STEP 2 2018 -- Pure Mathematics

STEP 2 2018 — Section A (Pure Mathematics)

Exam: STEP 2  |  Year: 2018  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	代数 (Algebra)	Challenging	倒数根代入, 因式分解, 判别式分析
2	分析 (Analysis)	Challenging	凹函数定义, Jensen不等式, 二阶导数判别, 对数变换
3	微积分 (Calculus)	Challenging	链式求导, 旋转对称性条件, 积分换元
4	三角函数 (Trigonometry)	Standard	和差化积公式, 因式分解, 余弦相等条件
5	级数与积分 (Series and Integration)	Challenging	二项展开, 逐项积分, 级数展开, 换元法
6	数论 (Number Theory)	Challenging	素数判定, 阶乘的素因子分析, 分情况讨论, 不等式估计
7	向量 (Vectors)	Challenging	向量共线条件, 比例关系, 面积比, 相似三角形
8	微分方程 (Differential Equations)	Challenging	变量代换, 分离变量法, 积分技巧, 渐近线分析, 图形比较

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Question 1

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

Problem

1 Show that, if $k$ is a root of the quartic equation

$x^4 + ax^3 + bx^2 + ax + 1 = 0 , \qquad \text{(*)}$

then $k^{-1}$ is a root.

You are now given that $a$ and $b$ in $(*)$ are both real and are such that the roots are all real.

(i) Write down all the values of $a$ and $b$ for which $(*)$ has only one distinct root.

(ii) Given that $(*)$ has exactly three distinct roots, show that either $b = 2a - 2$ or $b = -2a - 2$.

(iii) Solve $(*)$ in the case $b = 2a - 2$, giving your solutions in terms of $a$.

Given that $a$ and $b$ are both real and that the roots of $(*)$ are all real, find necessary and sufficient conditions, in terms of $a$ and $b$, for $(*)$ to have exactly three distinct real roots.

Hint

The first result can be shown by substituting $k^{-1}$ into the quartic expression.

It then follows for part (i) that the only way to achieve one distinct root is for that root to be either 1 or -1. In either case the factorised form of the quartic can then be considered to find the values of $a$ and $b$.

Similarly, for three distinct roots, there must be one pair $(k, k^{-1})$ along with either 1 or -1. Substitution of 1 or -1 into the quartic then leads to the required relationships.

For part (iii) note that the case $b = 2a - 2$ corresponds to the case where there is a root of -1 and it can be seen that it must be a repeated root. The other factor is therefore a quadratic, which can then be solved.

Finally, the conditions for there to be three roots can be found by considering the discriminant of the quadratic (and the corresponding one for the other case). It is also necessary to confirm that this quadratic does not repeat the root of either 1 or -1 depending on which case is being considered.

Model Solution

Show that, if $k$ is a root of the quartic equation, then $k^{-1}$ is a root.

Suppose $k$ is a root of $(*)$, so $k^4 + ak^3 + bk^2 + ak + 1 = 0.$ Note that $k \neq 0$ (since substituting $x = 0$ gives $1 \neq 0$). Substitute $x = k^{-1}$: $(k^{-1})^4 + a(k^{-1})^3 + b(k^{-1})^2 + a(k^{-1}) + 1 = \frac{1}{k^4} + \frac{a}{k^3} + \frac{b}{k^2} + \frac{a}{k} + 1.$ Multiplying through by $k^4$: $k^4 \left(\frac{1}{k^4} + \frac{a}{k^3} + \frac{b}{k^2} + \frac{a}{k} + 1\right) = 1 + ak + bk^2 + ak^3 + k^4 = k^4 + ak^3 + bk^2 + ak + 1 = 0.$ Since $k^4 \neq 0$, the expression equals 0, so $k^{-1}$ is a root.

The polynomial is palindromic: coefficients read the same forwards and backwards. Roots come in reciprocal pairs. Note $x = 1$ and $x = -1$ are each their own reciprocal.

Part (i): All four roots are real and equal to a single value $r$. Then $r = 1/r$, so $r = \pm 1$.

Case 1: $r = 1$. Then $(x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$, giving $a = -4$, $b = 6$.

Case 2: $r = -1$. Then $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$, giving $a = 4$, $b = 6$.

Part (ii): Exactly three distinct roots means one reciprocal pair collapses: $k = 1/k$, so $k = \pm 1$. Either $x = -1$ or $x = 1$ is a repeated root.

Case 1: $x = -1$ is a root. Substituting: $1 - a + b - a + 1 = 2 - 2a + b = 0$, so $b = 2a - 2$.

Case 2: $x = 1$ is a root. Substituting: $1 + a + b + a + 1 = 2 + 2a + b = 0$, so $b = -2a - 2$.

Part (iii): When $b = 2a - 2$, $x = -1$ is a root. Factor as $(x+1)^2(x^2 + px + q)$.

Expanding: $x^4 + (p+2)x^3 + (q+2p+1)x^2 + (2q+p)x + q$. Matching coefficients: $q = 1$, $p = a-2$. Check: $x^2$ coefficient $= 1 + 2(a-2) + 1 = 2a-2$ ✓; $x^1$ coefficient $= 2 + (a-2) = a$ ✓.

