STEP2 2010 -- Pure Mathematics

STEP2 2010 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2010  |  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	微积分 (Calculus)	Standard	曲率公式, 链式求导, 三角函数求导
2	三角函数 (Trigonometry)	Challenging	三倍角公式, 换元法, 不定积分与定积分
3	数列与级数 (Sequences and Series)	Standard	特征方程, 待定系数法, 等比级数求和
4	积分 (Integration)	Challenging	换元法, 积分对称性, King property
5	向量 (Vectors)	Challenging	向量点积, 方向余弦, 锥面方程
6	立体几何 (Solid Geometry)	Standard	三维坐标几何, 勾股定理, 二面角计算, 内切球半径公式
7	代数 (Algebra)	Challenging	导数求极值, 代换法解三次方程, 韦达定理, 不等式分析
8	积分 (Integration)	Challenging	指数函数积分, 三角函数积分, 无穷等比级数求和, 曲线交点

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Question 1

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

Problem

1 Let $P$ be a given point on a given curve $C$. The osculating circle to $C$ at $P$ is defined to be the circle that satisfies the following two conditions at $P$: it touches $C$; and the rate of change of its gradient is equal to the rate of change of the gradient of $C$.

Find the centre and radius of the osculating circle to the curve $y = 1 - x + \tan x$ at the point on the curve with $x$-coordinate $\frac{1}{4}\pi$.

Hint

1 When two curves meet they share common coordinates; when they “touch” they also share a common gradient. In the case of the osculating circle, they also have a common curvature at the point of contact. Since curvature (a further maths topic) is a function of both $\frac{\text{dy}}{\text{dx}}$ and $\frac{\text{d}^2\text{y}}{\text{dx}^2}$, the question merely states that $C$ and its osculating circle at $P$ have equal rates of change of gradient. It makes sense then to differentiate twice both the equation for $C$ and that for a circle, with equation of the form $(x - a)^2 + (y - b)^2 = r^2$, and then equate them when $x = \frac{1}{4}\pi$. The three resulting equations in the three unknowns $a$, $b$ and $r$ then simply need to be solved simultaneously.

For $y = 1 - x + \tan x$, $\frac{\text{dy}}{\text{dx}} = -1 + \sec^2 x$ and $\frac{\text{d}^2\text{y}}{\text{dx}^2} = 2 \sec^2 x \tan x$.

For $(x - a)^2 + (y - b)^2 = r^2$, $2(x - a) + 2(y - b) \frac{\text{dy}}{\text{dx}} = 0$ and $2 + 2(y - b) \frac{\text{d}^2\text{y}}{\text{dx}^2} + 2 \left( \frac{\text{dy}}{\text{dx}} \right)^2 = 0$.

When $x = \frac{1}{4}\pi$, $y = 2 - \frac{1}{4}\pi$ and so $(\frac{1}{4}\pi - a)^2 + (2 - \frac{1}{4}\pi - b)^2 = r^2$;

$\frac{\text{dy}}{\text{dx}} = -\frac{(x - a)}{(y - b)} = 1$ then gives a relationship between $a$ and $b$;

and $\frac{\text{d}^2\text{y}}{\text{dx}^2} = 4 = -\frac{4}{2(y - b)}$ gives the value of $b$.

Working back then gives $a$ and $r$.

Answers: The osculating circle to $C$ at $P$ has centre $(\frac{1}{4}\pi - \frac{1}{2}, \frac{5}{2} - \frac{1}{4}\pi)$ and radius $\frac{1}{\sqrt{2}}$.

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Model Solution

Step 1: Evaluate the curve and its derivatives at $x = \frac{\pi}{4}$.

For $y = 1 - x + \tan x$:

$\frac{dy}{dx} = -1 + \sec^2 x = \tan^2 x$

$\frac{d^2y}{dx^2} = 2\sec^2 x \tan x$

At $x = \frac{\pi}{4}$:

$y = 1 - \frac{\pi}{4} + \tan\frac{\pi}{4} = 2 - \frac{\pi}{4}$

$\frac{dy}{dx} = \tan^2\frac{\pi}{4} = 1$

$\frac{d^2y}{dx^2} = 2\sec^2\frac{\pi}{4}\tan\frac{\pi}{4} = 2 \cdot 2 \cdot 1 = 4$

Step 2: Differentiate the general circle equation.

Let the osculating circle have equation $(x - a)^2 + (y - b)^2 = r^2$.

Differentiating implicitly with respect to $x$:

$2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \qquad \text{(1)}$

Differentiating again:

$2 + 2(y - b)\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 = 0 \qquad \text{(2)}$

Step 3: Apply conditions at the point of contact.

At $x = \frac{\pi}{4}$, the circle passes through the point $\left(\frac{\pi}{4},\, 2 - \frac{\pi}{4}\right)$:

$\left(\frac{\pi}{4} - a\right)^2 + \left(2 - \frac{\pi}{4} - b\right)^2 = r^2 \qquad \text{(3)}$

The circle touches the curve, so their gradients are equal. From equation (1):

$\frac{dy}{dx} = -\frac{x - a}{y - b}$

Setting $\frac{dy}{dx} = 1$:

$-\frac{\frac{\pi}{4} - a}{2 - \frac{\pi}{4} - b} = 1$

$a - \frac{\pi}{4} = 2 - \frac{\pi}{4} - b$

$a + b = 2 \qquad \text{(4)}$

The rates of change of gradient are equal. From equation (2):

$\frac{d^2y}{dx^2} = -\frac{1 + \left(\frac{dy}{dx}\right)^2}{y - b}$

Setting $\frac{d^2y}{dx^2} = 4$:

$4 = -\frac{1 + 1}{2 - \frac{\pi}{4} - b} = -\frac{2}{2 - \frac{\pi}{4} - b}$

$2 - \frac{\pi}{4} - b = -\frac{1}{2}$

$b = \frac{5}{2} - \frac{\pi}{4}$

Step 4: Find $a$ and $r$.

From (4): $a = 2 - b = 2 - \frac{5}{2} + \frac{\pi}{4} = \frac{\pi}{4} - \frac{1}{2}$.

From (3):

$r^2 = \left(\frac{\pi}{4} - \frac{\pi}{4} + \frac{1}{2}\right)^2 + \left(2 - \frac{\pi}{4} - \frac{5}{2} + \frac{\pi}{4}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

$r = \frac{1}{\sqrt{2}}$

The osculating circle has centre $\left(\dfrac{\pi}{4} - \dfrac{1}{2},\; \dfrac{5}{2} - \dfrac{\pi}{4}\right)$ and radius $\dfrac{1}{\sqrt{2}}$.

Examiner Notes

This question was attempted by almost two-thirds of the candidature, with a mean mark of around $11 \frac{1}{2}$. Whilst most attempts were very successful, a lot of marks were lost by poorly structured working, where the candidate got themselves confused in some way or another. The only two common conceptual difficulties were the oversight of the equal gradients at the point of contact and the lack of a suitable circle equation to start working with. Apart from these, most candidates’ work went smoothly and successfully, although sign errors often cost them at least one of the final three answer marks.

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Question 2

Topic: 三角函数 (Trigonometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Prove that $\cos 3x = 4 \cos^3 x - 3 \cos x .$ Find and prove a similar result for $\sin 3x$ in terms of $\sin x$.

(i) Let $I(\alpha) = \int_{0}^{\alpha} (7 \sin x - 8 \sin^3 x) dx .$ Show that $I(\alpha) = -\frac{8}{3}c^3 + c + \frac{5}{3} ,$ where $c = \cos \alpha$. Write down one value of $c$ for which $I(\alpha) = 0$.

(ii) Useless Eustace believes that $\int \sin^n x \, dx = \frac{\sin^{n+1} x}{n + 1}$ for $n = 1, 2, 3, \dots$. Show that Eustace would obtain the correct value of $I(\beta)$, where $\cos \beta = -\frac{1}{6}$.

Find all values of $\alpha$ for which he would obtain the correct value of $I(\alpha)$.

Hint

2 The single-maths approach to the very first part is to use the standard trig. “Addition” formulae for sine and cosine, and then to use these results, twice, in (i); firstly, to rewrite $\sin^3 x$ in terms of $\sin 3x$ so that direct integration can be undertaken; then to express $\cos 3x$ in terms of $\cos^3 x$ in order to get the required “polynomial” in $\cos x$. Using the given “misunderstanding” in (ii) then leads to a second such polynomial which, when equated to the first, gives an equation for which a couple of roots have already been flagged. Unfortunately, the several versions of the question that were tried, in order to help candidates, ultimately led to the inadvertent disappearance of the interval $0$ to $\pi$ in which answers had originally been intended. This meant that there was a little bit more work to be done at the end than was initially planned.

$\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x = (2c^2 - 1)c - 2sc.s = (2c^2 - 1)c - 2c(1 - c^2)$ $= 4c^3 - 3c$.

