STEP2 2015 -- Pure Mathematics

STEP2 2015 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2015  |  Questions: Q1—Q8  |  Total marks per question: 20

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

Overview

Q	Topic	Difficulty	Key Techniques
1	微积分与级数 (Calculus and Series)	Standard	求导判断单调性, 利用对数差分的裂项求和, 泰勒展开不等式
2	三角学与几何 (Trigonometry and Geometry)	Standard	正弦定理, 和角/倍角公式, 三角形中的坐标与距离计算
3	组合数学与数列 (Combinatorics and Sequences)	Challenging	三角不等式, 分类计数, 数学归纳法, 利用递推关系
4	函数与图像 (Functions and Graphs)	Standard	反函数与复合函数图像变换, 驻点分析, 渐近线判断, 周期性平移
5	三角恒等式与级数 (Trigonometric Identities and Series)	Challenging	数学归纳法, tan 加法公式, arctan 的值域讨论, 反三角函数级数
6	积分 (Integration)	Challenging	三角恒等式证明,定积分换元法,利用对称性简化积分,分部积分思想
7	解析几何 (Analytic Geometry)	Challenging	圆的方程,勾股定理,距离公式,不等式推导,y²≥0 条件
8	向量几何 (Vector Geometry)	Challenging	向量线性组合,相似三角形比例,共线判定（向量平行）,中点条件

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Question 1

Topic: 微积分与级数 (Calculus and Series)  |  Difficulty: Standard  |  Marks: 20

Problem

1 (i) By use of calculus, show that $x - \ln(1 + x)$ is positive for all positive $x$. Use this result to show that $\sum_{k=1}^{n} \frac{1}{k} > \ln(n + 1) .$

(ii) By considering $x + \ln(1 - x)$, show that $\sum_{k=1}^{\infty} \frac{1}{k^2} < 1 + \ln 2 .$

Hint

For the first result, show that the gradient of the function is positive for all positive values of $x$ (by differentiating) and also that $f(0) \ge 0$. Once this result has been established sum a set of the terms, using x = rac{1}{k}, note that \ln(1 + rac{1}{k}) can be written as $\ln(k + 1) - \ln(k)$ and then the required result follows.

For the second part, first show that $x + \ln(1 - x)$ is negative for $0 < x < 1$ and then use the substitution x = rac{1}{k^2}, noting that \ln(1 - rac{1}{k^2}) can be written as $\ln(k - 1) - 2\ln(k) + \ln(k + 1)$. Deal with the sum starting with $k = 2$ and then add the initial 1 afterwards.

Model Solution

Part (i)

Let $f(x) = x - \ln(1 + x)$.

Differentiating:

$f'(x) = 1 - \frac{1}{1 + x} = \frac{x}{1 + x}$

For $x > 0$, both $x > 0$ and $1 + x > 0$, so $f'(x) > 0$. Since $f$ is strictly increasing on $(0, \infty)$ and

$f(0) = 0 - \ln 1 = 0,$

we conclude $f(x) > 0$ for all $x > 0$. That is,

$x > \ln(1 + x) \qquad \text{for all } x > 0. \qquad \cdots (\ast)$

Now set $x = \dfrac{1}{k}$ for $k = 1, 2, \ldots, n$. Since $\dfrac{1}{k} > 0$, applying $(\ast)$ gives

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

Summing from $k = 1$ to $n$:

$\sum_{k=1}^{n} \frac{1}{k} > \sum_{k=1}^{n} \bigl[\ln(k + 1) - \ln k\bigr].$

The right-hand side is a telescoping sum:

$\sum_{k=1}^{n} \bigl[\ln(k + 1) - \ln k\bigr] = (\ln 2 - \ln 1) + (\ln 3 - \ln 2) + \cdots + \bigl[\ln(n + 1) - \ln n\bigr] = \ln(n + 1).$

Therefore

$\sum_{k=1}^{n} \frac{1}{k} > \ln(n + 1). \qquad \text{(Q.E.D.)}$

Part (ii)

Let $g(x) = x + \ln(1 - x)$.

Differentiating:

$g'(x) = 1 - \frac{1}{1 - x} = \frac{(1 - x) - 1}{1 - x} = \frac{-x}{1 - x}$

For $0 < x < 1$, we have $-x < 0$ and $1 - x > 0$, so $g'(x) < 0$. Since $g$ is strictly decreasing on $(0, 1)$ and

$g(0) = 0 + \ln 1 = 0,$

we conclude $g(x) < 0$ for all $0 < x < 1$. That is,

$x < -\ln(1 - x) \qquad \text{for all } 0 < x < 1. \qquad \cdots (\ast\ast)$

Now set $x = \dfrac{1}{k^2}$ for $k \geqslant 2$. Since $0 < \dfrac{1}{k^2} < 1$ for $k \geqslant 2$, applying $(\ast\ast)$ gives

$\frac{1}{k^2} < -\ln\!\left(1 - \frac{1}{k^2}\right).$

We compute the logarithm:

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

Summing from $k = 2$ to $n$:

$\sum_{k=2}^{n} \frac{1}{k^2} < -\sum_{k=2}^{n} \bigl[\ln(k - 1) + \ln(k + 1) - 2\ln k\bigr].$

The right-hand side telescopes. Writing out the terms:

$\sum_{k=2}^{n} \bigl[\ln(k - 1) + \ln(k + 1) - 2\ln k\bigr]$ $= \sum_{k=2}^{n} \ln(k - 1) + \sum_{k=2}^{n} \ln(k + 1) - 2\sum_{k=2}^{n} \ln k.$

• First sum: $\ln 1 + \ln 2 + \cdots + \ln(n - 1)$.
• Second sum: $\ln 3 + \ln 4 + \cdots + \ln(n + 1)$.
• Third sum (with factor 2): $2\ln 2 + 2\ln 3 + \cdots + 2\ln n$.

Combining all terms:

• $\ln 1 = 0$.
• $\ln 2$ appears once (first sum) minus $2\ln 2$ (third sum) $= -\ln 2$.
• For $3 \leqslant j \leqslant n - 1$: $\ln j$ appears once in the first sum, once in the second sum, and $-2$ times in the third sum, giving coefficient $1 + 1 - 2 = 0$.
• $\ln n$ appears once (second sum) minus $2$ times (third sum) $= -\ln n$.
• $\ln(n + 1)$ appears once (second sum).

So the total is $-\ln 2 - \ln n + \ln(n + 1) = \ln\!\left(\dfrac{n + 1}{2n}\right)$.

Therefore:

$\sum_{k=2}^{n} \frac{1}{k^2} < -\ln\!\left(\frac{n + 1}{2n}\right) = \ln\!\left(\frac{2n}{n + 1}\right).$

Adding the $k = 1$ term ($\frac{1}{1^2} = 1$):

$\sum_{k=1}^{n} \frac{1}{k^2} < 1 + \ln\!\left(\frac{2n}{n + 1}\right).$

As $n \to \infty$, $\dfrac{2n}{n + 1} \to 2$, so $\ln\!\left(\dfrac{2n}{n + 1}\right) \to \ln 2$. Since the partial sums are increasing and bounded above, the infinite series converges and

$\sum_{k=1}^{\infty} \frac{1}{k^2} \leqslant 1 + \ln 2.$

To obtain the strict inequality, note that each term $\frac{1}{k^2}$ satisfies the strict inequality $\frac{1}{k^2} < -\ln\!\left(1 - \frac{1}{k^2}\right)$ for $k \geqslant 2$, so the partial sums are strictly less than $1 + \ln\!\left(\frac{2n}{n + 1}\right)$ for every finite $n$. Since the partial sums are strictly less than a sequence converging to $1 + \ln 2$, we conclude

$\sum_{k=1}^{\infty} \frac{1}{k^2} < 1 + \ln 2. \qquad \text{(Q.E.D.)}$

Examiner Notes

热门题但平均分偏低。常见错误：(1) 未同时验证梯度为正和 f(0)≥0；(2) 未按要求用第二部分结果而采用图解法；(3) 链式法则符号错误；(4) 求和公式写错。

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Question 2

Topic: 三角学与几何 (Trigonometry and Geometry)  |  Difficulty: Standard  |  Marks: 20

Problem

2 In the triangle $ABC$, angle $BAC = \alpha$ and angle $CBA = 2\alpha$, where $2\alpha$ is acute, and $BC = x$. Show that $AB = (3 - 4 \sin^2 \alpha)x$.

The point $D$ is the midpoint of $AB$ and the point $E$ is the foot of the perpendicular from $C$ to $AB$. Find an expression for $DE$ in terms of $x$.

