STEP2 2003 -- Pure Mathematics

STEP2 2003 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2003  |  Questions: Q1—Q7  |  Total marks per question: 20

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

Overview

Q	Topic	Difficulty	Key Techniques
1	线性方程组（Systems of Linear Equations）	Standard	参数讨论,线性方程组求解,约束优化
2	三角函数与反三角函数（Trigonometric and Inverse Trigonometric Functions）	Standard	辅助角公式,反三角函数恒等式
3	无理数与证明（Irrationality and Proof）	Challenging	反证法,数学归纳法,极限构造
4	解析几何（Analytic Geometry）	Standard	弓形面积公式,反余弦函数,坐标变换
5	向量（Vectors）	Standard	向量参数方程,中点公式,共面条件
6	函数迭代与积分（Function Iteration and Integration）	Challenging	函数迭代,绝对值函数图像,三角函数积分
7	积分与级数（Integration and Series）	Challenging	分部积分,反常积分,级数放缩

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Question 1

Topic: 线性方程组（Systems of Linear Equations）  |  Difficulty: Standard  |  Marks: 20

Problem

1 Consider the equations $ax - y - z = 3 ,$ $2ax - y - 3z = 7 ,$ $3ax - y - 5z = b ,$ where $a$ and $b$ are given constants.

(i) In the case $a = 0$, show that the equations have a solution if and only if $b = 11$.

(ii) In the case $a \neq 0$ and $b = 11$ show that the equations have a solution with $z = \lambda$ for any given number $\lambda$.

(iii) In the case $a = 2$ and $b = 11$ find the solution for which $x^2 + y^2 + z^2$ is least.

(iv) Find a value for $a$ for which there is a solution such that $x > 10^6$ and $y^2 + z^2 < 1$.

Model Solution

Part (i)

When $a = 0$, the equations become:

$-y - z = 3, \qquad -y - 3z = 7, \qquad -y - 5z = b.$

Subtracting the first from the second:

$(-y - 3z) - (-y - z) = 7 - 3 \implies -2z = 4 \implies z = -2.$

Substituting into the first equation: $-y - (-2) = 3 \implies -y + 2 = 3 \implies y = -1$.

Substituting $y = -1$ and $z = -2$ into the third equation:

$-(-1) - 5(-2) = b \implies 1 + 10 = b \implies b = 11.$

So the equations have a solution if and only if $b = 11$. $\quad \blacksquare$

Part (ii)

When $b = 11$, the system is:

$ax - y - z = 3, \qquad 2ax - y - 3z = 7, \qquad 3ax - y - 5z = 11.$

Subtracting the first equation from the second:

$ax - 2z = 4 \qquad \text{(...)}$

Subtracting the second from the third:

$ax - 2z = 4 \qquad \text{(...)}$

So the third equation is redundant (it gives the same condition as the first two combined). From $(*)$: $ax = 4 + 2z$. Since $a \neq 0$:

$x = \frac{4 + 2z}{a}.$

From the first equation: $y = ax - z - 3 = (4 + 2z) - z - 3 = 1 + z$.

Setting $z = \lambda$:

$x = \frac{4 + 2\lambda}{a}, \qquad y = 1 + \lambda, \qquad z = \lambda.$

This is a valid solution for any $\lambda$. $\quad \blacksquare$

Part (iii)

With $a = 2$ and $b = 11$, the solution from part (ii) gives:

$x = \frac{4 + 2\lambda}{2} = 2 + \lambda, \qquad y = 1 + \lambda, \qquad z = \lambda.$

We minimise $x^2 + y^2 + z^2$:

$x^2 + y^2 + z^2 = (2 + \lambda)^2 + (1 + \lambda)^2 + \lambda^2$

$= 4 + 4\lambda + \lambda^2 + 1 + 2\lambda + \lambda^2 + \lambda^2 = 3\lambda^2 + 6\lambda + 5.$

Completing the square:

$3\lambda^2 + 6\lambda + 5 = 3(\lambda^2 + 2\lambda) + 5 = 3(\lambda + 1)^2 - 3 + 5 = 3(\lambda + 1)^2 + 2.$

This is minimised when $\lambda = -1$, giving $x = 1$, $y = 0$, $z = -1$.

The minimum value of $x^2 + y^2 + z^2$ is $2$, and the solution is $(x, y, z) = (1, 0, -1)$.

Part (iv)

From part (ii), with $z = \lambda$, $a \neq 0$ and $b = 11$:

$x = \frac{4 + 2\lambda}{a}, \qquad y = 1 + \lambda, \qquad z = \lambda.$

We need $x > 10^6$ and $y^2 + z^2 < 1$. From $y^2 + z^2 < 1$:

$(1 + \lambda)^2 + \lambda^2 < 1 \implies 2\lambda^2 + 2\lambda + 1 < 1 \implies 2\lambda(\lambda + 1) < 0.$

So $-1 < \lambda < 0$. In particular, $4 + 2\lambda > 2$ (since $\lambda > -1$).

To make $x = \frac{4 + 2\lambda}{a} > 10^6$, we need $a$ to be small and positive (since $4 + 2\lambda > 0$).

