) = ae^{i\phi/2}\sin\tfrac{1}{2}\theta + be^{-i\theta/2}\sin\tfrac{1}{2}\phi \qquad \blacksquare$

If $\theta + \phi = 2\pi$: then $e^{i(\theta+\phi)} = 1$, so

$SR(z) = z + ae^{i\phi}(1 - e^{i\theta}) + b(1 - e^{i\phi})$

Since $\theta = 2\pi - \phi$, we have $e^{i\theta} = e^{-i\phi}$, so

$ae^{i\phi}(1 - e^{-i\phi}) = ae^{i\phi} - a$

$b(1 - e^{i\phi}) = b - be^{i\phi}$

Adding: $(ae^{i\phi} - a) + (b - be^{i\phi}) = (a - b)e^{i\phi} - (a - b) = (a - b)(e^{i\phi} - 1)$

$SR(z) = z + (a - b)(e^{i\phi} - 1)$

This is a translation by $(a - b)(e^{i\phi} - 1)$. $\qquad \text{(...)}$

Part (iii)

Case 1: $\theta + \phi = 2n\pi$ for some integer $n$.

By the same derivation as above (with $a \leftrightarrow b$, $\theta \leftrightarrow \phi$), $RS$ is a translation by

$(b - a)(e^{i\theta} - 1)$

while $SR$ is a translation by $(a - b)(e^{i\phi} - 1)$.

$RS = SR$ requires $(b-a)(e^{i\theta}-1) = (a-b)(e^{i\phi}-1)$, i.e.,

$(b - a)(e^{i\theta} - 1) + (b - a)(e^{i\phi} - 1) = 0$

$(b - a)(e^{i\theta} + e^{i\phi} - 2) = 0$

So either $a = b$, or $e^{i\theta} + e^{i\phi} = 2$.

Since $|e^{i\theta} + e^{i\phi}| \leq |e^{i\theta}| + |e^{i\phi}| = 2$, equality holds only when $e^{i\theta} = e^{i\phi}$, so $e^{i\theta} = e^{i\phi} = 1$, meaning $\theta$ and $\phi$ are both multiples of $2\pi$.

Case 2: $\theta + \phi \neq 2n\pi$ for any integer $n$.

Both $RS$ and $SR$ are rotations by $\theta + \phi$ (which is not a multiple of $2\pi$), so they are equal if and only if they share the same centre. Setting their centres equal:

$ae^{i\phi}(1 - e^{i\theta}) + b(1 - e^{i\phi}) = be^{i\theta}(1 - e^{i\phi}) + a(1 - e^{i\theta})$

$a\bigl[e^{i\phi}(1 - e^{i\theta}) - (1 - e^{i\theta})\bigr] = b\bigl[e^{i\theta}(1 - e^{i\phi}) - (1 - e^{i\phi})\bigr]$

$a(1 - e^{i\theta})(e^{i\phi} - 1) = b(1 - e^{i\phi})(e^{i\theta} - 1)$

Since $(e^{i\phi} - 1) = -(1 - e^{i\phi})$ and $(e^{i\theta} - 1) = -(1 - e^{i\theta})$, both sides equal

$-a(1 - e^{i\theta})(1 - e^{i\phi}) \quad \text{and} \quad -b(1 - e^{i\theta})(1 - e^{i\phi})$

respectively. So the condition is

$(a - b)(1 - e^{i\theta})(1 - e^{i\phi}) = 0$

Hence $a = b$, or $e^{i\theta} = 1$ (i.e., $\theta$ is a multiple of $2\pi$), or $e^{i\phi} = 1$ (i.e., $\phi$ is a multiple of $2\pi$).

Conclusion. Combining both cases: $RS = SR$ if and only if at least one of the following holds:

• $a = b$ (the two rotations share the same centre),
• $\theta$ is a multiple of $2\pi$ ($R$ is the identity),
• $\phi$ is a multiple of $2\pi$ ($S$ is the identity).

$\qquad \text{(...)}$

Examiner Notes

This was the least popular pure question being attempted by only just over a quarter of candidates, and was the least successful of all the questions scoring 5/20. Most candidates gave up after part (i), and some made much more of this first result, not being very succinct. Most could write down SR without difficulty, but then did not spot an easy way to move beyond this. The standard of algebra displayed was in general poor, in particular moving between complex and trigonometric forms.

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Question 3

Topic: 代数 (Algebra)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Let $\alpha, \beta, \gamma$ and $\delta$ be the roots of the quartic equation

$x^4 + px^3 + qx^2 + rx + s = 0 .$

You are given that, for any such equation, $\alpha\beta + \gamma\delta$, $\alpha\gamma + \beta\delta$ and $\alpha\delta + \beta\gamma$ satisfy a cubic equation of the form

$y^3 + Ay^2 + (pr - 4s)y + (4qs - p^2s - r^2) = 0 .$

Determine $A$.

Now consider the quartic equation given by $p = 0, q = 3, r = -6$ and $s = 10$.

(i) Find the value of $\alpha\beta + \gamma\delta$, given that it is the largest root of the corresponding cubic equation.

(ii) Hence, using the values of $q$ and $s$, find the value of $(\alpha + \beta)(\gamma + \delta)$ and the value of $\alpha\beta$ given that $\alpha\beta > \gamma\delta$.

(iii) Using these results, and the values of $p$ and $r$, solve the quartic equation.

Hint

Stem: $\alpha\beta+\gamma\delta + \alpha\gamma+\beta\delta + \alpha\delta+\beta\gamma = q$ (from Vieta’s), and the sum of the three roots of the cubic is $-A$, so $A = -q$.

For $p=0, q=3, r=-6, s=10$: cubic is $y^3 - 3y^2 - 40y + 84 = 0$.

(i) Factoring: $(y-2)(y-7)(y+6) = 0$. Largest root is $7$, so $\alpha\beta + \gamma\delta = 7$.

(ii) $(\alpha+\beta)(\gamma+\delta) = \alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta = q - (\alpha\beta+\gamma\delta) = 3-7 = -4$.

Since $(\alpha+\beta)+(\gamma+\delta) = -p = 0$, and $\alpha\beta\gamma\delta = s = 10$, with $\alpha\beta > \gamma\delta$:

$\alpha\beta$ and $\gamma\delta$ satisfy $t^2 - 7t + 10 = 0$, giving $\alpha\beta = 5$, $\gamma\delta = 2$.

(iii) $\alpha+\beta = -2$ (since $\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) = 6$ gives the sign), $\gamma+\delta = 2$.

$\alpha, \beta$ are roots of $t^2+2t+5=0$: $\alpha,\beta = -1 \pm 2i$.

$\gamma, \delta$ are roots of $t^2-2t+2=0$: $\gamma,\delta = 1 \pm i$.

The four roots of the quartic are $1 \pm i, -1 \pm 2i$.

Model Solution

Determining $A$

By Vieta’s formulas for the quartic $x^4 + px^3 + qx^2 + rx + s = 0$ with roots $\alpha, \beta, \gamma, \delta$:

$\alpha + \beta + \gamma + \delta = -p$

$\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = q$

$\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -r$

$\alpha\beta\gamma\delta = s$

The three roots of the cubic are $y_1 = \alpha\beta + \gamma\delta$, $y_2 = \alpha\gamma + \beta\delta$, $y_3 = \alpha\delta + \beta\gamma$. Their sum is:

$y_1 + y_2 + y_3 = \alpha\beta + \gamma\delta + \alpha\gamma + \beta\delta + \alpha\delta + \beta\gamma = q$

From the cubic $y^3 + Ay^2 + (pr - 4s)y + (4qs - p^2s - r^2) = 0$, the sum of the roots is $-A$. Therefore:

$q = -A \implies A = -q$

Part (i)

Substituting $p = 0$, $q = 3$, $r = -6$, $s = 10$ into the cubic:

$pr - 4s = 0 \cdot (-6) - 4(10) = -40$

$4qs - p^2 s - r^2 = 4(3)(10) - 0 - (-6)^2 = 120 - 36 = 84$

The cubic equation is:

$y^3 - 3y^2 - 40y + 84 = 0$

Testing $y = 2$: $8 - 12 - 80 + 84 = 0$. So $(y - 2)$ is a factor. Dividing:

$y^3 - 3y^2 - 40y + 84 = (y - 2)(y^2 - y - 42) = (y - 2)(y - 7)(y + 6) = 0$

The three roots are $y = -6$, $y = 2$, $y = 7$.