So the quartic factors as $(x+1)^2(x^2 + (a-2)x + 1) = 0$. By the quadratic formula: $x = \frac{(2-a) \pm \sqrt{a^2 - 4a}}{2}.$

Final part: Exactly three distinct real roots requires:

Case 1: $b = 2a - 2$. Quadratic $x^2 + (a-2)x + 1 = 0$ needs discriminant $> 0$: $a^2 - 4a > 0$, so $a < 0$ or $a > 4$. Also $a \neq 0$ (would give two distinct roots) and $a \neq 4$ (would give only one distinct root).

Case 2: $b = -2a - 2$. Quadratic $x^2 + (a+2)x + 1 = 0$ needs discriminant $> 0$: $a^2 + 4a > 0$, so $a < -4$ or $a > 0$. Also $a \neq 0$ and $a \neq -4$.

Necessary and sufficient conditions: ($b = 2a - 2$ with $a < 0$ or $a > 4$) or ($b = -2a - 2$ with $a < -4$ or $a > 0$).

Examiner Notes

This was the most popular question of the paper, attempted by 93% of candidates. While many candidates scored well on this question there were very few who achieved full marks and a significant number who scored 0 for this question.

The first two parts of the questions were generally done well, with the majority of marks being lost for arithmetic errors or from just considering one of the two cases. In some cases, candidates produced the reasoning for part (ii) as their answer for part (i). In these cases, the candidates generally repeated the work as their answer for part (ii). There were a number of solutions to part (ii) that were overcomplicated, involving consideration of the factorised form and comparison of coefficients.

Part (iii) was also very well completed by those who attempted it, with the main cause for loss of marks being algebraic errors. A number of errors were seen in the final part of the question, particularly not considering the second case or not realising that the quadratic would have a different discriminant in this case and a failure to check that the roots of the quadratic did not duplicate the repeated root already found in this case.

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Question 2

Topic: 分析 (Analysis)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 A function $f(x)$ is said to be concave for $a < x < b$ if

$t f(x_1) + (1 - t) f(x_2) \leqslant f(t x_1 + (1 - t) x_2) ,$

for $a < x_1 < b$, $a < x_2 < b$ and $0 \leqslant t \leqslant 1$.

Illustrate this definition by means of a sketch, showing the chord joining the points $(x_1, f(x_1))$ and $(x_2, f(x_2))$, in the case $x_1 < x_2$ and $f(x_1) < f(x_2)$.

Explain why a function $f(x)$ satisfying $f''(x) < 0$ for $a < x < b$ is concave for $a < x < b$.

(i) By choosing $t, x_1$ and $x_2$ suitably, show that, if $f(x)$ is concave for $a < x < b$, then

$f\left(\frac{u + v + w}{3}\right) \geqslant \frac{f(u) + f(v) + f(w)}{3} ,$

for $a < u < b$, $a < v < b$ and $a < w < b$.

(ii) Show that, if $A, B$ and $C$ are the angles of a triangle, then

$\sin A + \sin B + \sin C \leqslant \frac{3\sqrt{3}}{2} .$

(iii) By considering $\ln(\sin x)$, show that, if $A, B$ and $C$ are the angles of a triangle, then

$\sin A \times \sin B \times \sin C \leqslant \frac{3\sqrt{3}}{8} .$

Hint

The point with $x$-coordinate $tx_1 + (1 - t)x_2$ is a point within the range $(x_1, x_2)$ (for $0 < t < 1$), and $tf(x_1) + (1 - t)f(x_2)$ is the $y$-coordinate of a point on the chord joining the points $(x_1, f(x_1))$ and $(x_2, f(x_2))$. Therefore, for any position between the two endpoints on the curve, the inequality is comparing the point on the curve with the point on the chord joining the two points. The sketch therefore needs to show the chord entirely below the curve. The final part of the introductory section can be shown either with a proof by contradiction or by arguing that $f''(x) < 0$ means that the gradient is always decreasing within the interval.

Part (i) requires choosing values for $x_1$ and $x_2$ so that the inequality can be applied and then applying this process multiple times to reach the required result. In each case the choice of $x_1$ and $x_2$ need to be made so that they lie within the range for which applying the inequality is valid.

Parts (ii) and (iii) both follow from the result of part (i), but it is important to check that the function being used is concave in the relevant range which needs to be stated clearly. In part (ii) the result follows immediately, whereas the final part requires some manipulation of logarithms to reach the final form of the relationship.

Model Solution

Sketch: Draw a concave-down curve. Mark $P_1 = (x_1, f(x_1))$ and $P_2 = (x_2, f(x_2))$ with $x_1 < x_2$, $f(x_1) < f(x_2)$. The chord joining them lies entirely below the curve. For any $t \in [0,1]$, the point $(tx_1 + (1-t)x_2, tf(x_1) + (1-t)f(x_2))$ lies on the chord, while $(tx_1 + (1-t)x_2, f(tx_1 + (1-t)x_2))$ lies on the curve above.