For example, choose $\lambda = -\frac{1}{2}$ (which satisfies $-1 < \lambda < 0$). Then $y = \frac{1}{2}$, $z = -\frac{1}{2}$, so $y^2 + z^2 = \frac{1}{2} < 1$. And $x = \frac{3}{a}$, which exceeds $10^6$ when $a < \frac{3}{10^6} = 3 \times 10^{-6}$.

So any $a$ with $0 < a < 3 \times 10^{-6}$ works. For instance, $a = 10^{-6}$ is a valid choice.

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Question 2

Topic: 三角函数与反三角函数（Trigonometric and Inverse Trigonometric Functions）  |  Difficulty: Standard  |  Marks: 20

Problem

2 Write down a value of $\theta$ in the interval $\pi/4 < \theta < \pi/2$ that satisfies the equation $4 \cos \theta + 2\sqrt{3} \sin \theta = 5 .$ Hence, or otherwise, show that $\pi = 3 \arccos(5/\sqrt{28}) + 3 \arctan(\sqrt{3}/2) .$ Show that $\pi = 4 \arcsin(7\sqrt{2}/10) - 4 \arctan(3/4) .$

Model Solution

Finding θ

We use the auxiliary angle form. Write

$4\cos\theta + 2\sqrt{3}\sin\theta = R\cos(\theta - \alpha)$

where $R\cos\alpha = 4$ and $R\sin\alpha = 2\sqrt{3}$. Then

$R = \sqrt{16 + 12} = \sqrt{28} = 2\sqrt{7}, \qquad \tan\alpha = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}.$

So $\alpha = \arctan(\sqrt{3}/2)$. The equation becomes

$\sqrt{28}\cos(\theta - \alpha) = 5 \implies \cos(\theta - \alpha) = \frac{5}{\sqrt{28}}.$

Hence $\theta - \alpha = \pm\arccos(5/\sqrt{28})$, giving $\theta = \alpha \pm \arccos(5/\sqrt{28})$.

Taking the positive sign: $\theta = \arctan(\sqrt{3}/2) + \arccos(5/\sqrt{28})$.

Numerically: $\alpha \approx 40.9°$ and $\arccos(5/\sqrt{28}) \approx 19.1°$, so $\theta \approx 60° = \pi/3$.

We verify that $\theta = \pi/3$ lies in $(\pi/4, \pi/2)$ and satisfies the equation:

$4\cos\frac{\pi}{3} + 2\sqrt{3}\sin\frac{\pi}{3} = 4\cdot\frac{1}{2} + 2\sqrt{3}\cdot\frac{\sqrt{3}}{2} = 2 + 3 = 5. \quad \checkmark$

So $\theta = \pi/3$.

Show that $\pi = 3\arccos(5/\sqrt{28}) + 3\arctan(\sqrt{3}/2)$

From the auxiliary angle form with $\theta = \pi/3$:

$\sqrt{28}\cos\!\left(\frac{\pi}{3} - \alpha\right) = 5 \implies \cos\!\left(\frac{\pi}{3} - \alpha\right) = \frac{5}{\sqrt{28}}.$

Since $\pi/3 - \alpha \approx 19.1° \in (0, \pi/2)$, we have

$\frac{\pi}{3} - \alpha = \arccos\!\left(\frac{5}{\sqrt{28}}\right).$

Rearranging:

$\frac{\pi}{3} = \arccos\!\left(\frac{5}{\sqrt{28}}\right) + \alpha = \arccos\!\left(\frac{5}{\sqrt{28}}\right) + \arctan\!\left(\frac{\sqrt{3}}{2}\right).$

Multiplying both sides by 3:

$\pi = 3\arccos\!\left(\frac{5}{\sqrt{28}}\right) + 3\arctan\!\left(\frac{\sqrt{3}}{2}\right). \qquad \blacksquare$

Show that $\pi = 4\arcsin(7\sqrt{2}/10) - 4\arctan(3/4)$

Let $\phi = \arcsin(7\sqrt{2}/10)$ and $\beta = \arctan(3/4)$. We show that $\phi - \beta = \pi/4$.

From $\sin\phi = 7\sqrt{2}/10$, since $\phi$ is acute:

$\cos\phi = \sqrt{1 - \frac{98}{100}} = \sqrt{\frac{1}{50}} = \frac{\sqrt{2}}{10}.$

From $\tan\beta = 3/4$ (a 3-4-5 right triangle):

$\sin\beta = \frac{3}{5}, \qquad \cos\beta = \frac{4}{5}.$

Computing $\sin(\phi - \beta)$:

$\sin(\phi - \beta) = \sin\phi\cos\beta - \cos\phi\sin\beta = \frac{7\sqrt{2}}{10}\cdot\frac{4}{5} - \frac{\sqrt{2}}{10}\cdot\frac{3}{5} = \frac{28\sqrt{2} - 3\sqrt{2}}{50} = \frac{25\sqrt{2}}{50} = \frac{\sqrt{2}}{2}.$

Since $0 < \beta < \phi < \pi/2$, we have $0 < \phi - \beta < \pi/2$. The unique angle in $(0, \pi/2)$ with sine $\sqrt{2}/2$ is $\pi/4$. Therefore

$\phi - \beta = \frac{\pi}{4}.$

Multiplying by 4:

$4\phi - 4\beta = \pi \implies \pi = 4\arcsin\!\left(\frac{7\sqrt{2}}{10}\right) - 4\arctan\!\left(\frac{3}{4}\right). \qquad \blacksquare$

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Question 3

Topic: 无理数与证明（Irrationality and Proof）  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Prove that the cube root of any irrational number is an irrational number.

