STEP3 2018 -- Pure Mathematics

STEP3 2018 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2018  |  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	函数与曲线 (Functions and Curves)	Standard	求导求驻点, Vieta公式, 判别式约束, 曲线渐近线分析, 不等式证明
2	递推序列与数学归纳法 (Sequences and Proof by Induction)	Standard	数学归纳法, 乘积求导法则, 递推关系推导, 消元法
3	微分方程 (Differential Equations)	Challenging	待定系数法, 两次积分, 分情况讨论, 欧拉型方程
4	双曲线与参数方程 (Hyperbola and Parametric Equations)	Challenging	参数方程求切线, 联立方程求交点, 三角恒等式化简, 垂直切线条件
5	均值不等式 (AM-GM Inequality)	Challenging	AM-GM不等式, 辅助函数法, 求导判断单调性, 数学归纳法, 可逆推导
6	复数 (Complex Numbers)	Hard	复数共轭运算,单位圆上复数的性质 (aa*=1),共线条件的复数表示,线性组合消元
7	三角学 (Trigonometry)	Challenging	棣莫弗定理,二项式定理展开,韦达定理（根与系数关系）,小角近似不等式夹逼,极限
8	积分 (Integration)	Challenging	变量替换 y=x^{-1},部分分式分解,裂项求和（伸缩级数）,周期函数性质,分数部分函数的分段处理

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Question 1

Topic: 函数与曲线 (Functions and Curves)  |  Difficulty: Standard  |  Marks: 20

Problem

1 (i) The function $f$ is given by $f(\beta) = \beta - \frac{1}{\beta} - \frac{1}{\beta^2} \quad (\beta \neq 0) .$ Find the stationary point of the curve $y = f(\beta)$ and sketch the curve. Sketch also the curve $y = g(\beta)$, where $g(\beta) = \beta + \frac{3}{\beta} - \frac{1}{\beta^2} \quad (\beta \neq 0) .$ (ii) Let $u$ and $v$ be the roots of the equation $x^2 + \alpha x + \beta = 0 ,$ where $\beta \neq 0$. Obtain expressions in terms of $\alpha$ and $\beta$ for $u + v + \frac{1}{uv}$ and $\frac{1}{u} + \frac{1}{v} + uv$. (iii) Given that $u + v + \frac{1}{uv} = -1$, and that $u$ and $v$ are real, show that $\frac{1}{u} + \frac{1}{v} + uv \leqslant -1$. (iv) Given instead that $u + v + \frac{1}{uv} = 3$, and that $u$ and $v$ are real, find the greatest value of $\frac{1}{u} + \frac{1}{v} + uv$.

Hint

Differentiating $f(\beta)$ and setting equal to zero yields a cubic equation with a single real root, the quadratic factor being demonstrated to be non-zero either by completing the square or considering the discriminant. The stationary point is thus $(-1, -1)$ and with the asymptote $y = \beta$ the sketch results.

The same strategy for $g(\beta)$ yields a cubic equation with three real roots, two being coincident, and thus a maximum $(-2, -\frac{15}{4})$ and a point of inflection $(1, 3)$.

Employing Vieta’s formulae gives $u + v + \frac{1}{uv} = -\alpha + \frac{1}{\beta}$ and $\frac{1}{u} + \frac{1}{v} + uv = \frac{-\alpha}{\beta} + \beta$. The first condition of part (iii) enables an expression for $\alpha$ to be obtained in terms of $\beta$, and so the subject of the required inequality is $f(\beta)$. When the discriminant condition is employed to impose the reality of $u$ and $v$, the resulting cubic inequality has a quadratic factor which should be demonstrated to be positive (similarly to part (i)) and hence $\beta \le 1$ which by reference to part (i), gives the requested result. Part (iv) follows the same strategy as (iii) except that it makes use of $g(\beta)$ and the reality condition cubic inequality has a squared linear factor which allows $\beta = 1$ as well as $\beta \le \frac{1}{4}$, and so by reference to the sketch, the greatest value is 3. The reader might like to consider what effect it would have on parts (iii) and (iv) if the extra condition $u \neq v$ were to be imposed.

Model Solution

Part (i)

Differentiate $f(\beta) = \beta - \beta^{-1} - \beta^{-2}$: $f'(\beta) = 1 + \beta^{-2} + 2\beta^{-3}$

Setting $f'(\beta) = 0$ and multiplying by $\beta^3$: $\beta^3 + \beta + 2 = 0$

Testing $\beta = -1$: $(-1)^3 + (-1) + 2 = 0$, so $(\beta + 1)$ is a factor. Dividing: $\beta^3 + \beta + 2 = (\beta + 1)(\beta^2 - \beta + 2)$

The discriminant of $\beta^2 - \beta + 2$ is $1 - 8 = -7 < 0$, so $\beta^2 - \beta + 2 > 0$ for all real $\beta$. Hence $\beta = -1$ is the only real root, giving the unique stationary point.

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

The stationary point is $(-1, -1)$.

As $\beta \to \pm\infty$, $f(\beta) \sim \beta$, so $y = \beta$ is an oblique asymptote. As $\beta \to 0^{\pm}$, $f(\beta) \to -\infty$ (dominated by $-\beta^{-2}$).

For $g(\beta) = \beta + 3\beta^{-1} - \beta^{-2}$: $g'(\beta) = 1 - 3\beta^{-2} + 2\beta^{-3}$

Setting $g'(\beta) = 0$ and multiplying by $\beta^3$: $\beta^3 - 3\beta + 2 = 0$

Testing $\beta = 1$: $1 - 3 + 2 = 0$, so $(\beta - 1)$ is a factor. Dividing: $\beta^3 - 3\beta + 2 = (\beta - 1)(\beta^2 + \beta - 2) = (\beta - 1)^2(\beta + 2)$

So $\beta = 1$ (double root) and $\beta = -2$.

$g(-2) = -2 + \frac{3}{-2} - \frac{1}{4} = -\frac{8}{4} - \frac{6}{4} - \frac{1}{4} = -\frac{15}{4}$ $g(1) = 1 + 3 - 1 = 3$

To classify, write $g'(\beta) = \frac{(\beta - 1)^2(\beta + 2)}{\beta^3}$. Near $\beta = -2$: $(\beta - 1)^2 > 0$, $(\beta + 2)$ changes sign from $-$ to $+$, and $\beta^3 < 0$, so $g'$ changes from $+$ to $-$: this is a maximum. At $\beta = 1$: $g'$ does not change sign (squared factor), so this is a point of inflection.

The sketch of $g$ has asymptote $y = \beta$, a maximum at $(-2, -\tfrac{15}{4})$, and a horizontal point of inflection at $(1, 3)$.

Part (ii)

By Vieta’s formulae for $x^2 + \alpha x + \beta = 0$: $u + v = -\alpha$ and $uv = \beta$.

$u + v + \frac{1}{uv} = -\alpha + \frac{1}{\beta}$

$\frac{1}{u} + \frac{1}{v} + uv = \frac{u + v}{uv} + uv = \frac{-\alpha}{\beta} + \beta$

Part (iii)

Given $u + v + \frac{1}{uv} = -1$: $-\alpha + \frac{1}{\beta} = -1 \implies \alpha = 1 + \frac{1}{\beta}$

Substituting into the expression from (ii): $\frac{1}{u} + \frac{1}{v} + uv = \frac{-\alpha}{\beta} + \beta = -\frac{1}{\beta}\!\left(1 + \frac{1}{\beta}\right) + \beta = \beta - \frac{1}{\beta} - \frac{1}{\beta^2} = f(\beta)$

For $u$ and $v$ to be real, the discriminant must satisfy $\alpha^2 - 4\beta \geqslant 0$: $\left(1 + \frac{1}{\beta}\right)^2 - 4\beta \geqslant 0$

Expanding and multiplying by $\beta^2 > 0$: $\beta^2 + 2\beta + 1 - 4\beta^3 \geqslant 0$ $-4\beta^3 + \beta^2 + 2\beta + 1 \geqslant 0$

Testing $\beta = 1$: $-4 + 1 + 2 + 1 = 0$, so $(\beta - 1)$ is a factor. Dividing: $-4\beta^3 + \beta^2 + 2\beta + 1 = -(\beta - 1)(4\beta^2 + 3\beta + 1)$

The discriminant of $4\beta^2 + 3\beta + 1$ is $9 - 16 = -7 < 0$, so $4\beta^2 + 3\beta + 1 > 0$ for all real $\beta$.

Therefore $-(\beta - 1)(4\beta^2 + 3\beta + 1) \geqslant 0$ gives $\beta \leqslant 1$ (with $\beta \neq 0$).

Now examine $f(\beta) + 1$: $f(\beta) + 1 = \beta - \frac{1}{\beta} - \frac{1}{\beta^2} + 1 = \frac{\beta^3 + \beta^2 - \beta - 1}{\beta^2}$

Factor the numerator: $\beta^3 + \beta^2 - \beta - 1 = \beta^2(\beta + 1) - (\beta + 1) = (\beta + 1)(\beta^2 - 1) = (\beta + 1)^2(\beta - 1)$

So $f(\beta) + 1 = \dfrac{(\beta + 1)^2(\beta - 1)}{\beta^2}$.

