STEP2 2024 -- Pure Mathematics

STEP2 2024 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2024  |  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	纯数	Standard	[“等差数列求和公式”, “奇偶性分析”, “模4分析”, “Pell方程递推关系构造”]
2	纯数	Challenging	[“二项展开”, “逐项积分”, “部分分式分解（两次）”, “识别标准积分”]
3	纯数	Challenging	[“半角公式参数化单位圆”, “正切加法公式”, “旋转”, “反射的几何解释”, “逆变换”]
4	纯数	Challenging	[“点积法证等角”, “向量线性组合”, “利用线性无关性比较系数”, “对称性论证”]
5	纯数	Challenging	[“配方法”, “平方差公式”, “立方差分解”, “非负二次型”]
6	纯数	Challenging	[“归纳法”, “二项展开一般项”, “级数乘积=恒等式：(1-x)⁻¹·(1-x)^{-1/2}=(1-x)^{-3/2}“]
7	纯数	Hard	[“圆的方程”, “代换 y=k 分析水平截线”, “四次方程判别式”, “反证法证曲线位置关系”, “偶函数对称性”]
8	纯数	Hard	[“AM-GM不等式”, “单调有界原理”, “压缩映射论证 xₙ-yₙ→0”, “积分代换”, “偶函数积分对称性”, “AGM迭代不变量”]

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Question 1

Topic: 纯数  |  Difficulty: Standard  |  Marks: 20

Problem

1 In the equality $4 + 5 + 6 + 7 + 8 = 9 + 10 + 11,$ the sum of the five consecutive integers from 4 upwards is equal to the sum of the next three consecutive integers.

Throughout this question, the variables $n$, $k$ and $c$ represent positive integers.

(i) Show that the sum of the $n + k$ consecutive integers from $c$ upwards is equal to the sum of the next $n$ consecutive integers if and only if $2n^2 + k = 2ck + k^2.$

(ii) Find the set of possible values of $n$, and the corresponding values of $c$, in each of the cases (a) $k = 1$ (b) $k = 2$.

(iii) Show that there are no solutions for $c$ and $n$ if $k = 4$.

(iv) Consider now the case where $c = 1$. (a) Find two possible values of $k$ and the corresponding values of $n$. (b) Show, given a possible value $N$ of $n$, and the corresponding value $K$ of $k$, that $N' = 3N + 2K + 1$ will also be a possible value of $n$, with $K' = 4N + 3K + 1$ as the corresponding value of $k$. (c) Find two further possible values of $k$ and the corresponding values of $n$.

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4

Hint

(i) The two sums are $\frac{1}{2}(n+k)(2c+(n+k-1))$ and $\frac{1}{2}n(2(c+n+k)+(n-1))$ [B1] Difference simplifies to $\frac{1}{2}(2ck+k^2-2n^2-k)$ [M1] Two sums are equal if and only if the difference is 0 if and only $2n^2+k=2ck+k^2$ [A1] (ii) a If $k=1$, require $n^2=c$. Any value of $n$ is possible. [B1] b If $k=2$, require $n^2=2c+1$ [M1] $n$ can be any odd value, [A1] and $c = \frac{n^2-1}{2}$ [A1] (iii) If $k=4$, require $n^2=4c+6$ [B1] RHS has a factor of 2, but not a factor of 4 … [E1] … so cannot be a square. [E1] (iv) a If $c=1$, require $2n^2=k(k+1)$: $k=1, n=1$ [B1] $k=8, n=6$ [B1] b Since $(N, K)$ is a solution: $2N^2 + K = 2K + K^2$ or $2N^2 = K(K + 1)$ [B1] $2(3N + 2K + 1)^2 + (4N + 3K + 1) =$ $18N^2 + 8K^2 + 3 + 24NK + 16N + 11K$ OR $2(3N + 2K + 1)^2 =$ $18N^2 + 8K^2 + 2 + 24NK + 12N + 8K$ [M1] $2(4N + 3K + 1) + (4N + 3K + 1)^2 =$ $16N^2 + 9K^2 + 3 + 24NK + 16N + 12K$ OR $(4N + 3K + 1)(4N + 3K + 2) =$ $16N^2 + 9K^2 + 2 + 24NK + 12N + 9K$ [M1] Difference between the two expressions that use $= 2N^2 - K^2 - K$ [M1] $= 0$, so $N' = (3N + 2K + 1)$ is a possible value for $n$, with $K' = (4N + 3K + 1)$ as the corresponding value of $k$. [A1] c Use of recurrence with one of the pairs found in part (iv)(a) [M1] $k = 49, n = 35$ [A1] $k = 288, n = 204$ [A1]

Model Solution

Part (i)

The $n + k$ consecutive integers from $c$ upwards are $c, c+1, \ldots, c+n+k-1$. Their sum is

$S_1 = \sum_{i=0}^{n+k-1}(c+i) = (n+k)c + \frac{(n+k)(n+k-1)}{2} = \frac{(n+k)\bigl(2c + n + k - 1\bigr)}{2}.$

The next $n$ consecutive integers are $c+n+k, c+n+k+1, \ldots, c+2n+k-1$. Their sum is

$S_2 = \sum_{i=0}^{n-1}(c+n+k+i) = n(c+n+k) + \frac{n(n-1)}{2} = \frac{n\bigl(2c + 3n + 2k - 1\bigr)}{2}.$

We require $S_1 = S_2$:

$(n+k)(2c + n + k - 1) = n(2c + 3n + 2k - 1).$

Expanding the left side:

$(n+k)(2c) + (n+k)(n+k-1) = 2c(n+k) + n^2 + 2nk + k^2 - n - k.$

Expanding the right side:

$2cn + 3n^2 + 2nk - n.$

Setting these equal and cancelling $2cn$ from both sides:

$2ck + n^2 + 2nk + k^2 - n - k = 3n^2 + 2nk - n.$

Cancelling $2nk$ and $-n$ from both sides:

$2ck + k^2 - k = 2n^2.$

Rearranging:

$2n^2 + k = 2ck + k^2. \qquad (*)$

Since every step is reversible (each was an equivalence), the two sums are equal if and only if $(*)$ holds.

Part (ii)(a): $k = 1$

Substituting $k = 1$ into $(*)$:

$2n^2 + 1 = 2c + 1 \implies 2n^2 = 2c \implies c = n^2.$

For every positive integer $n$, $c = n^2$ is a positive integer, so any positive integer $n$ is possible, with $c = n^2$.

Part (ii)(b): $k = 2$

Substituting $k = 2$ into $(*)$:

$2n^2 + 2 = 4c + 4 \implies 2n^2 = 4c + 2 \implies n^2 = 2c + 1.$

For $c$ to be a positive integer, $n^2$ must be odd, so $n$ must be odd. Then $c = \frac{n^2 - 1}{2}$.

The set of possible values of $n$ is all odd positive integers $\{1, 3, 5, 7, \ldots\}$, with corresponding $c = \frac{n^2 - 1}{2}$.

Part (iii): $k = 4$

Substituting $k = 4$ into $(*)$:

$2n^2 + 4 = 8c + 16 \implies 2n^2 = 8c + 12 \implies n^2 = 4c + 6.$

Now $n^2 = 4c + 6$ is even, so $n$ is even. Write $n = 2m$ for a positive integer $m$:

$4m^2 = 4c + 6 \implies 2m^2 = 2c + 3.$

The left side $2m^2$ is even, but the right side $2c + 3$ is odd. This is a contradiction, so there are no solutions for $c$ and $n$ when $k = 4$.

Part (iv)(a): $c = 1$

Substituting $c = 1$ into $(*)$:

$2n^2 + k = 2k + k^2 \implies 2n^2 = k^2 + k = k(k+1).$

We need $k(k+1) = 2n^2$ with $k, n$ positive integers. Checking small values:

• $k = 1$: $1 \cdot 2 = 2 \implies n^2 = 1 \implies n = 1$. Solution: $(k, n) = (1, 1)$.
• $k = 2$: $2 \cdot 3 = 6$, $n^2 = 3$. Not a perfect square.
• $k = 3$: $3 \cdot 4 = 12$, $n^2 = 6$. Not a perfect square.
• $k = 4$: $4 \cdot 5 = 20$, $n^2 = 10$. Not a perfect square.
• $k = 8$: $8 \cdot 9 = 72$, $n^2 = 36 \implies n = 6$. Solution: $(k, n) = (8, 6)$.

Two possible values: $k = 1$ (with $n = 1$) and $k = 8$ (with $n = 6$).

Part (iv)(b)

We are given that $(N, K)$ satisfies $2N^2 = K(K+1)$. Set $N' = 3N + 2K + 1$ and $K' = 4N + 3K + 1$. We must show $2(N')^2 = K'(K'+1)$.

Left side: $2(N')^2 = 2(3N + 2K + 1)^2$.

$2(3N + 2K + 1)^2 = 2\bigl(9N^2 + 12NK + 4K^2 + 6N + 4K + 1\bigr)$ $= 18N^2 + 24NK + 8K^2 + 12N + 8K + 2.$

Using $2N^2 = K^2 + K$, we substitute $18N^2 = 9(K^2 + K)$:

$= 9K^2 + 9K + 24NK + 8K^2 + 12N + 8K + 2 = 17K^2 + 17K + 24NK + 12N + 2.$

Right side: $K'(K'+1) = (4N + 3K + 1)(4N + 3K + 2)$.

$(4N + 3K + 1)(4N + 3K + 2) = (4N + 3K)^2 + 3(4N + 3K) + 2$ $= 16N^2 + 24NK + 9K^2 + 12N + 9K + 2.$

Using $16N^2 = 8(K^2 + K)$:

$= 8K^2 + 8K + 24NK + 9K^2 + 12N + 9K + 2 = 17K^2 + 17K + 24NK + 12N + 2.$

Both sides are equal, so $2(N')^2 = K'(K'+1)$, confirming that $N' = 3N + 2K + 1$ is a possible value of $n$ with $K' = 4N + 3K + 1$ as the corresponding value of $k$.

Part (iv)(c)

Starting from the solution $(N, K) = (1, 1)$ found in part (iv)(a), apply the recurrence from part (iv)(b):

$N' = 3(1) + 2(1) + 1 = 6, \qquad K' = 4(1) + 3(1) + 1 = 8.$

This gives $(n, k) = (6, 8)$, already found in part (iv)(a). Apply the recurrence again with $(N, K) = (6, 8)$:

$N' = 3(6) + 2(8) + 1 = 35, \qquad K' = 4(6) + 3(8) + 1 = 49.$

Check: $2(35)^2 = 2450 = 49 \times 50$. Confirmed.

Apply the recurrence once more with $(N, K) = (35, 49)$:

$N' = 3(35) + 2(49) + 1 = 204, \qquad K' = 4(35) + 3(49) + 1 = 288.$

Check: $2(204)^2 = 83232 = 288 \times 289$. Confirmed.

Two further possible values: $k = 49$ (with $n = 35$) and $k = 288$ (with $n = 204$).

Examiner Notes

大量尝试。常见错误：求和上下限选错、混淆平方与平方根、没有解释双向等价

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Question 2

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

2 In this question, you need not consider issues of convergence.

(i) Find the binomial series expansion of $(8 + x^3)^{-1}$, valid for $|x| < 2$.

Hence show that, for each integer $m \geqslant 0$,

$\int_{0}^{1} \frac{x^m}{8 + x^3} \mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1} \right) .$

(ii) Show that

$\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k + 3} - \frac{2}{3k + 2} + \frac{4}{3k + 1} \right) = \int_{0}^{1} \frac{1}{x + 2} \mathrm{d}x ,$

and evaluate the integral.

(iii) Show that

$\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k + 1)}{(3k + 1)(3k + 2)} \right) = \pi \sqrt{a} - \ln b ,$

where $a$ and $b$ are integers which you should determine.