Let $u_n = 5^{1/(3^n)}$. Given that $\sqrt[3]{5}$ is an irrational number, prove by induction that $u_n$ is an irrational number for every positive integer $n$.

Hence, or otherwise, give an example of an infinite sequence of irrational numbers which converges to a given integer $m$.

[An irrational number is a number that cannot be expressed as the ratio of two integers.]

Model Solution

Part 1: Cube root of irrational is irrational

We prove the contrapositive: if $\sqrt[3]{x}$ is rational, then $x$ is rational.

Suppose $\sqrt[3]{x}$ is rational. Then there exist integers $p, q$ with $q \neq 0$ such that

$\sqrt[3]{x} = \frac{p}{q} .$

Cubing both sides:

$x = \frac{p^3}{q^3} .$

Since $p$ and $q$ are integers, $p^3$ and $q^3$ are integers, and $q^3 \neq 0$. Therefore $x$ is rational.

By contrapositive, if $x$ is irrational, then $\sqrt[3]{x}$ is irrational. $\blacksquare$

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Part 2: Induction that $u_n = 5^{1/3^n}$ is irrational

Base case ($n = 1$): $u_1 = 5^{1/3} = \sqrt[3]{5}$, which is given to be irrational.

Inductive step: Suppose $u_k = 5^{1/3^k}$ is irrational for some positive integer $k$. We show $u_{k+1}$ is irrational.

Observe that

$u_{k+1}^3 = \left(5^{1/3^{k+1}}\right)^3 = 5^{3/3^{k+1}} = 5^{1/3^k} = u_k .$

So $u_{k+1} = \sqrt[3]{u_k}$. By the induction hypothesis, $u_k$ is irrational. By Part 1, the cube root of an irrational number is irrational. Therefore $u_{k+1}$ is irrational.

By induction, $u_n$ is irrational for every positive integer $n$. $\blacksquare$

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Part 3: An infinite sequence of irrationals converging to integer $m$

Consider the sequence

$a_n = (m - 1) + 5^{1/3^n} \quad \text{for } n = 1, 2, 3, \ldots$

Each $a_n$ is irrational: if $a_n$ were rational, then $5^{1/3^n} = a_n - (m-1)$ would be rational (since the difference of two rationals is rational), contradicting Part 2.

As $n \to \infty$, $3^n \to \infty$, so $\frac{1}{3^n} \to 0$ and $5^{1/3^n} \to 5^0 = 1$. Therefore

$a_n \to (m - 1) + 1 = m .$

Thus $\{a_n\}$ is an infinite sequence of irrational numbers converging to $m$. $\blacksquare$

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Question 4

Topic: 解析几何（Analytic Geometry）  |  Difficulty: Standard  |  Marks: 20

Problem

4 The line $y = d$, where $d > 0$, intersects the circle $x^2 + y^2 = R^2$ at $G$ and $H$. Show that the area of the minor segment $GH$ is equal to

$R^2 \arccos \left( \frac{d}{R} \right) - d \sqrt{R^2 - d^2} . \qquad \text{(*)}$

In the following cases, the given line intersects the given circle. Determine how, in each case, the expression ($*$) should be modified to give the area of the minor segment.

(i) Line: $y = c$; circle: $(x - a)^2 + (y - b)^2 = R^2$.

(ii) Line: $y = mx + c$; circle: $x^2 + y^2 = R^2$.

(iii) Line: $y = mx + c$; circle: $(x - a)^2 + (y - b)^2 = R^2$.

Model Solution

Main derivation

The line $y = d$ meets $x^2 + y^2 = R^2$ at points $G$ and $H$ where $x^2 = R^2 - d^2$, so $G = (-\sqrt{R^2 - d^2},\; d)$ and $H = (\sqrt{R^2 - d^2},\; d)$.

The minor segment is the cap above $y = d$. Its area equals the area of the circular sector subtended by arc $GH$ minus the area of triangle $OGH$ (where $O$ is the origin).

Sector angle: The angle $\theta$ that $OH$ makes with the positive $y$-axis satisfies

$\cos \theta = \frac{d}{R}, \qquad \text{so} \qquad \theta = \arccos\left(\frac{d}{R}\right).$

The full angle subtended by arc $GH$ at the centre is $2\theta = 2\arccos(d/R)$.