For $\beta \leqslant 1$ and $\beta \neq 0$: $(\beta + 1)^2 \geqslant 0$, $(\beta - 1) \leqslant 0$, and $\beta^2 > 0$. Hence $f(\beta) + 1 \leqslant 0$, i.e. $\frac{1}{u} + \frac{1}{v} + uv \leqslant -1 \qquad \blacksquare$

Part (iv)

Given $u + v + \frac{1}{uv} = 3$: $-\alpha + \frac{1}{\beta} = 3 \implies \alpha = \frac{1}{\beta} - 3$

Substituting into the expression from (ii): $\frac{1}{u} + \frac{1}{v} + uv = -\frac{1}{\beta}\!\left(\frac{1}{\beta} - 3\right) + \beta = \beta + \frac{3}{\beta} - \frac{1}{\beta^2} = g(\beta)$

For $u$ and $v$ to be real, $\alpha^2 - 4\beta \geqslant 0$: $\left(\frac{1}{\beta} - 3\right)^2 - 4\beta \geqslant 0$

Expanding and multiplying by $\beta^2 > 0$: $1 - 6\beta + 9\beta^2 - 4\beta^3 \geqslant 0$

Testing $\beta = 1$: $1 - 6 + 9 - 4 = 0$, so $(\beta - 1)$ is a factor. Dividing: $-4\beta^3 + 9\beta^2 - 6\beta + 1 = -(\beta - 1)(4\beta^2 - 5\beta + 1) = -(\beta - 1)^2(4\beta - 1)$

So $-(\beta - 1)^2(4\beta - 1) \geqslant 0$. Since $(\beta - 1)^2 \geqslant 0$, this requires $4\beta - 1 \leqslant 0$, i.e. $\beta \leqslant \tfrac{1}{4}$. The case $\beta = 1$ also satisfies the inequality (with equality).

So the allowed values are $\beta \leqslant \tfrac{1}{4}$ (with $\beta \neq 0$) or $\beta = 1$.

From the sketch of $g$ in part (i): on $(-\infty, 0)$, $g$ has a maximum of $-\tfrac{15}{4}$ at $\beta = -2$. On $(0, \tfrac{1}{4}]$, $g$ is increasing (since $g' > 0$ for $0 < \beta < 1$) with $g(\tfrac{1}{4}) = \tfrac{1}{4} + 12 - 16 = -\tfrac{15}{4}$. At $\beta = 1$: $g(1) = 3$.

Therefore the greatest value is $3$, achieved when $\beta = 1$ (and $\alpha = -2$, giving $u = v = 1$).

Examiner Notes

As usual, this was the most popular question to be attempted with more than 93% of candidates doing so. However, scoring for it was only moderately good with a mean below 9/20. Most successfully differentiated and obtained a value for β from the cubic but ignored considering whether this was the only stationary point. Sketchs frequently did not display the asymptote; some that did showed the negative branch of the curve touching rather than intersecting the asymptote at the maximum. Many did not appreciate that to sketch the second curve in part (i) it was not sufficient to just offer a drawing without the working; the horizontal point of inflection and asymptote were frequent casualties. Part (ii) was straightforward for most. Many recognised that part (iii) made use of the first function f(β), provided that they used the condition to substitute for α. However, their justification suffered from ignoring the reality condition and using specious arguments, as a consequence. Part (iv) followed a similar trend to part (iii), except using g(β), and only differing in that of those that did apply the reality condition, quite a few overlooked β = 1 as a solution of the cubic inequality, and so their final answer was wrong.

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Question 2

Topic: 递推序列与数学归纳法 (Sequences and Proof by Induction)  |  Difficulty: Standard  |  Marks: 20

Problem

2 The sequence of functions $y_0, y_1, y_2, \dots$ is defined by $y_0 = 1$ and, for $n \geqslant 1$, $y_n = (-1)^n \frac{1}{z} \frac{\mathrm{d}^n z}{\mathrm{d}x^n} ,$ where $z = \mathrm{e}^{-x^2}$. (i) Show that $\frac{\mathrm{d}y_n}{\mathrm{d}x} = 2xy_n - y_{n+1}$ for $n \geqslant 1$. (ii) Prove by induction that, for $n \geqslant 1$, $y_{n+1} = 2xy_n - 2ny_{n-1} .$ Deduce that, for $n \geqslant 1$, $y_{n+1}^2 - y_n y_{n+2} = 2n(y_n^2 - y_{n-1} y_{n+1}) + 2y_n^2 .$ (iii) Hence show that $y_n^2 - y_{n-1} y_{n+1} > 0$ for $n \geqslant 1$.

Hint

(i) is obtained by differentiating the defined function $y_n$ using the product rule and a little tidying up of $z$ and its differential. The inductive step in the proof of part (ii) can be established by differentiating the assumed case and then removing the three differentials using the result of part (i); the base case is established by obtaining $y_1 = 2x$ and $y_2 = -2 + 4x^2$ from the original definition. The deduction in (ii) is most simply obtained by eliminating $x$ between the result for $y_{n+1}$ just obtained and the similar one for $y_{n+2}$. Part (iii) is obtained by using the deduction of part (ii) to establish the desired induction, the base case being obtained using the same results used for the base case in part (ii)‘s induction.

Model Solution

Preliminary. We compute the first few $y_n$ from the definition. With $z = e^{-x^2}$:

$z' = -2xe^{-x^2} = -2xz$

$y_1 = (-1)^1 \frac{z'}{z} = -\frac{-2xz}{z} = 2x$

$z'' = -2z - 2xz' = -2z + 4x^2z = (4x^2 - 2)z$

$y_2 = (-1)^2 \frac{z''}{z} = 4x^2 - 2$

Part (i)

For $n \geqslant 1$, write $z^{(n)} = \dfrac{\mathrm{d}^n z}{\mathrm{d}x^n}$. By definition, $y_n = (-1)^n \dfrac{z^{(n)}}{z}$.

Differentiate using the product rule:

$\frac{\mathrm{d}y_n}{\mathrm{d}x} = (-1)^n \frac{\mathrm{d}}{\mathrm{d}x}\!\left(\frac{z^{(n)}}{z}\right) = (-1)^n \frac{z^{(n+1)} \cdot z - z^{(n)} \cdot z'}{z^2}$

Since $z' = -2xz$:

$= (-1)^n \frac{z^{(n+1)} z + 2xz \cdot z^{(n)}}{z^2} = (-1)^n \frac{z^{(n+1)}}{z} + 2x \cdot (-1)^n \frac{z^{(n)}}{z}$

Now $(-1)^n \frac{z^{(n)}}{z} = y_n$ and $(-1)^n \frac{z^{(n+1)}}{z} = -(-1)^{n+1} \frac{z^{(n+1)}}{z} = -y_{n+1}$.

Therefore:

$\frac{\mathrm{d}y_n}{\mathrm{d}x} = 2xy_n - y_{n+1} \qquad \text{(*)}$

Part (ii)

Claim: $y_{n+1} = 2xy_n - 2ny_{n-1}$ for $n \geqslant 1$.

Base case ($n = 1$): We need $y_2 = 2xy_1 - 2y_0$. We have $y_0 = 1$, $y_1 = 2x$, $y_2 = 4x^2 - 2$. Check: $2x(2x) - 2(1) = 4x^2 - 2 = y_2$. $\checkmark$

Inductive step. Suppose $y_{k+1} = 2xy_k - 2ky_{k-1}$ for some $k \geqslant 1$. We prove $y_{k+2} = 2xy_{k+1} - 2(k+1)y_k$.

From (*): $y_{k+2} = 2xy_{k+1} - \dfrac{\mathrm{d}y_{k+1}}{\mathrm{d}x}$.

Differentiate the inductive hypothesis:

$\frac{\mathrm{d}y_{k+1}}{\mathrm{d}x} = 2y_k + 2x\frac{\mathrm{d}y_k}{\mathrm{d}x} - 2k\frac{\mathrm{d}y_{k-1}}{\mathrm{d}x}$

Apply () to $\dfrac{\mathrm{d}y_k}{\mathrm{d}x}$ and $\dfrac{\mathrm{d}y_{k-1}}{\mathrm{d}x}$ (valid since $k \geqslant 1$ and $k-1 \geqslant 0$; for $k = 1$, $y_0 = 1$ gives $\frac{\mathrm{d}y_0}{\mathrm{d}x} = 0 = 2x \cdot 1 - 2x = 2xy_0 - y_1$, so () holds for $n = 0$ as well):

$\frac{\mathrm{d}y_{k+1}}{\mathrm{d}x} = 2y_k + 2x(2xy_k - y_{k+1}) - 2k(2xy_{k-1} - y_k)$

$= 2y_k + 4x^2y_k - 2xy_{k+1} - 4kxy_{k-1} + 2ky_k$

From the inductive hypothesis, $2ky_{k-1} = 2xy_k - y_{k+1}$, so $4kxy_{k-1} = 2x(2xy_k - y_{k+1}) = 4x^2y_k - 2xy_{k+1}$:

$= 2y_k + 4x^2y_k - 2xy_{k+1} - 4x^2y_k + 2xy_{k+1} + 2ky_k = (2 + 2k)y_k = 2(k+1)y_k$

Therefore $y_{k+2} = 2xy_{k+1} - 2(k+1)y_k$, completing the induction.