The point $F$ lies on the perpendicular bisector of $AB$ and is a distance $x$ from $C$. The points $F$ and $B$ lie on the same side of the line through $A$ and $C$. Show that the line $FC$ trisects the angle $ACB$.

Hint

As with all geometric questions a good diagram of the information given makes the solution to this question much easier to reach. The first result in this question follows from an application of the sine rule with applications of the relevant formulae for $\sin(A + B)$ and the double angle formulae. From a diagram of the triangle it should then be an easy application of trigonometry to show that DE = rac{x}{2}. There are a number of different methods for establishing that $FC$ trisects the angle $ACB$ – one method is to show that \sin(ngle FCE) = rac{1}{2}, following which it is relatively straightforward to work out the sizes of angles $ACB$ and $ACF$ in terms of lpha and show that they must satisfy the correct relationship.

Model Solution

Showing $AB = (3 - 4\sin^2\alpha)\,x$:

In triangle $ABC$, the angles are $A = \alpha$, $B = 2\alpha$, so $C = \pi - 3\alpha$.

By the sine rule:

$\frac{BC}{\sin A} = \frac{AB}{\sin C} \implies \frac{x}{\sin\alpha} = \frac{AB}{\sin(\pi - 3\alpha)} = \frac{AB}{\sin 3\alpha}$

So $AB = \dfrac{x\sin 3\alpha}{\sin\alpha}$.

Expanding $\sin 3\alpha = \sin(2\alpha + \alpha)$:

$\sin 3\alpha = \sin 2\alpha\cos\alpha + \cos 2\alpha\sin\alpha = 2\sin\alpha\cos^2\alpha + (1 - 2\sin^2\alpha)\sin\alpha$ $= \sin\alpha\bigl[2\cos^2\alpha + 1 - 2\sin^2\alpha\bigr] = \sin\alpha\bigl[2(1 - \sin^2\alpha) + 1 - 2\sin^2\alpha\bigr] = \sin\alpha(3 - 4\sin^2\alpha).$

Therefore:

$AB = \frac{x \cdot \sin\alpha(3 - 4\sin^2\alpha)}{\sin\alpha} = (3 - 4\sin^2\alpha)\,x. \qquad \text{(Q.E.D.)}$

Finding $DE$:

Set up coordinates with $A$ at the origin and $AB$ along the $x$-axis. Then:

• $A = (0,\, 0)$, $\quad B = \bigl((3 - 4\sin^2\alpha)\,x,\; 0\bigr)$.

Since angle $BAC = \alpha$, the ray $AC$ makes angle $\alpha$ with the $x$-axis. By the sine rule, $AC = \dfrac{x\sin 2\alpha}{\sin\alpha} = 2x\cos\alpha$. So:

$C = (2x\cos^2\alpha,\; 2x\cos\alpha\sin\alpha) = (2x\cos^2\alpha,\; x\sin 2\alpha).$

$D$ is the midpoint of $AB$: $\quad D = \left(\dfrac{(3 - 4\sin^2\alpha)\,x}{2},\; 0\right)$.

$E$ is the foot of the perpendicular from $C$ to $AB$ (the $x$-axis): $\quad E = (2x\cos^2\alpha,\; 0)$.

So:

$DE = \left|2x\cos^2\alpha - \frac{(3 - 4\sin^2\alpha)\,x}{2}\right|.$

Using $\cos^2\alpha = 1 - \sin^2\alpha$:

$2\cos^2\alpha = 2 - 2\sin^2\alpha, \qquad 3 - 4\sin^2\alpha = 4\cos^2\alpha - 1.$

$DE = \left|x(2 - 2\sin^2\alpha) - \frac{x(4\cos^2\alpha - 1)}{2}\right| = \left|\frac{x(4 - 4\sin^2\alpha - 4\cos^2\alpha + 1)}{2}\right|$

$= \left|\frac{x(5 - 4\sin^2\alpha - 4\cos^2\alpha)}{2}\right| = \left|\frac{x(5 - 4)}{2}\right| = \frac{x}{2}.$

Therefore $DE = \dfrac{x}{2}$. $\qquad \text{(Q.E.D.)}$

Showing $FC$ trisects angle $ACB$:

The perpendicular bisector of $AB$ is the vertical line $x = \dfrac{(3 - 4\sin^2\alpha)\,x}{2}$. So $F = \left(\dfrac{(3 - 4\sin^2\alpha)\,x}{2},\; y\right)$ for some $y$.

The condition $FC = x$ gives:

$\left(\frac{(3 - 4\sin^2\alpha)\,x}{2} - 2x\cos^2\alpha\right)^2 + (y - x\sin 2\alpha)^2 = x^2.$

The horizontal difference simplifies: $\dfrac{(4\cos^2\alpha - 1)\,x}{2} - 2x\cos^2\alpha = -\dfrac{x}{2}$.

So $\dfrac{x^2}{4} + (y - x\sin 2\alpha)^2 = x^2$, giving $(y - x\sin 2\alpha)^2 = \dfrac{3x^2}{4}$, hence

$y = x\sin 2\alpha \pm \frac{\sqrt{3}}{2}\,x.$

To determine the sign, we use the condition that $F$ and $B$ lie on the same side of line $AC$. Line $AC$ passes through the origin in direction $(\cos\alpha,\, \sin\alpha)$, with equation $\sin\alpha \cdot X - \cos\alpha \cdot Y = 0$.

For $B = ((3 - 4\sin^2\alpha)\,x,\; 0)$: the value is $\sin\alpha \cdot (3 - 4\sin^2\alpha)\,x > 0$, so $B$ is on the positive side.

For $F$: the value is $\sin\alpha \cdot \dfrac{(3 - 4\sin^2\alpha)\,x}{2} - \cos\alpha \cdot y$. Substituting and simplifying, this equals $\dfrac{x}{2}(\sqrt{3}\cos\alpha - \sin\alpha)$ when $y = x\sin 2\alpha - \dfrac{\sqrt{3}}{2}\,x$ (the "$-$" sign), and $-\dfrac{x}{2}(\sin\alpha + \sqrt{3}\cos\alpha) < 0$ when $y = x\sin 2\alpha + \dfrac{\sqrt{3}}{2}\,x$ (the "$+$" sign).

Since $\sqrt{3}\cos\alpha > \sin\alpha$ for $0 < \alpha < \pi/4$ (as $\tan\alpha < 1 < \sqrt{3}$), we take the "$-$" sign:

$F = \left(\frac{(3 - 4\sin^2\alpha)\,x}{2},\; x\sin 2\alpha - \frac{\sqrt{3}}{2}\,x\right).$

Now compute the vector $\overrightarrow{CF} = F - C$:

$\overrightarrow{CF} = \left(-\frac{x}{2},\; -\frac{\sqrt{3}}{2}\,x\right), \qquad |\overrightarrow{CF}| = x\sqrt{\frac{1}{4} + \frac{3}{4}} = x. \quad \checkmark$

We compute $\angle ACF$ using the dot product. The vector $\overrightarrow{CA} = A - C = (-2x\cos^2\alpha,\; -x\sin 2\alpha)$, with $|\overrightarrow{CA}| = 2x\cos\alpha$.

$\overrightarrow{CA} \cdot \overrightarrow{CF} = (-2x\cos^2\alpha)\!\left(-\frac{x}{2}\right) + (-x\sin 2\alpha)\!\left(-\frac{\sqrt{3}}{2}\,x\right)$

$= x^2\cos^2\alpha + \frac{\sqrt{3}}{2}\,x^2\sin 2\alpha = x^2\cos^2\alpha + \sqrt{3}\,x^2\sin\alpha\cos\alpha = x^2\cos\alpha(\cos\alpha + \sqrt{3}\sin\alpha).$

Therefore:

$\cos(\angle ACF) = \frac{x^2\cos\alpha(\cos\alpha + \sqrt{3}\sin\alpha)}{2x\cos\alpha \cdot x} = \frac{\cos\alpha + \sqrt{3}\sin\alpha}{2}.$

Using the identity $\cos\alpha + \sqrt{3}\sin\alpha = 2\sin\!\left(\alpha + \dfrac{\pi}{6}\right) = 2\cos\!\left(\dfrac{\pi}{3} - \alpha\right)$:

$\cos(\angle ACF) = \cos\!\left(\frac{\pi}{3} - \alpha\right).$

Since $0 < \alpha < \pi/4$, we have $\pi/3 - \alpha > 0$, and $\angle ACF$ lies in $(0, \pi)$. So:

$\angle ACF = \frac{\pi}{3} - \alpha.$

Since angle $ACB = \pi - 3\alpha$:

$\angle FCB = \angle ACB - \angle ACF = (\pi - 3\alpha) - \left(\frac{\pi}{3} - \alpha\right) = \frac{2\pi}{3} - 2\alpha = 2\!\left(\frac{\pi}{3} - \alpha\right) = 2\,\angle ACF.$

So $\angle ACF : \angle FCB = 1 : 2$, meaning $FC$ divides angle $ACB$ in the ratio $1 : 2$. Since $\angle ACB = 3 \cdot \angle ACF$, the line $FC$ trisects angle $ACB$. $\qquad \text{(Q.E.D.)}$

Examiner Notes

平均分最高。解法多样但解释常不够清晰；建议画大图，注意图中角度标注不要颠倒。

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Question 3

Topic: 组合数学与数列 (Combinatorics and Sequences)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Three rods have lengths $a$, $b$ and $c$, where $a < b < c$. The three rods can be made into a triangle (possibly of zero area) if $a + b \geqslant c$.