$\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x = 2sc.c + (1 - 2s^2)s = 2s(1 - s^2) + s(1 - 2s^2)$ $= 3s - 4s^3$

(i) $I(\alpha) = \int\limits_0^\alpha (7 \sin x - 8 \sin^3 x) \text{dx} = \int\limits_0^\alpha (\sin x + 2 \sin 3x) \text{dx} = [-\cos x - \frac{2}{3} \cos 3x]_0^\alpha$ $= -\cos \alpha - \frac{2}{3} (4 \cos^3 \alpha - 3 \cos \alpha) + 1 + \frac{2}{3} = -\frac{8}{3} c^3 + c + \frac{5}{3}$ and $I(\alpha) = 0$ when $c = 1$ ($\alpha = 0$)

(ii) $J(\alpha) = \left[ \frac{7}{2} \sin^2 x - \frac{8}{4} \sin^4 x \right]_0^\alpha = \frac{7}{2} (1 - \cos^2 \alpha) - 2(1 - \cos^2 \alpha)^2 = -2c^4 + \frac{1}{2}c^2 + \frac{3}{2}$

$I(\alpha) = J(\alpha) \Rightarrow 0 = 12c^4 - 16c^3 - 3c^2 + 6c + 1 = (c - 1)^2 (2c + 1)(6c + 1)$

Thus $\cos \alpha = 1, \alpha = 0$; $\cos \alpha = -\frac{1}{2}, \alpha = \frac{2}{3}\pi$; and $\cos \alpha = -\frac{1}{6}, \alpha = \pi - \cos^{-1}(\frac{1}{6})$.

Answers: $\alpha = 2n\pi, 2n\pi \pm \frac{2}{3}\pi, (2n + 1)\pi \pm \cos^{-1} \frac{1}{6}$

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Model Solution

Proof of $\cos 3x = 4\cos^3 x - 3\cos x$:

$\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x$

Using $\cos 2x = 2\cos^2 x - 1$ and $\sin 2x = 2\sin x \cos x$:

$= (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x$

$= 2\cos^3 x - \cos x - 2\sin^2 x \cos x$

$= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x$

$= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x$

Proof of $\sin 3x = 3\sin x - 4\sin^3 x$:

$\sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x$

Using $\sin 2x = 2\sin x \cos x$ and $\cos 2x = 1 - 2\sin^2 x$:

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

$= 2\sin x \cos^2 x + \sin x - 2\sin^3 x$

$= 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x$

$= 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x = 3\sin x - 4\sin^3 x$

Part (i)

From the identity just proved, $\sin 3x = 3\sin x - 4\sin^3 x$, so:

$4\sin^3 x = 3\sin x - \sin 3x$

$8\sin^3 x = 6\sin x - 2\sin 3x$

$7\sin x - 8\sin^3 x = 7\sin x - 6\sin x + 2\sin 3x = \sin x + 2\sin 3x$

Therefore:

$I(\alpha) = \int_0^{\alpha} (\sin x + 2\sin 3x)\, dx = \left[-\cos x - \frac{2}{3}\cos 3x\right]_0^{\alpha}$

$= -\cos\alpha - \frac{2}{3}\cos 3\alpha + 1 + \frac{2}{3}$

$= -\cos\alpha - \frac{2}{3}(4\cos^3\alpha - 3\cos\alpha) + \frac{5}{3}$

$= -\cos\alpha - \frac{8}{3}\cos^3\alpha + 2\cos\alpha + \frac{5}{3}$

$= -\frac{8}{3}\cos^3\alpha + \cos\alpha + \frac{5}{3}$

Setting $c = \cos\alpha$:

$I(\alpha) = -\frac{8}{3}c^3 + c + \frac{5}{3}$

When $I(\alpha) = 0$: substituting $c = 1$ gives $-\frac{8}{3} + 1 + \frac{5}{3} = 0$. So $c = 1$ (i.e. $\alpha = 0$).

Part (ii)

Eustace uses $\int \sin^n x\, dx = \frac{\sin^{n+1} x}{n+1}$ for all $n = 1, 2, 3, \ldots$. So for $n = 1$:

$\int \sin x\, dx = \frac{\sin^2 x}{2} \quad \text{(Eustace's version, replacing the correct } {-\cos x}\text{)}$

and for $n = 3$:

$\int \sin^3 x\, dx = \frac{\sin^4 x}{4}$

Eustace’s evaluation of $I(\alpha)$:

$J(\alpha) = \left[\frac{7}{2}\sin^2 x - 8 \cdot \frac{\sin^4 x}{4}\right]_0^{\alpha} = \frac{7}{2}\sin^2\alpha - 2\sin^4\alpha$

Since $\sin^2\alpha = 1 - c^2$ where $c = \cos\alpha$:

$J(\alpha) = \frac{7}{2}(1 - c^2) - 2(1 - c^2)^2$

$= \frac{7}{2} - \frac{7}{2}c^2 - 2(1 - 2c^2 + c^4)$

$= \frac{7}{2} - \frac{7}{2}c^2 - 2 + 4c^2 - 2c^4$

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

Showing Eustace gets the correct value when $\cos\beta = -\frac{1}{6}$:

We need $I(\beta) = J(\beta)$, i.e. $I(\beta) - J(\beta) = 0$:

$I(\alpha) - J(\alpha) = \left(-\frac{8}{3}c^3 + c + \frac{5}{3}\right) - \left(-2c^4 + \frac{1}{2}c^2 + \frac{3}{2}\right)$

$= 2c^4 - \frac{8}{3}c^3 - \frac{1}{2}c^2 + c + \frac{1}{6}$

Multiplying by 6:

$12c^4 - 16c^3 - 3c^2 + 6c + 1 = 0$

We know from part (i) that $c = 1$ is a root (since $I(0) = J(0) = 0$). Dividing by $(c - 1)$:

$12c^4 - 16c^3 - 3c^2 + 6c + 1 = (c - 1)(12c^3 - 4c^2 - 7c - 1)$

Testing $c = 1$ again: $12 - 4 - 7 - 1 = 0$, so $(c - 1)$ is a repeated factor:

$(c - 1)^2(12c^2 + 8c + 1) = 0$

Factorising the quadratic: $12c^2 + 8c + 1 = (2c + 1)(6c + 1)$.

Therefore:

$12c^4 - 16c^3 - 3c^2 + 6c + 1 = (c - 1)^2(2c + 1)(6c + 1) = 0$

The roots are $c = 1$, $c = -\frac{1}{2}$, and $c = -\frac{1}{6}$.

When $c = -\frac{1}{6}$, the factor $(6c + 1) = 0$, confirming that Eustace obtains the correct value of $I(\beta)$ when $\cos\beta = -\frac{1}{6}$.

Finding all values of $\alpha$:

Setting $I(\alpha) = J(\alpha)$ gives $\cos\alpha = 1$, $\cos\alpha = -\frac{1}{2}$, or $\cos\alpha = -\frac{1}{6}$.

• $\cos\alpha = 1$: $\alpha = 2n\pi$
• $\cos\alpha = -\frac{1}{2}$: $\alpha = 2n\pi \pm \frac{2\pi}{3}$
• $\cos\alpha = -\frac{1}{6}$: $\alpha = 2n\pi \pm \arccos\!\left(-\frac{1}{6}\right)$, equivalently $\alpha = (2n + 1)\pi \pm \arccos\frac{1}{6}$

where $n$ is any integer.

Examiner Notes

This was the most popular question on the paper, drawing an attempt from almst every candidate. There were several proofs of the initial trigonometric identities using de Moivre’s Theorem but most settled for the more standard cosine and sine of $(2x + x)$. Personally, I was against the inclusion of the given answer of $\cos^{-1} (-\frac{1}{6})$ in (ii) as it led to what struck me as an unwelcome dichotomy of approaches. Most candidates opted to verify that the two polynomials in “c” that arose gave the same numerical answer, and this working was not entirely straightforward – in the event, lots of candidates failed to show the markers that they had done the working correctly for both expressions – whereas my original intention had been that they should collect terms up into a single polynomial equation and factorise it by first spotting the (repeated) factor $(c - 1)$ hinted at in (i).

There was one important mathematical oversight that many candidates made during this question, and it was due to not reading the question sufficiently carefully. The wording of the question in (ii) clearly states that Eustace’s misunderstanding of the integration of powers of the sine function was for $n = 1, 2, 3, \dots$. Unfortunately, rather a lot of candidates thought that he would then have integrated $\sin x$ (i.e. the case $n = 1$) correctly as $-\cos x$. We concocted a mark-scheme for this eventuality which allowed candidates ‘follow-through’ for 6 out of the 10 marks allocated here, but the self-imposed penalty of four marks could not be avoided as it was just no longer possible to get, for instance, the given answer.

Finally, there is a bit of an apology to make: at some final stage of the printing process, the bit of the question that identified $\alpha$ as lying in the range 0 to $\pi$ got removed; this left candidates having to think about general solutions rather than just the two decently small ones that had been looked-for when the question was first written. Nevertheless, not only was this the most popular question for number of attempts, it was also the most successful for candidates with a man score of almost 15.

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Question 3

Topic: 数列与级数 (Sequences and Series)  |  Difficulty: Standard  |  Marks: 20

Problem

3 The first four terms of a sequence are given by $F_0 = 0$, $F_1 = 1$, $F_2 = 1$ and $F_3 = 2$. The general term is given by $F_n = a\lambda^n + b\mu^n , \qquad \text{(*)}$ where $a, b, \lambda$ and $\mu$ are independent of $n$, and $a$ is positive.

