STEP3 2014 -- Pure Mathematics

STEP3 2014 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2014  |  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	代数与级数 (Algebra and Series)	Challenging	对称多项式, 对数级数展开, 系数比较, 反例构造
2	积分 (Integration)	Standard	双曲函数恒等式, 代换积分法, 分式分解, 定积分计算
3	解析几何与优化 (Coordinate Geometry and Optimisation)	Challenging	抛物线参数方程, 法线方程, 距离优化, 分类讨论
4	积分不等式 (Integral Inequalities)	Hard	完全微分, 积分恒等式, $\sec^2 x = 1 + \tan^2 x$, 边值条件, 积分因子
5	复数与几何 (Complex Numbers and Geometry)	Challenging	复数表示向量, 平行四边形判定, 正方形判定, 旋转乘 $i$, 充分必要条件
6	分析 (Analysis)	Hard	正函数积分保号性,二次积分法,凸函数判定,三角双曲函数构造
7	几何 (Geometry)	Hard	圆的弦性质,向量表示,反证法,相似三角形
8	级数与数论 (Series and Number Theory)	Hard	分块求和估计,几何级数界,数字组合计数,调和级数发散性

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Question 1

Topic: 代数与级数 (Algebra and Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and let

$(1 + ax)(1 + bx)(1 + cx) = 1 + qx^2 + rx^3$

for all real $x$. Show that $q = bc + ca + ab$ and $r = abc$.

(i) Show that the coefficient of $x^n$ in the series expansion (in ascending powers of $x$) of $\ln(1 + qx^2 + rx^3)$ is $(-1)^{n+1}S_n$ where

$S_n = \frac{a^n + b^n + c^n}{n}, \quad (n \geqslant 1).$

(ii) Find, in terms of $q$ and $r$, the coefficients of $x^2$, $x^3$ and $x^5$ in the series expansion (in ascending powers of $x$) of $\ln(1 + qx^2 + rx^3)$ and hence show that $S_2S_3 = S_5$.

(iii) Show that $S_2S_5 = S_7$.

(iv) Give a proof of, or find a counterexample to, the claim that $S_2S_7 = S_9$.

Hint

The stem results are obtained through algebraic expansion and equating coefficients. Using the expression $(1 + ax)(1 + bx)(1 + cx)$ for $1 + qx^2 + rx^3$, manipulating the logarithm of the product, and the series expansions for expressions like $\ln(1 + ax)$ yields the displayed result. In parts (ii), (iii), and (iv), it is simplest to find $S_2 = -q$, $S_3 = r$, $S_5 = -qr$, $S_7 = q^2r$, and $S_9 = \frac{r^3}{3} - q^3r$ by expanding the series for $\ln(1 + (qx^2 + rx^3))$, and choosing a counter-example, selecting a, b and c so that $r \neq 0$.

Model Solution

Stem result. Expanding the left-hand side:

$$(1+ax)(1+bx)(1+cx) = (1+ax)\bigl(1 + (b+c)x + bcx^2\bigr)$$ $$= 1 + (b+c)x + bcx^2 + ax + a(b+c)x^2 + abcx^3$$ $$= 1 + (a+b+c)x + (ab+bc+ca)x^2 + abc\,x^3.$$

Since $a+b+c = 0$, the coefficient of $x$ vanishes, giving $q = ab+bc+ca$ and $r = abc$. $\square$

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Part (i) Since $1 + qx^2 + rx^3 = (1+ax)(1+bx)(1+cx)$, we have

$$\ln(1+qx^2+rx^3) = \ln(1+ax) + \ln(1+bx) + \ln(1+cx).$$

Using the standard expansion $\ln(1+t) = \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\,t^k$ for $|t| < 1$:

$$\ln(1+ax) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\,a^k x^k,$$

and similarly for $b$ and $c$. Summing all three:

$$\ln(1+qx^2+rx^3) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\,(a^k+b^k+c^k)\,x^k.$$

The coefficient of $x^n$ is therefore

$$\frac{(-1)^{n+1}}{n}\,(a^n+b^n+c^n) = (-1)^{n+1}\,S_n.$$

$\square$

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Part (ii) We expand $\ln(1 + qx^2 + rx^3)$ using $\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$ with $u = qx^2 + rx^3$.

Coefficient of $x^2$: only the term $u = qx^2 + rx^3$ contributes, giving $q$.

Coefficient of $x^3$: only $u$ contributes (via $rx^3$), giving $r$.

Coefficient of $x^5$: from $u$ we get nothing (max power is $x^3$); from $-\frac{u^2}{2}$:

$$u^2 = (qx^2+rx^3)^2 = q^2x^4 + 2qrx^5 + r^2x^6,$$

so the $x^5$ contribution is $-\frac{1}{2}\cdot 2qr = -qr$. From $\frac{u^3}{3}$:

$$u^3 = (qx^2+rx^3)^3 = q^3x^6 + 3q^2rx^7 + \cdots,$$

so the $x^5$ contribution is $0$. Higher terms also give no $x^5$. Therefore the coefficient of $x^5$ is $-qr$.

From Part (i), the coefficient of $x^n$ is $(-1)^{n+1}S_n$. Matching:

$$\text{coeff of } x^2 = q = (-1)^3 S_2 = -S_2 \implies S_2 = -q,$$ $$\text{coeff of } x^3 = r = (-1)^4 S_3 = S_3 \implies S_3 = r,$$ $$\text{coeff of } x^5 = -qr = (-1)^6 S_5 = S_5 \implies S_5 = -qr.$$

Therefore $S_2 S_3 = (-q)(r) = -qr = S_5$. $\square$

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Part (iii) We need $S_7$, so we find the coefficient of $x^7$ in $\ln(1+qx^2+rx^3)$.

From $u$: no $x^7$ term (max power $x^3$).

From $-\frac{u^2}{2}$: $u^2 = q^2x^4 + 2qrx^5 + r^2x^6$, so no $x^7$.

From $\frac{u^3}{3}$:

$$u^3 = (qx^2+rx^3)^3 = q^3x^6 + 3q^2rx^7 + 3qr^2x^8 + r^3x^9,$$

so the $x^7$ contribution is $\frac{1}{3}\cdot 3q^2r = q^2r$.

From $-\frac{u^4}{4}$: $u^4$ starts at $x^8$, so no $x^7$.

Higher terms also give no $x^7$. Therefore the coefficient of $x^7$ is $q^2r$, giving

$$(-1)^8 S_7 = q^2r \implies S_7 = q^2r.$$

Hence $S_2 S_5 = (-q)(-qr) = q^2r = S_7$. $\square$

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Part (iv) We need $S_9$, so we find the coefficient of $x^9$ in $\ln(1+qx^2+rx^3)$.

From $u$: no $x^9$.

From $-\frac{u^2}{2}$: $u^2 = q^2x^4 + 2qrx^5 + r^2x^6$, so no $x^9$.

From $\frac{u^3}{3}$: $u^3 = q^3x^6 + 3q^2rx^7 + 3qr^2x^8 + r^3x^9$, giving $\frac{1}{3}r^3$.

From $-\frac{u^4}{4}$: $u^4 = (qx^2+rx^3)^4 = q^4x^8 + 4q^3rx^9 + \cdots$, giving $-\frac{1}{4}\cdot 4q^3r = -q^3r$.

From $\frac{u^5}{5}$: $u^5$ starts at $x^{10}$, so no $x^9$.

Higher terms also contribute nothing. Therefore the coefficient of $x^9$ is $\frac{r^3}{3} - q^3r$, giving

$$(-1)^{10} S_9 = \frac{r^3}{3} - q^3r \implies S_9 = \frac{r^3}{3} - q^3r.$$

Now $S_2 S_7 = (-q)(q^2r) = -q^3r$. The claim $S_2 S_7 = S_9$ requires

$$-q^3r = \frac{r^3}{3} - q^3r \implies \frac{r^3}{3} = 0 \implies r = 0.$$

Since the claim must hold for all valid $a, b, c$, it fails whenever $r \neq 0$. A concrete counterexample: take $a = 1, b = 1, c = -2$ (so $a+b+c=0$). Then $q = 1\cdot 1 + 1\cdot(-2) + (-2)\cdot 1 = -3$ and $r = 1\cdot 1\cdot(-2) = -2$.

$$S_2 = -q = 3, \quad S_7 = q^2r = 9\cdot(-2) = -18, \quad S_9 = \frac{(-2)^3}{3} - (-3)^3(-2) = -\frac{8}{3} - 54 = -\frac{170}{3}.$$ $$S_2 S_7 = 3\cdot(-18) = -54 \neq -\frac{170}{3} = S_9.$$

The claim is false. $\square$

Examiner Notes

This was the most popular question on the paper, being attempted by approximately 14 out of every 15 candidates. It was the second most successfully attempted with a mean score of half marks. The stem of the question caused no problems, but a common mistake in part (i) was to attempt derivatives to obtain the desired result. Most candidates came unstuck in part (ii), making it much more difficult for themselves by attempting to work with expressions in a, b, and c rather than using the log series working with q and r, and as a result making sign errors, putting part (iii) beyond reach, and although they could find counterexamples for the claim in part (iv), they did so without the clear direction that working with the expressions in q and r would have made obvious.