Let $T_n$ be the number of triangles that can be made with three rods chosen from $n$ rods of lengths $1, 2, 3, \dots, n$ (where $n \geqslant 3$). Show that $T_8 - T_7 = 2 + 4 + 6$ and evaluate $T_8 - T_6$. Write down expressions for $T_{2m} - T_{2m-1}$ and $T_{2m} - T_{2m-2}$.

Prove by induction that $T_{2m} = \frac{1}{6} m(m - 1)(4m + 1)$, and find the corresponding result for an odd number of rods.

Hint

For the first part note that $T_8 - T_7$ can be interpreted as the triangles that can be made using the rod of length 8 and two other, shorter rods. These can then be counted by noting that there are 6 possibilities if the length 7 rod is used, 4 possibilities if the length 6 (but not the length 7) rod is used and 2 possibilities if the length 5 (but not 6 or 7) rod is used. It is clear that at least one rod longer than length 4 must be used. To evaluate $T_8 - T_6$ note that it is equal to $(T_8 - T_7) + (T_7 - T_6)$ and then evaluate $T_7 - T_6$ in a similar manner to $T_8 - T_7$. Similar reasoning easily gives formulae for $T_{2m} - T_{2m-1}$ and $T_{2m} - T_{2m-2}$.

For the induction, the rule for $T_{2m} - T_{2m-2}$ deduced in the previous part can be used to show the inductive step, while the easiest way to show the base case is to list the possibilities. The easiest way to establish the result for an odd number of rods is to use the formula for $T_{2m} - T_{2m-1}$ and the formula for $T_{2m}$ that was just proven.

Model Solution

Counting triangles with longest side $n$:

The difference $T_n - T_{n-1}$ counts the number of triangles using the rod of length $n$ (the longest) together with two shorter rods $a < b < n$ satisfying $a + b \geq n$. For a given $b$ (with $b < n$), we need $a \geq n - b$ and $a < b$, so the number of valid choices for $a$ is $(b - 1) - (n - b) + 1 = 2b - n$, provided $2b - n \geq 1$, i.e.\ $b \geq \lceil \tfrac{n+1}{2} \rceil$.

Showing $T_8 - T_7 = 2 + 4 + 6$:

For $n = 8$, the middle rod $b$ ranges from $5$ to $7$:

• $b = 7$: $a$ ranges from $8 - 7 = 1$ to $6$, giving $6$ choices.
• $b = 6$: $a$ ranges from $8 - 6 = 2$ to $5$, giving $4$ choices.
• $b = 5$: $a$ ranges from $8 - 5 = 3$ to $4$, giving $2$ choices.

Hence $T_8 - T_7 = 6 + 4 + 2 = 2 + 4 + 6 = 12$.

Evaluating $T_8 - T_6$:

By the same reasoning with $n = 7$, the middle rod $b$ ranges from $4$ to $6$:

• $b = 6$: $a$ ranges from $1$ to $5$, giving $5$ choices.
• $b = 5$: $a$ ranges from $2$ to $4$, giving $3$ choices.
• $b = 4$: $a = 3$, giving $1$ choice.

So $T_7 - T_6 = 5 + 3 + 1 = 9$, and

$T_8 - T_6 = (T_8 - T_7) + (T_7 - T_6) = 12 + 9 = 21.$

General formula for $T_n - T_{n-1}$:

For even $n = 2m$: $b$ ranges from $m + 1$ to $2m - 1$, with $2b - 2m$ choices each. Setting $k = b - m$:

$T_{2m} - T_{2m-1} = \sum_{k=1}^{m-1} 2k = m(m - 1).$

For odd $n = 2m + 1$: $b$ ranges from $m + 1$ to $2m$, with $2b - (2m+1)$ choices each. Setting $k = b - m$:

$T_{2m+1} - T_{2m} = \sum_{k=1}^{m} (2k - 1) = m^2.$

Therefore, summarising the key differences:

$T_{2m} - T_{2m-1} = m(m-1), \qquad T_{2m-1} - T_{2m-2} = (m-1)^2.$

Combining these two:

$T_{2m} - T_{2m-2} = m(m-1) + (m-1)^2 = (m-1)[m + (m-1)] = (m-1)(2m - 1).$

Induction proof that $T_{2m} = \frac{1}{6}m(m-1)(4m+1)$:

Base case ($m = 2$): The possible triples from $\{1, 2, 3, 4\}$ are: $(1,2,3)$ with $1+2=3 \geq 3$; $(1,3,4)$ with $1+3=4 \geq 4$; $(2,3,4)$ with $2+3=5 \geq 4$. Triples $(1,2,4)$ fails ($1+2 < 4$). So $T_4 = 3 = \frac{1}{6}(2)(1)(9)$. The base case holds.

Inductive step: Assume $T_{2(m-1)} = \frac{1}{6}(m-1)(m-2)(4m-3)$ for some $m \geq 3$. Then:

$T_{2m} = T_{2(m-1)} + (T_{2m} - T_{2m-2}) = \frac{1}{6}(m-1)(m-2)(4m-3) + (m-1)(2m-1).$

Factor out $\frac{1}{6}(m-1)$:

$T_{2m} = \frac{1}{6}(m-1)\left[(m-2)(4m-3) + 6(2m-1)\right].$

Expanding the bracket:

$(m-2)(4m-3) + 6(2m-1) = 4m^2 - 11m + 6 + 12m - 6 = 4m^2 + m = m(4m+1).$

Therefore $T_{2m} = \frac{1}{6}(m-1) \cdot m(4m+1) = \frac{1}{6}m(m-1)(4m+1)$, completing the induction.

Odd number of rods:

Using $T_{2m-1} = T_{2m} - m(m-1)$:

$T_{2m-1} = \frac{1}{6}m(m-1)(4m+1) - m(m-1) = m(m-1)\left[\frac{4m+1}{6} - 1\right] = m(m-1) \cdot \frac{4m-5}{6}.$

So:

$T_{2m-1} = \frac{1}{6}m(m-1)(4m - 5).$

Verification: $T_3 = \frac{1}{6}(2)(1)(3) = 1$; $T_5 = \frac{1}{6}(3)(2)(7) = 7$; $T_7 = \frac{1}{6}(4)(3)(11) = 22$. These agree with direct counting.

Examiner Notes

整体表现好，满分人数多。常见问题：(1) 未将 $T_8-T_7$ 理解为包含特定杆的三角形数而逐一枚举；(2) 归纳法基础情况未穷举验证；(3) 奇数情况部分未尝试或错误使用归纳法。

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Question 4

Topic: 函数与图像 (Functions and Graphs)  |  Difficulty: Standard  |  Marks: 20

Problem

4 (i) The continuous function $f$ is defined by

$\tan f(x) = x \quad (-\infty < x < \infty)$

and $f(0) = \pi$. Sketch the curve $y = f(x)$.

(ii) The continuous function $g$ is defined by

$\tan g(x) = \frac{x}{1 + x^2} \quad (-\infty < x < \infty)$

and $g(0) = \pi$. Sketch the curves $y = \frac{x}{1 + x^2}$ and $y = g(x)$.

(iii) The continuous function $h$ is defined by $h(0) = \pi$ and

$\tan h(x) = \frac{x}{1 - x^2} \quad (x \neq \pm 1).$

(The values of $h(x)$ at $x = \pm 1$ are such that $h(x)$ is continuous at these points.) Sketch the curves $y = \frac{x}{1 - x^2}$ and $y = h(x)$.

Hint

For the first part, note that the graph of rctan x satisfies the requirement of being continuous, but does not satisfy $f(0) = \pi$. Since $an(x + \pi) = an x$, a translation of the graph of y = rctan x vertically by a distance of $\pi$ gives the required graph.

It should be clear that the graph of y = rac{x}{1+x^2} has no vertical asymptotes, approaches the $x$-axis as $x o \pm\infty$ and passes through the origin. Identifying the stationary points should be the next task after which a graph should be easy to sketch. The graph of $y = g(x)$ should then be easy to sketch by considering the fact that $f(x)$ is an increasing function and $g(x)$ is obtained by composing the two functions already sketched.