The largest root is $y = 7$, so:

$\alpha\beta + \gamma\delta = 7$

Part (ii)

We have:

$(\alpha + \beta)(\gamma + \delta) = \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta$

Since $\alpha\beta + \gamma\delta + \alpha\gamma + \beta\delta + \alpha\delta + \beta\gamma = q = 3$:

$(\alpha + \beta)(\gamma + \delta) = 3 - (\alpha\beta + \gamma\delta) = 3 - 7 = -4$

Now, $\alpha\beta + \gamma\delta = 7$ and $\alpha\beta \cdot \gamma\delta = \alpha\beta\gamma\delta = s = 10$.

So $\alpha\beta$ and $\gamma\delta$ are the roots of the quadratic:

$t^2 - 7t + 10 = 0 \implies (t - 5)(t - 2) = 0$

Since $\alpha\beta > \gamma\delta$:

$\alpha\beta = 5, \qquad \gamma\delta = 2$

Part (iii)

From $p = 0$: $\alpha + \beta + \gamma + \delta = 0$.

Let $u = \alpha + \beta$ and $v = \gamma + \delta$. Then $u + v = 0$ and $uv = -4$.

From Vieta’s third relation $\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -r = 6$:

$\alpha\beta(\gamma + \delta) + \gamma\delta(\alpha + \beta) = 5v + 2u = 6$

Since $v = -u$:

$5(-u) + 2u = 6 \implies -3u = 6 \implies u = -2$

So $v = 2$, giving $\alpha + \beta = -2$ and $\gamma + \delta = 2$.

The pair $\alpha, \beta$ are roots of:

$t^2 - (\alpha + \beta)t + \alpha\beta = t^2 + 2t + 5 = 0$

$t = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$

The pair $\gamma, \delta$ are roots of:

$t^2 - (\gamma + \delta)t + \gamma\delta = t^2 - 2t + 2 = 0$

$t = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i$

The four roots of the quartic are $1 + i$, $1 - i$, $-1 + 2i$, $-1 - 2i$.

Verification: $(1+i)(1-i)(-1+2i)(-1-2i) = 2 \cdot 5 = 10 = s$. ✓

Examiner Notes

The second most popular question at just over 70%, the success rate was about half marks in common with a number of other questions, with the majority earning either 16 and above, or 4 and below. A common mistake was omitting the minus sign in the first step to obtain A which resulted in candidates being unable to progress further. If the cubic equation was correctly found, then candidates tended to score all the marks as far as part (iii). A few candidates obtaining the correct results in (iii) then stated that the answers could not be complex, which was, of course, false.

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Question 4

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

Problem

4 For any function $f$ satisfying $f(x) > 0$, we define the geometric mean, $F$, by

$F(y) = e^{\frac{1}{y} \int_0^y \ln f(x) \, dx} \quad (y > 0) .$

(i) The function $f$ satisfies $f(x) > 0$ and $a$ is a positive number with $a \neq 1$. Prove that

$F(y) = a^{\frac{1}{y} \int_0^y \log_a f(x) \, dx} .$

(ii) The functions $f$ and $g$ satisfy $f(x) > 0$ and $g(x) > 0$, and the function $h$ is defined by $h(x) = f(x)g(x)$. Their geometric means are $F, G$ and $H$, respectively. Show that $H(y) = F(y)G(y)$.

(iii) Prove that, for any positive number $b$, the geometric mean of $b^x$ is $\sqrt{b^y}$.

(iv) Prove that, if $f(x) > 0$ and the geometric mean of $f(x)$ is $\sqrt{f(y)}$, then $f(x) = b^x$ for some positive number $b$.

Hint

(i) Let $\log_a f(x) = z$, then $f(x) = a^z = e^{z\ln a}$, so $\ln f(x) = \ln a \cdot \log_a f(x)$.

Substituting: $F(y) = e^{\frac{1}{y}\int_0^y \ln a \cdot \log_a f(x)\,dx} = a^{\frac{1}{y}\int_0^y \log_a f(x)\,dx}$.

(ii) $H(y) = e^{\frac{1}{y}\int_0^y \ln(f(x)g(x))\,dx} = e^{\frac{1}{y}\int_0^y (\ln f(x)+\ln g(x))\,dx} = F(y)G(y)$.

(iii) With $f(x)=b^x$: $F(y) = e^{\frac{1}{y}\int_0^y x\ln b\,dx} = e^{\frac{1}{y}\ln b \cdot \frac{y^2}{2}} = e^{\frac{y}{2}\ln b} = \sqrt{b^y}$.

(Also check $b=1$: $F(y) = e^0 = 1 = \sqrt{1^y}$.)

(iv) Setting $F(y) = \sqrt{f(y)}$: $\frac{1}{y}\int_0^y \ln f(x)\,dx = \frac{1}{2}\ln f(y)$.

Differentiating w.r.t. $y$: $\frac{f'(y)}{f(y)\ln f(y)} = \frac{1}{y}$.

Integrating: $\ln\ln f(y) = \ln y + c = \ln(ky)$, so $\ln f(y) = ky$, giving $f(y) = e^{ky} = b^y$.

Model Solution

Part (i)

Recall the change of base formula for logarithms: for any $a > 0$ with $a \neq 1$ and any $f(x) > 0$,

$\ln f(x) = \ln a \cdot \log_a f(x)$

This follows since if $\log_a f(x) = t$, then $a^t = f(x)$, so $t\ln a = \ln f(x)$.

Substituting into the definition of $F$:

$F(y) = e^{\frac{1}{y}\int_0^y \ln f(x)\,dx} = e^{\frac{1}{y}\int_0^y \ln a \cdot \log_a f(x)\,dx} = e^{\ln a \cdot \frac{1}{y}\int_0^y \log_a f(x)\,dx}$

Since $e^{\ln a \cdot k} = (e^{\ln a})^k = a^k$:

$F(y) = a^{\frac{1}{y}\int_0^y \log_a f(x)\,dx}$

Part (ii)

Since $h(x) = f(x)g(x)$ with $f(x) > 0$ and $g(x) > 0$:

$H(y) = e^{\frac{1}{y}\int_0^y \ln h(x)\,dx} = e^{\frac{1}{y}\int_0^y \ln\bigl(f(x)g(x)\bigr)\,dx}$

Using $\ln(fg) = \ln f + \ln g$:

$H(y) = e^{\frac{1}{y}\int_0^y \bigl[\ln f(x) + \ln g(x)\bigr]\,dx} = e^{\frac{1}{y}\int_0^y \ln f(x)\,dx \;+\; \frac{1}{y}\int_0^y \ln g(x)\,dx}$

$= e^{\frac{1}{y}\int_0^y \ln f(x)\,dx} \cdot e^{\frac{1}{y}\int_0^y \ln g(x)\,dx} = F(y)\,G(y)$

Part (iii)

Let $f(x) = b^x$ where $b > 0$. We compute:

$F(y) = e^{\frac{1}{y}\int_0^y \ln(b^x)\,dx}$

Since $\ln(b^x) = x\ln b$:

$F(y) = e^{\frac{1}{y}\int_0^y x\ln b\,dx} = e^{\frac{\ln b}{y}\int_0^y x\,dx} = e^{\frac{\ln b}{y} \cdot \frac{y^2}{2}} = e^{\frac{y\ln b}{2}}$

Therefore:

$F(y) = (e^{\ln b})^{y/2} = b^{y/2} = \sqrt{b^y}$

For the special case $b = 1$: $f(x) = 1$ for all $x$, so $\ln f(x) = 0$ and $F(y) = e^0 = 1 = \sqrt{1^y}$. ✓

Hence for every positive number $b$, the geometric mean of $b^x$ is $\sqrt{b^y}$.