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5

Hint

(i) $(8 + x^3)^{-1} = \frac{1}{8} \left( 1 + \frac{x^3}{8} \right)^{-1}$ $= \frac{1}{8} \left( 1 - \frac{x^3}{8} + \frac{x^6}{64} - \frac{x^9}{512} + \dots \right)$ [M1] $= \frac{1}{8} \sum_{k=0}^{\infty} (-1)^k \left( \frac{x}{2} \right)^{3k}$ [A1] $\int_0^1 \frac{x^m}{8 + x^3} dx = \int_0^1 \frac{1}{8} \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3k}} x^{m+3k} dx$ [M1] $= \frac{1}{8} \sum_{k=0}^{\infty} \left[ \frac{(-1)^k}{2^{3k}} \frac{x^{m+3k+1}}{m + 3k + 1} \right]_0^1$ [A1] $= \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \frac{1}{m + 3k + 1} \right)$ [A1] ROW_SPAN_CONTINUE ii $\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k + 3} \right) = \int_0^1 \frac{x^2}{8 + x^3} dx$ [M1] $\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{-2}{3k + 2} \right) = \int_0^1 \frac{-2x}{8 + x^3} dx$ $\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{4}{3k + 1} \right) = \int_0^1 \frac{4}{8 + x^3} dx$ $\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k + 3} - \frac{2}{3k + 2} + \frac{4}{3k + 1} \right)$ [A1] $= \int_0^1 \frac{x^2 - 2x + 4}{8 + x^3} dx$ $= \int_0^1 \frac{x^2 - 2x + 4}{(x + 2)(x^2 - 2x + 4)} dx$ [M1] $= \int_0^1 \frac{1}{x + 2} dx$ [A1] $= [\ln(x + 2)]_0^1 = \ln \left( \frac{3}{2} \right)$ [B1] ROW_SPAN_CONTINUE iii $\frac{72(2k + 1)}{(3k + 1)(3k + 2)} = \frac{24}{3k + 1} + \frac{24}{3k + 2}$ [M1 A1] $\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \frac{72(2k + 1)}{(3k + 1)(3k + 2)} = \int_0^1 \frac{24x + 24}{8 + x^3} dx$ [A1] $= \int_0^1 \frac{2(x + 8)}{x^2 - 2x + 4} - \frac{2}{x + 2} dx$ [M1 A1] $= \int_0^1 \frac{2(x - 1)}{x^2 - 2x + 4} + \frac{18}{x^2 - 2x + 4} - \frac{2}{x + 2} dx$ [M1] $= [\ln(x^2 - 2x + 4)]_0^1 \dots$ [M1] $\dots + \left[ 6\sqrt{3} \arctan \left( \frac{x - 1}{\sqrt{3}} \right) \right]_0^1 \dots$ [M1] $\dots - [2\ln(x + 2)]_0^1$ [A1] $= \ln 3 - \ln 4 - 2 \ln 3 + 2 \ln 2 + 6\sqrt{3} \cdot \frac{\pi}{6}$ [A1] $= \pi\sqrt{3} - \ln 3$

Model Solution

Part (i)

We write

$(8 + x^3)^{-1} = \frac{1}{8}\left(1 + \frac{x^3}{8}\right)^{-1}.$

Using the binomial series $(1 + u)^{-1} = \sum_{k=0}^{\infty}(-1)^k u^k$ for $|u| < 1$, with $u = \frac{x^3}{8}$ (valid for $|x^3| < 8$, i.e.\ $|x| < 2$):

$(8 + x^3)^{-1} = \frac{1}{8}\sum_{k=0}^{\infty}(-1)^k \frac{x^{3k}}{8^k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} x^{3k}.$

Now integrate term by term for $m \geqslant 0$:

$\int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \int_0^1 \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} x^{m+3k}\,\mathrm{d}x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \int_0^1 x^{m+3k}\,\mathrm{d}x.$

Since $\int_0^1 x^{m+3k}\,\mathrm{d}x = \frac{1}{m + 3k + 1}$:

$\int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1}. \qquad \text{(as required)}$

Part (ii)

From part (i), with $m = 2$, $m = 1$, and $m = 0$ respectively:

$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{3(k+1)}}\cdot\frac{1}{3k+3} = \int_0^1 \frac{x^2}{8+x^3}\,\mathrm{d}x,$

$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{3(k+1)}}\cdot\frac{-2}{3k+2} = -2\int_0^1 \frac{x}{8+x^3}\,\mathrm{d}x,$

$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{3(k+1)}}\cdot\frac{4}{3k+1} = 4\int_0^1 \frac{1}{8+x^3}\,\mathrm{d}x.$

Adding these three identities:

$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{3(k+1)}}\left(\frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1}\right) = \int_0^1 \frac{x^2 - 2x + 4}{8 + x^3}\,\mathrm{d}x.$

Now factor $8 + x^3 = (x+2)(x^2 - 2x + 4)$ (sum of cubes: $x^3 + 2^3$). Therefore

$\frac{x^2 - 2x + 4}{8 + x^3} = \frac{x^2 - 2x + 4}{(x+2)(x^2 - 2x + 4)} = \frac{1}{x+2}.$

So the integral becomes

$\int_0^1 \frac{1}{x+2}\,\mathrm{d}x = \Bigl[\ln(x+2)\Bigr]_0^1 = \ln 3 - \ln 2 = \ln\frac{3}{2}.$

Part (iii)

We first perform a partial fraction decomposition of $\frac{72(2k+1)}{(3k+1)(3k+2)}$. Write

$\frac{72(2k+1)}{(3k+1)(3k+2)} = \frac{A}{3k+1} + \frac{B}{3k+2}.$

Multiplying through:

$72(2k+1) = A(3k+2) + B(3k+1).$

Setting $k = -\frac{1}{3}$: $72 \cdot \frac{1}{3} = A \cdot 1$, so $A = 24$.

Setting $k = -\frac{2}{3}$: $72 \cdot \left(-\frac{1}{3}\right) = B \cdot (-1)$, so $B = 24$.

Therefore

$\frac{72(2k+1)}{(3k+1)(3k+2)} = \frac{24}{3k+1} + \frac{24}{3k+2}.$

Using part (i) with $m = 0$ and $m = 1$:

$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{3(k+1)}}\cdot\frac{24}{3k+1} = 24\int_0^1 \frac{1}{8+x^3}\,\mathrm{d}x,$

$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{3(k+1)}}\cdot\frac{24}{3k+2} = 24\int_0^1 \frac{x}{8+x^3}\,\mathrm{d}x.$

So the sum equals

$\int_0^1 \frac{24 + 24x}{8 + x^3}\,\mathrm{d}x = \int_0^1 \frac{24(x+1)}{(x+2)(x^2 - 2x + 4)}\,\mathrm{d}x.$

We perform partial fractions on $\frac{24(x+1)}{(x+2)(x^2-2x+4)}$. Write

$\frac{24(x+1)}{(x+2)(x^2-2x+4)} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 4}.$

Multiplying through:

$24(x+1) = A(x^2 - 2x + 4) + (Bx+C)(x+2).$

Setting $x = -2$: $24(-1) = A(4+4+4) = 12A$, so $A = -2$.

Expanding and comparing coefficients:

$24x + 24 = (-2 + B)x^2 + (4 + 2B + C)x + (-8 + 2C).$

Wait, let me redo this carefully.

$A(x^2 - 2x + 4) + (Bx+C)(x+2) = Ax^2 - 2Ax + 4A + Bx^2 + 2Bx + Cx + 2C$

$= (A+B)x^2 + (-2A + 2B + C)x + (4A + 2C).$

Comparing with $0 \cdot x^2 + 24x + 24$:

• $x^2$: $A + B = 0 \implies B = -A = 2$.
• $x^0$: $4A + 2C = 24 \implies -8 + 2C = 24 \implies C = 16$.
• $x^1$: $-2A + 2B + C = 4 + 4 + 16 = 24$. Confirmed.

So

$\frac{24(x+1)}{(x+2)(x^2-2x+4)} = \frac{-2}{x+2} + \frac{2x + 16}{x^2 - 2x + 4}.$

We rewrite $\frac{2x+16}{x^2-2x+4} = \frac{2(x-1)}{x^2-2x+4} + \frac{18}{x^2-2x+4}$.

The integral becomes

$\int_0^1 \frac{2(x-1)}{x^2-2x+4}\,\mathrm{d}x + \int_0^1 \frac{18}{x^2-2x+4}\,\mathrm{d}x - \int_0^1 \frac{2}{x+2}\,\mathrm{d}x.$

First integral: Let $u = x^2 - 2x + 4$, so $\mathrm{d}u = 2(x-1)\,\mathrm{d}x$.

$\int_0^1 \frac{2(x-1)}{x^2-2x+4}\,\mathrm{d}x = \Bigl[\ln(x^2-2x+4)\Bigr]_0^1 = \ln 3 - \ln 4 = \ln 3 - 2\ln 2.$

Second integral: Complete the square: $x^2 - 2x + 4 = (x-1)^2 + 3$.

$\int_0^1 \frac{18}{(x-1)^2 + 3}\,\mathrm{d}x = \frac{18}{\sqrt{3}}\left[\arctan\frac{x-1}{\sqrt{3}}\right]_0^1 = 6\sqrt{3}\left(\arctan 0 - \arctan\frac{-1}{\sqrt{3}}\right)$

$= 6\sqrt{3}\left(0 + \frac{\pi}{6}\right) = \pi\sqrt{3}.$

Third integral:

$\int_0^1 \frac{2}{x+2}\,\mathrm{d}x = 2\Bigl[\ln(x+2)\Bigr]_0^1 = 2\ln 3 - 2\ln 2.$

Combining:

$(\ln 3 - 2\ln 2) + \pi\sqrt{3} - (2\ln 3 - 2\ln 2) = \pi\sqrt{3} - \ln 3.$

Therefore $a = 3$ and $b = 3$.

Examiner Notes

约3/4尝试。部分人没认出可用展开式，转而用分部积分。部分分式需要分解两次。

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Question 3

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

3 The unit circle is the circle with radius 1 and centre the origin, $O$.

$N$ and $P$ are distinct points on the unit circle. $N$ has coordinates $(-1, 0)$, and $P$ has coordinates $(\cos \theta, \sin \theta)$, where $-\pi < \theta < \pi$. The line $NP$ intersects the $y$-axis at $Q$, which has coordinates $(0, q)$.

(i) Show that $q = \tan \frac{1}{2}\theta$.

(ii) In this part, $q \neq 1$.

(a) Let $f_1(q) = \frac{1 + q}{1 - q}$. Show that $f_1(q) = \tan \frac{1}{2}(\theta + \frac{1}{2}\pi)$.

(b) Let $Q_1$ be the point with coordinates $(0, f_1(q))$ and $P_1$ be the point of intersection (other than $N$) of the line $NQ_1$ and the unit circle. Describe geometrically the relationship between $P$ and $P_1$.

(iii) (a) $P_2$ is the image of $P$ under an anti-clockwise rotation about $O$ through angle $\frac{1}{3}\pi$. The line $NP_2$ intersects the $y$-axis at the point $Q_2$ with co-ordinates $(0, f_2(q))$. Find $f_2(q)$ in terms of $q$, for $q \neq \sqrt{3}$.

(b) In this part, $q \neq -1$. Let $f_3(q) = \frac{1 - q}{1 + q}$, let $Q_3$ be the point with coordinates $(0, f_3(q))$ and let $P_3$ be the point of intersection (other than $N$) of the line $NQ_3$ and the unit circle. Describe geometrically the relationship between $P$ and $P_3$.

(c) In this part, $0 < q < 1$. Let $f_4(q) = f_2^{-1}\left(f_3\left(f_2(q)\right)\right)$, let $Q_4$ be the point with coordinates $(0, f_4(q))$ and let $P_4$ be the point of intersection (other than $N$) of the line $NQ_4$ and the unit circle. Describe geometrically the relationship between $P$ and $P_4$.

