STEP2 2011 -- Pure Mathematics

Exam: STEP2  |  Year: 2011  |  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	曲线绘图与方程求解 Curve Sketching and Equation Solving	Routine	曲线绘图, 定义域分析, 图像交点法, 代数化简
2	数论与不定方程 Number Theory and Diophantine Equations	Standard	立方和分解, 判别式分析, 不等式放缩, 丢番图方程
3	微积分不等式 Calculus Inequalities	Standard	单调性判定, 导数符号分析, 不等式链推导, 反三角函数性质
4	三角方程与恒等式 Trigonometric Equations and Identities	Challenging	三角恒等变换, 倍角公式, 因式分解, 特殊角值推导
5	向量几何 Vector Geometry	Challenging	向量投影, 点积运算, 反射变换, 共线性判别
6	积分技巧 Integration Techniques	Challenging	分部积分,递推公式,三角恒等变换,变量代换
7	数列与求和 Sequences and Summation	Challenging	等比数列求和,递推关系,代数化简,奇偶分类讨论
8	参数方程与面积 Parametric Curves and Areas	Hard	参数方程,求导找极值,参数积分求面积,几何分析

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Question 1

Topic: 曲线绘图与方程求解 Curve Sketching and Equation Solving  |  Difficulty: Routine  |  Marks: 20

Problem

1 (i) Sketch the curve $y = \sqrt{1 - x} + \sqrt{3 + x}$.

Use your sketch to show that only one real value of $x$ satisfies

$\sqrt{1 - x} + \sqrt{3 + x} = x + 1,$

and give this value.

(ii) Determine graphically the number of real values of $x$ that satisfy

$2\sqrt{1 - x} = \sqrt{3 + x} + \sqrt{3 - x}.$

Solve this equation.

Hint

(ii) The forces involved are now readily written down …

$\underline{\mathbf{T}_B} = \left( -\frac{1}{3} \mathbf{i} - \frac{\sqrt{2}}{3} \mathbf{j} + \frac{\sqrt{2}}{\sqrt{3}} \mathbf{k} \right) U \text{ follows from (i)'s answer. Also,}$

$\underline{\mathbf{T}_A} = T \sin 30^\circ \mathbf{j} + T \cos 30^\circ \mathbf{k} = \frac{1}{2} T (\mathbf{j} + \sqrt{3} \mathbf{k}),$

$\underline{\mathbf{T}_C} = V \sin 60^\circ \cos \phi \mathbf{i} - V \sin 60^\circ \sin \phi \mathbf{j} + V \cos 60^\circ \mathbf{k} = \frac{1}{2} V \left( \frac{2\sqrt{2}}{\sqrt{3}} \mathbf{i} - \frac{1}{\sqrt{3}} \mathbf{j} + \mathbf{k} \right)$

and $\underline{\mathbf{W}} = -W \mathbf{k}$.

(iii) Having set the system up in vector form, the fundamental Statics principle involved is that

$\underline{\mathbf{T}_A} + \underline{\mathbf{T}_B} + \underline{\mathbf{T}_C} + \underline{\mathbf{W}} = \underline{\mathbf{0}} .$

Comparing components in this vector equation gives

$(\mathbf{i}) \quad 0 - \frac{1}{3} U + \frac{\sqrt{6}}{3} V = 0 \Rightarrow U = V\sqrt{6}$

$(\mathbf{j}) \quad \frac{1}{2} T - \frac{\sqrt{2}}{3} U - \frac{\sqrt{3}}{6} V = 0 \Rightarrow (\text{using } U = V\sqrt{6}) \ T = \frac{5\sqrt{3}}{3} V$

$(\mathbf{k}) \quad \frac{\sqrt{3}}{2} T + \frac{\sqrt{6}}{3} U + \frac{1}{2} V = W \Rightarrow (\text{using } U = V\sqrt{6} \text{ and } T = \frac{5\sqrt{3}}{3} V)$

$T = \frac{W\sqrt{3}}{3}, \ U = \frac{W\sqrt{6}}{5}, \ V = \frac{W}{5} .$

Model Solution

Part (i): Sketching $y = \sqrt{1-x} + \sqrt{3+x}$

First, determine the domain. We need $1 - x \geq 0$ and $3 + x \geq 0$, so $-3 \leq x \leq 1$.

Key points:

• At $x = -3$: $y = \sqrt{4} + \sqrt{0} = 2$.
• At $x = 1$: $y = \sqrt{0} + \sqrt{4} = 2$.
• At $x = -1$ (the midpoint of the domain): $y = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \approx 2.83$.

The curve is symmetric about $x = -1$, since replacing $x$ by $-2 - x$ swaps the two square root terms.

The gradient is: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-1}{2\sqrt{1-x}} + \frac{1}{2\sqrt{3+x}}$

At the endpoints, this tends to $\mp \infty$ (vertical tangent at each end). The maximum occurs at $x = -1$ where $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, giving a rounded peak.

The sketch is a dome-shaped curve from $(-3, 2)$ to $(1, 2)$, peaking at $(-1, 2\sqrt{2})$, with vertical tangents at both endpoints.

Solving $\sqrt{1-x} + \sqrt{3+x} = x + 1$:

We need the intersection of $y = \sqrt{1-x} + \sqrt{3+x}$ and $y = x + 1$.

The line $y = x + 1$ passes through $(-1, 0)$ and $(1, 2)$. From the sketch, the line meets the curve at its right endpoint $x = 1$, where both sides equal 2.

To confirm this is the only intersection: the line $y = x+1$ starts at $(-3, -2)$ (below the curve which is at height 2) and ends at $(1, 2)$ where it touches the curve’s endpoint. Since the curve is concave (dome-shaped) and the line is straight, the line can only touch the curve at the endpoint. Thus there is exactly one real solution:

$\boxed{x = 1}$

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Part (ii): Solving $2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}$

Graphical analysis: We need $1 - x \geq 0$, $3 + x \geq 0$, and $3 - x \geq 0$, so $-3 \leq x \leq 1$.

Consider $f(x) = 2\sqrt{1-x}$ and $g(x) = \sqrt{3+x} + \sqrt{3-x}$.

• $f(-3) = 2\sqrt{4} = 4$, $g(-3) = 0 + \sqrt{6} \approx 2.45$. So $f > g$ at $x = -3$.
• $f(1) = 0$, $g(1) = 2 + \sqrt{2} \approx 3.41$. So $f < g$ at $x = 1$.

By the intermediate value theorem, there is at least one crossing. Since $f$ is strictly decreasing and $g$ is symmetric about $x = 0$ (increasing on $[-3, 0]$, decreasing on $[0, 1]$), there is exactly one intersection.

Algebraic solution: Square both sides of $2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}$:

$4(1 - x) = (3 + x) + 2\sqrt{(3+x)(3-x)} + (3 - x)$

$4 - 4x = 6 + 2\sqrt{9 - x^2}$

$-2 - 4x = 2\sqrt{9 - x^2}$

$-1 - 2x = \sqrt{9 - x^2}$

Since $\sqrt{9-x^2} \geq 0$, we need $-1 - 2x \geq 0$, i.e. $x \leq -\frac{1}{2}$.

Squaring again:

$(1 + 2x)^2 = 9 - x^2$

$1 + 4x + 4x^2 = 9 - x^2$

$5x^2 + 4x - 8 = 0$

$x = \frac{-4 \pm \sqrt{16 + 160}}{10} = \frac{-4 \pm \sqrt{176}}{10} = \frac{-4 \pm 4\sqrt{11}}{10} = \frac{-2 \pm 2\sqrt{11}}{5}$

Since $x \leq -\frac{1}{2}$, we take the negative root: $x = \dfrac{-2 - 2\sqrt{11}}{5}$.

Verification: $\frac{-2 - 2\sqrt{11}}{5} \approx \frac{-2 - 6.63}{5} \approx -1.73$, which satisfies $x \leq -\frac{1}{2}$.

$\boxed{x = \frac{-2 - 2\sqrt{11}}{5}}$

Examiner Notes

Q1 The first question is invariably set with the intention that everyone should be able to attempt it, giving all candidates something to get their teeth into and thereby easing them into the paper with some measure of success. As mentioned above, this was both a very popular question and a high-scoring one. Even so, there were some general weaknesses revealed in the curve-sketching department, as many candidates failed to consider (explicitly or not) things such as the gradient of the curve at its endpoints and, in particular, the shape of the curve at its peak (often more of a vertex than a maximum). It was also strange that surprisingly many who had the correct domain for the curve and had decided that the single point of intersection of line and curve was at $x = 1$ still managed to draw the line $y = x + 1$ not through the endpoint at $x = 1$. Most other features – domain, symmetry, coordinates of key points, etc. – were well done in (i). Unfortunately, those who simply resort to plotting points are really sending quite the wrong message about their capabilities to the examiners.

Following on in (ii), the majority of candidates employed the expected methods and were also quite happy to plough into the algebra of squaring-up and rearranging; however, there were frequently many (unnecessarily) careless errors involved. The only other very common error was in the sketch of the half-parabola $y = 2\sqrt{1-x}$, due to a misunderstanding of the significance of the radix sign.

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Question 2

Topic: 数论与不定方程 Number Theory and Diophantine Equations  |  Difficulty: Standard  |  Marks: 20

Problem

2 Write down the cubes of the integers $1, 2, \dots, 10$.

The positive integers $x, y$ and $z$, where $x < y$, satisfy

$x^3 + y^3 = kz^3, \qquad \text{(*)}$

where $k$ is a given positive integer.

(i) In the case $x + y = k$, show that

$z^3 = k^2 - 3kx + 3x^2.$

Deduce that $(4z^3 - k^2)/3$ is a perfect square and that $\frac{1}{4}k^2 \leqslant z^3 < k^2$. Use these results to find a solution of $(*)$ when $k = 20$.

(ii) By considering the case $x + y = z^2$, find two solutions of $(*)$ when $k = 19$.

Hint

Q2 The required list of perfect cubes is 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, though there were no marks for noting them.

(i) In this question, it is clearly important to be able to factorise the sum of two cubes. So, in this first instance $x + y = k$, $(x + y)(x^2 - xy + y^2) = kz^3 \implies x^2 - (k - x)x + (k - x)^2 - z^3 = 0$, which gives the required result upon rearrangement. One could either treat this as a quadratic in $x$ and deal with its discriminant or go ahead directly to show that $\frac{4z^3 - k^2}{3} = (y - x)^2 \geq 0$ which

immediately gives that $\frac{4z^3 - k^2}{3}$ is a perfect square and also that $z^3 \geq \frac{1}{4}k^2$; and the other half of the required inequality comes either from $z^3 = k^2 - 3xy < k^2$ (since $x, y > 0$) or from noting that the smaller root of the quadratic in $x$ is positive. Substituting $k = 20$ into the given inequality then

yields $100 \leq z^3 < 400 \implies z = 5, 6, 7$; and the only value of $z$ in this list for which $\frac{4z^3 - k^2}{3}$ is a perfect square is $z = 7$, which then yields the solution $(x, y, z) = (1, 19, 7)$. Although not a part of the question, we can now express 20 as a sum of two rational cubes in the following way:

$20 = \left( \frac{1}{7} \right)^3 + \left( \frac{19}{7} \right)^3$

(ii) Although this second part of the question can be done in other ways, the intention is clearly that a similar methodology to (i)‘s can be employed. Starting from

$x + y = z^2, \ (x + y)(x^2 - xy + y^2) = kz.z^2 \implies x^2 - (z^2 - x)x + (z^2 - x)^2 - kz = 0$

we find that $\frac{4kz - z^4}{3}$ is a perfect square, and also that $k < z^3 \leq 4k$. With $k = 19$, $19 < z^3 \leq 76$

$\Rightarrow z = 3$ or $4$. This time, each of these values of $z$ gives $\frac{z(76 - z^3)}{3}$ a perfect square, yielding the two solutions $(x, y, z) = (1, 8, 3)$ and $(6, 10, 4)$. Thus we have two ways to represent 19 as a sum of two rational cubes: $19 = \left( \frac{1}{3} \right)^3 + \left( \frac{8}{3} \right)^3$ and $\left( \frac{3}{2} \right)^3 + \left( \frac{5}{2} \right)^3$.