Part (iv)

Suppose $f(x) > 0$ and $F(y) = \sqrt{f(y)}$ for all $y > 0$. By definition:

$e^{\frac{1}{y}\int_0^y \ln f(x)\,dx} = f(y)^{1/2}$

Taking the natural logarithm of both sides:

$\frac{1}{y}\int_0^y \ln f(x)\,dx = \frac{1}{2}\ln f(y)$

Multiplying through by $y$:

$\int_0^y \ln f(x)\,dx = \frac{y}{2}\ln f(y) \qquad (\star)$

The left side is differentiable by the Fundamental Theorem of Calculus (since $\ln f$ is continuous). As $\frac{y}{2}$ is smooth and nonzero for $y > 0$, this forces $\ln f(y)$ to be differentiable, so $f$ is differentiable.

Differentiating $(\star)$ with respect to $y$:

$\ln f(y) = \frac{1}{2}\ln f(y) + \frac{y}{2}\cdot\frac{f'(y)}{f(y)}$

$\frac{1}{2}\ln f(y) = \frac{y}{2}\cdot\frac{f'(y)}{f(y)}$

$\ln f(y) = y\cdot\frac{f'(y)}{f(y)}$

Set $g(y) = \ln f(y)$, so $g'(y) = \frac{f'(y)}{f(y)}$. The equation becomes:

$g(y) = y\,g'(y)$

Rearranging: $y\,g'(y) - g(y) = 0$, which gives:

$\frac{d}{dy}\!\left(\frac{g(y)}{y}\right) = \frac{y\,g'(y) - g(y)}{y^2} = 0$

Therefore $\frac{g(y)}{y} = k$ for some constant $k$, so $g(y) = ky$ for all $y > 0$.

Returning to $f$:

$\ln f(y) = ky \implies f(y) = e^{ky} = b^y$

where $b = e^k > 0$.

Verification: by part (iii), the geometric mean of $b^x$ is $\sqrt{b^y} = \sqrt{f(y)}$, confirming consistency. ✓

Examiner Notes

Three fifths of the candidates attempted question 4 with a marginally better success rate than question 3. A significant proportion of candidates struggled with changing base for part (i), but almost all completed (ii) successfully. A common strategy for part (iii) was to use the result of part (i) but very few remembered to check for b = 1. There were very few successful attempts for part (iv); many tried integration by parts, but rarely successfully.

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Question 5

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

Problem

5 The point with cartesian coordinates $(x, y)$ lies on a curve with polar equation $r = \text{f}(\theta)$. Find an expression for $\frac{\text{d}y}{\text{d}x}$ in terms of $\text{f}(\theta)$, $\text{f}'(\theta)$ and $\tan \theta$.

Two curves, with polar equations $r = \text{f}(\theta)$ and $r = \text{g}(\theta)$, meet at right angles. Show that where they meet $\text{f}'(\theta)\text{g}'(\theta) + \text{f}(\theta)\text{g}(\theta) = 0.$

The curve $C$ has polar equation $r = \text{f}(\theta)$ and passes through the point given by $r = 4$, $\theta = -\frac{1}{2}\pi$. For each positive value of $a$, the curve with polar equation $r = a(1 + \sin \theta)$ meets $C$ at right angles. Find $\text{f}(\theta)$.

Sketch on a single diagram the three curves with polar equations $r = 1 + \sin \theta$, $r = 4(1 + \sin \theta)$ and $r = \text{f}(\theta)$.

Hint

$y = r\sin\theta$, $x = r\cos\theta$. Differentiating w.r.t. $\theta$:

$\frac{dy}{dx} = \frac{f\cos\theta + f'\sin\theta}{-f\sin\theta + f'\cos\theta} = \frac{f + f'\tan\theta}{-f\tan\theta + f'}$

Orthogonality condition: product of gradients = $-1$. Setting:

$\frac{f+f'\tan\theta}{-f\tan\theta+f'} \cdot \frac{g+g'\tan\theta}{-g\tan\theta+g'} = -1$

Expanding and simplifying: $(fg+f'g')\sec^2\theta = 0$, hence $fg+f'g'=0$.

With $g(\theta) = a(1+\sin\theta)$, $g'(\theta) = a\cos\theta$:

$f'a\cos\theta + fa(1+\sin\theta) = 0 \implies \frac{f'}{f} = -\frac{1+\sin\theta}{\cos\theta} = -\sec\theta - \tan\theta$

Integrating: $\ln f = -\ln(\sec\theta+\tan\theta) + \ln\cos\theta + c = \ln\frac{k\cos^2\theta}{1+\sin\theta} = \ln(k(1-\sin\theta))$.

Using $f(-\pi/2) = 4$: $4 = k(1-(-1)) = 2k$, so $k=2$.

$f(\theta) = 2(1-\sin\theta)$

The sketch shows: $r=1+\sin\theta$ (small cardioid), $r=4(1+\sin\theta)$ (large cardioid), and $r=2(1-\sin\theta)$ (cardioid reflected in the $x$-axis, passing through the origin at $\theta=\pi/2$).

Model Solution

Finding $\dfrac{dy}{dx}$

For a curve with polar equation $r = \text{f}(\theta)$, the Cartesian coordinates are:

$x = \text{f}(\theta)\cos\theta, \qquad y = \text{f}(\theta)\sin\theta$

Differentiating with respect to $\theta$:

$\frac{dx}{d\theta} = \text{f}'(\theta)\cos\theta - \text{f}(\theta)\sin\theta$

$\frac{dy}{d\theta} = \text{f}'(\theta)\sin\theta + \text{f}(\theta)\cos\theta$

Therefore:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\text{f}'(\theta)\sin\theta + \text{f}(\theta)\cos\theta}{\text{f}'(\theta)\cos\theta - \text{f}(\theta)\sin\theta}$

Dividing numerator and denominator by $\cos\theta$:

$\frac{dy}{dx} = \frac{\text{f}(\theta) + \text{f}'(\theta)\tan\theta}{\text{f}'(\theta) - \text{f}(\theta)\tan\theta} \qquad \text{(*)}$

Orthogonality condition

For the curve $r = \text{g}(\theta)$, by the same argument:

$\frac{dy}{dx}\bigg|_{\text{g}} = \frac{\text{g}(\theta) + \text{g}'(\theta)\tan\theta}{\text{g}'(\theta) - \text{g}(\theta)\tan\theta}$

If the two curves meet at right angles, the product of their gradients equals $-1$:

$\frac{\text{f} + \text{f}'\tan\theta}{\text{f}' - \text{f}\tan\theta} \cdot \frac{\text{g} + \text{g}'\tan\theta}{\text{g}' - \text{g}\tan\theta} = -1$

Cross-multiplying:

$(\text{f} + \text{f}'\tan\theta)(\text{g} + \text{g}'\tan\theta) = -(\text{f}' - \text{f}\tan\theta)(\text{g}' - \text{g}\tan\theta)$

Expanding the left side:

$\text{fg} + \text{fg}'\tan\theta + \text{f}'\text{g}\tan\theta + \text{f}'\text{g}'\tan^2\theta$