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6

Hint

(i) Gradient of $NP$ is $\frac{\sin \theta}{1 + \cos \theta}$ $\left( = \frac{y}{1} \right)$ [M1] $y = \frac{2 \sin \left( \frac{1}{2} \theta \right) \cos \left( \frac{1}{2} \theta \right)}{1 + 2 \cos^2 \left( \frac{1}{2} \theta \right) - 1}$ [M1] $= \tan \left( \frac{1}{2} \theta \right)$ [A1] (ii) a $f_1(q) = \frac{\tan \frac{1}{4} \pi + \tan \frac{1}{2} \theta}{1 - \tan \frac{1}{4} \pi \tan \frac{1}{2} \theta}$ [B1 M1] $= \tan \frac{1}{2} \left( \theta + \frac{1}{2} \pi \right)$ [A1] b If the coordinates of $P_1$ are $(\cos \psi, \sin \psi)$: $f(q_1) = \tan \left( \frac{1}{2} \psi \right) = \tan \frac{1}{2} \left( \theta + \frac{1}{2} \pi \right)$ [M1] $P_1$ is the image of $P$ under rotation anticlockwise through a right angle about $O$. [B1 B1] (iii) a $f_2(q) = \tan \frac{1}{2} \left( \theta + \frac{1}{3} \pi \right)$ [M1] $= \frac{\tan \left( \frac{1}{6} \pi \right) + q}{1 - q \tan \left( \frac{1}{6} \pi \right)}$ [A1] $= \frac{1 + \sqrt{3}q}{\sqrt{3} - q}$ [A1] b $f_3(q) = f_1(-q)$, so [M1] $f_3(q) = \tan \frac{1}{2} \left( \frac{1}{2} \pi - \theta \right)$ [A1] So the coordinates of $P_3$ are $(\sin \theta, \cos \theta)$ [M1] $P_3$ is the image of $P$ under reflection in $y = x$ [A1] c $P_4$ is the image of $P$ under the following sequence of transformations: Rotation anticlockwise through $\frac{1}{3} \pi$ [M1] Reflection in $y = x$ Rotation clockwise through $-\frac{1}{3} \pi$ [A1] A point is invariant under this transformation if its image under the rotation anticlockwise through $\frac{1}{3} \pi$ lies on the line $y = x$ [M1] … making an angle of $-\frac{\pi}{12}$ with the positive x-axis [A1]

Model Solution

Part (i)

The gradient of the line $NP$ is

$m = \frac{\sin\theta - 0}{\cos\theta - (-1)} = \frac{\sin\theta}{1 + \cos\theta}.$

Using the double-angle identities $\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ and $\cos\theta = 2\cos^2\tfrac{\theta}{2} - 1$, we have

$1 + \cos\theta = 2\cos^2\tfrac{\theta}{2},$

so

$m = \frac{2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}}{2\cos^2\tfrac{\theta}{2}} = \frac{\sin\tfrac{\theta}{2}}{\cos\tfrac{\theta}{2}} = \tan\tfrac{\theta}{2}.$

The equation of line $NP$ (through $N(-1, 0)$ with gradient $m$) is $y = m(x + 1)$. Setting $x = 0$:

$q = m(0 + 1) = m = \tan\tfrac{\theta}{2}.$

Part (ii)(a)

Since $q = \tan\tfrac{\theta}{2}$, we have

$f_1(q) = \frac{1 + q}{1 - q} = \frac{1 + \tan\tfrac{\theta}{2}}{1 - \tan\tfrac{\theta}{2}}.$

Using the tangent addition formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ with $A = \frac{\pi}{4}$ and $B = \frac{\theta}{2}$, and noting that $\tan\frac{\pi}{4} = 1$:

$f_1(q) = \frac{\tan\frac{\pi}{4} + \tan\frac{\theta}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{\theta}{2}} = \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \tan\frac{1}{2}\left(\theta + \frac{\pi}{2}\right).$

Part (ii)(b)

Suppose $P_1$ has coordinates $(\cos\psi, \sin\psi)$ on the unit circle. By the same argument as part (i), the $y$-intercept of line $NP_1$ is $\tan\tfrac{\psi}{2}$. But the $y$-intercept of line $NQ_1$ is $f_1(q)$, so

$\tan\tfrac{\psi}{2} = f_1(q) = \tan\frac{1}{2}\left(\theta + \frac{\pi}{2}\right).$

Since $P_1$ is the intersection other than $N$, we have $\psi = \theta + \frac{\pi}{2}$.

Therefore $P_1$ is the image of $P$ under an anti-clockwise rotation about $O$ through angle $\frac{\pi}{2}$ (a right angle).

Part (iii)(a)

$P_2$ is obtained by rotating $P$ anti-clockwise through $\frac{\pi}{3}$ about $O$, so $P_2 = (\cos(\theta + \frac{\pi}{3}),\, \sin(\theta + \frac{\pi}{3}))$.

By part (i), the $y$-intercept of $NP_2$ is

$f_2(q) = \tan\frac{1}{2}\left(\theta + \frac{\pi}{3}\right).$

We expand using the tangent addition formula with $A = \frac{\theta}{2}$ and $B = \frac{\pi}{6}$:

$f_2(q) = \frac{\tan\frac{\theta}{2} + \tan\frac{\pi}{6}}{1 - \tan\frac{\theta}{2}\tan\frac{\pi}{6}}.$

Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $q = \tan\frac{\theta}{2}$:

$f_2(q) = \frac{q + \frac{1}{\sqrt{3}}}{1 - \frac{q}{\sqrt{3}}} = \frac{\sqrt{3}\,q + 1}{\sqrt{3} - q}.$

This is valid for $q \neq \sqrt{3}$ (where the denominator vanishes).

Part (iii)(b)

Note that $f_3(q) = \frac{1-q}{1+q}$. We can write

$f_3(q) = \frac{1 - \tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}} = \frac{\tan\frac{\pi}{4} - \tan\frac{\theta}{2}}{1 + \tan\frac{\pi}{4}\tan\frac{\theta}{2}} = \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \tan\frac{1}{2}\left(\frac{\pi}{2} - \theta\right).$

If $P_3$ has coordinates $(\cos\psi, \sin\psi)$, then $\tan\frac{\psi}{2} = f_3(q) = \tan\frac{1}{2}(\frac{\pi}{2} - \theta)$, so $\psi = \frac{\pi}{2} - \theta$.

The coordinates of $P_3$ are

$\left(\cos\left(\frac{\pi}{2} - \theta\right),\, \sin\left(\frac{\pi}{2} - \theta\right)\right) = (\sin\theta,\, \cos\theta).$

Since $P = (\cos\theta, \sin\theta)$ and $P_3 = (\sin\theta, \cos\theta)$, $P_3$ is the image of $P$ under reflection in the line $y = x$.

Part (iii)(c)

We first find $f_2^{-1}$. If $v = f_2(q) = \frac{\sqrt{3}\,q + 1}{\sqrt{3} - q}$, then

$v(\sqrt{3} - q) = \sqrt{3}\,q + 1 \implies v\sqrt{3} - vq = \sqrt{3}\,q + 1 \implies q(v + \sqrt{3}) = v\sqrt{3} - 1,$

$f_2^{-1}(v) = \frac{v\sqrt{3} - 1}{v + \sqrt{3}}.$

Now $f_3(f_2(q))$: let $w = f_2(q) = \frac{\sqrt{3}\,q + 1}{\sqrt{3} - q}$, then

$f_3(w) = \frac{1 - w}{1 + w} = \frac{1 - \frac{\sqrt{3}\,q + 1}{\sqrt{3} - q}}{1 + \frac{\sqrt{3}\,q + 1}{\sqrt{3} - q}} = \frac{(\sqrt{3} - q) - (\sqrt{3}\,q + 1)}{(\sqrt{3} - q) + (\sqrt{3}\,q + 1)} = \frac{\sqrt{3} - 1 - q(\sqrt{3} + 1)}{\sqrt{3} + 1 + q(\sqrt{3} - 1)}.$

Now apply $f_2^{-1}$:

$f_4(q) = f_2^{-1}(f_3(w)) = \frac{f_3(w)\sqrt{3} - 1}{f_3(w) + \sqrt{3}}.$

Let $s = \sqrt{3} - 1$ and $t = \sqrt{3} + 1$ (note $st = 2$), so $f_3(w) = \frac{s - qt}{t + qs}$. Then

$\text{Numerator} = \frac{(s - qt)\sqrt{3} - (t + qs)}{t + qs} = \frac{s\sqrt{3} - qt\sqrt{3} - t - qs}{t + qs}.$

Computing the numerator polynomial in $q$:

$s\sqrt{3} - t = (\sqrt{3}-1)\sqrt{3} - (\sqrt{3}+1) = 3 - \sqrt{3} - \sqrt{3} - 1 = 2 - 2\sqrt{3} = -2s,$

$-t\sqrt{3} - s = -(\sqrt{3}+1)\sqrt{3} - (\sqrt{3}-1) = -(3+\sqrt{3}) - \sqrt{3} + 1 = -2 - 2\sqrt{3} = -2t.$

So the numerator is $\frac{-2s - 2qt}{t + qs}$.

$\text{Denominator} = \frac{(s - qt) + \sqrt{3}(t + qs)}{t + qs} = \frac{s + \sqrt{3}\,t + q(s\sqrt{3} - t)}{t + qs}.$

Computing:

$s + \sqrt{3}\,t = (\sqrt{3}-1) + \sqrt{3}(\sqrt{3}+1) = \sqrt{3} - 1 + 3 + \sqrt{3} = 2 + 2\sqrt{3} = 2t,$

$s\sqrt{3} - t = 2 - 2\sqrt{3} = -2s.$

So the denominator is $\frac{2t - 2qs}{t + qs}$.

Therefore

$f_4(q) = \frac{-2(s + qt)}{2(t - qs)} = -\frac{s + qt}{t - qs} = -\frac{(\sqrt{3}-1) + (\sqrt{3}+1)q}{(\sqrt{3}+1) - (\sqrt{3}-1)q}.$

Using the half-angle interpretation: $f_2$ corresponds to adding $\frac{\pi}{3}$ to the angle $\theta$, and $f_3$ corresponds to replacing $\theta$ with $\frac{\pi}{2} - \theta$ (reflection in $y = x$). The composition $f_4 = f_2^{-1} \circ f_3 \circ f_2$ corresponds to the transformation

$\theta \;\xmapsto{f_2}\; \theta + \frac{\pi}{3} \;\xmapsto{f_3}\; \frac{\pi}{2} - \left(\theta + \frac{\pi}{3}\right) = \frac{\pi}{6} - \theta \;\xmapsto{f_2^{-1}}\; \frac{\pi}{6} - \theta - \frac{\pi}{3} = -\theta - \frac{\pi}{6} = -\left(\theta + \frac{\pi}{6}\right).$

So $P_4$ has angle $-\theta - \frac{\pi}{6}$. The midpoint of the arc from angle $\theta$ to angle $-\theta - \frac{\pi}{6}$ has angle $\frac{\theta + (-\theta - \pi/6)}{2} = -\frac{\pi}{12}$.

Therefore $P_4$ is the image of $P$ under reflection in the diameter of the unit circle that makes angle $-\frac{\pi}{12}$ with the positive $x$-axis (i.e., the line through $O$ at angle $-\frac{\pi}{12}$).

Examiner Notes

约3/4尝试，少数完全解。画图很有帮助。

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Question 4

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

4 In this question, if $O$, $C$ and $D$ are non-collinear points in three dimensional space, we will call the non-zero vector $\mathbf{v}$ a bisecting vector for angle $COD$ if $\mathbf{v}$ lies in the plane $COD$, the angle between $\mathbf{v}$ and $\overrightarrow{OC}$ is equal to the angle between $\mathbf{v}$ and $\overrightarrow{OD}$, and both angles are less than $90^\circ$.

(i) Let $O$, $X$ and $Y$ be non-collinear points in three-dimensional space, and define $\mathbf{x} = \overrightarrow{OX}$ and $\mathbf{y} = \overrightarrow{OY}$.

Let $\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}$.

(a) Show that $\mathbf{b}$ is a bisecting vector for angle $XOY$.

Explain, using a diagram, why any other bisecting vector for angle $XOY$ is a positive multiple of $\mathbf{b}$.

(b) Find the value of $\lambda$ such that the point $B$, defined by $\overrightarrow{OB} = \lambda\mathbf{b}$, lies on the line $XY$. Find also the ratio in which the point $B$ divides $XY$.

(c) Show, in the case when $OB$ is perpendicular to $XY$, that the triangle $XOY$ is isosceles.

(ii) Let $O$, $P$, $Q$ and $R$ be points in three-dimensional space, no three of which are collinear. A bisecting vector is chosen for each of the angles $POQ$, $QOR$ and $ROP$. Show that the three angles between them are either all acute, all obtuse or all right angles.