(i) Show that $\lambda^2 + \lambda\mu + \mu^2 = 2$, and find the values of $\lambda, \mu, a$ and $b$.

(ii) Use $(\ast)$ to evaluate $F_6$.

(iii) Evaluate $\displaystyle \sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} .$

Hint

3 You don’t have to have too wide an experience of mathematics to be able to recognise the Fibonacci Numbers in a modest disguise here. (However, this is of little help here, as you should be looking to follow the guidance of the question.) In (i), you are clearly intended to begin by substituting $n = 0, 1, 2$ and $3$, in turn, into the given formula for $F_n$, using the four given terms of the sequence. You now have four equations in four unknowns, and the given result in (i) is intended to help you make progress; with (ii) having you check the formula in a further case. In the final part, you should split the summation into two parts, each of which is an infinite geometric progression.

(i) $F_0 = 0 \Rightarrow 0 = a + b$ or $b = -a$. Then $F_1 = 1 \Rightarrow 1 = a(\lambda - \mu)$. $[F_2 = 1 \Rightarrow 1 = a(\lambda^2 - \mu^2) \Rightarrow \lambda + \mu = 1 \text{ is needed later}]$

and $F_3 = 2 \Rightarrow 2 = a(\lambda^3 - \mu^3) = a(\lambda - \mu)(\lambda^2 + \lambda\mu + \mu^2)$ by the difference of two cubes

$= 1 \cdot (\lambda^2 + \lambda\mu + \mu^2) \Rightarrow \lambda^2 + \lambda\mu + \mu^2 = 2$

Then, using any two suitable eqns., e.g. any two of $\lambda\mu = -1$, $\lambda - \mu = \frac{1}{a}$ and $\lambda + \mu = 1$, and

solving simultaneously gives $a = \frac{1}{\sqrt{5}}$, $b = -\frac{1}{\sqrt{5}}$, $\lambda = \frac{1}{2}(1 + \sqrt{5})$, $\mu = \frac{1}{2}(1 - \sqrt{5})$.

(ii) Using the formula $F_n = a \lambda^n + b \mu^n = \frac{1}{2^n \sqrt{5}} \left\{ (1 + \sqrt{5})^n - (1 - \sqrt{5})^n \right\}$ with $n = 6$; the Binomial Theorem gives $(1 + \sqrt{5})^6 = 1 + 6\sqrt{5} + 15 \cdot 5 + 20 \cdot 5\sqrt{5} + 15 \cdot 5^2 + 6 \cdot 5^2\sqrt{5} + 5^3 = 576 + 256\sqrt{5}$.

Similarly, $(1 - \sqrt{5})^6 = 576 - 256\sqrt{5}$ so that $F_6 = \frac{1}{2^6 \sqrt{5}} \{ 512\sqrt{5} \} = 8$.

(iii) $\sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} = \frac{a}{2} \sum_{n=0}^{\infty} \left( \frac{\lambda}{2} \right)^n - \frac{a}{2} \sum_{n=0}^{\infty} \left( \frac{\mu}{2} \right)^n = \frac{1}{2\sqrt{5}} \left( \frac{1}{1 - \frac{1}{4}(1 + \sqrt{5})} \right) - \frac{1}{2\sqrt{5}} \left( \frac{1}{1 - \frac{1}{4}(1 - \sqrt{5})} \right)$ using the

$S\infty$ formula for the two GPs;

$= \frac{1}{2\sqrt{5}} \left( \frac{4}{3 - \sqrt{5}} \right) - \frac{1}{2\sqrt{5}} \left( \frac{4}{3 + \sqrt{5}} \right)$.

Rationalising denominators then yields $\frac{2}{\sqrt{5}} \left( \frac{3 + \sqrt{5}}{9 - 5} \right) - \frac{2}{\sqrt{5}} \left( \frac{3 - \sqrt{5}}{9 - 5} \right) = \frac{2}{\sqrt{5}} \left( \frac{2\sqrt{5}}{4} \right) = 1$.

Model Solution

Part (i)

Substituting $n = 0, 1, 2, 3$ into $F_n = a\lambda^n + b\mu^n$:

$F_0 = 0 \implies a + b = 0 \implies b = -a \qquad \text{(*)}$

$F_1 = 1 \implies a\lambda + b\mu = 1 \implies a(\lambda - \mu) = 1 \qquad \text{(**)}$

$F_2 = 1 \implies a\lambda^2 + b\mu^2 = 1 \implies a(\lambda^2 - \mu^2) = 1 \qquad \text{(***)}$

$F_3 = 2 \implies a\lambda^3 + b\mu^3 = 2 \implies a(\lambda^3 - \mu^3) = 2 \qquad \text{(****)}$

Dividing $(***)$ by $(**)$:

$\frac{a(\lambda^2 - \mu^2)}{a(\lambda - \mu)} = \frac{1}{1} \implies \lambda + \mu = 1 \qquad \text{(1)}$

Now using the difference of two cubes factorisation on $(****)$:

$a(\lambda^3 - \mu^3) = a(\lambda - \mu)(\lambda^2 + \lambda\mu + \mu^2) = 2$

From $(**)$, $a(\lambda - \mu) = 1$, so:

$1 \cdot (\lambda^2 + \lambda\mu + \mu^2) = 2 \implies \lambda^2 + \lambda\mu + \mu^2 = 2 \qquad \checkmark$

Now we find the values. From $(1)$: $(\lambda + \mu)^2 = 1$, so $\lambda^2 + 2\lambda\mu + \mu^2 = 1$.

Subtracting $\lambda^2 + \lambda\mu + \mu^2 = 2$ from this:

$\lambda\mu = 1 - 2 = -1 \qquad \text{(2)}$

From $(1)$ and $(2)$, $\lambda$ and $\mu$ are roots of $t^2 - t - 1 = 0$:

$t = \frac{1 \pm \sqrt{5}}{2}$

Since $a > 0$ and $a(\lambda - \mu) = 1$, we need $\lambda > \mu$, so:

$\lambda = \frac{1 + \sqrt{5}}{2}, \qquad \mu = \frac{1 - \sqrt{5}}{2}$

From $(**)$: $a = \dfrac{1}{\lambda - \mu} = \dfrac{1}{\sqrt{5}}$, and from $(*)$: $b = -\dfrac{1}{\sqrt{5}}$.

Part (ii)

Using $F_n = \dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^n - \left(\dfrac{1-\sqrt{5}}{2}\right)^n\right]$ with $n = 6$.

We compute $(1 + \sqrt{5})^6$ using the Binomial Theorem:

$(1 + \sqrt{5})^6 = 1 + 6\sqrt{5} + 15 \cdot 5 + 20 \cdot 5\sqrt{5} + 15 \cdot 25 + 6 \cdot 25\sqrt{5} + 125$

$= 1 + 6\sqrt{5} + 75 + 100\sqrt{5} + 375 + 150\sqrt{5} + 125 = 576 + 256\sqrt{5}$

Similarly, replacing $\sqrt{5}$ with $-\sqrt{5}$:

$(1 - \sqrt{5})^6 = 576 - 256\sqrt{5}$

Therefore:

$F_6 = \frac{1}{\sqrt{5} \cdot 2^6}\left[(1+\sqrt{5})^6 - (1-\sqrt{5})^6\right] = \frac{1}{64\sqrt{5}}\left[576 + 256\sqrt{5} - 576 + 256\sqrt{5}\right]$

$= \frac{512\sqrt{5}}{64\sqrt{5}} = 8$

Part (iii)

$\sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} = \sum_{n=0}^{\infty} \frac{a\lambda^n + b\mu^n}{2^{n+1}} = \frac{a}{2}\sum_{n=0}^{\infty}\left(\frac{\lambda}{2}\right)^n + \frac{b}{2}\sum_{n=0}^{\infty}\left(\frac{\mu}{2}\right)^n$

Both are geometric series. We check convergence: $|\lambda/2| = \frac{1+\sqrt{5}}{4} \approx 0.809 < 1$ and $|\mu/2| = \frac{\sqrt{5}-1}{4} \approx 0.309 < 1$, so both converge.

$= \frac{1}{2\sqrt{5}} \cdot \frac{1}{1 - \frac{\lambda}{2}} - \frac{1}{2\sqrt{5}} \cdot \frac{1}{1 - \frac{\mu}{2}}$

Computing each denominator:

$1 - \frac{\lambda}{2} = 1 - \frac{1+\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{4}$

$1 - \frac{\mu}{2} = 1 - \frac{1-\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{4}$

So:

$\sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} = \frac{1}{2\sqrt{5}} \cdot \frac{4}{3-\sqrt{5}} - \frac{1}{2\sqrt{5}} \cdot \frac{4}{3+\sqrt{5}}$

Rationalising each fraction:

$\frac{4}{3-\sqrt{5}} = \frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} = \frac{4(3+\sqrt{5})}{9-5} = 3+\sqrt{5}$

$\frac{4}{3+\sqrt{5}} = \frac{4(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} = \frac{4(3-\sqrt{5})}{4} = 3-\sqrt{5}$

Therefore:

$\sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} = \frac{1}{2\sqrt{5}}\left[(3+\sqrt{5}) - (3-\sqrt{5})\right] = \frac{1}{2\sqrt{5}} \cdot 2\sqrt{5} = 1$

Examiner Notes

One doesn’t need to be too devoted a mathematician to recognise the Fibonacci numbers in this question, and many candidates clearly recognised this sequence. However, they were still required to answer the question in the way specified by the wording on the paper and a lot of attempts foundered at part (ii). This was the second most frequently attempted question, yet drew the second worst marks, averaging just over 8. Most attempts got little further than (i), and many foundered even here due to a lack of appreciation of the difference of two cubes factorisation. Things clearly got much worse in (ii) when far too many folks seemed incapable of attempting a binomial expansion of $(1 + \sqrt{5})^6$; many who did manage a decent stab at this then repeated the work for $(1 - \sqrt{5})^6$. Very few sorted this out correctly and, as a result, there were relatively few stabs at part (iii).