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Question 2

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

Problem

2 (i) Show, by means of the substitution $u = \cosh x$, that

$\int \frac{\sinh x}{\cosh 2x} \, dx = \frac{1}{2\sqrt{2}} \ln \left| \frac{\sqrt{2} \cosh x - 1}{\sqrt{2} \cosh x + 1} \right| + C.$

(ii) Use a similar substitution to find an expression for

$\int \frac{\cosh x}{\cosh 2x} \, dx.$

(iii) Using parts (i) and (ii) above, show that

$\int_0^1 \frac{1}{1 + u^4} \, du = \frac{\pi + 2 \ln(\sqrt{2} + 1)}{4\sqrt{2}}.$

Hint

The first part is solved using the given method, the formula $\cosh 2x = 2 \cosh^2 x - 1$, and then employing partial fractions or the standard form quoted in the formula book. The second part requires the substitution, $u = \sinh x$, the formula $\cosh 2x = 1 + 2 \sinh^2 x$, and a standard form to give $\frac{\sqrt{2}}{2} \tan^{-1} \sqrt{2} u + c$. The third part can be approached by making the substitution $u = e^x$ and division of the resulting fraction in the numerator and denominator by $e^{2x}$ to give half the difference of the integrals in the first two parts. Alternatively, a similar style of working with the substitution $u = e^{-x}$ results in a sum instead of a difference.

Model Solution

Part (i) Let $u = \cosh x$, so $du = \sinh x \, dx$. Using the double-angle identity $\cosh 2x = 2\cosh^2 x - 1 = 2u^2 - 1$:

$$\int \frac{\sinh x}{\cosh 2x} \, dx = \int \frac{du}{2u^2 - 1}.$$

We use partial fractions. Factor $2u^2 - 1 = (\sqrt{2}\,u - 1)(\sqrt{2}\,u + 1)$, so

$$\frac{1}{2u^2 - 1} = \frac{1}{(\sqrt{2}\,u-1)(\sqrt{2}\,u+1)} = \frac{A}{\sqrt{2}\,u - 1} + \frac{B}{\sqrt{2}\,u + 1}.$$

Multiplying through: $1 = A(\sqrt{2}\,u+1) + B(\sqrt{2}\,u-1)$.

Setting $u = \frac{1}{\sqrt{2}}$: $1 = 2A$, so $A = \frac{1}{2}$.

Setting $u = -\frac{1}{\sqrt{2}}$: $1 = -2B$, so $B = -\frac{1}{2}$.

Therefore

$$\int \frac{du}{2u^2 - 1} = \frac{1}{2}\int \frac{du}{\sqrt{2}\,u - 1} - \frac{1}{2}\int \frac{du}{\sqrt{2}\,u + 1}$$ $$= \frac{1}{2\sqrt{2}}\ln|\sqrt{2}\,u - 1| - \frac{1}{2\sqrt{2}}\ln|\sqrt{2}\,u + 1| + C$$ $$= \frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}\,u - 1}{\sqrt{2}\,u + 1}\right| + C.$$

Substituting back $u = \cosh x$:

$$\int \frac{\sinh x}{\cosh 2x} \, dx = \frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}\cosh x - 1}{\sqrt{2}\cosh x + 1}\right| + C.$$

$\square$

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Part (ii) Let $u = \sinh x$, so $du = \cosh x \, dx$. Using $\cosh 2x = 1 + 2\sinh^2 x = 1 + 2u^2$:

$$\int \frac{\cosh x}{\cosh 2x} \, dx = \int \frac{du}{1 + 2u^2}.$$

This is a standard arctangent integral. Setting $v = \sqrt{2}\,u$, $dv = \sqrt{2}\,du$:

$$\int \frac{du}{1 + 2u^2} = \int \frac{dv/\sqrt{2}}{1 + v^2} = \frac{1}{\sqrt{2}}\arctan v + C = \frac{1}{\sqrt{2}}\arctan(\sqrt{2}\,\sinh x) + C.$$

$\square$

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Part (iii) We evaluate $I = \displaystyle\int_0^1 \frac{du}{1 + u^4}$.

Factor $1 + u^4 = (u^2 + \sqrt{2}\,u + 1)(u^2 - \sqrt{2}\,u + 1)$. We use partial fractions:

$$\frac{1}{1+u^4} = \frac{Au + B}{u^2 + \sqrt{2}\,u + 1} + \frac{Cu + D}{u^2 - \sqrt{2}\,u + 1}.$$

Multiplying through:

$$1 = (Au+B)(u^2 - \sqrt{2}\,u + 1) + (Cu+D)(u^2 + \sqrt{2}\,u + 1).$$

Expanding and collecting powers of $u$:

$$u^3: \quad A + C = 0 \implies C = -A,$$ $$u^2: \quad -\sqrt{2}\,A + B + \sqrt{2}\,C + D = 0 \implies -2\sqrt{2}\,A + B + D = 0,$$ $$u^1: \quad A - \sqrt{2}\,B + C + \sqrt{2}\,D = 0 \implies \sqrt{2}(D - B) = 0 \implies D = B,$$ $$u^0: \quad B + D = 1 \implies 2B = 1 \implies B = D = \tfrac{1}{2}.$$

From $u^2$: $-2\sqrt{2}\,A + 1 = 0$, so $A = \frac{1}{2\sqrt{2}}$, and $C = -\frac{1}{2\sqrt{2}}$.

Therefore

$$I = \frac{1}{2\sqrt{2}}\int_0^1 \frac{u}{u^2+\sqrt{2}\,u+1}\,du - \frac{1}{2\sqrt{2}}\int_0^1 \frac{u}{u^2-\sqrt{2}\,u+1}\,du + \frac{1}{2}\int_0^1 \frac{du}{u^2+\sqrt{2}\,u+1} + \frac{1}{2}\int_0^1 \frac{du}{u^2-\sqrt{2}\,u+1}.$$

We evaluate each integral separately.

Logarithmic integrals. Write $u = \frac{1}{2}(2u + \sqrt{2}) - \frac{\sqrt{2}}{2}$ in the first, and $u = \frac{1}{2}(2u - \sqrt{2}) + \frac{\sqrt{2}}{2}$ in the second:

$$\int \frac{u}{u^2+\sqrt{2}\,u+1}\,du = \frac{1}{2}\ln(u^2+\sqrt{2}\,u+1) - \frac{\sqrt{2}}{2}\int \frac{du}{u^2+\sqrt{2}\,u+1},$$ $$\int \frac{u}{u^2-\sqrt{2}\,u+1}\,du = \frac{1}{2}\ln(u^2-\sqrt{2}\,u+1) + \frac{\sqrt{2}}{2}\int \frac{du}{u^2-\sqrt{2}\,u+1}.$$

Substituting back, the $\frac{\sqrt{2}}{2}$-type terms cancel with parts of the $\frac{1}{2}\int \frac{du}{u^2 \pm \sqrt{2}u + 1}$ contributions, and we are left with:

$$I = \frac{1}{4\sqrt{2}}\left[\ln(u^2+\sqrt{2}\,u+1) - \ln(u^2-\sqrt{2}\,u+1)\right]_0^1 + \frac{1}{2}\int_0^1 \frac{du}{u^2+\sqrt{2}\,u+1} + \frac{1}{2}\int_0^1 \frac{du}{u^2-\sqrt{2}\,u+1}.$$

Arctangent integrals. Complete the square:

$$u^2 + \sqrt{2}\,u + 1 = \left(u + \frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}, \qquad u^2 - \sqrt{2}\,u + 1 = \left(u - \frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}.$$

Using $\displaystyle\int \frac{dw}{w^2 + a^2} = \frac{1}{a}\arctan\frac{w}{a}$ with $a = \frac{1}{\sqrt{2}}$:

$$\int_0^1 \frac{du}{u^2+\sqrt{2}\,u+1} = \left[\sqrt{2}\,\arctan\!\left(\sqrt{2}\,u + 1\right)\right]_0^1 = \sqrt{2}\left(\arctan(\sqrt{2}+1) - \arctan 1\right),$$ $$\int_0^1 \frac{du}{u^2-\sqrt{2}\,u+1} = \left[\sqrt{2}\,\arctan\!\left(\sqrt{2}\,u - 1\right)\right]_0^1 = \sqrt{2}\left(\arctan(\sqrt{2}-1) + \arctan 1\right).$$

Adding these two:

$$\arctan(\sqrt{2}+1) + \arctan(\sqrt{2}-1) = \frac{\pi}{2},$$

since $(\sqrt{2}+1)(\sqrt{2}-1) = 1$, and the identity $\arctan p + \arctan q = \frac{\pi}{2}$ holds when $pq = 1$ and $p > 0$. Therefore the sum of the arctangent integrals is $\sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi\sqrt{2}}{2}$.

Logarithmic part. At $u = 1$: $u^2 + \sqrt{2}\,u + 1 = 2 + \sqrt{2}$ and $u^2 - \sqrt{2}\,u + 1 = 2 - \sqrt{2}$. At $u = 0$: both equal $1$. So

$$\frac{1}{4\sqrt{2}}\ln\frac{2+\sqrt{2}}{2-\sqrt{2}}.$$

Rationalise:

$$\frac{2+\sqrt{2}}{2-\sqrt{2}} \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{(2+\sqrt{2})^2}{4-2} = \frac{6+4\sqrt{2}}{2} = 3+2\sqrt{2} = (1+\sqrt{2})^2.$$

So the logarithmic part is $\frac{1}{4\sqrt{2}} \cdot 2\ln(1+\sqrt{2}) = \frac{\ln(1+\sqrt{2})}{2\sqrt{2}}$.