Expanding the right side:

$-(\text{f}'\text{g}' - \text{f}'\text{g}\tan\theta - \text{fg}'\tan\theta + \text{fg}\tan^2\theta)$

Bringing everything to one side:

$\text{fg} + \text{fg}'\tan\theta + \text{f}'\text{g}\tan\theta + \text{f}'\text{g}'\tan^2\theta + \text{f}'\text{g}' - \text{f}'\text{g}\tan\theta - \text{fg}'\tan\theta + \text{fg}\tan^2\theta = 0$

The terms $\text{fg}'\tan\theta$ and $\text{f}'\text{g}\tan\theta$ cancel in pairs:

$\text{fg}(1 + \tan^2\theta) + \text{f}'\text{g}'(1 + \tan^2\theta) = 0$

$(\text{f}'\text{g}' + \text{fg})\sec^2\theta = 0$

Since $\sec^2\theta \neq 0$:

$\text{f}'(\theta)\text{g}'(\theta) + \text{f}(\theta)\text{g}(\theta) = 0 \qquad \text{(**)}$

Finding $\text{f}(\theta)$

We need $\text{f}(\theta)$ such that $r = \text{f}(\theta)$ meets $r = a(1 + \sin\theta)$ at right angles for every $a > 0$.

Set $\text{g}(\theta) = a(1 + \sin\theta)$, so $\text{g}'(\theta) = a\cos\theta$. Substituting into $(**)$:

$\text{f}'(\theta) \cdot a\cos\theta + \text{f}(\theta) \cdot a(1 + \sin\theta) = 0$

Since $a \neq 0$:

$\text{f}'(\theta)\cos\theta + \text{f}(\theta)(1 + \sin\theta) = 0$

$\frac{\text{f}'(\theta)}{\text{f}(\theta)} = -\frac{1 + \sin\theta}{\cos\theta}$

This is a separable ODE. Integrating both sides with respect to $\theta$:

$\ln\text{f}(\theta) = -\int \frac{1 + \sin\theta}{\cos\theta}\, d\theta = -\int \sec\theta\, d\theta - \int \tan\theta\, d\theta$

$= -\ln|\sec\theta + \tan\theta| + \ln|\cos\theta| + c$

Combining the logarithms:

$\ln\text{f}(\theta) = \ln\frac{k\cos\theta}{\sec\theta + \tan\theta} \qquad \text{where } k = e^c > 0$

Simplifying the fraction:

$\frac{\cos\theta}{\sec\theta + \tan\theta} = \frac{\cos\theta}{\frac{1 + \sin\theta}{\cos\theta}} = \frac{\cos^2\theta}{1 + \sin\theta} = \frac{1 - \sin^2\theta}{1 + \sin\theta} = 1 - \sin\theta$

Therefore:

$\text{f}(\theta) = k(1 - \sin\theta)$

Applying the condition $\text{f}(-\frac{1}{2}\pi) = 4$:

$4 = k(1 - \sin(-\tfrac{1}{2}\pi)) = k(1 + 1) = 2k$

So $k = 2$, giving:

$\text{f}(\theta) = 2(1 - \sin\theta)$

Sketch

The three curves are:

• $r = 1 + \sin\theta$: a cardioid with cusp at the origin (at $\theta = -\frac{1}{2}\pi$) and maximum $r = 2$ at $\theta = \frac{1}{2}\pi$.
• $r = 4(1 + \sin\theta)$: a cardioid with cusp at the origin and maximum $r = 8$ at $\theta = \frac{1}{2}\pi$.
• $r = 2(1 - \sin\theta)$: a cardioid reflected in the $x$-axis, with cusp at the origin (at $\theta = \frac{1}{2}\pi$) and maximum $r = 4$ at $\theta = -\frac{1}{2}\pi$.

All three cardioids pass through the origin. The first two are nested (same orientation, different sizes) with their dimples pointing downward. The third is oriented upward with its dimple pointing upward, crossing the first two at right angles at every intersection.

Examiner Notes

Very slightly more popular than question 4, the marks scored were on average 1 less per attempt. Most found dy/dx successfully, though a significant minority swapped x and y. In this case, they could still obtain the displayed equation successfully, but in both categories, there were frequent sign errors when differentiating trigonometric functions. Most then attempted using the displayed result to find f(theta), either by separating variables or using an integrating factor and got as far as f(theta) = (k cos^2 theta / (1 + sin theta)) but then more than half got stuck. Most plotted the two given curves relatively correctly, but then a substantial number used guesswork having not previously obtained C correctly.

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Question 6

Topic: 积分与换元 (Integration and Substitution)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 In this question, you are not permitted to use any properties of trigonometric functions or inverse trigonometric functions.

The function $\text{T}$ is defined for $x > 0$ by $\text{T}(x) = \int_{0}^{x} \frac{1}{1 + u^2} \, \text{d}u \, ,$ and $\text{T}_{\infty} = \int_{0}^{\infty} \frac{1}{1 + u^2} \, \text{d}u$ (which has a finite value).

(i) By making an appropriate substitution in the integral for $\text{T}(x)$, show that $\text{T}(x) = \text{T}_{\infty} - \text{T}(x^{-1}) \, .$

(ii) Let $v = \frac{u + a}{1 - au}$, where $a$ is a constant. Verify that, for $u \neq a^{-1}$, $\frac{\text{d}v}{\text{d}u} = \frac{1 + v^2}{1 + u^2} \, .$

Hence show that, for $a > 0$ and $x < \frac{1}{a}$, $\text{T}(x) = \text{T}\left(\frac{x + a}{1 - ax}\right) - \text{T}(a) \, .$

Deduce that $\text{T}(x^{-1}) = 2\text{T}_{\infty} - \text{T}\left(\frac{x + a}{1 - ax}\right) - \text{T}(a^{-1})$ and hence that, for $b > 0$ and $y > \frac{1}{b}$, $\text{T}(y) = 2\text{T}_{\infty} - \text{T}\left(\frac{y + b}{by - 1}\right) - \text{T}(b) \, .$

(iii) Use the above results to show that $\text{T}(\sqrt{3}) = \frac{2}{3}\text{T}_{\infty}$ and $\text{T}(\sqrt{2} - 1) = \frac{1}{4}\text{T}_{\infty}$.

Hint

(i) $T(x) = \int_0^x \frac{1}{1+u^2}\,du$.

Let $u = v^{-1}$, $\frac{du}{dv} = -v^{-2}$. Then:

$T(x) = \int_{\infty}^{x^{-1}} \frac{1}{1+v^{-2}} \cdot (-v^{-2})\,dv = \int_{x^{-1}}^{\infty} \frac{1}{v^2+1}\,dv = \int_0^{\infty} \frac{1}{1+u^2}\,du - \int_0^{x^{-1}} \frac{1}{1+u^2}\,du$

$T(x) = T_{\infty} - T(x^{-1})$

(ii) $v = \frac{u+a}{1-au} \Leftrightarrow a = \frac{v-u}{1+uv}$.

Differentiating implicitly: $\frac{dv}{du} = \frac{1+v^2}{1+u^2}$.

Alternatively, directly: $\frac{dv}{du} = \frac{(1-au)+a(u+a)}{(1-au)^2} = \frac{1+a^2}{(1-au)^2} = \frac{(1-au)^2+(u+a)^2}{(1-au)^2(1+u^2)} = \frac{1+v^2}{1+u^2}$.

Applying this substitution: $T(x) = \int_a^{\frac{x+a}{1-ax}} \frac{1}{1+v^2}\,dv = T\!\left(\frac{x+a}{1-ax}\right) - T(a)$.

Using $T(x) = T_{\infty} - T(x^{-1})$ and $T(a) = T_{\infty} - T(a^{-1})$:

$T(x^{-1}) = 2T_{\infty} - T\!\left(\frac{x+a}{1-ax}\right) - T(a^{-1})$.

