STEP2 2012 -- Pure Mathematics

STEP2 2012 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2012  |  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	级数展开 Series Expansion	Challenging	二项式级数展开, 生成函数乘积, 分类讨论系数
2	多项式方程 Polynomial Equations	Challenging	多项式次数比较, 函数方程求解, 待定系数法
3	积分 Integration	Challenging	变量代换, 反常积分计算, 三角代换x=tanθ
4	级数与不等式 Series and Inequalities	Challenging	对数级数展开, 不等式放缩, 极限与e的定义
5	曲线描绘 Curve Sketching	Standard	有理函数曲线描绘, 渐近线分析, 驻点求解
6	几何 Geometry	Challenging	余弦定理, 面积分解, 代数化简
7	向量与几何 Vectors and Geometry	Challenging	向量运算, 圆的性质, 代数化简
8	数列与递推关系 Sequences and Recurrence Relations	Challenging	递推关系, 数学归纳法, 不等式证明

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Question 1

Topic: 级数展开 Series Expansion  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Write down the general term in the expansion in powers of $x$ of $(1 - x^6)^{-2}$.

(i) Find the coefficient of $x^{24}$ in the expansion in powers of $x$ of

$(1 - x^6)^{-2}(1 - x^3)^{-1} .$

Obtain also, and simplify, formulae for the coefficient of $x^n$ in the different cases that arise.

(ii) Show that the coefficient of $x^{24}$ in the expansion in powers of $x$ of

$(1 - x^6)^{-2}(1 - x^3)^{-1}(1 - x)^{-1}$

is 55, and find the coefficients of $x^{25}$ and $x^{66}$.

Hint

1. Therefore the LHS is not constant so $n = 1$ and $p(x)$

is linear. Setting $p(x) = ax + b \implies p(p(x)) = a(ax + b) + b = a^2x + (a + 1)b$ and

$p(p(p(x))) = a[a^2x + (a + 1)b] + b = a^3x + (a^2 + a + 1)b$.

Then $a^3x + (a^2 + a + 1)b - 3ax - 3b + 2x \equiv 0 \implies (a^3 - 3a + 2)x + (a^2 + a - 2)b \equiv 0$

$\implies (a - 1)(a^2 + a - 2)x + (a^2 + a - 2)b \equiv 0$

$\implies (a^2 + a - 2)[(a - 1)x + b] \equiv 0$

$\implies (a + 2)(a - 1)[(a - 1)x + b] \equiv 0$

We have, then, that $a = -2$ or $1$. In either case, $b$ takes any (arbitrary) value and the solutions are thus $p_1(x) = -2x + b$ and $p_2(x) = x + b$.

(ii) $\text{Deg}[\text{RHS}] = 4$ while $\text{Deg}[\text{LHS}] \leq \max(n^2, 2n, n)$, so it follows that $n = 2$ and $p(x)$ is quadratic.

Setting $p(x) = ax^2 + bx + c$, we have

$2p(p(x)) = 2a(ax^2 + bx + c)^2 + 2b(ax^2 + bx + c) + 2c$

$= 2a \{ a^2x^4 + 2abx^3 + 2acx^2 + b^2x^2 + 2bcx + c^2 \} + 2b(ax^2 + bx + c) + 2c$

$3(p(x))^2 = 3[a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2]$ and $-4p(x) = -4ax^2 - 4bx - 4c$.

Thus, $\text{LHS} = (2a^3 + 3a^2)x^4 + (4a^2b + 6ab)x^3 + (2ab^2 + 4a^2c + 2ab + 3b^2 + 6ac - 4a)x^2$

$+ \ (4abc + 2b^2 + 6bc - 4b)x + (2ac^2 + 2bc + 2c + 3c^2 - 4c)$,

while the $\text{RHS} = x^4$.

Equating terms gives

$x^4) \quad 2a^3 + 3a^2 - 1 = 0 \implies (a + 1)^2(2a - 1) \implies a = -1$ or $\frac{1}{2}$

$x^3) \quad 2ab(2a + 3) = 0 \implies b = 0$

$x^2) \quad 2a(2ac + 3c - 2) = 0 \implies c = 2$ when $a = -1$; i.e. $p_1(x) = -x^2 + 2$

OR $c = \frac{1}{2}$ when $a = \frac{1}{2}$; i.e. $p_2(x) = \frac{1}{2}(x^2 + 1)$.

Note that there are two sets of conditions yet to be used, so the results obtained need to be checked (visibly) for consistency:

$x^1) \quad 2b(2ac + b + 3c - 2) = 0$ checks and $x^0) \quad c(2ac + 3c - 2) = 0$ checks also.

Question 3

It helps greatly to begin with, to note that if $t = \sqrt{x^2 + 1} + x$, then $\frac{1}{t} = \sqrt{x^2 + 1} - x$. These then give the

result $x = \frac{1}{2}t - \frac{1}{2}t^{-1}$, from which we find $\frac{dx}{dt} = \frac{1}{2} + \frac{1}{2}t^{-2}$ and (changing the limits) $x : (0, \infty) \to t : (1, \infty)$,

so that $\int_0^\infty f(\sqrt{x^2 + 1} + x) \, dx = \int_1^\infty f(t) \times \frac{1}{2} \left( 1 + \frac{1}{t^2} \right) \, dt = \frac{1}{2} \int_1^\infty f(x) \left( 1 + \frac{1}{x^2} \right) \, dx$, as required.

For the first integral, $I_1 = \int_0^\infty \frac{1}{(\sqrt{x^2 + 1} + x)^2} \, dx$, we are using $f(x) = \frac{1}{x^2}$ in the result established initially.

Then $I_1 = \frac{1}{2} \int_1^\infty \left( 1 + \frac{1}{x^2} \right) \cdot \frac{1}{x^2} \, dx = \frac{1}{2} \int_1^\infty (x^{-2} + x^{-4}) \, dx = \frac{1}{2} \left[ -\frac{1}{x} - \frac{1}{3x^3} \right]_1^\infty = \frac{1}{2}(0 + 1 + \frac{1}{3}) = \frac{2}{3}$.

In the case of the second integral, the substitution $x = \tan \theta \Rightarrow dx = \sec^2 \theta \, d\theta$. Also $\sqrt{1 + x^2} = \sec \theta$ and the required change of limits yields $(0, \frac{1}{2}\pi) \to (0, \infty)$. We then have

$\begin{aligned} I_2 &= \int_0^{\frac{1}{2}\pi} \frac{1}{(1 + \sin \theta)^3} \, d\theta = \int_0^{\frac{1}{2}\pi} \left( \frac{\sec \theta}{\sec \theta + \tan \theta} \right)^3 \, d\theta \quad \text{[Note the importance of changing to sec and tan]} \\ &= \int_0^{\frac{1}{2}\pi} \frac{\sec \theta}{(\sec \theta + \tan \theta)^3} \cdot \sec^2 \theta \, d\theta = \int_0^\infty \frac{\sqrt{x^2 + 1}}{(\sqrt{x^2 + 1} + x)^3} \, dx . \end{aligned}$

We now note, matching this up with the initial result, that we are using $f(t) = \frac{\frac{1}{2} \left( t + \frac{1}{t} \right)}{t^3} = \frac{t^2 + 1}{2t^4}$, so that

$I_2 = \frac{1}{2} \int_1^\infty \left( \frac{t^2 + 1}{t^2} \right) \left( \frac{t^2 + 1}{2t^4} \right) \, dt = \frac{1}{4} \int_1^\infty (t^{-2} + 2t^{-4} + t^{-6}) \, dt = \frac{1}{4} \left[ -\frac{1}{t} - \frac{2}{3t^3} - \frac{1}{5t^5} \right]_1^\infty = \frac{1}{4} (0 + 1 + \frac{2}{3} + \frac{1}{5}) = \frac{7}{15} .$

Question 4

(i) This first result is easily established: For $n, k > 1$, $n^{k+1} > n^k$ and $k + 1 > k$ so $(k + 1) \times n^{k+1} > k \times n^k$ $\Rightarrow \frac{1}{(k + 1)n^{k+1}} < \frac{1}{kn^k}$ (since all terms are positive).

$\begin{aligned} \text{Then } \ln \left( 1 + \frac{1}{n} \right) &= \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \frac{1}{5n^5} - \dots \quad \text{(a result which is valid since } 0 < \frac{1}{n} < 1) \\ &= \frac{1}{n} - \left( \frac{1}{2n^2} - \frac{1}{3n^3} \right) - \left( \frac{1}{4n^4} - \frac{1}{5n^5} \right) - \dots < \frac{1}{n} \text{ since each bracketed term is positive, using} \end{aligned}$ A1

the previous result. Exponentiating then gives $1 + \frac{1}{n} < e^{\frac{1}{n}} \Rightarrow \left( 1 + \frac{1}{n} \right)^n < e$.

(ii) A bit of preliminary log. work enables us to use the $\ln(1 + x)$ result on

$\begin{aligned} \ln \left( \frac{2y + 1}{2y - 1} \right) &= \ln \left( 1 + \frac{1}{2y} \right) - \ln \left( 1 - \frac{1}{2y} \right) = \left( \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \frac{1}{4(2y)^4} + \frac{1}{5(2y)^5} - \dots \right) \\ &\qquad\qquad\qquad\qquad\qquad\qquad - \left( -\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \frac{1}{4(2y)^4} - \frac{1}{5(2y)^5} - \dots \right) \\ &= 2 \left( \frac{1}{2y} + \frac{1}{3(2y)^3} + \frac{1}{5(2y)^5} + \dots \right) > \frac{1}{y} \quad \text{(since all terms after the first are positive).} \end{aligned}$

Again, note that we should justify that the series is valid for $0 < \frac{1}{2y} < 1$ i.e. $y > \frac{1}{2}$ in order to justify the

use of the given series. It then follows that $\ln \left( \frac{2y + 1}{2y - 1} \right)^y > 1$, and setting $y = n + \frac{1}{2}$ (the crucial final step)

gives $\ln \left( \frac{2n + 2}{2n} \right)^{n + \frac{1}{2}} > 1 \Rightarrow \left( 1 + \frac{1}{n} \right)^{n + \frac{1}{2}} > e$.

(iii) This final part only required a fairly informal argument, but the details still required a little bit of care in order to avoid being too vague.

As $n \to \infty$, $\left( 1 + \frac{1}{n} \right)^{n + \frac{1}{2}} = \left( 1 + \frac{1}{n} \right)^n \times \left( 1 + \frac{1}{n} \right)^{\frac{1}{2}} \to \left( 1 + \frac{1}{n} \right)^n \times 1 + \to \left( 1 + \frac{1}{n} \right)^n$ from above and e is squeezed

into the same limit from both above and below.

Question 5

With any curve-sketching question of this kind, it is important to grasp those features that are important and ignore those that aren’t. For instance, throughout this question, the position of the $y$-axis is entirely immaterial: it could be drawn through any branch of the curves in question or, indeed, appear as an

asymptote. So the usually key detail of the $y$-intercept, at $\left( 0, \frac{1}{a^2 - 1} \right)$ in part (i), does not help decide

what the function is up to. The asymptotes, turning points (clearly important in part (ii) since they are specifically requested), and any symmetries are important. The other key features to decide upon are the “short-term” (when $x$ is small) and the “long-term” (as $x \to \pm \infty$) behaviours.

In (i), there are vertical asymptotes at $x = a - 1$ and $x = a + 1$; while the $x$-axis is a horizontal asymptote. There is symmetry in the line $x = a$ (a consequence of which is the maximum TP in the “middle” branch) and the “long-term” behaviour of the curve is that it ultimately resembles the graph of $y = \frac{1}{x^2}$.

(ii) Differentiating the function in (ii) gives

$g'(x) = \frac{-2}{\left[ (x - a)^2 - 1 \right]^2 \left[ (x - b)^2 - 1 \right]^2} \left\{ (x - b) \left[ (x - a)^2 - 1 \right] + (x - a) \left[ (x - b)^2 - 1 \right] \right\}$

and setting the numerator $= 0 \implies (x - a)(x - b) [x - a + x - b] + [x - a + x - b] = 0$. Factorising yields

$(2x - a - b)(x^2 - (a + b)x + (ab - 1)) = 0$, so that $x = \frac{1}{2}(a + b)$ or $\frac{a + b \pm \sqrt{(a + b)^2 - 4ab + 4}}{2}$.

In the first case, where $b > a + 2$ (i.e. $a + 1 < b - 1$), there are five branches of the curve, with 4 vertical asymptotes: $x = a \pm 1$ and $x = b \pm 1$. As the function changes sign as it “crosses” each asymptote, and the “long-term” behaviour is still to resemble $y = \frac{1}{x^2}$, these branches alternate above and below the $x$-axis,

with symmetry in $x = \frac{1}{2}(a + b)$.

In the second case, where $b = a + 2$ (i.e. $a + 1 = b - 1$), the very middle section has collapsed, leaving only the four branches, but the curve is otherwise essentially unchanged from the previous case.

Question 6

A quick diagram helps here, leading to the important observation, from the GCSE geometry result “opposite angles of a cyclic quad. add to $180^\circ$”, that $\angle BCD = 180^\circ - \theta$. Then, using the Cosine Rule twice (and noting that $\cos(180^\circ - \theta) = -\cos\theta$):

in $\Delta BAD$: $BD^2 = a^2 + d^2 - 2ad \cos\theta$

in $\Delta BCD$: $BD^2 = b^2 + c^2 + 2bc \cos\theta$

Equating for $BD^2$ and re-arranging gives $\cos\theta = \frac{a^2 - b^2 - c^2 + d^2}{2(ad + bc)}$

Next, the well-known formula for triangle area, $\Delta = \frac{1}{2}ab \sin C$, twice, gives $Q = \frac{1}{2}ad \sin \theta + \frac{1}{2}bc \sin \theta$,

since $\sin(\pi - \theta) = \sin \theta$. Rearranging then gives $\sin \theta = \frac{2Q}{ad + bc}$ or $\frac{4Q}{2(ad + bc)}$.