Deduction. We eliminate $x$ from $y_{n+1} = 2xy_n - 2ny_{n-1}$ and $y_{n+2} = 2xy_{n+1} - 2(n+1)y_n$.

From the first equation: $2x = \dfrac{y_{n+1} + 2ny_{n-1}}{y_n}$ (we work formally; the final identity holds regardless).

Substitute into the second:

$y_{n+2} = \frac{y_{n+1} + 2ny_{n-1}}{y_n} \cdot y_{n+1} - 2(n+1)y_n$

$y_n y_{n+2} = y_{n+1}^2 + 2ny_{n-1}y_{n+1} - 2(n+1)y_n^2$

Rearranging:

$y_{n+1}^2 - y_n y_{n+2} = 2(n+1)y_n^2 - 2ny_{n-1}y_{n+1} = 2n(y_n^2 - y_{n-1}y_{n+1}) + 2y_n^2$

$y_{n+1}^2 - y_n y_{n+2} = 2n(y_n^2 - y_{n-1}y_{n+1}) + 2y_n^2 \qquad \text{(**)}$

Part (iii)

Let $S_n = y_n^2 - y_{n-1}y_{n+1}$. We prove $S_n > 0$ for all $n \geqslant 1$ by induction.

From (**): $S_{n+1} = 2nS_n + 2y_n^2$.

Base case ($n = 1$):

$S_1 = y_1^2 - y_0 y_2 = (2x)^2 - 1 \cdot (4x^2 - 2) = 4x^2 - 4x^2 + 2 = 2 > 0. \; \checkmark$

Inductive step. Suppose $S_k > 0$ for some $k \geqslant 1$. Then

$S_{k+1} = 2kS_k + 2y_k^2.$

Since $k \geqslant 1$, $S_k > 0$, and $y_k^2 \geqslant 0$:

$S_{k+1} \geqslant 2 \cdot 1 \cdot S_k > 0.$

By induction, $S_n = y_n^2 - y_{n-1}y_{n+1} > 0$ for all $n \geqslant 1$. $\qquad \blacksquare$

Examiner Notes

With about 89% attempting this, it was the second most popular question, and with a mean score of nearly 11/20, the second most successful. Part (i) was generally well-handled, and although most scored some marks on the proof by induction in part (ii), candidates often struggled to complete it, many of them because they attempted to use the original definition of the functions, which rarely led to success, rather than employ the result of part (i). Some candidates noticed that the proof by induction in part (ii) was equivalent to proving dy_n/dx = 2ny_{n-1} by induction. This gave them a simpler base case but did not significantly simplify the inductive step. For the deduction in part (ii), it was very common for the first result of part (ii) to be squared, leading to pages of algebra, although they were often then successful. Part (iii) was surprisingly poorly attempted as few candidates realised that a proof by induction (or equivalent) was required.

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Question 3

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

Problem

3 Show that the second-order differential equation

$x^2y'' + (1 - 2p)x y' + (p^2 - q^2) y = f(x) ,$

where $p$ and $q$ are constants, can be written in the form

$x^a (x^b (x^c y)')' = f(x) , \qquad \text{(*)}$

where $a$, $b$ and $c$ are constants.

(i) Use ($*$) to derive the general solution of the equation

$x^2y'' + (1 - 2p)xy' + (p^2 - q^2)y = 0$

in the different cases that arise according to the values of $p$ and $q$.

(ii) Use ($*$) to derive the general solution of the equation

$x^2y'' + (1 - 2p)xy' + p^2y = x^n$

in the different cases that arise according to the values of $p$ and $n$.

Hint

Differentiating and equating coefficients generates three simultaneous equations, two of which are in $b$ and $c$ only, and can be solved by substituting for one of these variables, then giving $a$ from the third; $a = 1 + p \mp q$, $b = 1 \pm 2q$, $c = \mp q - p$. Part (i) makes use of the form obtained in the stem which can be then integrated twice, with a minor algebraic rearrangement after each integration, with different cases arising as $b = 1$ or not; the solutions are

$y = Ax^{p+q} + Bx^{p-q}$ if $q \neq 0$, and $y = Ax^p \ln x + Bx^p$ if $q = 0$. Part (ii) proceeds similarly, except that $a$, $b$, and $c$ are simplified as $q = 0$. However, two cases arise again and so

$y = \frac{x^n}{(n-p)^2} + Ax^p \ln x + Bx^p \text{ if } n \neq p, \text{ and } y = x^n \frac{(\ln x)^2}{2} + Ax^n \ln x + Bx^n \text{ if } n = p.$

Model Solution

Showing the equation can be written in the required form

Let $u = x^c y$. Then $u' = cx^{c-1}y + x^c y'$, and

$x^b u' = cx^{b+c-1}y + x^{b+c}y'.$

Differentiating again:

$(x^b u')' = c(b+c-1)x^{b+c-2}y + (b+2c)x^{b+c-1}y' + x^{b+c}y''.$

Multiplying by $x^a$:

$x^a(x^b(x^c y)')' = x^{a+b+c}y'' + (a+2c)x^{a+b+c-1}y' + c(b+c-1)x^{a+b+c-2}y.$

Comparing with $x^2y'' + (1-2p)xy' + (p^2 - q^2)y$:

$a + b + c = 2, \qquad a + 2c = 1 - 2p, \qquad c(b+c-1) = p^2 - q^2.$

From the first two equations: $b = 1 + 2p - 2c$. Substituting into the third:

$c(1 + 2p - 2c + c - 1) = c(2p - c) = p^2 - q^2$

$\Longrightarrow \quad c^2 - 2pc + p^2 - q^2 = 0 \quad \Longrightarrow \quad (c - p)^2 = q^2$

$\Longrightarrow \quad c = p \mp q.$

Taking $c = p - q$ (the alternative $c = p + q$ yields the same solutions to the ODE):

$a = 1 + q, \qquad b = 1 + 2q, \qquad c = p - q. \qquad \square$

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Part (i)

With these values of $a, b, c$, the equation $x^a(x^b(x^c y)')' = 0$ becomes

$x^{1+q}\!\left(x^{1+2q}\!\left(x^{p-q}y\right)'\right)' = 0.$

Since $x^{1+q} \neq 0$ for $x > 0$, the first integration gives

$x^{1+2q}\!\left(x^{p-q}y\right)' = A_1 \quad \Longrightarrow \quad \left(x^{p-q}y\right)' = A_1 x^{-1-2q}.$

Case $q \neq 0$: Integrating again:

$x^{p-q}y = \frac{A_1}{-2q}x^{-2q} + B_1 \quad \Longrightarrow \quad y = Ax^{p+q} + Bx^{p-q}.$

Case $q = 0$: Here $a = 1+p$, $b = 1$, $c = -p$, so after the first integration $(x^{-p}y)' = A_1 x^{-1}$. Integrating:

$x^{-p}y = A_1 \ln x + B_1 \quad \Longrightarrow \quad y = Ax^p \ln x + Bx^p.$

$\boxed{y = Ax^{p+q} + Bx^{p-q} \;\text{ if } q \neq 0, \qquad y = Ax^p \ln x + Bx^p \;\text{ if } q = 0.}$

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Part (ii)

With $q = 0$: $a = 1+p$, $b = 2p-1$, $c = -p$. The equation becomes

$x^{1+p}\!\left(x^{2p-1}\!\left(x^{-p}y\right)'\right)' = x^n.$

Case $n \neq p$: To find the particular integral, try $y_p = Kx^n$ in the original equation $x^2y'' + (1-2p)xy' + p^2y = x^n$:

$K\big[n(n-1) + n(1-2p) + p^2\big]x^n = x^n \quad \Longrightarrow \quad K(n - p)^2 = 1 \quad \Longrightarrow \quad K = \frac{1}{(n-p)^2}.$

(To verify: $n(n-1) + n - 2np + p^2 = n^2 - 2np + p^2 = (n-p)^2$.)

The general solution is the particular integral plus the homogeneous solution from part (i):

$y = \frac{x^n}{(n-p)^2} + Ax^p \ln x + Bx^p \qquad (n \neq p).$

Case $n = p$: The equation is $x^2y'' + (1-2p)xy' + p^2y = x^p$. Using the factored form and dividing by $x^{1+p}$:

$\left(x^{2p-1}\!\left(x^{-p}y\right)'\right)' = x^{-1}.$

First integration: $x^{2p-1}(x^{-p}y)' = \ln x + A$.

For the particular integral, try $y_p = Cx^p(\ln x)^2$. Then:

$y_p' = Cpx^{p-1}(\ln x)^2 + 2Cx^{p-1}\ln x$

$y_p'' = Cp(p-1)x^{p-2}(\ln x)^2 + 2C(2p-1)x^{p-2}\ln x + 2Cx^{p-2}$

Substituting into $x^2y'' + (1-2p)xy' + p^2y$:

• Coefficient of $x^p(\ln x)^2$: $C[p(p-1) + p(1-2p) + p^2] = C[p^2 - p + p - 2p^2 + p^2] = 0$.
• Coefficient of $x^p \ln x$: $C[2(2p-1) + 2(1-2p)] = C[4p - 2 + 2 - 4p] = 0$.
• Coefficient of $x^p$: $2C = 1$, giving $C = \tfrac{1}{2}$.