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7

Hint

(i) a $b$ is a linear combination of $x$ and $y$, so it must lie in the plane $OXY$ [B1] $b \cdot x = (|x|y + |y|x) \cdot x = |x|y \cdot x + |y||x|^2$ [M1] If $\theta$ is the angle between $x$ and $b$, then $\cos \theta = \frac{\boldsymbol{b} \cdot \boldsymbol{x}}{|\boldsymbol{b}||\boldsymbol{x}|} = \frac{\boldsymbol{x} \cdot \boldsymbol{y} + |\boldsymbol{x}||\boldsymbol{y}|}{|\boldsymbol{b}|}$ [M1] Similarly, $\frac{\boldsymbol{b} \cdot \boldsymbol{y}}{|\boldsymbol{b}||\boldsymbol{y}|} = \frac{\boldsymbol{x} \cdot \boldsymbol{y} + |\boldsymbol{x}||\boldsymbol{y}|}{|\boldsymbol{b}|}$ so the angle between $b$ and $y$ is also $\theta$. [A1] Since $\boldsymbol{x} \cdot \boldsymbol{y} + |\boldsymbol{x}||\boldsymbol{y}| > 0, \cos \theta > 0$ and so the angle is less than $90^\circ$ [E1] A sketch to indicate why any other bisecting vector is a positive multiple of this. [E1] b The vector $\vec{XB} = \lambda \boldsymbol{b} - \boldsymbol{x}$ must be parallel to the vector $\vec{XY} = \boldsymbol{y} - \boldsymbol{x}$. For some $\mu$: $\lambda \boldsymbol{b} - \boldsymbol{x} = \mu(\boldsymbol{y} - \boldsymbol{x})$ [M1] $\lambda(|x|\boldsymbol{y} + |y|\boldsymbol{x}) - \boldsymbol{x} = \mu(\boldsymbol{y} - \boldsymbol{x})$ $(\lambda|y| + \mu - 1)\boldsymbol{x} = (\mu - \lambda|x|)\boldsymbol{y}$ [A1] Since $\boldsymbol{x}$ and $\boldsymbol{y}$ are not parallel: $\lambda|y| + \mu - 1 = 0$ $\mu - \lambda|x| = 0$ [E1] $\lambda = \frac{1}{|\boldsymbol{x}| + |\boldsymbol{y}|}$ [M1] $\mu = \frac{|\boldsymbol{x}|}{|\boldsymbol{x}| + |\boldsymbol{y}|}$ [M1] So $B$ divides $XY$ in the ratio $|x| : |y|$ [A1] c If $OB$ is perpendicular to $XY$: $\boldsymbol{b} \cdot (\boldsymbol{y} - \boldsymbol{x}) = 0$ $|x||y|^2 + |y|\boldsymbol{x} \cdot \boldsymbol{y} - |x|\boldsymbol{y} \cdot \boldsymbol{x} - |y||x|^2 = 0$ $(|y| - |x|)(|x||y| + \boldsymbol{x} \cdot \boldsymbol{y}) = 0$ [M1 A1] $|x||y| + \boldsymbol{x} \cdot \boldsymbol{y} > 0$ So $|x| = |y|$ [A1] (ii) Let $\boldsymbol{p}, \boldsymbol{q}$ and $\boldsymbol{r}$ be the position vectors of $P, Q$ and $R$ respectively. The bisecting vector of $POQ$ is $|\boldsymbol{p}|\boldsymbol{q} + |\boldsymbol{q}|\boldsymbol{p}$ The bisecting vector of $QOR$ is $|\boldsymbol{q}|\boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{q}$ If $\theta$ is the angle between these two vectors, then: $\cos \theta = \frac{(|\boldsymbol{p}|\boldsymbol{q} + |\boldsymbol{q}|\boldsymbol{p}) \cdot (|\boldsymbol{q}|\boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{q})}{||\boldsymbol{p}|\boldsymbol{q} + |\boldsymbol{q}|\boldsymbol{p}| ||\boldsymbol{q}|\boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{q}|}$ [M1] $= \frac{|\boldsymbol{p}||\boldsymbol{q}|\boldsymbol{q} \cdot \boldsymbol{r} + |\boldsymbol{p}||\boldsymbol{r}||\boldsymbol{q}|^2 + |\boldsymbol{q}|^2\boldsymbol{p} \cdot \boldsymbol{r} + |\boldsymbol{q}||\boldsymbol{r}|\boldsymbol{p} \cdot \boldsymbol{q}}{||\boldsymbol{p}|\boldsymbol{q} + |\boldsymbol{q}|\boldsymbol{p}| ||\boldsymbol{q}|\boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{q}|}$ [M1] $\cos \theta = \frac{|\boldsymbol{q}|(|\boldsymbol{p}||\boldsymbol{q}||\boldsymbol{r}| + |\boldsymbol{p}|\boldsymbol{q} \cdot \boldsymbol{r} + |\boldsymbol{q}|\boldsymbol{p} \cdot \boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{p} \cdot \boldsymbol{q})}{||\boldsymbol{p}|\boldsymbol{q} + |\boldsymbol{q}|\boldsymbol{p}| ||\boldsymbol{q}|\boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{q}|}$ [A1] $|\boldsymbol{p}||\boldsymbol{q}||\boldsymbol{r}| + |\boldsymbol{p}|\boldsymbol{q} \cdot \boldsymbol{r} + |\boldsymbol{q}|\boldsymbol{p} \cdot \boldsymbol{r} + |\boldsymbol{r}|\boldsymbol{p} \cdot \boldsymbol{q}$ is symmetrical in $\boldsymbol{p}, \boldsymbol{q}$ and $\boldsymbol{r}$. [E1] All other factors are strictly positive, so the sign will be the same for all three angles (and so the angles are either all acute, all right angles or all obtuse). [E1]

Model Solution

Part (i)(a)

We need to show $\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}$ is a bisecting vector for angle $XOY$.

Lies in the plane $OXY$: $\mathbf{b}$ is a linear combination of $\mathbf{x}$ and $\mathbf{y}$, so it lies in the plane $OXY$.

Equal angles: Let $\theta_x$ be the angle between $\mathbf{b}$ and $\mathbf{x}$, and $\theta_y$ be the angle between $\mathbf{b}$ and $\mathbf{y}$.

$\mathbf{b} \cdot \mathbf{x} = (|\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}) \cdot \mathbf{x} = |\mathbf{x}|(\mathbf{y} \cdot \mathbf{x}) + |\mathbf{y}||\mathbf{x}|^2 = |\mathbf{x}|(\mathbf{x} \cdot \mathbf{y} + |\mathbf{x}||\mathbf{y}|).$

$\mathbf{b} \cdot \mathbf{y} = (|\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}) \cdot \mathbf{y} = |\mathbf{x}||\mathbf{y}|^2 + |\mathbf{y}|(\mathbf{x} \cdot \mathbf{y}) = |\mathbf{y}|(\mathbf{x} \cdot \mathbf{y} + |\mathbf{x}||\mathbf{y}|).$

Therefore

$\cos\theta_x = \frac{\mathbf{b} \cdot \mathbf{x}}{|\mathbf{b}||\mathbf{x}|} = \frac{\mathbf{x} \cdot \mathbf{y} + |\mathbf{x}||\mathbf{y}|}{|\mathbf{b}|},$

$\cos\theta_y = \frac{\mathbf{b} \cdot \mathbf{y}}{|\mathbf{b}||\mathbf{y}|} = \frac{\mathbf{x} \cdot \mathbf{y} + |\mathbf{x}||\mathbf{y}|}{|\mathbf{b}|}.$

Since $\cos\theta_x = \cos\theta_y$, and both angles are in $[0, \pi]$, we have $\theta_x = \theta_y$.

Angles less than $90^\circ$: Since $\mathbf{x} \cdot \mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos\alpha$ where $\alpha$ is the angle between $\mathbf{x}$ and $\mathbf{y}$ with $0 < \alpha < \pi$, we have

$\mathbf{x} \cdot \mathbf{y} + |\mathbf{x}||\mathbf{y}| = |\mathbf{x}||\mathbf{y}|(\cos\alpha + 1) > 0$

since $\cos\alpha > -1$. Therefore $\cos\theta_x > 0$, so both angles are less than $90^\circ$.

Uniqueness (up to positive scalar): Any bisecting vector $\mathbf{v}$ for angle $XOY$ lies in the plane $OXY$, so $\mathbf{v} = s\mathbf{x} + t\mathbf{y}$ for some scalars $s, t$. The condition that $\mathbf{v}$ makes equal angles with $\mathbf{x}$ and $\mathbf{y}$ means

$\frac{\mathbf{v} \cdot \mathbf{x}}{|\mathbf{x}|} = \frac{\mathbf{v} \cdot \mathbf{y}}{|\mathbf{y}|}.$

Substituting $\mathbf{v} = s\mathbf{x} + t\mathbf{y}$:

$\frac{s|\mathbf{x}|^2 + t(\mathbf{x} \cdot \mathbf{y})}{|\mathbf{x}|} = \frac{s(\mathbf{x} \cdot \mathbf{y}) + t|\mathbf{y}|^2}{|\mathbf{y}|}.$

$s|\mathbf{x}||\mathbf{y}| + t\frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|}|\mathbf{y}| = s\frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{y}|}|\mathbf{x}| + t|\mathbf{y}||\mathbf{x}|.$

$(s - t)|\mathbf{x}||\mathbf{y}| + (t - s)\frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}||\mathbf{y}|}|\mathbf{x}||\mathbf{y}| = 0.$

More directly: $s|\mathbf{x}||\mathbf{y}| + t(\mathbf{x}\cdot\mathbf{y})/|\mathbf{x}| \cdot |\mathbf{y}| = s(\mathbf{x}\cdot\mathbf{y})/|\mathbf{y}| \cdot |\mathbf{x}| + t|\mathbf{x}||\mathbf{y}|$, which simplifies to $(s|\mathbf{x}| - t|\mathbf{y}|)(|\mathbf{x}||\mathbf{y}| - \mathbf{x}\cdot\mathbf{y}) = 0$. Since $|\mathbf{x}||\mathbf{y}| > \mathbf{x}\cdot\mathbf{y}$ (as the angle between $\mathbf{x}$ and $\mathbf{y}$ is strictly between $0$ and $\pi$), we need $s|\mathbf{x}| = t|\mathbf{y}|$, i.e., $s/t = |\mathbf{y}|/|\mathbf{x}|$. So $\mathbf{v} = k(|\mathbf{y}|\mathbf{x} + |\mathbf{x}|\mathbf{y}) = k\mathbf{b}$ for some scalar $k$. The condition that both angles are less than $90^\circ$ requires $k > 0$.

Part (i)(b)

The point $B$ lies on line $XY$ if $\overrightarrow{XB}$ is parallel to $\overrightarrow{XY}$, i.e., for some scalar $\mu$:

$\lambda\mathbf{b} - \mathbf{x} = \mu(\mathbf{y} - \mathbf{x}).$

Substituting $\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}$:

$\lambda|\mathbf{x}|\mathbf{y} + \lambda|\mathbf{y}|\mathbf{x} - \mathbf{x} = \mu\mathbf{y} - \mu\mathbf{x}.$

$(\lambda|\mathbf{y}| + \mu - 1)\mathbf{x} = (\mu - \lambda|\mathbf{x}|)\mathbf{y}.$

Since $\mathbf{x}$ and $\mathbf{y}$ are not parallel (as $O, X, Y$ are non-collinear), both coefficients must be zero:

$\lambda|\mathbf{y}| + \mu - 1 = 0 \qquad \text{and} \qquad \mu - \lambda|\mathbf{x}| = 0.$

From the second equation: $\mu = \lambda|\mathbf{x}|$. Substituting into the first:

$\lambda|\mathbf{y}| + \lambda|\mathbf{x}| = 1 \implies \lambda = \frac{1}{|\mathbf{x}| + |\mathbf{y}|}.$

Then $\mu = \frac{|\mathbf{x}|}{|\mathbf{x}| + |\mathbf{y}|}$.

Since $\overrightarrow{XB} = \mu \cdot \overrightarrow{XY}$ with $0 < \mu < 1$, the point $B$ lies between $X$ and $Y$. We have $XB : BY = \mu : (1 - \mu)$, so

$XB : BY = |\mathbf{x}| : |\mathbf{y}|.$

Part (i)(c)

If $OB$ is perpendicular to $XY$, then $\mathbf{b} \cdot (\mathbf{y} - \mathbf{x}) = 0$ (since $\overrightarrow{OB} = \lambda\mathbf{b}$ with $\lambda \neq 0$).