Combining:

$$I = \frac{\ln(1+\sqrt{2})}{2\sqrt{2}} + \frac{1}{2} \cdot \frac{\pi\sqrt{2}}{2} = \frac{\ln(1+\sqrt{2})}{2\sqrt{2}} + \frac{\pi}{2\sqrt{2}} = \frac{\pi + 2\ln(\sqrt{2}+1)}{4\sqrt{2}}.$$

$\square$

Examiner Notes

This was only marginally less popular than question 1, but was the most successfully attempted with a mean of two thirds marks. Most that attempted the question were able to do the first two parts easily, but could not find a suitable substitution to do the last part. In about a tenth of the attempts, a helpful substitution was made in part (iii) which then usually resulted in successful completion of the question. Modulus signs were often ignored, or could not be distinguished from usual parentheses, and the arbitrary constant, even though it appeared in the result for part (i), was frequently overlooked. A few did not use the correct formulae for $\cosh 2x$, instead resorting to the trigon versions. A handful of candidates attempted partial fractions in the last part having correctly factorised the quartic, but this did not use the previous parts as instructed.

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Question 3

Topic: 解析几何与优化 (Coordinate Geometry and Optimisation)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) The line $L$ has equation $y = mx + c$, where $m > 0$ and $c > 0$. Show that, in the case $mc > a > 0$, the shortest distance between $L$ and the parabola $y^2 = 4ax$ is

$\frac{mc - a}{m\sqrt{m^2 + 1}}.$

           What is the shortest distance in the case that $mc \leqslant a$?

     **(ii)**     Find the shortest distance between the point $(p, 0)$, where $p > 0$, and the parabola $y^2 = 4ax$, where $a > 0$, in the different cases that arise according to the value of $p/a$. [*You may wish to use the parametric coordinates $(at^2, 2at)$ of points on the parabola.*]

           Hence find the shortest distance between the circle $(x - p)^2 + y^2 = b^2$, where $p > 0$ and $b > 0$, and the parabola $y^2 = 4ax$, where $a > 0$, in the different cases that arise according to the values of $p, a$ and $b$.

Hint

(i) Given that the shortest distance between the line and the parabola will be zero if they meet, investigating the solution of the equations simultaneously, and the discriminant of the resulting quadratic equation, the first result of the question is the case that they do not meet. The closest approach is the perpendicular distance of the point on the parabola where the tangent is parallel to the line, so using the standard parametric form, it is the perpendicular distance of $(\frac{a}{m^2}, \frac{2a}{m})$ from $y = mx + c$, giving the required result with care being taken over the sign of the numerator bearing in mind the inequalities.

(ii) The shortest distance of a point on the axis from the parabola, is either the distance from the vertex to the point, or the distance along one of the normals (which are symmetrically situated) which is not the axis. If the normal at $(at^2, 2at)$ passes through $(p, 0)$, then $p = 2a + at^2$. From this it can be simply shown that shortest distance is $p$ if $\frac{p}{a} < 2$, and is $2\sqrt{a(p - a)}$ if $\frac{p}{a} \geq 2$.

Then for the circle, the results follow simply, that the shortest distance will be $p - b$ if $p > b$, and 0 otherwise if $\frac{p}{a} < 2$, and $2\sqrt{a(p - a)} - b$ if $4a(p - a) > b^2$ or 0 otherwise if $\frac{p}{a} \geq 2$.

Model Solution

Part (i)

Substituting $y = mx + c$ into $y^2 = 4ax$:

$(mx + c)^2 = 4ax$

$m^2 x^2 + 2mcx + c^2 = 4ax$

$m^2 x^2 + (2mc - 4a)x + c^2 = 0.$

This is a quadratic in $x$. The discriminant is

$\Delta = (2mc - 4a)^2 - 4m^2 c^2 = 4(mc - 2a)^2 - 4m^2 c^2$

$= 4(m^2 c^2 - 4amc + 4a^2 - m^2 c^2) = 4(4a^2 - 4amc) = 16a(a - mc).$

When $mc > a > 0$, we have $\Delta < 0$, so the line and parabola do not meet. The shortest distance between them is therefore strictly positive.

The shortest distance from $L$ to the parabola occurs at a point on the parabola where the tangent is parallel to $L$ (i.e., has slope $m$). Using the parametric form $(at^2, 2at)$ on the parabola:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}.$

Setting $\frac{1}{t} = m$ gives $t = \frac{1}{m}$, so the point is $\left(\frac{a}{m^2},\, \frac{2a}{m}\right)$.

The perpendicular distance from a point $(x_0, y_0)$ to the line $y = mx + c$, i.e.\ $mx - y + c = 0$, is $\frac{|mx_0 - y_0 + c|}{\sqrt{m^2 + 1}}$.

At the point $\left(\frac{a}{m^2},\, \frac{2a}{m}\right)$:

$d = \frac{\left|m \cdot \frac{a}{m^2} - \frac{2a}{m} + c\right|}{\sqrt{m^2 + 1}} = \frac{\left|\frac{a}{m} - \frac{2a}{m} + c\right|}{\sqrt{m^2 + 1}} = \frac{\left|c - \frac{a}{m}\right|}{\sqrt{m^2 + 1}} = \frac{|mc - a|}{m\sqrt{m^2 + 1}}.$

Since $mc > a > 0$, we have $mc - a > 0$, so $|mc - a| = mc - a$, giving

$d = \frac{mc - a}{m\sqrt{m^2 + 1}}.$

When $mc \leqslant a$, the discriminant $\Delta \geqslant 0$, so the line and parabola intersect (or are tangent). The shortest distance in this case is 0.

Part (ii)

We seek the shortest distance from the point $(p, 0)$ to the parabola $y^2 = 4ax$.

Using parametric coordinates $(at^2, 2at)$ on the parabola, the distance squared is

$d^2(t) = (at^2 - p)^2 + (2at)^2 = (at^2 - p)^2 + 4a^2 t^2.$

The normal to the parabola at parameter $t$ has slope $-t$, and its equation is

$y - 2at = -t(x - at^2).$

If the normal passes through $(p, 0)$:

$0 - 2at = -t(p - at^2) = -tp + at^3$

$-2at + tp - at^3 = 0$

$t(p - 2a - at^2) = 0.$

So either $t = 0$ (the normal at the vertex, which is the $x$-axis itself), or

$at^2 = p - 2a \quad \implies \quad t^2 = \frac{p - 2a}{a}.$

This has real solutions for $t$ if and only if $p \geqslant 2a$.

Case 1: $p < 2a$. No normal from $(p, 0)$ meets the parabola at a point other than the vertex. Since $p > 0$ and the vertex is at the origin, the closest point is the vertex itself, and the shortest distance is $p$.

Case 2: $p \geqslant 2a$. The normal through $(p, 0)$ meets the parabola at $t^2 = \frac{p - 2a}{a}$. The distance squared at this point is

$d^2 = \left(a \cdot \frac{p - 2a}{a} - p\right)^2 + 4a^2 \cdot \frac{p - 2a}{a}$

$= (p - 2a - p)^2 + 4a(p - 2a) = 4a^2 + 4a(p - 2a) = 4a(p - a).$

So $d = 2\sqrt{a(p - a)}$. We verify this is less than $p$ when $p > 2a$: indeed $4a(p-a) < p^2 \iff 4ap - 4a^2 < p^2 \iff (p - 2a)^2 > 0$, which holds for $p \neq 2a$.

Therefore:

$\text{Shortest distance} = \begin{cases} p & \text{if } p < 2a, \\[4pt] 2\sqrt{a(p - a)} & \text{if } p \geqslant 2a. \end{cases}$

Distance from the circle to the parabola.

The circle $(x - p)^2 + y^2 = b^2$ has centre $(p, 0)$ and radius $b$. The shortest distance from the circle to the parabola equals the shortest distance from the centre $(p, 0)$ to the parabola, minus $b$ (provided this is positive; otherwise the circle and parabola meet, and the distance is $0$).

Case 1: $p < 2a$. The shortest distance from $(p, 0)$ to the parabola is $p$.

• If $p > b$: the shortest distance from the circle to the parabola is $p - b$.
• If $p \leqslant b$: the shortest distance is $0$.

Case 2: $p \geqslant 2a$. The shortest distance from $(p, 0)$ to the parabola is $2\sqrt{a(p - a)}$.

• If $4a(p - a) > b^2$, i.e.\ $2\sqrt{a(p - a)} > b$: the shortest distance from the circle to the parabola is $2\sqrt{a(p - a)} - b$.
• If $4a(p - a) \leqslant b^2$: the shortest distance is $0$.

Examiner Notes

About half of the candidates attempted this, but it was the second least successfully attempted with a mean score just below a quarter marks. Most managed the first result, with those not doing so falling foul of various basic algebraic errors. The second result of part (i) was often answered with no justification. The second part was poorly done with a variety of approaches attempted such as obtaining a distance function, and then using completing the square or differentiation, or investigating the intersections of the circle and parabola. Few considered the geometry of the parabola and its normal which would have yielded the results fairly simply.

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Question 4

Topic: 积分不等式 (Integral Inequalities)  |  Difficulty: Hard  |  Marks: 20

Problem

4 (i) Let

$I = \int_0^1 ((y')^2 - y^2) \, dx \quad \text{and} \quad I_1 = \int_0^1 (y' + y \tan x)^2 \, dx,$

           where $y$ is a given function of $x$ satisfying $y = 0$ at $x = 1$. Show that $I - I_1 = 0$ and deduce that $I \geqslant 0$. Show further that $I = 0$ only if $y = 0$ for all $x$ ($0 \leqslant x \leqslant 1$).