Sector area:

$A_{\text{sector}} = \frac{1}{2} R^2 \cdot 2\theta = R^2 \arccos\left(\frac{d}{R}\right).$

Triangle area: Triangle $OGH$ has base $GH = 2\sqrt{R^2 - d^2}$ and height $d$ (the perpendicular distance from $O$ to line $GH$):

$A_{\triangle} = \frac{1}{2} \cdot 2\sqrt{R^2 - d^2} \cdot d = d\sqrt{R^2 - d^2} .$

Segment area:

$A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = R^2 \arccos\left(\frac{d}{R}\right) - d\sqrt{R^2 - d^2} \qquad \text{as required.} \quad \blacksquare$

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Key observation for parts (i)—(iii)

Formula (*) depends only on two geometric quantities:

• $R$: the radius of the circle;
• $d$: the perpendicular distance from the centre of the circle to the chord (line).

In each case below, we compute the perpendicular distance $\delta$ from the circle’s centre to the given line and substitute $d \to \delta$ in (*).

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Part (i)

Line: $y = c$; circle: $(x - a)^2 + (y - b)^2 = R^2$.

The circle has centre $(a, b)$ and radius $R$. The line $y = c$ is horizontal, so the perpendicular distance from the centre to the line is

$\delta = |b - c| .$

Substituting $d \to |b - c|$ in (*):

$\boxed{R^2 \arccos\left(\frac{|b - c|}{R}\right) - |b - c|\sqrt{R^2 - (b - c)^2}} .$

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Part (ii)

Line: $y = mx + c$, i.e.\ $mx - y + c = 0$; circle: $x^2 + y^2 = R^2$ (centre at origin).

The perpendicular distance from the origin to the line $mx - y + c = 0$ is

$\delta = \frac{|m \cdot 0 - 0 + c|}{\sqrt{m^2 + 1}} = \frac{|c|}{\sqrt{m^2 + 1}} .$

Substituting $d \to \dfrac{|c|}{\sqrt{m^2 + 1}}$ in (*):

$R^2 \arccos\left(\frac{|c|}{R\sqrt{m^2 + 1}}\right) - \frac{|c|}{\sqrt{m^2 + 1}} \sqrt{R^2 - \frac{c^2}{m^2 + 1}} .$

Simplify the square root:

$\sqrt{R^2 - \frac{c^2}{m^2 + 1}} = \sqrt{\frac{R^2(m^2 + 1) - c^2}{m^2 + 1}} = \frac{\sqrt{R^2(m^2 + 1) - c^2}}{\sqrt{m^2 + 1}} .$

So the second term becomes

$\frac{|c|}{\sqrt{m^2 + 1}} \cdot \frac{\sqrt{R^2(m^2 + 1) - c^2}}{\sqrt{m^2 + 1}} = \frac{|c|\sqrt{R^2(m^2 + 1) - c^2}}{m^2 + 1} .$

The modified formula is:

$\boxed{R^2 \arccos\left(\frac{|c|}{R\sqrt{m^2 + 1}}\right) - \frac{|c|\sqrt{R^2(m^2 + 1) - c^2}}{m^2 + 1}} .$

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Part (iii)

Line: $y = mx + c$, i.e.\ $mx - y + c = 0$; circle: $(x - a)^2 + (y - b)^2 = R^2$ (centre $(a, b)$).

The perpendicular distance from $(a, b)$ to the line $mx - y + c = 0$ is

$\delta = \frac{|ma - b + c|}{\sqrt{m^2 + 1}} .$

Substituting $d \to \dfrac{|ma - b + c|}{\sqrt{m^2 + 1}}$ in (*):

$R^2 \arccos\left(\frac{|ma - b + c|}{R\sqrt{m^2 + 1}}\right) - \frac{|ma - b + c|}{\sqrt{m^2 + 1}} \sqrt{R^2 - \frac{(ma - b + c)^2}{m^2 + 1}} .$

Simplify the square root as before:

$\sqrt{R^2 - \frac{(ma - b + c)^2}{m^2 + 1}} = \frac{\sqrt{R^2(m^2 + 1) - (ma - b + c)^2}}{\sqrt{m^2 + 1}} .$

The modified formula is:

$\boxed{R^2 \arccos\left(\frac{|ma - b + c|}{R\sqrt{m^2 + 1}}\right) - \frac{|ma - b + c|\sqrt{R^2(m^2 + 1) - (ma - b + c)^2}}{m^2 + 1}} .$

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Summary: In every case, the modification is to replace $d$ in formula (*) with the perpendicular distance $\delta$ from the centre of the circle to the line:

Case	$\delta$
Original	$d$
(i)	$\lvert b - c \rvert$
(ii)	$\dfrac{\lvert c \rvert}{\sqrt{m^2 + 1}}$
(iii)	$\dfrac{\lvert ma - b + c \rvert}{\sqrt{m^2 + 1}}$

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Question 5

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

Problem

5 The position vectors of the points $A$, $B$ and $P$ with respect to an origin $O$ are $a\mathbf{i}$, $b\mathbf{j}$ and $l\mathbf{i} + m\mathbf{j} + n\mathbf{k}$, respectively, where $a$, $b$, and $n$ are all non-zero. The points $E$, $F$, $G$ and $H$ are the midpoints of $OA$, $BP$, $OB$ and $AP$, respectively. Show that the lines $EF$ and $GH$ intersect.