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Question 4

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

Problem

4 (i) Let $I = \int_{0}^{a} \frac{f(x)}{f(x) + f(a - x)} \, dx .$ Use a substitution to show that $I = \int_{0}^{a} \frac{f(a - x)}{f(x) + f(a - x)} \, dx$ and hence evaluate $I$ in terms of $a$. Use this result to evaluate the integrals $\int_{0}^{1} \frac{\ln(x + 1)}{\ln(2 + x - x^2)} \, dx \quad \text{and} \quad \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin(x + \frac{\pi}{4})} \, dx .$

(ii) Evaluate $\int_{\frac{1}{2}}^{2} \frac{\sin x}{x(\sin x + \sin \frac{1}{x})} \, dx .$

Hint

4 Hopefully, the obvious choice is $y = a - x$ for the initial substitution and, as with any given result, you should make every effort to be clear in your working to establish it. Thereafter, the two integrals that follow in (i) use this result with differing functions and for different choices of the upper limit $a$. Since this may be thought an obvious way to proceed, it is (again) important that your working is clear in identifying the roles of $f(x)$ and $f(a - x)$ in each case. In part (ii), however, it is not the first result that is to be used, but rather the process that yielded it. The required substitution should, again, be obvious, and then you should be trying to mimic the first process in this second situation.

(i) Using the substn. $y = a - x$, $dy = -dx$ and $(0, a) \rightarrow (a, 0)$ so that

$\int_{0}^{a} \frac{f(x)}{f(x) + f(a - x)} dx = \int_{a}^{0} \frac{f(a - y)}{f(a - y) + f(y)} \cdot -dy = \int_{0}^{a} \frac{f(a - y)}{f(a - y) + f(y)} dy$

$= \int_{0}^{a} \frac{f(a - x)}{f(x) + f(a - x)} dx \text{, since the } x/y \text{ interchange here is nothing more than a re-labelling.}$

$\text{Then } 2I = \int_{0}^{a} \frac{f(x) + f(a - x)}{f(x) + f(a - x)} dx = \int_{0}^{a} 1.dx = [x]_{0}^{a} = a \Rightarrow I = \frac{1}{2}a.$

For $f(x) = \ln(1 + x)$, $\ln(2 + x - x^2) = \ln[(1 + x)(2 - x)] = \ln(1 + x) + \ln(2 - x)$

and $\ln(2 - x) = \ln(1 + [1 - x]) = f(a - x)$ with $a = 1$ so that $\int_{0}^{1} \frac{f(x)}{f(x) + f(1 - x)} dx = \frac{1}{2}$.

$\int_{0}^{\pi/2} \frac{\sin x}{\sin(x + \frac{1}{4}\pi)} dx = \int_{0}^{\pi/2} \frac{\sin x}{\sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}}} dx = \sqrt{2} \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \sin(\frac{1}{2}\pi - x)} dx = \frac{1}{4}\pi\sqrt{2}.$

(ii) For $u = \frac{1}{x}$, $du = -\frac{1}{x^2} dx$ and $(\frac{1}{2}, 2) \rightarrow (2, \frac{1}{2})$.

$\text{Then } \int_{0.5}^{2} \frac{1}{x} \cdot \frac{\sin x}{(\sin x + \sin(\frac{1}{x}))} dx = \int_{0.5}^{2} \frac{1}{x^2} \cdot \frac{x \sin x}{(\sin x + \sin(\frac{1}{x}))} dx = \int_{2}^{0.5} \frac{\frac{1}{u} \cdot \sin(\frac{1}{u})}{(\sin(\frac{1}{u}) + \sin u)} \cdot -du$

$= \int_{0.5}^{2} \frac{1}{u} \cdot \frac{\sin(\frac{1}{u})}{(\sin u + \sin(\frac{1}{u}))} du \quad \text{or} \quad \int_{0.5}^{2} \frac{1}{x} \cdot \frac{\sin(\frac{1}{x})}{(\sin x + \sin(\frac{1}{x}))} dx$

$\text{Adding then gives } 2I = \int_{0.5}^{2} \frac{1}{x} dx = [\ln x]_{0.5}^{2} = 2 \ln 2 \Rightarrow I = \ln 2.$

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Model Solution

Part (i) — Establishing the result

Let $I = \displaystyle\int_{0}^{a} \frac{f(x)}{f(x) + f(a - x)} \, dx$.

Substitute $y = a - x$, so $dy = -dx$. When $x = 0$, $y = a$; when $x = a$, $y = 0$.

$I = \int_{a}^{0} \frac{f(a - y)}{f(a - y) + f(y)} \cdot (-dy) = \int_{0}^{a} \frac{f(a - y)}{f(a - y) + f(y)} \, dy$

Relabelling $y$ as $x$ (a dummy variable):

$I = \int_{0}^{a} \frac{f(a - x)}{f(x) + f(a - x)} \, dx \qquad \checkmark$

Adding the two expressions for $I$:

$2I = \int_{0}^{a} \frac{f(x) + f(a - x)}{f(x) + f(a - x)} \, dx = \int_{0}^{a} 1 \, dx = a$

$\therefore \quad I = \frac{a}{2}$

First application: Evaluate $\displaystyle\int_{0}^{1} \frac{\ln(x + 1)}{\ln(2 + x - x^2)} \, dx$.

Note that $2 + x - x^2 = (1 + x)(2 - x)$, so:

$\ln(2 + x - x^2) = \ln(1 + x) + \ln(2 - x)$

Let $f(x) = \ln(1 + x)$ with $a = 1$. Then:

$f(a - x) = f(1 - x) = \ln(1 + (1 - x)) = \ln(2 - x)$

So the integrand is $\dfrac{f(x)}{f(x) + f(1 - x)}$, and by the result:

$\int_{0}^{1} \frac{\ln(x + 1)}{\ln(2 + x - x^2)} \, dx = \frac{1}{2}$

Second application: Evaluate $\displaystyle\int_{0}^{\pi/2} \frac{\sin x}{\sin(x + \frac{\pi}{4})} \, dx$.

Expanding the denominator using the addition formula:

$\sin\!\left(x + \frac{\pi}{4}\right) = \sin x \cos\frac{\pi}{4} + \cos x \sin\frac{\pi}{4} = \frac{\sin x + \cos x}{\sqrt{2}}$

So the integral becomes:

$\sqrt{2}\int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$

Let $f(x) = \sin x$ with $a = \pi/2$. Then $f(a - x) = \sin(\pi/2 - x) = \cos x$, and $f(x) + f(a - x) = \sin x + \cos x$.

By the result:

$\sqrt{2}\int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx = \sqrt{2} \cdot \frac{\pi/2}{2} = \frac{\pi\sqrt{2}}{4}$

Part (ii)

We evaluate $\displaystyle I = \int_{1/2}^{2} \frac{\sin x}{x(\sin x + \sin \frac{1}{x})} \, dx$.

This can be written as $I = \displaystyle\int_{1/2}^{2} \frac{1}{x} \cdot \frac{\sin x}{\sin x + \sin(1/x)} \, dx$.

Substitute $u = 1/x$, so $x = 1/u$ and $dx = -1/u^2 \, du$. When $x = 1/2$, $u = 2$; when $x = 2$, $u = 1/2$.

$I = \int_{2}^{1/2} u \cdot \frac{\sin(1/u)}{\sin(1/u) + \sin u} \cdot \left(-\frac{1}{u^2}\right) du = \int_{1/2}^{2} \frac{1}{u} \cdot \frac{\sin(1/u)}{\sin u + \sin(1/u)} \, du$

Relabelling $u$ as $x$:

$I = \int_{1/2}^{2} \frac{1}{x} \cdot \frac{\sin(1/x)}{\sin x + \sin(1/x)} \, dx$

Adding this to the original expression for $I$:

$2I = \int_{1/2}^{2} \frac{1}{x} \cdot \frac{\sin x + \sin(1/x)}{\sin x + \sin(1/x)} \, dx = \int_{1/2}^{2} \frac{1}{x} \, dx = \Big[\ln x\Big]_{1/2}^{2} = \ln 2 - \ln\frac{1}{2} = 2\ln 2$

$\therefore \quad I = \ln 2$

Examiner Notes

This question received about the same number of “hits” as Q1 and came out with an average mark only fractionally lower. For the majority, the introductory work was successfully completed along with the rest of (i), although a lot of candidates’ working was very unclear in the first integral, involving logarithms. One or two marks were commonly lost as the correct answer of $\frac{1}{2}$ could easily have been guessed from the initial result, and the working produced by the candidates failed to convince markers that it had been obtained legitimately otherwise. The fault was often little more than failing either to identify the relevant "$f(x)$" or to show it implicitly by careful presentation of the working of the log. function.