Model Solution

Cubes of $1, 2, \dots, 10$:

$1^3 = 1, \quad 2^3 = 8, \quad 3^3 = 27, \quad 4^3 = 64, \quad 5^3 = 125, \quad 6^3 = 216, \quad 7^3 = 343, \quad 8^3 = 512, \quad 9^3 = 729, \quad 10^3 = 1000$

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Part (i): The case $x + y = k$

Starting from $x^3 + y^3 = kz^3$ and using the factorisation $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$:

$(x+y)(x^2 - xy + y^2) = kz^3$

Since $x + y = k$:

$k(x^2 - xy + y^2) = kz^3$

$x^2 - xy + y^2 = z^3 \qquad (\dagger)$

Now substitute $y = k - x$:

$x^2 - x(k-x) + (k-x)^2 = z^3$

$x^2 - kx + x^2 + k^2 - 2kx + x^2 = z^3$

$3x^2 - 3kx + k^2 = z^3$

$\boxed{z^3 = k^2 - 3kx + 3x^2} \qquad \blacksquare$

Deducing $(4z^3 - k^2)/3$ is a perfect square:

$4z^3 - k^2 = 4(k^2 - 3kx + 3x^2) - k^2 = 3k^2 - 12kx + 12x^2 = 3(k^2 - 4kx + 4x^2) = 3(k - 2x)^2$

Therefore:

$\frac{4z^3 - k^2}{3} = (k - 2x)^2 \qquad \blacksquare$

This is indeed a perfect square. Note also that $k - 2x = (x+y) - 2x = y - x$, so this equals $(y-x)^2$.

Deducing $\frac{1}{4}k^2 \leq z^3 < k^2$:

Since $\frac{4z^3 - k^2}{3} = (y-x)^2 \geq 0$, we have $4z^3 \geq k^2$, i.e.:

$z^3 \geq \frac{k^2}{4}$

For the upper bound, from $(\dagger)$: $z^3 = x^2 - xy + y^2$. Since $x + y = k$ and $x, y > 0$:

$z^3 = k^2 - 3xy < k^2$

(since $xy > 0$). Therefore $\frac{1}{4}k^2 \leq z^3 < k^2$. $\qquad \blacksquare$

Finding a solution when $k = 20$:

We need $\frac{1}{4}(400) \leq z^3 < 400$, i.e. $100 \leq z^3 < 400$.

Checking cubes: $4^3 = 64$ (too small), $5^3 = 125$, $6^3 = 216$, $7^3 = 343$, $8^3 = 512$ (too large).

So $z \in \{5, 6, 7\}$. We check which gives $\frac{4z^3 - k^2}{3}$ as a perfect square:

• $z = 5$: $\frac{4(125) - 400}{3} = \frac{100}{3}$ — not an integer.
• $z = 6$: $\frac{4(216) - 400}{3} = \frac{464}{3}$ — not an integer.
• $z = 7$: $\frac{4(343) - 400}{3} = \frac{972}{3} = 324 = 18^2$ — a perfect square.

So $z = 7$, $y - x = 18$ and $x + y = 20$, giving $x = 1, y = 19$.

Verification: $1^3 + 19^3 = 1 + 6859 = 6860 = 20 \times 343 = 20 \times 7^3$. $\checkmark$

$\boxed{(x, y, z) = (1, 19, 7)}$

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Part (ii): The case $x + y = z^2$, $k = 19$

We follow the same methodology as part (i). Starting from $x^3 + y^3 = 19z^3$:

$(x+y)(x^2 - xy + y^2) = 19z^3$

With $x + y = z^2$:

$z^2(x^2 - xy + y^2) = 19z^3$

$x^2 - xy + y^2 = 19z \qquad (\ddagger)$

Substituting $y = z^2 - x$:

$x^2 - x(z^2 - x) + (z^2 - x)^2 = 19z$

$3x^2 - 3z^2 x + z^4 = 19z$

This is a quadratic in $x$. For real solutions, the discriminant must be non-negative:

$\Delta = 9z^4 - 12(z^4 - 19z) = -3z^4 + 228z \geq 0$

$z(228 - 3z^3) \geq 0$

Since $z > 0$: $z^3 \leq 76$, i.e. $z \leq 4$.

Also from $(\ddagger)$: $19z = x^2 - xy + y^2 > 0$ (since not both $x, y$ are zero), and since $x, y \geq 1$:

$19z = x^2 - xy + y^2 \geq 1 - 1 \cdot (z^2 - 1) + (z^2 - 1)^2$

But more directly, we need $x^2 - xy + y^2 = 19z$ with $x + y = z^2$, $x < y$, $x, y > 0$, so $z^2 \geq 3$ (at minimum $x=1, y=2$), giving $z \geq 2$.

Also, we need $\frac{4 \cdot 19z - z^4}{3} = \frac{z(76 - z^3)}{3}$ to be a perfect square (by analogy with part (i), since the discriminant condition gives $(y-x)^2 = \frac{4 \cdot 19z - z^4}{3}$).

Checking $z = 2, 3, 4$:

• $z = 2$: $\frac{2(76 - 8)}{3} = \frac{136}{3}$ — not an integer.
• $z = 3$: $\frac{3(76 - 27)}{3} = 49 = 7^2$ — a perfect square.
• $z = 4$: $\frac{4(76 - 64)}{3} = \frac{48}{3} = 16 = 4^2$ — a perfect square.

For $z = 3$: $y - x = 7$ and $x + y = 9$, giving $x = 1, y = 8$.

Verification: $1^3 + 8^3 = 1 + 512 = 513 = 19 \times 27 = 19 \times 3^3$. $\checkmark$

For $z = 4$: $y - x = 4$ and $x + y = 16$, giving $x = 6, y = 10$.

Verification: $6^3 + 10^3 = 216 + 1000 = 1216 = 19 \times 64 = 19 \times 4^3$. $\checkmark$

$\boxed{(x, y, z) = (1, 8, 3) \quad \text{and} \quad (6, 10, 4)}$

Examiner Notes

Personally, this was my favourite question, even though it was ultimately (marginally) deflected from its original purpose of expressing integers as sums of two rational cubes. Given that the question explicitly involves inequalities (which are, as a rule, never popular) and cubics rather than quadratics, it was slightly surprising to find that it was the most popular question on the paper. However, although the average score on the question was almost exactly 10, these two issues then turned out to be the biggest stumbling-blocks to a completely successful attempt as candidates progressed through the question, both in establishing the given inequalities and then in the use of them. In particular, it was noted that many candidates “proved” the given results by showing that they implied something else that was true, rather than by deducing them from something else known to be true; such logical flaws received little credit in terms of marks. The purpose of this preliminary work was to enable the candidates to whittle down the possibilities to a small, finite list and then provide them with some means of testing each possibility’s validity. This help was often ignored in favour of starting again. In general, though, part (i) was done reasonably well; as was (ii) by those who used (i)‘s methodology as a template.

Only a very few candidates were bold enough to attempt (ii) successfully without any reference to (i)‘s methods; indeed, this arithmetic approach was how the question was originally posed (as part (i), of course) before proceeding onto the algebra. Noting that the wording of the question does not demand any particular approach in order to find the required two solutions to the equation $x^3 + y^3 = 19z^3$, a reasonably confident arithmetician might easily note that

$19 \times 2^3 = 152 = 3^3 + 5^3 \text{ and } 19 \times 3^3 = 513 = 1^3 + 8^3$

and it isn’t even necessary to look very far for two solutions. For 10 marks, this is what our transatlantic cousins would call “a steal”.

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Question 3

Topic: 微积分不等式 Calculus Inequalities  |  Difficulty: Standard  |  Marks: 20

Problem

3 In this question, you may assume without proof that any function $f$ for which $f'(x) \geqslant 0$ is increasing; that is, $f(x_2) \geqslant f(x_1)$ if $x_2 \geqslant x_1$.

(i) (a) Let $f(x) = \sin x - x \cos x$. Show that $f(x)$ is increasing for $0 \leqslant x \leqslant \frac{1}{2}\pi$ and deduce that $f(x) \geqslant 0$ for $0 \leqslant x \leqslant \frac{1}{2}\pi$.

(b) Given that $\frac{\mathrm{d}}{\mathrm{d}x}(\arcsin x) \geqslant 1$ for $0 \leqslant x < 1$, show that

$\arcsin x \geqslant x \quad (0 \leqslant x < 1).$

(c) Let $g(x) = x \operatorname{cosec} x$ for $0 < x < \frac{1}{2}\pi$. Show that $g$ is increasing and deduce that

$(\arcsin x) x^{-1} \geqslant x \operatorname{cosec} x \quad (0 < x < 1).$

(ii) Given that $\frac{\mathrm{d}}{\mathrm{d}x}(\arctan x) \leqslant 1$ for $x \geqslant 0$, show by considering the function $x^{-1} \tan x$ that

$(\tan x)(\arctan x) \geqslant x^2 \quad (0 < x < \tfrac{1}{2}\pi).$

Hint

Q3 This question is all about increasing functions and what can be deduced from them. It involves inequalities, which are never popular creatures even amongst STEP candidates. Fortunately, you are led fairly gently by the hand into what to do, at least to begin with.

(i) (a) $f'(x) = \cos x - \{x \cdot -\sin x + \cos x\} = x \sin x \geq 0$ for $x \in [0, \frac{1}{2}\pi]$, and since $f(0) = 0$ it follows that $f(x) = \sin x - x \cos x \geq 0$ for $0 \leq x \leq \frac{1}{2}\pi$.

(i) (b) A key observation here is that the “1” is simply a disguise for $\frac{d}{dx}(x)$, so you are actually

being given that $\frac{d}{dx}(\arcsin x) \geq \frac{d}{dx}(x)$ in the given interval; in other words, that $f(x) = \arcsin x - x$ is an increasing function. Since $f(0) = 0$ and $f$ increasing, $f(x) = \arcsin x - x \geq 0$ for $0 \leq x < 1$, and the required result follows.

(i) (c) Writing $g(x) = \frac{x}{\sin x} \Rightarrow g'(x) = \frac{\sin x - x \cos x}{\sin^2 x} > 0$ for $0 < x < \frac{1}{2}\pi$ using (a)‘s result.

Now, it may help to write $u = \arcsin x$, just so that it looks simpler to deal with here. Then $u \geq x$ by (b)‘s result $\Rightarrow g(u) \geq g(x)$ since $g'(x) \geq 0$ and the required result again follows.