Why $f'' < 0$ implies concavity: Define $g(t) = f(tx_1 + (1-t)x_2)$. Then $g''(t) = (x_1 - x_2)^2 f''(tx_1 + (1-t)x_2) < 0$ since $(x_1-x_2)^2 > 0$ and $f'' < 0$. A function with $g'' < 0$ lies above its chord (the tangent lines lie above the curve, so the chord does too). Hence $g(t) \geq (1-t)g(0) + tg(1)$, which is the concavity inequality.

Part (i): Apply the concavity inequality twice.

Step 1. Take $x_1 = u$, $x_2 = v$, $t = 1/2$: $\frac{f(u) + f(v)}{2} \leqslant f\!\left(\frac{u+v}{2}\right). \qquad \text{(1)}$

Step 2. Take $x_1 = \frac{u+v}{2}$, $x_2 = w$, $t = \frac{2}{3}$: $\frac{2}{3}f\!\left(\frac{u+v}{2}\right) + \frac{1}{3}f(w) \leqslant f\!\left(\frac{u+v+w}{3}\right). \qquad \text{(2)}$

Step 3. Substitute (1) into (2): $\frac{2}{3} \cdot \frac{f(u)+f(v)}{2} + \frac{1}{3}f(w) = \frac{f(u)+f(v)+f(w)}{3} \leqslant f\!\left(\frac{u+v+w}{3}\right).$

Part (ii): Verify $f(x) = \sin x$ is concave on $(0,\pi)$: $f''(x) = -\sin x < 0$ for $0 < x < \pi$. ✓

Since $A + B + C = \pi$ and $0 < A, B, C < \pi$, apply part (i): $\frac{\sin A + \sin B + \sin C}{3} \leqslant \sin\frac{A+B+C}{3} = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}.$

Hence $\sin A + \sin B + \sin C \leqslant \frac{3\sqrt{3}}{2}$.

Part (iii): Verify $f(x) = \ln(\sin x)$ is concave on $(0,\pi)$: $f'(x) = \cot x, \quad f''(x) = -\csc^2 x = -\frac{1}{\sin^2 x} < 0. \checkmark$

Apply part (i): $\frac{\ln(\sin A) + \ln(\sin B) + \ln(\sin C)}{3} \leqslant \ln\!\left(\sin\frac{\pi}{3}\right) = \ln\frac{\sqrt{3}}{2}.$

The left side is $\frac{1}{3}\ln(\sin A \cdot \sin B \cdot \sin C)$. Since $\ln$ is increasing: $(\sin A \cdot \sin B \cdot \sin C)^{1/3} \leqslant \frac{\sqrt{3}}{2}.$

Cubing: $\sin A \cdot \sin B \cdot \sin C \leqslant \frac{3\sqrt{3}}{8}$.

Examiner Notes

This was the least popular of the pure questions on the paper, attempted by only 35% of the candidates. Very few candidates were able to score full marks and a fairly high proportion scored 0 on this question.

In the first part of the question, a significant number of candidates sketched convex functions and so had sketches that did not match the inequality. Most candidates were able to identify many of the key points required for the final section of this introductory part, but many could not explain it clearly enough for full marks in this section.

Most candidates made good attempts at the next section, but again the presentations often lacked enough clarity about how the inequalities were being linked together to be awarded full marks. In some cases, the choices for $x_1$ and $x_2$ were not within the interval $(x_1, x_2)$ and so these candidates lost marks.

Candidates who attempted to complete the final parts of the question were generally confident about how the results from the first part were to be applied, but a significant number omitted to demonstrate that the function being used was concave and so were not able to achieve full marks here.

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Question 3

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

Problem

3 (i) Let $f(x) = \frac{1}{1 + \tan x}$ for $0 \le x < \frac{1}{2}\pi$. Show that $f'(x) = -\frac{1}{1 + \sin 2x}$ and hence find the range of $f'(x)$. Sketch the curve $y = f(x)$.

(ii) The function $g(x)$ is continuous for $-1 \le x \le 1$. Show that the curve $y = g(x)$ has rotational symmetry of order 2 about the point $(a, b)$ on the curve if and only if $g(x) + g(2a - x) = 2b .$ Given that the curve $y = g(x)$ passes through the origin and has rotational symmetry of order 2 about the origin, write down the value of $\int_{-1}^{1} g(x) \, dx .$

(iii) Show that the curve $y = \frac{1}{1 + \tan^k x}$, where $k$ is a positive constant and $0 < x < \frac{1}{2}\pi$, has rotational symmetry of order 2 about a certain point (which you should specify) and evaluate $\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \tan^k x} \, dx .$

Hint

The differentiation for the first part of the question can be achieved by applying the chain rule. Consideration of the sine function within the interval then allows the range to be determined. Consideration of the gradient function then allows the graph to be sketched - it can be seen from the symmetry of the sine function that $f'(x) = f' \left( \frac{1}{2} \pi \right) - x$, which means that the graph must have rotational symmetry about the point where $x = \frac{1}{4} \pi$.

A sketch of a rotationally symmetric function is a helpful way of demonstrating the second part as it allows the distances that must be equal to be identified clearly. It is important to show clearly that the result works in both the if and only if directions. For the final question in this part, note that the sections of the graph above the axis must exactly match those below the axis, so the area must be 0.