The general solution is

$y = \frac{x^p(\ln x)^2}{2} + Ax^p \ln x + Bx^p \qquad (n = p).$

Examiner Notes

Only 65% attempted this, and it was the second weakest of the Pure questions with a mean score of about 7/20. It was imperative for candidates to demonstrate high levels of algebraic accuracy to score highly. Most successfully differentiated the initial expression, and equated coefficients but then failed to solve explicitly for a, b and c in terms of p and q (or demonstrate that such a, b and c existed). Candidates without these explicit expressions then often failed to spot one of the main strands of the question, integration of x^n in the two cases n = -1, and n ≠ -1; some considered superfluous other cases. Some candidates fell at the final hurdle having used correct methods but then did not express their solutions in terms of p and q. Also, some were thrown by the two possible sets of solutions for a, b and c, successful candidates realising that these gave the same solutions to the differential equations.

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Question 4

Topic: 双曲线与参数方程 (Hyperbola and Parametric Equations)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The point $P(a \sec \theta, b \tan \theta)$ lies on the hyperbola

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ,$

where $a > b > 0$. Show that the equation of the tangent to the hyperbola at $P$ can be written as

$bx - ay \sin \theta = ab \cos \theta .$

(i) This tangent meets the lines $\frac{x}{a} = \frac{y}{b}$ and $\frac{x}{a} = -\frac{y}{b}$ at $S$ and $T$, respectively.

How is the mid-point of $ST$ related to $P$?

(ii) The point $Q(a \sec \phi, b \tan \phi)$ also lies on the hyperbola and the tangents to the hyperbola at $P$ and $Q$ are perpendicular. These two tangents intersect at $(x, y)$.

Obtain expressions for $x^2$ and $y^2$ in terms of $a$, $\theta$ and $\phi$.

Hence, or otherwise, show that $x^2 + y^2 = a^2 - b^2$.

Hint

The equation of the tangent to the hyperbola at P is found in the usual way, having obtained the gradient through differentiation. In part (i) the points S and T can be found by solving simultaneously the equations for each of the given lines with that for the tangent and their midpoint is found to be P, using a Pythagorean trigonometry result to simplify the common denominators. Solving simultaneous equations using the equations of the two tangents gives $x^2 = \left[ a \frac{(\sin \varphi \cos \theta - \sin \theta \cos \varphi)}{(\sin \varphi - \sin \theta)} \right]^2$ and $y^2 = -a^2 \sin \theta \sin \varphi \left[ \frac{(\cos \theta - \cos \varphi)}{(\sin \varphi - \sin \theta)} \right]^2$, having eliminated $b$ from the latter expression using the condition that the tangents are perpendicular; that same condition and the knowledge that $a > b$ not only justifies that such tangents exist, but also that the denominator of the two results found is non-zero. Adding the two results, expanding brackets in the numerator, and then removing any cos squared terms in favour of sin squared terms leaves an expression which cancels with the denominator, and the resulting simple expression can then be seen to yield the desired result.

Whilst not within the remit of the question posed, an elegant method of obtaining the result of part (ii) is to consider the solution of the hyperbola equation with the equation of a general straight line through a point on the hyperbola and another point. Solving simultaneously for (say) the $x$-coordinate of the meeting of the curve and line and imposing that the solutions to the quadratic must be coincident for a tangent yields a quadratic equation for the gradients, and as the tangents are perpendicular, the product of those gradients is -1 giving the desired result without recourse to trigonometry.

Model Solution

Stem: Finding the tangent equation

Implicit differentiation of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ gives $\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$.

At $P(a\sec\theta, b\tan\theta)$:

$m = \frac{b^2 \cdot a\sec\theta}{a^2 \cdot b\tan\theta} = \frac{b\sec\theta}{a\tan\theta} = \frac{b}{a\sin\theta}.$

Tangent equation using point-slope form:

$y - b\tan\theta = \frac{b}{a\sin\theta}(x - a\sec\theta)$

$a\sin\theta \cdot y - ab\sin\theta\tan\theta = bx - ab\sec\theta$

$a\sin\theta \cdot y - ab \cdot \frac{\sin^2\theta}{\cos\theta} = bx - \frac{ab}{\cos\theta}$

Multiply through by $\cos\theta$:

$a\sin\theta\cos\theta \cdot y - ab\sin^2\theta = bx\cos\theta - ab$

$bx\cos\theta - a\sin\theta\cos\theta \cdot y = ab(1 - \sin^2\theta) = ab\cos^2\theta$

Dividing by $\cos\theta \neq 0$:

$bx - ay\sin\theta = ab\cos\theta. \qquad \square$

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Part (i)

The tangent at $P$ is $bx - ay\sin\theta = ab\cos\theta$.

Finding $S$: On the line $y = \frac{bx}{a}$. Substituting:

$bx - bx\sin\theta = ab\cos\theta \quad \Longrightarrow \quad bx(1 - \sin\theta) = ab\cos\theta$

$\Longrightarrow \quad x_S = \frac{a\cos\theta}{1 - \sin\theta}, \qquad y_S = \frac{b\cos\theta}{1 - \sin\theta}.$

Finding $T$: On the line $y = -\frac{bx}{a}$. Substituting:

$bx + bx\sin\theta = ab\cos\theta \quad \Longrightarrow \quad bx(1 + \sin\theta) = ab\cos\theta$

$\Longrightarrow \quad x_T = \frac{a\cos\theta}{1 + \sin\theta}, \qquad y_T = \frac{-b\cos\theta}{1 + \sin\theta}.$

Mid-point of $ST$:

$x_M = \frac{1}{2}\left(\frac{a\cos\theta}{1 - \sin\theta} + \frac{a\cos\theta}{1 + \sin\theta}\right) = \frac{a\cos\theta}{2} \cdot \frac{(1+\sin\theta) + (1-\sin\theta)}{1 - \sin^2\theta} = \frac{a\cos\theta}{\cos^2\theta} = a\sec\theta$

$y_M = \frac{1}{2}\left(\frac{b\cos\theta}{1 - \sin\theta} - \frac{b\cos\theta}{1 + \sin\theta}\right) = \frac{b\cos\theta}{2} \cdot \frac{(1+\sin\theta) - (1-\sin\theta)}{\cos^2\theta} = \frac{b\sin\theta}{\cos\theta} = b\tan\theta$

The mid-point of $ST$ is $P$ itself.

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Part (ii)

Tangent at $P$: $\;bx - ay\sin\theta = ab\cos\theta$. Tangent at $Q$: $\;bx - ay\sin\phi = ab\cos\phi$.

Finding the intersection $(x, y)$: Subtracting:

$ay(\sin\phi - \sin\theta) = ab(\cos\theta - \cos\phi)$

$\Longrightarrow \quad y = \frac{b(\cos\theta - \cos\phi)}{\sin\phi - \sin\theta}$

(valid when $\sin\phi \neq \sin\theta$; justified below).

Substituting back into the tangent at $P$:

$bx = ab\cos\theta + \frac{ab\sin\theta(\cos\theta - \cos\phi)}{\sin\phi - \sin\theta} = ab\left(\frac{\cos\theta(\sin\phi - \sin\theta) + \sin\theta(\cos\theta - \cos\phi)}{\sin\phi - \sin\theta}\right)$

$= ab \cdot \frac{\sin\phi\cos\theta - \sin\theta\cos\phi}{\sin\phi - \sin\theta} = \frac{ab\sin(\phi - \theta)}{\sin\phi - \sin\theta}$

$\Longrightarrow \quad x = \frac{a\sin(\phi - \theta)}{\sin\phi - \sin\theta}.$

Perpendicularity condition: Product of tangent slopes is $-1$:

$\frac{b}{a\sin\theta} \cdot \frac{b}{a\sin\phi} = -1 \quad \Longrightarrow \quad \sin\theta\sin\phi = -\frac{b^2}{a^2}.$

Since $a > b > 0$, we have $0 < \frac{b^2}{a^2} < 1$, so $\sin\theta$ and $\sin\phi$ have opposite signs, guaranteeing $\sin\phi \neq \sin\theta$.

Expressions for $x^2$ and $y^2$:

$x^2 = \frac{a^2\sin^2(\phi - \theta)}{(\sin\phi - \sin\theta)^2}, \qquad y^2 = \frac{b^2(\cos\theta - \cos\phi)^2}{(\sin\phi - \sin\theta)^2}.$

Computing $x^2 + y^2$: Using $b^2 = -a^2\sin\theta\sin\phi$:

$x^2 + y^2 = \frac{a^2\sin^2(\phi - \theta) - a^2\sin\theta\sin\phi \cdot (\cos\theta - \cos\phi)^2}{(\sin\phi - \sin\theta)^2}.$

Expanding $\sin^2(\phi - \theta) = (\sin\phi\cos\theta - \cos\phi\sin\theta)^2 = \sin^2\phi\cos^2\theta - 2\sin\theta\sin\phi\cos\theta\cos\phi + \cos^2\phi\sin^2\theta$.