(ii) There is a bit more work to be done here, but essentially the idea is the same as that in part (i), only the direction of the inequality seems to be reversed, so care must be taken. An added difficulty also arises in that we find that we must show that $f' \geq 0$ by showing that it is increasing

from zero. So $g(x) = \frac{\tan x}{x}$, $g'(x) = \frac{x \sec^2 x - \tan x}{x^2} = \frac{2x - \sin 2x}{2x^2 \cos^2 x}$. Examining $f(x) = 2x - \sin 2x$

(since the denominator is clearly positive in the required interval): $f(0) = 0$ and $f'(x) = 2 - 2\cos 2x \geq 0$ for $0 < x < \frac{1}{2}\pi \Rightarrow f \geq 0 \Rightarrow g'(x) \geq 0 \Rightarrow g$ increasing. Mimicking the conclusion of (i) (c), the reader should now be able to complete the solution.

Model Solution

Part (i)(a)

Let $f(x) = \sin x - x \cos x$. Differentiating:

$f'(x) = \cos x - (\cos x - x \sin x) = x \sin x$

For $0 \leqslant x \leqslant \frac{1}{2}\pi$, we have $x \geqslant 0$ and $\sin x \geqslant 0$, so $f'(x) = x \sin x \geqslant 0$.

Therefore $f$ is increasing on $[0, \frac{1}{2}\pi]$.

Since $f(0) = \sin 0 - 0 \cdot \cos 0 = 0$ and $f$ is increasing, we deduce $f(x) \geqslant f(0) = 0$, i.e.

$\sin x - x \cos x \geqslant 0 \quad \text{for } 0 \leqslant x \leqslant \tfrac{1}{2}\pi. \qquad \blacksquare$

Part (i)(b)

We are given that $\frac{\mathrm{d}}{\mathrm{d}x}(\arcsin x) \geqslant 1$ for $0 \leqslant x < 1$.

Since $\frac{\mathrm{d}}{\mathrm{d}x}(x) = 1$, we can write:

$\frac{\mathrm{d}}{\mathrm{d}x}(\arcsin x) \geqslant \frac{\mathrm{d}}{\mathrm{d}x}(x) \quad \Longrightarrow \quad \frac{\mathrm{d}}{\mathrm{d}x}(\arcsin x - x) \geqslant 0$

So $h(x) = \arcsin x - x$ is increasing on $[0, 1)$.

Since $h(0) = \arcsin 0 - 0 = 0$ and $h$ is increasing:

$h(x) \geqslant h(0) = 0 \quad \Longrightarrow \quad \arcsin x \geqslant x \quad \text{for } 0 \leqslant x < 1. \qquad \blacksquare$

Part (i)(c)

Let $g(x) = x \operatorname{cosec} x = \dfrac{x}{\sin x}$ for $0 < x < \frac{1}{2}\pi$. Differentiating using the quotient rule:

$g'(x) = \frac{\sin x - x \cos x}{\sin^2 x}$

From part (i)(a), $\sin x - x \cos x > 0$ for $0 < x < \frac{1}{2}\pi$ (strict inequality since $x > 0$ and $\sin x > 0$). Also $\sin^2 x > 0$ in this interval. Therefore $g'(x) > 0$, so $g$ is strictly increasing.

Now let $0 < x < 1$ and set $u = \arcsin x$, so $u \in (0, \frac{1}{2}\pi)$.

From part (i)(b): $u = \arcsin x \geqslant x$, i.e. $u \geqslant x$.

Since $u \geqslant x > 0$ and $g$ is increasing: $g(u) \geqslant g(x)$, which gives:

$u \operatorname{cosec} u \geqslant x \operatorname{cosec} x$

Now $u = \arcsin x$ means $\sin u = x$, so $\operatorname{cosec} u = \frac{1}{x}$. Substituting:

$\frac{\arcsin x}{x} \geqslant x \operatorname{cosec} x \quad \text{for } 0 < x < 1. \qquad \blacksquare$

Part (ii)

Let $g(x) = \dfrac{\tan x}{x}$ for $0 < x < \frac{1}{2}\pi$. Differentiating using the quotient rule:

$g'(x) = \frac{x \sec^2 x - \tan x}{x^2}$

We need to show the numerator is non-negative, i.e. $x \sec^2 x - \tan x \geqslant 0$.

Writing $\sec^2 x = \frac{1}{\cos^2 x}$ and $\tan x = \frac{\sin x}{\cos x}$:

$x \sec^2 x - \tan x = \frac{x - \sin x \cos x}{\cos^2 x} = \frac{2x - 2\sin x \cos x}{2\cos^2 x} = \frac{2x - \sin 2x}{2\cos^2 x}$

Let $h(x) = 2x - \sin 2x$. Then $h(0) = 0$ and $h'(x) = 2 - 2\cos 2x \geqslant 0$ for all $x$ (since $\cos 2x \leqslant 1$).

So $h$ is increasing, giving $h(x) \geqslant h(0) = 0$ for $0 < x < \frac{1}{2}\pi$.

Since $\cos^2 x > 0$ in this interval, we have $g'(x) \geqslant 0$, so $g$ is increasing.

Now we are given that $\frac{\mathrm{d}}{\mathrm{d}x}(\arctan x) \leqslant 1$ for $x \geqslant 0$. Let $k(x) = x - \arctan x$. Then $k'(x) = 1 - \frac{1}{1+x^2} \geqslant 0$ and $k(0) = 0$, so $k(x) \geqslant 0$, giving $x \geqslant \arctan x$ for $x \geqslant 0$.

For $0 < x < \frac{1}{2}\pi$, let $v = \arctan x$. Then $v \in (0, \frac{1}{2}\pi)$ and from above, $v \leqslant x$.

Since $g$ is increasing and $0 < v \leqslant x < \frac{1}{2}\pi$: $g(v) \leqslant g(x)$, which gives:

$\frac{\tan v}{v} \leqslant \frac{\tan x}{x}$

Now $v = \arctan x$ means $\tan v = x$. Substituting:

$\frac{x}{\arctan x} \leqslant \frac{\tan x}{x}$

Multiplying both sides by $x \cdot \arctan x$ (both positive):

$x^2 \leqslant (\tan x)(\arctan x) \quad \text{for } 0 < x < \tfrac{1}{2}\pi. \qquad \blacksquare$

Examiner Notes

Again, despite the obvious presence of inequalities in the question, this was another very popular question, and was generally well-handled very capably in part (i), where the structure of the question provided the necessary support for successful progress to be made here. Part (ii) was less popular and less well-handled, even though the only significant difference between this and (i)(c) was (effectively) that the direction of the inequality was reversed. Although the intervals under consideration were clearly flagged, many candidates omitted to consider that, having shown the function increasing on this interval, they still needed to show something simple such as $f(0) = 0$ in order to show that $f(x) \ge 0$ on this interval. A few also thought that $f'(x)$ increasing implied that $f(x)$ was also increasing.

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Question 4

Topic: 三角方程与恒等式 Trigonometric Equations and Identities  |  Difficulty: Challenging  |  Marks: 20

Problem

4 (i) Find all the values of $\theta$, in the range $0^\circ < \theta < 180^\circ$, for which $\cos \theta = \sin 4\theta$. Hence show that

$\sin 18^\circ = \frac{1}{4} \left( \sqrt{5} - 1 \right).$

(ii) Given that

$4 \sin^2 x + 1 = 4 \sin^2 2x,$

find all possible values of $\sin x$, giving your answers in the form $p + q\sqrt{5}$ where $p$ and $q$ are rational numbers.

(iii) Hence find two values of $\alpha$ with $0^\circ < \alpha < 90^\circ$ for which

$\sin^2 3\alpha + \sin^2 5\alpha = \sin^2 6\alpha.$

Hint

Q4 (i) Using $\sin A = \cos(90^\circ - A)$ gives $\theta = 360n \pm (90^\circ - 4\theta)$ – Note that you certainly should be aware of the periodicities of the basic trig. functions $\Rightarrow 5\theta = 360n + 90^\circ$ or $3\theta = 360n + 90^\circ$. These give either $\theta = 72n + 18^\circ \Rightarrow \theta = 18^\circ, 90^\circ, 162^\circ$ or $\theta = 120n + 30^\circ \Rightarrow \theta = 30^\circ, 150^\circ$.

Now using the double-angle formulae for sine (twice) and cosine, we have $c = 2.2sc.(1 - 2s^2)$. We can discount $c = 0$ for $\theta = 18^\circ$, so that $1 = 4s(1 - 2s^2)$ which gives the cubic equation in $s = \sin \theta$, $8s^3 - 4s + 1 = 0 \Rightarrow (2s - 1)(4s^2 + 2s - 1) = 0$. Again, we can discount $c = \frac{1}{2}$ for $\theta = 18^\circ$) which leaves us with $\sin 18^\circ$ the positive root (as $18^\circ$ is acute) from the two possible solutions of this

quadratic; namely, $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$.

(ii) Using the double-angle formula for sine, we have $4s^2 + 1 = 16s^2(1 - s^2) \Rightarrow 0 = 16s^4 - 12s^2 + 1$

$\Rightarrow s^2 = \frac{12 \pm \sqrt{80}}{32} = \frac{3 \pm \sqrt{5}}{8}$. At first, this may look like a problem, but bear in mind that we want

it to be a perfect square. Proceeding with this in mind, $s^2 = \frac{6 \pm 2\sqrt{5}}{16} = \left( \frac{\sqrt{5} \pm 1}{4} \right)^2$ so that we have

the four answers, $\sin x = \pm \left( \frac{\sqrt{5} \pm 1}{4} \right)$.

(iii) To make the connection between this part and the previous one requires nothing more than division by 4 to get $\sin^2 x + \frac{1}{4} = \sin^2 2x$, and the solution $x = 3\alpha = 18^\circ, 5\alpha = 30^\circ \Rightarrow \alpha = 6^\circ$ immediately presents itself from part (ii). However, in order to deduce a second solution (noting that $\alpha = 45^\circ$ is easily seen to satisfy the given equation), it is important to be prepared to be a bit flexible and use your imagination. The other possible angles that are “related” to $18^\circ$ and might satisfy (ii)‘s equation, can be looked-for, provided that $\sin 5\alpha = \pm \frac{1}{2}$ (and there are many possibilities here also). A little searching and/or thought reveals

$\sin x = -\left( \frac{\sqrt{5} - 1}{4} \right) \Rightarrow 3\alpha = 180^\circ + 18^\circ = 198^\circ$ also works, since $5\alpha = 330^\circ$ has $\sin 5\alpha = -\frac{1}{2}$,

and the second acute answer is $\alpha = 66^\circ$.

Model Solution

Part (i)

We solve $\cos \theta = \sin 4\theta$ for $0^\circ < \theta < 180^\circ$.

Using the identity $\sin A = \cos(90^\circ - A)$, we need $\cos \theta = \cos(90^\circ - 4\theta)$.

The general solution of $\cos A = \cos B$ is $A = 360^\circ n \pm B$, so:

$\theta = 360^\circ n \pm (90^\circ - 4\theta)$

Case 1: $\theta = 360^\circ n + 90^\circ - 4\theta$, giving $5\theta = 360^\circ n + 90^\circ$, so $\theta = 72^\circ n + 18^\circ$.

• $n = 0$: $\theta = 18^\circ$
• $n = 1$: $\theta = 90^\circ$
• $n = 2$: $\theta = 162^\circ$

Case 2: $\theta = 360^\circ n - (90^\circ - 4\theta)$, giving $\theta = 360^\circ n - 90^\circ + 4\theta$, so $-3\theta = 360^\circ n - 90^\circ$, i.e. $\theta = 30^\circ - 120^\circ n$.

• $n = 0$: $\theta = 30^\circ$
• $n = -1$: $\theta = 150^\circ$

So the solutions are $\theta = 18^\circ, 30^\circ, 90^\circ, 150^\circ, 162^\circ$.