     **(ii)**     Let

$J = \int_0^1 ((y')^2 - a^2y^2) \, dx,$

           where $a$ is a given positive constant and $y$ is a given function of $x$, not identically zero, satisfying $y = 0$ at $x = 1$. By considering an integral of the form

$\int_0^1 (y' + ay \tan bx)^2 \, dx,$

           where $b$ is suitably chosen, show that $J \geqslant 0$. You should state the range of values of $a$, in the form $a < k$, for which your proof is valid.

           In the case $a = k$, find a function $y$ (not everywhere zero) such that $J = 0$.

Hint

Expanding the bracket in the integral $I_1$, and employing $\sec^2 x = 1 + \tan^2 x$ yields $I$ plus the integral of a perfect differential which can be evaluated simply. For $I = 0$, $y' + y \tan x = 0$ over the interval which can be solved using an integrating factor and then the condition $y = 0, x = 1$ enables the arbitrary constant to be evaluated giving the required result. In part (ii), similar working can be undertaken with the integral which is to be considered, given $b = a$. The argument requires no discontinuity in the interval so $a < \frac{\pi}{2}$. The function $y = \cos ax$ can be shown to meet the requirement.

Model Solution

Part (i)

We expand $I_1$:

$I_1 = \int_0^1 (y' + y\tan x)^2 \, dx = \int_0^1 \left[(y')^2 + 2yy'\tan x + y^2 \tan^2 x\right] dx.$

We evaluate the middle integral $\int_0^1 2yy'\tan x \, dx$ by parts. Let $u = \tan x$ and $dv = 2yy' \, dx = d(y^2)$, so $du = \sec^2 x \, dx$ and $v = y^2$. Then

$\int_0^1 2yy'\tan x \, dx = \left[y^2 \tan x\right]_0^1 - \int_0^1 y^2 \sec^2 x \, dx.$

At $x = 1$: $y^2 \tan 1$. At $x = 0$: $y^2 \tan 0 = 0$. But $y = 0$ at $x = 1$, so $y^2 \tan 1 = 0$. Therefore

$\int_0^1 2yy'\tan x \, dx = -\int_0^1 y^2 \sec^2 x \, dx.$

Substituting back:

$I_1 = \int_0^1 (y')^2 \, dx - \int_0^1 y^2 \sec^2 x \, dx + \int_0^1 y^2 \tan^2 x \, dx$

$= \int_0^1 (y')^2 \, dx + \int_0^1 y^2 (\tan^2 x - \sec^2 x) \, dx.$

Using $\sec^2 x = 1 + \tan^2 x$, we get $\tan^2 x - \sec^2 x = -1$. Therefore

$I_1 = \int_0^1 (y')^2 \, dx - \int_0^1 y^2 \, dx = I.$

Hence $I - I_1 = 0$.

Since $I = I_1$ and $I_1 = \int_0^1 (y' + y\tan x)^2 \, dx$, and the integrand is a perfect square, we have $I_1 \geqslant 0$, so $I \geqslant 0$.

If $I = 0$, then $I_1 = 0$, which requires $(y' + y\tan x)^2 = 0$ for all $x \in [0, 1]$, i.e.

$y' + y\tan x = 0.$

This is a separable ODE. Dividing by $y$ (assuming $y \neq 0$):

$\frac{y'}{y} = -\tan x \quad \implies \quad \ln|y| = \ln|\cos x| + C \quad \implies \quad y = A\cos x$

for some constant $A$. The boundary condition $y(1) = 0$ requires $A\cos 1 = 0$. Since $\cos 1 \neq 0$ (as $1$ radian $< \frac{\pi}{2}$), we need $A = 0$, so $y = 0$ for all $x \in [0, 1]$.

Part (ii)

We consider

$I_2 = \int_0^1 (y' + ay\tan bx)^2 \, dx$

where $b$ is to be chosen. Expanding:

$I_2 = \int_0^1 (y')^2 \, dx + 2a\int_0^1 yy'\tan bx \, dx + a^2\int_0^1 y^2 \tan^2 bx \, dx.$

We evaluate the middle term by parts with $u = \tan bx$ and $dv = 2yy' \, dx = d(y^2)$:

$2a\int_0^1 yy'\tan bx \, dx = a\left[y^2 \tan bx\right]_0^1 - ab\int_0^1 y^2 \sec^2 bx \, dx.$

At $x = 0$: $y^2 \tan 0 = 0$. At $x = 1$: $y = 0$, so $y^2 \tan b = 0$ provided $\tan b$ is finite, i.e.\ $b \neq \frac{\pi}{2} + n\pi$. Therefore

$2a\int_0^1 yy'\tan bx \, dx = -ab\int_0^1 y^2 \sec^2 bx \, dx.$

Substituting:

$I_2 = \int_0^1 (y')^2 \, dx + \int_0^1 y^2 (a^2\tan^2 bx - ab\sec^2 bx) \, dx.$

Using $\sec^2 bx = 1 + \tan^2 bx$:

$a^2\tan^2 bx - ab\sec^2 bx = a^2\tan^2 bx - ab(1 + \tan^2 bx) = (a^2 - ab)\tan^2 bx - ab.$

We want $I_2 = J + (\text{non-negative terms})$, where $J = \int_0^1 ((y')^2 - a^2 y^2) \, dx$. Comparing:

$I_2 = \int_0^1 (y')^2 \, dx + \int_0^1 y^2\left[(a^2 - ab)\tan^2 bx - ab\right] dx.$

For this to equal $J + \int_0^1 (\text{something})^2 \, dx$, we need $-ab = -a^2$, giving $b = a$.

With $b = a$: $a^2 - ab = 0$, so

$I_2 = \int_0^1 (y')^2 \, dx - a^2\int_0^1 y^2 \, dx = J.$

Therefore $J = I_2 = \int_0^1 (y' + ay\tan ax)^2 \, dx \geqslant 0$.

Validity: This proof requires $\tan ax$ to be continuous on $[0, 1]$, so $ax \neq \frac{\pi}{2}$ for any $x \in [0, 1]$, i.e.\ $a < \frac{\pi}{2}$. So the proof is valid for $a < \frac{\pi}{2}$, giving $k = \frac{\pi}{2}$.

The case $a = k = \frac{\pi}{2}$. We need a function $y$, not everywhere zero, with $y(1) = 0$ and $J = 0$. Since $J = \int_0^1 ((y')^2 - \frac{\pi^2}{4} y^2) \, dx$, consider $y = \cos\frac{\pi x}{2}$. Then $y(1) = \cos\frac{\pi}{2} = 0$ and

$y' = -\frac{\pi}{2}\sin\frac{\pi x}{2}, \qquad (y')^2 = \frac{\pi^2}{4}\sin^2\frac{\pi x}{2}, \qquad \frac{\pi^2}{4}y^2 = \frac{\pi^2}{4}\cos^2\frac{\pi x}{2}.$

Therefore

$J = \frac{\pi^2}{4}\int_0^1 \left(\sin^2\frac{\pi x}{2} - \cos^2\frac{\pi x}{2}\right) dx = -\frac{\pi^2}{4}\int_0^1 \cos(\pi x) \, dx = -\frac{\pi^2}{4}\left[\frac{\sin \pi x}{\pi}\right]_0^1 = 0.$

So $y = \cos\frac{\pi x}{2}$ gives $J = 0$.

Examiner Notes

Two thirds of the candidature attempted this but with only moderate success earning just a third of the marks. The very first result was frequently obtained although some fell at the first hurdle through not appreciating that they needed to use $\sec^2 x = 1 + \tan^2 x$, or else that there was then an exact differential. The second result in part (i) was “only if” whereas many read it, or answered it, as “if”. In part (ii), most spotted $b = a$. There were many inappropriate functions suggested for the last part of the question, many which ignored the requirement that $y = 0, x = 1$.

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Question 5

Topic: 复数与几何 (Complex Numbers and Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 A quadrilateral drawn in the complex plane has vertices $A, B, C$ and $D$, labelled anticlockwise. These vertices are represented, respectively, by the complex numbers $a, b, c$ and $d$. Show that $ABCD$ is a parallelogram (defined as a quadrilateral in which opposite sides are parallel and equal in length) if and only if $a + c = b + d$. Show further that, in this case, $ABCD$ is a square if and only if $i(a - c) = b - d$.

Let $PQRS$ be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less than $180^\circ$. Squares with centres $X, Y, Z$ and $T$ are constructed externally to the quadrilateral on the sides $PQ, QR, RS$ and $SP$, respectively.

(i) If $P$ and $Q$ are represented by the complex numbers $p$ and $q$, respectively, show that $X$ can be represented by $\frac{1}{2}(p(1 + i) + q(1 - i)) .$

(ii) Show that $XYZT$ is a square if and only if $PQRS$ is a parallelogram.

Hint

ABCD is a parallelogram if and only if $\vec{AB} = \vec{DC}$ which yields the required result. To be a square as well, angle $ABC = 90^\circ$, and $|AB| = |BC|$, so $c - b = i(b - a)$. Treating the two results as simultaneous equations to be solved for $a$ and $c$ in terms of $b$ and $d$, the second result of the stem can be shown with reversible logic. For part (i) the same logic can be used for $PXQ$ as just used for $ABC$. From the stem, XYZT is a square if and only if $i(x - z) = y - t$, and $x + z = y + t$ and given the working for X in part (i), these can be shown to be true treating Y, Z, and T similarly.

Model Solution

Stem: Parallelogram condition

$ABCD$ is a parallelogram if and only if $\vec{AB} = \vec{DC}$, i.e.