To sketch the graph of y = rac{x}{1-x^2} first note that there must be two vertical asymptotes. Once stationary points have been checked for it should be straightforward to complete the sketch. In this case, the asymptotes need to be considered to deduce the shape of the graph for $y = h(x)$ as the composition with $f(x)$ will lead to discontinuities. Noting again that $an(x + \pi) = an x$ the discontinuities can be resolved by translating sections of that graph vertically by a distance of $\pi$.

Model Solution

Part (i)

Since $\tan(x + \pi) = \tan x$ for all $x$, and $\arctan x$ is the continuous inverse of $\tan$ on $(-\frac{\pi}{2}, \frac{\pi}{2})$, the function satisfying $\tan f(x) = x$ and $f(0) = \pi$ is:

$f(x) = \arctan x + \pi.$

Graph features:

• Continuous and strictly increasing on $(-\infty, \infty)$.
• Passes through $(0, \pi)$.
• Horizontal asymptotes: $y = \frac{\pi}{2}$ as $x \to -\infty$, and $y = \frac{3\pi}{2}$ as $x \to +\infty$.
• This is the standard graph of $y = \arctan x$ shifted upward by $\pi$.

Part (ii)

First sketch $y = \frac{x}{1+x^2}$.

Let $u(x) = \frac{x}{1+x^2}$. Key features:

• $u(0) = 0$; $u$ is odd, so the graph is symmetric about the origin.
• $\lim_{x \to \pm\infty} u(x) = 0$ (horizontal asymptote $y = 0$).
• No vertical asymptotes (denominator $1 + x^2 > 0$ for all $x$).

To find stationary points, differentiate:

$u'(x) = \frac{(1+x^2) - x \cdot 2x}{(1+x^2)^2} = \frac{1 - x^2}{(1+x^2)^2}.$

Setting $u'(x) = 0$ gives $x = \pm 1$.

• $u(1) = \frac{1}{2}$ (maximum), $u(-1) = -\frac{1}{2}$ (minimum).

The graph rises from $0$ at $x = -\infty$, reaches a minimum of $-\frac{1}{2}$ at $x = -1$, passes through the origin, reaches a maximum of $\frac{1}{2}$ at $x = 1$, then returns to $0$.

Now sketch $y = g(x)$.

Since $g(0) = \pi$ and $\tan g(x) = u(x)$, we have $g(x) = \arctan u(x) + \pi$. Because $f(x) = \arctan x + \pi$ is strictly increasing, composing with $u$ preserves the qualitative shape:

• $g(0) = \pi$, $g(\pm 1) = \pi \pm \arctan \tfrac{1}{2}$.
• As $x \to \pm\infty$, $u(x) \to 0$, so $g(x) \to \pi$.
• Since $-\frac{1}{2} \leq u(x) \leq \frac{1}{2}$, the range of $g$ is $[\pi - \arctan \frac{1}{2},\; \pi + \arctan \frac{1}{2}] \approx [2.68,\; 3.46]$.
• No vertical asymptotes; $g$ is continuous and bounded.

The graph of $y = g(x)$ has the same “bump” shape as $y = u(x)$, centred on $y = \pi$ instead of $y = 0$, with a maximum of $\pi + \arctan \frac{1}{2}$ at $x = 1$ and a minimum of $\pi - \arctan \frac{1}{2}$ at $x = -1$.

Part (iii)

First sketch $y = \frac{x}{1-x^2}$.

Let $v(x) = \frac{x}{1-x^2}$. Key features:

• $v(0) = 0$; $v$ is odd.
• Vertical asymptotes at $x = \pm 1$ (denominator vanishes).
• $\lim_{x \to \pm\infty} v(x) = 0$ (horizontal asymptote $y = 0$).

Differentiate:

$v'(x) = \frac{(1-x^2) - x(-2x)}{(1-x^2)^2} = \frac{1 + x^2}{(1-x^2)^2} > 0 \quad \text{for all } x \neq \pm 1.$

So $v$ is strictly increasing on each of the three intervals $(-\infty, -1)$, $(-1, 1)$, $(1, \infty)$. No stationary points.

On $(-1, 1)$: $v$ increases from $-\infty$ to $+\infty$ (crossing the origin). On $(-\infty, -1)$: $v$ increases from $0^+$ to $+\infty$. On $(1, \infty)$: $v$ increases from $-\infty$ to $0^-$.

Now sketch $y = h(x)$.

Since $v(x)$ takes all real values on $(-1, 1)$, the composition $h(x) = \arctan v(x) + k\pi$ (for suitable integer $k$) has discontinuities at $x = \pm 1$ unless we choose different shifts on each interval. We require $h$ to be continuous everywhere.

On $(-1, 1)$: $v$ ranges over $(-\infty, \infty)$, so $\arctan v(x)$ ranges over $(-\frac{\pi}{2}, \frac{\pi}{2})$. With $h(0) = \pi$, the correct branch is:

$h(x) = \arctan v(x) + \pi \qquad \text{on } (-1, 1).$

This gives $h \to \frac{\pi}{2}$ as $x \to -1^+$ and $h \to \frac{3\pi}{2}$ as $x \to 1^-$.

On $(-\infty, -1)$: $v(x) > 0$, so $\arctan v(x) \in (0, \frac{\pi}{2})$. With the basic branch $\arctan v(x)$, we get $h \to \frac{\pi}{2}$ as $x \to -1^-$, matching the left-limit from $(-1, 1)$. So:

$h(x) = \arctan v(x) \qquad \text{on } (-\infty, -1).$

On $(1, \infty)$: $v(x) < 0$, so $\arctan v(x) \in (-\frac{\pi}{2}, 0)$. We need $h \to \frac{3\pi}{2}$ as $x \to 1^+$, so we shift by $2\pi$:

$h(x) = \arctan v(x) + 2\pi \qquad \text{on } (1, \infty).$

Summary of $h$:

$h(x) = \begin{cases} \arctan\dfrac{x}{1-x^2}, & x < -1, \\[6pt] \arctan\dfrac{x}{1-x^2} + \pi, & -1 < x < 1, \\[6pt] \arctan\dfrac{x}{1-x^2} + 2\pi, & x > 1. \end{cases}$

Continuity checks:

• At $x = -1$: as $x \to -1^-$, $v(x) \to +\infty$, so $\arctan v(x) \to \frac{\pi}{2}$, giving $h \to \frac{\pi}{2}$. As $x \to -1^+$, $v(x) \to -\infty$, so $\arctan v(x) \to -\frac{\pi}{2}$, giving $h \to -\frac{\pi}{2} + \pi = \frac{\pi}{2}$. Continuous. ✓
• At $x = 1$: as $x \to 1^-$, $v(x) \to +\infty$, so $\arctan v(x) \to \frac{\pi}{2}$, giving $h \to \frac{\pi}{2} + \pi = \frac{3\pi}{2}$. As $x \to 1^+$, $v(x) \to -\infty$, so $\arctan v(x) \to -\frac{\pi}{2}$, giving $h \to -\frac{\pi}{2} + 2\pi = \frac{3\pi}{2}$. Continuous. ✓

Graph features of $y = h(x)$:

• Passes through $(0, \pi)$.
• On $(-\infty, -1)$: increases from $0$ (as $x \to -\infty$, since $\frac{x}{1-x^2} \to 0^+$) to $\frac{\pi}{2}$ (at $x = -1$).
• On $(-1, 1)$: increases from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, passing through $(0, \pi)$.
• On $(1, \infty)$: increases from $\frac{3\pi}{2}$ (at $x = 1$) to $2\pi$ (as $x \to +\infty$, since $\frac{x}{1-x^2} \to 0^-$).
• Horizontal asymptotes: $y = 0$ as $x \to -\infty$, and $y = 2\pi$ as $x \to +\infty$.
• $h$ is strictly increasing and continuous on all of $\mathbb{R}$, with the steepest gradient near $x = \pm 1$.

Examiner Notes

普遍表现较好。常见错误：(1) 画出不连续的图像（尽管声称连续）；(2) 图像上标注点标错；(3) 用图像变换推理时出错；(4) 说要微分却做了积分。

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Question 5

Topic: 三角恒等式与级数 (Trigonometric Identities and Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 In this question, the arctan function satisfies $0 \leqslant \arctan x < \frac{1}{2}\pi$ for $x \geqslant 0$.