The excellent part (ii) required candidates to mimic the method used to find the opening result rather than repeat its use in a new case, and this was only accessible to those with that extra bit of insight or determination.

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Question 5

Topic: 向量 (Vectors)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The points $A$ and $B$ have position vectors $\mathbf{i} + \mathbf{j} + \mathbf{k}$ and $5\mathbf{i} - \mathbf{j} - \mathbf{k}$, respectively, relative to the origin $O$. Find $\cos 2\alpha$, where $2\alpha$ is the angle $\angle AOB$.

(i) The line $L_1$ has equation $\mathbf{r} = \lambda(m\mathbf{i} + n\mathbf{j} + p\mathbf{k})$. Given that $L_1$ is inclined equally to $OA$ and to $OB$, determine a relationship between $m$, $n$ and $p$. Find also values of $m$, $n$ and $p$ for which $L_1$ is the angle bisector of $\angle AOB$.

(ii) The line $L_2$ has equation $\mathbf{r} = \mu(u\mathbf{i} + v\mathbf{j} + w\mathbf{k})$. Given that $L_2$ is inclined at an angle $\alpha$ to $OA$, where $2\alpha = \angle AOB$, determine a relationship between $u$, $v$ and $w$. Hence describe the surface with Cartesian equation $x^2 + y^2 + z^2 = 2(yz + zx + xy)$.

Hint

5 The opener here is a standard bit of A-level maths using the scalar product, and the following parts use this method, but with a bit of additional imagination needed. In 3-dimensions, there are infinitely lines inclined at a given angle to another, specified line, and this is the key idea of the final part of the question. Leading up to that, in (i), you need only realise that a line equally inclined to two specified (non-skew) lines must lie in the plane that bisects them (and is perpendicular to the plane that contains, in this case, the points $O$, $A$ and $B$). One might argue that the vector treatment of “planes” is further maths work, but these ideas are simple geometric ones.

$\cos 2\alpha = \frac{(1, 1, 1) \bullet (5, -1, -1)}{\sqrt{3} \cdot \sqrt{27}} = \frac{1}{3}$

** (i) ** $l_1$ equally inclined to $OA$ and $OB$ iff $\frac{(m, n, p) \bullet (1, 1, 1)}{\sqrt{m^2 + n^2 + p^2} \cdot \sqrt{3}} = \frac{(m, n, p) \bullet (5, -1, -1)}{\sqrt{m^2 + n^2 + p^2} \cdot \sqrt{27}}$

i.e. $3(m + n + p) = 5m - n - p$ or $m = 2(n + p)$.

For $l_1$ to be the angle bisector, we also require (e.g.) $\frac{m + n + p}{\sqrt{m^2 + n^2 + p^2} \cdot \sqrt{3}} = \cos \alpha$, where

$\cos 2\alpha = 2 \cos^2 \alpha - 1 = \frac{1}{3} \implies \cos \alpha = \frac{\sqrt{2}}{\sqrt{3}}$, so that $m + n + p = \sqrt{m^2 + n^2 + p^2} \cdot \sqrt{2}$.

Squaring both sides: $m^2 + n^2 + p^2 + 2mn + 2np + 2pm = 2(m^2 + n^2 + p^2)$

$\implies 2mn + 2np + 2pm = m^2 + n^2 + p^2$

Setting $m = 2n + 2p$ (or equivalent) then gives $2np + (2n + 2p)^2 = (2n + 2p)^2 + n^2 + p^2$

which gives $(n - p)^2 = 0 \implies p = n$, $m = 4n$.

Thus $\begin{pmatrix} m \\ n \\ p \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$, or any non-zero multiple will suffice.

** (ii) ** If you used the above method then you already have this relationship; namely,

$2uv + 2vw + 2wu = u^2 + v^2 + w^2$

Thus, $2xy + 2yz + 2zx = x^2 + y^2 + z^2$ gives all lines inclined at an angle $\cos^{-1} \frac{\sqrt{2}}{\sqrt{3}}$ to $OA$ and

hence describes the surface which is a double-cone, vertex at $O$, having central axis $OA$.

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6

Although it seems that 3-dimensional problems are not popular, this is actually a very, very easy question indeed and requires little more than identifying an appropriate right-angled triangle and using some basic trig. and/or Pythagoras. There are thus so many ways in which one can approach the three parts to this question that it is difficult to put forward just the one.

** (i) ** Taking the midpoint of $AB$ as the origin, $O$, with the $x$-axis along $AB$ and the $y$-axis along $OC$, we have a cartesian coordinate system to help us organise our thoughts.

Then $A = (-\frac{1}{2}, 0, 0)$, $B = (\frac{1}{2}, 0, 0)$,

$C = (0, \frac{\sqrt{3}}{2}, 0)$ by trig. or Pythagoras, and

$P = (0, \frac{\sqrt{3}}{6}, 0)$. The standard distance formula

then gives $PA$ (or $PB$) = $\frac{\sqrt{3}}{3}$ and $PD = \frac{\sqrt{6}}{3}$ or $\sqrt{\frac{2}{3}}$.

** (ii) ** The angle between adjacent faces is (e.g.) $\angle DOC = \cos^{-1} \left( \frac{\frac{1}{6}\sqrt{3}}{\frac{1}{2}\sqrt{3}} \right)$ in right-angled triangle

$DOP$, which gives the required answer, $\cos^{-1} \frac{1}{3}$.

(iii)

The centre, $S$, of the inscribed sphere must, by symmetry, lie on $PD$, equidistant from each vertex.

By Pythagoras, $x^2 = \frac{1}{12} + \left( \frac{\sqrt{6}}{9} - 2\frac{\sqrt{6}}{3}x + x^2 \right) \Rightarrow x = \frac{\sqrt{6}}{4}$. Then $r = x \sin(90^\circ - \text{(ii)}) = \frac{1}{3}x = \frac{\sqrt{6}}{12}$.

Alternatively, if you know that the sphere’s centre is at the centre of mass of the tetrahedron, the point ($S$) with position vector $\frac{1}{4}(\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d})$, then the answer

is just $\frac{1}{4} DP = \frac{\sqrt{6}}{12}$.

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Model Solution

Preamble: Finding $\cos 2\alpha$

We have $\overrightarrow{OA} = \mathbf{i} + \mathbf{j} + \mathbf{k}$ and $\overrightarrow{OB} = 5\mathbf{i} - \mathbf{j} - \mathbf{k}$.

$|\overrightarrow{OA}| = \sqrt{1 + 1 + 1} = \sqrt{3}, \qquad |\overrightarrow{OB}| = \sqrt{25 + 1 + 1} = 3\sqrt{3}$

$\overrightarrow{OA} \cdot \overrightarrow{OB} = 5 - 1 - 1 = 3$

$\cos 2\alpha = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}|\,|\overrightarrow{OB}|} = \frac{3}{\sqrt{3} \cdot 3\sqrt{3}} = \frac{3}{9} = \frac{1}{3}$

Part (i)

The line $L_1$ has direction vector $(m, n, p)$. For $L_1$ to be equally inclined to $OA$ and $OB$, the cosine of the angle with $OA$ must equal the cosine of the angle with $OB$:

$\frac{(m, n, p) \cdot (1, 1, 1)}{\sqrt{m^2 + n^2 + p^2} \cdot \sqrt{3}} = \frac{(m, n, p) \cdot (5, -1, -1)}{\sqrt{m^2 + n^2 + p^2} \cdot \sqrt{27}}$

Cancelling $\sqrt{m^2 + n^2 + p^2}$ and multiplying both sides by $3\sqrt{3}$:

$3(m + n + p) = 5m - n - p$

$3m + 3n + 3p = 5m - n - p$

$4n + 4p = 2m$

$m = 2(n + p)$

This is the required relationship between $m$, $n$ and $p$.

For $L_1$ to be the angle bisector, the angle between $L_1$ and $OA$ must be exactly $\alpha$. From $\cos 2\alpha = 2\cos^2\alpha - 1 = \frac{1}{3}$:

$2\cos^2\alpha = \frac{4}{3} \implies \cos\alpha = \sqrt{\frac{2}{3}}$

So we additionally require:

$\frac{m + n + p}{\sqrt{m^2 + n^2 + p^2} \cdot \sqrt{3}} = \sqrt{\frac{2}{3}} \implies \frac{m + n + p}{\sqrt{m^2 + n^2 + p^2}} = \sqrt{2}$

Squaring both sides:

$m^2 + n^2 + p^2 + 2mn + 2np + 2pm = 2(m^2 + n^2 + p^2)$

$2mn + 2np + 2pm = m^2 + n^2 + p^2 \qquad (\dagger)$

Substitute $m = 2n + 2p$ into $(\dagger)$. Left-hand side:

$2(2n+2p)n + 2np + 2(2n+2p)p = 4n^2 + 4np + 2np + 4np + 4p^2 = 4n^2 + 10np + 4p^2$

Right-hand side:

$(2n+2p)^2 + n^2 + p^2 = 4n^2 + 8np + 4p^2 + n^2 + p^2 = 5n^2 + 8np + 5p^2$

Setting equal:

$4n^2 + 10np + 4p^2 = 5n^2 + 8np + 5p^2$

$0 = n^2 - 2np + p^2 = (n - p)^2$

So $n = p$, giving $m = 4n$. Therefore $(m, n, p) = (4, 1, 1)$, or any non-zero scalar multiple.