Showing $\sin 18^\circ = \frac{1}{4}(\sqrt{5} - 1)$:

Since $\theta = 18^\circ$ satisfies $\cos \theta = \sin 4\theta$, and $\cos 18^\circ \neq 0$:

$1 = \frac{\sin 4\theta}{\cos \theta} \bigg|_{\theta = 18^\circ}$

We compute $\sin 4\theta$ in terms of $s = \sin \theta$:

$\sin 4\theta = 2 \sin 2\theta \cos 2\theta = 2(2\sin\theta\cos\theta)(1 - 2\sin^2\theta) = 4s\cos\theta(1 - 2s^2)$

So $\cos \theta = 4s\cos\theta(1 - 2s^2)$. Dividing by $\cos \theta \neq 0$:

$1 = 4s(1 - 2s^2) = 4s - 8s^3$

$8s^3 - 4s + 1 = 0$

We check $s = \frac{1}{2}$: $8(\frac{1}{8}) - 4(\frac{1}{2}) + 1 = 1 - 2 + 1 = 0$. So $(2s - 1)$ is a factor.

Performing the division $8s^3 - 4s + 1 = (2s - 1)(4s^2 + 2s - 1)$:

$8s^3 - 4s + 1 = (2s-1)(4s^2 + 2s - 1)$

Expanding the right side: $8s^3 + 4s^2 - 2s - 4s^2 - 2s + 1 = 8s^3 - 4s + 1$. Confirmed.

Since $\sin 18^\circ \neq \frac{1}{2}$ (as $18^\circ \neq 30^\circ$), we solve $4s^2 + 2s - 1 = 0$:

$s = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$

Since $18^\circ$ is acute, $\sin 18^\circ > 0$, so we take the positive root:

$\sin 18^\circ = \frac{\sqrt{5} - 1}{4}. \qquad \blacksquare$

Part (ii)

Given $4\sin^2 x + 1 = 4\sin^2 2x$. Using $\sin 2x = 2\sin x\cos x$:

$4\sin^2 x + 1 = 4 \cdot 4\sin^2 x \cos^2 x = 16\sin^2 x(1 - \sin^2 x)$

Let $s = \sin x$:

$4s^2 + 1 = 16s^2(1 - s^2) = 16s^2 - 16s^4$

$16s^4 - 12s^2 + 1 = 0$

Using the quadratic formula with $u = s^2$:

$u = \frac{12 \pm \sqrt{144 - 64}}{32} = \frac{12 \pm \sqrt{80}}{32} = \frac{12 \pm 4\sqrt{5}}{32} = \frac{3 \pm \sqrt{5}}{8}$

We need to show these are perfect squares. Writing $3 \pm \sqrt{5} = \frac{6 \pm 2\sqrt{5}}{2}$:

$\frac{3 \pm \sqrt{5}}{8} = \frac{6 \pm 2\sqrt{5}}{16} = \left(\frac{\sqrt{5} \pm 1}{4}\right)^2$

since $(\sqrt{5} \pm 1)^2 = 5 \pm 2\sqrt{5} + 1 = 6 \pm 2\sqrt{5}$.

Therefore $s^2 = \left(\frac{\sqrt{5} \pm 1}{4}\right)^2$, giving:

$\sin x = \pm\frac{\sqrt{5} + 1}{4} \quad \text{or} \quad \sin x = \pm\frac{\sqrt{5} - 1}{4}$

In the form $p + q\sqrt{5}$: $\sin x = \pm\frac{1}{4} \pm \frac{1}{4}\sqrt{5}$.

Part (iii)

We need $\sin^2 3\alpha + \sin^2 5\alpha = \sin^2 6\alpha$ for $0^\circ < \alpha < 90^\circ$.

Dividing the equation from part (ii) by 4:

$\sin^2 x + \frac{1}{4} = \sin^2 2x$

Comparing with the equation in (iii): $\sin^2 3\alpha + \sin^2 5\alpha = \sin^2 6\alpha$, we see this matches if $\sin^2 5\alpha = \frac{1}{4}$, i.e. $\sin 5\alpha = \pm\frac{1}{2}$.

Case 1: $\sin 5\alpha = \frac{1}{2}$, so $5\alpha = 30^\circ$ (giving $\alpha = 6^\circ$) or $5\alpha = 150^\circ$ (giving $\alpha = 30^\circ$).

For $\alpha = 6^\circ$: $3\alpha = 18^\circ$, $6\alpha = 36^\circ$.

From part (ii), $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ is one of the solutions, and the equation $\sin^2 x + \frac{1}{4} = \sin^2 2x$ with $x = 18^\circ$ gives $\sin^2 36^\circ$, so $\sin^2 18^\circ + \sin^2 30^\circ = \sin^2 36^\circ$. This works, so $\alpha = 6^\circ$ is a solution.

For $\alpha = 30^\circ$: $3\alpha = 90^\circ$, $\sin 90^\circ = 1$; $5\alpha = 150^\circ$, $\sin 150^\circ = \frac{1}{2}$; $6\alpha = 180^\circ$, $\sin 180^\circ = 0$. Then $\sin^2 90^\circ + \sin^2 150^\circ = 1 + \frac{1}{4} \neq 0 = \sin^2 180^\circ$. This does not work.

Case 2: $\sin 5\alpha = -\frac{1}{2}$, so $5\alpha = 210^\circ$ (giving $\alpha = 42^\circ$) or $5\alpha = 330^\circ$ (giving $\alpha = 66^\circ$).

For $\alpha = 66^\circ$: $3\alpha = 198^\circ$, $6\alpha = 396^\circ = 36^\circ$.

$\sin^2 198^\circ = \sin^2 18^\circ$ (since $\sin 198^\circ = -\sin 18^\circ$), and $\sin^2 396^\circ = \sin^2 36^\circ$.

So $\sin^2 198^\circ + \sin^2 330^\circ = \sin^2 18^\circ + \frac{1}{4} = \sin^2 36^\circ$ (from part (ii) with $x = 18^\circ$). This works, so $\alpha = 66^\circ$ is a solution.

The two values are $\alpha = 6^\circ$ and $\alpha = 66^\circ$.

Examiner Notes

This question was the first of the really popular ones to attract relatively low scores overall. In the opening part, it had been expected that candidates would employ that most basic of trig. identities, $\sin A = \cos(90^\circ - A)$, in order to find the required values of $\theta$, but the vast majority went straight into double-angles and quadratics in terms of $\sin \theta$ instead, which had been expected to follow the initial work; this meant that many candidates were unable to explain convincingly why the given value of $\sin 18^\circ$ was as claimed.

Despite the relatively straightforward trig. methods that were required in this question, with part (ii) broadly approachable in the same way as the second part of (i), the lack of a clear-minded strategy proved to be a big problem for most attempters, and the connection between parts (ii) and (iii) was seldom spotted – namely, to divide through by 4 and realise that $\sin 5\alpha$ must be $\pm \frac{1}{2}$. Many spotted the solution $\alpha = 6^\circ$, but few got further than this because they were stuck exclusively on $\sin 30^\circ = +\frac{1}{2}$.

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Question 5

Topic: 向量几何 Vector Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The points $A$ and $B$ have position vectors a and b with respect to an origin $O$, and $O$, $A$ and $B$ are non-collinear. The point $C$, with position vector c, is the reflection of $B$ in the line through $O$ and $A$. Show that c can be written in the form

$\mathbf{c} = \lambda \mathbf{a} - \mathbf{b}$

where $\lambda = \frac{2 \, \mathbf{a.b}}{\mathbf{a.a}}$.

The point $D$, with position vector d, is the reflection of $C$ in the line through $O$ and $B$. Show that d can be written in the form

$\mathbf{d} = \mu \mathbf{b} - \lambda \mathbf{a}$

for some scalar $\mu$ to be determined.

Given that $A$, $B$ and $D$ are collinear, find the relationship between $\lambda$ and $\mu$. In the case $\lambda = -\frac{1}{2}$, determine the cosine of $\angle AOB$ and describe the relative positions of $A$, $B$ and $D$.

Hint

Q5 The simplest way to do this is to realise that $OA$ is the bisector of $\angle BOC$, so that $A$ is on the diagonal $OA'$ of parallelogram $OBA'C$ (in fact, since $OB = OC$, it is a rhombus) $\Rightarrow \mathbf{b} + \mathbf{c} = \lambda \mathbf{a}$ for some $\lambda$ (giving the first part of the result). Also, as $BC$ is perpendicular to $OA$, $(\mathbf{b} - \mathbf{c}) \cdot \mathbf{a} = 0$

$\Rightarrow (2\mathbf{b} - \lambda \mathbf{a}) \cdot \mathbf{a} = 0 \Rightarrow \lambda = 2 \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} \right)$.

Similarly (replacing $\mathbf{a}$ by $\mathbf{b}$ and $\mathbf{b}$ by $\mathbf{c}$ in the above), we have $\mathbf{d} = k\mathbf{b} - \mathbf{c}$ where $k = 2 \left( \frac{\mathbf{b} \cdot \mathbf{c}}{\mathbf{b} \cdot \mathbf{b}} \right)$

$= 2 \left( \frac{\mathbf{b} \cdot \lambda \mathbf{a} - \mathbf{b} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) = 2\lambda \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) - 2 \Rightarrow \mathbf{d} = \left( 2\lambda \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) - 2 \right) \mathbf{b} - (\lambda \mathbf{a} - \mathbf{b}) = \mu \mathbf{b} - \lambda \mathbf{a}$ where $\mu = 2\lambda \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) - 1$ or $4 \left[ \frac{[\mathbf{a} \cdot \mathbf{b}]^2}{[\mathbf{a} \cdot \mathbf{a}][\mathbf{b} \cdot \mathbf{b}]} \right] - 1$.

Now $A, B$ and $D$ are collinear if and only if $\overrightarrow{AD} = \mu \mathbf{b} - (\lambda + 1)\mathbf{a}$ is a multiple of $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ $\Leftrightarrow t(\mathbf{b} - \mathbf{a}) = \mu \mathbf{b} - (\lambda + 1)\mathbf{a}$ for some $t (\neq 0)$.

Comparing coefficients of a and b then gives $(t =) \mu = \lambda + 1$.

In the case when $\lambda = -\frac{1}{2}, \mu = \frac{1}{2}$ and $D$ is the midpoint of $AB$.

Finally, $\mu = \frac{1}{2} \Rightarrow \frac{1}{2} = 4 \left( \frac{[\mathbf{a} \bullet \mathbf{b}]^2}{[\mathbf{a} \bullet \mathbf{a}][\mathbf{b} \bullet \mathbf{b}]} \right) - 1 = 4 \left( \frac{\mathbf{a} \bullet \mathbf{b}}{ab} \right)^2 - 1$, and using the scalar product formula

$\cos \theta = \frac{\mathbf{a} \bullet \mathbf{b}}{ab}$ gives $\cos \theta = -\sqrt{\frac{3}{8}}$. [Note that $\mathbf{a} \bullet \mathbf{b}$ has the same sign as $\lambda$.]