Setting $y = x^{-1}$, $b = a^{-1}$ (so $y > 1/b$ when $x < 1/a$):

$T(y) = 2T_{\infty} - T\!\left(\frac{y+b}{by-1}\right) - T(b)$.

(iii) For $T(\sqrt{3})$: apply the final result of (ii) with $y = b = \sqrt{3}$:

$T(\sqrt{3}) = 2T_{\infty} - T(\sqrt{3}) - T(\sqrt{3})$, so $3T(\sqrt{3}) = 2T_{\infty}$, giving $T(\sqrt{3}) = \frac{2}{3}T_{\infty}$.

For $T(\sqrt{2}-1)$: use $T(x) = T\!\left(\frac{x+a}{1-ax}\right) - T(a)$ with $x = \sqrt{2}-1$, $a = 1$:

$T(\sqrt{2}-1) = T(\sqrt{2}+1) - T(1)$.

From $T(x) = T_{\infty} - T(x^{-1})$ with $x=1$: $T(1) = \frac{1}{2}T_{\infty}$.

And $T(\sqrt{2}+1) = T_{\infty} - T(\sqrt{2}-1)$.

So $T(\sqrt{2}-1) = T_{\infty} - T(\sqrt{2}-1) - \frac{1}{2}T_{\infty}$, giving $T(\sqrt{2}-1) = \frac{1}{4}T_{\infty}$.

Model Solution

Part (i)

We wish to show that $\text{T}(x) = \text{T}_{\infty} - \text{T}(x^{-1})$ for $x > 0$.

In the integral $\text{T}(x) = \int_0^x \frac{1}{1+u^2}\,du$, substitute $u = t^{-1}$, so $du = -t^{-2}\,dt$.

When $u = 0$, $t = \infty$; when $u = x$, $t = x^{-1}$.

$\text{T}(x) = \int_{\infty}^{x^{-1}} \frac{1}{1+t^{-2}} \cdot (-t^{-2})\,dt = \int_{\infty}^{x^{-1}} \frac{-t^{-2}}{1+t^{-2}}\,dt$

Multiplying numerator and denominator by $t^2$:

$= \int_{\infty}^{x^{-1}} \frac{-1}{t^2+1}\,dt = \int_{x^{-1}}^{\infty} \frac{1}{1+t^2}\,dt$

Splitting the integral:

$= \int_0^{\infty} \frac{1}{1+t^2}\,dt - \int_0^{x^{-1}} \frac{1}{1+t^2}\,dt = \text{T}_{\infty} - \text{T}(x^{-1})$

Hence $\text{T}(x) = \text{T}_{\infty} - \text{T}(x^{-1})$. $\qquad \text{(R1)}$

Part (ii)

Verification. Let $v = \frac{u+a}{1-au}$. Differentiating using the quotient rule:

$\frac{dv}{du} = \frac{(1-au) \cdot 1 - (u+a)(-a)}{(1-au)^2} = \frac{1 - au + au + a^2}{(1-au)^2} = \frac{1+a^2}{(1-au)^2}$

Now we compute $\frac{1+v^2}{1+u^2}$:

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

Expanding the numerator:

$(1-au)^2 + (u+a)^2 = 1 - 2au + a^2u^2 + u^2 + 2au + a^2 = 1 + a^2 + a^2u^2 + u^2 = (1+a^2)(1+u^2)$

Therefore:

$\frac{1+v^2}{1+u^2} = \frac{(1+a^2)(1+u^2)}{(1-au)^2(1+u^2)} = \frac{1+a^2}{(1-au)^2} = \frac{dv}{du}$

Hence $\frac{dv}{du} = \frac{1+v^2}{1+u^2}$. $\qquad \text{(R2)}$

Showing $\text{T}(x) = \text{T}\!\left(\frac{x+a}{1-ax}\right) - \text{T}(a)$.

From (R2), $\frac{dv}{1+v^2} = \frac{du}{1+u^2}$. In the integral $\text{T}(x) = \int_0^x \frac{1}{1+u^2}\,du$, substituting $v = \frac{u+a}{1-au}$:

When $u = 0$: $v = a$. When $u = x$: $v = \frac{x+a}{1-ax}$ (valid since $x < 1/a$ ensures $1-ax > 0$).

$\text{T}(x) = \int_0^x \frac{1}{1+u^2}\,du = \int_a^{\frac{x+a}{1-ax}} \frac{1}{1+v^2}\,dv$

$= \int_0^{\frac{x+a}{1-ax}} \frac{1}{1+v^2}\,dv - \int_0^a \frac{1}{1+v^2}\,dv = \text{T}\!\left(\frac{x+a}{1-ax}\right) - \text{T}(a) \qquad \text{(R3)}$

Deducing $\text{T}(x^{-1}) = 2\text{T}_{\infty} - \text{T}\!\left(\frac{x+a}{1-ax}\right) - \text{T}(a^{-1})$.

From (R1): $\text{T}(x) = \text{T}_{\infty} - \text{T}(x^{-1})$, so $\text{T}(x^{-1}) = \text{T}_{\infty} - \text{T}(x)$.

Substituting (R3) for $\text{T}(x)$:

$\text{T}(x^{-1}) = \text{T}_{\infty} - \left[\text{T}\!\left(\frac{x+a}{1-ax}\right) - \text{T}(a)\right]$

$= \text{T}_{\infty} - \text{T}\!\left(\frac{x+a}{1-ax}\right) + \text{T}(a)$

From (R1) with $x$ replaced by $a$: $\text{T}(a) = \text{T}_{\infty} - \text{T}(a^{-1})$.

$\text{T}(x^{-1}) = \text{T}_{\infty} - \text{T}\!\left(\frac{x+a}{1-ax}\right) + \text{T}_{\infty} - \text{T}(a^{-1})$

$= 2\text{T}_{\infty} - \text{T}\!\left(\frac{x+a}{1-ax}\right) - \text{T}(a^{-1}) \qquad \text{(R4)}$

Showing $\text{T}(y) = 2\text{T}_{\infty} - \text{T}\!\left(\frac{y+b}{by-1}\right) - \text{T}(b)$.

In (R4), set $y = x^{-1}$ and $b = a^{-1}$ (so $x = y^{-1}$ and $a = b^{-1}$). The condition $x < 1/a$ becomes $y^{-1} < b$, i.e. $y > 1/b$.

$\frac{x+a}{1-ax} = \frac{y^{-1} + b^{-1}}{1 - b^{-1}y^{-1}} = \frac{\frac{b+y}{by}}{\frac{by-1}{by}} = \frac{y+b}{by-1}$

Substituting into (R4):

$\text{T}(y) = 2\text{T}_{\infty} - \text{T}\!\left(\frac{y+b}{by-1}\right) - \text{T}(b) \qquad \text{(R5)}$

for $b > 0$ and $y > 1/b$.

Part (iii)

Showing $\text{T}(\sqrt{3}) = \frac{2}{3}\text{T}_{\infty}$.

Apply (R5) with $y = b = \sqrt{3}$. We check: $b > 0$ and $y = \sqrt{3} > 1/\sqrt{3}$, so the conditions are satisfied.

$\frac{y+b}{by-1} = \frac{\sqrt{3}+\sqrt{3}}{3-1} = \frac{2\sqrt{3}}{2} = \sqrt{3}$

Substituting into (R5):

$\text{T}(\sqrt{3}) = 2\text{T}_{\infty} - \text{T}(\sqrt{3}) - \text{T}(\sqrt{3})$

$3\text{T}(\sqrt{3}) = 2\text{T}_{\infty}$

$\text{T}(\sqrt{3}) = \frac{2}{3}\text{T}_{\infty} \qquad \text{(R6)}$

Showing $\text{T}(\sqrt{2}-1) = \frac{1}{4}\text{T}_{\infty}$.