Use of $\sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \frac{16Q^2}{4(ad + bc)^2} + \frac{(a^2 - b^2 - c^2 + d^2)^2}{4(ad + bc)^2} = 1$ and this then gives the printed

result, $16Q^2 = 4(ad + bc)^2 - (a^2 - b^2 - c^2 + d^2)^2$.

Then, $16Q^2 = (2ad + 2bc - a^2 + b^2 + c^2 - d^2)(2ad + 2bc + a^2 - b^2 - c^2 + d^2)$ by the difference-of-two-squares factorisation

$= ([b + c]^2 - [a - d]^2)([a + d]^2 - [b - c]^2)$

$= ([b + c] - [a - d])([b + c] + [a - d])([a + d] - [b - c])([a + d] + [b - c])$

using the difference-of-two-squares factorisation in each large bracket

$= (b + c + d - a)(a + b + c - d)(a + c + d - b)(a + b + d - c)$.

Splitting the 16 into four 2’s (one per bracket) and using $2s = a + b + c + d$

$\Rightarrow Q^2 = \frac{(2s - 2a)}{2} \frac{(2s - 2b)}{2} \frac{(2s - 2c)}{2} \frac{(2s - 2d)}{2} = (s - a)(s - b)(s - c)(s - d)$.

Finally, for a triangle (guaranteed cyclic), letting $d \rightarrow 0$ (Or $s - d \rightarrow s$ Or let $D \rightarrow A$), we get the result known as Heron’s Formula: $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.

Question 7

Many of you will know that this point $G$, used here, is the centroid of the triangle, and has position vector $\mathbf{g} = \frac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3)$.

Then $\vec{GX_1} = \mathbf{x}_1 - \mathbf{g} = \frac{1}{3}(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3)$ and so $\vec{GY_1} = -\frac{1}{3}\lambda_1(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3)$, where $\lambda_1 > 0$.

Also $\vec{OY_1} = \vec{OG} + \vec{GY_1} = \frac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3) - \frac{1}{3}\lambda_1(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3) = \frac{1}{3}([1 - 2\lambda_1]\mathbf{x}_1 + 1 + \lambda_1)$, the first printed result.

The really critical observation here is that the circle centre $O$, radius 1 has equation $|\mathbf{x}|^2 = 1$ or $\mathbf{x} \cdot \mathbf{x} = 1$, where $\mathbf{x}$ can be the p.v. of any point on the circle.

Thus, since $\vec{OY_1} \cdot \vec{OY_1} = 1$, we have

$1 = \frac{1}{9} \left\{ (1 - 2\lambda_1)^2 + 2(1 + \lambda_1)^2 + 2(1 - 2\lambda_1)(1 + \lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + 2(1 + \lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right\}$

$\Rightarrow 9 = 1 - 4\lambda_1 + 4\lambda_1^2 + 2 + 4\lambda_1 + 2\lambda_1^2 + 2(1 - 2\lambda_1)(1 + \lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + 2(1 + \lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3$

$\Rightarrow 0 = -3(1 - \lambda_1)(1 + \lambda_1) + (1 - 2\lambda_1)(1 + \lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + (1 + \lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3$

As $\lambda_1 > 0$, $0 = -3(1 - \lambda_1) + (1 - 2\lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + (1 + \lambda_1)\mathbf{x}_2 \cdot \mathbf{x}_3$

$\Rightarrow 0 = -3 + 3\lambda_1 + (\mathbf{x}_1 \cdot \mathbf{x}_2 + \mathbf{x}_2 \cdot \mathbf{x}_3 + \mathbf{x}_3 \cdot \mathbf{x}_1) + \lambda_1(\mathbf{x}_2 \cdot \mathbf{x}_3) - 2\lambda_1(\mathbf{x}_1 \cdot \mathbf{x}_2 + \mathbf{x}_1 \cdot \mathbf{x}_3)$

$\Rightarrow \lambda_1 = \frac{3 - (\alpha + \beta + \gamma)}{3 + \alpha - 2\beta - 2\gamma}$, using $\alpha = \mathbf{x}_2 \cdot \mathbf{x}_3$, $\beta = \mathbf{x}_3 \cdot \mathbf{x}_1$ and $\gamma = \mathbf{x}_1 \cdot \mathbf{x}_2$.

Similarly, $\lambda_2 = \frac{3 - (\alpha + \beta + \gamma)}{3 + \beta - 2\alpha - 2\gamma}$ and $\lambda_3 = \frac{3 - (\alpha + \beta + \gamma)}{3 + \gamma - 2\alpha - 2\beta}$.

Using $\frac{GX_i}{GY_i} = \frac{1}{\lambda_i}$ ($i = 1, 2, 3$), $\frac{GX_1}{GY_1} + \frac{GX_2}{GY_2} + \frac{GX_3}{GY_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \frac{9 + (\alpha + \beta + \gamma) - 4(\alpha + \beta + \gamma)}{3 - (\alpha + \beta + \gamma)}$ $= \frac{9 - 3(\alpha + \beta + \gamma)}{3 - (\alpha + \beta + \gamma)} = 3$.

[Interestingly, this result generalises to $n$ points on a circle: $\sum_{i=1}^{n} \frac{GX_i}{GY_i} = n$.]

Question 8

$\beta - \alpha > q$ ($q > 0$) $\Rightarrow \beta^2 - 2\alpha\beta + \alpha^2 > q^2 \Rightarrow \alpha^2 + \beta^2 - q^2 > 2\alpha\beta \Rightarrow \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} > 2 \Rightarrow$ the opening result, $\frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} - 2 > 0$.

$u_{n+1} = \frac{{u_n}^2 - q^2}{u_{n-1}}$ etc. $\Rightarrow {u_n}^2 - u_{n+1}u_{n-1} = q^2 = {u_{n+1}}^2 - u_{n+2}u_n$ (since the result is true at all stages) and equating for $q^2 \Rightarrow u_n(u_n + u_{n+2}) = u_{n+1}(u_{n-1} + u_{n+1})$.

Now this gives $\frac{u_n + u_{n+2}}{u_{n+1}} = \frac{u_{n-1} + u_{n+1}}{u_n}$ which $\Rightarrow \frac{u_{n-1} + u_{n+1}}{u_n}$ is constant (independent of $n$). Calling this constant $p$ gives $u_{n+1} - pu_n + u_{n-1} = 0$, as required. In order to determine $p$, we only need to use the fact that $p = \frac{u_{n-1} + u_{n+1}}{u_n}$ for all $n$, so we choose the first few terms to work with.

$u_2 = \frac{\beta^2 - q^2}{\alpha} \text{ and } p = \frac{u_0 + u_2}{u_1} = \frac{\alpha + \frac{\beta^2 - q^2}{\alpha}}{\beta} = \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta}.$

Alternatively, $u_2 = \gamma = \frac{\beta^2 - q^2}{\alpha} = p\beta - \alpha \Leftrightarrow p = \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta}$

and $u_3 = \frac{\gamma^2 - q^2}{\beta} = p\gamma - \beta \Leftrightarrow p = \frac{\gamma^2 + \beta^2 - q^2}{\beta\gamma} = \frac{\left(\frac{\beta^2 - q^2}{\alpha}\right)^2 + \beta^2 - q^2}{\beta\left(\frac{\beta^2 - q^2}{\alpha}\right)}$ $= \frac{(\beta^2 - q^2)^2 + \alpha^2(\beta^2 - q^2)}{\alpha\beta(\beta^2 - q^2)} = \frac{\beta^2 - q^2 + \alpha^2}{\alpha\beta}$

since $\beta^2 - q^2 \neq 0$ as $u_2$ non-zero (given). Since $p$ is consistent for any chosen $\alpha, \beta$, the proof follows inductively on any two consecutive terms of the sequence.

Finally, on to the given cases.

If $\beta > \alpha + q$, $u_{n+1} - u_n = (p - 1)u_n - u_{n-1} = \left(\frac{\beta^2 + \alpha^2 - q^2}{\alpha\beta} - 1\right)u_n - u_{n-1}$ $> (2 - 1)u_n - u_{n-1}$ by the initial result $> u_n - u_{n-1}$

Hence, if $u_n - u_{n-1} > 0$ then so is $u_{n+1} - u_n$. Since $\beta > \alpha$, $u_2 - u_1 > 0$ and proof follows inductively.

If $\beta = \alpha + q$ then $p = 2$ and $u_{n+1} - u_n = u_n - u_{n-1}$ so that the sequence is an AP.

Also, $u_0 = \alpha$, $u_1 = \alpha + q$, $u_2 = \alpha + 2q, \dots \Rightarrow$ the common difference is $q$ (and we still have a strictly increasing sequence, since $q > 0$ given).

Question 9

In the standard way, we use the constant-acceleration formulae to get

$x = ut \cos \alpha \text{ and } y = 2h - ut \sin \alpha - \frac{1}{2} gt^2 .$

When $x = a$, $t = \frac{a}{u \cos \alpha}$. Substituting this into the equation for $y \Rightarrow y = 2h - a \tan \alpha - \frac{ga^2}{2u^2} \sec^2 \alpha$.

As $y > h$ at this point (the ball, assuming it to be “a particle”, is above the net), we get

$h - a \tan \alpha > \frac{ga^2}{2u^2} \sec^2 \alpha \Rightarrow \frac{1}{u^2} < \frac{2(h - a \tan \alpha)}{ga^2 \sec^2 \alpha}, \text{ as required.}$

For the next part, we set $y = 0$ in $y = 2h - ut \sin \alpha - \frac{1}{2} gt^2$ and solve as a quadratic in $t$ to get

$t = \frac{-2u \sin \alpha + \sqrt{4u^2 \sin^2 \alpha + 16gh}}{2g} \dots \text{ (the positive root is required).}$

Setting $x = (u \cos \alpha)t$ and noting that $x < b$, $u \cos \alpha \left( \frac{\sqrt{u^2 \sin^2 \alpha + 4gh} - u \sin \alpha}{g} \right) < b$

$\Rightarrow \sqrt{u^2 \sin^2 \alpha + 4gh} < \frac{bg}{u \cos \alpha} + u \sin \alpha .$

There are several ways to proceed from here, but this is (perhaps) the most straightforward.

Squaring $\Rightarrow u^2 \sin^2 \alpha + 4gh < \frac{b^2 g^2 \sec^2 \alpha}{u^2} + 2bg \tan \alpha + u^2 \sin^2 \alpha$

Cancelling $u^2 \sin^2 \alpha$ both sides & dividing by $g \Rightarrow 4h < \frac{b^2 g^2 \sec^2 \alpha}{u^2} + 2b \tan \alpha$

Re-arranging for $\frac{1}{u^2} \Rightarrow \frac{2(2h - b \tan \alpha)}{b^2 g \sec^2 \alpha} < \frac{1}{u^2}$

Using the first result, $\frac{1}{u^2} < \frac{2(h - a \tan \alpha)}{a^2 g \sec^2 \alpha}$, in here $\Rightarrow \frac{2(2h - b \tan \alpha)}{b^2 g \sec^2 \alpha} < \frac{2(h - a \tan \alpha)}{a^2 g \sec^2 \alpha}$

Re-arranging for $\tan \alpha \Rightarrow ab(b - a) \tan \alpha < h(b^2 - 2a^2)$, which leads to the required final answer

$\tan \alpha < \frac{h(b^2 - 2a^2)}{ab(b - a)}$. However, it is necessary (since we might otherwise be dividing by a quantity that

could be negative) to explain that $b > a$ (we are now on the other side of the net to the projection point) else the direction of the inequality would reverse.

Question 10

As with many statics problems, a good diagram is essential to successful progress. Then there are relatively few mechanical principles to be applied … resolving (twice), taking moments, and the standard “Friction Law”. It is, of course, also important to get the angles right.

Taking moments about $M$: $R_1 a \sin \phi = R_2 a \sin \phi + F_1 a \cos \phi + F_2 a \cos \phi$

Using the Friction Law: $F_1 = \mu R_1$ and $F_2 = \mu R_2$

Dividing by $\cos \phi$ and re-arranging $R_1 \tan \phi = R_2 \tan \phi + \mu R_1 + \mu R_2$ $\Rightarrow (R_1 - R_2) \tan \phi = \mu (R_1 + R_2)$

For the second part, it seems likely that we will have to resolve twice (not having yet used this particular set of tools), though we could take moments about some other point in place of one resolution. There is also the question of which directions to resolve in – here, it should be clear very quickly that “horizontally and vertically” will only yield some very messy results.

Moments about $O$: $\mu (R_1 - R_2) r = W r \sin \phi \sin \theta$ Resolving // $AB$: $(R_1 - R_2) \cos \phi + \mu (R_1 + R_2) \sin \phi = W \sin \theta$ (Give one A1 here if all correct apart from a $-$ sign) Resolving $\perp^r AB$: $(R_1 + R_2) \sin \phi - \mu (R_1 - R_2) \cos \phi = W \cos \theta$

Note that only two of these are actually required, but it may be easier to write them all down first and then decide which two are best used.