The numerator (omitting the overall $a^2$ factor) is:

$\sin^2\phi\cos^2\theta - 2\sin\theta\sin\phi\cos\theta\cos\phi + \cos^2\phi\sin^2\theta - \sin\theta\sin\phi(\cos^2\theta - 2\cos\theta\cos\phi + \cos^2\phi)$

$= \cos^2\theta(\sin^2\phi - \sin\theta\sin\phi) + \cos^2\phi(\sin^2\theta - \sin\theta\sin\phi)$

$= \sin\phi(\sin\phi - \sin\theta)\cos^2\theta + \sin\theta(\sin\theta - \sin\phi)\cos^2\phi$

$= (\sin\phi - \sin\theta)\big[\sin\phi\cos^2\theta - \sin\theta\cos^2\phi\big].$

Now expand $\sin\phi\cos^2\theta - \sin\theta\cos^2\phi$:

$= \sin\phi(1 - \sin^2\theta) - \sin\theta(1 - \sin^2\phi) = (\sin\phi - \sin\theta) + \sin\theta\sin\phi(\sin\phi - \sin\theta)$

$= (\sin\phi - \sin\theta)(1 + \sin\theta\sin\phi).$

Therefore the numerator is $a^2(\sin\phi - \sin\theta)^2(1 + \sin\theta\sin\phi)$, giving

$x^2 + y^2 = a^2(1 + \sin\theta\sin\phi) = a^2\!\left(1 - \frac{b^2}{a^2}\right) = a^2 - b^2. \qquad \square$

Examiner Notes

The third most popular question being attempted by just short of three quarters of the candidature, it was however the most successfully attempted with a mean score of not quite 12/20. The stem was usually correctly attempted either using parametric or implicit differentiation. Simultaneous equations were sensibly attempted for part (i), but sometimes they confused the two pairs of equations and as a result got the wrong answer. Some solutions elegantly achieved the correct result having found just one of the coordinates and arguing that as it lay on the tangent, it had to be the point P. Part (ii) was quite often abandoned partway through, giving up after obtaining x^2 and y^2 in the face of the algebra, although some forgot to answer the question at this point even though they had employed simultaneous equations to obtain x and y. Few managed to conclude the question, and it was very rare indeed that the non-zero nature of the denominator (sin φ - sin θ) was justified.

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Question 5

Topic: 均值不等式 (AM-GM Inequality)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The real numbers $a_1, a_2, a_3, \dots$ are all positive. For each positive integer $n$, $A_n$ and $G_n$ are defined by $A_n = \frac{a_1 + a_2 + \dots + a_n}{n} \quad \text{and} \quad G_n = (a_1 a_2 \dots a_n)^{1/n} .$

(i) Show that, for any given positive integer $k$, $(k + 1)(A_{k+1} - G_{k+1}) \geqslant k(A_k - G_k)$ if and only if $\lambda_k^{k+1} - (k + 1)\lambda_k + k \geqslant 0 ,$ where $\lambda_k = \left( \frac{a_{k+1}}{G_k} \right)^{\frac{1}{k+1}} .$

(ii) Let $f(x) = x^{k+1} - (k + 1)x + k ,$ where $x > 0$ and $k$ is a positive integer. Show that $f(x) \geqslant 0$ and that $f(x) = 0$ if and only if $x = 1$.

(iii) Deduce that:

(a) $A_n \geqslant G_n$ for all $n$;

(b) if $A_n = G_n$ for some $n$, then $a_1 = a_2 = \dots = a_n$.

Hint

As with many inequalities, rather than proving that one expression is greater than or equal to another, it is easier to subtract and produce a single expression that one requires to show is greater than or equal to zero. Substituting for the $A$s, simplifying and dividing by $G_k$ ($G_k > 0$) gets most of the way to obtaining the first result, having demonstrated that $\frac{G_{k+1}}{G_k} = \lambda_k$ en route, and observing the reversibility of the argument throughout. Part (ii) can be simply shown using differentiation to find and justify that there is a single stationary point for $f(x)$, that it is a minimum (zero) and that it occurs for $f(x)$. Part (iii) (a) can be deduced using part (i), observing that the condition for (i) is met by use of part (ii). There are several different but equivalent arguments that can be used for part (iii) (b). Assuming $A_k = G_k$ for some $k$, then by part (a), and (i), it can be shown that $A_{k-1} = G_{k-1}$, but also that $\lambda_{k-1} = 1$ and so $a_k = G_{k-1}$. Thus if $A_n = G_n$, the argument just employed obtains $A_k = G_k$ and $a_k = G_{k-1}$ for all $k$, for $k = 1$ to $n$. The final step of the argument follows simply.

Model Solution

Part (i)

We express $A_{k+1}$ and $G_{k+1}$ in terms of $A_k$, $G_k$ and $a_{k+1}$.

$A_{k+1} = \frac{a_1 + a_2 + \dots + a_k + a_{k+1}}{k+1} = \frac{kA_k + a_{k+1}}{k+1}$

$G_{k+1} = (G_k^k \cdot a_{k+1})^{1/(k+1)} = G_k \left(\frac{a_{k+1}}{G_k}\right)^{1/(k+1)} = G_k \lambda_k$

Substituting into the inequality $(k+1)(A_{k+1} - G_{k+1}) \geqslant k(A_k - G_k)$:

$kA_k + a_{k+1} - (k+1)G_k\lambda_k \geqslant kA_k - kG_k$

$a_{k+1} - (k+1)G_k\lambda_k + kG_k \geqslant 0$

Since $a_{k+1} = G_k \lambda_k^{k+1}$ (from the definition of $\lambda_k$):

$G_k\lambda_k^{k+1} - (k+1)G_k\lambda_k + kG_k \geqslant 0$

$G_k\!\left(\lambda_k^{k+1} - (k+1)\lambda_k + k\right) \geqslant 0$

Since all $a_i > 0$, we have $G_k > 0$, so this is equivalent to

$\lambda_k^{k+1} - (k+1)\lambda_k + k \geqslant 0.$

Every step is reversible (division by $G_k > 0$ and substitution $a_{k+1} = G_k\lambda_k^{k+1}$ work in both directions), so the two inequalities are equivalent.

Part (ii)

Let $f(x) = x^{k+1} - (k+1)x + k$ for $x > 0$, where $k$ is a positive integer.

$f'(x) = (k+1)x^k - (k+1) = (k+1)(x^k - 1)$

Setting $f'(x) = 0$: since $k+1 \neq 0$, we need $x^k = 1$, giving $x = 1$ (the only real positive root).

• For $0 < x < 1$: $x^k < 1$, so $f'(x) < 0$ (decreasing).
• For $x > 1$: $x^k > 1$, so $f'(x) > 0$ (increasing).

Hence $x = 1$ is the unique stationary point on $(0, \infty)$ and is a global minimum.

$f(1) = 1 - (k+1) + k = 0$

Therefore $f(x) \geqslant 0$ for all $x > 0$, and $f(x) = 0$ if and only if $x = 1$.

Part (iii)(a)

We prove $A_n \geqslant G_n$ for all $n$ by induction.

Base case ($n = 1$): $A_1 = a_1 = G_1$, so $A_1 \geqslant G_1$ holds.

Inductive step: Suppose $A_k \geqslant G_k$ for some $k \geqslant 1$. By part (ii), $f(\lambda_k) \geqslant 0$. By part (i):

$(k+1)(A_{k+1} - G_{k+1}) \geqslant k(A_k - G_k) \geqslant 0$

where the second inequality uses the inductive hypothesis. Since $k+1 > 0$, we obtain $A_{k+1} \geqslant G_{k+1}$.

By induction, $A_n \geqslant G_n$ for all positive integers $n$.

Part (iii)(b)

Suppose $A_n = G_n$ for some $n \geqslant 1$. We prove $a_1 = a_2 = \dots = a_n$.

From parts (i) and (ii), for each $1 \leqslant k \leqslant n-1$:

$(k+1)(A_{k+1} - G_{k+1}) \geqslant k(A_k - G_k)$

with equality if and only if $\lambda_k = 1$.

Since $A_n = G_n$, the left side at $k = n-1$ is zero:

$0 = (n-1)(A_{n-1} - G_{n-1})$

so $A_{n-1} = G_{n-1}$ and $\lambda_{n-1} = 1$.

$\lambda_{n-1} = 1$ gives $\left(\frac{a_n}{G_{n-1}}\right)^{1/n} = 1$, hence $a_n = G_{n-1}$.

Since $A_{n-1} = G_{n-1}$, we have $a_n = A_{n-1}$.

Repeating: $A_{n-1} = G_{n-1}$ at $k = n-2$ gives $A_{n-2} = G_{n-2}$ and $a_{n-1} = G_{n-2} = A_{n-2}$, and so on.

Continuing down to $k = 1$: at each step, $A_k = G_k$ and $a_{k+1} = G_k$.

In particular, for $k = 1$: $A_1 = G_1 = a_1$ and $a_2 = G_1 = a_1$.

Now $A_2 = \frac{a_1 + a_2}{2} = a_1$ and $a_3 = G_2 = A_2 = a_1$.

By induction on the remaining indices, $a_k = a_1$ for all $k = 1, 2, \dots, n$.

Therefore $a_1 = a_2 = \dots = a_n$.