Let $D$ be the point with position vector $d\mathbf{k}$, where $d$ is non-zero, and let $S$ be the point of intersection of $EF$ and $GH$. The point $T$ is such that the mid-point of $DT$ is $S$. Find the position vector of $T$ and hence find $d$ in terms of $n$ if $T$ lies in the plane $OAB$.

Model Solution

Position vectors of the midpoints

$E = \frac{a}{2}\,\mathbf{i}, \quad F = \frac{l}{2}\,\mathbf{i} + \frac{b + m}{2}\,\mathbf{j} + \frac{n}{2}\,\mathbf{k}, \quad G = \frac{b}{2}\,\mathbf{j}, \quad H = \frac{a + l}{2}\,\mathbf{i} + \frac{m}{2}\,\mathbf{j} + \frac{n}{2}\,\mathbf{k}.$

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Showing that $EF$ and $GH$ intersect

We first observe that

$E + F = \frac{a + l}{2}\,\mathbf{i} + \frac{b + m}{2}\,\mathbf{j} + \frac{n}{2}\,\mathbf{k} = G + H.$

Therefore the midpoint of segment $EF$ equals the midpoint of segment $GH$:

$M = \frac{E + F}{2} = \frac{G + H}{2} = \frac{a + l}{4}\,\mathbf{i} + \frac{b + m}{4}\,\mathbf{j} + \frac{n}{4}\,\mathbf{k}.$

Since $M$ is the midpoint of segment $EF$, it lies on line $EF$ (at parameter $t = 1/2$ in $\mathbf{r} = E + t(F - E)$). Similarly, $M$ lies on line $GH$ (at parameter $s = 1/2$ in $\mathbf{r} = G + s(H - G)$). Hence both lines pass through $M$.

To confirm the lines are distinct (not the same line), we check that the direction vectors $\overrightarrow{EF}$ and $\overrightarrow{GH}$ are not parallel:

$\overrightarrow{EF} = \frac{l}{2}\,\mathbf{i} + \frac{b + m - a}{2}\,\mathbf{j} + \frac{n}{2}\,\mathbf{k}, \qquad \overrightarrow{GH} = \frac{a + l}{2}\,\mathbf{i} + \frac{m - b}{2}\,\mathbf{j} + \frac{n}{2}\,\mathbf{k}.$

If $\overrightarrow{EF} = \lambda\,\overrightarrow{GH}$, comparing $\mathbf{k}$-components gives $\lambda = 1$ (since $n \neq 0$). Then $\mathbf{i}$-components give $l = a + l$, i.e.\ $a = 0$, contradicting $a \neq 0$. So the lines are distinct.

Therefore $EF$ and $GH$ intersect at the unique point

$S = \frac{a + l}{4}\,\mathbf{i} + \frac{b + m}{4}\,\mathbf{j} + \frac{n}{4}\,\mathbf{k}. \qquad \blacksquare$

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Finding $T$ and $d$

Since $S$ is the midpoint of $DT$ with $D = d\mathbf{k}$:

$S = \frac{D + T}{2} \implies T = 2S - D = \frac{a + l}{2}\,\mathbf{i} + \frac{b + m}{2}\,\mathbf{j} + \frac{n}{2}\,\mathbf{k} - d\,\mathbf{k},$

$T = \frac{a + l}{2}\,\mathbf{i} + \frac{b + m}{2}\,\mathbf{j} + \left(\frac{n}{2} - d\right)\mathbf{k}.$

The plane $OAB$ is the $xy$-plane (since $O$, $A$, $B$ all have zero $\mathbf{k}$-component). For $T$ to lie in this plane, its $\mathbf{k}$-component must be zero:

$\frac{n}{2} - d = 0 \implies d = \frac{n}{2}.$

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Question 6

Topic: 函数迭代与积分（Function Iteration and Integration）  |  Difficulty: Challenging  |  Marks: 20

Problem

6 The function $f$ is defined by

$f(x) = |x - 1| ,$

where the domain is $\mathbf{R}$, the set of all real numbers. The function $g_n = f^n$, with domain $\mathbf{R}$, so for example $g_3(x) = f(f(f(x)))$. In separate diagrams, sketch graphs of $g_1$, $g_2$, $g_3$ and $g_4$.

The function $h$ is defined by

$h(x) = \left| \sin \frac{\pi x}{2} \right| ,$

where the domain is $\mathbf{R}$. Show that if $n$ is even,

$\int_0^n (h(x) - g_n(x)) \ dx = \frac{2n}{\pi} - \frac{n}{2} .$

Model Solution

Sketches of $g_1, g_2, g_3, g_4$

Each $g_n$ is piecewise linear with slopes $\pm 1$ on every unit interval. The key pattern is:

$g_n(k) = \begin{cases} 0 & \text{if } n + k \text{ is even}, \\ 1 & \text{if } n + k \text{ is odd}, \end{cases} \qquad \text{for every integer } k \geq 0.$

This means each $g_n$ is a zigzag of unit triangles on $[0, n]$.

Graph of $g_1(x) = |x - 1|$: V-shape with vertex at $(1, 0)$. Passes through $(0, 1)$; slope $-1$ for $x < 1$, slope $+1$ for $x > 1$.