For part (iii), begin by showing that the equation from part (ii) holds for this function. Once the rotational symmetry has been demonstrated it follows that the area of any interval with the centre of rotation in the centre will be equal to the area of a rectangle over the same interval passing through the centre of rotation.

Model Solution

Part (i): $f(x) = (1 + \tan x)^{-1}$.

$f'(x) = -(1 + \tan x)^{-2} \sec^2 x = -\frac{\sec^2 x}{(1 + \tan x)^2} = -\frac{1 + \tan^2 x}{(1 + \tan x)^2}.$

Since $\sin 2x = \frac{2\tan x}{1 + \tan^2 x}$: $1 + \sin 2x = \frac{1 + \tan^2 x + 2\tan x}{1 + \tan^2 x} = \frac{(1 + \tan x)^2}{1 + \tan^2 x}.$

Therefore $-\frac{1}{1 + \sin 2x} = -\frac{1 + \tan^2 x}{(1 + \tan x)^2} = f'(x)$. ✓

Range: For $0 < x < \pi/2$, $0 < 2x < \pi$, so $\sin 2x \in (0, 1]$. Thus $1 + \sin 2x \in (1, 2]$, giving $f'(x) \in [-1, -1/2)$.

Sketch: $f(0) = 1$, $f(x) \to 0$ as $x \to \pi/2^-$. Strictly decreasing (since $f' < 0$). Flattest at $x = \pi/4$ where $f = 1/2$.

Part (ii): Rotational symmetry of order 2 about $(a, b)$ means a $180°$ rotation maps the curve to itself. Point $(x, g(x))$ maps to $(2a - x, 2b - g(x))$. For this to lie on the curve: $g(2a - x) = 2b - g(x)$, i.e. $g(x) + g(2a - x) = 2b$. The argument works in both directions. ✓

With $a = b = 0$: $g(x) + g(-x) = 0$, so $g$ is odd. Hence $\int_{-1}^{1} g(x)\,dx = 0$.

Part (iii): Compute $f(\pi/2 - x)$: $f\!\left(\frac{\pi}{2} - x\right) = \frac{1}{1 + \cot^k x} = \frac{\tan^k x}{1 + \tan^k x}.$

So $f(x) + f(\pi/2 - x) = \frac{1}{1 + \tan^k x} + \frac{\tan^k x}{1 + \tan^k x} = 1$.

This confirms rotational symmetry about $(\pi/4, 1/2)$.

Evaluating the integral: $\pi/4$ is the midpoint of $[\pi/6, \pi/3]$. By the symmetry: $\int_{\pi/6}^{\pi/3} f(x)\,dx = \int_{\pi/6}^{\pi/3} [1 - f(x)]\,dx,$ so $2\int_{\pi/6}^{\pi/3} f(x)\,dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$, giving $\int_{\pi/6}^{\pi/3} f(x)\,dx = \frac{\pi}{12}$ (independent of $k$).

Examiner Notes

The first part of this question was well attempted in general, although some solutions required more care to be taken to check whether or not the extreme values are included within the range or not.

Many solutions to part (ii) did not generally include very clear explanations of the method and candidates often did not make it clear that they had demonstrated the result in both directions.

For the final part, candidates often did not find the “certain point” referred to in the question and instead tried to work with a general point $(a, b)$. Some candidates also attempted to calculate the integral directly rather than making use of the fact that the graph has rotational symmetry.

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Question 4

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

Problem

4 In this question, you may use the following identity without proof: $\cos A + \cos B = 2 \cos \frac{1}{2}(A + B) \cos \frac{1}{2}(A - B) .$

(i) Given that $0 \le x \le 2\pi$, find all the values of $x$ that satisfy the equation $\cos x + 3 \cos 2x + 3 \cos 3x + \cos 4x = 0 .$

(ii) Given that $0 \le x \le \pi$ and $0 \le y \le \pi$ and that $\cos(x + y) + \cos(x - y) - \cos 2x = 1 ,$ show that either $x = y$ or $x$ takes one specific value which you should find.

(iii) Given that $0 \le x \le \pi$ and $0 \le y \le \pi$, find the values of $x$ and $y$ that satisfy the equation $\cos x + \cos y - \cos(x + y) = \frac{3}{2} .$

Hint

For the first part, note that two of the terms in the left-hand side of the equation have a coefficient of 1 and two have a coefficient of 3. Applying the given identity to each of these pairs gives a common factor of $\cos \frac{5x}{2}$. The equation can therefore be factorised and then another application of the given identity will allow the full set of roots to be found.

The identity given at the start of the question can be applied to the first two terms of the left-hand side of the equation in part (ii) and the double angle formula can be applied to the $\cos 2x$. This then leads to an equation that can easily be factorised to show the required result. The range of possible values needs to be considered when considering the case where $\cos x = \cos y$.

For the final part a similar process to part (ii) can be used to create a quadratic function of $\cos \frac{1}{2}(x + y)$. Completing the square or considering a discriminant then allows the solutions to be found.

Model Solution

Part (i): Group: $(\cos x + \cos 4x) + 3(\cos 2x + \cos 3x) = 0$.