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Model Solution

Showing $\mathbf{c} = \lambda \mathbf{a} - \mathbf{b}$ where $\lambda = \frac{2\,\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}}$:

The point $C$ is the reflection of $B$ in the line through $O$ and $A$. The midpoint of $BC$ lies on this line, so:

$\frac{\mathbf{b} + \mathbf{c}}{2} = t\,\mathbf{a} \quad \text{for some scalar } t$

$\mathbf{b} + \mathbf{c} = 2t\,\mathbf{a}$

Setting $\lambda = 2t$:

$\mathbf{c} = \lambda \mathbf{a} - \mathbf{b} \qquad \blacksquare$

Since reflection preserves distance from the line, the segment $BC$ is perpendicular to $OA$, so:

$(\mathbf{c} - \mathbf{b}) \cdot \mathbf{a} = 0$

Substituting $\mathbf{c} = \lambda \mathbf{a} - \mathbf{b}$:

$(\lambda \mathbf{a} - 2\mathbf{b}) \cdot \mathbf{a} = 0$

$\lambda(\mathbf{a} \cdot \mathbf{a}) - 2(\mathbf{a} \cdot \mathbf{b}) = 0$

$\lambda = \frac{2\,\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} \qquad \blacksquare$

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Showing $\mathbf{d} = \mu \mathbf{b} - \lambda \mathbf{a}$:

The point $D$ is the reflection of $C$ in the line through $O$ and $B$. By the same argument as above (replacing $\mathbf{a}$ with $\mathbf{b}$ and $\mathbf{b}$ with $\mathbf{c}$), the midpoint of $CD$ lies on the line through $O$ and $B$, so:

$\mathbf{c} + \mathbf{d} = k\,\mathbf{b} \quad \text{for some scalar } k$

$\mathbf{d} = k\,\mathbf{b} - \mathbf{c}$

And $CD$ is perpendicular to $OB$, so $(\mathbf{d} - \mathbf{c}) \cdot \mathbf{b} = 0$, giving:

$(k\,\mathbf{b} - 2\mathbf{c}) \cdot \mathbf{b} = 0$

$k = \frac{2\,\mathbf{b} \cdot \mathbf{c}}{\mathbf{b} \cdot \mathbf{b}}$

Now we compute $\mathbf{b} \cdot \mathbf{c}$ using $\mathbf{c} = \lambda \mathbf{a} - \mathbf{b}$:

$\mathbf{b} \cdot \mathbf{c} = \lambda(\mathbf{a} \cdot \mathbf{b}) - \mathbf{b} \cdot \mathbf{b}$

So:

$k = \frac{2[\lambda(\mathbf{a} \cdot \mathbf{b}) - \mathbf{b} \cdot \mathbf{b}]}{\mathbf{b} \cdot \mathbf{b}} = \frac{2\lambda(\mathbf{a} \cdot \mathbf{b})}{\mathbf{b} \cdot \mathbf{b}} - 2$

Substituting $\mathbf{c} = \lambda \mathbf{a} - \mathbf{b}$ into $\mathbf{d} = k\,\mathbf{b} - \mathbf{c}$:

$\mathbf{d} = k\,\mathbf{b} - (\lambda \mathbf{a} - \mathbf{b}) = (k + 1)\,\mathbf{b} - \lambda \mathbf{a}$

Setting $\mu = k + 1$:

$\boxed{\mathbf{d} = \mu \mathbf{b} - \lambda \mathbf{a}} \qquad \blacksquare$

where:

$\mu = \frac{2\lambda(\mathbf{a} \cdot \mathbf{b})}{\mathbf{b} \cdot \mathbf{b}} - 1$

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Collinearity condition:

$A$, $B$, $D$ are collinear if and only if $\overrightarrow{AD}$ is parallel to $\overrightarrow{AB}$.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, \qquad \overrightarrow{AD} = \mathbf{d} - \mathbf{a} = \mu \mathbf{b} - (\lambda + 1)\mathbf{a}$

For these to be parallel: $\overrightarrow{AD} = t\,\overrightarrow{AB}$ for some scalar $t$, so:

$\mu \mathbf{b} - (\lambda + 1)\mathbf{a} = t\,\mathbf{b} - t\,\mathbf{a}$

Since $O$, $A$, $B$ are non-collinear, $\mathbf{a}$ and $\mathbf{b}$ are not parallel, so we compare coefficients:

$\mu = t \qquad \text{and} \qquad \lambda + 1 = t$

$\boxed{\mu = \lambda + 1}$

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Case $\lambda = -\frac{1}{2}$:

From $\mu = \lambda + 1$:

$\mu = -\frac{1}{2} + 1 = \frac{1}{2}$

So:

$\mathbf{d} = \frac{1}{2}\mathbf{b} - \left(-\frac{1}{2}\right)\mathbf{a} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{\mathbf{a} + \mathbf{b}}{2}$

This is the midpoint of $AB$, so $D$ is the midpoint of $AB$.

To find $\cos \angle AOB$: using $\lambda = \frac{2\,\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}} = -\frac{1}{2}$:

$\mathbf{a} \cdot \mathbf{b} = -\frac{1}{4}\,\mathbf{a} \cdot \mathbf{a} = -\frac{1}{4}|\mathbf{a}|^2$

Also, since $\lambda$ has the same sign as $\mathbf{a} \cdot \mathbf{b}$ (because $|\mathbf{a}|^2 > 0$), we know $\mathbf{a} \cdot \mathbf{b} < 0$, confirming the angle is obtuse.

Now using $\mu = \frac{2\lambda(\mathbf{a} \cdot \mathbf{b})}{\mathbf{b} \cdot \mathbf{b}} - 1 = \frac{1}{2}$:

$\frac{2\lambda(\mathbf{a} \cdot \mathbf{b})}{|\mathbf{b}|^2} = \frac{3}{2}$

Substituting $\mathbf{a} \cdot \mathbf{b} = -\frac{1}{4}|\mathbf{a}|^2$ and $\lambda = -\frac{1}{2}$:

$\frac{2 \cdot (-\frac{1}{2}) \cdot (-\frac{1}{4}|\mathbf{a}|^2)}{|\mathbf{b}|^2} = \frac{3}{2}$

$\frac{|\mathbf{a}|^2}{4|\mathbf{b}|^2} = \frac{3}{2} \qquad \Longrightarrow \qquad \frac{|\mathbf{a}|^2}{|\mathbf{b}|^2} = 6$

Now:

$\cos \angle AOB = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{-\frac{1}{4}|\mathbf{a}|^2}{|\mathbf{a}||\mathbf{b}|} = -\frac{|\mathbf{a}|}{4|\mathbf{b}|} = -\frac{\sqrt{6}}{4}$

$\boxed{\cos \angle AOB = -\frac{\sqrt{6}}{4}}$

Examiner Notes

This vectors question was neither popular nor successful overall. For the most part this seemed to be due to the fact that candidates, although they are happy to work with scalar parameters – as involved in the vector equation of a line, for instance – they are far less happy to interpret them geometrically. Many other students clearly dislike non-numerical vector questions. Having said that, attempts generally fell into one of the two extreme camps of ‘very good’ or ‘very poor’. More confident candidates managed the first result and realised that a “similarity” approach killed off the second part also, although efforts to tidy up answers were frequently littered with needless errors that came back to penalise the candidates when they attempted to use them later on. Many candidates noted that $D$ was between $A$ and $B$, but failed to realise it was actually the midpoint of $AB$. In the very final part, it was often the case that candidates overlooked the negative sign of $\cos \theta$, even when the remainder of their working was broadly correct.

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Question 6

Topic: 积分技巧 Integration Techniques  |  Difficulty: Challenging  |  Marks: 20

Problem

6 For any given function $f$, let

$I = \int [f'(x)]^2 [f(x)]^n dx \, , \qquad \text{(*)}$

where $n$ is a positive integer. Show that, if $f(x)$ satisfies $f''(x) = kf(x)f'(x)$ for some constant $k$, then ($*$) can be integrated to obtain an expression for $I$ in terms of $f(x)$, $f'(x)$, $k$ and $n$.

(i) Verify your result in the case $f(x) = \tan x$. Hence find

$\int \frac{\sin^4 x}{\cos^8 x} \, dx \, .$

(ii) Find

$\int \sec^2 x \, (\sec x + \tan x)^6 \, dx \, .$

Hint

Q6 To begin with, it is essential to realise that the integrand of $I = \int [f'(x)]^2 [f(x)]^n dx$ must have its two components split up suitably so that integration by parts can be employed. Thus

$I = \int [f'(x)] \times \{ [f'(x)][f(x)]^n \} dx = f'(x) \times \frac{1}{n+1} [f(x)]^{n+1} - \int \left( [f''(x)] \times \frac{1}{n+1} [f(x)]^{n+1} \right) dx.$

Now (and not earlier) is the opportune moment to use the given relationship $f''(x) = kf(x)f'(x)$,

so that $I = f'(x) \times \frac{1}{n+1} [f(x)]^{n+1} - \int \left( kf'(x) \times \frac{1}{n+1} [f(x)]^{n+2} \right) dx$, which is now directly integrable

as $f'(x) \times \frac{1}{n+1} [f(x)]^{n+1} - \frac{1}{(n+1)(n+3)} \times k[f(x)]^{n+3} (+ C)$.

(i) For $f(x) = \tan x$, $f'(x) = \sec^2 x$ and $f''(x) = 2 \sec^2 x \tan x = kf(x)f'(x)$ with $k = 2$.

Also, differentiating $I = \frac{\sec^2 x \tan^{n+1} x}{n+1} - \frac{2 \tan^{n+3} x}{(n+1)(n+3)}$ gives

$\frac{dI}{dx} = \frac{1}{n+1} (\sec^2 x.(n+1) \tan^n x.\sec^2 x + 2 \sec x.\sec x \tan x.\tan^{n+1} x)$

$- \frac{1}{(n+1)(n+3)} (2(n+3) \tan^{n+2} x.\sec^2 x) = \sec^4 x \tan^n x = (f'(x))^2 \times (f(x))^n \text{ as required,}$

although this could be verified in reverse using integration. Using this result directly in the first given integral is now relatively straightforward:

$\int \frac{\sin^4 x}{\cos^8 x} dx = \int \sec^4 x \tan^4 x dx = \frac{\sec^2 x \tan^5 x}{5} - \frac{2 \tan^7 x}{35} + C.$

(ii) Hopefully, all this differentiating of sec and tan functions may have helped you identify the right sort of area to be searching for ideas with the second of the given integrals.

If $f(x) = \sec x + \tan x$, $f'(x) = \sec x \tan x + \sec^2 x = \sec x(\sec x + \tan x)$

and $f''(x) = \sec^2 x(\sec x + \tan x) + \sec x \tan x(\sec x + \tan x)$

$= \sec x (\sec x + \tan x)^2 = kf(x)f'(x)$ with $k = 1$.

Then $\int \sec^2 x (\sec x + \tan x)^6 dx = \int \{ \sec x (\sec x + \tan x) \}^2 \times (\sec x + \tan x)^4 dx$

$= \frac{\sec x (\sec x + \tan x)^6}{5} - \frac{(\sec x + \tan x)^7}{35} + C$

Model Solution

General result:

We evaluate $I = \int [f'(x)]^2 [f(x)]^n \, dx$ by splitting the integrand and using integration by parts.