Graph of $g_2(x)$: Zigzag on $[0, 2]$. Passes through $(0, 0)$, $(1, 1)$, $(2, 0)$. A tent (inverted V) shape: rises from $(0, 0)$ to $(1, 1)$, then falls to $(2, 0)$. For $x > 2$, continues with slope $+1$.

Graph of $g_3(x)$: Zigzag on $[0, 3]$. Passes through $(0, 1)$, $(1, 0)$, $(2, 1)$, $(3, 0)$. Alternating V-shapes: falls from $(0, 1)$ to $(1, 0)$, rises to $(2, 1)$, falls to $(3, 0)$. For $x > 3$, continues with slope $+1$.

Graph of $g_4(x)$: Zigzag on $[0, 4]$. Passes through $(0, 0)$, $(1, 1)$, $(2, 0)$, $(3, 1)$, $(4, 0)$. Repeating tents: rises from $(0, 0)$ to $(1, 1)$, falls to $(2, 0)$, rises to $(3, 1)$, falls to $(4, 0)$. For $x > 4$, continues with slope $+1$.

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Proof of the integral identity

We establish two results and then combine them.

Step 1: $\int_0^n g_n(x) \, dx = \frac{n}{2}$ for even $n$.

We prove by induction on $n$ that $g_n$ satisfies:

(a) $g_n$ is piecewise linear with slope $\pm 1$ on every interval $[k, k+1]$, $k \in \{0, 1, \ldots, n-1\}$;

(b) $g_n(k) = 0$ if $n + k$ is even, and $g_n(k) = 1$ if $n + k$ is odd, for each integer $k \in \{0, 1, \ldots, n\}$.

Base case $n = 1$: $g_1(x) = |x - 1|$. At integers: $g_1(0) = 1$ (since $1 + 0 = 1$ is odd), $g_1(1) = 0$ (since $1 + 1 = 2$ is even). Slope is $-1$ on $[0, 1]$, $+1$ on $[1, \infty)$. Both (a) and (b) hold.

Inductive step: Assume (a) and (b) hold for $g_n$. We show they hold for $g_{n+1} = f \circ g_n$.

For (a): On each interval $[k, k+1]$, $g_n$ is linear with slope $\pm 1$, so $g_n$ takes values in $[0, 1] \subseteq [0, \infty)$. On this range, $f(y) = |y - 1| = 1 - y$, which is linear in $y$. The composition of two linear functions is linear. Since $g_n$ has slope $\pm 1$ and $f$ has slope $-1$ on $[0, 1]$, the composition $g_{n+1}$ has slope $\pm 1$.

For (b): At $k = 0$: $g_n(0)$ has parity $n \bmod 2$ by (b). So $g_n(0) = 0$ if $n$ is even, $g_n(0) = 1$ if $n$ is odd. Then $g_{n+1}(0) = |g_n(0) - 1|$, giving $g_{n+1}(0) = 1$ if $n$ is even, $g_{n+1}(0) = 0$ if $n$ is odd. In both cases, $g_{n+1}(0)$ has parity $(n+1) \bmod 2$.

At $k = 1$: By (b), $g_n(1) = 0$ when $n+1$ is even (i.e.\ $n$ odd), and $g_n(1) = 1$ when $n+1$ is odd (i.e.\ $n$ even). In fact, from the base case: $g_1(1) = 0$, and each subsequent application of $f$ flips between 0 and 1. So $g_n(1) = 0$ when $n$ is odd, $g_n(1) = 1$ when $n$ is even. Then $g_{n+1}(1) = |g_n(1) - 1|$: if $n$ is odd, $g_{n+1}(1) = 1$; if $n$ is even, $g_{n+1}(1) = 0$. In both cases, $g_{n+1}(1)$ has parity $(n + 2) \bmod 2 = n \bmod 2$, which matches $(n+1)+1 \bmod 2$. ✓

At $k \geq 2$: By (b), $g_n(k) \in \{0, 1\}$ with parity $(n+k) \bmod 2$. Then $g_{n+1}(k) = |g_n(k) - 1| \in \{0, 1\}$, which flips the value. If $g_n(k) = 0$ then $g_{n+1}(k) = 1$; if $g_n(k) = 1$ then $g_{n+1}(k) = 0$. So $g_{n+1}(k)$ has parity $1 - ((n+k) \bmod 2) = (n+k+1) \bmod 2 = ((n+1)+k) \bmod 2$. ✓

This completes the induction.

Computing the integral: On each interval $[k, k+1]$ with $0 \leq k \leq n - 1$, $g_n$ is linear connecting the points $(k, g_n(k))$ and $(k+1, g_n(k+1))$. By property (b), these endpoints are $\{0, 1\}$ in some order, so $g_n(k) + g_n(k+1) = 1$. The area under a linear function on a unit interval is the average of the endpoint values:

$\int_k^{k+1} g_n(x) \, dx = \frac{g_n(k) + g_n(k+1)}{2} \cdot 1 = \frac{1}{2}.$

Summing over all $n$ intervals:

$\int_0^n g_n(x) \, dx = \sum_{k=0}^{n-1} \frac{1}{2} = \frac{n}{2}. \qquad (\star)$

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Step 2: $\int_0^n h(x) \, dx = \frac{2n}{\pi}$ for even $n$.