Apply the identity: $\cos x + \cos 4x = 2\cos\frac{5x}{2}\cos\frac{3x}{2}, \quad \cos 2x + \cos 3x = 2\cos\frac{5x}{2}\cos\frac{x}{2}.$

Factor: $2\cos\frac{5x}{2}\left(\cos\frac{3x}{2} + 3\cos\frac{x}{2}\right) = 0$.

Simplify: $\cos\frac{3x}{2} + 3\cos\frac{x}{2} = \cos\frac{3x}{2} + \cos\frac{x}{2} + 2\cos\frac{x}{2} = 2\cos x\cos\frac{x}{2} + 2\cos\frac{x}{2} = 2\cos\frac{x}{2}(\cos x + 1)$.

Equation becomes: $4\cos\frac{5x}{2}\cos\frac{x}{2}(\cos x + 1) = 0$.

Case 1: $\cos\frac{5x}{2} = 0 \Rightarrow x = \frac{\pi}{5} + \frac{2n\pi}{5}$, giving $x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}$.

Case 2: $\cos\frac{x}{2} = 0 \Rightarrow x = \pi$ (already found).

Case 3: $\cos x = -1 \Rightarrow x = \pi$ (already found).

Solutions: $x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}$.

Part (ii): Apply identity with $A = x+y$, $B = x-y$: $\cos(x+y) + \cos(x-y) = 2\cos x\cos y$.

Equation: $2\cos x\cos y - \cos 2x = 1$. Use $\cos 2x = 2\cos^2 x - 1$: $2\cos x\cos y - 2\cos^2 x + 1 = 1 \implies 2\cos x(\cos y - \cos x) = 0.$

Case 1: $\cos x = 0 \Rightarrow x = \pi/2$. Case 2: $\cos y = \cos x$. Since $\cos$ is strictly decreasing on $[0, \pi]$, $x = y$.

Therefore either $x = y$ or $x = \pi/2$.

Part (iii): Let $s = (x+y)/2$, $d = (x-y)/2$. Then $\cos x + \cos y = 2\cos s\cos d$ and $\cos(x+y) = 2\cos^2 s - 1$.

Substituting: $2\cos s\cos d - 2\cos^2 s + 1 = 3/2$, so $2\cos s(\cos d - \cos s) = 1/2$.

Since $\cos d \leq 1$: $\frac{1}{2} \leq 2\cos s(1 - \cos s)$. Let $u = \cos s$: $4u^2 - 4u + 1 \leq 0$, i.e. $(2u-1)^2 \leq 0$.

Since a square is $\geq 0$, we need $(2u-1)^2 = 0$, so $u = 1/2$, meaning $\cos s = 1/2$, $s = \pi/3$. Equality requires $\cos d = 1$, so $d = 0$.

Therefore $x = y = \pi/3$. Verification: $\cos(\pi/3) + \cos(\pi/3) - \cos(2\pi/3) = 1/2 + 1/2 + 1/2 = 3/2$. ✓

Examiner Notes

This was the second most popular question on the paper. The average mark scored by candidates attempting this question was the highest of all the pure questions.

The first two parts of the question were generally well done with candidates showing confidence in applying the given identity and then factorising and solving the resulting equation. A sizeable minority struggled to deal with the given range for $x$ and so were unable to find all the solutions. In part (ii) many candidates were able to explain clearly why $\cos x = \cos y$ leads to $x = y$.

Part (iii), however, proved to be difficult for the majority of candidates. A small number of candidates did manage to find the quadratic equation and most of these were able to proceed and complete the question fully. Most candidates did not score very many marks in this section.

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Question 5

Topic: 级数与积分 (Series and Integration)  |  Difficulty: Challenging  |  Marks: 20

Problem

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

(i) Write down the binomial expansion, for $|x| < 1$, of $\frac{1}{1 + x}$ and deduce that

$\ln(1 + x) = - \sum_{n=1}^{\infty} \frac{(-x)^n}{n}$

for $|x| < 1$.

(ii) Write down the series expansion in powers of $x$ of $e^{-ax}$. Use this expansion to show that

$\int_{0}^{\infty} \frac{(1 - e^{-ax}) e^{-x}}{x} dx = \ln(1 + a) \quad (|a| < 1) .$

(iii) Deduce the value of

$\int_{0}^{1} \frac{x^p - x^q}{\ln x} dx \quad (|p| < 1, |q| < 1) .$

Hint

For the first part of the question note that $\ln(1 + x)$ can be obtained by integrating $(1 + x)^{-1}$ and so the required expansion can be found by integrating the binomial expansion term by term. Note also, that the integration produces a constant, which needs to be shown to be 0.

In part (ii), the series expansion of $e^{ax}$ can be obtained by adjusting the series expansion of $e^x$. To evaluate the integral, substitute the series expansion for the $e^{ax}$, but leave the $e^{-x}$ unchanged. The integration can then be completed term by term.

For part (iii) note that a substitution of $u = -\ln x$ will transform the integral into one that can be expressed in terms of the integral in part (ii), which then allows the result to follow.

Model Solution

Part (i): Binomial expansion: $\frac{1}{1+x} = \sum_{n=0}^{\infty} (-x)^n$ for $|x| < 1$.

Integrate from $0$ to $x$: $\ln(1+x) = \int_0^x \frac{1}{1+t}\,dt = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} = -\sum_{n=1}^{\infty} \frac{(-x)^n}{n}.$

Check: $\ln(1+0) = 0$ ✓.