Dividing these last two eqns. $\Rightarrow \tan \theta = \frac{(R_1 - R_2) \cos \phi + \mu (R_1 + R_2) \sin \phi}{(R_1 + R_2) \sin \phi - \mu (R_1 - R_2) \cos \phi}$

Using first result, $\mu (R_1 + R_2) = (R_1 - R_2) \tan \phi \Rightarrow \tan \theta = \frac{(R_1 - R_2) \cos \phi + (R_1 - R_2) \tan \phi \sin \phi}{(R_1 - R_2) \frac{\tan \phi}{\mu} \sin \phi - \mu (R_1 - R_2) \cos \phi}$

$\Rightarrow \tan \theta = \frac{\cos \phi + \tan \phi \sin \phi}{\frac{\tan \phi}{\mu} \sin \phi - \mu \cos \phi}$. (There is no need to note that $R_1 \neq R_2$ for then the rod would hve to be

positioned symmetrically in the cylinder.)

Multiplying throughout by $\mu \cos \phi \Rightarrow \tan \theta = \frac{\mu (\cos^2 \phi + \sin^2 \phi)}{\sin^2 \phi - \mu^2 \cos^2 \phi} = \frac{\mu}{1 - \cos^2 \phi - \mu^2 \cos^2 \phi}$ and, using

$\cos \phi = \frac{a}{r}$ gives $\tan \theta = \frac{\mu}{1 - \frac{a^2}{r^2} - \mu^2 \left( \frac{a^2}{r^2} \right)} = \frac{\mu r^2}{r^2 - a^2 (1 + \mu^2)}$.

Finally, $\tan \lambda = \mu = \left( \frac{R_1 - R_2}{R_1 + R_2} \right) \tan \phi$, from the first result, $< \tan \phi \Rightarrow \lambda < \phi$.

Question 11

Again, a diagram is really useful for helping put ones thoughts in order; also, we are going to have to consider what is going on generally (and not just “pattern-spot” our way up the line).


Using the principle of Conservation of Linear Momentum,

$\underline{\text{CLM} \rightarrow} \quad m u + M V_{i-1} = (M + im) V_i \quad (\text{NB } V_0 = 0) \quad \text{leads to}$ $V_1 = \frac{u}{k+1}, \ V_2 = \frac{2u}{k+1+2}, \ V_3 = \frac{3u}{k+1+2+3}, \dots, V_n = \frac{nu}{k + \frac{1}{2}n(n+1)} = \frac{2nu}{2k + n(n+1)}.$

Alternatively, $\underline{\text{CLM} \rightarrow \text{for all particles}}$ gives $mu + 2m\left(\frac{u}{2}\right) + 3m\left(\frac{u}{3}\right) + \dots + nm\left(\frac{u}{n}\right) = \left(k + \frac{1}{2}n(n+1)\right)mV$,

and rearranging for $V = V_n$ yields $V_n = \frac{2nu}{2k + n(n+1)}$.

The last collision occurs when $V_n \geq \frac{u}{n+1}$, i.e. $\frac{2nu}{N(N+1) + n(n+1)} \geq \frac{u}{n+1}$

$\Rightarrow 2n(n+1) \geq N(N+1) + n(n+1) \Rightarrow n(n+1) \geq N(N+1) \Rightarrow$ there are $N$ collisions.

Now, the total KE of all the $P_i$‘s is $\sum_{i=1}^{N} \frac{1}{2}(i\,m)\left(\frac{u}{i}\right)^2 = \frac{1}{2}mu^2 \sum_{i=1}^{N} \frac{1}{i}$.

The final KE of the block is $\frac{1}{2}N(N+1)mV_N^2 = \frac{1}{2}N(N+1)m\left(\frac{u}{N+1}\right)^2 = \frac{1}{2}mu^2 \left(\frac{N}{N+1}\right)$.

Therefore, the loss in KE is the difference: $\frac{1}{2}mu^2 \sum_{i=1}^{N} \frac{1}{i} - \frac{1}{2}mu^2 \left(\frac{N}{N+1}\right)$.

Since $\frac{N}{N+1} = 1 - \frac{1}{N+1}$, the loss in KE is $\frac{1}{2}mu^2 \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{N} - 1 + \frac{1}{N+1}\right) = \frac{1}{2}mu^2 \sum_{i=2}^{N+1} \left(\frac{1}{i}\right)$.

Question 12

This can be broken down into more (four) separate cases, but there is no need to:

P(light on) = $p \times \frac{3}{4} \times \frac{1}{2} + (1 - p) \times \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}(1 + 2p)$, and then the conditional probability

$\text{P(Hall | on)} = \frac{\frac{1}{8}(1 - p)}{\frac{1}{8}(1 + 2p)} = \frac{(1 - p)}{(1 + 2p)} .$

To make progress with this next part of the question, it is important to recognise the underlying binomial distribution, and that each day represents one such (Bernoulli) trial. We are thus dealing with $\text{B}(7, p_1)$,

where $p_1 = \frac{(1 - p)}{(1 + 2p)}$ is the previously given answer.

For the modal value to be 3, we must have $\text{P}(2) < \text{P}(3) < \text{P}(4)$; that is,

$\binom{7}{2}(p_1)^2(1 - p_1)^5 < \binom{7}{3}(p_1)^3(1 - p_1)^4 \quad \text{and} \quad \binom{7}{4}(p_1)^4(1 - p_1)^3 < \binom{7}{3}(p_1)^3(1 - p_1)^4 .$

Using $p_1 = \frac{(1 - p)}{(1 + 2p)}$ gives

$21 \left( \frac{1 - p}{1 + 2p} \right)^2 \left( \frac{3p}{1 + 2p} \right)^5 < 35 \left( \frac{1 - p}{1 + 2p} \right)^3 \left( \frac{3p}{1 + 2p} \right)^4 \implies 3(3p) < 5(1 - p) \implies p < \frac{5}{14}$

and

$35 \left( \frac{1 - p}{1 + 2p} \right)^4 \left( \frac{3p}{1 + 2p} \right)^3 < 35 \left( \frac{1 - p}{1 + 2p} \right)^3 \left( \frac{3p}{1 + 2p} \right)^4 \implies (1 - p) < (3p) \implies p > \frac{1}{4} .$

Question 13

Working with the distribution $\text{P}_o(\lambda = k\pi y^2)$, $\text{P(no supermarkets)} = e^{-k\pi y^2}$ and $\text{P}(Y < y) = 1 - e^{-k\pi y^2}$. Differentiating w.r.t. $y$ to find the pdf of $Y \implies f(y) = 2k\pi y e^{-k\pi y^2}$, as given. Then

$\text{E}(Y) = \int_{0}^{\infty} 2k\pi y^2 e^{-k\pi y^2} dy . \text{ Using Integration by Parts and writing } 2k\pi y^2 e^{-k\pi y^2} \text{ as } y \left( 2k\pi y e^{-k\pi y^2} \right)$

gives $\text{E}(Y) = \left[ y \left( e^{-k\pi y^2} \right) \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-k\pi y^2} dy = 0 + \int_{0}^{\infty} e^{-k\pi y^2} dy$. It is useful (but not essential) to use the

simplifying substitution $x = y\sqrt{2k\pi}$ at this stage to get $\frac{1}{\sqrt{2k\pi}} \int_{0}^{\infty} e^{-\frac{1}{2}x^2} dx = \frac{1}{\sqrt{2k\pi}} \sqrt{\frac{\pi}{2}} = \frac{1}{2\sqrt{k}}$ (by the

given result, relating to the standard normal distribution’s pdf, at the very beginning of the question).

Next, $\text{E}(Y^2) = \int_{0}^{\infty} 2k\pi y^3 e^{-k\pi y^2} dy$, and using Integration by Parts and, in a similar way to earlier,

writing $2k\pi y^3 e^{-k\pi y^2}$ as $y^2 \left( 2k\pi y e^{-k\pi y^2} \right)$, $\text{E}(Y^2) = \left[ y^2 \left( e^{-k\pi y^2} \right) \right]_{0}^{\infty} + \int_{0}^{\infty} 2y e^{-k\pi y^2} dy$

$= 0 + \frac{1}{k\pi} \int_{0}^{\infty} 2k\pi y e^{-k\pi y^2} = \frac{-1}{k\pi} \left[ e^{-k\pi y^2} \right]_{0}^{\infty} \text{ (using a previous result, or by substitution) } = \frac{1}{k\pi}$

$\implies \text{Var}(Y) = \frac{1}{k\pi} - \frac{1}{4k} = \frac{4 - \pi}{4k\pi} , \text{ the given answer, as required.}$

STEP 3 2012 Hints and Solutions

Model Solution

General term of $(1 - x^6)^{-2}$.

By the generalized binomial theorem, $(1 - t)^{-2} = \sum_{k=0}^{\infty}(k+1)t^k$ for $|t| < 1$. Setting $t = x^6$:

$(1 - x^6)^{-2} = \sum_{k=0}^{\infty}(k+1)x^{6k}.$

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Part (i): Coefficient of $x^{24}$ in $(1 - x^6)^{-2}(1 - x^3)^{-1}$.

Since $(1 - x^3)^{-1} = \sum_{j=0}^{\infty} x^{3j}$, the product is

$(1 - x^6)^{-2}(1 - x^3)^{-1} = \sum_{k=0}^{\infty}\sum_{j=0}^{\infty}(k+1)x^{6k+3j}.$

For $x^{24}$, we need $6k + 3j = 24$, i.e. $j = 8 - 2k$. The constraint $j \ge 0$ gives $k = 0, 1, 2, 3, 4$.

$[x^{24}] = \sum_{k=0}^{4}(k+1) = 1 + 2 + 3 + 4 + 5 = 15.$

General formula. For the coefficient of $x^n$, we need $6k + 3j = n$, so $j = \frac{n - 6k}{3}$. Since $3 \mid 6k$ always, we require $3 \mid n$; otherwise the coefficient is $0$.

If $3 \mid n$, write $n = 3m$. Then $j = m - 2k$ with $0 \le k \le \lfloor m/2 \rfloor$.

$[x^n] = \sum_{k=0}^{\lfloor m/2 \rfloor}(k+1) = \frac{(\lfloor m/2 \rfloor + 1)(\lfloor m/2 \rfloor + 2)}{2}.$

Explicitly, writing $r = \lfloor n/6 \rfloor$:

$[x^n] = \begin{cases} \dfrac{(r+1)(r+2)}{2} & \text{if } 3 \mid n, \\[6pt] 0 & \text{if } 3 \nmid n. \end{cases}$

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Part (ii): Coefficient of $x^{24}$ in $(1 - x^6)^{-2}(1 - x^3)^{-1}(1 - x)^{-1}$.

With $(1 - x)^{-1} = \sum_{i=0}^{\infty} x^i$, the coefficient of $x^n$ in the triple product is

$\sum_{\substack{k, j, i \ge 0 \\ 6k + 3j + i = n}} (k+1).$

For each pair $(k, j)$ with $6k + 3j \le n$, there is exactly one $i = n - 6k - 3j \ge 0$.

For $n = 24$: the valid pairs are $k = 0, 1, \ldots, 4$ with $j = 0, 1, \ldots, 8 - 2k$.

$k = 0: \ 9 \text{ terms}, \quad k = 1: \ 7 \text{ terms}, \quad k = 2: \ 5 \text{ terms}, \quad k = 3: \ 3 \text{ terms}, \quad k = 4: \ 1 \text{ term}.$

$[x^{24}] = 1 \times 9 + 2 \times 7 + 3 \times 5 + 4 \times 3 + 5 \times 1 = 9 + 14 + 15 + 12 + 5 = 55. \qquad \blacksquare$

Coefficient of $x^{25}$. The coefficient of $x^n$ in $(1-x)^{-1}g(x)$ is $\sum_{\ell=0}^{n}[x^\ell]g(x)$. Since $[x^\ell](1-x^6)^{-2}(1-x^3)^{-1} = 0$ when $3 \nmid \ell$:

$[x^{25}] = \sum_{\ell=0}^{25}[x^\ell]g(x) = \sum_{\ell=0}^{24}[x^\ell]g(x) = [x^{24}] = 55.$

(The term $\ell = 25$ vanishes since $3 \nmid 25$.)

Coefficient of $x^{66}$. Similarly,

$[x^{66}] = \sum_{m=0}^{22}[x^{3m}]g(x).$

Using the formula $[x^{3m}]g(x) = \frac{(\lfloor m/2 \rfloor + 1)(\lfloor m/2 \rfloor + 2)}{2}$, we split into even and odd $m$:

Even $m = 0, 2, 4, \ldots, 22$ (12 terms, with $m = 2s$, $s = 0, 1, \ldots, 11$): $\displaystyle\sum_{s=0}^{11}\frac{(s+1)(s+2)}{2} = \frac{12 \cdot 13 \cdot 14}{6} = 364.$

Odd $m = 1, 3, 5, \ldots, 21$ (11 terms, with $m = 2s+1$, $s = 0, 1, \ldots, 10$): $\displaystyle\sum_{s=0}^{10}\frac{(s+1)(s+2)}{2} = \frac{11 \cdot 12 \cdot 13}{6} = 286.$

$[x^{66}] = 364 + 286 = 650.$

Examiner Notes

Question 1

In spite of the printing error at the start of the question, two thirds of the candidates attempted this question. Most candidates earned a quarter of the marks by obtaining $z$ in terms of $y$ in part (i), and then went no further. Some candidates realised the significance of the first line of the question that the expression given was an exact differential, and those that did frequently then scored highly. Some candidates found their own way through having obtained $z$ in terms of $y$ in part (i), then making $y$ the subject substituted back to find a second order differential equation for $z$, which they then solved and hence completed the solution to each part.