Apply (R3) with $x = \sqrt{2}-1$ and $a = 1$. We check: $a = 1 > 0$ and $x = \sqrt{2}-1 \approx 0.414 < 1 = 1/a$, so the conditions are satisfied.

$\frac{x+a}{1-ax} = \frac{(\sqrt{2}-1)+1}{1-1 \cdot (\sqrt{2}-1)} = \frac{\sqrt{2}}{2-\sqrt{2}}$

Rationalising:

$\frac{\sqrt{2}}{2-\sqrt{2}} \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}(2+\sqrt{2})}{4-2} = \frac{2\sqrt{2}+2}{2} = \sqrt{2}+1$

From (R3):

$\text{T}(\sqrt{2}-1) = \text{T}(\sqrt{2}+1) - \text{T}(1) \qquad \text{(R7)}$

From (R1) with $x = 1$: $\text{T}(1) = \text{T}_{\infty} - \text{T}(1)$, so $2\text{T}(1) = \text{T}_{\infty}$, giving $\text{T}(1) = \frac{1}{2}\text{T}_{\infty}$.

From (R1) with $x = \sqrt{2}+1$: $\text{T}(\sqrt{2}+1) = \text{T}_{\infty} - \text{T}\!\left(\frac{1}{\sqrt{2}+1}\right) = \text{T}_{\infty} - \text{T}(\sqrt{2}-1)$

since $\frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.

Substituting into (R7):

$\text{T}(\sqrt{2}-1) = \text{T}_{\infty} - \text{T}(\sqrt{2}-1) - \frac{1}{2}\text{T}_{\infty}$

$2\text{T}(\sqrt{2}-1) = \frac{1}{2}\text{T}_{\infty}$

$\text{T}(\sqrt{2}-1) = \frac{1}{4}\text{T}_{\infty} \qquad \text{(R8)}$

Examiner Notes

Attempted by four fifths of candidates; success rate only slightly less than Q4. As every part required obtaining a given result, had to be marked strictly on presentation. Surprising problems with changing variables in first part - candidates did not clearly understand dummy variables, others integrated with respect to constants. Despite ban on trigonometric functions, some still tried to use tangent. The two results in (iii), especially the second, were testing but found very hard, and previous inapplicable results were used ignoring conditions given as inequalities.

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Question 7

Topic: 解析几何：椭圆与切线 (Analytic Geometry: Ellipse and Tangents)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Show that the point $T$ with coordinates $\left( \frac{a(1 - t^2)}{1 + t^2} , \frac{2bt}{1 + t^2} \right) \qquad \text{(*)}$ (where $a$ and $b$ are non-zero) lies on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 .$

(i) The line $L$ is the tangent to the ellipse at $T$. The point $(X, Y)$ lies on $L$, and $X^2 \neq a^2$. Show that $(a + X)bt^2 - 2aYt + b(a - X) = 0 .$ Deduce that if $a^2Y^2 > (a^2 - X^2)b^2$, then there are two distinct lines through $(X, Y)$ that are tangents to the ellipse. Interpret this result geometrically. Show, by means of a sketch, that the result holds also if $X^2 = a^2$.

(ii) The distinct points $P$ and $Q$ are given by $(*)$, with $t = p$ and $t = q$, respectively. The tangents to the ellipse at $P$ and $Q$ meet at the point with coordinates $(X, Y)$, where $X^2 \neq a^2$. Show that $(a + X)pq = a - X$ and find an expression for $p + q$ in terms of $a, b, X$ and $Y$. Given that the tangents meet the $y$-axis at points $(0, y_1)$ and $(0, y_2)$, where $y_1 + y_2 = 2b$, show that $\frac{X^2}{a^2} + \frac{Y}{b} = 1 .$

Hint

Stem: substituting $T$‘s coordinates into LHS of ellipse equation:

$\frac{a^2(1-t^2)^2/(1+t^2)^2}{a^2} + \frac{4b^2t^2/(1+t^2)^2}{b^2} = \frac{(1-t^2)^2+4t^2}{(1+t^2)^2} = 1$.

(i) Implicit differentiation of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{dy}{dx} = -\frac{b^2x}{a^2y} = -\frac{b(1-t^2)}{2at}$.

Tangent $L$ at $T$: $y - \frac{2bt}{1+t^2} = -\frac{b(1-t^2)}{2at}\left(x - \frac{a(1-t^2)}{1+t^2}\right)$.

Simplifying and substituting $(X,Y)$ on $L$:

$0 = (a+X)bt^2 - 2aYt + b(a-X)$.

For two distinct tangents, discriminant $> 0$:

$4a^2Y^2 > 4b^2(a^2-X^2)$, i.e. $a^2Y^2 > (a^2-X^2)b^2$.

This means $\frac{X^2}{a^2} + \frac{Y^2}{b^2} > 1$, so $(X,Y)$ lies outside the ellipse.

(ii) $p$ and $q$ are roots of the quadratic in $t$. By Vieta’s formulas:

$p+q = \frac{2aY}{(a+X)b}$, $pq = \frac{a-X}{a+X}$, so $(a+X)pq = a-X$.

For tangent at $P$ ($t=p$) meeting $y$-axis at $(0,y_1)$: $bp^2 - 2py_1 + b = 0$.

Similarly for $Q$ ($t=q$): $bq^2 - 2qy_2 + b = 0$.

With $y_1+y_2 = 2b$: $\frac{p^2+1}{p} + \frac{q^2+1}{q} = 4$, i.e. $p+q + \frac{p+q}{pq} = 4$.

Substituting Vieta’s expressions: $\frac{2aY}{a+X} + \frac{2aY}{a-X} = 4b$, which simplifies to $\frac{X^2}{a^2} + \frac{Y}{b} = 1$.

Model Solution

Showing $T$ lies on the ellipse.

Substituting $x = \frac{a(1-t^2)}{1+t^2}$ and $y = \frac{2bt}{1+t^2}$ into the left side of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{a^2} \cdot \frac{a^2(1-t^2)^2}{(1+t^2)^2} + \frac{1}{b^2} \cdot \frac{4b^2t^2}{(1+t^2)^2} = \frac{(1-t^2)^2 + 4t^2}{(1+t^2)^2}$

Expanding the numerator:

$(1-t^2)^2 + 4t^2 = 1 - 2t^2 + t^4 + 4t^2 = 1 + 2t^2 + t^4 = (1+t^2)^2$

So $\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{(1+t^2)^2}{(1+t^2)^2} = 1$, confirming $T$ lies on the ellipse.