Examiner Notes

A little under half the candidates attempted this, scoring marginally less on average than on question 1. Some candidates used the arithmetic mean/geometric mean inequality which was what the question was proving in part (iii), and so were heavily penalised as their arguments were thus circular. Part (i) was generally well done, with most justifying that their steps were reversible to obtain ‘if and only if’. Marks were sometimes lost when justifying that G_k > 0 was not used to legitimise division by G_k and also that inequalities did not change sign. Part (ii) was well done too, though many candidates did not justify the ‘only if’, although some tried to use part (i) to prove (ii), which could not succeed. Part (iii) expressly required deduction, and only deduction, so those who had learned another proof of the inequality and just copied it out could not be rewarded. Many just used the results of (i) and (ii) without justifying why they could be used and some were imprecise with their induction arguments.

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Question 6

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

Problem

6 (i) The distinct points $A$, $Q$ and $C$ lie on a straight line in the Argand diagram, and represent the distinct complex numbers $a$, $q$ and $c$, respectively. Show that $\frac{q - a}{c - a}$ is real and hence that $(c - a)(q^* - a^*) = (c^* - a^*)(q - a)$.

Given that $aa^* = cc^* = 1$, show further that

$q + acq^* = a + c .$

(ii) The distinct points $A$, $B$, $C$ and $D$ lie, in anticlockwise order, on the circle of unit radius with centre at the origin (so that, for example, $aa^* = 1$). The lines $AC$ and $BD$ meet at $Q$. Show that

$(ac - bd)q^* = (a + c) - (b + d) ,$

where $b$ and $d$ are complex numbers represented by the points $B$ and $D$ respectively, and show further that

$(ac - bd)(q + q^*) = (a - b)(1 + cd) + (c - d)(1 + ab) .$

(iii) The lines $AB$ and $CD$ meet at $P$, which represents the complex number $p$. Given that $p$ is real, show that $p(1 + ab) = a + b$. Given further that $ac - bd \neq 0$, show that

$p(q + q^*) = 2 .$

Hint

第一问是简单的基础结论，有多种证法。将表达式与其共轭相等（利用实数条件），运用共轭的和、积等性质并代数整理即得(i)的第二个结果。利用单位圆性质替换 a 和 c 的共轭，经代数整理（包括除以可证明非零的 c-a）得到(i)的最终结果。利用 Q 在 AC 上且结合(i)的结果，类似地利用 Q 在 BD 上得到两个方程，线性组合得到(ii)中关于 q* 的方程。类似线性组合也可得 q 的方程，与已有结果相加整理得(ii)的第二个结果。(iii)第一个结果利用(i)的结论，P 在 AB 上且 p 为实数。类似地对 P 在 CD 上得到类似结果。将(ii)的最终结果乘以 p，利用刚得到的两个表达式化简，再除以 ac-bd 完成证明。本题实质上是关于极点与极线的问题。

Model Solution

Part (i)

Since $A$, $Q$, $C$ are distinct and collinear, the vector from $A$ to $Q$ is a real scalar multiple of the vector from $A$ to $C$: there exists $t \in \mathbb{R}$ such that $q - a = t(c - a)$, so

$\frac{q - a}{c - a} = t \in \mathbb{R}.$

A complex number is real if and only if it equals its own conjugate:

$\frac{q - a}{c - a} = \left(\frac{q - a}{c - a}\right)^* = \frac{q^* - a^*}{c^* - a^*}.$

Cross-multiplying:

$(c - a)(q^* - a^*) = (c^* - a^*)(q - a). \qquad \text{(*)}$

Now suppose $aa^* = cc^* = 1$, so $a^* = 1/a$ and $c^* = 1/c$. From $q = a + t(c-a)$ with $t$ real, conjugating:

$q^* = a^* + t(c^* - a^*) = \frac{1}{a} + t\!\left(\frac{1}{c} - \frac{1}{a}\right).$

Multiplying by $ac$:

$acq^* = c + t(a - c).$

Adding to $q = a + t(c-a)$:

$q + acq^* = [a + t(c-a)] + [c + t(a-c)] = a + c + \underbrace{t(c-a+a-c)}_{0} = a + c.$

Hence $q + acq^* = a + c$.

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Part (ii)

Since $Q$ lies on line $AC$ and all four points lie on the unit circle, part (i) gives:

$q + acq^* = a + c. \qquad \text{(1)}$

Since $Q$ also lies on line $BD$:

$q + bdq^* = b + d. \qquad \text{(2)}$

Finding $q^*$: subtract (2) from (1):

$(ac - bd)q^* = (a + c) - (b + d). \qquad \text{(3)}$

Finding $q + q^*$: add (1) and (2):

$2q + (ac + bd)q^* = (a + c) + (b + d). \qquad \text{(4)}$

From (3): $q^* = \dfrac{(a+c) - (b+d)}{ac - bd}$. Substitute into (4):

$2q = (a+c) + (b+d) - (ac + bd) \cdot \frac{(a+c) - (b+d)}{ac - bd}$

$= \frac{[(a+c) + (b+d)](ac - bd) - (ac + bd)[(a+c) - (b+d)]}{ac - bd}.$

Expanding the numerator. Write it as $[(a+c) + (b+d)](ac - bd) - [(a+c) - (b+d)](ac + bd)$:

$(a+c)[(ac-bd) - (ac+bd)] + (b+d)[(ac-bd) + (ac+bd)]$

$= (a+c)(-2bd) + (b+d)(2ac)$

$= 2[ac(b+d) - bd(a+c)].$

So $2q = \dfrac{2[ac(b+d) - bd(a+c)]}{ac - bd}$, giving

$q = \frac{ac(b+d) - bd(a+c)}{ac - bd}. \qquad \text{(5)}$

Now compute $q + q^*$ from (3) and (5):

$q + q^* = \frac{ac(b+d) - bd(a+c) + (a+c) - (b+d)}{ac - bd}$

$= \frac{(a+c)(1 - bd) + (b+d)(ac - 1)}{ac - bd}. \qquad \text{(6)}$

We verify that the numerator equals $(a-b)(1+cd) + (c-d)(1+ab)$ by expanding:

$(a-b)(1+cd) + (c-d)(1+ab) = a + acd - b - bcd + c + abc - d - abd$

$= (a+c) + ac(b+d) - (b+d) - bd(a+c)$

$= (a+c)(1-bd) + (b+d)(ac-1). \qquad \checkmark$

Therefore:

$(ac - bd)(q + q^*) = (a - b)(1 + cd) + (c - d)(1 + ab). \qquad \text{(6')}$

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Part (iii)

Since $P$ lies on line $AB$ and $p$ is real, part (i) gives $p + abp^* = a + b$. With $p^* = p$:

$p(1 + ab) = a + b. \qquad \text{(7)}$

Similarly, $P$ lies on line $CD$:

$p(1 + cd) = c + d. \qquad \text{(8)}$

Multiply equation (6’) by $p$:

$p(ac - bd)(q + q^*) = p(a - b)(1 + cd) + p(c - d)(1 + ab).$

Using (7) and (8):

$p(a-b)(1+cd) = (a-b) \cdot p(1+cd) = (a-b)(c+d),$

$p(c-d)(1+ab) = (c-d) \cdot p(1+ab) = (c-d)(a+b).$

Adding:

$(a-b)(c+d) + (c-d)(a+b) = ac + ad - bc - bd + ac + bc - ad - bd = 2(ac - bd).$

Therefore $p(ac - bd)(q + q^*) = 2(ac - bd)$. Since $ac - bd \neq 0$:

$p(q + q^*) = 2. \qquad \text{(9)}$

Examiner Notes

最不受欢迎的纯数题，尝试率不足40%，平均分仅略高于5/20。第一问有多种成功解法，但常见错误是将 q 和 c 视为 a 的标量倍数，且 * 运算令部分考生困惑。进入(ii)的考生起步尚可，但利用单位圆条件的考生推导最终方程更为轻松。能坚持到(iii)的考生通常表现不错，但在除以各种因子时偶有未说明非零的扣分。

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Question 7

Topic: 三角学 (Trigonometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 (i) Use De Moivre’s theorem to show that, if $\sin \theta \neq 0$, then $\frac{(\cot \theta + \mathrm{i})^{2n+1} - (\cot \theta - \mathrm{i})^{2n+1}}{2\mathrm{i}} = \frac{\sin(2n+1)\theta}{\sin^{2n+1}\theta},$ for any positive integer $n$. Deduce that the solutions of the equation $\binom{2n+1}{1}x^n - \binom{2n+1}{3}x^{n-1} + \dots + (-1)^n = 0$ are $x = \cot^2 \left( \frac{m\pi}{2n+1} \right)$ where $m = 1, 2, \dots, n$.