Part (ii)

A line through the origin with direction $(u, v, w)$ makes angle $\alpha$ with $OA$ when:

$\frac{u + v + w}{\sqrt{u^2 + v^2 + w^2} \cdot \sqrt{3}} = \sqrt{\frac{2}{3}}$

By the same algebra as in part (i), squaring and simplifying yields:

$u^2 + v^2 + w^2 = 2(uv + vw + wu)$

This is the required relationship between $u$, $v$ and $w$.

For the surface $x^2 + y^2 + z^2 = 2(yz + zx + xy)$: a point $(x, y, z) \neq \mathbf{0}$ lies on this surface if and only if its position vector satisfies the condition above, meaning the line from the origin to $(x, y, z)$ makes angle $\alpha = \frac{1}{2}\cos^{-1}\!\frac{1}{3}$ with $OA$.

The set of all lines through a fixed point making a fixed angle with a given line is a double cone. Thus the surface is a double cone with vertex at the origin, axis along $OA$, and half-angle $\alpha = \frac{1}{2}\cos^{-1}\!\frac{1}{3}$.

Examiner Notes

This was the least popular question on the paper and attracted the lowest average score of about 7. This is partly explained by the way that, like Q3 and Q6 particularly, it drew a lot of attempts from desperate weaker students who started, only to give up before too long (in order, presumably, to try yet another question in some hit-and-miss approach, scrambling for odd marks here and there). Of those who persevered, there were plenty of marks to be had. Little more was required than the use of the scalar product, a careful application of algebra, and a modest grasp of the geometrical implications of what the working represented.

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Question 6

Topic: 立体几何 (Solid Geometry)  |  Difficulty: Standard  |  Marks: 20

Problem

6 Each edge of the tetrahedron $ABCD$ has unit length. The face $ABC$ is horizontal, and $P$ is the point in $ABC$ that is vertically below $D$.

(i) Find the length of $PD$.

(ii) Show that the cosine of the angle between adjacent faces of the tetrahedron is $1/3$.

(iii) Find the radius of the largest sphere that can fit inside the tetrahedron.

Hint

Although it seems that 3-dimensional problems are not popular, this is actually a very, very easy question indeed and requires little more than identifying an appropriate right-angled triangle and using some basic trig. and/or Pythagoras. There are thus so many ways in which one can approach the three parts to this question that it is difficult to put forward just the one.

** (i) ** Taking the midpoint of $AB$ as the origin, $O$, with the $x$-axis along $AB$ and the $y$-axis along $OC$, we have a cartesian coordinate system to help us organise our thoughts.

Then $A = (-\frac{1}{2}, 0, 0)$, $B = (\frac{1}{2}, 0, 0)$,

$C = (0, \frac{\sqrt{3}}{2}, 0)$ by trig. or Pythagoras, and

$P = (0, \frac{\sqrt{3}}{6}, 0)$. The standard distance formula

then gives $PA$ (or $PB$) = $\frac{\sqrt{3}}{3}$ and $PD = \frac{\sqrt{6}}{3}$ or $\sqrt{\frac{2}{3}}$.

** (ii) ** The angle between adjacent faces is (e.g.) $\angle DOC = \cos^{-1} \left( \frac{\frac{1}{6}\sqrt{3}}{\frac{1}{2}\sqrt{3}} \right)$ in right-angled triangle

$DOP$, which gives the required answer, $\cos^{-1} \frac{1}{3}$.

(iii) The centre, $S$, of the inscribed sphere must, by symmetry, lie on $PD$, equidistant from each vertex.

By Pythagoras, $x^2 = \frac{1}{12} + \left( \frac{\sqrt{6}}{9} - 2\frac{\sqrt{6}}{3}x + x^2 \right) \Rightarrow x = \frac{\sqrt{6}}{4}$. Then $r = x \sin(90^\circ - \text{(ii)}) = \frac{1}{3}x = \frac{\sqrt{6}}{12}$.

Alternatively, if you know that the sphere’s centre is at the centre of mass of the tetrahedron, the point ($S$) with position vector $\frac{1}{4}(\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d})$, then the answer

is just $\frac{1}{4} DP = \frac{\sqrt{6}}{12}$.

Model Solution

Part (i)

Let $M$ be the midpoint of $AB$. We set up coordinates with $M$ at the origin, the $x$-axis along $AB$, and the $y$-axis along $MC$.

Since $ABC$ is an equilateral triangle with side 1:

$A = \left(-\tfrac{1}{2}, 0, 0\right), \quad B = \left(\tfrac{1}{2}, 0, 0\right), \quad C = \left(0, \tfrac{\sqrt{3}}{2}, 0\right)$

The centroid $P$ of triangle $ABC$ lies at the average of the three vertices:

$P = \left(\frac{-\frac{1}{2} + \frac{1}{2} + 0}{3},\; \frac{0 + 0 + \frac{\sqrt{3}}{2}}{3},\; 0\right) = \left(0,\; \frac{\sqrt{3}}{6},\; 0\right)$

Since $D$ is vertically above $P$, we have $D = \left(0, \frac{\sqrt{3}}{6}, h\right)$ for some $h > 0$.

Using $|DA| = 1$:

$\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{6}\right)^2 + h^2 = 1$

$\frac{1}{4} + \frac{3}{36} + h^2 = 1$

$\frac{1}{4} + \frac{1}{12} + h^2 = 1$

$\frac{4}{12} + h^2 = 1 \implies h^2 = \frac{2}{3}$

$PD = h = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$

Part (ii)

Let $M'$ be the midpoint of $BC$. The dihedral angle between faces $ABC$ and $DBC$ along edge $BC$ equals the angle $\angle M'PD$ in the triangle $DM'P$, since:

• $DM'$ is the median of equilateral triangle $DBC$, so $DM' \perp BC$
• $PM'$ lies in the base $ABC$ and $PM' \perp BC$ (since $P$ is the centroid and $M'$ is the midpoint of $BC$, the line $PM'$ is along the median from $A$ to $BC$, which is perpendicular to $BC$)

Computing the lengths. First, $PM'$: since $P$ is the centroid and $M'$ is the midpoint of $BC$, the distance $PM'$ is one-third of the median from $A$ to $BC$:

$PM' = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}$

Next, $DM'$ is the height of equilateral triangle $DBC$:

$DM' = \frac{\sqrt{3}}{2}$

In right-angled triangle $DPM'$ (right-angled at $P$ since $DP \perp$ the base):

$\tan(\angle DM'P) = \frac{DP}{PM'} = \frac{\sqrt{6}/3}{\sqrt{3}/6} = \frac{\sqrt{6}}{3} \cdot \frac{6}{\sqrt{3}} = 2\sqrt{2}$

So $\sec^2(\angle DM'P) = 1 + 8 = 9$, giving $\cos(\angle DM'P) = \frac{1}{3}$.

The cosine of the angle between adjacent faces is $\frac{1}{3}$. $\checkmark$

Part (iii)

The centre $S$ of the inscribed sphere lies, by symmetry, on the altitude $PD$. Let $SP = r$, so $S$ is at distance $r$ from the base $ABC$, and $SD = PD - r = \frac{\sqrt{6}}{3} - r$.

For the sphere to be tangent to face $DBC$, the perpendicular distance from $S$ to the plane $DBC$ must equal $r$. Since $DP$ is perpendicular to the base $ABC$, and the dihedral angle between the base and face $DBC$ is $\alpha$ (where $\cos\alpha = \frac{1}{3}$), the angle between $DP$ and face $DBC$ is $90^\circ - \alpha$. Therefore the perpendicular distance from $S$ to face $DBC$ is:

$SD \cdot \sin(90^\circ - \alpha) = SD \cdot \cos\alpha = \frac{SD}{3}$

Setting this equal to $r$:

$r = \frac{1}{3}\left(\frac{\sqrt{6}}{3} - r\right)$

$3r = \frac{\sqrt{6}}{3} - r$

$4r = \frac{\sqrt{6}}{3}$

$r = \frac{\sqrt{6}}{12}$

Alternative method: For a regular tetrahedron, the inscribed sphere’s centre coincides with the centroid, which lies on each altitude at a ratio of $1 : 3$ from the base. Since $PD = \frac{\sqrt{6}}{3}$:

$r = \frac{1}{4} \cdot PD = \frac{\sqrt{6}}{12}$

Examiner Notes

Of the pure maths questions on the paper, only Q5 and this one attracted attempts from under half the candidature; this despite the fact that it is obviously (to the trained eye, at least) the easiest question on the paper. Parts (i) and (ii) require nothing more than GCSE trigonometry, and (iii) can be done in one line if one knows a little bit about geometric centres of 3-d shapes. Clearly 3-dimensional objects, and the associated trig., are sufficiently daunting to have put most folks off either completely or early on in the proceedings, and the average mark scored here was under 10.