(i) Let

$S_n = \sum_{m=1}^{n} \arctan \left( \frac{1}{2m^2} \right),$

for $n = 1, 2, 3, \dots$. Prove by induction that

$\tan S_n = \frac{n}{n + 1}.$

Prove also that

$S_n = \arctan \frac{n}{n + 1}.$

(ii) In a triangle $ABC$, the lengths of the sides $AB$ and $BC$ are $4n^2$ and $4n^4 - 1$, respectively, and the angle at $B$ is a right angle. Let angle $BCA = 2\alpha_n$. Show that

$\sum_{n=1}^{\infty} \alpha_n = \frac{1}{4}\pi.$

Hint

The initial proof by induction is a straightforward application of the $an(A + B)$ formula. The final part of section (i) requires recognition that there are many possible values of $x$ to give a particular value of $an x$, but only one of them is the value that would be obtained by applying the rctan function. The result can therefore be shown by establishing that the difference between consecutive terms of the sequence is never more than $\pi$.

For the second part of the question a diagram of the triangle and application of the $an 2A$ formula shows that the value of lpha_n must be of the form used in the first part of the question. All that remains is then to show that the limit of the sum must give the required value.

Model Solution

Part (i)

We prove $\tan S_n = \dfrac{n}{n+1}$ by induction.

Base case ($n = 1$): $S_1 = \arctan\dfrac{1}{2}$, so $\tan S_1 = \dfrac{1}{2} = \dfrac{1}{1+1}$. ✓

Inductive step: Assume $\tan S_k = \dfrac{k}{k+1}$ for some $k \geqslant 1$. Then

$S_{k+1} = S_k + \arctan\frac{1}{2(k+1)^2}.$

Using the tangent addition formula:

$\tan S_{k+1} = \frac{\tan S_k + \dfrac{1}{2(k+1)^2}}{1 - \tan S_k \cdot \dfrac{1}{2(k+1)^2}} = \frac{\dfrac{k}{k+1} + \dfrac{1}{2(k+1)^2}}{1 - \dfrac{k}{(k+1) \cdot 2(k+1)^2}}.$

For the numerator:

$\frac{k}{k+1} + \frac{1}{2(k+1)^2} = \frac{2k(k+1) + 1}{2(k+1)^2} = \frac{2k^2 + 2k + 1}{2(k+1)^2}.$

For the denominator:

$1 - \frac{k}{2(k+1)^3} = \frac{2(k+1)^3 - k}{2(k+1)^3} = \frac{2k^3 + 6k^2 + 6k + 2 - k}{2(k+1)^3} = \frac{2k^3 + 6k^2 + 5k + 2}{2(k+1)^3}.$

Therefore:

$\tan S_{k+1} = \frac{2k^2 + 2k + 1}{2(k+1)^2} \times \frac{2(k+1)^3}{2k^3 + 6k^2 + 5k + 2} = \frac{(2k^2 + 2k + 1)(k+1)}{2k^3 + 6k^2 + 5k + 2}.$

Factorising the denominator: $2k^3 + 6k^2 + 5k + 2 = (k+1)(2k^2 + 4k + 2) = 2(k+1)^3$.

$\tan S_{k+1} = \frac{(2k^2+2k+1)(k+1)}{2(k+1)^3} = \frac{2k^2+2k+1}{2(k+1)^2}.$

Hmm, let me recheck. We need $\tan S_{k+1} = \frac{k+1}{k+2}$.

Actually, let me recompute the denominator more carefully. We have:

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

Now $2(k+1)^3 = 2(k^3 + 3k^2 + 3k + 1) = 2k^3 + 6k^2 + 6k + 2$, so the denominator is

$\frac{2k^3 + 6k^2 + 5k + 2}{2(k+1)^3}.$

We can check that $(k+2)(2k^2 + 2k + 1) = 2k^3 + 2k^2 + k + 4k^2 + 4k + 2 = 2k^3 + 6k^2 + 5k + 2$. ✓

Therefore:

$\tan S_{k+1} = \frac{(2k^2+2k+1)(k+1)}{(2k^2+2k+1)(k+2)} \cdot \frac{1}{1} = \frac{k+1}{k+2}.$

This completes the induction. $\qquad \text{(Q.E.D.)}$

Proof that $S_n = \arctan\dfrac{n}{n+1}$:

Since $\dfrac{1}{2m^2} > 0$ for each $m$, and $\arctan$ maps $[0, \infty)$ to $[0, \tfrac{\pi}{2})$, each term $\arctan\dfrac{1}{2m^2}$ lies in $(0, \tfrac{\pi}{2})$. Therefore

$0 < S_n < \sum_{m=1}^{n} \frac{\pi}{2} \qquad \text{(not useful as stated)}.$

More precisely, $S_n$ is a sum of positive terms, so $0 < S_n$. Also, since $\arctan x < x$ for $x > 0$ (from Q1(i) style reasoning), $S_n < \sum_{m=1}^n \frac{1}{2m^2} < \sum_{m=1}^\infty \frac{1}{2m^2} = \frac{\pi^2}{12} < \frac{\pi}{2}$. Hence $S_n \in (0, \frac{\pi}{2})$.

In the interval $[0, \frac{\pi}{2})$, the tangent function is strictly increasing and therefore injective. We have shown $\tan S_n = \frac{n}{n+1}$, and since $\arctan\frac{n}{n+1}$ is the unique value in $[0, \frac{\pi}{2})$ with this tangent, we conclude

$S_n = \arctan\frac{n}{n+1}. \qquad \text{(Q.E.D.)}$

Part (ii)

In right-angled triangle $ABC$ with the right angle at $B$, we have $AB = 4n^2$, $BC = 4n^4 - 1$, and angle $BCA = 2\alpha_n$.

$\tan(2\alpha_n) = \frac{AB}{BC} = \frac{4n^2}{4n^4 - 1} = \frac{4n^2}{(2n^2-1)(2n^2+1)}.$

Let $t = \tan\alpha_n$. The double angle formula gives $\tan(2\alpha_n) = \dfrac{2t}{1-t^2}$, so:

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

This gives $2t(2n^2-1)(2n^2+1) = 4n^2(1-t^2)$, i.e.

$2t(4n^4 - 1) = 4n^2 - 4n^2 t^2.$

Rearranging: $4n^2 t^2 + 2(4n^4-1)t - 4n^2 = 0$, i.e. $2n^2 t^2 + (4n^4-1)t - 2n^2 = 0$.

Using the quadratic formula or factoring: we try $t = \dfrac{1}{2n^2}$:

$2n^2 \cdot \frac{1}{4n^4} + (4n^4-1)\cdot\frac{1}{2n^2} - 2n^2 = \frac{1}{2n^2} + \frac{4n^4-1}{2n^2} - 2n^2 = \frac{4n^4}{2n^2} - 2n^2 = 2n^2 - 2n^2 = 0. \quad \checkmark$

Since $t > 0$ (as $0 < 2\alpha_n < \frac{\pi}{2}$), we take $\tan\alpha_n = \dfrac{1}{2n^2}$.

Therefore $\alpha_n = \arctan\dfrac{1}{2n^2}$, and from part (i):

$\sum_{n=1}^{N} \alpha_n = \sum_{n=1}^{N} \arctan\frac{1}{2n^2} = S_N = \arctan\frac{N}{N+1}.$

Taking the limit:

$\sum_{n=1}^{\infty} \alpha_n = \lim_{N\to\infty} \arctan\frac{N}{N+1} = \arctan 1 = \frac{\pi}{4}. \qquad \text{(Q.E.D.)}$

Examiner Notes

平均分偏低。多数人能完成归纳证明 $an S_n = n/(n+1)$，但绝大多数未意识到还需进一步证明 S_n = rctan(n/(n+1))（需讨论值域）。第二部分若识别出与第一部分的关联则表现较好。

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Question 6

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

Problem

6 (i) Show that $\sec^2 \left( \frac{1}{4}\pi - \frac{1}{2}x \right) = \frac{2}{1 + \sin x} .$ Hence integrate $\frac{1}{1 + \sin x}$ with respect to $x$.

(ii) By means of the substitution $y = \pi - x$, show that $\int_{0}^{\pi} xf(\sin x) \, dx = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx ,$ where $f$ is any function for which these integrals exist.

Hence evaluate $\int_{0}^{\pi} \frac{x}{1 + \sin x} \, dx .$

(iii) Evaluate $\int_{0}^{\pi} \frac{2x^3 - 3\pi x^2}{(1 + \sin x)^2} \, dx .$

Hint

The first part of the question requires use of the $\cos(A + B)$ formula. Following this the integral should be easy to evaluate given that $\int \sec^2 x \, dx = \tan x + c$. In the second part, apply the substitution and note that the limits of the integral are reversed, which is equivalent to multiplying by -1. Following this a simple rearrangement (noting that the variable that the integration is taken over can be changed from $y$ to $x$) should establish the required result. The integral at the end of this part can then be evaluated simply by applying this result along with the integral evaluated in part (i).