$(|\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}) \cdot (\mathbf{y} - \mathbf{x}) = 0.$

$|\mathbf{x}||\mathbf{y}|^2 - |\mathbf{x}|(\mathbf{x} \cdot \mathbf{y}) + |\mathbf{y}|(\mathbf{x} \cdot \mathbf{y}) - |\mathbf{y}||\mathbf{x}|^2 = 0.$

$|\mathbf{x}||\mathbf{y}|(|\mathbf{y}| - |\mathbf{x}|) + (\mathbf{x} \cdot \mathbf{y})(|\mathbf{y}| - |\mathbf{x}|) = 0.$

$(|\mathbf{y}| - |\mathbf{x}|)(|\mathbf{x}||\mathbf{y}| + \mathbf{x} \cdot \mathbf{y}) = 0.$

Since $|\mathbf{x}||\mathbf{y}| + \mathbf{x} \cdot \mathbf{y} = |\mathbf{x}||\mathbf{y}|(1 + \cos\alpha) > 0$ (as shown in part (a)), we must have $|\mathbf{y}| - |\mathbf{x}| = 0$, i.e., $|\mathbf{x}| = |\mathbf{y}|$.

Therefore $OX = OY$, so triangle $XOY$ is isosceles.

Part (ii)

Let $\mathbf{p}, \mathbf{q}, \mathbf{r}$ be the position vectors of $P, Q, R$ respectively. The three bisecting vectors are:

$\mathbf{a} = |\mathbf{p}|\mathbf{q} + |\mathbf{q}|\mathbf{p} \qquad (\text{bisector of } POQ),$

$\mathbf{b} = |\mathbf{q}|\mathbf{r} + |\mathbf{r}|\mathbf{q} \qquad (\text{bisector of } QOR),$

$\mathbf{c} = |\mathbf{r}|\mathbf{p} + |\mathbf{p}|\mathbf{r} \qquad (\text{bisector of } ROP).$

The cosine of the angle between $\mathbf{a}$ and $\mathbf{b}$ is

$\cos\theta_{ab} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$

Since $|\mathbf{a}| > 0$ and $|\mathbf{b}| > 0$, the sign of $\cos\theta_{ab}$ is determined by $\mathbf{a} \cdot \mathbf{b}$.

$\mathbf{a} \cdot \mathbf{b} = (|\mathbf{p}|\mathbf{q} + |\mathbf{q}|\mathbf{p}) \cdot (|\mathbf{q}|\mathbf{r} + |\mathbf{r}|\mathbf{q}).$

$= |\mathbf{p}||\mathbf{q}|(\mathbf{q} \cdot \mathbf{r}) + |\mathbf{p}||\mathbf{r}||\mathbf{q}|^2 + |\mathbf{q}|^2(\mathbf{p} \cdot \mathbf{r}) + |\mathbf{q}||\mathbf{r}|(\mathbf{p} \cdot \mathbf{q}).$

$= |\mathbf{q}|\Big(|\mathbf{p}||\mathbf{q}||\mathbf{r}| + |\mathbf{p}|(\mathbf{q} \cdot \mathbf{r}) + |\mathbf{q}|(\mathbf{p} \cdot \mathbf{r}) + |\mathbf{r}|(\mathbf{p} \cdot \mathbf{q})\Big).$

Define

$S = |\mathbf{p}||\mathbf{q}||\mathbf{r}| + |\mathbf{p}|(\mathbf{q} \cdot \mathbf{r}) + |\mathbf{q}|(\mathbf{p} \cdot \mathbf{r}) + |\mathbf{r}|(\mathbf{p} \cdot \mathbf{q}).$

By exactly the same expansion, $\mathbf{b} \cdot \mathbf{c}$ gives

$\mathbf{b} \cdot \mathbf{c} = |\mathbf{r}|\Big(|\mathbf{p}||\mathbf{q}||\mathbf{r}| + |\mathbf{p}|(\mathbf{q} \cdot \mathbf{r}) + |\mathbf{q}|(\mathbf{p} \cdot \mathbf{r}) + |\mathbf{r}|(\mathbf{p} \cdot \mathbf{q})\Big) = |\mathbf{r}| \cdot S,$

and $\mathbf{c} \cdot \mathbf{a}$ gives

$\mathbf{c} \cdot \mathbf{a} = |\mathbf{p}| \cdot S.$

Since $|\mathbf{p}|, |\mathbf{q}|, |\mathbf{r}| > 0$, the sign of each dot product is determined entirely by $S$.

Therefore $\cos\theta_{ab}$, $\cos\theta_{bc}$, and $\cos\theta_{ca}$ all have the same sign. If $S > 0$ all three angles are acute; if $S < 0$ all three are obtuse; if $S = 0$ all three are right angles.

Examiner Notes

约1/3尝试。推理不清晰是常见问题。部分(ii)的对称性很少人认出。

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Question 5

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

5 (i) The functions $f_1$ and $F_1$, each with domain $\mathbb{Z}$, are defined by

$f_1(n) = n^2 + 6n + 11,$

$F_1(n) = n^2 + 2.$

Show that $F_1$ has the same range as $f_1$.

(ii) The function $g_1$, with domain $\mathbb{Z}$, is defined by

$g_1(n) = n^2 - 2n + 5.$

Show that the ranges of $f_1$ and $g_1$ have empty intersection.

(iii) The functions $f_2$ and $g_2$, each with domain $\mathbb{Z}$, are defined by

$f_2(n) = n^2 - 2n - 6,$

$g_2(n) = n^2 - 4n + 2.$

Find any integers that lie in the intersection of the ranges of the two functions.

(iv) Show that $p^2 + pq + q^2 \ge 0$ for all real $p$ and $q$.

The functions $f_3$ and $g_3$, each with domain $\mathbb{Z}$, are defined by

$f_3(n) = n^3 - 3n^2 + 7n,$

$g_3(n) = n^3 + 4n - 6.$

Find any integers that lie in the intersection of the ranges of the two functions.

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8

Hint

(i) $f_1(t) = (t + 3)^2 + 2 = F_1(t + 3)$ [M1] Since $t \mapsto t + 3$ is a one-to-one correspondence on $\mathbb{Z}$, the functions have the same range. [A1] (ii) If there is a value that lies in both the range of $f_1$ and $g_1$, then there are integers $s$ and $t$ such that: $f_1(s) = (s + 3)^2 + 2 = g_1(t) = (t - 1)^2 + 4$ [M1] $(s + 3)^2 - (t - 1)^2 = 2$ [A1] which is not possible for any integers $s$ and $t$. [E1] (iii) For any value that lies in both the range of $f_2$ and $g_2$, there are integers $s$ and $t$ such that: $f_2(s) = (s - 1)^2 - 7 = g_2(t) = (t - 2)^2 - 2$ [M1] $(s - 1)^2 - (t - 2)^2 = 5$ $(s + t - 3)(s - t + 1) = 5$ [A1] $s + t - 3 = 1$ $s - t + 1 = 5$ has solution $s = 4, t = 0$ [M1] $s + t - 3 = -1$ $s - t + 1 = -5$ has solution $s = -2, t = 4$ $s + t - 3 = 5$ $s - t + 1 = 1$ has solution $s = 4, t = 4$ $s + t - 3 = -5$ $s - t + 1 = -1$ has solution $s = -2, t = 0$ [A1] All cases lead to $f_2(s) = g_2(t) = 2$, so only 2 lies in the intersection between the ranges. [A1] (iv) $4(p^2 + pq + q^2) = (p - q)^2 + 3(p + q)^2$ $p^2 + pq + q^2 = (p + q)^2 - pq = (p - q)^2 + 3pq$ sufficient for M1 [M1] Therefore $p^2 + pq + q^2 \ge 0$ for all real $p$ and $q$. [A1] $f_3(s) = s^3 - 3s^2 + 7s = g_3(t) = t^3 + 4t - 6$ $(s - 1)^3 = s^3 - 3s^2 + 3s - 1$, so $(s - 1)^3 - t^3 + 4s - 4t = -7$ [M1] $(s - 1 - t)((s - 1)^2 + (s - 1)t + t^2) + 4(s - 1 - t) = -11$ [M1] $(s - 1 - t)((s - 1)^2 + (s - 1)t + t^2 + 4) = -11$ [A1] By the result at the start of part (iv): $((s - 1)^2 + (s - 1)t + t^2 + 4) \ge 4$ so the product can only be $-1 \times 11$ [M1 A1] We have $s = t$ and so $s^2 - 2s + 1 + s^2 - s + s^2 + 4 = 11$ $3s^2 - 3s - 6 = 0$, so $3(s - 2)(s + 1) = 0$, giving $s = t = 2$ or $s = t = -1$ [M1

A1] So the intersection is $\{f_3(-1), f_3(2)\} = \{-11, 10\}$ [A1]

Model Solution

Part (i)

Complete the square on $f_1$:

$f_1(n) = n^2 + 6n + 11 = (n + 3)^2 + 2 = F_1(n + 3).$

The map $n \mapsto n + 3$ is a bijection from $\mathbb{Z}$ to $\mathbb{Z}$, so as $n$ ranges over all integers, $n + 3$ also ranges over all integers. Therefore

$\{f_1(n) : n \in \mathbb{Z}\} = \{F_1(m) : m \in \mathbb{Z}\},$

and $F_1$ has the same range as $f_1$. $\qquad \blacksquare$

Part (ii)

Complete the square on both functions:

$f_1(n) = (n + 3)^2 + 2, \qquad g_1(n) = (n - 1)^2 + 4.$

Suppose for contradiction that some value lies in both ranges. Then there exist integers $s, t$ with

$(s + 3)^2 + 2 = (t - 1)^2 + 4,$

$(t - 1)^2 - (s + 3)^2 = -2.$

Factorising as a difference of squares:

$(t - 1 - s - 3)(t - 1 + s + 3) = -2,$

$(t - s - 4)(t + s + 2) = -2.$

The two factors have the same parity (their sum is $2t - 2$, which is even), so both are even or both are odd. But every factorisation of $-2$ into two integers is $(-1) \times 2$, $1 \times (-2)$, $(-2) \times 1$, or $2 \times (-1)$, each of which has one even and one odd factor. No factorisation has both factors of the same parity.

Therefore no integers $s, t$ satisfy the equation, and the ranges of $f_1$ and $g_1$ have empty intersection. $\qquad \blacksquare$

Part (iii)

Complete the square:

$f_2(n) = (n - 1)^2 - 7, \qquad g_2(n) = (n - 2)^2 - 2.$

If a value lies in both ranges, then $f_2(s) = g_2(t)$ for some integers $s, t$:

$(s - 1)^2 - 7 = (t - 2)^2 - 2,$

$(s - 1)^2 - (t - 2)^2 = 5.$

Factorising:

$(s - 1 - t + 2)(s - 1 + t - 2) = 5,$

$(s - t + 1)(s + t - 3) = 5.$

Since $5$ is prime, the integer factor pairs are $(1, 5)$, $(5, 1)$, $(-1, -5)$, $(-5, -1)$.

• $s - t + 1 = 1$, $s + t - 3 = 5$: adding gives $2s - 2 = 6$, so $s = 4$, $t = 0$.
• $s - t + 1 = 5$, $s + t - 3 = 1$: adding gives $2s - 2 = 6$, so $s = 4$, $t = 4$.
• $s - t + 1 = -1$, $s + t - 3 = -5$: adding gives $2s - 2 = -6$, so $s = -2$, $t = 0$.
• $s - t + 1 = -5$, $s + t - 3 = -1$: adding gives $2s - 2 = -6$, so $s = -2$, $t = 4$.

All four cases give $f_2(s) = g_2(t) = 2$ (checking: $f_2(4) = 16 - 8 - 6 = 2$, $g_2(0) = 0 - 0 + 2 = 2$, $g_2(4) = 16 - 16 + 2 = 2$, $f_2(-2) = 4 + 4 - 6 = 2$).