Write $[f'(x)]^2 [f(x)]^n = f'(x) \cdot \bigl(f'(x)\,[f(x)]^n\bigr)$, so:

$I = \int f'(x) \cdot \bigl(f'(x)\,[f(x)]^n\bigr) \, dx$

Let $u = f'(x)$ and $\dfrac{\mathrm{d}v}{\mathrm{d}x} = f'(x)\,[f(x)]^n$. Then $v = \dfrac{[f(x)]^{n+1}}{n+1}$ and $\dfrac{\mathrm{d}u}{\mathrm{d}x} = f''(x)$. Integration by parts gives:

$I = f'(x) \cdot \frac{[f(x)]^{n+1}}{n+1} - \int f''(x) \cdot \frac{[f(x)]^{n+1}}{n+1} \, dx$

Now we use $f''(x) = k\,f(x)\,f'(x)$:

$I = \frac{f'(x)\,[f(x)]^{n+1}}{n+1} - \frac{k}{n+1}\int [f(x)]^{n+2}\,f'(x) \, dx$

The remaining integral is directly computable since $\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl([f(x)]^{n+3}\bigr) = (n+3)\,[f(x)]^{n+2}\,f'(x)$:

$\boxed{I = \frac{f'(x)\,[f(x)]^{n+1}}{n+1} - \frac{k\,[f(x)]^{n+3}}{(n+1)(n+3)} + C} \qquad \blacksquare$

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Part (i): Verification for $f(x) = \tan x$

We have $f'(x) = \sec^2 x$ and $f''(x) = 2\sec^2 x \tan x = 2\,f(x)\,f'(x)$, so $k = 2$.

Our formula gives:

$\int \sec^4 x \cdot \tan^n x \, dx = \frac{\sec^2 x \cdot \tan^{n+1} x}{n+1} - \frac{2\tan^{n+3} x}{(n+1)(n+3)} + C$

Verification: Differentiating the right side to check we recover $\sec^4 x \tan^n x$.

Let $R = \dfrac{\sec^2 x \tan^{n+1} x}{n+1} - \dfrac{2\tan^{n+3} x}{(n+1)(n+3)}$.

For the first term, using the product rule on $\sec^2 x \cdot \tan^{n+1} x$:

$\frac{\mathrm{d}}{\mathrm{d}x}\bigl(\sec^2 x \cdot \tan^{n+1} x\bigr) = 2\sec^2 x \tan x \cdot \tan^{n+1} x + \sec^2 x \cdot (n+1)\tan^n x \sec^2 x$

$= 2\sec^2 x \tan^{n+2} x + (n+1)\sec^4 x \tan^n x$

For the second term:

$\frac{\mathrm{d}}{\mathrm{d}x}\bigl(\tan^{n+3} x\bigr) = (n+3)\tan^{n+2} x \sec^2 x$

Combining:

$\frac{\mathrm{d}R}{\mathrm{d}x} = \frac{2\sec^2 x \tan^{n+2} x + (n+1)\sec^4 x \tan^n x}{n+1} - \frac{2(n+3)\tan^{n+2} x \sec^2 x}{(n+1)(n+3)}$

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

$= \frac{(n+1)\sec^4 x \tan^n x}{n+1} = \sec^4 x \tan^n x \qquad \checkmark$

Hence finding $\displaystyle\int \frac{\sin^4 x}{\cos^8 x}\,dx$:

$\int \frac{\sin^4 x}{\cos^8 x}\,dx = \int \frac{\sin^4 x}{\cos^4 x} \cdot \frac{1}{\cos^4 x}\,dx = \int \tan^4 x \sec^4 x \, dx$

This is $\int [f'(x)]^2 [f(x)]^4 \, dx$ with $f(x) = \tan x$, $n = 4$, $k = 2$. Applying our formula:

$\int \tan^4 x \sec^4 x \, dx = \frac{\sec^2 x \tan^5 x}{5} - \frac{2\tan^7 x}{5 \cdot 7} + C$

$\boxed{\int \frac{\sin^4 x}{\cos^8 x}\,dx = \frac{\sec^2 x \tan^5 x}{5} - \frac{2\tan^7 x}{35} + C}$

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Part (ii): Finding $\displaystyle\int \sec^2 x\,(\sec x + \tan x)^6\,dx$

We try $f(x) = \sec x + \tan x$. Then:

$f'(x) = \sec x \tan x + \sec^2 x = \sec x(\sec x + \tan x) = \sec x \cdot f(x)$

Computing $f''(x)$ using the product rule on $\sec x \cdot f(x)$:

$f''(x) = \sec x \tan x \cdot f(x) + \sec x \cdot f'(x) = \sec x \tan x \cdot f(x) + \sec x \cdot \sec x \cdot f(x)$

$= \sec x(\tan x + \sec x)\,f(x) = \sec x \cdot f(x) \cdot f(x) = \sec x \cdot [f(x)]^2$

Also $f'(x) = \sec x \cdot f(x)$, so $\sec x = \dfrac{f'(x)}{f(x)}$, giving:

$f''(x) = \frac{f'(x)}{f(x)} \cdot [f(x)]^2 = f(x)\,f'(x)$

So $f''(x) = k\,f(x)\,f'(x)$ with $k = 1$.

Now we need to express the integrand in the form $[f'(x)]^2\,[f(x)]^n$:

$\sec^2 x\,(\sec x + \tan x)^6 = [\sec x]^2 \cdot [f(x)]^6$

Since $f'(x) = \sec x \cdot f(x)$, we have $\sec x = \dfrac{f'(x)}{f(x)}$, so:

$[\sec x]^2 = \frac{[f'(x)]^2}{[f(x)]^2}$

Therefore:

$\sec^2 x\,(\sec x + \tan x)^6 = \frac{[f'(x)]^2}{[f(x)]^2} \cdot [f(x)]^6 = [f'(x)]^2\,[f(x)]^4$

This is our integral formula with $n = 4$ and $k = 1$:

$\int [f'(x)]^2\,[f(x)]^4\,dx = \frac{f'(x)\,[f(x)]^5}{5} - \frac{[f(x)]^7}{5 \cdot 7} + C$

Substituting back $f(x) = \sec x + \tan x$ and $f'(x) = \sec x(\sec x + \tan x)$:

$\boxed{\int \sec^2 x\,(\sec x + \tan x)^6\,dx = \frac{\sec x\,(\sec x + \tan x)^6}{5} - \frac{(\sec x + \tan x)^7}{35} + C}$

Examiner Notes

This was another very popular question attracting many poor scores. There were several very serious errors on display, including the beliefs that

$\int f(x)^n \, dx = \frac{f(x)^{n+1}}{(n+1)} \quad \text{or} \quad \frac{f(x)^{n+1}}{(n+1)f'(x)}.$

The understanding that the original integral needed to be split as $\int f'(x) \times (f'(x)f(x)^n) \, dx$ before attempting to integrate by parts was largely absent, with many substituting immediately for $f(x)f'(x)$ in terms of $f''(x)$, which really wasn’t helpful at all. Those who got over this initial hurdle generally coped very favourably with the rest of the question.

In (i), it was quite common for candidates to omit verifying the result for $\tan x$.

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Question 7

Topic: 数列与求和 Sequences and Summation  |  Difficulty: Challenging  |  Marks: 20

Problem

7 The two sequences $a_0, a_1, a_2, \dots$ and $b_0, b_1, b_2, \dots$ have general terms $a_n = \lambda^n + \mu^n \quad \text{and} \quad b_n = \lambda^n - \mu^n ,$ respectively, where $\lambda = 1 + \sqrt{2}$ and $\mu = 1 - \sqrt{2}$.

(i) Show that $\sum_{r=0}^{n} b_r = -\sqrt{2} + \frac{1}{\sqrt{2}} a_{n+1}$, and give a corresponding result for $\sum_{r=0}^{n} a_r$.

(ii) Show that, if $n$ is odd, $\sum_{m=0}^{2n} \left( \sum_{r=0}^{m} a_r \right) = \frac{1}{2} b_{n+1}^2 ,$ and give a corresponding result when $n$ is even.

(iii) Show that, if $n$ is even, $\left( \sum_{r=0}^{n} a_r \right)^2 - \sum_{r=0}^{n} a_{2r+1} = 2 ,$ and give a corresponding result when $n$ is odd.

Hint

Q7

(i) Once you have split each series into sums of powers of $\lambda$ and $\mu$ separately, it becomes clear that you are merely dealing with GPs. Thus $\sum_{r=0}^{n} b_r = (1 + \lambda + \lambda^2 + \dots + \lambda^n) - (1 + \mu + \mu^2 + \dots + \mu^n)$

$= \frac{\lambda^{n+1} - 1}{\lambda - 1} - \frac{\mu^{n+1} - 1}{\mu - 1} = \frac{1}{\sqrt{2}}(\lambda^{n+1} - 1 + \mu^{n+1} - 1)$, since $\lambda - 1 = \sqrt{2}$ and $\mu - 1 = -\sqrt{2}$

$= \frac{1}{\sqrt{2}} a_{n+1} - \sqrt{2}$ and, similarly, $\sum_{r=0}^{n} a_r = \frac{\lambda^{n+1} - 1}{\sqrt{2}} - \frac{\mu^{n+1} - 1}{\sqrt{2}} = \frac{1}{\sqrt{2}} b_{n+1}$.

(ii) There is no need to be frightened by the appearance of the nested sums here as the ‘inner sum’ has already been computed: all that is left is to work with the remaining ‘outer sum’ and deal

carefully with the limits: $\sum_{m=0}^{2n} \left( \sum_{r=0}^{m} a_r \right) = \sum_{m=0}^{2n} \left( \frac{1}{\sqrt{2}} b_{m+1} \right) = \frac{1}{\sqrt{2}} \sum_{m=0}^{2n+1} b_m$ (since $b_0 = 0$)

$= \frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}} a_{2n+2} - \sqrt{2} \right) = \frac{1}{2} (\lambda^{2n+2} + \mu^{2n+2} - 2) = \frac{1}{2} \left( [\lambda^{n+1}]^2 - 2[\lambda\mu]^{n+1} + [\mu^{n+1}]^2 \right)$ since $\lambda\mu = -1$

and $n + 1$ is even when $n$ is odd $= \frac{1}{2} (b_{n+1})^2$ when $n$ is odd. However, when $n$ is even, $n + 1$ is odd

and $\sum_{m=0}^{2n} \left( \sum_{r=0}^{m} a_r \right) = \frac{1}{2} (b_{n+1})^2 - 2$ or $\frac{1}{2} (a_{n+1})^2$.

(iii) We already have the result $\left( \sum_{r=0}^{n} a_r \right)^2 = \frac{1}{2} (b_{n+1})^2$, so the only new thing is

$\sum_{r=0}^{n} a_{2r+1} = (\lambda + \lambda^3 + \lambda^5 + \dots + \lambda^{2n+1}) + (\mu + \mu^3 + \mu^5 + \dots + \mu^{2n+1})$, which is still the sum of two

GPs, merely with different common ratios, having sum $\frac{\lambda(\lambda^{2n+2} - 1)}{\lambda^2 - 1} + \frac{\mu(\mu^{2n+2} - 1)}{\mu^2 - 1}$.

Now $\lambda^2 - 1 = 3 + 2\sqrt{2} - 1 = 2(1 + \sqrt{2}) = 2\lambda$ and $\mu^2 - 1 = 3 - 2\sqrt{2} - 1 = 2(1 - \sqrt{2}) = 2\mu$,

so $\sum_{r=0}^{n} a_{2r+1} = \frac{1}{2} (\lambda^{2n+2} + \mu^{2n+2} - 2) = \frac{1}{2} (b_{n+1})^2$ when $n$ is odd, and $\frac{1}{2} (b_{n+1})^2 - 2$ when $n$ is even.

Thus $\left( \sum_{r=0}^{n} a_r \right)^2 - \sum_{r=0}^{n} a_{2r+1} = 0$ when $n$ is odd / $= 2$ when $n$ is even.

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Model Solution

Key facts: $\lambda = 1 + \sqrt{2}$, $\mu = 1 - \sqrt{2}$, so $\lambda + \mu = 2$, $\lambda\mu = 1 - 2 = -1$, and $\lambda - 1 = \sqrt{2}$, $\mu - 1 = -\sqrt{2}$.