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Question 2

Topic: 多项式方程 Polynomial Equations  |  Difficulty: Challenging  |  Marks: 20

Problem

2 If $p(x)$ and $q(x)$ are polynomials of degree $m$ and $n$, respectively, what is the degree of $p(q(x))$?

(i) The polynomial $p(x)$ satisfies

$p(p(p(x))) - 3p(x) = -2x$

for all $x$. Explain carefully why $p(x)$ must be of degree 1, and find all polynomials that satisfy this equation.

(ii) Find all polynomials that satisfy

$2p(p(x)) + 3[p(x)]^2 - 4p(x) = x^4$

for all $x$.

Hint

Firstly, $p(q(x))$ has degree $mn$.

(i) $\text{Deg}[p(x)] = n \implies \text{Deg}[p(p(x))] = n^2 \ \& \ \text{Deg}[p(p(p(x)))] = n^3$.

$\text{Deg}[\text{LHS}] \leq \max(n^3, n)$ while RHS is of degree 1. Therefore the LHS is not constant so $n = 1$ and $p(x)$

is linear. Setting $p(x) = ax + b \implies p(p(x)) = a(ax + b) + b = a^2x + (a + 1)b$ and

$p(p(p(x))) = a[a^2x + (a + 1)b] + b = a^3x + (a^2 + a + 1)b$.

Then $a^3x + (a^2 + a + 1)b - 3ax - 3b + 2x \equiv 0 \implies (a^3 - 3a + 2)x + (a^2 + a - 2)b \equiv 0$

$\implies (a - 1)(a^2 + a - 2)x + (a^2 + a - 2)b \equiv 0$

$\implies (a^2 + a - 2)[(a - 1)x + b] \equiv 0$

$\implies (a + 2)(a - 1)[(a - 1)x + b] \equiv 0$

We have, then, that $a = -2$ or $1$. In either case, $b$ takes any (arbitrary) value and the solutions are thus $p_1(x) = -2x + b$ and $p_2(x) = x + b$.

(ii) $\text{Deg}[\text{RHS}] = 4$ while $\text{Deg}[\text{LHS}] \leq \max(n^2, 2n, n)$, so it follows that $n = 2$ and $p(x)$ is quadratic.

Setting $p(x) = ax^2 + bx + c$, we have

$2p(p(x)) = 2a(ax^2 + bx + c)^2 + 2b(ax^2 + bx + c) + 2c$

$= 2a \{ a^2x^4 + 2abx^3 + 2acx^2 + b^2x^2 + 2bcx + c^2 \} + 2b(ax^2 + bx + c) + 2c$

$3(p(x))^2 = 3[a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2]$ and $-4p(x) = -4ax^2 - 4bx - 4c$.

Thus, $\text{LHS} = (2a^3 + 3a^2)x^4 + (4a^2b + 6ab)x^3 + (2ab^2 + 4a^2c + 2ab + 3b^2 + 6ac - 4a)x^2$

$+ \ (4abc + 2b^2 + 6bc - 4b)x + (2ac^2 + 2bc + 2c + 3c^2 - 4c)$,

while the $\text{RHS} = x^4$.

Equating terms gives

$x^4) \quad 2a^3 + 3a^2 - 1 = 0 \implies (a + 1)^2(2a - 1) \implies a = -1$ or $\frac{1}{2}$

$x^3) \quad 2ab(2a + 3) = 0 \implies b = 0$

$x^2) \quad 2a(2ac + 3c - 2) = 0 \implies c = 2$ when $a = -1$; i.e. $p_1(x) = -x^2 + 2$

OR $c = \frac{1}{2}$ when $a = \frac{1}{2}$; i.e. $p_2(x) = \frac{1}{2}(x^2 + 1)$.

Note that there are two sets of conditions yet to be used, so the results obtained need to be checked (visibly) for consistency:

$x^1) \quad 2b(2ac + b + 3c - 2) = 0$ checks and $x^0) \quad c(2ac + 3c - 2) = 0$ checks also.

Model Solution

Stem. If $p(x)$ has degree $m$ and $q(x)$ has degree $n$, then $p(q(x))$ has degree $mn$.

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Part (i): $p(p(p(x))) - 3p(x) = -2x$.

Let $\deg p = n$. Then $\deg p(p(p(x))) = n^3$ and $\deg 3p(x) = n$.

If $n \ge 2$: $\deg(\text{LHS}) = \max(n^3, n) = n^3 \ge 8$, but $\deg(\text{RHS}) = 1$. Contradiction.

If $n = 0$: $p(x) = c$ (constant), giving $c - 3c = -2x$, i.e. $-2c = -2x$ for all $x$. Contradiction.

So $n = 1$ and $p(x) = ax + b$. Computing the iterates:

$p(p(x)) = a(ax + b) + b = a^2 x + (a + 1)b$

$p(p(p(x))) = a[a^2 x + (a+1)b] + b = a^3 x + (a^2 + a + 1)b$

Substituting into the equation:

$a^3 x + (a^2 + a + 1)b - 3ax - 3b = -2x$

$(a^3 - 3a + 2)x + (a^2 + a - 2)b = 0 \quad \text{for all } x.$

For the $x$-coefficient: $a^3 - 3a + 2 = 0$. Testing $a = 1$: $1 - 3 + 2 = 0$. Factoring out $(a - 1)$:

$a^3 - 3a + 2 = (a - 1)(a^2 + a - 2) = (a - 1)^2(a + 2) = 0.$

So $a = 1$ or $a = -2$.

For the constant term: $(a^2 + a - 2)b = (a + 2)(a - 1)b = 0$.

• $a = 1$: $(3)(0)b = 0$ is satisfied for all $b$.
• $a = -2$: $(0)(-3)b = 0$ is satisfied for all $b$.

The solutions are $p(x) = x + b$ and $p(x) = -2x + b$ for any constant $b$.

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Part (ii): $2p(p(x)) + 3[p(x)]^2 - 4p(x) = x^4$.

Let $\deg p = n$. Then $\deg p(p(x)) = n^2$, $\deg [p(x)]^2 = 2n$, $\deg p(x) = n$.

For $n \ge 3$: $n^2 \ge 9 > 4$, so $\deg(\text{LHS}) > 4$. For $n = 1$: $\max(1, 2, 1) = 2 \ne 4$.

So $n = 2$ and $p(x) = ax^2 + bx + c$ with $a \ne 0$.

Computing each term. First, $(ax^2 + bx + c)^2 = a^2x^4 + 2abx^3 + (b^2 + 2ac)x^2 + 2bcx + c^2$.

$2p(p(x)) = 2a(ax^2 + bx + c)^2 + 2b(ax^2 + bx + c) + 2c$

$= 2a^3x^4 + 4a^2bx^3 + 2(ab^2 + 2a^2c + ab)x^2 + 2(2abc + b^2)x + 2(ac^2 + bc + c)$

$3[p(x)]^2 = 3a^2x^4 + 6abx^3 + 3(b^2 + 2ac)x^2 + 6bcx + 3c^2$

$-4p(x) = -4ax^2 - 4bx - 4c$

Equating coefficients. Comparing LHS with $x^4$:

$x^4)$ $\quad 2a^3 + 3a^2 = 1$, i.e. $2a^3 + 3a^2 - 1 = 0$. Testing $a = -1$: $-2 + 3 - 1 = 0$. Factoring:

$2a^3 + 3a^2 - 1 = (a + 1)(2a^2 + a - 1) = (a + 1)^2(2a - 1) = 0.$

So $a = -1$ or $a = \frac{1}{2}$.

$x^3)$ $\quad 4a^2b + 6ab = 2ab(2a + 3) = 0$. Since $a \ne 0$ and $2a + 3 \ne 0$ for $a = -1$ or $\frac{1}{2}$, we get $b = 0$.

$x^2)$ $\quad$ With $b = 0$: $4a^2c + 6ac - 4a = 0$, i.e. $2a(2ac + 3c - 2) = 0$. Since $a \ne 0$:

$c(2a + 3) = 2 \implies c = \frac{2}{2a + 3}.$

For $a = -1$: $c = \frac{2}{1} = 2$, giving $p_1(x) = -x^2 + 2$.

For $a = \frac{1}{2}$: $c = \frac{2}{4} = \frac{1}{2}$, giving $p_2(x) = \frac{1}{2}(x^2 + 1)$.

$x^1)$ $\quad$ With $b = 0$: $0 = 0$. Consistent.

$x^0)$ $\quad$ With $b = 0$: $2ac^2 + 3c^2 - 2c = c(2ac + 3c - 2) = 0$. Since $2ac + 3c - 2 = 0$ from the $x^2$ equation, this is satisfied.

The solutions are $p(x) = -x^2 + 2$ and $p(x) = \frac{1}{2}(x^2 + 1)$.

Examiner Notes

2. Most did very well with the stem, though a few were unable to obtain a proper second order equation. Those that attempted part (i) were usually successful. The non-trivial exponential calculations in part (ii) caused problems for some making computation mistakes whilst others were totally on top of this. Part (iii) tested the candidates on two levels, interpreting the sigma notation correctly, and recognising and using the geometric series. Some managed this excellently.

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Question 3

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

Problem

3 Show that, for any function f (for which the integrals exist), $\int_{0}^{\infty} f(x + \sqrt{1 + x^2}) \, dx = \frac{1}{2} \int_{1}^{\infty} \left( 1 + \frac{1}{t^2} \right) f(t) \, dt \, .$ Hence evaluate $\int_{0}^{\infty} \frac{1}{2x^2 + 1 + 2x\sqrt{x^2 + 1}} \, dx \, ,$ and, using the substitution $x = \tan \theta$, $\int_{0}^{\frac{1}{2}\pi} \frac{1}{(1 + \sin \theta)^3} \, d\theta \, .$

Hint

It helps greatly to begin with, to note that if $t = \sqrt{x^2 + 1} + x$, then $\frac{1}{t} = \sqrt{x^2 + 1} - x$. These then give the

result $x = \frac{1}{2}t - \frac{1}{2}t^{-1}$, from which we find $\frac{dx}{dt} = \frac{1}{2} + \frac{1}{2}t^{-2}$ and (changing the limits) $x : (0, \infty) \to t : (1, \infty)$,

so that $\int_0^\infty f(\sqrt{x^2 + 1} + x) \, dx = \int_1^\infty f(t) \times \frac{1}{2} \left( 1 + \frac{1}{t^2} \right) \, dt = \frac{1}{2} \int_1^\infty f(x) \left( 1 + \frac{1}{x^2} \right) \, dx$, as required.

For the first integral, $I_1 = \int_0^\infty \frac{1}{(\sqrt{x^2 + 1} + x)^2} \, dx$, we are using $f(x) = \frac{1}{x^2}$ in the result established initially.

Then $I_1 = \frac{1}{2} \int_1^\infty \left( 1 + \frac{1}{x^2} \right) \cdot \frac{1}{x^2} \, dx = \frac{1}{2} \int_1^\infty (x^{-2} + x^{-4}) \, dx = \frac{1}{2} \left[ -\frac{1}{x} - \frac{1}{3x^3} \right]_1^\infty = \frac{1}{2}(0 + 1 + \frac{1}{3}) = \frac{2}{3}$.

In the case of the second integral, the substitution $x = \tan \theta \Rightarrow dx = \sec^2 \theta \, d\theta$. Also $\sqrt{1 + x^2} = \sec \theta$ and the required change of limits yields $(0, \frac{1}{2}\pi) \to (0, \infty)$. We then have

$\begin{aligned} I_2 &= \int_0^{\frac{1}{2}\pi} \frac{1}{(1 + \sin \theta)^3} \, d\theta = \int_0^{\frac{1}{2}\pi} \left( \frac{\sec \theta}{\sec \theta + \tan \theta} \right)^3 \, d\theta \quad \text{[Note the importance of changing to sec and tan]} \\ &= \int_0^{\frac{1}{2}\pi} \frac{\sec \theta}{(\sec \theta + \tan \theta)^3} \cdot \sec^2 \theta \, d\theta = \int_0^\infty \frac{\sqrt{x^2 + 1}}{(\sqrt{x^2 + 1} + x)^3} \, dx . \end{aligned}$

We now note, matching this up with the initial result, that we are using $f(t) = \frac{\frac{1}{2} \left( t + \frac{1}{t} \right)}{t^3} = \frac{t^2 + 1}{2t^4}$, so that

$I_2 = \frac{1}{2} \int_1^\infty \left( \frac{t^2 + 1}{t^2} \right) \left( \frac{t^2 + 1}{2t^4} \right) \, dt = \frac{1}{4} \int_1^\infty (t^{-2} + 2t^{-4} + t^{-6}) \, dt = \frac{1}{4} \left[ -\frac{1}{t} - \frac{2}{3t^3} - \frac{1}{5t^5} \right]_1^\infty = \frac{1}{4} (0 + 1 + \frac{2}{3} + \frac{1}{5}) = \frac{7}{15} .$

Model Solution

Step 1: Establish the substitution.