$b - a = c - d.$

Rearranging: $a + c = b + d$. $\qquad \text{(shown)}$

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Stem: Square condition (given parallelogram)

Suppose $ABCD$ is a parallelogram, so $a + c = b + d$. Then $ABCD$ is a square if and only if adjacent sides are perpendicular and equal in length, i.e. $\vec{BC} = i\,\vec{AB}$ (rotation by $90°$):

$c - b = i(b - a). \qquad (\star)$

We show that $(\star)$ together with $a + c = b + d$ is equivalent to $i(a - c) = b - d$.

($\Rightarrow$) From $a + c = b + d$: $d = a + c - b$, so

$b - d = 2b - a - c.$

From $(\star)$: $c = b + i(b - a) = b(1 + i) - ia$, giving

$a - c = a(1 + i) - b(1 + i) = (1 + i)(a - b).$

Therefore $i(a - c) = i(1 + i)(a - b) = (i - 1)(a - b)$. Meanwhile

$b - d = 2b - a - [b(1+i) - ia] = b(1 - i) - a(1 - i) = -(1 - i)(a - b) = (i - 1)(a - b).$

So $i(a - c) = b - d$.

($\Leftarrow$) Suppose $a + c = b + d$ and $i(a - c) = b - d$. Let $m = \frac{a+c}{2} = \frac{b+d}{2}$ and set $u = a - m$, $v = b - m$. Then $c = m - u$, $d = m - v$, and $a - c = 2u$, $b - d = 2v$. The condition $i(a-c) = b-d$ gives $v = iu$. Now

$\vec{AB} = b - a = v - u = u(i - 1), \qquad \vec{BC} = c - b = -u(1 + i).$

$\frac{\vec{BC}}{\vec{AB}} = \frac{-u(1+i)}{u(i-1)} = \frac{-(1+i)}{i-1}.$

Multiplying numerator and denominator by $\overline{(i-1)} = -(1+i)$:

$\frac{-(1+i)}{i-1} \cdot \frac{-(1+i)}{-(1+i)} = \frac{(1+i)^2}{|i-1|^2} = \frac{2i}{2} = i.$

Since $\vec{BC} = i\,\vec{AB}$, the sides $AB$ and $BC$ are perpendicular and equal in length. Combined with the parallelogram condition, $ABCD$ is a square. $\qquad \square$

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Part (i)

The square on side $PQ$ is constructed externally. Its four vertices, in order, are

$P, \quad Q, \quad Q + R, \quad P + R,$

where $R = -i(q - p)$ is the vector $\vec{PQ}$ rotated $90°$ clockwise (toward the exterior of the anticlockwise quadrilateral).

The centre $X$ is the average of the four vertices:

$X = \frac{1}{4}\bigl[p + q + (q + R) + (p + R)\bigr] = \frac{2p + 2q + 2R}{4} = \frac{p + q + R}{2}.$

Substituting $R = -i(q - p)$:

$X = \frac{p + q - i(q - p)}{2} = \frac{p(1 + i) + q(1 - i)}{2}. \qquad \text{(shown)} \quad \square$

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Part (ii)

By the same construction, the centres of the external squares on the four sides are:

$x = \tfrac{1}{2}\bigl(p(1+i) + q(1-i)\bigr), \quad y = \tfrac{1}{2}\bigl(q(1+i) + r(1-i)\bigr),$

$z = \tfrac{1}{2}\bigl(r(1+i) + s(1-i)\bigr), \quad t = \tfrac{1}{2}\bigl(s(1+i) + p(1-i)\bigr).$

From the stem results, $XYZT$ is a square if and only if both:

• (a) $x + z = y + t$ (parallelogram condition), and
• (b) $i(x - z) = y - t$ (right angle + equal sides condition).

Condition (b) — perpendicularity:

$x - z = \tfrac{1}{2}\bigl[p(1+i) + q(1-i) - r(1+i) - s(1-i)\bigr].$

Multiplying by $i$ and using $(1+i)i = i - 1$, $(1-i)i = 1 + i$:

$i(x - z) = \tfrac{1}{2}\bigl[p(i-1) + q(1+i) - r(i-1) - s(1+i)\bigr].$

Since $-r(i-1) = r(1-i)$:

$i(x - z) = \tfrac{1}{2}\bigl[p(i-1) + q(1+i) + r(1-i) - s(1+i)\bigr].$

Now compute directly:

$y - t = \tfrac{1}{2}\bigl[q(1+i) + r(1-i) - s(1+i) - p(1-i)\bigr].$

Since $-p(1-i) = p(i - 1)$, the two expressions are identical. Therefore $i(x - z) = y - t$ holds identically for all $p, q, r, s$.

Condition (a) — parallelogram condition:

$x + z = \tfrac{1}{2}\bigl[(p+r)(1+i) + (q+s)(1-i)\bigr],$

$y + t = \tfrac{1}{2}\bigl[(q+s)(1+i) + (p+r)(1-i)\bigr].$

Setting $u = p + r$, $v = q + s$:

$u(1+i) + v(1-i) = v(1+i) + u(1-i)$

$(u - v)(1+i) - (u - v)(1-i) = 0$

$(u - v)\bigl[(1+i) - (1-i)\bigr] = 0$

$2i(u - v) = 0 \implies u = v.$

So $x + z = y + t$ if and only if $p + r = q + s$, which is the condition for $PQRS$ to be a parallelogram.

Conclusion: Condition (b) is always satisfied. Therefore $XYZT$ is a square if and only if condition (a) holds, i.e. if and only if $PQRS$ is a parallelogram. $\qquad \square$

Examiner Notes

This was a moderately popular question attempted by half the candidates, with some success, scoring a little below half marks. There were some basic problems exposed in this question such as the differences between a vector and its length, the negative of a vector and the vector, and the meaning of “if and only if” resulting in things being shown in one direction only throughout the question. Part (i) was generally well done, but in part (ii), it was commonly forgotten that there were two conditions for XYZT to be a square. Approaches using real and imaginary parts (breaking into components) were not very successful.

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Question 6

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

Problem

6 Starting from the result that $h(t) > 0 \text{ for } 0 < t < x \implies \int_{0}^{x} h(t) \, dt > 0 ,$ show that, if $f''(t) > 0$ for $0 < t < x_0$ and $f(0) = f'(0) = 0$, then $f(t) > 0$ for $0 < t < x_0$.

(i) Show that, for $0 < x < \frac{1}{2}\pi$, $\cos x \cosh x < 1 .$

(ii) Show that, for $0 < x < \frac{1}{2}\pi$, $\frac{1}{\cosh x} < \frac{\sin x}{x} < \frac{x}{\sinh x} .$

Hint

Starting from $f''(t) > 0$ for $0 < t < x_0$, integrate from $0$ to $t_0$ (where $0 < t_0 < x_0$): $\int_0^{t_0} f''(t) \, dt > 0 \implies f'(t_0) - f'(0) > 0 \implies f'(t_0) > 0$ since $f'(0) = 0$. Repeat the argument with $f'(t)$: integrate from $0$ to $t_0$: $\int_0^{t_0} f'(t) \, dt > 0 \implies f(t_0) - f(0) > 0 \implies f(t_0) > 0$ since $f(0) = 0$.

(i) Choose $f(x) = 1 - \cos x \cosh x$. Then $f(0) = 0$ and $f'(x) = \sin x \cosh x - \cos x \sinh x$, so $f'(0) = 0$. Now $f''(x) = 2\sin x \sinh x > 0 \quad \text{for } 0 < x < \pi.$ By the stem result, $f(x) > 0$ for $0 < x < \pi$, so $1 - \cos x \cosh x > 0$, i.e. $\cos x \cosh x < 1$ for $0 < x < \pi$, and in particular for $0 < x < \frac{1}{2}\pi$.

(ii) Choose $g(x) = x^2 - \sin x \sinh x$. Then $g(0) = 0$ and $g'(x) = 2x - \cos x \sinh x - \sin x \cosh x$, so $g'(0) = 0$. Now $g''(x) = 2 - 2\cos x \cosh x = 2f(x) > 0 \quad \text{for } 0 < x < \pi.$ By the stem result, $g(x) > 0$ for $0 < x < \pi$, so $x^2 > \sin x \sinh x$, giving $\frac{x}{\sinh x} > \frac{\sin x}{x}$ for $0 < x < \frac{1}{2}\pi$.

Choose $h(x) = \sin x \cosh x - x$. Then $h(0) = 0$ and $h'(x) = \cos x \cosh x + \sin x \sinh x - 1$, so $h'(0) = 0$. Now $h''(x) = 2\sin x \cosh x > 0 \quad \text{for } 0 < x < \pi.$ By the stem result, $h(x) > 0$ for $0 < x < \pi$, so $\sin x \cosh x > x$, giving $\frac{\sin x}{x} > \frac{1}{\cosh x}$ for $0 < x < \frac{1}{2}\pi$. Combining: $\frac{1}{\cosh x} < \frac{\sin x}{x} < \frac{x}{\sinh x}$.

Model Solution

Stem result

Suppose $f''(t) > 0$ for $0 < t < x_0$ and $f(0) = f'(0) = 0$. We apply the given result twice.

First application. Take $h(t) = f''(t)$. Since $f''(t) > 0$ for $0 < t < x_0$, for any $t_0$ with $0 < t_0 < x_0$:

$\int_0^{t_0} f''(t) \, dt > 0 \implies f'(t_0) - f'(0) > 0 \implies f'(t_0) > 0,$

using $f'(0) = 0$.