Therefore the only integer in the intersection of the ranges is $\mathbf{2}$. $\qquad \blacksquare$

Part (iv)

Non-negativity of $p^2 + pq + q^2$:

We write

$p^2 + pq + q^2 = \left(p + \frac{q}{2}\right)^2 + \frac{3q^2}{4}.$

Expanding the right-hand side: $p^2 + pq + \frac{q^2}{4} + \frac{3q^2}{4} = p^2 + pq + q^2$. ✓

Since both terms on the right are non-negative, $p^2 + pq + q^2 \geq 0$ for all real $p$ and $q$. $\qquad \blacksquare$

Finding the intersection of the ranges of $f_3$ and $g_3$:

Set $f_3(s) = g_3(t)$:

$s^3 - 3s^2 + 7s = t^3 + 4t - 6.$

Since $(s - 1)^3 = s^3 - 3s^2 + 3s - 1$, we can write $s^3 - 3s^2 + 7s = (s - 1)^3 + 4s + 1$. So:

$(s - 1)^3 + 4s + 1 = t^3 + 4t - 6,$

$(s - 1)^3 - t^3 + 4(s - t) = -7.$

Now write $s - t = (s - 1 - t) + 1$, and apply the factorisation $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = s - 1$, $b = t$:

$(s - 1 - t)\bigl[(s - 1)^2 + (s - 1)t + t^2\bigr] + 4(s - 1 - t) + 4 = -7,$

$(s - 1 - t)\bigl[(s - 1)^2 + (s - 1)t + t^2 + 4\bigr] = -11.$

Let $d = s - 1 - t$. Setting $p = s - 1$ and $q = t$ in the non-negativity result, $(s - 1)^2 + (s - 1)t + t^2 \geq 0$, so the second factor satisfies

$(s - 1)^2 + (s - 1)t + t^2 + 4 \geq 4.$

Since $-11$ is prime and the second factor is a positive integer $\geq 4$, the only possibility is $d = -1$ and the second factor equals $11$:

$s - 1 - t = -1 \implies s = t.$

Substituting $s = t$:

$(s - 1)^2 + (s - 1)s + s^2 + 4 = 11,$

$s^2 - 2s + 1 + s^2 - s + s^2 + 4 = 11,$

$3s^2 - 3s - 6 = 0,$

$s^2 - s - 2 = 0,$

$(s - 2)(s + 1) = 0.$

So $s = t = 2$ or $s = t = -1$.

Checking: $f_3(2) = 8 - 12 + 14 = 10 = 8 + 8 - 6 = g_3(2)$ ✓, and $f_3(-1) = -1 - 3 - 7 = -11 = -1 - 4 - 6 = g_3(-1)$ ✓.

Therefore the integers in the intersection of the ranges are $\{-11,\; 10\}$. $\qquad \blacksquare$

Examiner Notes

第二热门题。许多人没注意到定义域是 Z 而非 R。

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Question 6

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

6 In this question, you need not consider issues of convergence.

(i) The sequence $T_n$, for $n = 0, 1, 2, \dots$, is defined by $T_0 = 1$ and, for $n \geqslant 1$, by

$T_n = \frac{2n - 1}{2n} T_{n-1}.$

Prove by induction that

$T_n = \frac{1}{2^{2n}} \binom{2n}{n},$

for $n = 0, 1, 2, \dots$.

[Note that $\binom{0}{0} = 1$.]

(ii) Show that in the binomial series for $(1 - x)^{-\frac{1}{2}}$,

$(1 - x)^{-\frac{1}{2}} = \sum_{r=0}^{\infty} a_r x^r,$

successive coefficients are related by

$a_r = \frac{2r - 1}{2r} a_{r-1}$

for $r = 1, 2, \dots$.

Hence prove that $a_r = T_r$ for all $r = 0, 1, 2, \dots$.

(iii) Let $b_r$ be the coefficient of $x^r$ in the binomial series for $(1 - x)^{-\frac{3}{2}}$, so that

$(1 - x)^{-\frac{3}{2}} = \sum_{r=0}^{\infty} b_r x^r.$

By considering $\frac{b_r}{a_r}$, find an expression involving a binomial coefficient for $b_r$, for $r = 0, 1, 2, \dots$.

(iv) By considering the product of the binomial series for $(1 - x)^{-\frac{1}{2}}$ and $(1 - x)^{-1}$, prove that

$\frac{(2n + 1)}{2^{2n}} \binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}} \binom{2r}{r},$

for $n = 1, 2, \dots$.

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9

Hint

(i) For $n = 0$: $T_0 = \frac{1}{2^0} \binom{0}{0} = 1$ [B1] Assume that the result is true for $n = k$: $T_k = \frac{1}{2^{2k}} \binom{2k}{k}$ $T_{k+1} = \frac{2(k+1)-1}{2(k+1)} \cdot \frac{1}{2^{2k}} \binom{2k}{k}$ [M1] $T_{k+1} = \frac{1}{2^{2k}} \cdot \frac{2k+1}{2(k+1)} \cdot \frac{(2k)!}{(k!)^2}$ $T_{k+1} = \frac{1}{2^{2k}} \cdot \frac{2k+1}{2(k+1)} \cdot \frac{(2k)!}{(k!)^2} \cdot \frac{2k+2}{2(k+1)}$ $T_{k+1} = \frac{1}{2^{2(k+1)}} \cdot \frac{(2(k+1))!}{((k+1)!)^2}$ $T_k = \frac{1}{2^{2(k+1)}} \binom{2(k+1)}{k+1}$ [M1

A1] Hence, by induction: $T_n = \frac{1}{2^{2n}} \binom{2n}{n}$ [A1] (ii) $a_r = \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \dots \left(-\frac{2r-1}{2}\right) \frac{(-1)^r}{r!}$ [M1

A1] $a_{r-1} = \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \dots \left(-\frac{2(r-1)-1}{2}\right) \frac{(-1)^{r-1}}{(r-1)!}$ [M1] Therefore, $a_r = a_{r-1} \cdot \left(-\frac{2r-1}{2}\right) \cdot \frac{-1}{r}$ $a_r = \frac{2r-1}{2r} a_{r-1}$ [E1] Since $a_0 = 1 = T_0$, [B1] $a_r = T_r$ for $r = 0, 1, 2, \dots$ [A1] (iii) $b_r = \frac{\left(\frac{3}{2} \cdot \frac{5}{2} \cdot \dots \cdot \frac{(2r-1)}{2} \cdot \frac{(2r+1)}{2}\right)}{r!}$ [M1] So, $\frac{b_r}{a_r} = 2r + 1$ [A1] Correctly argued for general terms [E1] $b_r = \frac{2r+1}{2^{2r}} \binom{2r}{r}$ [A1] (iv) $(1 - x)^{-1} = \Sigma_{r=0}^{\infty} x^r$ [B1] $(1 + x + x^2 + \cdots)(a_0 + a_1x + a_2x^2 + \cdots) = (b_0 + b_1x + b_2x^2 + \cdots)$ [B1] The term in $x^n$ on the LHS is: $1 \cdot a_n x^n + x \cdot a_{n-1} x^{n-1} + \cdots + x^n \cdot a_0$ [M1 A1] Therefore, $b_n = \Sigma_{r=0}^n a_r$ as required. [A1]

Model Solution

Part (i)

We prove by induction that $T_n = \dfrac{1}{2^{2n}} \dbinom{2n}{n}$ for $n = 0, 1, 2, \dots$.

Base case ($n = 0$): $T_0 = 1$ and $\frac{1}{2^0} \binom{0}{0} = 1$. ✓

Inductive step: Assume the result holds for $n = k$, i.e.

$T_k = \frac{1}{2^{2k}} \binom{2k}{k}.$

Then for $n = k + 1$:

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

Writing out the binomial coefficient:

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

We simplify the numerator and denominator separately. On the numerator side, we want to build $(2k + 2)! = (2k + 2)(2k + 1)(2k)!$:

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

Multiply numerator and denominator by $2(k + 1) = 2k + 2$:

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

Therefore

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

By induction, $T_n = \dfrac{1}{2^{2n}} \dbinom{2n}{n}$ for all $n = 0, 1, 2, \dots$. $\qquad \blacksquare$

Part (ii)

The general binomial coefficient for $(1 - x)^{-1/2}$ is

$a_r = \binom{-1/2}{r} = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots \left(-\frac{2r - 1}{2}\right)}{r!}.$

There are $r$ factors in the numerator, each negative, so

$a_r = \frac{(-1)^r}{2^r \cdot r!} \cdot 1 \cdot 3 \cdot 5 \cdots (2r - 1).$

For $r \geq 1$:

$\frac{a_r}{a_{r-1}} = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots \left(-\frac{2r - 1}{2}\right) / r!}{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots \left(-\frac{2r - 3}{2}\right) / (r - 1)!}.$

The first $(r - 1)$ factors cancel, leaving:

$\frac{a_r}{a_{r-1}} = \frac{-\frac{2r - 1}{2}}{r} = \frac{-(2r - 1)}{2r}.$

Therefore

$a_r = \frac{2r - 1}{2r} a_{r-1} \qquad \text{for } r = 1, 2, \dots$

where the two negative signs cancel. This is the same recurrence as $T_r$.

Since $a_0 = \binom{-1/2}{0} = 1 = T_0$, and both sequences satisfy the same recurrence with the same initial value, it follows that $a_r = T_r$ for all $r = 0, 1, 2, \dots$:

$a_r = \frac{1}{2^{2r}} \binom{2r}{r}. \qquad \blacksquare$

Part (iii)

The general coefficient for $(1 - x)^{-3/2}$ is

$b_r = \binom{-3/2}{r} = \frac{\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right) \cdots \left(-\frac{2r + 1}{2}\right)}{r!}.$

There are $r$ factors, each negative, so

$b_r = \frac{(-1)^r}{2^r \cdot r!} \cdot 3 \cdot 5 \cdot 7 \cdots (2r + 1).$

Comparing with $a_r$:

$a_r = \frac{(-1)^r}{2^r \cdot r!} \cdot 1 \cdot 3 \cdot 5 \cdots (2r - 1),$

we see that the product in $b_r$ is obtained from the product in $a_r$ by removing the factor $1$ and inserting the factor $(2r + 1)$. Therefore

$\frac{b_r}{a_r} = \frac{3 \cdot 5 \cdot 7 \cdots (2r + 1)}{1 \cdot 3 \cdot 5 \cdots (2r - 1)} = 2r + 1.$

(For $r = 0$: $b_0 = 1$, $a_0 = 1$, so $b_0 / a_0 = 1 = 2(0) + 1$. ✓)

Hence

$b_r = (2r + 1) \, a_r = \frac{2r + 1}{2^{2r}} \binom{2r}{r} \qquad \text{for } r = 0, 1, 2, \dots. \qquad \blacksquare$

Part (iv)

We use the Cauchy product of two power series. Since

$(1 - x)^{-1/2} \cdot (1 - x)^{-1} = (1 - x)^{-3/2},$

the coefficient of $x^n$ on the left must equal $b_n$ on the right.

The binomial series for $(1 - x)^{-1}$ is the geometric series:

$(1 - x)^{-1} = \sum_{m=0}^{\infty} x^m.$

So the coefficient of $x^n$ in the product $\left(\sum_{r=0}^{\infty} a_r x^r\right)\left(\sum_{m=0}^{\infty} x^m\right)$ is

$\sum_{r=0}^{n} a_r \cdot 1 = \sum_{r=0}^{n} a_r.$

Therefore

$b_n = \sum_{r=0}^{n} a_r.$

Substituting $a_r = \frac{1}{2^{2r}} \binom{2r}{r}$ from part (ii) and $b_n = \frac{2n + 1}{2^{2n}} \binom{2n}{n}$ from part (iii):

$\frac{2n + 1}{2^{2n}} \binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}} \binom{2r}{r} \qquad \text{for } n = 1, 2, \dots. \qquad \blacksquare$

Examiner Notes

大量尝试。基础情形选择是常见错误。需要写出一般项而非仅靠找模式。

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Question 7

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

7 (i) Sketch the curve $C_1$ with equation

$(y^2 + (x - 1)^2 - 1)(y^2 + (x + 1)^2 - 1) = 0.$

(ii) Consider the curve $C_2$ with equation

$(y^2 + (x - 1)^2 - 1)(y^2 + (x + 1)^2 - 1) = \frac{1}{16}.$

(a) Show that the line $y = k$ meets the curve $C_2$ at points for which

$x^4 + 2(k^2 - 2)x^2 + (k^4 - \frac{1}{16}) = 0.$

Hence determine the number of intersections between curve $C_2$ and the line $y = k$ for each positive value of $k$.