Let $t = x + \sqrt{1 + x^2}$. We note a useful companion identity: multiplying by the conjugate gives

$\frac{1}{t} = \frac{1}{x + \sqrt{1 + x^2}} = \frac{\sqrt{1 + x^2} - x}{(1 + x^2) - x^2} = \sqrt{1 + x^2} - x.$

Adding and subtracting these two expressions:

$t + \frac{1}{t} = 2\sqrt{1 + x^2}, \qquad t - \frac{1}{t} = 2x.$

So $x = \frac{1}{2}\left(t - \frac{1}{t}\right)$ and differentiating: $\frac{dx}{dt} = \frac{1}{2}\left(1 + \frac{1}{t^2}\right)$.

When $x = 0$: $t = 0 + \sqrt{1} = 1$. When $x \to \infty$: $t \to \infty$. So the limits transform as $x \in (0, \infty) \to t \in (1, \infty)$.

Therefore:

$\int_0^\infty f\!\left(x + \sqrt{1 + x^2}\right) dx = \int_1^\infty f(t) \cdot \frac{1}{2}\!\left(1 + \frac{1}{t^2}\right) dt = \frac{1}{2}\int_1^\infty \left(1 + \frac{1}{t^2}\right) f(t)\, dt. \qquad \blacksquare$

Step 2: Evaluate the first integral.

We need $I_1 = \int_0^\infty \frac{1}{2x^2 + 1 + 2x\sqrt{x^2 + 1}}\, dx$.

Notice that $2x^2 + 1 + 2x\sqrt{x^2 + 1} = (x + \sqrt{1 + x^2})^2$, since

$(x + \sqrt{1 + x^2})^2 = x^2 + 2x\sqrt{1 + x^2} + (1 + x^2) = 2x^2 + 1 + 2x\sqrt{1 + x^2}.$

So $I_1 = \int_0^\infty \frac{1}{(x + \sqrt{1 + x^2})^2}\, dx$, which has the form $\int_0^\infty f(x + \sqrt{1 + x^2})\, dx$ with $f(u) = \frac{1}{u^2}$.

Applying the result from Step 1:

$I_1 = \frac{1}{2}\int_1^\infty \left(1 + \frac{1}{t^2}\right) \cdot \frac{1}{t^2}\, dt = \frac{1}{2}\int_1^\infty \left(t^{-2} + t^{-4}\right) dt$

$= \frac{1}{2}\left[-\frac{1}{t} - \frac{1}{3t^3}\right]_1^\infty = \frac{1}{2}\left(0 - \left(-1 - \frac{1}{3}\right)\right) = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}.$

Step 3: Evaluate the second integral using $x = \tan\theta$.

We need $I_2 = \int_0^{\frac{1}{2}\pi} \frac{1}{(1 + \sin\theta)^3}\, d\theta$.

Setting $x = \tan\theta$, so $dx = \sec^2\theta\, d\theta$ and $\sqrt{1 + x^2} = \sec\theta$ (for $0 \le \theta < \frac{\pi}{2}$). Also $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{x}{\sqrt{1 + x^2}}$. The limits transform as $\theta \in (0, \frac{\pi}{2}) \to x \in (0, \infty)$.

Write the integrand by multiplying numerator and denominator by $\sec^3\theta$:

$\frac{1}{(1 + \sin\theta)^3} = \frac{\sec^3\theta}{(\sec\theta + \tan\theta)^3} = \frac{\sec\theta \cdot \sec^2\theta}{(\sec\theta + \tan\theta)^3}.$

So:

$I_2 = \int_0^{\frac{\pi}{2}} \frac{\sec\theta \cdot \sec^2\theta}{(\sec\theta + \tan\theta)^3}\, d\theta = \int_0^\infty \frac{\sqrt{1 + x^2}}{(\sqrt{1 + x^2} + x)^3}\, dx.$

This has the form $\int_0^\infty f(x + \sqrt{1 + x^2})\, dx$ where we need to identify $f$. We have $x + \sqrt{1 + x^2} = t$ and $\sqrt{1 + x^2} = \frac{1}{2}(t + t^{-1})$, so

$\frac{\sqrt{1 + x^2}}{(\sqrt{1 + x^2} + x)^3} = \frac{\frac{1}{2}(t + t^{-1})}{t^3} = \frac{t^2 + 1}{2t^4}.$

Thus $f(t) = \frac{t^2 + 1}{2t^4}$, and applying the result:

$I_2 = \frac{1}{2}\int_1^\infty \left(1 + \frac{1}{t^2}\right) \cdot \frac{t^2 + 1}{2t^4}\, dt = \frac{1}{4}\int_1^\infty \frac{(t^2 + 1)^2}{t^6}\, dt = \frac{1}{4}\int_1^\infty \left(t^{-2} + 2t^{-4} + t^{-6}\right) dt$

$= \frac{1}{4}\left[-\frac{1}{t} - \frac{2}{3t^3} - \frac{1}{5t^5}\right]_1^\infty = \frac{1}{4}\left(0 - \left(-1 - \frac{2}{3} - \frac{1}{5}\right)\right) = \frac{1}{4}\left(1 + \frac{2}{3} + \frac{1}{5}\right) = \frac{1}{4} \cdot \frac{15 + 10 + 3}{15} = \frac{7}{15}.$

Examiner Notes

3. Lots of attempts relied on manipulating series for $e$, and would have struggled had the first two results not been given, and even so, there were varying levels of success and conviction. This approach fell apart in the this part with the cubic term. Some candidates used a generating function method successfully with $an(x) = \sum_{n=1}^{\infty} \frac{n+1}{n!} x^n$. However, whilst this worked well for part (i), it got very nasty for part (ii). There were lots of sign errors with the log series in part (ii), having begun well with partial fractions.

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Question 4

Topic: 级数与不等式 Series and Inequalities  |  Difficulty: Challenging  |  Marks: 20

Problem

4 In this question, you may assume that the infinite series $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots + (-1)^{n+1} \frac{x^n}{n} + \dots$ is valid for $|x| < 1$.

(i) Let $n$ be an integer greater than 1. Show that, for any positive integer $k$, $\frac{1}{(k + 1)n^{k+1}} < \frac{1}{kn^k} \, .$ Hence show that $\ln \left( 1 + \frac{1}{n} \right) < \frac{1}{n} \, .$ Deduce that $\left( 1 + \frac{1}{n} \right)^n < e \, .$

(ii) Show, using an expansion in powers of $\frac{1}{y} \, ,$ that $\ln \left( \frac{2y + 1}{2y - 1} \right) > \frac{1}{y}$ for $y > \frac{1}{2} \, .$ Deduce that, for any positive integer $n$, $e < \left( 1 + \frac{1}{n} \right)^{n + \frac{1}{2}} \, .$

(iii) Use parts (i) and (ii) to show that as $n \to \infty$ $\left( 1 + \frac{1}{n} \right)^n \to e \, .$

Hint

(i) This first result is easily established: For $n, k > 1$, $n^{k+1} > n^k$ and $k + 1 > k$ so $(k + 1) \times n^{k+1} > k \times n^k$ $\Rightarrow \frac{1}{(k + 1)n^{k+1}} < \frac{1}{kn^k}$ (since all terms are positive).

$\begin{aligned} \text{Then } \ln \left( 1 + \frac{1}{n} \right) &= \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \frac{1}{5n^5} - \dots \quad \text{(a result which is valid since } 0 < \frac{1}{n} < 1) \\ &= \frac{1}{n} - \left( \frac{1}{2n^2} - \frac{1}{3n^3} \right) - \left( \frac{1}{4n^4} - \frac{1}{5n^5} \right) - \dots < \frac{1}{n} \text{ since each bracketed term is positive, using} \end{aligned}$ A1

the previous result. Exponentiating then gives $1 + \frac{1}{n} < e^{\frac{1}{n}} \Rightarrow \left( 1 + \frac{1}{n} \right)^n < e$.

(ii) A bit of preliminary log. work enables us to use the $\ln(1 + x)$ result on

$\begin{aligned} \ln \left( \frac{2y + 1}{2y - 1} \right) &= \ln \left( 1 + \frac{1}{2y} \right) - \ln \left( 1 - \frac{1}{2y} \right) = \left( \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \frac{1}{4(2y)^4} + \frac{1}{5(2y)^5} - \dots \right) \\ &\qquad\qquad\qquad\qquad\qquad\qquad - \left( -\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \frac{1}{4(2y)^4} - \frac{1}{5(2y)^5} - \dots \right) \\ &= 2 \left( \frac{1}{2y} + \frac{1}{3(2y)^3} + \frac{1}{5(2y)^5} + \dots \right) > \frac{1}{y} \quad \text{(since all terms after the first are positive).} \end{aligned}$

Again, note that we should justify that the series is valid for $0 < \frac{1}{2y} < 1$ i.e. $y > \frac{1}{2}$ in order to justify the

use of the given series. It then follows that $\ln \left( \frac{2y + 1}{2y - 1} \right)^y > 1$, and setting $y = n + \frac{1}{2}$ (the crucial final step)

gives $\ln \left( \frac{2n + 2}{2n} \right)^{n + \frac{1}{2}} > 1 \Rightarrow \left( 1 + \frac{1}{n} \right)^{n + \frac{1}{2}} > e$.

(iii) This final part only required a fairly informal argument, but the details still required a little bit of care in order to avoid being too vague.

As $n \to \infty$, $\left( 1 + \frac{1}{n} \right)^{n + \frac{1}{2}} = \left( 1 + \frac{1}{n} \right)^n \times \left( 1 + \frac{1}{n} \right)^{\frac{1}{2}} \to \left( 1 + \frac{1}{n} \right)^n \times 1 + \to \left( 1 + \frac{1}{n} \right)^n$ from above and e is squeezed

into the same limit from both above and below.

Model Solution

Part (i)

Step 1: Show the inequality $\frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}$.

Since $n > 1$ and $k \ge 1$, we have $n^{k+1} = n \cdot n^k > n^k$ (as $n > 1$) and $k + 1 > k$. Both sides are positive, so:

$(k+1) \cdot n^{k+1} > k \cdot n^k \implies \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}. \qquad \blacksquare$

Step 2: Show $\ln\!\left(1 + \frac{1}{n}\right) < \frac{1}{n}$.

Since $n > 1$, we have $0 < \frac{1}{n} < 1$, so the series for $\ln(1+x)$ is valid with $x = \frac{1}{n}$:

$\ln\!\left(1 + \frac{1}{n}\right) = \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \frac{1}{5n^5} - \cdots$

Group consecutive pairs of terms (after the first):

$= \frac{1}{n} - \left(\frac{1}{2n^2} - \frac{1}{3n^3}\right) - \left(\frac{1}{4n^4} - \frac{1}{5n^5}\right) - \cdots$

Each bracketed pair has the form $\frac{1}{kn^k} - \frac{1}{(k+1)n^{k+1}}$, which is positive by the inequality just proved. Therefore

$\ln\!\left(1 + \frac{1}{n}\right) < \frac{1}{n}. \qquad \blacksquare$

Step 3: Deduce $\left(1 + \frac{1}{n}\right)^n < e$.

From Step 2, $\ln\!\left(1 + \frac{1}{n}\right) < \frac{1}{n}$. Multiplying both sides by $n$:

$n \ln\!\left(1 + \frac{1}{n}\right) < 1 \implies \ln\!\left(1 + \frac{1}{n}\right)^n < 1 = \ln e.$

Since $\ln$ is strictly increasing, exponentiating gives

$\left(1 + \frac{1}{n}\right)^n < e. \qquad \blacksquare$

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Part (ii)

Step 1: Expand $\ln\!\left(\frac{2y+1}{2y-1}\right)$ in powers of $\frac{1}{y}$.

For $y > \frac{1}{2}$, we have $0 < \frac{1}{2y} < 1$, so the series for $\ln(1+x)$ is valid with $x = \frac{1}{2y}$:

$\ln\!\left(\frac{2y+1}{2y-1}\right) = \ln\!\left(1 + \frac{1}{2y}\right) - \ln\!\left(1 - \frac{1}{2y}\right).$

We use the series $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ and $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots$ (valid for $|x| < 1$), with $x = \frac{1}{2y}$:

$\ln\!\left(1 + \frac{1}{2y}\right) = \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \frac{1}{4(2y)^4} + \cdots$

$\ln\!\left(1 - \frac{1}{2y}\right) = -\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \frac{1}{4(2y)^4} - \cdots$

Subtracting (the even-power terms cancel, the odd-power terms double):

$\ln\!\left(\frac{2y+1}{2y-1}\right) = 2\left(\frac{1}{2y} + \frac{1}{3(2y)^3} + \frac{1}{5(2y)^5} + \cdots\right).$

Every term after the first is positive, so

$\ln\!\left(\frac{2y+1}{2y-1}\right) > 2 \cdot \frac{1}{2y} = \frac{1}{y}. \qquad \blacksquare$

Step 2: Deduce $e < \left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}$.

From Step 1 with $y > \frac{1}{2}$:

$\ln\!\left(\frac{2y+1}{2y-1}\right) > \frac{1}{y} \implies y \ln\!\left(\frac{2y+1}{2y-1}\right) > 1 = \ln e \implies \left(\frac{2y+1}{2y-1}\right)^y > e.$

Now set $y = n + \frac{1}{2}$ (which is $> \frac{1}{2}$ for any positive integer $n$). Then

$\frac{2y+1}{2y-1} = \frac{2(n + \frac{1}{2}) + 1}{2(n + \frac{1}{2}) - 1} = \frac{2n + 2}{2n} = \frac{n+1}{n} = 1 + \frac{1}{n}.$

So $\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}} > e$, i.e.