Part (ii): $e^{-ax} = \sum_{n=0}^{\infty} \frac{(-ax)^n}{n!}$, so $1 - e^{-ax} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} a^n x^n}{n!}$.

Divide by $x$ and multiply by $e^{-x}$: $\frac{(1-e^{-ax})e^{-x}}{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} a^n}{n!} x^{n-1} e^{-x}.$

Integrate term by term. Using $\int_0^{\infty} x^{n-1} e^{-x}\,dx = (n-1)!$ (by repeated integration by parts): $\int_0^{\infty} \frac{(1-e^{-ax})e^{-x}}{x}\,dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} a^n}{n!} \cdot (n-1)! = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} a^n}{n} = \ln(1+a).$

Part (iii): Substitute $x = e^{-t}$, so $\ln x = -t$, $dx = -e^{-t}\,dt$: $I = \int_0^1 \frac{x^p - x^q}{\ln x}\,dx = \int_0^{\infty} \frac{(e^{-qt} - e^{-pt})e^{-t}}{t}\,dt.$

Using part (ii): $I = \int_0^{\infty} \frac{(1-e^{-pt})e^{-t}}{t}\,dt - \int_0^{\infty} \frac{(1-e^{-qt})e^{-t}}{t}\,dt = \ln(1+p) - \ln(1+q) = \ln\!\left(\frac{1+p}{1+q}\right).$

Examiner Notes

Of the pure questions this was the one that attracted the poorest responses in general, with a significant proportion of attempts scoring 0.

In the first part of the question the majority of candidates did not include the constant of integration and so did not produce a fully justified solution. The expansion and substitution in the second part of the question was done well in general, although many candidates attempted to expand the $e^{-x}$ as well as the $(1 - e^{ax})$, after which they were unable to complete the integral. In many of these cases there were then unjustified jumps to the series expansion found in part (i).

Answers to part (iii) were generally much better than part (ii), although some substitution errors were seen. Most of the marks lost in this part were because candidates failed to spot the connection with the previous part. It is worth noting that many candidates did not attempt part (iii), having failed to complete part (ii) successfully. It is likely that some additional marks could have been scored by these candidates had they attempted this final part.

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Question 6

Topic: 数论 (Number Theory)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 (i) Find all pairs of positive integers $(n, p)$, where $p$ is a prime number, that satisfy

$n! + 5 = p .$

(ii) In this part of the question you may use the following two theorems:

1. For $n \geqslant 7$, $1! \times 3! \times \dots \times (2n - 1)! > (4n)!$.

2. For every positive integer $n$, there is a prime number between $2n$ and $4n$.

Find all pairs of positive integers $(n, m)$ that satisfy

$1! \times 3! \times \dots \times (2n - 1)! = m! .$

Hint

For the first part, note that 5 will certainly be a factor of the left-hand side in any case where n >= 5. It then remains to check the other cases one at a time.

For part (ii), first explain how the two theorems show that there will not be any solutions if n >= 7, by showing that m > 4n and so there must be a prime factor of the right-hand side that cannot exist in the product on the left-hand side.

The remaining cases then need to be checked one at a time, noting that the individual numbers within the product can be split into their prime factorisation and then combined differently to form the right-hand side.

Model Solution

Part (i): Check small $n$:

$n$	$n! + 5$	Prime?
1	6 = 2×3	No
2	7	Yes
3	11	Yes
4	29	Yes
5	125 = 5³	No

For $n \geq 5$: $5 \mid n!$, so $5 \mid (n! + 5)$ and $n! + 5 > 5$, hence composite.

Solutions: $(n,p) = (2,7), (3,11), (4,29)$.

Part (ii): Check small $n$:

• $n=1$: $1! = 1 = 1!$, so $m = 1$. ✓
• $n=2$: $1! \times 3! = 6 = 3!$, so $m = 3$. ✓
• $n=3$: $1! \times 3! \times 5! = 720 = 6!$, so $m = 6$. ✓
• $n=4$: $1! \times 3! \times 5! \times 7! = 3{,}628{,}800 = 10!$, so $m = 10$. ✓

No solutions for $n \geq 5$:

Case $n = 5$: By Theorem 2, there is a prime $p$ with $10 < p < 20$; take $p = 11$. Since $P(5) = 10! \times 9! > 10!$, we have $m \geq 11$, so $11 \mid m!$. But $11 > 9 = 2(5)-1$, so $11 \nmid P(5)$. Contradiction.

Case $n = 6$: Prime $p = 13$ (between 12 and 24). $P(6) = P(5) \times 11! > 12!$, so $m \geq 13$, $13 \mid m!$. But $13 > 11 = 2(6)-1$, so $13 \nmid P(6)$. Contradiction.

Case $n \geq 7$: By Theorem 1, $P(n) > (4n)!$, so $m > 4n$. By Theorem 2, prime $p$ with $2n < p < 4n$ exists. Then $p \mid m!$ but $p > 2n-1$ means $p \nmid P(n)$. Contradiction.

Solutions: $(n,m) = (1,1), (2,3), (3,6), (4,10)$.