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Question 7

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

Problem

7 (i) By considering the positions of its turning points, show that the curve with equation $y = x^3 - 3qx - q(1 + q) ,$ where $q > 0$ and $q \neq 1$, crosses the $x$-axis once only.

(ii) Given that $x$ satisfies the cubic equation $x^3 - 3qx - q(1 + q) = 0 ,$ and that $x = u + q/u ,$ obtain a quadratic equation satisfied by $u^3$. Hence find the real root of the cubic equation in the case $q > 0, q \neq 1$.

(iii) The quadratic equation $t^2 - pt + q = 0$ has roots $\alpha$ and $\beta$. Show that $\alpha^3 + \beta^3 = p^3 - 3qp .$ It is given that one of these roots is the square of the other. By considering the expression $(\alpha^2 - \beta)(\beta^2 - \alpha)$, find a relationship between $p$ and $q$. Given further that $q > 0, q \neq 1$ and $p$ is real, determine the value of $p$ in terms of $q$.

Hint

7 The first two parts of the question begin, helpfully, by saying exactly what to consider in order to proceed, and the material should certainly appear to be routine enough to make these parts very accessible. Where things are going in (iii) may not immediately be obvious but, presumably, there is a purpose to (i) and (ii) which should become clear in (iii).

(i) $y = x^3 - 3qx - q(1 + q) \Rightarrow \frac{dy}{dx} = 3(x^2 - q) = 0$ for $x = \pm\sqrt{q}$.

When $x = +\sqrt{q}$, $y = -q(\sqrt{q} + 1)^2 < 0$ since $q > 0$

When $x = -\sqrt{q}$, $y = -q(\sqrt{q} - 1)^2 < 0$ since $q > 0$ and $q \neq 1$

Since both TPs below x-axis, the curve crosses the x-axis once only (possibly with sketch)

(ii) $x = u + \frac{q}{u} \Rightarrow x^3 = u^3 + 3uq + 3\frac{q^2}{u} + \frac{q^3}{u^3}$

$0 = x^3 - 3qx - q(1 + q) = u^3 + 3uq + 3\frac{q^2}{u} + \frac{q^3}{u^3} - 3qu - 3\frac{q^2}{u} - q - q^2$

$\Rightarrow u^3 + \frac{q^3}{u^3} - q(1 + q) = 0$ or $(u^3)^2 - q(1 + q)(u^3) + q^3 = 0$

$u^3 = \frac{q(1 + q) \pm \sqrt{q^2(1 + q)^2 - 4q^3}}{2} = \frac{q}{2} \left\{ 1 + q \pm \sqrt{1 + 2q + q^2 - 4q} \right\}$

$= \frac{q}{2} \left\{ 1 + q \pm \sqrt{(1 - q)^2} \right\} = \frac{q}{2} \{ 1 + q \pm (1 - q) \} = q$ or $q^2$

giving $u = q^{\frac{1}{3}}$ or $q^{\frac{2}{3}}$ and $x = q^{\frac{1}{3}} + q^{\frac{2}{3}}$

(iii) $\alpha + \beta = p$, $\alpha\beta = q \Rightarrow \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = p^3 - 3qp$.

One root is the square of the other $\Leftrightarrow \alpha = \beta^2$ or $\beta = \alpha^2 \Leftrightarrow 0 = (\alpha^2 - \beta)(\alpha - \beta^2)$. Then $0 = (\alpha^2 - \beta)(\alpha - \beta^2) = \alpha^3 + \beta^3 - \alpha\beta - (\alpha\beta)^2 = p^3 - 3qp - q(1 + q)$

$\Leftrightarrow p = q^{\frac{1}{3}} + q^{\frac{2}{3}}$.

Model Solution

Part (i)

We have $y = x^3 - 3qx - q(1+q)$, so

$\frac{dy}{dx} = 3x^2 - 3q = 3(x^2 - q).$

Setting $\frac{dy}{dx} = 0$ gives turning points at $x = \pm\sqrt{q}$.

At $x = \sqrt{q}$:

$y = q\sqrt{q} - 3q\sqrt{q} - q - q^2 = -2q\sqrt{q} - q - q^2 = -q(2\sqrt{q} + 1 + q) = -q(\sqrt{q} + 1)^2.$

Since $q > 0$, we have $-q(\sqrt{q}+1)^2 < 0$.

At $x = -\sqrt{q}$:

$y = -q\sqrt{q} + 3q\sqrt{q} - q - q^2 = 2q\sqrt{q} - q - q^2 = -q(-2\sqrt{q} + 1 + q) = -q(\sqrt{q} - 1)^2.$

Since $q > 0$ and $q \neq 1$, we have $(\sqrt{q}-1)^2 > 0$, so $-q(\sqrt{q}-1)^2 < 0$.

Both turning points lie below the $x$-axis. Since the leading coefficient of $x^3$ is positive, $y \to -\infty$ as $x \to -\infty$ and $y \to +\infty$ as $x \to +\infty$. The curve therefore crosses the $x$-axis exactly once (to the right of $x = \sqrt{q}$).

Part (ii)

Substituting $x = u + q/u$ into $x^3 - 3qx - q(1+q) = 0$:

First, compute $x^3$:

$x^3 = \left(u + \frac{q}{u}\right)^3 = u^3 + 3u^2 \cdot \frac{q}{u} + 3u \cdot \frac{q^2}{u^2} + \frac{q^3}{u^3} = u^3 + 3qu + \frac{3q^2}{u} + \frac{q^3}{u^3}.$

Next, compute $3qx$:

$3qx = 3q\!\left(u + \frac{q}{u}\right) = 3qu + \frac{3q^2}{u}.$

Substituting into the cubic:

$u^3 + 3qu + \frac{3q^2}{u} + \frac{q^3}{u^3} - 3qu - \frac{3q^2}{u} - q(1+q) = 0.$

The $3qu$ and $3q^2/u$ terms cancel, leaving:

$u^3 + \frac{q^3}{u^3} - q(1+q) = 0.$

Multiplying through by $u^3$:

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

This is a quadratic in $u^3$. By the quadratic formula:

$u^3 = \frac{q(1+q) \pm \sqrt{q^2(1+q)^2 - 4q^3}}{2}.$

The discriminant simplifies as:

$q^2(1+q)^2 - 4q^3 = q^2\!\left((1+q)^2 - 4q\right) = q^2(1 - 2q + q^2) = q^2(1-q)^2.$

So $\sqrt{q^2(1-q)^2} = q|1-q|$. Since $q > 0$ and $q \neq 1$:

$u^3 = \frac{q(1+q) \pm q|1-q|}{2}.$

If $0 < q < 1$: $|1-q| = 1-q$, giving $u^3 = \frac{q(1+q) \pm q(1-q)}{2}$, so $u^3 = q$ or $u^3 = q^2$.

If $q > 1$: $|1-q| = q-1$, giving $u^3 = \frac{q(1+q) \pm q(q-1)}{2}$, so $u^3 = q^2$ or $u^3 = q$.

In either case, $u^3 = q$ or $u^3 = q^2$, yielding $u = q^{1/3}$ or $u = q^{2/3}$.

Since $x = u + q/u$, taking $u = q^{1/3}$ gives $x = q^{1/3} + q^{2/3}$. (Taking $u = q^{2/3}$ gives the same value of $x$.)

The real root of the cubic is $x = q^{1/3} + q^{2/3}$.

Part (iii)

By Vieta’s relations: $\alpha + \beta = p$ and $\alpha\beta = q$.

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = p^3 - 3qp.$

Now suppose one root is the square of the other, i.e., $\alpha = \beta^2$ or $\beta = \alpha^2$. This is equivalent to

$(\alpha^2 - \beta)(\beta^2 - \alpha) = 0.$

Expanding:

$(\alpha^2 - \beta)(\beta^2 - \alpha) = \alpha^2\beta^2 - \alpha^3 - \beta^3 + \alpha\beta = (\alpha\beta)^2 + \alpha\beta - (\alpha^3 + \beta^3).$

Setting this to zero and substituting $\alpha\beta = q$ and $\alpha^3 + \beta^3 = p^3 - 3qp$:

$q^2 + q - (p^3 - 3qp) = 0,$

$p^3 - 3qp - q(1+q) = 0. \qquad \text{(...)}$

This is exactly the cubic equation from part (ii) with $p$ in place of $x$. Since $q > 0$ and $q \neq 1$, part (ii) gives the unique real root:

$p = q^{1/3} + q^{2/3}.$

Examiner Notes

Q7 This proved to be the second most popular question on the paper, both by choice and by success. I imagine that its helpful structure probably contributed significantly to both. Part of the problem is that there are ways to do this using methods not on single maths specifications, so it was necessary to be quite specific. Nonetheless, there were still areas where marks were commonly lost; in (i), candidates were required to show that both TPs lie below the $x$-axis and, while one of the $y$-coordinates was obviously negative (being the sum of three negative terms), the other one was only obviously so by completing the square. The problems found by candidates, even in the first case, just highlights the widespread difficulty found by students when dealing with inequalities.