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Part (i)

We find the tangent to the ellipse at $T$. Differentiating $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ implicitly with respect to $x$:

$\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$

At $T$, with $x_T = \frac{a(1-t^2)}{1+t^2}$ and $y_T = \frac{2bt}{1+t^2}$:

$\frac{dy}{dx}\bigg|_T = -\frac{b^2 \cdot \frac{a(1-t^2)}{1+t^2}}{a^2 \cdot \frac{2bt}{1+t^2}} = -\frac{b(1-t^2)}{2at}$

The tangent line $L$ at $T$ is:

$y - \frac{2bt}{1+t^2} = -\frac{b(1-t^2)}{2at}\left(x - \frac{a(1-t^2)}{1+t^2}\right)$

Since $(X, Y)$ lies on $L$:

$Y - \frac{2bt}{1+t^2} = -\frac{b(1-t^2)}{2at}\left(X - \frac{a(1-t^2)}{1+t^2}\right)$

Multiplying both sides by $2at(1+t^2)$:

$2at(1+t^2)Y - 4abt^2 = -b(1-t^2)\left[X(1+t^2) - a(1-t^2)\right]$

Expanding the right side:

$2at(1+t^2)Y - 4abt^2 = -bX(1-t^4) + ab(1-t^2)^2$

Bringing everything to one side and expanding:

$2aYt + 2aYt^3 - 4abt^2 + bX - bXt^4 - ab + 2abt^2 - abt^4 = 0$

Collecting by powers of $t$:

$-b(X+a)t^4 + 2aYt^3 - 2abt^2 + 2aYt - b(a-X) = 0$

We can factor out $(1+t^2)$. Performing polynomial division:

$-b(X+a)t^4 + 2aYt^3 - 2abt^2 + 2aYt - b(a-X) = (1+t^2)\left[-b(X+a)t^2 + 2aYt - b(a-X)\right]$

Verification: expanding the right side gives $-b(X+a)t^4 + 2aYt^3 - b(X+a)t^2 - b(a-X)t^2 + 2aYt - b(a-X)$, and $-b(X+a)t^2 - b(a-X)t^2 = -b[(X+a)+(a-X)]t^2 = -2abt^2$. ✓

Since $1 + t^2 \neq 0$ for real $t$, we require:

$-b(X+a)t^2 + 2aYt - b(a-X) = 0$

Multiplying by $-1$:

$(a+X)bt^2 - 2aYt + b(a-X) = 0 \qquad \text{(*)}$

Deduction. For a given point $(X, Y)$ with $X^2 \neq a^2$, the values of $t$ corresponding to tangents through $(X,Y)$ satisfy the quadratic $(*)$. Two distinct tangent lines exist when $(*)$ has two distinct real roots, i.e. when the discriminant is positive:

$\Delta = (-2aY)^2 - 4 \cdot b(a+X) \cdot b(a-X) = 4a^2Y^2 - 4b^2(a^2 - X^2) > 0$

$a^2Y^2 > b^2(a^2 - X^2)$

This condition is equivalent to $\frac{X^2}{a^2} + \frac{Y^2}{b^2} > 1$, meaning the point $(X, Y)$ lies outside the ellipse. Geometrically, from any exterior point there are exactly two tangent lines to the ellipse.

When $X^2 = a^2$. If $X = a$, then $(*)$ becomes $2abt^2 - 2aYt = 0$, i.e. $t(bt - Y) = 0$, giving $t = 0$ (the tangent at $(a, 0)$, which is the vertical line $x = a$) and $t = Y/b$ (another tangent). If $Y \neq 0$, these are distinct. Similarly for $X = -a$. Since the point $(a, Y)$ with $Y \neq 0$ lies outside the ellipse, the result holds.

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Part (ii)

The parameter values $p$ and $q$ for the two tangent points are the two roots of $(*)$:

$(a+X)bt^2 - 2aYt + b(a-X) = 0$

By Vieta’s formulas:

$p + q = \frac{2aY}{(a+X)b} \qquad \text{and} \qquad pq = \frac{b(a-X)}{b(a+X)} = \frac{a-X}{a+X}$

From the product:

$(a+X)pq = a - X$

and from the sum:

$p + q = \frac{2aY}{(a+X)b}$

Finding the $y$-intercepts. The tangent at $T(t)$ has equation:

$y - \frac{2bt}{1+t^2} = -\frac{b(1-t^2)}{2at}\left(x - \frac{a(1-t^2)}{1+t^2}\right)$

Setting $x = 0$ and writing $y_1$ for the intercept when $t = p$:

$y_1 = \frac{2bp}{1+p^2} + \frac{b(1-p^2)}{2ap} \cdot \frac{a(1-p^2)}{1+p^2} = \frac{2bp}{1+p^2} + \frac{b(1-p^2)^2}{2p(1+p^2)}$

$= \frac{b}{2p(1+p^2)}\left[4p^2 + (1-p^2)^2\right] = \frac{b(1+p^2)^2}{2p(1+p^2)} = \frac{b(1+p^2)}{2p}$

Similarly $y_2 = \frac{b(1+q^2)}{2q}$.

Applying the condition $y_1 + y_2 = 2b$:

$\frac{b(1+p^2)}{2p} + \frac{b(1+q^2)}{2q} = 2b$

Dividing by $b$ and multiplying by $2pq$:

$q(1+p^2) + p(1+q^2) = 4pq$

$q + qp^2 + p + pq^2 = 4pq$

$(p+q) + pq(p+q) = 4pq$

$(p+q)(1 + pq) = 4pq$

Substituting $pq = \frac{a-X}{a+X}$ and $p+q = \frac{2aY}{(a+X)b}$:

$\frac{2aY}{(a+X)b} \cdot \left(1 + \frac{a-X}{a+X}\right) = 4 \cdot \frac{a-X}{a+X}$

$\frac{2aY}{(a+X)b} \cdot \frac{a+X+a-X}{a+X} = \frac{4(a-X)}{a+X}$

$\frac{2aY}{(a+X)b} \cdot \frac{2a}{a+X} = \frac{4(a-X)}{a+X}$

$\frac{4a^2 Y}{b(a+X)^2} = \frac{4(a-X)}{a+X}$

$\frac{a^2 Y}{b(a+X)} = a - X$

$a^2 Y = b(a-X)(a+X) = b(a^2 - X^2)$

$\frac{Y}{b} = \frac{a^2 - X^2}{a^2} = 1 - \frac{X^2}{a^2}$

$\frac{X^2}{a^2} + \frac{Y}{b} = 1$

Examiner Notes

Popularity between Q4 and Q5, mean score about 8/20 making it one of the least successful pure questions. Most did stem correctly then scored about half marks on (i) before stopping. Common slip: differentiating ellipse equation implicitly with RHS = 1 instead of 0. Geometric interpretation in (i) frequently omitted. Few continued to (ii), though some courted disaster by labelling coefficients of quadratic in (i) as a, b, and c.

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Question 8

Topic: 求和与三角恒等式 (Summation and Trigonometric Identities)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Prove that, for any numbers $a_1, a_2, \dots$, and $b_1, b_2, \dots$, and for $n \geqslant 1$, $\sum_{m=1}^{n} a_m(b_{m+1} - b_m) = a_{n+1}b_{n+1} - a_1b_1 - \sum_{m=1}^{n} b_{m+1}(a_{m+1} - a_m) .$

(i) By setting $b_m = \sin mx$, show that $\sum_{m=1}^{n} \cos(m + \tfrac{1}{2})x = \tfrac{1}{2} \big( \sin(n + 1)x - \sin x \big) \operatorname{cosec} \tfrac{1}{2}x .$ Note: $\sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right)$.

(ii) Show that $\sum_{m=1}^{n} m \sin mx = (p \sin(n + 1)x + q \sin nx) \operatorname{cosec}^2 \tfrac{1}{2}x ,$ where $p$ and $q$ are to be determined in terms of $n$. Note: $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$; $2 \cos A \sin B = \sin(A + B) - \sin(A - B)$.

Hint

Stem: expanding and telescoping:

$\sum_{m=1}^{n} a_m(b_{m+1}-b_m) + \sum_{m=1}^{n} b_{m+1}(a_{m+1}-a_m) = \sum_{m=1}^{n} (-a_mb_m + b_{m+1}a_{m+1}) = a_{n+1}b_{n+1} - a_1b_1$.

Hence $\sum_{m=1}^{n} a_m(b_{m+1}-b_m) = a_{n+1}b_{n+1} - a_1b_1 - \sum_{m=1}^{n} b_{m+1}(a_{m+1}-a_m)$.

(i) Let $a_m = 1$, $b_m = \sin mx$. Then $b_{m+1}-b_m = \sin(m+1)x - \sin mx$.