Examiner Notes

Solutions on this question either scored very well or very poorly, depending on the quality of explanation provided by candidates in their solutions. In the weakest cases the only marks that were awarded were for finding some of the particular cases.

In the first part of the question candidates were generally able to find the cases that satisfied the equation, but many of the explanations that there are no solutions if n >= 5 were not sufficiently well produced to receive full marks.

In the second part of the question many students just restated the theorems without explaining the reasoning that followed from them. The cases for small values of n were generally found, but some candidates struggled to find the pair (4,10). Some candidates also did not attempt to explain why the cases n = 5 and n = 6 did not produce solutions.

There were some attempts to calculate the values of large factorials in this question. Candidates should be aware that such an approach will not be the correct method with which to tackle the questions.

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Question 7

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

Problem

7 The points $O$, $A$ and $B$ are the vertices of an acute-angled triangle. The points $M$ and $N$ lie on the sides $OA$ and $OB$ respectively, and the lines $AN$ and $BM$ intersect at $Q$. The position vector of $A$ with respect to $O$ is a, and the position vectors of the other points are labelled similarly.

Given that $|MQ| = \mu|QB|$, and that $|NQ| = \nu|QA|$, where $\mu$ and $\nu$ are positive and $\mu\nu < 1$, show that

$\mathbf{m} = \frac{(1 + \mu)\nu}{1 + \nu} \mathbf{a} .$

The point $L$ lies on the side $OB$, and $|OL| = \lambda|OB|$. Given that $ML$ is parallel to $AN$, express $\lambda$ in terms of $\mu$ and $\nu$.

What is the geometrical significance of the condition $\mu\nu < 1$?

Hint

It is very useful to draw a diagram to represent the situation described at the start of this question. Defining the vectors m and n as scalar multiples of a and b (using two new unknowns) allows the position vector of Q to be written in two different ways. Since the vectors a and b are not parallel, the coefficients of these vectors can be equated and this then leads to the correct expression for m.

A similar process then leads to an expression for the position vector of L in terms of a and b, but since L lies on OB, the coefficient of a must be 0.

It then follows that μν < 1 means that L lies on the segment OB.

Model Solution

Set $\mathbf{m} = s\mathbf{a}$ (since $M$ lies on $OA$) and $\mathbf{n} = t\mathbf{b}$ (since $N$ lies on $OB$).

$Q$ lies on $BM$: $\mathbf{q} = \frac{1}{1+\mu}\mathbf{m} + \frac{\mu}{1+\mu}\mathbf{b}$ (since $|MQ| = \mu|QB|$, so $Q$ divides $MB$ in ratio $\mu:1$).

$Q$ lies on $AN$: $\mathbf{q} = \frac{\nu}{1+\nu}\mathbf{a} + \frac{1}{1+\nu}\mathbf{n}$ (since $|NQ| = \nu|QA|$, so $Q$ divides $NA$ in ratio $\nu:1$).

From BM: $\mathbf{q} = \frac{s}{1+\mu}\mathbf{a} + \frac{\mu}{1+\mu}\mathbf{b}$. From AN: $\mathbf{q} = \frac{\nu}{1+\nu}\mathbf{a} + \frac{t}{1+\nu}\mathbf{b}$.

Since $\mathbf{a}$ and $\mathbf{b}$ are not parallel, equate coefficients:

• $\mathbf{a}$: $\frac{s}{1+\mu} = \frac{\nu}{1+\nu}$, so $s = \frac{(1+\mu)\nu}{1+\nu}$.
• $\mathbf{b}$: $\frac{\mu}{1+\mu} = \frac{t}{1+\nu}$, so $t = \frac{\mu(1+\nu)}{1+\mu}$.

Therefore $\mathbf{m} = \frac{(1+\mu)\nu}{1+\nu}\mathbf{a}$. ✓

Finding $\lambda$: $\mathbf{l} = \lambda\mathbf{b}$. $ML \parallel AN$ means $\mathbf{l} - \mathbf{m} \parallel \mathbf{n} - \mathbf{a}$.

$\mathbf{l} - \mathbf{m} = \lambda\mathbf{b} - s\mathbf{a}$ and $\mathbf{n} - \mathbf{a} = t\mathbf{b} - \mathbf{a}$.

Parallel: $\frac{\lambda}{t} = \frac{-s}{-1}$, so $\lambda = st = \frac{(1+\mu)\nu}{1+\nu} \cdot \frac{\mu(1+\nu)}{1+\mu} = \mu\nu$.

Geometric significance: $\lambda = \mu\nu$. Since $L$ lies on segment $OB$, we need $0 < \lambda < 1$, i.e. $\mu\nu < 1$.

Examiner Notes

This was an unpopular question among the pure questions, with slightly less than two fifths of the candidates attempting it. Many of the responses did not progress far through the problem, although there were some excellent solutions seen. As a result, attempts at this question generally received either very few or most of the marks.

A range of different methods were used, such as consideration of ratios of areas, or parallel lines to the diagram and the use of similar triangles.

The most common difficulties arose from candidates not recognising that m was parallel to a and n was parallel to b.