In the final part of the question it is tempting to make repeated applications of the result proven in part (ii). However, this is not valid as it would require the use of a function satisfying $f(\sin x) = x$, which is not possible on the interval over which the integral is defined. Instead, application of a similar substitution to part (ii) to $\int_0^\pi x^3 f(\sin x) \, dx$ will simplify to allow this integral to be evaluated based on the integration of $\frac{1}{(1+\sin x)^2}$. An application of the result from part (ii) will also be required.

Model Solution

Part (i)

We show $\sec^2\!\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right) = \dfrac{2}{1+\sin x}$.

Using the cosine addition formula:

$\cos\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = \cos\frac{\pi}{4}\cos\frac{x}{2} + \sin\frac{\pi}{4}\sin\frac{x}{2} = \frac{1}{\sqrt{2}}\!\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right).$

Squaring:

$\cos^2\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1}{2}\!\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 = \frac{1}{2}\!\left(\cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} + \sin^2\frac{x}{2}\right) = \frac{1}{2}(1 + \sin x).$

Therefore

$\sec^2\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1}{\cos^2\!\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)} = \frac{2}{1 + \sin x}. \qquad \text{(Q.E.D.)}$

To integrate $\dfrac{1}{1+\sin x}$, note that

$\frac{d}{dx}\tan\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = -\frac{1}{2}\sec^2\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = -\frac{1}{1+\sin x}.$

Therefore

$\int \frac{1}{1+\sin x}\,dx = -\tan\!\left(\frac{\pi}{4} - \frac{x}{2}\right) + C.$

Part (ii)

Let $I = \displaystyle\int_0^\pi x\,f(\sin x)\,dx$. Substituting $y = \pi - x$, so $x = \pi - y$ and $dx = -dy$:

$I = \int_{\pi}^{0}(\pi - y)\,f\!\big(\sin(\pi - y)\big)(-dy) = \int_0^\pi (\pi - y)\,f(\sin y)\,dy,$

using $\sin(\pi - y) = \sin y$. Expanding:

$I = \pi\int_0^\pi f(\sin y)\,dy - \int_0^\pi y\,f(\sin y)\,dy = \pi\int_0^\pi f(\sin x)\,dx - I.$

Rearranging: $2I = \pi\displaystyle\int_0^\pi f(\sin x)\,dx$, giving

$\int_0^\pi x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\,dx. \qquad \text{(Q.E.D.)}$

Evaluation. Apply this with $f(\sin x) = \dfrac{1}{1+\sin x}$:

$\int_0^\pi \frac{x}{1+\sin x}\,dx = \frac{\pi}{2}\int_0^\pi \frac{1}{1+\sin x}\,dx.$

From part (i):

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

Therefore

$\int_0^\pi \frac{x}{1+\sin x}\,dx = \frac{\pi}{2}\cdot 2 = \pi.$

Part (iii)

Let $I_0 = \displaystyle\int_0^\pi \frac{dx}{(1+\sin x)^2}$ and $I_k = \displaystyle\int_0^\pi \frac{x^k}{(1+\sin x)^2}\,dx$ for $k = 1, 2, 3$. We need $2I_3 - 3\pi I_2$.

Step 1: Derive relations via the substitution $y = \pi - x$.

For $I_3$, substitute $y = \pi - x$:

$I_3 = \int_0^\pi \frac{(\pi - x)^3}{(1+\sin x)^2}\,dx = \int_0^\pi \frac{\pi^3 - 3\pi^2 x + 3\pi x^2 - x^3}{(1+\sin x)^2}\,dx = \pi^3 I_0 - 3\pi^2 I_1 + 3\pi I_2 - I_3.$

So $2I_3 = \pi^3 I_0 - 3\pi^2 I_1 + 3\pi I_2$. $\qquad \cdots (*)$

For $I_2$, the same substitution gives:

$I_2 = \int_0^\pi \frac{(\pi - x)^2}{(1+\sin x)^2}\,dx = \pi^2 I_0 - 2\pi I_1 + I_2.$

So $0 = \pi^2 I_0 - 2\pi I_1$, giving $I_1 = \dfrac{\pi}{2}\,I_0$. $\qquad \cdots (**)$

This is also the result from part (ii) applied to $f(\sin x) = \dfrac{1}{(1+\sin x)^2}$.

Step 2: Compute $2I_3 - 3\pi I_2$.

From $(*)$:

$2I_3 - 3\pi I_2 = \pi^3 I_0 - 3\pi^2 I_1 + 3\pi I_2 - 3\pi I_2 = \pi^3 I_0 - 3\pi^2 I_1.$

Substituting $(**)$:

$2I_3 - 3\pi I_2 = \pi^3 I_0 - 3\pi^2 \cdot \frac{\pi}{2}\,I_0 = \pi^3 I_0 - \frac{3\pi^3}{2}\,I_0 = -\frac{\pi^3}{2}\,I_0.$

Step 3: Evaluate $I_0 = \displaystyle\int_0^\pi \frac{dx}{(1+\sin x)^2}$.

Use $t = \tan\dfrac{x}{2}$, so $\sin x = \dfrac{2t}{1+t^2}$, $dx = \dfrac{2\,dt}{1+t^2}$, with limits $x = 0 \to t = 0$, $x = \pi \to t = 1$.

$1 + \sin x = \frac{(1+t)^2}{1+t^2}, \qquad (1+\sin x)^2 = \frac{(1+t)^4}{(1+t^2)^2}.$

$I_0 = \int_0^1 \frac{(1+t^2)^2}{(1+t)^4}\cdot\frac{2\,dt}{1+t^2} = 2\int_0^1 \frac{1+t^2}{(1+t)^4}\,dt.$

Write $1+t^2 = (1+t)^2 - 2(1+t) + 2$:

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

Integrating:

$\int_0^1 \frac{dt}{(1+t)^2} = \left[-\frac{1}{1+t}\right]_0^1 = -\frac{1}{2} + 1 = \frac{1}{2},$

$\int_0^1 \frac{dt}{(1+t)^3} = \left[-\frac{1}{2(1+t)^2}\right]_0^1 = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8},$

$\int_0^1 \frac{dt}{(1+t)^4} = \left[-\frac{1}{3(1+t)^3}\right]_0^1 = -\frac{1}{24} + \frac{1}{3} = \frac{7}{24}.$

Therefore

$I_0 = 2\left(\frac{1}{2} - \frac{6}{8} + \frac{14}{24}\right) = 2\left(\frac{12 - 18 + 14}{24}\right) = 2\cdot\frac{8}{24} = \frac{2}{3}.$

Step 4: Final answer.

$\int_0^\pi \frac{2x^3 - 3\pi x^2}{(1+\sin x)^2}\,dx = 2I_3 - 3\pi I_2 = -\frac{\pi^3}{2}\cdot\frac{2}{3} = -\frac{\pi^3}{3}.$

Examiner Notes

This was the most popular question on the paper with over 1000 attempts made. The first section did not present significant difficulties to candidates and the integration was generally well completed, although occasionally with an error in the factor. The second part proved difficult for a number of candidates who failed to change the variable in the integral correctly, or in some cases did not change the variable in every position that it occurred. Other candidates did not apply a correct result for dealing with the trigonometric functions involved or did not clearly show how the required result was reached as the solution jumped through several steps to a statement of the result asked for in the question. There were very few successful attempts at the final part of the question, but they did include a variety of methods for evaluating the integral once the substitution had been made.

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Question 7

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

Problem

7 A circle $C$ is said to be bisected by a curve $X$ if $X$ meets $C$ in exactly two points and these points are diametrically opposite each other on $C$.

(i) Let $C$ be the circle of radius $a$ in the $x$-$y$ plane with centre at the origin. Show, by giving its equation, that it is possible to find a circle of given radius $r$ that bisects $C$ provided $r > a$. Show that no circle of radius $r$ bisects $C$ if $r \leqslant a$.

(ii) Let $C_1$ and $C_2$ be circles with centres at $(-d, 0)$ and $(d, 0)$ and radii $a_1$ and $a_2$, respectively, where $d > a_1$ and $d > a_2$. Let $D$ be a circle of radius $r$ that bisects both $C_1$ and $C_2$. Show that the $x$-coordinate of the centre of $D$ is $\frac{a_2^2 - a_1^2}{4d}$.