$e < \left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}. \qquad \blacksquare$

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Part (iii)

From parts (i) and (ii), for every positive integer $n$:

$\left(1 + \frac{1}{n}\right)^n < e < \left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}.$

Write the right-hand expression as

$\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}} = \left(1 + \frac{1}{n}\right)^n \cdot \left(1 + \frac{1}{n}\right)^{\frac{1}{2}}.$

As $n \to \infty$, $\left(1 + \frac{1}{n}\right)^{\frac{1}{2}} \to 1$, so

$\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}} = \left(1 + \frac{1}{n}\right)^n \cdot \left(1 + \frac{1}{n}\right)^{\frac{1}{2}} \to \left(1 + \frac{1}{n}\right)^n \cdot 1.$

More precisely, let $a_n = \left(1 + \frac{1}{n}\right)^n$. Then $a_n < e$ for all $n$ (from part (i)), and $e < a_n \cdot \left(1 + \frac{1}{n}\right)^{\frac{1}{2}}$ (from part (ii)), which gives

$\frac{e}{\left(1 + \frac{1}{n}\right)^{1/2}} < a_n < e.$

As $n \to \infty$, the left side $\to \frac{e}{1} = e$. By the squeeze theorem,

$\left(1 + \frac{1}{n}\right)^n \to e. \qquad \blacksquare$

Examiner Notes

Just over 70% of the candidates attempted this, with marginally less success than question 3. Lots of attempts relied on manipulating series for $e$, and would have struggled had the first two results not been given, and even so, there were varying levels of success and conviction. This approach fell apart in the this part with the cubic term. Some candidates used a generating function method successfully with $an(x) = \sum_{n=1}^{\infty} \frac{n+1}{n!} x^n$. However, whilst this worked well for part (i), it got very nasty for part (ii). There were lots of sign errors with the log series in part (ii), having begun well with partial fractions.

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Question 5

Topic: 曲线描绘 Curve Sketching  |  Difficulty: Standard  |  Marks: 20

Problem

5 (i) Sketch the curve $y = \text{f}(x)$, where

$\text{f}(x) = \frac{1}{(x - a)^2 - 1} \quad (x \neq a \pm 1),$

and $a$ is a constant.

(ii) The function $\text{g}(x)$ is defined by

$\text{g}(x) = \frac{1}{((x - a)^2 - 1)((x - b)^2 - 1)} \quad (x \neq a \pm 1, \ x \neq b \pm 1),$

where $a$ and $b$ are constants, and $b > a$. Sketch the curves $y = \text{g}(x)$ in the two cases $b > a + 2$ and $b = a + 2$, finding the values of $x$ at the stationary points.

Hint

With any curve-sketching question of this kind, it is important to grasp those features that are important and ignore those that aren’t. For instance, throughout this question, the position of the $y$-axis is entirely immaterial: it could be drawn through any branch of the curves in question or, indeed, appear as an

asymptote. So the usually key detail of the $y$-intercept, at $\left( 0, \frac{1}{a^2 - 1} \right)$ in part (i), does not help decide

what the function is up to. The asymptotes, turning points (clearly important in part (ii) since they are specifically requested), and any symmetries are important. The other key features to decide upon are the “short-term” (when $x$ is small) and the “long-term” (as $x \to \pm \infty$) behaviours.

In (i), there are vertical asymptotes at $x = a - 1$ and $x = a + 1$; while the $x$-axis is a horizontal asymptote. There is symmetry in the line $x = a$ (a consequence of which is the maximum TP in the “middle” branch) and the “long-term” behaviour of the curve is that it ultimately resembles the graph of $y = \frac{1}{x^2}$.

(ii) Differentiating the function in (ii) gives

$g'(x) = \frac{-2}{\left[ (x - a)^2 - 1 \right]^2 \left[ (x - b)^2 - 1 \right]^2} \left\{ (x - b) \left[ (x - a)^2 - 1 \right] + (x - a) \left[ (x - b)^2 - 1 \right] \right\}$

and setting the numerator $= 0 \implies (x - a)(x - b) [x - a + x - b] + [x - a + x - b] = 0$. Factorising yields

$(2x - a - b)(x^2 - (a + b)x + (ab - 1)) = 0$, so that $x = \frac{1}{2}(a + b)$ or $\frac{a + b \pm \sqrt{(a + b)^2 - 4ab + 4}}{2}$.

In the first case, where $b > a + 2$ (i.e. $a + 1 < b - 1$), there are five branches of the curve, with 4 vertical asymptotes: $x = a \pm 1$ and $x = b \pm 1$. As the function changes sign as it “crosses” each asymptote, and the “long-term” behaviour is still to resemble $y = \frac{1}{x^2}$, these branches alternate above and below the $x$-axis,

with symmetry in $x = \frac{1}{2}(a + b)$.

In the second case, where $b = a + 2$ (i.e. $a + 1 = b - 1$), the very middle section has collapsed, leaving only the four branches, but the curve is otherwise essentially unchanged from the previous case.

Model Solution

Part (i)

We analyse $f(x) = \dfrac{1}{(x - a)^2 - 1}$.

Domain. The denominator vanishes when $(x - a)^2 = 1$, i.e.\ $x = a - 1$ or $x = a + 1$. These are excluded from the domain.

Vertical asymptotes. As $x \to (a \pm 1)^{\pm}$, the denominator tends to $0$, so $f(x) \to \pm\infty$. Specifically:

• As $x \to (a-1)^-$: $(x-a)^2 - 1 \to 0^+$, so $f(x) \to +\infty$.
• As $x \to (a-1)^+$: $(x-a)^2 - 1 \to 0^-$, so $f(x) \to -\infty$.
• As $x \to (a+1)^-$: $(x-a)^2 - 1 \to 0^-$, so $f(x) \to -\infty$.
• As $x \to (a+1)^+$: $(x-a)^2 - 1 \to 0^+$, so $f(x) \to +\infty$.

Horizontal asymptote. As $x \to \pm\infty$, $(x-a)^2 - 1 \to +\infty$, so $f(x) \to 0^+$. The $x$-axis is a horizontal asymptote.

Sign of $f(x)$. For $x < a - 1$ or $x > a + 1$: $(x-a)^2 > 1$, so $f(x) > 0$. For $a - 1 < x < a + 1$: $(x-a)^2 < 1$, so $f(x) < 0$.

Symmetry. $f(a + h) = \dfrac{1}{h^2 - 1} = f(a - h)$, so the curve is symmetric about $x = a$.

Stationary point. Differentiating:

$f'(x) = \frac{-2(x - a)}{[(x - a)^2 - 1]^2}$

Setting $f'(x) = 0$ gives $x = a$. At $x = a$: $f(a) = \dfrac{1}{0 - 1} = -1$. This is the only stationary point, and it is a local minimum in the middle branch (since $f(x) < 0$ on $(a-1, a+1)$ with $f(x) \to -\infty$ at both endpoints).

Sketch summary for part (i): Three branches — the outer two branches are positive (above the $x$-axis), rising from the horizontal asymptote to $+\infty$ at the vertical asymptotes; the middle branch is negative (below the $x$-axis), dropping from $-\infty$ to the minimum $f(a) = -1$ and back up to $-\infty$.

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Part (ii)

We have $g(x) = \dfrac{1}{[(x - a)^2 - 1][(x - b)^2 - 1]}$ with $b > a$.

Finding the stationary points. Write $g(x) = [(x-a)^2 - 1]^{-1} \cdot [(x-b)^2 - 1]^{-1}$ and differentiate using the product rule:

$g'(x) = \frac{-2(x-a)}{[(x-a)^2-1]^2} \cdot \frac{1}{[(x-b)^2-1]} + \frac{1}{[(x-a)^2-1]} \cdot \frac{-2(x-b)}{[(x-b)^2-1]^2}$

$= \frac{-2\Big[(x-b)[(x-a)^2-1] + (x-a)[(x-b)^2-1]\Big]}{[(x-a)^2-1]^2[(x-b)^2-1]^2}$

Setting the numerator to zero: $(x - b)[(x - a)^2 - 1] + (x - a)[(x - b)^2 - 1] = 0$.

With $u = x - a$ and $v = x - b$, this becomes

$v(u^2 - 1) + u(v^2 - 1) = u^2 v - v + uv^2 - u = uv(u + v) - (u + v) = (u + v)(uv - 1) = 0.$

So either $u + v = 0$ or $uv = 1$:

Case 1: $u + v = 0$, i.e.\ $(x - a) + (x - b) = 0$, giving $x = \dfrac{a + b}{2}$.

Case 2: $uv = 1$, i.e.\ $(x - a)(x - b) = 1$, so $x^2 - (a + b)x + ab - 1 = 0$. By the quadratic formula:

$x = \frac{(a + b) \pm \sqrt{(a + b)^2 - 4(ab - 1)}}{2} = \frac{(a + b) \pm \sqrt{(a - b)^2 + 4}}{2}$

Since $(a - b)^2 + 4 > 0$, these roots are always real. The three stationary points are

$x = \frac{a + b - \sqrt{(a - b)^2 + 4}}{2}, \qquad x = \frac{a + b}{2}, \qquad x = \frac{a + b + \sqrt{(a - b)^2 + 4}}{2}.$

Case $b > a + 2$:

The four vertical asymptotes are at $x = a - 1,\, a + 1,\, b - 1,\, b + 1$. Since $b > a + 2$, we have $a + 1 < b - 1$, so the ordering is $a - 1 < a + 1 < b - 1 < b + 1$, giving five branches.

We verify the locations of the outer two stationary points. For $x_+ = \dfrac{a + b + \sqrt{(a-b)^2+4}}{2}$: we claim $x_+ > b + 1$, i.e.\ $\sqrt{(a-b)^2+4} > b - a + 2$. Squaring both sides (both positive): $(a-b)^2 + 4 > (b-a)^2 + 4(b-a) + 4$, which simplifies to $0 > -4(b-a)$, true since $b > a$. So $x_+$ lies to the right of $b + 1$.

By a symmetric argument, $x_- = \dfrac{a + b - \sqrt{(a-b)^2+4}}{2} < a - 1$, so $x_-$ lies to the left of $a - 1$.

The middle stationary point $\dfrac{a+b}{2}$ lies between $a + 1$ and $b - 1$ (since $b > a + 2$ implies $\frac{a+b}{2} > a + 1$ and $\frac{a+b}{2} < b - 1$).

Sketch for $b > a + 2$: Five branches, alternating above and below the $x$-axis (since $g$ changes sign across each vertical asymptote). The outer two branches and the middle branch are positive (above the axis), each having a local maximum at the respective stationary point. The two inner branches (between $a-1$ and $a+1$, and between $b-1$ and $b+1$) are negative, tending to $-\infty$ at their asymptote boundaries. The curve has symmetry about $x = \frac{a+b}{2}$ and approaches $y = 0$ as $x \to \pm\infty$.

Case $b = a + 2$:

Now $b - 1 = a + 1$, so the asymptotes at $x = a + 1$ and $x = b - 1$ coincide. There are three vertical asymptotes: $x = a - 1,\, a + 1,\, a + 3$, giving four branches instead of five. The middle positive branch (which existed between $a+1$ and $b-1$ in the previous case) has collapsed since that interval has zero width. The three stationary points are the same in form, but $\frac{a+b}{2} = a + 1$ now coincides with a vertical asymptote rather than being an interior stationary point. The remaining four branches are otherwise essentially unchanged: the outer two positive branches and the two inner negative branches, with the two negative branches now meeting at the common asymptote $x = a + 1$.

Examiner Notes

This was only very slightly more popular than question 4, though with the same level of success. A lot of candidates scored just the first 5 marks, getting as far as completing the simplification in part (i) (b), but then, being unable to apply it for the final result, and then making no progress with part (ii). The biggest problem was that candidates ignored the definitions given at the start of the question, most notably that “$a$ and $b$ are rational numbers”. The other common problem was that candidates chose a simple value for $\theta$ such as $\frac{\pi}{4}$ or $\frac{\pi}{3}$ rather than for $\cos \theta$ such as $\frac{4}{5}$. In part (ii), quite frequently, candidates substituted $x = p + \sqrt{2}q$, and $y = r + \sqrt{2}s$ and some then successfully found solutions. For part (ii) (c), a method using $\cosh \theta$ and $\sinh \theta$ was not unexpected, although the comparable one with $\sec \theta$ and $\tan \theta$ was quite commonly used too.

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Question 6

Topic: 几何 Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

6 A cyclic quadrilateral $ABCD$ has sides $AB$, $BC$, $CD$ and $DA$ of lengths $a$, $b$, $c$ and $d$, respectively. The area of the quadrilateral is $Q$, and angle $DAB$ is $\theta$.

Find an expression for $\cos \theta$ in terms of $a$, $b$, $c$ and $d$, and an expression for $\sin \theta$ in terms of $a$, $b$, $c$, $d$ and $Q$. Hence show that

$16Q^2 = 4(ad + bc)^2 - (a^2 + d^2 - b^2 - c^2)^2,$

and deduce that

$Q^2 = (s - a)(s - b)(s - c)(s - d),$

where $s = \frac{1}{2}(a + b + c + d)$.

Deduce a formula for the area of a triangle with sides of length $a$, $b$ and $c$.