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Question 8

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

Problem

8 The curves $C_1$ and $C_2$ are defined by $y = e^{-x} \quad (x > 0) \quad \text{and} \quad y = e^{-x} \sin x \quad (x > 0),$ respectively. Sketch roughly $C_1$ and $C_2$ on the same diagram.

Let $x_n$ denote the $x$-coordinate of the $n$th point of contact between the two curves, where $0 < x_1 < x_2 < \cdots$, and let $A_n$ denote the area of the region enclosed by the two curves between $x_n$ and $x_{n+1}$. Show that $A_n = \tfrac{1}{2}(e^{2\pi} - 1)e^{-(4n+1)\pi/2}$ and hence find $\displaystyle \sum_{n=1}^{\infty} A_n$.

Hint

8 When asked to draw sketches of graphs, it is important to note the key features. The first curve is a standard “exponential decay” curve; the second has the extra factor of $\sin x$. Now $\sin x$ oscillates between $-1$ and $1$, and introduces zeroes at intervals of $\pi$. Thus, $C_2$ oscillates between $C_1$ and $-C_1$, with zeroes every $\pi$ units along the $x$-axis. This sketch of the two curves should then make it clear that the $x_i$ that are then introduced are the $x$-coordinates of $C_2$’s maxima, when $\sin x = 1$. [It is important to be clear in your description of $x_n$ and $x_{n+1}$ in terms of $n$ as these are going to be substituted as limits into the area integrals that follow.] The integration required to find one representative area will involve the use of “parts”, and the final summation looks like it must be that of an infinite GP.

The curves meet each time $\sin x = 1$ when $x = 2n\pi + \frac{\pi}{2}$ ($n = 0, 1, 2, \dots$).

Thus $x_n = \frac{(4n-3)\pi}{2}$ and $x_{n+1} = \frac{(4n+1)\pi}{2}$.

$\int (e^{-x} \sin x) dx$ attempted by parts $= -e^{-x} \cdot \cos x - \int (e^{-x} \cdot \cos x) dx$ or $-e^{-x} \cdot \sin x - \int (e^{-x} \cdot \sin x) dx$

(depending on your choice of ‘$1^{\text{st}}$’ and ‘$2^{\text{nd}}$’ part) $= -e^{-x} \cdot \cos x - \left\{ e^{-x} \cdot \sin x + \int (e^{-x} \cdot \sin x) dx \right\}$.

Then $I = -e^{-x} (\cos x + \sin x) - I$ (by “looping”) $= -\frac{1}{2} e^{-x} (\cos x + \sin x)$

$A_n = \int_{x_n}^{x_{n+1}} (e^{-x} - e^{-x} \sin x) dx = \left[ -e^{-x} + \frac{1}{2} e^{-x} (\cos x + \sin x) \right]_{x_n}^{x_{n+1}}$ or $\left[ \frac{1}{2} e^{-x} (\cos x + \sin x - 2) \right]_{x_n}^{x_{n+1}}$

$= \frac{1}{2} e^{-\frac{1}{2}\pi(4n+1)} (0+1-2) - \frac{1}{2} e^{-\frac{1}{2}\pi(4n-3)} (0+1-2) = \frac{1}{2} e^{-\frac{1}{2}\pi(4n+1)} (-1+e^{2\pi})$

Note that $A_1 = \frac{1}{2} e^{-\frac{5}{2}\pi} (e^{2\pi} - 1)$ and $A_{n+1} = e^{-2\pi} A_n$ so that $\sum_{n=1}^{\infty} A_n = A_1 \left\{ 1 + (e^{-2\pi}) + (e^{-2\pi})^2 + \dots \right\}$

$= \frac{1}{2} e^{-\frac{5}{2}\pi} (e^{2\pi} - 1) \times \frac{1}{1 - e^{-2\pi}} = \frac{1}{2} e^{-\frac{5}{2}\pi} (e^{2\pi} - 1) \times \frac{e^{2\pi}}{e^{2\pi} - 1}$ (using the $S_{\infty}$ of a GP formula) $= \frac{1}{2} e^{-\frac{1}{2}\pi}$

Model Solution

Sketch

$C_1: y = e^{-x}$ is a standard exponential decay curve, starting at $(0, 1)$ and approaching $0$ as $x \to \infty$.

$C_2: y = e^{-x}\sin x$ oscillates between $e^{-x}$ and $-e^{-x}$, crossing the $x$-axis at $x = \pi, 2\pi, 3\pi, \ldots$, and touching $C_1$ at its maxima where $\sin x = 1$.

Finding the points of contact

The curves meet when $e^{-x} = e^{-x}\sin x$, i.e., $\sin x = 1$. For $x > 0$:

$x = \frac{\pi}{2},\; \frac{5\pi}{2},\; \frac{9\pi}{2},\; \ldots$

In general, $x_n = \frac{(4n-3)\pi}{2}$ for $n = 1, 2, 3, \ldots$

So $x_n = \frac{(4n-3)\pi}{2}$ and $x_{n+1} = \frac{(4n+1)\pi}{2}$.

Computing $\int e^{-x}\sin x\, dx$ by parts

Let $I = \int e^{-x}\sin x\, dx$. Using integration by parts with $u = \sin x$, $dv = e^{-x}dx$:

$I = -e^{-x}\sin x + \int e^{-x}\cos x\, dx$

Apply parts again to $\int e^{-x}\cos x\, dx$ with $u = \cos x$, $dv = e^{-x}dx$:

$\int e^{-x}\cos x\, dx = -e^{-x}\cos x - \int e^{-x}\sin x\, dx = -e^{-x}\cos x - I$

Substituting back:

$I = -e^{-x}\sin x - e^{-x}\cos x - I$

$2I = -e^{-x}(\sin x + \cos x)$

$I = -\frac{1}{2}e^{-x}(\sin x + \cos x)$

Computing $A_n$

Between $x_n$ and $x_{n+1}$, we have $\sin x < 1$ (since $\sin x = 1$ only at the endpoints), so $e^{-x}\sin x < e^{-x}$. The enclosed area is:

$A_n = \int_{x_n}^{x_{n+1}} \left(e^{-x} - e^{-x}\sin x\right) dx = \int_{x_n}^{x_{n+1}} e^{-x}\, dx - \int_{x_n}^{x_{n+1}} e^{-x}\sin x\, dx$

$= \Big[-e^{-x}\Big]_{x_n}^{x_{n+1}} - \left[-\frac{1}{2}e^{-x}(\sin x + \cos x)\right]_{x_n}^{x_{n+1}}$

$= \left[-e^{-x} + \frac{1}{2}e^{-x}(\sin x + \cos x)\right]_{x_n}^{x_{n+1}}$

$= \left[\frac{1}{2}e^{-x}(\sin x + \cos x - 2)\right]_{x_n}^{x_{n+1}}$

At $x_{n+1} = \frac{(4n+1)\pi}{2}$: $\sin x_{n+1} = \sin\!\left(2n\pi + \frac{\pi}{2}\right) = 1$ and $\cos x_{n+1} = 0$.

At $x_n = \frac{(4n-3)\pi}{2}$: $\sin x_n = \sin\!\left(2(n-1)\pi + \frac{\pi}{2}\right) = 1$ and $\cos x_n = 0$.

So at both limits, $\sin x + \cos x - 2 = 1 + 0 - 2 = -1$.

$A_n = \frac{1}{2}e^{-x_{n+1}}(-1) - \frac{1}{2}e^{-x_n}(-1) = \frac{1}{2}\left(e^{-x_n} - e^{-x_{n+1}}\right)$

Now $x_{n+1} - x_n = \frac{(4n+1)\pi}{2} - \frac{(4n-3)\pi}{2} = 2\pi$, so $e^{-x_n} = e^{-x_{n+1}} \cdot e^{2\pi}$.

$A_n = \frac{1}{2}e^{-x_{n+1}}\!\left(e^{2\pi} - 1\right)$

Since $x_{n+1} = \frac{(4n+1)\pi}{2}$:

$A_n = \frac{1}{2}\!\left(e^{2\pi} - 1\right)e^{-(4n+1)\pi/2} \qquad \checkmark$

Summing the series

The ratio between consecutive terms is:

$\frac{A_{n+1}}{A_n} = \frac{e^{-(4(n+1)+1)\pi/2}}{e^{-(4n+1)\pi/2}} = e^{-2\pi}$

So $\{A_n\}$ is a geometric series with first term $A_1 = \frac{1}{2}(e^{2\pi}-1)e^{-5\pi/2}$ and common ratio $e^{-2\pi} < 1$.

$\sum_{n=1}^{\infty} A_n = \frac{A_1}{1 - e^{-2\pi}} = \frac{\frac{1}{2}(e^{2\pi}-1)e^{-5\pi/2}}{1 - e^{-2\pi}}$

Since $1 - e^{-2\pi} = \frac{e^{2\pi} - 1}{e^{2\pi}}$:

$\sum_{n=1}^{\infty} A_n = \frac{1}{2}(e^{2\pi}-1)e^{-5\pi/2} \cdot \frac{e^{2\pi}}{e^{2\pi}-1} = \frac{1}{2}e^{-5\pi/2} \cdot e^{2\pi} = \frac{1}{2}e^{-\pi/2}$

Examiner Notes

Q8 This was the sixth most popular of the pure maths questions, with an average score of just under