Obtain an expression in terms of $d$, $r$, $a_1$ and $a_2$ for the $y$-coordinate of the centre of $D$, and deduce that $r$ must satisfy $16r^2d^2 \geqslant \left( 4d^2 + (a_2 - a_1)^2 \right) \left( 4d^2 + (a_2 + a_1)^2 \right) .$

Hint

For part (i) note that the lines joining the centres of the two circles and one of the points where the bisection occurs form a right-angled triangle, so the radius of the new circle can be calculated. To show that no such circle can exist when $r < a$ note that the diametrically opposite points on $C$ must be a distance of $2a$ apart, and no two points on a circle of radius $r$ can be that far apart. For the case $r = a$ note that the new circle would be the same as $C$ (and so would have more than two intersection points).

For part (ii) a similar method can be used to deduce the distances between the centre of the new circle and each of $C_1$ and $C_2$. From these distances equations can be formed relating the $x$ and $y$ coordinates of the centre of the new circle. It is then an easy task to eliminate the $y$-coordinate of the centre of the circle from the equations to get the given value of the $x$-coordinate.

The expression for $y$ can easily be found by substituting back into the equations obtained from the distance between the centres of two of the circles. Once this is done, note that $y^2 \geq 0$ to obtain the final inequality.

Model Solution

Part (i)

Let $C$ be the circle of radius $a$ centred at the origin $O$, and let $D$ be a circle of radius $r$ centred at point $P$ that bisects $C$. The two intersection points are diametrically opposite on $C$, so they lie on a diameter of $C$. Call them $A$ and $B$, where $\vec{OA} = a\hat{\mathbf{u}}$ and $\vec{OB} = -a\hat{\mathbf{u}}$ for some unit vector $\hat{\mathbf{u}}$.

Since $A$ and $B$ both lie on $D$, the centre $P$ is equidistant from $A$ and $B$. Therefore $P$ lies on the perpendicular bisector of $AB$, which passes through $O$ and is perpendicular to $\hat{\mathbf{u}}$. Hence $\vec{OP} \cdot \hat{\mathbf{u}} = 0$.

Since $A$ lies on $D$, we have $|PA| = r$:

$|\vec{OP} - a\hat{\mathbf{u}}|^2 = r^2$ $|\vec{OP}|^2 - 2a(\vec{OP} \cdot \hat{\mathbf{u}}) + a^2 = r^2$ $|\vec{OP}|^2 + a^2 = r^2 \qquad \text{(since } \vec{OP} \cdot \hat{\mathbf{u}} = 0\text{)}$

So $|\vec{OP}| = \sqrt{r^2 - a^2}$, which requires $r \geq a$.

If $r > a$: Choose $P$ at distance $\sqrt{r^2 - a^2}$ from $O$ in any direction perpendicular to $\hat{\mathbf{u}}$. For instance, with $\hat{\mathbf{u}} = \hat{\mathbf{j}}$, take $P = (\sqrt{r^2 - a^2},\, 0)$. Then $D$ has equation

$(x - \sqrt{r^2 - a^2})^2 + y^2 = r^2.$

The intersection with $C$: substituting the condition $\mathbf{x} \cdot \hat{\mathbf{u}} = 0$ (i.e. $y = 0$ for this choice) into $x^2 + y^2 = a^2$ gives exactly the two points $(0, a)$ and $(0, -a)$, which are diametrically opposite on $C$. So $D$ bisects $C$.

If $r = a$: Then $|\vec{OP}| = 0$, so $P = O$ and $D$ coincides with $C$. The circles meet at every point (not exactly two), so $D$ does not bisect $C$.

If $r < a$: Then $r^2 - a^2 < 0$, so no real position for $P$ exists. No circle of radius $r$ can bisect $C$.

Therefore a circle of radius $r$ bisects $C$ if and only if $r > a$. $\qquad \text{(Q.E.D.)}$

Part (ii)

Let $D$ have centre $P = (x, y)$ and radius $r$. Since $D$ bisects $C_1$ (centre $(-d, 0)$, radius $a_1$), by the same argument as part (i):

$(x + d)^2 + y^2 = r^2 - a_1^2 \qquad \cdots (1)$

Since $D$ bisects $C_2$ (centre $(d, 0)$, radius $a_2$):

$(x - d)^2 + y^2 = r^2 - a_2^2 \qquad \cdots (2)$

Subtracting (2) from (1):

$(x + d)^2 - (x - d)^2 = (r^2 - a_1^2) - (r^2 - a_2^2)$

Expanding the left side: $(x^2 + 2dx + d^2) - (x^2 - 2dx + d^2) = 4dx$.

$4dx = a_2^2 - a_1^2$

$x = \frac{a_2^2 - a_1^2}{4d} \qquad \text{(Q.E.D.)}$

To find $y$, substitute into equation (1):

$y^2 = r^2 - a_1^2 - (x + d)^2$

$x + d = \frac{a_2^2 - a_1^2}{4d} + d = \frac{a_2^2 - a_1^2 + 4d^2}{4d}$

$y^2 = r^2 - a_1^2 - \frac{(4d^2 + a_2^2 - a_1^2)^2}{16d^2} \qquad \cdots (*)$

Since $y^2 \geq 0$, we require $r^2 - a_1^2 \geq \dfrac{(4d^2 + a_2^2 - a_1^2)^2}{16d^2}$, i.e.

$16d^2(r^2 - a_1^2) \geq (4d^2 + a_2^2 - a_1^2)^2.$

We show that $16d^2(r^2 - a_1^2) - (4d^2 + a_2^2 - a_1^2)^2$ equals the right side of the desired inequality minus the left side. Expanding:

$16d^2 r^2 - 16d^2 a_1^2 - (4d^2 + a_2^2 - a_1^2)^2 \geq 0$

$16d^2 r^2 \geq 16d^2 a_1^2 + (4d^2 + a_2^2 - a_1^2)^2$

We claim that $16d^2 a_1^2 + (4d^2 + a_2^2 - a_1^2)^2 = (4d^2 + (a_2 - a_1)^2)(4d^2 + (a_2 + a_1)^2)$.

Expanding the right side with $S = a_1 + a_2$ and $D = a_2 - a_1$ (noting $SD = a_2^2 - a_1^2$ and $S^2 + D^2 = 2(a_1^2 + a_2^2)$):

$(4d^2 + D^2)(4d^2 + S^2) = 16d^4 + 4d^2(S^2 + D^2) + S^2 D^2$ $= 16d^4 + 8d^2(a_1^2 + a_2^2) + (a_2^2 - a_1^2)^2$

And the left side:

$16d^2 a_1^2 + (4d^2 + a_2^2 - a_1^2)^2 = 16d^2 a_1^2 + 16d^4 + 8d^2(a_2^2 - a_1^2) + (a_2^2 - a_1^2)^2$ $= 16d^4 + 8d^2 a_2^2 + (a_2^2 - a_1^2)^2 = 16d^4 + 8d^2(a_1^2 + a_2^2) + (a_2^2 - a_1^2)^2 \checkmark$

These are identical, so the condition $y^2 \geq 0$ becomes:

$16r^2 d^2 \geq (4d^2 + (a_2 - a_1)^2)(4d^2 + (a_2 + a_1)^2). \qquad \text{(Q.E.D.)}$

Examiner Notes

The first part of this question was generally well answered, although a significant number of answers did not give the equation of the new circle. The case in part (i) where the two circles have the same radius was often not considered and the explanations for there not being such a circle in some cases were often not sufficiently clear. A significant number of candidates made the incorrect assumption in the second part that the centres of the three circles must lie on a straight line or attempted this part of the question with incorrect methods, such as equating the equations of the two given circles. In the final part of the question not all candidates realised that y² must be positive and were unable to obtain the required inequality by any other means.

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Question 8

Topic: 向量几何 (Vector Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

The diagram above shows two non-overlapping circles $C_1$ and $C_2$ of different sizes. The lines $L$ and $L'$ are the two common tangents to $C_1$ and $C_2$ such that the two circles lie on the same side of each of the tangents. The lines $L$ and $L'$ intersect at the point $P$ which is called the focus of $C_1$ and $C_2$.

(i) Let $\mathbf{x}_1$ and $\mathbf{x}_2$ be the position vectors of the centres of $C_1$ and $C_2$, respectively. Show that the position vector of $P$ is

$\frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1 - r_2},$

where $r_1$ and $r_2$ are the radii of $C_1$ and $C_2$, respectively.

(ii) The circle $C_3$ does not overlap either $C_1$ or $C_2$ and its radius, $r_3$, satisfies $r_1 \neq r_3 \neq r_2$. The focus of $C_1$ and $C_3$ is $Q$, and the focus of $C_2$ and $C_3$ is $R$. Show that $P$, $Q$ and $R$ lie on the same straight line.

(iii) Find a condition on $r_1$, $r_2$ and $r_3$ for $Q$ to lie half-way between $P$ and $R$.