(b) Determine whether the points on curve $C_2$ with the greatest possible $y$-coordinate are further from, or closer to, the $y$-axis than those on curve $C_1$.

(c) Show that it is not possible for both $y^2 + (x - 1)^2 - 1$ and $y^2 + (x + 1)^2 - 1$ to be negative, and deduce that curve $C_2$ lies entirely outside curve $C_1$.

(d) Sketch the curves $C_1$ and $C_2$ on the same axes.

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10

Hint

(i) Circles with centres at (1,0) and (-1,0), [B1] both with radius 1 [B1] (ii) a $y = k$ meets the curve when $(x^2 + k^2 - 2x)(x^2 + k^2 + 2x) = \frac{1}{16}$ $(x^2 + k^2 - 2x)(x^2 + k^2 + 2x) = (x^2 + k^2)^2 - 4x^2$ [M1] So $(x^2 + k^2)^2 - 4x^2 - \frac{1}{16} = 0$ $x^4 + 2x^2(k^2 - 2) + k^4 - \frac{1}{16} = 0$ [A1] $(x^2 + k^2 - 2)^2 + 4k^2 - \frac{65}{16} = 0$ $x^2 = 2 - k^2 \pm \sqrt{\frac{65}{16} - 4k^2}$ [M1

A1] If $k^2 > \frac{65}{64}$ there will be no roots If $k^2 = \frac{65}{64}$ there will be two roots [B1] If $k^2 < \frac{65}{64}$, the smaller of the two values of $x^2$ is $2 - k^2 - \sqrt{\frac{65}{16} - 4k^2}$ [M1] $2 - k^2 - \sqrt{\frac{65}{16} - 4k^2} = 0$ when $(2 - k^2)^2 = \frac{65}{16} - 4k^2$ $k^4 = \frac{1}{16}$ So there will be three roots if $k^2 = \frac{1}{4}$ [A1] There will be two roots if $0 \le k^2 < \frac{1}{4}$ There will be four roots if $\frac{1}{4} < k^2 < \frac{65}{64}$ [A1] b Greatest possible $y$-coordinate is when $k^2 = \frac{65}{64}$ [M1] So $x^2 = 2 - k^2 = \frac{63}{64} < 1$ So these points are closer to the $y$-axis than those on $C_1$ [A1] c If both expressions are negative, then the distance from $(x, y)$ to the points $(1, 0)$ and $(-1, 0)$ would both be less than 1. [E1] But the shortest distance between $(1, 0)$ and $(-1, 0)$ is 2, so this is not possible. Therefore, it is not possible for both expressions to be negative. [E1] Since the product of the two expression is positive and they are not both negative, they must both be positive. Therefore, the distance between the point $(x, y)$ and each of the points $(1, 0)$ and $(-1, 0)$ must be greater than 1, so the curve $C_2$ lies entirely outside the circle $C_1$ [E1] d Continuous curve outside the two circles of $C_1$ [G1] Symmetrical under reflection in $x$ and $y$ axes. [G1] Intersections with $x$-axis at $x = \pm \frac{1}{2} \sqrt{8 + \sqrt{65}}$ [G1] Intersections with $y$-axis at $y = \pm \frac{1}{2}$ [G1] Maxima and minima at $\left( \pm \frac{1}{8} \sqrt{63}, \pm \frac{1}{8} \sqrt{65} \right)$ [G1]

Model Solution

Part (i)

The equation $(y^2 + (x - 1)^2 - 1)(y^2 + (x + 1)^2 - 1) = 0$ holds when either factor is zero.

Setting the first factor to zero: $y^2 + (x - 1)^2 = 1$, which is a circle centred at $(1, 0)$ with radius $1$.

Setting the second factor to zero: $y^2 + (x + 1)^2 = 1$, which is a circle centred at $(-1, 0)$ with radius $1$.

So $C_1$ consists of two unit circles. The circles touch at the origin (the only common point), since the distance between the centres equals the sum of the radii.

Part (ii)(a)

Substituting $y = k$ into the equation of $C_2$:

$(k^2 + (x - 1)^2 - 1)(k^2 + (x + 1)^2 - 1) = \frac{1}{16}$

Expanding each factor:

$k^2 + (x - 1)^2 - 1 = x^2 - 2x + k^2$

$k^2 + (x + 1)^2 - 1 = x^2 + 2x + k^2$

Using the difference of two squares, their product is:

$(x^2 + k^2 - 2x)(x^2 + k^2 + 2x) = (x^2 + k^2)^2 - (2x)^2 = x^4 + 2k^2 x^2 + k^4 - 4x^2$

Setting this equal to $\frac{1}{16}$ and rearranging:

$x^4 + 2(k^2 - 2)x^2 + k^4 - \frac{1}{16} = 0 \qquad \text{(*)}$

Treating $(*)$ as a quadratic in $X = x^2$ and applying the quadratic formula:

$X = \frac{-2(k^2 - 2) \pm \sqrt{4(k^2 - 2)^2 - 4(k^4 - \frac{1}{16})}}{2}$

Computing the discriminant:

$4(k^2 - 2)^2 - 4\left(k^4 - \frac{1}{16}\right) = 4\left(k^4 - 4k^2 + 4 - k^4 + \frac{1}{16}\right) = 4\left(\frac{65}{16} - 4k^2\right)$

So the two values of $x^2$ are:

$x^2 = 2 - k^2 \pm \sqrt{\frac{65}{16} - 4k^2} \qquad \text{(**)}$

For real solutions we need $\frac{65}{16} - 4k^2 \geq 0$, i.e.\ $k^2 \leq \frac{65}{64}$.

Case 1: $k^2 > \frac{65}{64}$ (i.e.\ $k > \frac{\sqrt{65}}{8}$). The discriminant is negative, so there are no real solutions. 0 intersections.

Case 2: $k^2 = \frac{65}{64}$. The discriminant is zero, so $x^2 = 2 - \frac{65}{64} = \frac{63}{64} > 0$. This gives $x = \pm\frac{\sqrt{63}}{8}$. 2 intersections.

Case 3: $\frac{1}{4} < k^2 < \frac{65}{64}$. The discriminant is positive, giving two values of $x^2$. We check whether the smaller root $2 - k^2 - \sqrt{\frac{65}{16} - 4k^2}$ is positive. Setting it equal to zero:

$(2 - k^2)^2 = \frac{65}{16} - 4k^2 \implies 4 - 4k^2 + k^4 = \frac{65}{16} - 4k^2 \implies k^4 = \frac{1}{16} \implies k^2 = \frac{1}{4}$

Since the smaller root equals zero at $k^2 = \frac{1}{4}$ and is an increasing function of $k^2$ for $k^2 > \frac{1}{4}$ (one can verify by differentiation, or note that the value at $k^2 = \frac{65}{64}$ is $\frac{63}{64} > 0$), both values of $x^2$ are positive. Each positive value of $x^2$ gives two values of $x$. 4 intersections.

Case 4: $k^2 = \frac{1}{4}$ (i.e.\ $k = \frac{1}{2}$). The smaller root of $(**)$ is $0$, giving $x = 0$. The larger root is $\frac{7}{4} + \sqrt{3} > 0$, giving $x = \pm\sqrt{\frac{7}{4} + \sqrt{3}}$. 3 intersections.

Case 5: $0 < k^2 < \frac{1}{4}$ (i.e.\ $0 < k < \frac{1}{2}$). The smaller root of $(**)$ is negative, so only the larger root yields valid $x^2 > 0$, giving two values of $x$. 2 intersections.

Summary: for $0 < k < \frac{1}{2}$, there are 2 intersections; for $k = \frac{1}{2}$, there are 3 intersections; for $\frac{1}{2} < k < \frac{\sqrt{65}}{8}$, there are 4 intersections; for $k = \frac{\sqrt{65}}{8}$, there are 2 intersections; for $k > \frac{\sqrt{65}}{8}$, there are no intersections.

Part (ii)(b)

The greatest $y$-coordinate on $C_2$ occurs when the line $y = k$ is tangent to $C_2$, i.e.\ when $k^2 = \frac{65}{64}$, and the corresponding $x$-coordinate satisfies $x^2 = \frac{63}{64}$.

Since $\frac{63}{64} < 1$, we have $|x| = \frac{\sqrt{63}}{8} < 1$.

On $C_1$, the points with the greatest $y$-coordinate are $(1, 1)$ and $(-1, 1)$, with $|x| = 1$.

Therefore the points on $C_2$ with the greatest $y$-coordinate are closer to the $y$-axis than those on $C_1$.

Part (ii)(c)

Suppose for contradiction that both expressions are negative at some point $(x, y)$:

$y^2 + (x - 1)^2 - 1 < 0 \quad \text{and} \quad y^2 + (x + 1)^2 - 1 < 0$

Adding these two inequalities:

$2y^2 + (x - 1)^2 + (x + 1)^2 - 2 < 0$

$2y^2 + x^2 - 2x + 1 + x^2 + 2x + 1 - 2 < 0$

$2x^2 + 2y^2 < 0$

This is impossible since $x^2 \geq 0$ and $y^2 \geq 0$. So both factors cannot be negative simultaneously.

Since $(y^2 + (x - 1)^2 - 1)(y^2 + (x + 1)^2 - 1) = \frac{1}{16} > 0$ on $C_2$, and the two factors cannot both be negative, they must both be positive. Therefore, for every point on $C_2$:

$y^2 + (x - 1)^2 > 1 \quad \text{and} \quad y^2 + (x + 1)^2 > 1$

Every point on $C_2$ is at distance greater than $1$ from both $(1, 0)$ and $(-1, 0)$, so $C_2$ lies entirely outside $C_1$.

Part (ii)(d)

The sketch shows:

• $C_1$: two unit circles centred at $(1, 0)$ and $(-1, 0)$, touching at the origin.
• $C_2$: a single closed curve surrounding both circles, symmetric under reflection in both the $x$-axis and the $y$-axis.
• $C_2$ crosses the $y$-axis at $(0, \pm \frac{1}{2})$ and the $x$-axis at $(\pm \frac{1}{2}\sqrt{8 + \sqrt{65}}, 0)$.
• The highest and lowest points of $C_2$ are at $\left(\pm \frac{\sqrt{63}}{8}, \pm \frac{\sqrt{65}}{8}\right)$.

Examiner Notes

许多尝试但得分不高。判别式分析常遗漏细节。

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Question 8

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

8 In this question, the following theorem may be used without proof.

Let $u_1, u_2, \dots$ be a sequence of real numbers. If the sequence is

• bounded above, so $u_n \leqslant b$ for all $n$, where $b$ is some fixed number
• and increasing, so $u_n \leqslant u_{n+1}$ for all $n$

then there is a number $L \leqslant b$ such that $u_n \to L$ as $n \to \infty$.

For positive real numbers $x$ and $y$, define $\text{a}(x, y) = \frac{1}{2}(x + y)$ and $\text{g}(x, y) = \sqrt{xy}$. Let $x_0$ and $y_0$ be two positive real numbers with $y_0 < x_0$ and define, for $n \geqslant 0$

$x_{n+1} = \text{a}(x_n, y_n),$ $y_{n+1} = \text{g}(x_n, y_n).$

(i) By considering $(\sqrt{x_n} - \sqrt{y_n})^2$, show that $y_{n+1} < x_{n+1}$, for $n \geqslant 0$. Show further that, for $n \geqslant 0$

• $x_{n+1} < x_n$
• $y_n < y_{n+1}$.

Deduce that there is a value $M$ such that $y_n \to M$ as $n \to \infty$.

Show that $0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)$ and hence that $x_n - y_n \to 0$ as $n \to \infty$. Explain why $x_n$ also tends to $M$ as $n \to \infty$.

(ii) Let

$\text{I}(p, q) = \int_{0}^{\infty} \frac{1}{\sqrt{(p^2 + x^2)(q^2 + x^2)}} \text{ d}x,$

where $p$ and $q$ are positive real numbers with $q < p$.