(ii) Hence show that $\sum_{m=1}^n \cot^2 \left( \frac{m\pi}{2n+1} \right) = \frac{n(2n-1)}{3}.$

(iii) Given that $0 < \sin \theta < \theta < \tan \theta$ for $0 < \theta < \frac{1}{2}\pi$, show that $\cot^2 \theta < \frac{1}{\theta^2} < 1 + \cot^2 \theta.$ Hence show that $\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}.$

Hint

将 cot θ 表示为 cos θ/sin θ 并运用棣莫弗定理得到第一个结果。将等式左边用二项式展开，按奇次幂项整理并除以 2i，化简为第二个结果的左边，当 (2n+1)θ = mπ（m=1,2,…,n）时等于零。(ii)通过考虑(i)中所得方程的根之和得到。(iii)的第一个结果由给定的小角不等式出发，注意取倒数和平方时的正负号，再利用适当的毕达哥拉斯三角恒等式。对 m=1 到 n 求和不等式，整理出所需求和对象，利用(i)中定义的 θ 值，得到上下界分别为 n(2n-1)π²/(3(2n+1)²) 和 n(2n+2)π²/(3(2n+1)²)，两者当 n→∞ 时均趋于 π²/6。

Model Solution

Part (i)

Write $\cot\theta = \frac{\cos\theta}{\sin\theta}$, so that

$\cot\theta + \mathrm{i} = \frac{\cos\theta + \mathrm{i}\sin\theta}{\sin\theta}, \qquad \cot\theta - \mathrm{i} = \frac{\cos\theta - \mathrm{i}\sin\theta}{\sin\theta}.$

Raising to the power $2n+1$:

$(\cot\theta + \mathrm{i})^{2n+1} = \frac{(\cos\theta + \mathrm{i}\sin\theta)^{2n+1}}{\sin^{2n+1}\theta}, \qquad (\cot\theta - \mathrm{i})^{2n+1} = \frac{(\cos\theta - \mathrm{i}\sin\theta)^{2n+1}}{\sin^{2n+1}\theta}.$

By De Moivre’s theorem, $(\cos\theta + \mathrm{i}\sin\theta)^{2n+1} = \cos(2n+1)\theta + \mathrm{i}\sin(2n+1)\theta$ and $(\cos\theta - \mathrm{i}\sin\theta)^{2n+1} = \cos(2n+1)\theta - \mathrm{i}\sin(2n+1)\theta$.

Subtracting:

$(\cot\theta + \mathrm{i})^{2n+1} - (\cot\theta - \mathrm{i})^{2n+1} = \frac{2\mathrm{i}\sin(2n+1)\theta}{\sin^{2n+1}\theta}.$

Dividing both sides by $2\mathrm{i}$:

$\frac{(\cot\theta + \mathrm{i})^{2n+1} - (\cot\theta - \mathrm{i})^{2n+1}}{2\mathrm{i}} = \frac{\sin(2n+1)\theta}{\sin^{2n+1}\theta}. \qquad \text{(*)}$

Now expand the left side using the binomial theorem. We have

$(\cot\theta + \mathrm{i})^{2n+1} = \sum_{r=0}^{2n+1} \binom{2n+1}{r} \cot^{2n+1-r}\theta \cdot \mathrm{i}^r,$

$(\cot\theta - \mathrm{i})^{2n+1} = \sum_{r=0}^{2n+1} \binom{2n+1}{r} \cot^{2n+1-r}\theta \cdot (-\mathrm{i})^r.$

Subtracting, the even-$r$ terms cancel and the odd-$r$ terms double:

$(\cot\theta + \mathrm{i})^{2n+1} - (\cot\theta - \mathrm{i})^{2n+1} = 2\sum_{\substack{r=1,3,5,\dots}}^{2n+1} \binom{2n+1}{r} \cot^{2n+1-r}\theta \cdot \mathrm{i}^r.$

Writing $r = 2s+1$ for $s = 0, 1, \dots, n$ and noting $\mathrm{i}^{2s+1} = \mathrm{i} \cdot (-1)^s$:

$= 2\mathrm{i} \sum_{s=0}^{n} (-1)^s \binom{2n+1}{2s+1} \cot^{2(n-s)}\theta.$

Dividing by $2\mathrm{i}$ and substituting into (*):

$\sum_{s=0}^{n} (-1)^s \binom{2n+1}{2s+1} \cot^{2(n-s)}\theta = \frac{\sin(2n+1)\theta}{\sin^{2n+1}\theta}. \qquad \text{(**)}$

Deduction. Setting the right side of (**) to zero requires $\sin(2n+1)\theta = 0$, i.e.\ $(2n+1)\theta = m\pi$ for integer $m$. For $0 < \theta < \pi/2$, this gives $\theta_m = \frac{m\pi}{2n+1}$ with $m = 1, 2, \dots, n$.

At these values, the left side of (**) is zero. Writing $x = \cot^2\theta$, the left side becomes

$\binom{2n+1}{1}x^n - \binom{2n+1}{3}x^{n-1} + \binom{2n+1}{5}x^{n-2} - \dots + (-1)^n.$

The values $x_m = \cot^2\theta_m = \cot^2\!\left(\frac{m\pi}{2n+1}\right)$ for $m = 1, 2, \dots, n$ are all distinct (since $\cot^2$ is strictly decreasing on $(0, \pi/2)$). This is a degree-$n$ polynomial with leading coefficient $\binom{2n+1}{1} = 2n+1 \neq 0$, so it has at most $n$ roots. We have found $n$ distinct roots, so these are all the solutions.

Part (ii)

The equation from part (i) can be written as

$\sum_{s=0}^{n} (-1)^s \binom{2n+1}{2s+1} x^{n-s} = 0,$

or equivalently, dividing by the leading coefficient $\binom{2n+1}{1} = 2n+1$:

$x^n - \frac{\binom{2n+1}{3}}{2n+1} x^{n-1} + \dots = 0.$

By Vieta’s formula, the sum of the roots equals $\frac{\binom{2n+1}{3}}{\binom{2n+1}{1}}$. We compute:

$\frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{\frac{(2n+1)(2n)(2n-1)}{6}}{2n+1} = \frac{2n(2n-1)}{6} = \frac{n(2n-1)}{3}.$

Therefore

$\sum_{m=1}^{n} \cot^2\!\left(\frac{m\pi}{2n+1}\right) = \frac{n(2n-1)}{3}. \qquad \text{(***)}$

Part (iii)

For $0 < \theta < \frac{\pi}{2}$, we are given $0 < \sin\theta < \theta < \tan\theta$.

Left inequality: From $\theta < \tan\theta$, dividing both sides by $\theta\tan\theta > 0$:

$\frac{1}{\tan\theta} < \frac{1}{\theta} \implies \cot\theta < \frac{1}{\theta}.$

Since both sides are positive, squaring preserves the inequality:

$\cot^2\theta < \frac{1}{\theta^2}. \qquad \text{(L)}$

Right inequality: From $\sin\theta < \theta$, dividing both sides by $\theta\sin\theta > 0$:

$\frac{1}{\theta} < \frac{1}{\sin\theta}.$

Squaring (both sides positive): $\frac{1}{\theta^2} < \frac{1}{\sin^2\theta} = 1 + \cot^2\theta.$

$\frac{1}{\theta^2} < 1 + \cot^2\theta. \qquad \text{(R)}$

Combining (L) and (R): $\cot^2\theta < \frac{1}{\theta^2} < 1 + \cot^2\theta$.

Evaluating the Basel sum. Set $\theta_m = \frac{m\pi}{2n+1}$ for $m = 1, 2, \dots, n$. Since $0 < \theta_m < \frac{\pi}{2}$, the double inequality applies. Summing over $m$:

$\sum_{m=1}^{n} \cot^2\theta_m < \sum_{m=1}^{n} \frac{1}{\theta_m^2} < \sum_{m=1}^{n} (1 + \cot^2\theta_m).$

The middle sum is

$\sum_{m=1}^{n} \frac{(2n+1)^2}{m^2\pi^2} = \frac{(2n+1)^2}{\pi^2} \sum_{m=1}^{n} \frac{1}{m^2}.$

Using (***) for both bounds:

$\frac{n(2n-1)}{3} < \frac{(2n+1)^2}{\pi^2} \sum_{m=1}^{n} \frac{1}{m^2} < n + \frac{n(2n-1)}{3} = \frac{n(2n+2)}{3}.$

Dividing through by $\frac{(2n+1)^2}{\pi^2}$:

$\frac{n(2n-1)\pi^2}{3(2n+1)^2} < \sum_{m=1}^{n} \frac{1}{m^2} < \frac{n(2n+2)\pi^2}{3(2n+1)^2}.$

As $n \to \infty$, we evaluate the limits. For the lower bound:

$\frac{n(2n-1)}{3(2n+1)^2} = \frac{2n^2 - n}{3(4n^2 + 4n + 1)} = \frac{2 - 1/n}{3(4 + 4/n + 1/n^2)} \longrightarrow \frac{2}{12} = \frac{1}{6}.$

For the upper bound:

$\frac{n(2n+2)}{3(2n+1)^2} = \frac{2n^2 + 2n}{3(4n^2 + 4n + 1)} = \frac{2 + 2/n}{3(4 + 4/n + 1/n^2)} \longrightarrow \frac{2}{12} = \frac{1}{6}.$

Both bounds converge to $\frac{\pi^2}{6}$. Since the partial sums $\sum_{m=1}^{n} \frac{1}{m^2}$ are bounded between two sequences both tending to $\frac{\pi^2}{6}$, by the squeeze theorem:

$\sum_{m=1}^{\infty} \frac{1}{m^2} = \frac{\pi^2}{6}.$

Examiner Notes

尝试率略高于 Q6，平均分 8/20。(i)的第一部分没有问题，但更进一步的考生较少；展开 (cot θ + i)^{2n+1} 的考生几乎都成功了，即使不总能处理 sin θ ≠ 0 的条件。尝试(ii)的考生大多得满分，但有人忘记将 C(2n+1,3) 除以 C(2n+1,1)。几乎所有尝试(iii)的考生都证明了第一个结果，但应用结果得出最终结论较难。常见错误是对 θ 而非 m 求和，且最后部分的代数运算常出错。

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Question 8

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

Problem

8 In this question, you should ignore issues of convergence.

(i) Let $I = \int_0^1 \frac{\mathrm{f}(x^{-1})}{1+x} \, \mathrm{d}x,$ where $\mathrm{f}(x)$ is a function for which the integral exists. Show that $I = \sum_{n=1}^\infty \int_n^{n+1} \frac{\mathrm{f}(y)}{y(1+y)} \, \mathrm{d}y$ and deduce that, if $\mathrm{f}(x) = \mathrm{f}(x+1)$ for all $x$, then $I = \int_0^1 \frac{\mathrm{f}(x)}{1+x} \, \mathrm{d}x.$

(ii) The fractional part, $\{x\}$, of a real number $x$ is defined to be $x - \lfloor x \rfloor$ where $\lfloor x \rfloor$ is the largest integer less than or equal to $x$. For example $\{3.2\} = 0.2$ and $\{3\} = 0$. Use the result of part (i) to evaluate $\int_0^1 \frac{\{x^{-1}\}}{1+x} \, \mathrm{d}x \quad \text{and} \quad \int_0^1 \frac{\{2x^{-1}\}}{1+x} \, \mathrm{d}x.$

Hint

(i)的第一个结果可通过将积分和解释为从1到∞的单一积分，并作换元 y=x^{-1} 来证明。推论可通过将积分拆为部分分式，在第一个分式的积分中用 y=x+1 换元，利用周期性条件，然后裂项求和化简为单一积分。(ii)的第一个积分直接使用(i)的结果，观察到在积分范围内 {x}=x，然后正常积分得 1-ln 2。第二个积分需观察到 {2(x+1)}={2x+2}={2x}，利用(i)的结果将积分拆为从0到1/2和从1/2到1的两个积分，分别处理 {2x} 的两个不同显式表达式；结果可化简为 2+ln(3/16) 或等价形式。

Model Solution

Part (i)

First result. In the integral $I = \int_0^1 \frac{f(x^{-1})}{1+x}\,dx$, substitute $y = x^{-1}$, so $x = y^{-1}$ and $dx = -y^{-2}\,dy$. When $x = 0^+$, $y \to \infty$; when $x = 1$, $y = 1$. Thus

$I = \int_{\infty}^{1} \frac{f(y)}{1 + y^{-1}} \left(-\frac{1}{y^2}\right) dy = \int_1^{\infty} \frac{f(y)}{y^2\left(1 + y^{-1}\right)}\,dy = \int_1^{\infty} \frac{f(y)}{y^2 \cdot \frac{y+1}{y}}\,dy = \int_1^{\infty} \frac{f(y)}{y(y+1)}\,dy.$

Splitting the integral over consecutive unit intervals:

$I = \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(y)}{y(y+1)}\,dy. \qquad \text{(*)}$

Deduction. Suppose $f(x) = f(x+1)$ for all $x$. Using partial fractions, $\frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}$, so each term in (*) becomes

$\int_n^{n+1} \frac{f(y)}{y(y+1)}\,dy = \int_n^{n+1} \frac{f(y)}{y}\,dy - \int_n^{n+1} \frac{f(y)}{y+1}\,dy.$

In the first integral, substitute $u = y - 1$ (so $y = u+1$, $dy = du$), with limits $u = n-1$ to $u = n$:

$\int_n^{n+1} \frac{f(y)}{y}\,dy = \int_{n-1}^{n} \frac{f(u+1)}{u+1}\,du = \int_{n-1}^{n} \frac{f(u)}{u+1}\,du,$

where the last step uses $f(u+1) = f(u)$. Therefore

$\int_n^{n+1} \frac{f(y)}{y(y+1)}\,dy = \int_{n-1}^{n} \frac{f(u)}{u+1}\,du - \int_n^{n+1} \frac{f(y)}{y+1}\,dy.$

Summing from $n = 1$ to $N$:

$\sum_{n=1}^{N} \int_n^{n+1} \frac{f(y)}{y(y+1)}\,dy = \sum_{n=1}^{N} \int_{n-1}^{n} \frac{f(u)}{u+1}\,du - \sum_{n=1}^{N} \int_n^{n+1} \frac{f(y)}{y+1}\,dy.$

The first sum telescopes to $\int_0^{N} \frac{f(u)}{u+1}\,du$. The second sum telescopes to $\int_1^{N+1} \frac{f(y)}{y+1}\,dy$. Therefore

$\sum_{n=1}^{N} \int_n^{n+1} \frac{f(y)}{y(y+1)}\,dy = \int_0^{N} \frac{f(u)}{u+1}\,du - \int_1^{N+1} \frac{f(u)}{u+1}\,du = \int_0^1 \frac{f(u)}{u+1}\,du - \int_N^{N+1} \frac{f(u)}{u+1}\,du.$

As $N \to \infty$, the last integral tends to zero (the integrand is $O(1/N)$ over an interval of length 1), so

$I = \int_0^1 \frac{f(x)}{1+x}\,dx.$

Part (ii)

First integral: $\int_0^1 \frac{\{x^{-1}\}}{1+x}\,dx$.

Take $f(x) = \{x\} = x - \lfloor x \rfloor$. Since $\{x+1\} = (x+1) - \lfloor x+1 \rfloor = (x+1) - (\lfloor x \rfloor + 1) = x - \lfloor x \rfloor = \{x\}$, the function $f$ is periodic with period 1. By part (i):

$\int_0^1 \frac{\{x^{-1}\}}{1+x}\,dx = \int_0^1 \frac{\{x\}}{1+x}\,dx.$

For $0 \leq x < 1$, $\lfloor x \rfloor = 0$, so $\{x\} = x$. Therefore

$\int_0^1 \frac{x}{1+x}\,dx = \int_0^1 \left(1 - \frac{1}{1+x}\right) dx = \Big[x - \ln(1+x)\Big]_0^1 = 1 - \ln 2.$

Second integral: $\int_0^1 \frac{\{2x^{-1}\}}{1+x}\,dx$.

Take $f(x) = \{2x\}$. Since $\{2(x+1)\} = \{2x + 2\} = \{2x\}$, this function is periodic with period 1. By part (i):

$\int_0^1 \frac{\{2x^{-1}\}}{1+x}\,dx = \int_0^1 \frac{\{2x\}}{1+x}\,dx.$

For $0 \leq x < \frac{1}{2}$: $\lfloor 2x \rfloor = 0$, so $\{2x\} = 2x$. For $\frac{1}{2} \leq x < 1$: $\lfloor 2x \rfloor = 1$, so $\{2x\} = 2x - 1$.

Split the integral:

$\int_0^1 \frac{\{2x\}}{1+x}\,dx = \int_0^{1/2} \frac{2x}{1+x}\,dx + \int_{1/2}^{1} \frac{2x-1}{1+x}\,dx.$

First piece:

$\int_0^{1/2} \frac{2x}{1+x}\,dx = 2\int_0^{1/2} \left(1 - \frac{1}{1+x}\right) dx = 2\Big[x - \ln(1+x)\Big]_0^{1/2} = 2\left(\frac{1}{2} - \ln\frac{3}{2}\right) = 1 - 2\ln\frac{3}{2}.$

Second piece: Write $\frac{2x - 1}{1+x} = \frac{2(x+1) - 3}{1+x} = 2 - \frac{3}{1+x}$.

$\int_{1/2}^{1} \left(2 - \frac{3}{1+x}\right) dx = \Big[2x - 3\ln(1+x)\Big]_{1/2}^{1} = (2 - 3\ln 2) - (1 - 3\ln\tfrac{3}{2}) = 1 - 3\ln 2 + 3\ln\frac{3}{2}.$

Combining:

$\int_0^1 \frac{\{2x\}}{1+x}\,dx = (1 - 2\ln\tfrac{3}{2}) + (1 - 3\ln 2 + 3\ln\tfrac{3}{2}) = 2 + \ln\frac{3}{2} - 3\ln 2.$

Simplifying: $\ln\frac{3}{2} = \ln 3 - \ln 2$, so

$2 + (\ln 3 - \ln 2) - 3\ln 2 = 2 + \ln 3 - 4\ln 2 = 2 + \ln\frac{3}{16}.$

Therefore

$\int_0^1 \frac{\{x^{-1}\}}{1+x}\,dx = 1 - \ln 2, \qquad \int_0^1 \frac{\{2x^{-1}\}}{1+x}\,dx = 2 + \ln\frac{3}{16}.$

Examiner Notes

第五受欢迎的题目，尝试率为八分之五，平均分略高于9/20，为第四成功的题目。(i)的第一个结果通常做得很好，但第二个结果通常只限于发现可以使用部分分式的考生。(ii)的第一个积分通常能正确完成。但第二个积分出现各种错误，如未能看出或说明第一部分如何使用以及未化简答案。周期性的性质以及在不同区间上积分不同表达式的需要是常见的绊脚石。