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Question 8

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

Problem

8 (i) Use the substitution $v = \sqrt{y}$ to solve the differential equation

$\frac{dy}{dt} = \alpha y^{\frac{1}{2}} - \beta y \quad (y \geqslant 0, \enspace t \geqslant 0),$

where $\alpha$ and $\beta$ are positive constants. Find the non-constant solution $y_1(x)$ that satisfies $y_1(0) = 0$.

(ii) Solve the differential equation

$\frac{dy}{dt} = \alpha y^{\frac{2}{3}} - \beta y \quad (y \geqslant 0, \enspace t \geqslant 0),$

where $\alpha$ and $\beta$ are positive constants. Find the non-constant solution $y_2(x)$ that satisfies $y_2(0) = 0$.

(iii) In the case $\alpha = \beta$, sketch $y_1(x)$ and $y_2(x)$ on the same axes, indicating clearly which is $y_1(x)$ and which is $y_2(x)$. You should explain how you determined the positions of the curves relative to each other.

Hint

Making the substitution given reduces the differential equation into one for which it is easy to separate the variables. The two sides can then be integrated to find a general solution to the equation and then the boundary condition can be applied to find the required solution.

The differential equation in part (ii) is similar to the one from part (i), so a similar substitution should work (using a cube root rather than square root). The same process can then be followed as in part (i) to solve this differential equation.

In part (iii), the information that α = β can be used to simplify the equations being considered and then it can be seen that the two curves will both approach an asymptote at y = 1. We also know that the curves both pass through the origin and the differential equations show that the curves should both have gradient 0 at the origin. All that remains is to deduce the relative positions of the two curves by considering the behaviour of the exponential function.

Model Solution

Part (i): $v = \sqrt{y}$, so $y = v^2$, $dy/dt = 2v\,dv/dt$. Substituting: $2v\frac{dv}{dt} = \alpha v - \beta v^2.$ For $v \neq 0$: $2\frac{dv}{dt} = \alpha - \beta v$. Separate: $\frac{2\,dv}{\alpha - \beta v} = dt.$ Integrate: $-\frac{2}{\beta}\ln|\alpha - \beta v| = t + C$.

At $t = 0$, $y = 0$ so $v = 0$: $C = -\frac{2}{\beta}\ln\alpha$.

$-\frac{2}{\beta}\ln\frac{\alpha - \beta v}{\alpha} = t \implies \frac{\alpha - \beta v}{\alpha} = e^{-\beta t/2} \implies v = \frac{\alpha}{\beta}(1 - e^{-\beta t/2}).$

$y_1(t) = \frac{\alpha^2}{\beta^2}(1 - e^{-\beta t/2})^2.$

Part (ii): $v = y^{1/3}$, so $y = v^3$, $dy/dt = 3v^2\,dv/dt$. Substituting: $3v^2\frac{dv}{dt} = \alpha v^2 - \beta v^3.$ For $v \neq 0$: $3\frac{dv}{dt} = \alpha - \beta v$. Separate and integrate: $-\frac{3}{\beta}\ln|\alpha - \beta v| = t + C.$ At $t = 0$, $v = 0$: $C = -\frac{3}{\beta}\ln\alpha$.

$v = \frac{\alpha}{\beta}(1 - e^{-\beta t/3}).$

$y_2(t) = \frac{\alpha^3}{\beta^3}(1 - e^{-\beta t/3})^3.$

Part (iii): When $\alpha = \beta$: $y_1 = (1 - e^{-t/2})^2, \quad y_2 = (1 - e^{-t/3})^3.$

Both $\to 1$ as $t \to \infty$. Both $= 0$ at $t = 0$ with $dy/dt = 0$.

For $t > 0$: since $-t/3 > -t/2$, we have $e^{-t/3} > e^{-t/2}$, so $1 - e^{-t/3} < 1 - e^{-t/2}$. Let $a = 1 - e^{-t/3}$ and $b = 1 - e^{-t/2}$, with $0 < a < b < 1$. Then $a^3 < a^2 < b^2$ (since cubing a number in $(0,1)$ makes it smaller). Wait, we need $a^3 < b^2$. Since $a < b$: $a^3 < ab^2 < b^3 < b^2$ (the last inequality since $b < 1$). Actually $b^3 < b^2$ since $b < 1$. And $a^3 < b^3 < b^2$. So $y_2 < y_1$ for all $t > 0$.

Therefore $y_1$ lies above $y_2$.

Examiner Notes

This was the third most popular question on the paper and a number of very good responses were seen from candidates.

Most candidates were able to apply the given substitution to the differential equation and then separated the variables successfully. Candidates adopted a range of approaches to solving the differential equation, such as use of an integrating factor or a solution by finding a complementary function and then a particular integral. Success was seen with all of these methods. Some candidates omitted the constant of integration and then were unable to reach the correct answer.

In the second part of the question most candidates were able to spot a suitable substitution and proceeded to solve the differential equation successfully. The best candidates spotted the similarity to part (i) and therefore saved some time on this part by modifying the answer to (i) rather than working through all of the steps again.

The final part of the question was the least well answered. Although most candidates realised that y_1(x) > y_2(x), many did not justify this or substituted a value to check rather than demonstrating that it was true for all values. Many candidates also failed to recognise that the gradient should be 0 at the origin for both curves.