Hint

A quick diagram helps here, leading to the important observation, from the GCSE geometry result “opposite angles of a cyclic quad. add to $180^\circ$”, that $\angle BCD = 180^\circ - \theta$. Then, using the Cosine Rule twice (and noting that $\cos(180^\circ - \theta) = -\cos\theta$):

in $\Delta BAD$: $BD^2 = a^2 + d^2 - 2ad \cos\theta$

in $\Delta BCD$: $BD^2 = b^2 + c^2 + 2bc \cos\theta$

Equating for $BD^2$ and re-arranging gives $\cos\theta = \frac{a^2 - b^2 - c^2 + d^2}{2(ad + bc)}$

Next, the well-known formula for triangle area, $\Delta = \frac{1}{2}ab \sin C$, twice, gives $Q = \frac{1}{2}ad \sin \theta + \frac{1}{2}bc \sin \theta$,

since $\sin(\pi - \theta) = \sin \theta$. Rearranging then gives $\sin \theta = \frac{2Q}{ad + bc}$ or $\frac{4Q}{2(ad + bc)}$.

Use of $\sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \frac{16Q^2}{4(ad + bc)^2} + \frac{(a^2 - b^2 - c^2 + d^2)^2}{4(ad + bc)^2} = 1$ and this then gives the printed

result, $16Q^2 = 4(ad + bc)^2 - (a^2 - b^2 - c^2 + d^2)^2$.

Then, $16Q^2 = (2ad + 2bc - a^2 + b^2 + c^2 - d^2)(2ad + 2bc + a^2 - b^2 - c^2 + d^2)$ by the difference-of-two-squares factorisation

$= ([b + c]^2 - [a - d]^2)([a + d]^2 - [b - c]^2)$

$= ([b + c] - [a - d])([b + c] + [a - d])([a + d] - [b - c])([a + d] + [b - c])$

using the difference-of-two-squares factorisation in each large bracket

$= (b + c + d - a)(a + b + c - d)(a + c + d - b)(a + b + d - c)$.

Splitting the 16 into four 2’s (one per bracket) and using $2s = a + b + c + d$

$\Rightarrow Q^2 = \frac{(2s - 2a)}{2} \frac{(2s - 2b)}{2} \frac{(2s - 2c)}{2} \frac{(2s - 2d)}{2} = (s - a)(s - b)(s - c)(s - d)$.

Finally, for a triangle (guaranteed cyclic), letting $d \rightarrow 0$ (Or $s - d \rightarrow s$ Or let $D \rightarrow A$), we get the result known as Heron’s Formula: $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.

Model Solution

Finding $\cos\theta$.

In the cyclic quadrilateral $ABCD$, opposite angles sum to $180^\circ$, so $\angle BCD = 180^\circ - \theta$.

Applying the cosine rule to triangle $BAD$:

$BD^2 = a^2 + d^2 - 2ad\cos\theta. \qquad (1)$

Applying the cosine rule to triangle $BCD$ (and noting $\cos(180^\circ - \theta) = -\cos\theta$):

$BD^2 = b^2 + c^2 - 2bc\cos(180^\circ - \theta) = b^2 + c^2 + 2bc\cos\theta. \qquad (2)$

Equating $(1)$ and $(2)$:

$a^2 + d^2 - 2ad\cos\theta = b^2 + c^2 + 2bc\cos\theta$

$a^2 + d^2 - b^2 - c^2 = 2(ad + bc)\cos\theta$

$\cos\theta = \frac{a^2 + d^2 - b^2 - c^2}{2(ad + bc)}. \qquad (3)$

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Finding $\sin\theta$.

The area of $ABCD$ is the sum of the areas of triangles $BAD$ and $BCD$:

$Q = \tfrac{1}{2}ad\sin\theta + \tfrac{1}{2}bc\sin(180^\circ - \theta) = \tfrac{1}{2}ad\sin\theta + \tfrac{1}{2}bc\sin\theta = \tfrac{1}{2}(ad + bc)\sin\theta,$

since $\sin(180^\circ - \theta) = \sin\theta$. Rearranging:

$\sin\theta = \frac{2Q}{ad + bc}. \qquad (4)$

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Showing $16Q^2 = 4(ad + bc)^2 - (a^2 + d^2 - b^2 - c^2)^2$.

Using $\sin^2\theta + \cos^2\theta = 1$ with $(3)$ and $(4)$:

$\left(\frac{2Q}{ad + bc}\right)^2 + \left(\frac{a^2 + d^2 - b^2 - c^2}{2(ad + bc)}\right)^2 = 1$

$\frac{4Q^2}{(ad + bc)^2} + \frac{(a^2 + d^2 - b^2 - c^2)^2}{4(ad + bc)^2} = 1$

Multiplying through by $4(ad + bc)^2$:

$16Q^2 + (a^2 + d^2 - b^2 - c^2)^2 = 4(ad + bc)^2$

$16Q^2 = 4(ad + bc)^2 - (a^2 + d^2 - b^2 - c^2)^2. \qquad \blacksquare$

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Deducing $Q^2 = (s-a)(s-b)(s-c)(s-d)$.

We factorise $16Q^2$ as a difference of two squares:

$16Q^2 = [2(ad + bc) - (a^2 + d^2 - b^2 - c^2)][2(ad + bc) + (a^2 + d^2 - b^2 - c^2)].$

First factor:

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

Second factor:

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

Now each factor is again a difference of two squares:

$(b + c)^2 - (a - d)^2 = (b + c - a + d)(b + c + a - d),$

$(a + d)^2 - (b - c)^2 = (a + d - b + c)(a + d + b - c).$

Therefore:

$16Q^2 = (b + c + d - a)(a + b + c - d)(a + c + d - b)(a + b + d - c).$

With $s = \frac{1}{2}(a + b + c + d)$, we have $2s = a + b + c + d$, so:

$b + c + d - a = 2s - 2a = 2(s - a),$

and similarly for the other three factors. Substituting:

$16Q^2 = 2(s-a) \cdot 2(s-b) \cdot 2(s-c) \cdot 2(s-d) = 16(s-a)(s-b)(s-c)(s-d).$

Dividing by 16:

$Q^2 = (s - a)(s - b)(s - c)(s - d). \qquad \blacksquare$

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Deducing Heron’s formula for a triangle.

Every triangle is cyclic (it can be inscribed in a unique circle). A triangle with sides $a$, $b$, $c$ can be viewed as a degenerate cyclic quadrilateral $ABCD$ with $d = 0$ (vertex $D$ coincides with vertex $A$). Setting $d = 0$ in the formula above: $s = \frac{1}{2}(a + b + c)$ and

$Q^2 = (s - a)(s - b)(s - c)(s - 0) = s(s - a)(s - b)(s - c).$

So the area of a triangle with sides $a$, $b$, $c$ is

$\Delta = \sqrt{s(s - a)(s - b)(s - c)}, \qquad \text{where } s = \tfrac{1}{2}(a + b + c).$

This is Heron’s formula. $\qquad \blacksquare$

Examiner Notes

Two thirds attempted this, with less success than its three predecessors. Very few indeed scored full marks, for even those that mastered the question rarely sketched the last locus correctly, putting in a non-existent cusp. Most candidates managed the first part, good ones the second part too, and only the very best the third part. Quite a few assumed the roots were complex and then used complex conjugates, with varying success. Many candidates lost marks through careless arithmetic and algebraic errors. Given that most could do the first part, it was possible for candidates to score reasonably if they took care and took real parts and imaginary parts correctly.

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Question 7

Topic: 向量与几何 Vectors and Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Three distinct points, $X_1$, $X_2$ and $X_3$, with position vectors $\mathbf{x}_1$, $\mathbf{x}_2$ and $\mathbf{x}_3$ respectively, lie on a circle of radius 1 with its centre at the origin $O$. The point $G$ has position vector $\frac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3)$. The line through $X_1$ and $G$ meets the circle again at the point $Y_1$ and the points $Y_2$ and $Y_3$ are defined correspondingly.

Given that $\overrightarrow{GY_1} = -\lambda_1 \overrightarrow{GX_1}$, where $\lambda_1$ is a positive scalar, show that

$\overrightarrow{OY_1} = \frac{1}{3}((1 - 2\lambda_1)\mathbf{x}_1 + (1 + \lambda_1)(\mathbf{x}_2 + \mathbf{x}_3))$

and hence that

$\lambda_1 = \frac{3 - \alpha - \beta - \gamma}{3 + \alpha - 2\beta - 2\gamma},$

where $\alpha = \mathbf{x}_2 \cdot \mathbf{x}_3$, $\beta = \mathbf{x}_3 \cdot \mathbf{x}_1$ and $\gamma = \mathbf{x}_1 \cdot \mathbf{x}_2$.

Deduce that $\frac{GX_1}{GY_1} + \frac{GX_2}{GY_2} + \frac{GX_3}{GY_3} = 3$.

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Hint

Many of you will know that this point $G$, used here, is the centroid of the triangle, and has position vector $\mathbf{g} = \frac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3)$.

Then $\vec{GX_1} = \mathbf{x}_1 - \mathbf{g} = \frac{1}{3}(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3)$ and so $\vec{GY_1} = -\frac{1}{3}\lambda_1(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3)$, where $\lambda_1 > 0$.

Also $\vec{OY_1} = \vec{OG} + \vec{GY_1} = \frac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3) - \frac{1}{3}\lambda_1(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3) = \frac{1}{3}([1 - 2\lambda_1]\mathbf{x}_1 + 1 + \lambda_1)$, the first printed result.

The really critical observation here is that the circle centre $O$, radius 1 has equation $|\mathbf{x}|^2 = 1$ or $\mathbf{x} \cdot \mathbf{x} = 1$, where $\mathbf{x}$ can be the p.v. of any point on the circle.

Thus, since $\vec{OY_1} \cdot \vec{OY_1} = 1$, we have

$1 = \frac{1}{9} \left\{ (1 - 2\lambda_1)^2 + 2(1 + \lambda_1)^2 + 2(1 - 2\lambda_1)(1 + \lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + 2(1 + \lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right\}$

$\Rightarrow 9 = 1 - 4\lambda_1 + 4\lambda_1^2 + 2 + 4\lambda_1 + 2\lambda_1^2 + 2(1 - 2\lambda_1)(1 + \lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + 2(1 + \lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3$

$\Rightarrow 0 = -3(1 - \lambda_1)(1 + \lambda_1) + (1 - 2\lambda_1)(1 + \lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + (1 + \lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3$

As $\lambda_1 > 0$, $0 = -3(1 - \lambda_1) + (1 - 2\lambda_1)\mathbf{x}_1 \cdot (\mathbf{x}_2 + \mathbf{x}_3) + (1 + \lambda_1)\mathbf{x}_2 \cdot \mathbf{x}_3$

$\Rightarrow 0 = -3 + 3\lambda_1 + (\mathbf{x}_1 \cdot \mathbf{x}_2 + \mathbf{x}_2 \cdot \mathbf{x}_3 + \mathbf{x}_3 \cdot \mathbf{x}_1) + \lambda_1(\mathbf{x}_2 \cdot \mathbf{x}_3) - 2\lambda_1(\mathbf{x}_1 \cdot \mathbf{x}_2 + \mathbf{x}_1 \cdot \mathbf{x}_3)$

$\Rightarrow \lambda_1 = \frac{3 - (\alpha + \beta + \gamma)}{3 + \alpha - 2\beta - 2\gamma}$, using $\alpha = \mathbf{x}_2 \cdot \mathbf{x}_3$, $\beta = \mathbf{x}_3 \cdot \mathbf{x}_1$ and $\gamma = \mathbf{x}_1 \cdot \mathbf{x}_2$.

Similarly, $\lambda_2 = \frac{3 - (\alpha + \beta + \gamma)}{3 + \beta - 2\alpha - 2\gamma}$ and $\lambda_3 = \frac{3 - (\alpha + \beta + \gamma)}{3 + \gamma - 2\alpha - 2\beta}$.

Using $\frac{GX_i}{GY_i} = \frac{1}{\lambda_i}$ ($i = 1, 2, 3$), $\frac{GX_1}{GY_1} + \frac{GX_2}{GY_2} + \frac{GX_3}{GY_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \frac{9 + (\alpha + \beta + \gamma) - 4(\alpha + \beta + \gamma)}{3 - (\alpha + \beta + \gamma)}$ $= \frac{9 - 3(\alpha + \beta + \gamma)}{3 - (\alpha + \beta + \gamma)} = 3$.

[Interestingly, this result generalises to $n$ points on a circle: $\sum_{i=1}^{n} \frac{GX_i}{GY_i} = n$.]

Model Solution

Step 1: Show the position vector formula.

The point $G$ (centroid of the triangle) has position vector $\mathbf{g} = \frac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3)$.

$\overrightarrow{GX_1} = \mathbf{x}_1 - \mathbf{g} = \mathbf{x}_1 - \tfrac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3) = \tfrac{1}{3}(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3).$

Given $\overrightarrow{GY_1} = -\lambda_1 \overrightarrow{GX_1}$ (with $\lambda_1 > 0$, since $Y_1$ is on the opposite side of $G$ from $X_1$):

$\overrightarrow{OY_1} = \overrightarrow{OG} + \overrightarrow{GY_1} = \mathbf{g} - \lambda_1 \overrightarrow{GX_1}$

$= \tfrac{1}{3}(\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3) - \tfrac{\lambda_1}{3}(2\mathbf{x}_1 - \mathbf{x}_2 - \mathbf{x}_3)$

$= \tfrac{1}{3}\big[(1 - 2\lambda_1)\mathbf{x}_1 + (1 + \lambda_1)(\mathbf{x}_2 + \mathbf{x}_3)\big]. \qquad \blacksquare$

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Step 2: Find $\lambda_1$.