Since $h(x) = \left|\sin \frac{\pi x}{2}\right|$ has period 2, and $n$ is even:

$\int_0^n h(x) \, dx = \frac{n}{2} \int_0^2 h(x) \, dx.$

On $[0, 2]$: $\sin \frac{\pi x}{2} \geq 0$ for $x \in [0, 1]$ and $\sin \frac{\pi x}{2} \leq 0$ for $x \in [1, 2]$. So

$\int_0^2 h(x) \, dx = \int_0^1 \sin \frac{\pi x}{2} \, dx + \int_1^2 \left(-\sin \frac{\pi x}{2}\right) dx.$

For the first integral:

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

For the second integral:

$\int_1^2 \left(-\sin \frac{\pi x}{2}\right) dx = \left[\frac{2}{\pi} \cos \frac{\pi x}{2}\right]_1^2 = \frac{2}{\pi}\cos\pi - \frac{2}{\pi}\cos\frac{\pi}{2} = -\frac{2}{\pi} - 0 = \frac{2}{\pi}.$

Wait: $\frac{2}{\pi}\cos\pi = \frac{2}{\pi}(-1) = -\frac{2}{\pi}$. So the second integral is $-\frac{2}{\pi} - 0 = -\frac{2}{\pi}$? Let me recompute.

$\int_1^2 \left(-\sin \frac{\pi x}{2}\right) dx = \left[\frac{2}{\pi} \cos \frac{\pi x}{2}\right]_1^2 = \frac{2}{\pi}\cos\pi - \frac{2}{\pi}\cos\frac{\pi}{2} = -\frac{2}{\pi} - 0 = -\frac{2}{\pi}.$

But this should be positive since the integrand is non-negative on $[1, 2]$. Let me recheck the antiderivative.

$\frac{d}{dx}\left[\frac{2}{\pi}\cos\frac{\pi x}{2}\right] = \frac{2}{\pi} \cdot \left(-\sin\frac{\pi x}{2}\right) \cdot \frac{\pi}{2} = -\sin\frac{\pi x}{2}.$

So $\int(-\sin\frac{\pi x}{2})\,dx = \frac{2}{\pi}\cos\frac{\pi x}{2} + C$. ✓

Evaluating:

$\left[\frac{2}{\pi}\cos\frac{\pi x}{2}\right]_1^2 = \frac{2}{\pi}\cos\pi - \frac{2}{\pi}\cos\frac{\pi}{2} = \frac{2}{\pi}(-1) - \frac{2}{\pi}(0) = -\frac{2}{\pi}.$

This is negative, which contradicts the integrand being non-negative. Let me recheck whether $-\sin\frac{\pi x}{2} \geq 0$ on $[1, 2]$.

At $x = 1.5$: $\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2} > 0$, so $-\sin\frac{3\pi}{4} < 0$.

The issue is: $\sin\frac{\pi x}{2}$ on $[1, 2]$: at $x = 1$, $\sin\frac{\pi}{2} = 1 > 0$; at $x = 2$, $\sin\pi = 0$. So $\sin\frac{\pi x}{2} \geq 0$ on $[1, 2]$ as well! Therefore $h(x) = \sin\frac{\pi x}{2}$ on all of $[0, 2]$.

So:

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

Therefore:

$\int_0^n h(x) \, dx = \frac{n}{2} \cdot \frac{4}{\pi} = \frac{2n}{\pi}. \qquad (\star\star)$

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Combining the results

From $(\star)$ and $(\star\star)$:

$\int_0^n \bigl(h(x) - g_n(x)\bigr) \, dx = \frac{2n}{\pi} - \frac{n}{2}. \qquad \blacksquare$

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Question 7

Topic: 积分与级数（Integration and Series）  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Show that, if $n > 0$, then $\int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} dx = \frac{2}{n^2 e} .$

You may assume that $\frac{\ln x}{x} \to 0$ as $x \to \infty$.

Explain why, if $1 < a < b$, then $\int_{b}^{\infty} \frac{\ln x}{x^{n+1}} dx < \int_{a}^{\infty} \frac{\ln x}{x^{n+1}} dx .$

Deduce that $\sum_{n=1}^{N} \frac{1}{n^2} < \frac{e}{2} \int_{e^{1/N}}^{\infty} \frac{1 - x^{-N}}{x^2 - x} \ln x dx ,$

Model Solution

Part (i): Show that $\int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx = \frac{2}{n^2 e}$

We use integration by parts with

$u = \ln x, \qquad dv = x^{-(n+1)} \, dx ,$

so that

$du = \frac{1}{x} \, dx, \qquad v = -\frac{1}{n x^n} .$

Applying integration by parts:

$\int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx = \left[ -\frac{\ln x}{n x^n} \right]_{e^{1/n}}^{\infty} + \frac{1}{n} \int_{e^{1/n}}^{\infty} \frac{1}{x^{n+1}} \, dx \qquad \text{(*)}$

Evaluating the boundary term. As $x \to \infty$:

$\frac{\ln x}{x^n} = \frac{\ln x}{x} \cdot \frac{1}{x^{n-1}} .$

Since $n > 0$, for $n \geq 1$ the factor $\frac{1}{x^{n-1}}$ is bounded (by $1$ when $n = 1$, or tends to $0$ when $n > 1$), and we are given that $\frac{\ln x}{x} \to 0$. So $\frac{\ln x}{x^n} \to 0$ as $x \to \infty$.