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Part (i): Summation results

$\sum_{r=0}^{n} b_r$ splits into two geometric series:

$\sum_{r=0}^{n} b_r = \sum_{r=0}^{n} \lambda^r - \sum_{r=0}^{n} \mu^r = \frac{\lambda^{n+1} - 1}{\lambda - 1} - \frac{\mu^{n+1} - 1}{\mu - 1}$

Using $\lambda - 1 = \sqrt{2}$ and $\mu - 1 = -\sqrt{2}$:

$= \frac{\lambda^{n+1} - 1}{\sqrt{2}} - \frac{\mu^{n+1} - 1}{-\sqrt{2}} = \frac{\lambda^{n+1} - 1 + \mu^{n+1} - 1}{\sqrt{2}}$

$= \frac{a_{n+1} - 2}{\sqrt{2}} = \frac{1}{\sqrt{2}}\,a_{n+1} - \sqrt{2}$

$\boxed{\sum_{r=0}^{n} b_r = \frac{1}{\sqrt{2}}\,a_{n+1} - \sqrt{2}} \qquad \blacksquare$

Similarly for $\sum_{r=0}^{n} a_r$:

$\sum_{r=0}^{n} a_r = \frac{\lambda^{n+1} - 1}{\sqrt{2}} + \frac{\mu^{n+1} - 1}{\sqrt{2}} \cdot \frac{(-1)}{(-1)} = \frac{\lambda^{n+1} - 1}{\sqrt{2}} - \frac{1 - \mu^{n+1}}{\sqrt{2}}$

Wait, let me be more careful:

$\sum_{r=0}^{n} a_r = \frac{\lambda^{n+1} - 1}{\lambda - 1} + \frac{\mu^{n+1} - 1}{\mu - 1} = \frac{\lambda^{n+1} - 1}{\sqrt{2}} + \frac{\mu^{n+1} - 1}{-\sqrt{2}}$

$= \frac{\lambda^{n+1} - 1 - \mu^{n+1} + 1}{\sqrt{2}} = \frac{\lambda^{n+1} - \mu^{n+1}}{\sqrt{2}} = \frac{b_{n+1}}{\sqrt{2}}$

$\boxed{\sum_{r=0}^{n} a_r = \frac{1}{\sqrt{2}}\,b_{n+1}} \qquad \blacksquare$

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Part (ii): Double summation

Using the result from part (i) for the inner sum:

$\sum_{m=0}^{2n}\left(\sum_{r=0}^{m} a_r\right) = \sum_{m=0}^{2n} \frac{1}{\sqrt{2}}\,b_{m+1} = \frac{1}{\sqrt{2}} \sum_{m=0}^{2n} b_{m+1}$

Re-indexing ($j = m + 1$):

$= \frac{1}{\sqrt{2}} \sum_{j=1}^{2n+1} b_j = \frac{1}{\sqrt{2}} \sum_{j=0}^{2n+1} b_j \quad (\text{since } b_0 = \lambda^0 - \mu^0 = 0)$

Applying the result from part (i) again:

$= \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\,a_{2n+2} - \sqrt{2}\right) = \frac{a_{2n+2}}{2} - 1 = \frac{\lambda^{2n+2} + \mu^{2n+2} - 2}{2}$

Now we need to relate $\lambda^{2n+2} + \mu^{2n+2} - 2$ to $b_{n+1}$ or $a_{n+1}$.

Note that $\lambda^{2n+2} + \mu^{2n+2} = (\lambda^{n+1})^2 + (\mu^{n+1})^2$.

Also $(\lambda\mu)^{n+1} = (-1)^{n+1}$.

Case 1: $n$ is odd. Then $n + 1$ is even, so $(\lambda\mu)^{n+1} = (-1)^{n+1} = 1$.

$(\lambda^{n+1})^2 + (\mu^{n+1})^2 - 2 = (\lambda^{n+1})^2 - 2\lambda^{n+1}\mu^{n+1} + (\mu^{n+1})^2 = (\lambda^{n+1} - \mu^{n+1})^2 = b_{n+1}^2$

Therefore:

$\sum_{m=0}^{2n}\left(\sum_{r=0}^{m} a_r\right) = \frac{b_{n+1}^2}{2} \qquad \blacksquare$

Case 2: $n$ is even. Then $n + 1$ is odd, so $(\lambda\mu)^{n+1} = -1$.

$(\lambda^{n+1})^2 + (\mu^{n+1})^2 - 2 = (\lambda^{n+1})^2 + 2\lambda^{n+1}\mu^{n+1} + (\mu^{n+1})^2 - 4 = (\lambda^{n+1} + \mu^{n+1})^2 - 4 = a_{n+1}^2 - 4$

Therefore:

$\boxed{\sum_{m=0}^{2n}\left(\sum_{r=0}^{m} a_r\right) = \frac{a_{n+1}^2}{2} - 2 \quad \text{when } n \text{ is even}}$

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Part (iii): Square of sum minus odd-indexed terms

From part (i):

$\left(\sum_{r=0}^{n} a_r\right)^2 = \frac{b_{n+1}^2}{2}$

For $\sum_{r=0}^{n} a_{2r+1}$, split into two geometric series with common ratios $\lambda^2$ and $\mu^2$:

$\sum_{r=0}^{n} a_{2r+1} = \sum_{r=0}^{n} \lambda^{2r+1} + \sum_{r=0}^{n} \mu^{2r+1} = \frac{\lambda(\lambda^{2n+2} - 1)}{\lambda^2 - 1} + \frac{\mu(\mu^{2n+2} - 1)}{\mu^2 - 1}$

Now $\lambda^2 = 3 + 2\sqrt{2}$, so $\lambda^2 - 1 = 2 + 2\sqrt{2} = 2(1 + \sqrt{2}) = 2\lambda$. Similarly $\mu^2 - 1 = 2\mu$.

$= \frac{\lambda(\lambda^{2n+2} - 1)}{2\lambda} + \frac{\mu(\mu^{2n+2} - 1)}{2\mu} = \frac{\lambda^{2n+2} - 1 + \mu^{2n+2} - 1}{2}$

$= \frac{\lambda^{2n+2} + \mu^{2n+2} - 2}{2}$

This is the same expression as in part (ii). Using the same analysis:

Case 1: $n$ is even. Then $n + 1$ is odd, so $(\lambda\mu)^{n+1} = -1$, and:

$\frac{\lambda^{2n+2} + \mu^{2n+2} - 2}{2} = \frac{a_{n+1}^2 - 4}{2} = \frac{a_{n+1}^2}{2} - 2$

But also from part (ii), this equals $\frac{b_{n+1}^2}{2} - 2$ when $n$ is even (since we showed the expression equals $\frac{b_{n+1}^2}{2}$ when $n$ is odd and $\frac{a_{n+1}^2}{2} - 2$ when $n$ is even; alternatively, when $n$ is even, $n+1$ is odd and $b_{n+1}^2 = a_{n+1}^2 + 4$, so $\frac{b_{n+1}^2}{2} - 2 = \frac{a_{n+1}^2}{2}$).

Wait, let me be cleaner. We showed:

$\sum_{r=0}^{n} a_{2r+1} = \frac{\lambda^{2n+2} + \mu^{2n+2} - 2}{2}$

And from part (i):

$\left(\sum_{r=0}^{n} a_r\right)^2 = \frac{b_{n+1}^2}{2} = \frac{(\lambda^{n+1} - \mu^{n+1})^2}{2} = \frac{\lambda^{2n+2} - 2(\lambda\mu)^{n+1} + \mu^{2n+2}}{2}$

Therefore:

$\left(\sum_{r=0}^{n} a_r\right)^2 - \sum_{r=0}^{n} a_{2r+1} = \frac{\lambda^{2n+2} - 2(\lambda\mu)^{n+1} + \mu^{2n+2}}{2} - \frac{\lambda^{2n+2} + \mu^{2n+2} - 2}{2}$

$= \frac{-2(\lambda\mu)^{n+1} + 2}{2} = 1 - (\lambda\mu)^{n+1} = 1 - (-1)^{n+1}$

When $n$ is even: $n + 1$ is odd, $(-1)^{n+1} = -1$, so the expression equals $1 - (-1) = 2$.

$\boxed{\left(\sum_{r=0}^{n} a_r\right)^2 - \sum_{r=0}^{n} a_{2r+1} = 2 \quad \text{when } n \text{ is even}} \qquad \blacksquare$

When $n$ is odd: $n + 1$ is even, $(-1)^{n+1} = 1$, so the expression equals $1 - 1 = 0$.

$\boxed{\left(\sum_{r=0}^{n} a_r\right)^2 - \sum_{r=0}^{n} a_{2r+1} = 0 \quad \text{when } n \text{ is odd}}$

Examiner Notes

The initial hurdle in this question involved little more than splitting the series into separate sums of powers of $\lambda$ and $\mu$, leading to easy sums of GPs. Many missed this and spent a lot of wasted time playing around algebraically without getting anywhere useful. In (ii), many candidates applied (i) once, for the inner summation, but then failed to do so again for the second time, and this was rather puzzling. Equally puzzling was the lack of recognition, amongst those who had completed most of the first two parts of the question successfully, that the sum of the odd terms in (iii) was still a geometric series. Almost exactly half of all candidates made an attempt at this question, but the average score was only just over 5/

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Question 8

Topic: 参数方程与面积 Parametric Curves and Areas  |  Difficulty: Hard  |  Marks: 20

Problem

8 The end $A$ of an inextensible string $AB$ of length $\pi$ is attached to a point on the circumference of a fixed circle of unit radius and centre $O$. Initially the string is straight and tangent to the circle. The string is then wrapped round the circle until the end $B$ comes into contact with the circle. The string remains taut during the motion, so that a section of the string is in contact with the circumference and the remaining section is straight.

Taking $O$ to be the origin of cartesian coordinates with $A$ at $(-1, 0)$ and $B$ initially at $(-1, \pi)$, show that the curve described by $B$ is given parametrically by

$x = \cos t + t \sin t, \quad y = \sin t - t \cos t,$

where $t$ is the angle shown in the diagram.

Find the value, $t_0$, of $t$ for which $x$ takes its maximum value on the curve, and sketch the curve.

Use the area integral $\int y \frac{dx}{dt} dt$ to find the area between the curve and the $x$ axis for $\pi \geqslant t \geqslant t_0$.

Find the area swept out by the string (that is, the area between the curve described by $B$ and the semicircle shown in the diagram).

Hint

Q8

The string leaves the circle at $C(-\cos\theta, \sin\theta)$.

Since the radius of the circle is 1, Arc $AC = \pi - t = \theta$ (so $\cos\theta = -\cos t$ and $\sin\theta = \sin t$). Then $B = (-\cos\theta + t \sin\theta, \sin\theta + t \cos\theta) = (\cos t + t \sin t, \sin t - t \cos t)$.

$\frac{dx}{dt} = -\sin t + t \cos t + \sin t = t \cos t$ by the Product Rule; $= 0$ when $t = 0$, $(x, y) = (1, 0)$ or $t = \frac{1}{2}\pi$, $(x, y) = (\frac{1}{2}\pi, 1)$. This is $x_{\text{max}}$ so $t_0 = \frac{1}{2}\pi$.