Hint

The first part of the question follows from consideration of similar triangles in the diagram if the line through $P$ and the centres of the circles is added. For the second part, expressions can be written down for the position vectors of $Q$ and $R$ by noting that the same method as in part (i) will still apply. The vectors $\vec{PQ}$ and $\vec{QR}$ can then be compared to show that one is a multiple of the other.

For the final part of the question, note that $Q$ will lie halfway between $P$ and $R$ if $\vec{PQ} = \vec{QR}$.

Model Solution

Part (i)

Let $O_1$ and $O_2$ be the centres of $C_1$ and $C_2$ with position vectors $\mathbf{x}_1$ and $\mathbf{x}_2$, and radii $r_1$ and $r_2$ respectively (with $r_1 \neq r_2$).

Consider the line through $O_1$, $O_2$ and $P$. Drop perpendiculars from $O_1$ and $O_2$ to one of the common tangents $L$, meeting $L$ at $T_1$ and $T_2$ respectively. Then $O_1T_1 = r_1$ and $O_2T_2 = r_2$, with $O_1T_1 \perp L$ and $O_2T_2 \perp L$.

Since $O_1T_1 \parallel O_2T_2$ (both perpendicular to $L$), the triangles $\triangle PO_1T_1$ and $\triangle PO_2T_2$ are similar:

$\frac{PO_1}{PO_2} = \frac{O_1T_1}{O_2T_2} = \frac{r_1}{r_2}$

Let $P$ have position vector $\mathbf{p}$. Since $P$ lies on the line through $O_1$ and $O_2$ (on the side of the larger circle, beyond the larger centre), we can write:

$\mathbf{p} = \mathbf{x}_1 + t(\mathbf{x}_2 - \mathbf{x}_1) \quad \text{for some scalar } t.$

Then $PO_1 = |t| \cdot |\mathbf{x}_2 - \mathbf{x}_1|$ and $PO_2 = |1 - t| \cdot |\mathbf{x}_2 - \mathbf{x}_1|$.

From the ratio (taking $P$ beyond $O_2$ when $r_1 > r_2$, so $t > 1$):

$\frac{t}{t - 1} = \frac{r_1}{r_2} \implies r_2 t = r_1(t - 1) \implies t(r_2 - r_1) = -r_1 \implies t = \frac{r_1}{r_1 - r_2}$

Therefore:

$\mathbf{p} = \mathbf{x}_1 + \frac{r_1}{r_1 - r_2}(\mathbf{x}_2 - \mathbf{x}_1) = \frac{(r_1 - r_2)\mathbf{x}_1 + r_1\mathbf{x}_2 - r_1\mathbf{x}_1}{r_1 - r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1 - r_2}. \qquad \text{(Q.E.D.)}$

Part (ii)

By the same formula, the foci are:

$\mathbf{p} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1 - r_2}, \qquad \mathbf{q} = \frac{r_1\mathbf{x}_3 - r_3\mathbf{x}_1}{r_1 - r_3}, \qquad \mathbf{r} = \frac{r_2\mathbf{x}_3 - r_3\mathbf{x}_2}{r_2 - r_3}$

To show $P$, $Q$, $R$ are collinear, we show $\overrightarrow{PQ}$ is parallel to $\overrightarrow{QR}$.

Compute $\overrightarrow{PQ} = \mathbf{q} - \mathbf{p}$:

$\overrightarrow{PQ} = \frac{r_1\mathbf{x}_3 - r_3\mathbf{x}_1}{r_1 - r_3} - \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1 - r_2}$

With common denominator $(r_1 - r_2)(r_1 - r_3)$:

$= \frac{(r_1 - r_2)(r_1\mathbf{x}_3 - r_3\mathbf{x}_1) - (r_1 - r_3)(r_1\mathbf{x}_2 - r_2\mathbf{x}_1)}{(r_1 - r_2)(r_1 - r_3)}$

Expanding the numerator and collecting by $\mathbf{x}_1$, $\mathbf{x}_2$, $\mathbf{x}_3$:

• Coefficient of $\mathbf{x}_1$: $-(r_1 - r_2)r_3 + (r_1 - r_3)r_2 = -r_1 r_3 + r_2 r_3 + r_1 r_2 - r_2 r_3 = r_1(r_2 - r_3)$

• Coefficient of $\mathbf{x}_2$: $-(r_1 - r_3)r_1 = -r_1(r_1 - r_3)$

• Coefficient of $\mathbf{x}_3$: $(r_1 - r_2)r_1 = r_1(r_1 - r_2)$

So:

$\overrightarrow{PQ} = \frac{r_1}{(r_1 - r_2)(r_1 - r_3)}\left[(r_2 - r_3)\mathbf{x}_1 - (r_1 - r_3)\mathbf{x}_2 + (r_1 - r_2)\mathbf{x}_3\right]$

Compute $\overrightarrow{QR} = \mathbf{r} - \mathbf{q}$:

$\overrightarrow{QR} = \frac{r_2\mathbf{x}_3 - r_3\mathbf{x}_2}{r_2 - r_3} - \frac{r_1\mathbf{x}_3 - r_3\mathbf{x}_1}{r_1 - r_3}$

With common denominator $(r_1 - r_3)(r_2 - r_3)$:

$= \frac{(r_1 - r_3)(r_2\mathbf{x}_3 - r_3\mathbf{x}_2) - (r_2 - r_3)(r_1\mathbf{x}_3 - r_3\mathbf{x}_1)}{(r_1 - r_3)(r_2 - r_3)}$

Expanding and collecting:

• Coefficient of $\mathbf{x}_1$: $(r_2 - r_3)r_3 = r_3(r_2 - r_3)$

• Coefficient of $\mathbf{x}_2$: $-(r_1 - r_3)r_3 = -r_3(r_1 - r_3)$

• Coefficient of $\mathbf{x}_3$: $(r_1 - r_3)r_2 - (r_2 - r_3)r_1 = r_1 r_2 - r_2 r_3 - r_1 r_2 + r_1 r_3 = r_3(r_1 - r_2)$

So:

$\overrightarrow{QR} = \frac{r_3}{(r_1 - r_3)(r_2 - r_3)}\left[(r_2 - r_3)\mathbf{x}_1 - (r_1 - r_3)\mathbf{x}_2 + (r_1 - r_2)\mathbf{x}_3\right]$

Both $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ are scalar multiples of the same vector $\mathbf{v} = (r_2 - r_3)\mathbf{x}_1 - (r_1 - r_3)\mathbf{x}_2 + (r_1 - r_2)\mathbf{x}_3$, so they are parallel and $P$, $Q$, $R$ are collinear. $\qquad \text{(Q.E.D.)}$

In fact $\overrightarrow{PQ} = \dfrac{r_1(r_2 - r_3)}{r_3(r_1 - r_2)} \overrightarrow{QR}$.

Part (iii)

For $Q$ to lie halfway between $P$ and $R$, we need $\overrightarrow{PQ} = \overrightarrow{QR}$.

From part (ii), $\overrightarrow{PQ} = \lambda \overrightarrow{QR}$ where:

$\lambda = \frac{r_1/(r_1 - r_2)}{r_3/(r_2 - r_3)} = \frac{r_1(r_2 - r_3)}{r_3(r_1 - r_2)}$

Setting $\lambda = 1$:

$\frac{r_1(r_2 - r_3)}{r_3(r_1 - r_2)} = 1$

$r_1(r_2 - r_3) = r_3(r_1 - r_2)$

$r_1 r_2 - r_1 r_3 = r_1 r_3 - r_2 r_3$

$r_1 r_2 + r_2 r_3 = 2r_1 r_3$

$r_2(r_1 + r_3) = 2r_1 r_3$

$r_2 = \frac{2r_1 r_3}{r_1 + r_3}.$

This is the required condition: $r_2$ must be the harmonic mean of $r_1$ and $r_3$. $\qquad \text{(Q.E.D.)}$

We can verify this is consistent with $r_1 \neq r_2 \neq r_3$: for example $r_1 = 1$, $r_3 = 3$ gives $r_2 = \frac{6}{4} = \frac{3}{2}$, which satisfies all three being distinct.

Examiner Notes

This was one of the least attempted questions on the paper and the average score for the question was quite low. However, there were a number of very good answers to the question. Part (i) was answered correctly by the majority of candidates, but part (ii) was approached in a much more complicated manner than necessary by many candidates, attempting to work out the equation of the line rather than comparing vectors in its direction. Where the vectors were considered, solutions could have been made clearer by better grouping of the terms. A number of solutions referred to division of vectors rather than comparing coefficients. In the final part some candidates did not identify the simplest relationship between the vectors to ensure that Q lies halfway between P and R. Generally, more complicated relationships did not lead to correct solutions to this part of the question.