Since $Y_1$ lies on the unit circle centred at $O$, we have $\overrightarrow{OY_1} \cdot \overrightarrow{OY_1} = 1$. Also $|\mathbf{x}_i|^2 = 1$ for $i = 1, 2, 3$.

$9 = \big|(1 - 2\lambda_1)\mathbf{x}_1 + (1 + \lambda_1)(\mathbf{x}_2 + \mathbf{x}_3)\big|^2$

Expanding using $|\mathbf{x}_i|^2 = 1$, $\mathbf{x}_2 \cdot \mathbf{x}_3 = \alpha$, $\mathbf{x}_3 \cdot \mathbf{x}_1 = \beta$, $\mathbf{x}_1 \cdot \mathbf{x}_2 = \gamma$:

$9 = (1 - 2\lambda_1)^2 + 2(1 + \lambda_1)^2 + 2(1 - 2\lambda_1)(1 + \lambda_1)(\gamma + \beta) + 2(1 + \lambda_1)^2 \alpha$

The first two terms give:

$(1 - 2\lambda_1)^2 + 2(1 + \lambda_1)^2 = 1 - 4\lambda_1 + 4\lambda_1^2 + 2 + 4\lambda_1 + 2\lambda_1^2 = 3 + 6\lambda_1^2$

So:

$9 = 3 + 6\lambda_1^2 + 2(1 + \lambda_1)\big[(1 - 2\lambda_1)(\beta + \gamma) + (1 + \lambda_1)\alpha\big]$

$6(1 - \lambda_1^2) = 2(1 + \lambda_1)\big[(1 - 2\lambda_1)(\beta + \gamma) + (1 + \lambda_1)\alpha\big]$

Since $\lambda_1 > 0$, we have $1 + \lambda_1 > 0$ and can divide by $2(1 + \lambda_1)$:

$3(1 - \lambda_1) = (1 - 2\lambda_1)(\beta + \gamma) + (1 + \lambda_1)\alpha$

$3 - 3\lambda_1 = \beta + \gamma - 2\lambda_1(\beta + \gamma) + \alpha + \lambda_1 \alpha$

$3 - (\alpha + \beta + \gamma) = 3\lambda_1 + \lambda_1(\alpha - 2\beta - 2\gamma) = \lambda_1(3 + \alpha - 2\beta - 2\gamma)$

$\lambda_1 = \frac{3 - \alpha - \beta - \gamma}{3 + \alpha - 2\beta - 2\gamma}. \qquad \blacksquare$

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Step 3: Deduce $\frac{GX_1}{GY_1} + \frac{GX_2}{GY_2} + \frac{GX_3}{GY_3} = 3$.

By the same argument with cyclic permutation of $\alpha, \beta, \gamma$:

$\lambda_2 = \frac{3 - \alpha - \beta - \gamma}{3 + \beta - 2\alpha - 2\gamma}, \qquad \lambda_3 = \frac{3 - \alpha - \beta - \gamma}{3 + \gamma - 2\alpha - 2\beta}.$

Since $\overrightarrow{GY_i} = -\lambda_i \overrightarrow{GX_i}$ with $\lambda_i > 0$, we have $\frac{GX_i}{GY_i} = \frac{1}{\lambda_i}$.

Let $S = \alpha + \beta + \gamma$ and $D = 3 - S$ (the common numerator). Then:

$\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \frac{(3 + \alpha - 2\beta - 2\gamma) + (3 + \beta - 2\alpha - 2\gamma) + (3 + \gamma - 2\alpha - 2\beta)}{D}$

The numerator is:

$9 + (\alpha + \beta + \gamma) - 2(\beta + \gamma + \alpha + \gamma + \alpha + \beta) = 9 + S - 4S = 9 - 3S = 3D$

Therefore:

$\frac{GX_1}{GY_1} + \frac{GX_2}{GY_2} + \frac{GX_3}{GY_3} = \frac{3D}{D} = 3. \qquad \blacksquare$

Examiner Notes

Two thirds attempted this too, with marginally greater success than question 2. Most did very well with the stem, though a few were unable to obtain a proper second order equation. Those that attempted part (i) were usually successful. The non-trivial exponential calculations in part (ii) caused problems for some making computation mistakes whilst others were totally on top of this. Part (iii) tested the candidates on two levels, interpreting the sigma notation correctly, and recognising and using the geometric series. Some managed this excellently.

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Question 8

Topic: 数列与递推关系 Sequences and Recurrence Relations  |  Difficulty: Challenging  |  Marks: 20

Problem

8 The positive numbers $\alpha$, $\beta$ and $q$ satisfy $\beta - \alpha > q$. Show that

$\frac{\alpha^2 + \beta^2 - q^2}{\alpha \beta} - 2 > 0.$

The sequence $u_0, u_1, \dots$ is defined by $u_0 = \alpha$, $u_1 = \beta$ and

$u_{n+1} = \frac{u_n^2 - q^2}{u_{n-1}} \quad (n \geqslant 1),$

where $\alpha$, $\beta$ and $q$ are given positive numbers (and $\alpha$ and $\beta$ are such that no term in the sequence is zero). Prove that $u_n(u_n + u_{n+2}) = u_{n+1}(u_{n-1} + u_{n+1})$. Prove also that

$u_{n+1} - p u_n + u_{n-1} = 0$

for some number $p$ which you should express in terms of $\alpha$, $\beta$ and $q$.

Hence, or otherwise, show that if $\beta > \alpha + q$, the sequence is strictly increasing (that is, $u_{n+1} - u_n > 0$ for all $n$). Comment on the case $\beta = \alpha + q$.

Hint

$\beta - \alpha > q$ ($q > 0$) $\Rightarrow \beta^2 - 2\alpha\beta + \alpha^2 > q^2 \Rightarrow \alpha^2 + \beta^2 - q^2 > 2\alpha\beta \Rightarrow \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} > 2 \Rightarrow$ the opening result, $\frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} - 2 > 0$.

$u_{n+1} = \frac{{u_n}^2 - q^2}{u_{n-1}}$ etc. $\Rightarrow {u_n}^2 - u_{n+1}u_{n-1} = q^2 = {u_{n+1}}^2 - u_{n+2}u_n$ (since the result is true at all stages) and equating for $q^2 \Rightarrow u_n(u_n + u_{n+2}) = u_{n+1}(u_{n-1} + u_{n+1})$.

Now this gives $\frac{u_n + u_{n+2}}{u_{n+1}} = \frac{u_{n-1} + u_{n+1}}{u_n}$ which $\Rightarrow \frac{u_{n-1} + u_{n+1}}{u_n}$ is constant (independent of $n$). Calling this constant $p$ gives $u_{n+1} - pu_n + u_{n-1} = 0$, as required. In order to determine $p$, we only need to use the fact that $p = \frac{u_{n-1} + u_{n+1}}{u_n}$ for all $n$, so we choose the first few terms to work with.

$u_2 = \frac{\beta^2 - q^2}{\alpha} \text{ and } p = \frac{u_0 + u_2}{u_1} = \frac{\alpha + \frac{\beta^2 - q^2}{\alpha}}{\beta} = \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta}.$

Alternatively, $u_2 = \gamma = \frac{\beta^2 - q^2}{\alpha} = p\beta - \alpha \Leftrightarrow p = \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta}$

and $u_3 = \frac{\gamma^2 - q^2}{\beta} = p\gamma - \beta \Leftrightarrow p = \frac{\gamma^2 + \beta^2 - q^2}{\beta\gamma} = \frac{\left(\frac{\beta^2 - q^2}{\alpha}\right)^2 + \beta^2 - q^2}{\beta\left(\frac{\beta^2 - q^2}{\alpha}\right)}$ $= \frac{(\beta^2 - q^2)^2 + \alpha^2(\beta^2 - q^2)}{\alpha\beta(\beta^2 - q^2)} = \frac{\beta^2 - q^2 + \alpha^2}{\alpha\beta}$

since $\beta^2 - q^2 \neq 0$ as $u_2$ non-zero (given). Since $p$ is consistent for any chosen $\alpha, \beta$, the proof follows inductively on any two consecutive terms of the sequence.

Finally, on to the given cases.

If $\beta > \alpha + q$, $u_{n+1} - u_n = (p - 1)u_n - u_{n-1} = \left(\frac{\beta^2 + \alpha^2 - q^2}{\alpha\beta} - 1\right)u_n - u_{n-1}$ $> (2 - 1)u_n - u_{n-1}$ by the initial result $> u_n - u_{n-1}$

Hence, if $u_n - u_{n-1} > 0$ then so is $u_{n+1} - u_n$. Since $\beta > \alpha$, $u_2 - u_1 > 0$ and proof follows inductively.

If $\beta = \alpha + q$ then $p = 2$ and $u_{n+1} - u_n = u_n - u_{n-1}$ so that the sequence is an AP.

Also, $u_0 = \alpha$, $u_1 = \alpha + q$, $u_2 = \alpha + 2q, \dots \Rightarrow$ the common difference is $q$ (and we still have a strictly increasing sequence, since $q > 0$ given).

Model Solution

Step 1: Show $\frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} - 2 > 0$.

Since $\beta - \alpha > q > 0$, squaring (both sides positive) gives:

$(\beta - \alpha)^2 > q^2 \implies \beta^2 - 2\alpha\beta + \alpha^2 > q^2 \implies \alpha^2 + \beta^2 - q^2 > 2\alpha\beta.$

Dividing by $\alpha\beta > 0$:

$\frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} > 2, \qquad \text{i.e.} \quad \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} - 2 > 0. \qquad \blacksquare$

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Step 2: Prove $u_n(u_n + u_{n+2}) = u_{n+1}(u_{n-1} + u_{n+1})$.

From the recurrence $u_{n+1} = \frac{u_n^2 - q^2}{u_{n-1}}$, rearranging gives $u_n^2 - u_{n+1}u_{n-1} = q^2$.

This holds for all $n \ge 1$, so also $u_{n+1}^2 - u_{n+2}u_n = q^2$.

Equating the two expressions for $q^2$:

$u_n^2 - u_{n+1}u_{n-1} = u_{n+1}^2 - u_{n+2}u_n$

$u_n^2 + u_{n+2}u_n = u_{n+1}^2 + u_{n+1}u_{n-1}$

$u_n(u_n + u_{n+2}) = u_{n+1}(u_{n+1} + u_{n-1}). \qquad \blacksquare$

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Step 3: Prove $u_{n+1} - pu_n + u_{n-1} = 0$ for some constant $p$.

From Step 2, dividing by $u_n u_{n+1}$ (both positive since all terms are positive):

$\frac{u_n + u_{n+2}}{u_{n+1}} = \frac{u_{n-1} + u_{n+1}}{u_n}.$

This shows $\frac{u_{n-1} + u_{n+1}}{u_n}$ is independent of $n$. Call this constant $p$. Then:

$u_{n-1} + u_{n+1} = pu_n, \qquad \text{i.e.} \quad u_{n+1} - pu_n + u_{n-1} = 0. \qquad \blacksquare$

Finding $p$. Using $n = 1$: $p = \frac{u_0 + u_2}{u_1}$.

From the recurrence with $n = 1$: $u_2 = \frac{\beta^2 - q^2}{\alpha}$.

$p = \frac{\alpha + \frac{\beta^2 - q^2}{\alpha}}{\beta} = \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta}.$

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Step 4: Show the sequence is strictly increasing when $\beta > \alpha + q$.

From Step 1, $\beta - \alpha > q$ implies $p = \frac{\alpha^2 + \beta^2 - q^2}{\alpha\beta} > 2$.

From the linear recurrence: $u_{n+1} = pu_n - u_{n-1}$, so

$u_{n+1} - u_n = (p - 1)u_n - u_{n-1} > (2 - 1)u_n - u_{n-1} = u_n - u_{n-1}$

using $p > 2$ and $u_n > 0$.

Base case. $u_1 - u_0 = \beta - \alpha > 0$ (since $\beta > \alpha + q > \alpha$). Also $u_0 = \alpha > 0$ and $u_1 = \beta > 0$.

Inductive step. Suppose $u_n > u_{n-1} > 0$. Then:

$u_{n+1} - u_n > u_n - u_{n-1} > 0, \qquad \text{and} \qquad u_{n+1} = pu_n - u_{n-1} > 2u_n - u_{n-1} > u_n > 0.$

So $u_{n+1} > u_n > 0$, and by induction the sequence is strictly increasing. $\qquad \blacksquare$

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Step 5: Comment on the case $\beta = \alpha + q$.

If $\beta = \alpha + q$, then:

$p = \frac{\alpha^2 + (\alpha + q)^2 - q^2}{\alpha(\alpha + q)} = \frac{2\alpha^2 + 2\alpha q}{\alpha(\alpha + q)} = \frac{2\alpha(\alpha + q)}{\alpha(\alpha + q)} = 2.$

The recurrence becomes $u_{n+1} = 2u_n - u_{n-1}$, i.e. $u_{n+1} - u_n = u_n - u_{n-1}$, so the common difference is constant.

With $u_0 = \alpha$ and $u_1 = \alpha + q$, the common difference is $q$, giving:

$u_n = \alpha + nq.$

Since $q > 0$, the sequence is still strictly increasing (in fact, an arithmetic progression with common difference $q$).

Examiner Notes

This was the most popular question attempted by over 83% of candidates, and the third most successful with, on average, half marks being scored. Part (i) caused no problems, though some chose to obtain the result algebraically. Part (ii) was not well attempted, with a number stating the two values the expression can take but failing to do anything else or failing with the algebra. Part (iii) was generally fairly well done although frequently the details were not quite tied up fully.