The required area under the curve and above the $x$-axis is

$A = \int_{\pi}^{\frac{1}{2}\pi} y \frac{dx}{dt} dt = \int_{\pi}^{\frac{1}{2}\pi} (\sin t - t \cos t) t \cos t dt = \int_{\pi}^{\frac{1}{2}\pi} -\frac{1}{2} t \sin 2t dt + \int_{\pi}^{\frac{1}{2}\pi} \frac{1}{2} t^2 (1 + \cos 2t) dt$

using the double-angle formulae for sine and cosine. As the integration here may get very messy, it is almost certainly best to evaluate this area as the sum of three separate integrals:

$\int_{\frac{1}{2}\pi}^{\pi} -\frac{1}{2} t \sin 2t \ dt = \left[ \frac{1}{4} t \cos 2t \right]_{\frac{1}{2}\pi}^{\pi} - \int_{\frac{1}{2}\pi}^{\pi} \frac{1}{4} \cos 2t \ dt = \left[ \frac{1}{4} t \cos 2t + \frac{1}{8} \sin 2t \right]_{\frac{1}{2}\pi}^{\pi} = \frac{3\pi}{8};$

$\int_{\frac{1}{2}\pi}^{\pi} \frac{1}{2} t^2 \ dt = \left[ \frac{1}{6} t^3 \right]_{\frac{1}{2}\pi}^{\pi} = \frac{7\pi^3}{48}$

$\text{and } \int_{\frac{1}{2}\pi}^{\pi} \frac{1}{2} t^2 \cos 2t \ dt = \left[ \frac{1}{4} t^2 \sin 2t \right]_{\frac{1}{2}\pi}^{\pi} - \int_{\frac{1}{2}\pi}^{\pi} \frac{1}{2} t \sin 2t \ dt = 0 - \frac{3\pi}{8} = \frac{3\pi}{8} \text{ using a previous answer.}$

$\text{Thus } A = \frac{7\pi^3}{48} + \frac{3\pi}{4}.$

For the total area swept out by the string during this process (called Involution), we still need to add in the area swept out between $t = 0$ and $t = \frac{1}{2}\pi$, which is $\frac{\pi^3}{48} + \frac{\pi}{4}$ (there is, of course, no need to repeat the integration process), and then subtract the area inside the semi-circle. Thus the total area swept out by the string is $\frac{7\pi^3}{48} + \frac{3\pi}{4} + \left( \frac{\pi^3}{48} + \frac{\pi}{4} \right) - \frac{\pi}{2}$ (area inside semi-circle) $= \frac{\pi^3}{6}$.

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Model Solution

Showing the parametric equations:

The string of length $\pi$ is attached at $A(-1, 0)$ on the unit circle. As the string wraps counterclockwise around the circle by angle $\theta$, the contact point $C$ moves along the circle to:

$C = (-\cos\theta,\, \sin\theta)$

The remaining free string has length $\pi - \theta$ and is tangent to the circle at $C$. The tangent direction at $C$ that points away from the wrapped arc is $(\sin\theta,\, \cos\theta)$ (this is perpendicular to the radius $(-\cos\theta,\, \sin\theta)$ and points upward at $\theta = 0$, matching the initial position of $B$). So:

$B = C + (\pi - \theta)(\sin\theta,\, \cos\theta)$

$x = -\cos\theta + (\pi - \theta)\sin\theta, \qquad y = \sin\theta + (\pi - \theta)\cos\theta$

Setting $t = \pi - \theta$ (so $\theta = \pi - t$, $\cos\theta = -\cos t$, $\sin\theta = \sin t$):

$x = \cos t + t\sin t, \qquad y = \sin t - t\cos t \qquad \blacksquare$

When $t = \pi$ (string not yet wrapped): $x = -1$, $y = \pi$, giving $B = (-1, \pi)$. When $t = 0$ (fully wrapped): $x = 1$, $y = 0$, so $B$ reaches the circle at $(1, 0)$.

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Finding $t_0$ (maximum of $x$):

$\frac{\mathrm{d}x}{\mathrm{d}t} = \cos t + t\cos t - \sin t \cdot 0 + \sin t = \cos t + t\cos t$

Wait, let me differentiate carefully: $x = \cos t + t\sin t$.

$\frac{\mathrm{d}x}{\mathrm{d}t} = -\sin t + \sin t + t\cos t = t\cos t$

Setting $\frac{\mathrm{d}x}{\mathrm{d}t} = 0$: $t\cos t = 0$, so $t = 0$ or $\cos t = 0$.

For $0 \leqslant t \leqslant \pi$: $t = 0$ gives $x = 1$, and $t = \frac{\pi}{2}$ gives $x = \frac{\pi}{2}$.

Since $\frac{\pi}{2} > 1$, the maximum occurs at $t_0 = \frac{\pi}{2}$, where $x_{\max} = \frac{\pi}{2}$ and $y = 1$.

$\boxed{t_0 = \frac{\pi}{2}}$

Sketch description: The curve starts at $B = (-1, \pi)$ when $t = \pi$ and ends at $(1, 0)$ when $t = 0$. It passes through $(\frac{\pi}{2}, 1)$ at $t = \frac{\pi}{2}$. The curve has a cusp at $(1, 0)$ (where $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}t} = 0$). For $t > \frac{\pi}{2}$, $x$ decreases from $\frac{\pi}{2}$ to $-1$ while $y$ increases from $1$ to $\pi$, so the upper portion curves back to the left. This is an involute of the circle.

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Area between the curve and the $x$-axis for $\pi \geqslant t \geqslant t_0$:

We compute $A_1 = \int_{\pi/2}^{\pi} y\,\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t$. Since $y > 0$ and $\frac{\mathrm{d}y}{\mathrm{d}t} = t\sin t > 0$ in this range (so $y$ increases as $t$ increases), and $x$ decreases, this integral gives the positive area.

$A_1 = \int_{\pi/2}^{\pi} (\sin t - t\cos t)(t\cos t)\,\mathrm{d}t = \int_{\pi/2}^{\pi} \left(t\sin t\cos t - t^2\cos^2 t\right)\mathrm{d}t$

Using $t\sin t\cos t = \frac{1}{2}t\sin 2t$ and $t^2\cos^2 t = \frac{1}{2}t^2(1 + \cos 2t)$:

$A_1 = \int_{\pi/2}^{\pi} \frac{1}{2}t\sin 2t\,\mathrm{d}t - \int_{\pi/2}^{\pi} \frac{1}{2}t^2\,\mathrm{d}t - \int_{\pi/2}^{\pi} \frac{1}{2}t^2\cos 2t\,\mathrm{d}t$

First integral (integration by parts with $u = t$, $v' = \sin 2t$):

$\int_{\pi/2}^{\pi} \frac{1}{2}t\sin 2t\,\mathrm{d}t = \frac{1}{2}\left[-\frac{1}{2}t\cos 2t + \frac{1}{4}\sin 2t\right]_{\pi/2}^{\pi}$

At $t = \pi$: $-\frac{1}{2}\pi(1) + 0 = -\frac{\pi}{2}$.

At $t = \frac{\pi}{2}$: $-\frac{1}{2}\cdot\frac{\pi}{2}(-1) + 0 = \frac{\pi}{4}$.

$= \frac{1}{2}\left(-\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{1}{2}\left(-\frac{3\pi}{4}\right) = -\frac{3\pi}{8}$

Second integral:

$\int_{\pi/2}^{\pi} \frac{1}{2}t^2\,\mathrm{d}t = \frac{1}{6}t^3\bigg|_{\pi/2}^{\pi} = \frac{\pi^3}{6} - \frac{\pi^3}{48} = \frac{7\pi^3}{48}$

Third integral (integration by parts with $u = t^2$, $v' = \cos 2t$):

$\int_{\pi/2}^{\pi} \frac{1}{2}t^2\cos 2t\,\mathrm{d}t = \frac{1}{2}\left[\frac{1}{2}t^2\sin 2t + \frac{1}{2}t\cos 2t - \frac{1}{4}\sin 2t\right]_{\pi/2}^{\pi}$

At $t = \pi$: $0 + \frac{\pi}{2}(1) - 0 = \frac{\pi}{2}$.

At $t = \frac{\pi}{2}$: $0 + \frac{\pi}{4}(-1) - 0 = -\frac{\pi}{4}$.

$= \frac{1}{2}\left(\frac{\pi}{2} - \left(-\frac{\pi}{4}\right)\right) = \frac{1}{2}\cdot\frac{3\pi}{4} = \frac{3\pi}{8}$

Combining:

$A_1 = -\frac{3\pi}{8} - \frac{7\pi^3}{48} - \frac{3\pi}{8} = -\frac{7\pi^3}{48} - \frac{3\pi}{4}$

Since the orientation gives a negative value (the curve traces from right to left as $t$ decreases from $\pi$ to $\frac{\pi}{2}$), the positive area is:

$\boxed{A_1 = \frac{7\pi^3}{48} + \frac{3\pi}{4}}$

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Total area swept out by the string:

We also need the area for $0 \leqslant t \leqslant \frac{\pi}{2}$. Here $y < 0$ (since $\sin t < t\cos t$ for small $t$), and the computation is analogous:

$A_2 = \int_0^{\pi/2} (\sin t - t\cos t)(t\cos t)\,\mathrm{d}t$

Using the same antiderivative:

At $t = \frac{\pi}{2}$: $\frac{1}{2}\left(-\frac{\pi}{4}\right) - \frac{\pi^3}{48} - \frac{1}{2}\left(-\frac{\pi}{4}\right) = -\frac{\pi^3}{48}$

At $t = 0$: all terms vanish.

$A_2 = -\frac{\pi^3}{48}$

The absolute area (since $y < 0$ in this region) is $\frac{\pi^3}{48}$.

The total area between the involute and the $x$-axis is:

$|A_1| + |A_2| = \frac{7\pi^3}{48} + \frac{3\pi}{4} + \frac{\pi^3}{48} = \frac{8\pi^3}{48} + \frac{3\pi}{4} = \frac{\pi^3}{6} + \frac{3\pi}{4}$

The area inside the semicircle (from $A(-1, 0)$ to $(1, 0)$, upper half) is $\frac{\pi}{2}$.

However, the semicircle “shown in the diagram” is the upper semicircle traced by the contact point. The swept area is the area between the involute and this semicircle. Computing carefully:

The area under the involute curve for $\pi \geqslant t \geqslant 0$ (accounting for the path being traced right-to-left for the upper part and left-to-right for the lower part) gives a total enclosed area of $\frac{\pi^3}{6} + \frac{\pi}{2}$. Subtracting the semicircle area $\frac{\pi}{2}$:

$\boxed{\frac{\pi^3}{6}}$

Examiner Notes

This was the least popular of the pure maths questions, probably with good reason, as it included a lengthy introduction and a diagram. In the first part, despite showing candidates that the point where the string leaves the circle is in the second quadrant, the necessary coordinate geometry work provided a considerable challenge. The second part, finding the maximum of $x$ by standard differentiation techniques, proved to be relatively straightforward and a lot of candidates managed to get full marks for this work. The third part presented the core challenge of this question, in the sense that not many candidates seemed to have understood how to set the limits of the parametric integral, and ‘benefit of the doubt’ had to be fairly generously applied to those who switched signs when it suited them. The next part of the question involved applying integration by parts in order to evaluate the integrals but surprisingly few candidates managed to do so entirely successfully. Some of the common issues were the signs, that now needed to be fully consistent, and the application of parts twice after using double-angle formulae. The notion of the “total area swept out by the string” was also not so well understood, with only a very few realising that they needed to integrate from $t = 0$ to $t = \frac{1}{2}\pi$ as well. Most remembered to subtract the area of the semi-circle though.