At the lower limit $x = e^{1/n}$:

$-\frac{\ln(e^{1/n})}{n(e^{1/n})^n} = -\frac{1/n}{n \cdot e} = -\frac{1}{n^2 e} .$

Therefore:

$\left[ -\frac{\ln x}{n x^n} \right]_{e^{1/n}}^{\infty} = 0 - \left( -\frac{1}{n^2 e} \right) = \frac{1}{n^2 e} .$

Evaluating the remaining integral.

$\frac{1}{n} \int_{e^{1/n}}^{\infty} x^{-(n+1)} \, dx = \frac{1}{n} \left[ \frac{x^{-n}}{-n} \right]_{e^{1/n}}^{\infty} = \frac{1}{n} \left( 0 - \frac{(e^{1/n})^{-n}}{-n} \right) = \frac{1}{n} \cdot \frac{e^{-1}}{n} = \frac{1}{n^2 e} .$

Substituting both results into $(*)$:

$\int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx = \frac{1}{n^2 e} + \frac{1}{n^2 e} = \frac{2}{n^2 e} \qquad \blacksquare$

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Part (ii): Explain why $\int_{b}^{\infty} \frac{\ln x}{x^{n+1}} dx < \int_{a}^{\infty} \frac{\ln x}{x^{n+1}} dx$ when $1 < a < b$

For $x > 1$, we have $\ln x > 0$ and $x^{n+1} > 0$, so the integrand $\frac{\ln x}{x^{n+1}} > 0$.

Since $a < b$, we can split the integral from $a$:

$\int_{a}^{\infty} \frac{\ln x}{x^{n+1}} \, dx = \int_{a}^{b} \frac{\ln x}{x^{n+1}} \, dx + \int_{b}^{\infty} \frac{\ln x}{x^{n+1}} \, dx .$

The first integral on the right is strictly positive (integral of a strictly positive function over $[a, b]$ with $a < b$). Therefore:

$\int_{b}^{\infty} \frac{\ln x}{x^{n+1}} \, dx < \int_{a}^{\infty} \frac{\ln x}{x^{n+1}} \, dx \qquad \blacksquare$

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Part (iii): Deduce the series inequality

From part (i), each term satisfies

$\frac{1}{n^2} = \frac{e}{2} \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx .$

For each $n$ with $1 \leq n \leq N$, since $n \leq N$ we have $\frac{1}{N} \leq \frac{1}{n}$, hence $e^{1/N} \leq e^{1/n}$. Applying part (ii) with $a = e^{1/n}$ and $b = e^{1/N}$:

$\int_{e^{1/N}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx \leq \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx = \frac{2}{n^2 e} \qquad \text{(**)}$

Note: the inequality is strict whenever $n < N$ (i.e. $e^{1/N} < e^{1/n}$), and becomes an equality only when $n = N$.

Multiplying $(**)$ by $\frac{e}{2}$:

$\frac{e}{2} \int_{e^{1/N}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx \leq \frac{1}{n^2} .$

Summing over $n = 1, 2, \ldots, N$:

$\frac{e}{2} \sum_{n=1}^{N} \int_{e^{1/N}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx \leq \sum_{n=1}^{N} \frac{1}{n^2} .$

Since all integrals share the same limits, we combine them:

$\frac{e}{2} \int_{e^{1/N}}^{\infty} \left( \sum_{n=1}^{N} \frac{1}{x^{n+1}} \right) \ln x \, dx \leq \sum_{n=1}^{N} \frac{1}{n^2} .$

The inner sum is a geometric series with first term $\frac{1}{x^2}$ and common ratio $\frac{1}{x}$:

$\sum_{n=1}^{N} \frac{1}{x^{n+1}} = \frac{1}{x^2} + \frac{1}{x^3} + \cdots + \frac{1}{x^{N+1}} = \frac{1}{x^2} \cdot \frac{1 - (1/x)^N}{1 - 1/x} = \frac{1 - x^{-N}}{x^2(1 - x^{-1})} = \frac{1 - x^{-N}}{x^2 - x} .$

Substituting:

$\frac{e}{2} \int_{e^{1/N}}^{\infty} \frac{(1 - x^{-N}) \ln x}{x^2 - x} \, dx \leq \sum_{n=1}^{N} \frac{1}{n^2} .$

Strict inequality for $N \geq 2$. The bound $(**)$ is an equality only when $n = N$. For every $n$ with $1 \leq n < N$, the bound is strict. Since $N \geq 2$ ensures that $n = 1$ satisfies $n < N$, at least one strict inequality appears in the sum. Therefore:

$\sum_{n=1}^{N} \frac{1}{n^2} < \frac{e}{2} \int_{e^{1/N}}^{\infty} \frac{1 - x^{-N}}{x^2 - x} \ln x \, dx \qquad (N \geq 2) \qquad \blacksquare$