Second application. Take $h(t) = f'(t)$. Since $f'(t) > 0$ for $0 < t < x_0$ (just shown), for any $t_0$ with $0 < t_0 < x_0$:

$\int_0^{t_0} f'(t) \, dt > 0 \implies f(t_0) - f(0) > 0 \implies f(t_0) > 0,$

using $f(0) = 0$. Therefore $f(t) > 0$ for $0 < t < x_0$. $\qquad \square$

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Part (i)

Define $f(x) = 1 - \cos x \cosh x$. We verify the hypotheses of the stem result.

Check boundary conditions:

$f(0) = 1 - \cos 0 \cdot \cosh 0 = 1 - 1 = 0.$

$f'(x) = \sin x \cosh x - \cos x \sinh x \implies f'(0) = 0 - 0 = 0.$

Compute $f''$:

$f''(x) = \frac{d}{dx}[\sin x \cosh x - \cos x \sinh x]$

$= (\cos x \cosh x + \sin x \sinh x) - (-\sin x \sinh x + \cos x \cosh x)$

$= 2\sin x \sinh x.$

For $0 < x < \pi$: $\sin x > 0$ and $\sinh x > 0$, so $f''(x) = 2\sin x \sinh x > 0$.

Apply the stem result with $x_0 = \pi$: $f(x) > 0$ for $0 < x < \pi$, i.e.

$1 - \cos x \cosh x > 0 \implies \cos x \cosh x < 1.$

In particular, this holds for $0 < x < \frac{1}{2}\pi$. $\qquad \square$

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Part (ii)

We prove two inequalities separately.

Right inequality: $\dfrac{\sin x}{x} < \dfrac{x}{\sinh x}$ for $0 < x < \frac{1}{2}\pi$.

This is equivalent to $\sin x \cdot \sinh x < x^2$ (valid since $\sin x > 0$, $\sinh x > 0$, $x > 0$ in this range).

Define $g(x) = x^2 - \sin x \sinh x$. Check boundary conditions:

$g(0) = 0 - 0 = 0.$

$g'(x) = 2x - \cos x \sinh x - \sin x \cosh x \implies g'(0) = 0 - 0 - 0 = 0.$

Compute $g''$:

$g''(x) = 2 - (-\sin x \sinh x + \cos x \cosh x) - (\cos x \cosh x + \sin x \sinh x)$

$= 2 - 2\cos x \cosh x.$

By part (i), $\cos x \cosh x < 1$ for $0 < x < \pi$, so $g''(x) = 2(1 - \cos x \cosh x) > 0$ for $0 < x < \pi$.

Apply the stem result with $x_0 = \pi$: $g(x) > 0$ for $0 < x < \pi$, i.e. $x^2 > \sin x \sinh x$.

Dividing by $x \sinh x > 0$:

$\frac{x}{\sinh x} > \frac{\sin x}{x}. \qquad (\dagger)$

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Left inequality: $\dfrac{1}{\cosh x} < \dfrac{\sin x}{x}$ for $0 < x < \frac{1}{2}\pi$.

This is equivalent to $x < \sin x \cosh x$ (valid since $\cosh x > 0$, $x > 0$ in this range).

Define $h(x) = \sin x \cosh x - x$. Check boundary conditions:

$h(0) = 0 - 0 = 0.$

$h'(x) = \cos x \cosh x + \sin x \sinh x - 1 \implies h'(0) = 1 + 0 - 1 = 0.$

Compute $h''$:

$h''(x) = (-\sin x \cosh x + \cos x \sinh x) + (\cos x \sinh x + \sin x \cosh x)$

$= 2\cos x \sinh x.$

For $0 < x < \frac{1}{2}\pi$: $\cos x > 0$ and $\sinh x > 0$, so $h''(x) > 0$.

Apply the stem result with $x_0 = \frac{1}{2}\pi$: $h(x) > 0$ for $0 < x < \frac{1}{2}\pi$, i.e. $\sin x \cosh x > x$.

Dividing by $x \cosh x > 0$:

$\frac{\sin x}{x} > \frac{1}{\cosh x}. \qquad (\ddagger)$

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Combining $(\dagger)$ and $(\ddagger)$:

$\frac{1}{\cosh x} < \frac{\sin x}{x} < \frac{x}{\sinh x} \quad \text{for } 0 < x < \tfrac{1}{2}\pi. \qquad \square$

Examiner Notes

Many did not use the required starting point, instead resorting to monotonicity or drawing pictures (graphs) which were not proofs. In parts (i) especially and (ii) as well, candidates failed to use the result that $f(t) > 0$, cavalierly using $f(t) < 0$, or even $f(t) < 1$ without justification. Many made complicated choices of functions for (i) and (ii), and then got lost in their differentiations, and finally there was frequent lack of care to ensure that quantities dividing inequalities were positive.

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Question 7

Topic: 几何 (Geometry)  |  Difficulty: Hard  |  Marks: 20

Problem

7 The four distinct points $P_i$ ($i = 1, 2, 3, 4$) are the vertices, labelled anticlockwise, of a cyclic quadrilateral. The lines $P_1P_3$ and $P_2P_4$ intersect at $Q$.

(i) By considering the triangles $P_1QP_4$ and $P_2QP_3$ show that $(P_1Q)(QP_3) = (P_2Q)(QP_4)$.

(ii) Let $\mathbf{p}_i$ be the position vector of the point $P_i$ ($i = 1, 2, 3, 4$). Show that there exist numbers $a_i$, not all zero, such that

$\sum_{i=1}^{4} a_i = 0 \quad \text{and} \quad \sum_{i=1}^{4} a_i\mathbf{p}_i = \mathbf{0} . \qquad \text{(*)}$

(iii) Let $a_i$ ($i = 1, 2, 3, 4$) be any numbers, not all zero, that satisfy ($*$). Show that $a_1 + a_3 \neq 0$ and that the lines $P_1P_3$ and $P_2P_4$ intersect at the point with position vector

$\frac{a_1\mathbf{p}_1 + a_3\mathbf{p}_3}{a_1 + a_3} .$

Deduce that $a_1a_3(P_1P_3)^2 = a_2a_4(P_2P_4)^2$.

Hint

(i) Since $P_1P_2P_3P_4$ is cyclic, $\angle P_1P_4Q = \angle P_1P_2Q$ (angles in the same segment) and $\angle P_4P_1Q = \angle P_3P_2Q$ (angles in the same segment). Thus $\triangle P_1QP_4 \sim \triangle P_2QP_3$ (AA similarity). Therefore $\frac{P_1Q}{P_2Q} = \frac{QP_4}{QP_3}$, giving $(P_1Q)(QP_3) = (P_2Q)(QP_4)$.

(ii) Since $Q$ lies on $P_1P_3$, we can write $\mathbf{q} = \mathbf{p}_1 + \lambda(\mathbf{p}_3 - \mathbf{p}_1) = (1-\lambda)\mathbf{p}_1 + \lambda\mathbf{p}_3$ for some $\lambda$. Since $Q$ also lies on $P_2P_4$, we can write $\mathbf{q} = (1-\mu)\mathbf{p}_2 + \mu\mathbf{p}_4$ for some $\mu$. Equating: $(1-\lambda)\mathbf{p}_1 + \lambda\mathbf{p}_3 = (1-\mu)\mathbf{p}_2 + \mu\mathbf{p}_4$. Setting $a_1 = 1-\lambda$, $a_3 = \lambda$, $a_2 = -(1-\mu)$, $a_4 = -\mu$: then $a_1 + a_2 + a_3 + a_4 = (1-\lambda) - (1-\mu) + \lambda - \mu = 0$ and $a_1\mathbf{p}_1 + a_2\mathbf{p}_2 + a_3\mathbf{p}_3 + a_4\mathbf{p}_4 = \mathbf{0}$.

(iii) Suppose $a_1 + a_3 = 0$. Then $a_1 = -a_3$ and from (*): $a_1(\mathbf{p}_1 - \mathbf{p}_3) = -(a_2\mathbf{p}_2 + a_4\mathbf{p}_4)$. Also $a_2 + a_4 = 0$ so $a_4 = -a_2$, giving $a_1(\mathbf{p}_1 - \mathbf{p}_3) = -a_2(\mathbf{p}_2 - \mathbf{p}_4) = a_2(\mathbf{p}_4 - \mathbf{p}_2)$. Since $P_1P_3$ and $P_2P_4$ intersect at $Q$, the vectors $\mathbf{p}_3 - \mathbf{p}_1$ and $\mathbf{p}_4 - \mathbf{p}_2$ are not parallel, so $a_1 = a_2 = 0$, contradicting not all $a_i$ being zero. Thus $a_1 + a_3 \neq 0$.

From (*): $a_1\mathbf{p}_1 + a_3\mathbf{p}_3 = -(a_2\mathbf{p}_2 + a_4\mathbf{p}_4)$ and $a_1 + a_3 = -(a_2 + a_4)$. The point $\frac{a_1\mathbf{p}_1 + a_3\mathbf{p}_3}{a_1 + a_3} = \mathbf{p}_1 + \frac{a_3(\mathbf{p}_3 - \mathbf{p}_1)}{a_1 + a_3}$ lies on $P_1P_3$. Similarly $\frac{a_2\mathbf{p}_2 + a_4\mathbf{p}_4}{a_2 + a_4} = \mathbf{p}_2 + \frac{a_4(\mathbf{p}_4 - \mathbf{p}_2)}{a_2 + a_4}$ lies on $P_2P_4$. These are equal, so they represent $Q$.