Using the note: $\sin(m+1)x - \sin mx = 2\cos(m+\tfrac{1}{2})x \sin\tfrac{1}{2}x$.

RHS: $a_{n+1}b_{n+1} - a_1b_1 = \sin(n+1)x - \sin x$, and $\sum b_{m+1}(a_{m+1}-a_m) = 0$ since $a_m$ is constant.

So $2\sin\tfrac{1}{2}x \sum_{m=1}^{n} \cos(m+\tfrac{1}{2})x = \sin(n+1)x - \sin x$, giving $\sum_{m=1}^{n} \cos(m+\tfrac{1}{2})x = \tfrac{1}{2}(\sin(n+1)x - \sin x)\csc\tfrac{1}{2}x$.

(ii) Let $a_m = m$, $b_m = \sin(m-1)x - \sin mx$.

$b_{m+1}-b_m = (\sin mx - \sin(m+1)x) - (\sin(m-1)x - \sin mx) = 4\sin mx \sin^2\tfrac{1}{2}x$.

Using the stem and simplifying:

$4\sin^2\tfrac{1}{2}x \sum_{m=1}^{n} m\sin mx = (n+1)\sin nx - n\sin(n+1)x$.

Thus $\sum_{m=1}^{n} m\sin mx = (p\sin nx + q\sin(n+1)x)\csc^2\tfrac{1}{2}x$

where $p = -\tfrac{1}{4}n$ and $q = \tfrac{1}{4}(n+1)$.

Model Solution

Stem: Proving the summation identity.

We expand the left side by writing out the two sums together:

$\sum_{m=1}^{n} a_m(b_{m+1} - b_m) + \sum_{m=1}^{n} b_{m+1}(a_{m+1} - a_m)$

$= \sum_{m=1}^{n} \left[a_m b_{m+1} - a_m b_m + b_{m+1} a_{m+1} - b_{m+1} a_m\right]$

$= \sum_{m=1}^{n} \left[b_{m+1} a_{m+1} - a_m b_m\right]$

This is a telescoping sum. Writing out the terms:

$(a_2 b_2 - a_1 b_1) + (a_3 b_3 - a_2 b_2) + \cdots + (a_{n+1} b_{n+1} - a_n b_n) = a_{n+1} b_{n+1} - a_1 b_1$

Therefore:

$\sum_{m=1}^{n} a_m(b_{m+1} - b_m) + \sum_{m=1}^{n} b_{m+1}(a_{m+1} - a_m) = a_{n+1}b_{n+1} - a_1 b_1$

Rearranging:

$\sum_{m=1}^{n} a_m(b_{m+1} - b_m) = a_{n+1}b_{n+1} - a_1 b_1 - \sum_{m=1}^{n} b_{m+1}(a_{m+1} - a_m) \qquad \text{(*)}$

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Part (i)

Set $a_m = 1$ for all $m$ and $b_m = \sin mx$. Then:

• $a_{m+1} - a_m = 0$ for all $m$, so the rightmost sum in $(*)$ vanishes.
• $a_{n+1}b_{n+1} - a_1 b_1 = \sin(n+1)x - \sin x$.
• $b_{m+1} - b_m = \sin(m+1)x - \sin mx$.

By the given identity $\sin A - \sin B = 2\cos\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$:

$\sin(m+1)x - \sin mx = 2\cos\!\left(\frac{(m+1)x + mx}{2}\right)\sin\!\left(\frac{(m+1)x - mx}{2}\right) = 2\cos\!\left(m+\tfrac{1}{2}\right)x \cdot \sin\tfrac{1}{2}x$

Substituting into $(*)$:

$\sum_{m=1}^{n} 1 \cdot 2\cos\!\left(m+\tfrac{1}{2}\right)x \cdot \sin\tfrac{1}{2}x = \sin(n+1)x - \sin x$

$2\sin\tfrac{1}{2}x \sum_{m=1}^{n} \cos\!\left(m+\tfrac{1}{2}\right)x = \sin(n+1)x - \sin x$

$\sum_{m=1}^{n} \cos\!\left(m+\tfrac{1}{2}\right)x = \frac{\sin(n+1)x - \sin x}{2\sin\tfrac{1}{2}x} = \tfrac{1}{2}\bigl(\sin(n+1)x - \sin x\bigr)\operatorname{cosec}\tfrac{1}{2}x$

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Part (ii)

To obtain a sum involving $m\sin mx$, we need a clever choice of sequences. Set:

$a_m = m, \qquad b_m = \sin(m-1)x - \sin mx$

Then:

• $a_{m+1} - a_m = 1$.
• $b_{m+1} = \sin mx - \sin(m+1)x$.

Computing $b_{m+1} - b_m$:

$b_{m+1} - b_m = [\sin mx - \sin(m+1)x] - [\sin(m-1)x - \sin mx]$

$= 2\sin mx - \sin(m+1)x - \sin(m-1)x$

Using the sum-to-product identity:

$\sin(m+1)x + \sin(m-1)x = 2\sin mx \cos x$

So:

$b_{m+1} - b_m = 2\sin mx - 2\sin mx\cos x = 2\sin mx(1 - \cos x)$

Since $1 - \cos x = 2\sin^2\!\tfrac{1}{2}x$:

$b_{m+1} - b_m = 4\sin mx \sin^2\!\tfrac{1}{2}x$

Left side of $(*)$:

$\sum_{m=1}^{n} a_m(b_{m+1} - b_m) = \sum_{m=1}^{n} m \cdot 4\sin mx\sin^2\!\tfrac{1}{2}x = 4\sin^2\!\tfrac{1}{2}x \sum_{m=1}^{n} m\sin mx$

Right side of $(*)$: The boundary terms are:

$a_{n+1}b_{n+1} = (n+1)[\sin nx - \sin(n+1)x]$

$a_1 b_1 = 1 \cdot [\sin 0 - \sin x] = -\sin x$

$a_{n+1}b_{n+1} - a_1 b_1 = (n+1)\sin nx - (n+1)\sin(n+1)x + \sin x$

The remaining sum is:

$\sum_{m=1}^{n} b_{m+1}(a_{m+1} - a_m) = \sum_{m=1}^{n} [\sin mx - \sin(m+1)x] \cdot 1$

This telescopes:

$(\sin x - \sin 2x) + (\sin 2x - \sin 3x) + \cdots + (\sin nx - \sin(n+1)x) = \sin x - \sin(n+1)x$

Substituting into $(*)$:

$4\sin^2\!\tfrac{1}{2}x \sum_{m=1}^{n} m\sin mx = (n+1)\sin nx - (n+1)\sin(n+1)x + \sin x - [\sin x - \sin(n+1)x]$

$= (n+1)\sin nx - (n+1)\sin(n+1)x + \sin(n+1)x$

$= (n+1)\sin nx - n\sin(n+1)x$

Therefore:

$\sum_{m=1}^{n} m\sin mx = \frac{(n+1)\sin nx - n\sin(n+1)x}{4\sin^2\!\tfrac{1}{2}x}$

Since $\operatorname{cosec}^2\!\tfrac{1}{2}x = \frac{1}{\sin^2\!\tfrac{1}{2}x}$:

$\sum_{m=1}^{n} m\sin mx = \left[-\tfrac{1}{4}n\sin(n+1)x + \tfrac{1}{4}(n+1)\sin nx\right]\operatorname{cosec}^2\!\tfrac{1}{2}x$

Comparing with the form $(p\sin(n+1)x + q\sin nx)\operatorname{cosec}^2\!\tfrac{1}{2}x$:

$p = -\frac{n}{4}, \qquad q = \frac{n+1}{4}$

Examiner Notes

Attempted as many times as Q5, success rate about halfway between Q5 and Q7. Many attempts using induction which wasted time. Stem and part (i) generally well solved, but many could not spot the method to proceed with part (ii).