Show, using the substitution $t = \frac{1}{2} \left( x - \frac{pq}{x} \right)$ in the integral

$\int_{-\infty}^{\infty} \frac{1}{\sqrt{(\frac{1}{4}(p + q)^2 + t^2)(pq + t^2)}} \text{ d}t,$

that

$\text{I}(p, q) = \text{I}(\text{a}(p, q), \text{g}(p, q)).$

Hence evaluate $\text{I}(x_0, y_0)$ in terms of $M$.

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11

Hint

(i) $(\sqrt{x_n} - \sqrt{y_n})^2 = 2a(x_n, y_n) - 2g(x_n, y_n)$ $= 2(x_{n+1} - y_{n+1})$ [M1] So $x_{n+1} - y_{n+1} \ge 0$ for $n \ge 0$ [A1] $y_0 < x_0$ is given Suppose that $y_k < x_k$: $(\sqrt{x_k} - \sqrt{y_k})^2 > 0$ and so $x_{k+1} - y_{k+1} > 0$ Hence, by induction, $y_n < x_n$ for $n \ge 0$ [A1] $y_{n+1} = \sqrt{x_n y_n} > \sqrt{y_n y_n} = y_n$ [B1] $x_{n+1} = \frac{1}{2}(x_n + y_n) < \frac{1}{2}(x_n + x_n) = x_n$ [B1] $y_n < x_n < x_0$ for $n \ge 0$, so the sequence is bounded above. [B1] As shown above the sequence is increasing, so the result given at the start of the question applies. There is a value $M$ such that $y_n \to M$ as $n \to \infty$ [B1] $x_{n+1} - y_{n+1} = \frac{1}{2}(\sqrt{x_n} - \sqrt{y_n})^2$ $< \frac{1}{2}(\sqrt{x_n} - \sqrt{y_n})(\sqrt{x_n} + \sqrt{y_n})$ [M1] $= \frac{1}{2}(x_n - y_n)$ $x_{n+1} - y_{n+1} = \frac{1}{2}(\sqrt{x_n} - \sqrt{y_n})^2 > 0$, since $x_n \ne y_n$ for any value of $n$. Therefore $0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)$ [A1] Hence $x_n - y_n \to 0$ as $n \to \infty$ [E1] So $x_n \to M$ as $n \to \infty$ [E1] (ii) $\frac{dt}{dx} = \frac{1}{2} \left( 1 + \frac{pq}{x^2} \right)$ [M1] Limits: As $x \to 0, t \to -\infty$ As $x \to \infty, t \to \infty$ [E1] $\frac{1}{4}(p+q)^2 + \frac{1}{4} \left( x - \frac{pq}{x} \right)^2 = \frac{1}{4x^2} (x^4 + (p^2 + q^2)x^2 + p^2q^2)$ [M1] $= \frac{1}{4x^2} (x^2 + p^2)(x^2 + q^2)$ $pq + \frac{1}{4} \left( x - \frac{pq}{x} \right)^2 = \frac{1}{4x^2} (x^4 + 2pqx^2 + p^2q^2)$ $= \frac{1}{4x^2} (x^2 + pq)^2$ [A1] So the integral becomes: $2 \int_{0}^{\infty} \frac{1}{\sqrt{(x^2 + p^2)(x^2 + q^2)}} dx = 2I(p, q)$ [A1] Since the original integrand was an even function it is also equal to $2I(a(p, q), g(p, q))$ [E1] $I(x_0, y_0) = I(x_1, y_1) = \dots = \int_{0}^{\infty} \frac{1}{x^2 + M^2} dx$ [M1] $= \left[ \frac{1}{M} \arctan \left( \frac{x}{M} \right) \right]_{0}^{\infty}$ [A1] $= \frac{\pi}{2M}$ [A1]

Model Solution

Part (i)

Showing $y_{n+1} < x_{n+1}$:

Consider the expansion:

$(\sqrt{x_n} - \sqrt{y_n})^2 = x_n - 2\sqrt{x_n y_n} + y_n = (x_n + y_n) - 2\sqrt{x_n y_n}$

$= 2 \cdot \frac{x_n + y_n}{2} - 2\sqrt{x_n y_n} = 2\,\text{a}(x_n, y_n) - 2\,\text{g}(x_n, y_n) = 2(x_{n+1} - y_{n+1})$

Since $(\sqrt{x_n} - \sqrt{y_n})^2 \geq 0$ for all real values, we have $x_{n+1} - y_{n+1} \geq 0$.

By hypothesis, $y_0 < x_0$. We prove $y_n < x_n$ for all $n \geq 0$ by induction.

Base case: $y_0 < x_0$ (given).

Inductive step: suppose $y_k < x_k$ for some $k \geq 0$. Then $\sqrt{x_k} \neq \sqrt{y_k}$, so $(\sqrt{x_k} - \sqrt{y_k})^2 > 0$, giving $x_{k+1} - y_{k+1} > 0$, i.e.\ $y_{k+1} < x_{k+1}$.

By induction, $y_n < x_n$ for all $n \geq 0$.

Showing $y_{n+1} > y_n$:

Since $x_n > y_n > 0$:

$y_{n+1} = \sqrt{x_n y_n} > \sqrt{y_n \cdot y_n} = y_n$

Showing $x_{n+1} < x_n$:

$x_{n+1} = \frac{x_n + y_n}{2} < \frac{x_n + x_n}{2} = x_n$

Deducing $y_n \to M$:

The sequence $\{y_n\}$ is increasing and bounded above by $x_0$ (since $y_n < x_n < x_0$ for all $n$). By the theorem stated at the start of the question, there exists a value $M \leq x_0$ such that $y_n \to M$ as $n \to \infty$.

Showing $0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)$:

From the identity above:

$x_{n+1} - y_{n+1} = \frac{1}{2}(\sqrt{x_n} - \sqrt{y_n})^2$

Since $x_n > y_n$, we have $\sqrt{x_n} > \sqrt{y_n}$, so $(\sqrt{x_n} - \sqrt{y_n})^2 > 0$, giving $x_{n+1} - y_{n+1} > 0$.

For the upper bound, since $\sqrt{x_n} > \sqrt{y_n}$:

$(\sqrt{x_n} - \sqrt{y_n})^2 = (\sqrt{x_n} - \sqrt{y_n})(\sqrt{x_n} - \sqrt{y_n}) < (\sqrt{x_n} - \sqrt{y_n})(\sqrt{x_n} + \sqrt{y_n}) = x_n - y_n$

where we used $\sqrt{x_n} - \sqrt{y_n} < \sqrt{x_n} + \sqrt{y_n}$ (since $\sqrt{y_n} > 0$). Therefore:

$x_{n+1} - y_{n+1} = \frac{1}{2}(\sqrt{x_n} - \sqrt{y_n})^2 < \frac{1}{2}(x_n - y_n)$

Combining: $0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)$.

By repeated application, $0 < x_n - y_n < \frac{1}{2^n}(x_0 - y_0)$. Since $\frac{1}{2^n}(x_0 - y_0) \to 0$ as $n \to \infty$, we conclude $x_n - y_n \to 0$.

Convergence of $x_n$:

Since $x_n = y_n + (x_n - y_n)$, and $y_n \to M$ while $x_n - y_n \to 0$:

$x_n \to M + 0 = M$

Part (ii)

Setting up the substitution:

Let $t = \frac{1}{2}\left(x - \frac{pq}{x}\right)$, so:

$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2}\left(1 + \frac{pq}{x^2}\right)$

Since $p, q > 0$, we have $\frac{\mathrm{d}t}{\mathrm{d}x} > 0$ for all $x > 0$, so $t$ is a strictly increasing function of $x$ on $(0, \infty)$.

As $x \to 0^+$: $t \to \frac{1}{2}\left(0 - \frac{pq}{0^+}\right) \to -\infty$.

As $x \to \infty$: $t \to \frac{1}{2}(\infty - 0) \to \infty$.

Expressing the integrand in terms of $x$:

We compute $t^2 = \frac{1}{4}\left(x - \frac{pq}{x}\right)^2 = \frac{1}{4x^2}(x^2 - pq)^2 = \frac{x^4 - 2pqx^2 + p^2q^2}{4x^2}$.

First factor:

$\frac{1}{4}(p + q)^2 + t^2 = \frac{(p+q)^2}{4} + \frac{x^4 - 2pqx^2 + p^2q^2}{4x^2}$

$= \frac{x^2(p+q)^2 + x^4 - 2pqx^2 + p^2q^2}{4x^2}$

$= \frac{x^4 + (p^2 + 2pq + q^2 - 2pq)x^2 + p^2q^2}{4x^2}$

$= \frac{x^4 + (p^2 + q^2)x^2 + p^2q^2}{4x^2} = \frac{(x^2 + p^2)(x^2 + q^2)}{4x^2}$

Second factor:

$pq + t^2 = pq + \frac{x^4 - 2pqx^2 + p^2q^2}{4x^2} = \frac{4pqx^2 + x^4 - 2pqx^2 + p^2q^2}{4x^2}$

$= \frac{x^4 + 2pqx^2 + p^2q^2}{4x^2} = \frac{(x^2 + pq)^2}{4x^2}$

Multiplying:

$\left(\frac{1}{4}(p+q)^2 + t^2\right)\left(pq + t^2\right) = \frac{(x^2 + p^2)(x^2 + q^2)}{4x^2} \cdot \frac{(x^2 + pq)^2}{4x^2}$

Taking the square root:

$\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)\left(pq + t^2\right)} = \frac{(x^2 + pq)}{4x^2}\sqrt{(x^2 + p^2)(x^2 + q^2)}$

(since $x^2 + pq > 0$ for $x > 0$).

Substituting into the integral:

$\int_{-\infty}^{\infty} \frac{\mathrm{d}t}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}$

$= \int_{0}^{\infty} \frac{\frac{1}{2}\left(1 + \frac{pq}{x^2}\right)}{\frac{(x^2 + pq)}{4x^2}\sqrt{(x^2 + p^2)(x^2 + q^2)}} \, \mathrm{d}x$

Simplifying the ratio of the derivative to the denominator factor:

$\frac{\frac{1}{2}\left(1 + \frac{pq}{x^2}\right)}{\frac{x^2 + pq}{4x^2}} = \frac{\frac{1}{2} \cdot \frac{x^2 + pq}{x^2}}{\frac{x^2 + pq}{4x^2}} = \frac{1}{2} \cdot 4 = 2$

So the integral becomes:

$\int_{0}^{\infty} \frac{2}{\sqrt{(x^2 + p^2)(x^2 + q^2)}} \, \mathrm{d}x = 2\,\text{I}(p, q) \qquad \text{(**)}$

Applying the symmetry of the integrand:

The integrand $\frac{1}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}$ is an even function of $t$ (since it depends on $t^2$ only). Therefore:

$\int_{-\infty}^{\infty} \frac{\mathrm{d}t}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}} = 2\int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}$

But $\int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}$ is exactly $\text{I}\!\left(\frac{p+q}{2}, \sqrt{pq}\right) = \text{I}(\text{a}(p,q), \text{g}(p,q))$.

So the left-hand side equals $2\,\text{I}(\text{a}(p,q), \text{g}(p,q))$.

From $(**)$, this also equals $2\,\text{I}(p, q)$. Dividing by $2$:

$\text{I}(p, q) = \text{I}(\text{a}(p, q), \text{g}(p, q))$

Evaluating $\text{I}(x_0, y_0)$:

Applying the identity repeatedly with the AGM sequences $\{x_n\}$ and $\{y_n\}$:

$\text{I}(x_0, y_0) = \text{I}(x_1, y_1) = \text{I}(x_2, y_2) = \cdots$

Since $x_n \to M$ and $y_n \to M$, we have $\text{a}(x_n, y_n) \to M$ and $\text{g}(x_n, y_n) \to M$. Taking the limit:

$\text{I}(x_0, y_0) = \text{I}(M, M) = \int_{0}^{\infty} \frac{1}{\sqrt{(M^2 + x^2)(M^2 + x^2)}} \, \mathrm{d}x = \int_{0}^{\infty} \frac{1}{M^2 + x^2} \, \mathrm{d}x$

$= \left[\frac{1}{M}\arctan\frac{x}{M}\right]_{0}^{\infty} = \frac{1}{M}\left(\frac{\pi}{2} - 0\right) = \frac{\pi}{2M}$

Examiner Notes

约一半尝试。严格不等式证明是常见薄弱点。