For the final deduction: let $Q$ divide $P_1P_3$ in ratio $\lambda:(1-\lambda)$ and $P_2P_4$ in ratio $\mu:(1-\mu)$. From (i), $(P_1Q)(QP_3) = (P_2Q)(QP_4)$, so $\lambda(1-\lambda)(P_1P_3)^2 = \mu(1-\mu)(P_2P_4)^2$. From the representation, $\frac{a_3}{a_1+a_3} = \lambda$ and $\frac{a_1}{a_1+a_3} = 1-\lambda$, so $a_1a_3 = (a_1+a_3)^2 \lambda(1-\lambda)$. Similarly $a_2a_4 = (a_2+a_4)^2 \mu(1-\mu)$. Since $a_1+a_3 = -(a_2+a_4)$, we get $(a_1+a_3)^2 = (a_2+a_4)^2$, so $a_1a_3(P_1P_3)^2 = a_2a_4(P_2P_4)^2$.

Model Solution

Part (i)

Since $P_1P_2P_3P_4$ is a cyclic quadrilateral, we apply the inscribed angle theorem (angles in the same segment):

• $\angle P_1P_4Q = \angle P_1P_2Q$: both subtend arc $P_1P_4$ on the same side.
• $\angle P_4P_1Q = \angle P_3P_2Q$: both subtend arc $P_3P_4$ on the same side.

(Note that $Q$ lies on segment $P_1P_3$ between $P_1$ and $P_3$, and on segment $P_2P_4$ between $P_2$ and $P_4$, so these equalities hold as stated.)

Therefore $\triangle P_1QP_4 \sim \triangle P_2QP_3$ by AA similarity (two pairs of equal angles).

From the similarity, corresponding sides are proportional:

$\frac{P_1Q}{P_2Q} = \frac{QP_4}{QP_3}$

Cross-multiplying:

$(P_1Q)(QP_3) = (P_2Q)(QP_4) \qquad \blacksquare$

Part (ii)

Since $Q$ lies on line $P_1P_3$, we can write

$\mathbf{q} = (1-\lambda)\mathbf{p}_1 + \lambda\mathbf{p}_3$

for some real number $\lambda$. Since $Q$ also lies on line $P_2P_4$, we can write

$\mathbf{q} = (1-\mu)\mathbf{p}_2 + \mu\mathbf{p}_4$

for some real number $\mu$. Equating these two expressions:

$(1-\lambda)\mathbf{p}_1 + \lambda\mathbf{p}_3 = (1-\mu)\mathbf{p}_2 + \mu\mathbf{p}_4$

Rearranging:

$(1-\lambda)\mathbf{p}_1 - (1-\mu)\mathbf{p}_2 + \lambda\mathbf{p}_3 - \mu\mathbf{p}_4 = \mathbf{0}$

Set $a_1 = 1-\lambda$, $a_2 = -(1-\mu)$, $a_3 = \lambda$, $a_4 = -\mu$. Then:

$a_1 + a_2 + a_3 + a_4 = (1-\lambda) - (1-\mu) + \lambda - \mu = 0$

$a_1\mathbf{p}_1 + a_2\mathbf{p}_2 + a_3\mathbf{p}_3 + a_4\mathbf{p}_4 = \mathbf{0}$

Since $P_1 \neq P_3$ (distinct vertices), at least one of $a_1, a_3$ is non-zero, so the $a_i$ are not all zero. $\qquad \blacksquare$

Part (iii)

Showing $a_1 + a_3 \neq 0$.

Suppose for contradiction that $a_1 + a_3 = 0$, so $a_3 = -a_1$.

From $\sum a_i = 0$: $a_2 + a_4 = -(a_1 + a_3) = 0$, so $a_4 = -a_2$.

From $\sum a_i\mathbf{p}_i = \mathbf{0}$:

$a_1\mathbf{p}_1 + a_2\mathbf{p}_2 - a_1\mathbf{p}_3 - a_2\mathbf{p}_4 = \mathbf{0}$

$a_1(\mathbf{p}_1 - \mathbf{p}_3) + a_2(\mathbf{p}_2 - \mathbf{p}_4) = \mathbf{0}$

$a_1(\mathbf{p}_3 - \mathbf{p}_1) = a_2(\mathbf{p}_4 - \mathbf{p}_2)$

Since $P_1P_2P_3P_4$ is a convex quadrilateral, the diagonals $P_1P_3$ and $P_2P_4$ intersect and are therefore not parallel. Hence $\mathbf{p}_3 - \mathbf{p}_1$ and $\mathbf{p}_4 - \mathbf{p}_2$ are not scalar multiples of each other. The equation above forces $a_1 = 0$ and $a_2 = 0$, giving $a_3 = 0$ and $a_4 = 0$, which contradicts the requirement that not all $a_i$ are zero. Therefore $a_1 + a_3 \neq 0$. $\qquad \blacksquare$

Showing the intersection point.

Since $a_1 + a_3 \neq 0$ and $a_2 + a_4 = -(a_1 + a_3) \neq 0$, we rearrange $\sum a_i\mathbf{p}_i = \mathbf{0}$:

$a_1\mathbf{p}_1 + a_3\mathbf{p}_3 = -(a_2\mathbf{p}_2 + a_4\mathbf{p}_4) \qquad \text{and} \qquad a_1 + a_3 = -(a_2 + a_4)$

Define $\mathbf{q} = \dfrac{a_1\mathbf{p}_1 + a_3\mathbf{p}_3}{a_1 + a_3}$. Rewriting:

$\mathbf{q} = \mathbf{p}_1 + \frac{a_3}{a_1 + a_3}(\mathbf{p}_3 - \mathbf{p}_1)$

This is a point on line $P_1P_3$ (with parameter $\lambda = \frac{a_3}{a_1 + a_3}$).

Similarly, using $a_1 + a_3 = -(a_2 + a_4)$:

$\mathbf{q} = \frac{-(a_2\mathbf{p}_2 + a_4\mathbf{p}_4)}{-(a_2 + a_4)} = \frac{a_2\mathbf{p}_2 + a_4\mathbf{p}_4}{a_2 + a_4} = \mathbf{p}_2 + \frac{a_4}{a_2 + a_4}(\mathbf{p}_4 - \mathbf{p}_2)$

This is a point on line $P_2P_4$ (with parameter $\mu = \frac{a_4}{a_2 + a_4}$).

Since $\mathbf{q}$ lies on both lines $P_1P_3$ and $P_2P_4$, and these lines intersect at the unique point $Q$, we conclude $\mathbf{q} = \vec{OQ}$. $\qquad \blacksquare$

Deducing $a_1a_3(P_1P_3)^2 = a_2a_4(P_2P_4)^2$.

From the above, $Q$ divides $P_1P_3$ in ratio $\lambda:(1-\lambda)$ where $\lambda = \frac{a_3}{a_1+a_3}$, and divides $P_2P_4$ in ratio $\mu:(1-\mu)$ where $\mu = \frac{a_4}{a_2+a_4}$.

This gives:

$P_1Q = (1-\lambda)\,|P_1P_3|, \quad QP_3 = \lambda\,|P_1P_3|$

$P_2Q = (1-\mu)\,|P_2P_4|, \quad QP_4 = \mu\,|P_2P_4|$

Substituting into the result from part (i):

$\lambda(1-\lambda)(P_1P_3)^2 = \mu(1-\mu)(P_2P_4)^2 \qquad (\dagger)$

Now compute $\lambda(1-\lambda)$:

$\lambda(1-\lambda) = \frac{a_3}{a_1+a_3} \cdot \frac{a_1}{a_1+a_3} = \frac{a_1a_3}{(a_1+a_3)^2}$

Similarly:

$\mu(1-\mu) = \frac{a_4}{a_2+a_4} \cdot \frac{a_2}{a_2+a_4} = \frac{a_2a_4}{(a_2+a_4)^2}$

Substituting into $(\dagger)$:

$\frac{a_1a_3}{(a_1+a_3)^2}(P_1P_3)^2 = \frac{a_2a_4}{(a_2+a_4)^2}(P_2P_4)^2$

Since $a_1 + a_3 = -(a_2 + a_4)$, we have $(a_1+a_3)^2 = (a_2+a_4)^2$. Multiplying both sides by this common value:

$a_1a_3(P_1P_3)^2 = a_2a_4(P_2P_4)^2 \qquad \blacksquare$

Examiner Notes

Roughly two fifths of the candidates attempted this with a mean score of just over three marks making it the least well attempted question on the paper. Most could do part (i), which is GCSE material, but frustratingly quite a few stated that the triangles were similar with no justification. Part (ii) was by far the most poorly attempted part with a lot of hand-waving arguments. Part (iii) was done well by virtue of only the best candidates making it past part (i) with 75% of solutions containing good proofs by contradiction for the first result and the last two parts were pretty well done.

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Question 8

Topic: 级数与数论 (Series and Number Theory)  |  Difficulty: Hard  |  Marks: 20

Problem

8 The numbers $f(r)$ satisfy $f(r) > f(r + 1)$ for $r = 1, 2, \dots$. Show that, for any non-negative integer $n$,

$k^n(k - 1) f(k^{n+1}) \leqslant \sum_{r=k^n}^{k^{n+1}-1} f(r) \leqslant k^n(k - 1) f(k^n)$

where $k$ is an integer greater than 1.

(i) By taking $f(r) = 1/r$, show that

$\frac{N + 1}{2} \leqslant \sum_{r=1}^{2^{N+1}-1} \frac{1}{r} \leqslant N + 1 .$

Deduce that the sum $\sum_{r=1}^{\infty} \frac{1}{r}$ does not converge.

(ii) By taking $f(r) = 1/r^3$, show that

$\sum_{r=1}^{\infty} \frac{1}{r^3} \leqslant 1\frac{1}{3} .$

(iii) Let $S(n)$ be the set of positive integers less than $n$ which do not have a 2 in their decimal representation and let $\sigma(n)$ be the sum of the reciprocals of the numbers in $S(n)$, so for example $\sigma(5) = 1 + \frac{1}{3} + \frac{1}{4}$. Show that $S(1000)$ contains $9^3 - 1$ distinct numbers.

Show that $\sigma(n) < 80$ for all $n$.

Hint

Since $f(r) > f(r+1)$, for $k^n \leqslant r \leqslant k^{n+1}-1$ we have $f(k^{n+1}) \leqslant f(r) \leqslant f(k^n)$. The sum has $k^{n+1} - k^n = k^n(k-1)$ terms, so $k^n(k-1)f(k^{n+1}) \leqslant \sum_{r=k^n}^{k^{n+1}-1} f(r) \leqslant k^n(k-1)f(k^n).$

(i) With $f(r) = 1/r$ and $k = 2$: $2^n \cdot 1 \cdot \frac{1}{2^{n+1}} \leqslant \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} \leqslant 2^n \cdot 1 \cdot \frac{1}{2^n}$, i.e. $\frac{1}{2} \leqslant \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} \leqslant 1$. Summing from $n = 0$ to $N$: $\frac{N+1}{2} \leqslant \sum_{r=1}^{2^{N+1}-1} \frac{1}{r} \leqslant N+1$. Since $\lim_{N \to \infty} \frac{N+1}{2} = \infty$, the partial sums $\sum_{r=1}^{M} \frac{1}{r}$ are unbounded, so $\sum_{r=1}^{\infty} \frac{1}{r}$ does not converge.

(ii) With $f(r) = 1/r^3$ and $k = 2$: $2^n \cdot \frac{1}{2^{3(n+1)}} \leqslant \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r^3} \leqslant 2^n \cdot \frac{1}{2^{3n}}$, i.e. $\frac{1}{2^{2n+3}} \leqslant \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r^3} \leqslant \frac{1}{4^n}$. Summing from $n = 0$ to $\infty$: $\sum_{r=1}^{\infty} \frac{1}{r^3} \leqslant \sum_{n=0}^{\infty} \frac{1}{4^n} = \frac{1}{1 - 1/4} = \frac{4}{3} = 1\frac{1}{3}$.

(iii) The numbers in $S(1000)$ have at most 3 digits, none of which is 2. For 1-digit numbers: digits from $\{1,3,4,5,6,7,8,9\}$ gives 8 numbers. For 2-digit: first digit from $\{1,\dots,9\}\setminus\{2\} = 8$ choices, second from $\{0,\dots,9\}\setminus\{2\} = 9$ choices, giving $8 \times 9 = 72$. For 3-digit: $8 \times 9 \times 9 = 648$. Total: $8 + 72 + 648 = 728 = 9^3 - 1$.

For $\sigma(n) < 80$: use the stem with $k = 10$ and $f(r) = 0$ if $r$ contains a 2 in its decimal representation, $f(r) = 1/r$ otherwise. For each block $[10^n, 10^{n+1}-1]$, the numbers without digit 2 contribute at most $10^n \cdot 9 \cdot \frac{1}{10^n} = 9$ (by the stem upper bound, since there are at most $9 \times 10^{n-1} \times 10^n/(10^n)$ valid terms). More precisely, the number of $(n+1)$-digit integers with no digit 2 is $8 \times 9^n$, and by the upper bound each block contributes at most $8 \times 9^n / 10^n$. Summing: $\sigma(n) \leqslant \sum_{m=0}^{\infty} 8 \cdot (9/10)^m = 8 \cdot \frac{1}{1-9/10} = 80$. Since the bound is strict (the first block contributes strictly less than 8), $\sigma(n) < 80$.

Model Solution

Stem result. Since $f$ is strictly decreasing, for every $r$ with $k^n \leqslant r \leqslant k^{n+1}-1$:

$f(k^{n+1}) < f(r) \leqslant f(k^n)$

(The strict lower bound uses $f(r) > f(r+1)$; the upper bound uses $f(k^n) \geqslant f(r)$ since $r \geqslant k^n$.)

The number of terms in the sum is $(k^{n+1}-1) - k^n + 1 = k^{n+1} - k^n = k^n(k-1)$.

Multiplying the inequalities by this count:

$k^n(k-1)\,f(k^{n+1}) < \sum_{r=k^n}^{k^{n+1}-1} f(r) \leqslant k^n(k-1)\,f(k^n)$

In particular:

$k^n(k-1)\,f(k^{n+1}) \leqslant \sum_{r=k^n}^{k^{n+1}-1} f(r) \leqslant k^n(k-1)\,f(k^n) \qquad \blacksquare$

Part (i)

Take $f(r) = 1/r$ (which is strictly decreasing) and $k = 2$. The stem gives:

$2^n \cdot 1 \cdot \frac{1}{2^{n+1}} \leqslant \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} \leqslant 2^n \cdot 1 \cdot \frac{1}{2^n}$

$\frac{1}{2} \leqslant \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} \leqslant 1$

The intervals $[2^n, 2^{n+1}-1]$ for $n = 0, 1, \ldots, N$ partition the integers from $1$ to $2^{N+1}-1$. Summing:

$\sum_{n=0}^{N} \frac{1}{2} \leqslant \sum_{n=0}^{N} \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} \leqslant \sum_{n=0}^{N} 1$

$\frac{N+1}{2} \leqslant \sum_{r=1}^{2^{N+1}-1} \frac{1}{r} \leqslant N+1 \qquad \blacksquare$

Divergence of the harmonic series. Since $\sum_{r=1}^{2^{N+1}-1} \frac{1}{r} \geqslant \frac{N+1}{2}$ and $\frac{N+1}{2} \to \infty$ as $N \to \infty$, the partial sums $\sum_{r=1}^{M} \frac{1}{r}$ are unbounded. Therefore $\sum_{r=1}^{\infty} \frac{1}{r}$ does not converge. $\qquad \blacksquare$

Part (ii)

Take $f(r) = 1/r^3$ (strictly decreasing) and $k = 2$. The stem upper bound gives:

$\sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r^3} \leqslant 2^n(2-1) \cdot \frac{1}{(2^n)^3} = \frac{2^n}{2^{3n}} = \frac{1}{4^n}$

Since the intervals $[2^n, 2^{n+1}-1]$ for $n = 0, 1, 2, \ldots$ partition all positive integers:

$\sum_{r=1}^{\infty} \frac{1}{r^3} = \sum_{n=0}^{\infty} \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r^3} \leqslant \sum_{n=0}^{\infty} \frac{1}{4^n} = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3} = 1\tfrac{1}{3} \qquad \blacksquare$

Part (iii)

Counting $S(1000)$. The elements of $S(1000)$ are positive integers less than $1000$ whose decimal representation contains no digit $2$.

• 1-digit numbers: Digits from $\{1, 3, 4, 5, 6, 7, 8, 9\}$ (8 choices, excluding $0$ and $2$). Count: $8$.
• 2-digit numbers: First digit from $\{1, 3, 4, 5, 6, 7, 8, 9\}$ (8 choices); second digit from $\{0, 1, 3, 4, 5, 6, 7, 8, 9\}$ (9 choices). Count: $8 \times 9 = 72$.
• 3-digit numbers: First digit: 8 choices; second and third digits: 9 choices each. Count: $8 \times 9 \times 9 = 648$.

Total: $8 + 72 + 648 = 728$.

We check: $9^3 - 1 = 729 - 1 = 728$. $\qquad \blacksquare$

Showing $\sigma(n) < 80$ for all $n$.

We bound $\sigma(n)$ by summing over all positive integers without digit $2$ (which includes $S(n)$ as a subset, so $\sigma(n) \leqslant \sigma(\infty)$).

Group the positive integers without digit $2$ by their number of digits $d$:

• For $d = 1$: there are $8$ such numbers, each at least $1$. So the contribution is at most $8$.
• For $d \geqslant 2$: there are $8 \times 9^{d-1}$ such numbers, and each has $d$ digits so is at least $10^{d-1}$. The contribution is at most

$\frac{8 \times 9^{d-1}}{10^{d-1}} = 8\left(\frac{9}{10}\right)^{d-1}$

Summing over all $d \geqslant 1$:

$\sigma(\infty) \leqslant \sum_{d=1}^{\infty} 8\left(\frac{9}{10}\right)^{d-1} = 8 \cdot \frac{1}{1 - \frac{9}{10}} = 8 \times 10 = 80$

This bound is not tight: for $d = 1$, the actual sum $1 + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{9} \approx 2.83$ is strictly less than $8$. Therefore $\sigma(\infty) < 80$.

Since $\sigma(n) \leqslant \sigma(\infty) < 80$ for every $n$:

$\sigma(n) < 80 \qquad \blacksquare$

Examiner Notes

Just fewer than half the candidates attempted this scoring just over a third of the marks. Many managed all but part (iii) easily but few managed that last part, and most did not try it. In part (i), having correctly used the result from the stem, there was frequently not enough care taken in extending this to the full sum. A not infrequent error of logic was that $\sum_{r=1}^{2^{N+1}-1} 1/r < N + 1$ and $\lim_{N \to \infty} N + 1 = \infty$ somehow implies that $\sum_{r=1}^{\infty} 1/r$ does not converge.