STEP3 2011 -- Pure Mathematics

Exam: STEP3  |  Year: 2011  |  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	微分方程 Differential Equations	Standard	变量代换, 积分因子, 降阶法
2	代数 Algebra	Standard	有理根定理, 反证法, 整除性分析
3	代数 Algebra	Challenging	三次多项式分解, 复数单位根, 代数恒等式
4	分析 Analysis	Challenging	Young不等式, 积分性质, 反三角函数
5	几何与力学 Geometry and Mechanics	Hard	参数方程, 扫掠面积公式, 三角积分
6	积分 Integral Calculus	Challenging	变量代换, 双曲函数恒等式, artanh 与对数关系
7	数论与代数 Number Theory & Algebra	Challenging	数学归纳法, 二项展开, Pell 方程结构
8	复数 Complex Numbers	Challenging	Möbius 变换, 实虚部分离, 半角代换 tan(θ/2), 轨迹分析

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Question 1

Topic: 微分方程 Differential Equations  |  Difficulty: Standard  |  Marks: 20

Problem

1 (i) Find the general solution of the differential equation $\frac{\mathrm{d}u}{\mathrm{d}x} - \left( \frac{x + 2}{x + 1} \right) u = 0 .$

(ii) Show that substituting $y = ze^{-x}$ (where $z$ is a function of $x$) into the second order differential equation $(x + 1) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x \frac{\mathrm{d}y}{\mathrm{d}x} - y = 0 \qquad \text{(*)}$ leads to a first order differential equation for $\frac{\mathrm{d}z}{\mathrm{d}x}$. Find $z$ and hence show that the general solution of $(*)$ is $y = Ax + Be^{-x} ,$ where $A$ and $B$ are arbitrary constants.

(iii) Find the general solution of the differential equation $(x + 1) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x \frac{\mathrm{d}y}{\mathrm{d}x} - y = (x + 1)^2 .$

Hint

1. (i) The differential equation can be solved either by separating variables or using an integrating factor. In either case, $\int \left( \frac{x+2}{x+1} \right) dx$, or the negative of it is required, and this can be found either by re-writing $\left( \frac{x+2}{x+1} \right)$ as $1 + \frac{1}{x+1}$ or using the substitution, $v = x + 1$. Thus the solution is $u = k(x + 1)e^x$.

(ii) The substitution $y = ze^{-x}$ yields $\frac{dy}{dx} = z'e^{-x} - ze^{-x}$, and $\frac{d^2y}{dx^2} = z''e^{-x} - 2z'e^{-x} + ze^{-x}$.

Substituting these expressions in the differential equation and simplifying gives

$((x + 1)z'' - (x + 2)z')e^{-x} = 0$ which is effectively the first order differential equation from part (i) with $u = z'$.

So $z' = k(x + 1)e^x$, which is an exact differential (or integration by parts could be used), $z = kxe^x + c$ and so $y = Ax + Be^{-x}$ as required.

(iii) Part (ii)’s substitution gives $z'' - \frac{(x+2)}{(x+1)}z' = (x + 1)e^x$ which using the integrating factor from part (i) gives $\frac{e^{-x}}{x+1}z' = \int 1 dx = x + c$, and thus

$y = (x^2 + 1) + Ax + Be^{-x}$. Alternatively, the solution to part (ii) is the complementary function and a quadratic particular integral should be conjectured, which in view of the cf need only be $y = Cx^2 + D$, yielding the same result.

Model Solution

Part (i)

The equation is first-order linear. Separating variables:

$\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{x + 2}{x + 1} \, u$

$\int \frac{\mathrm{d}u}{u} = \int \frac{x + 2}{x + 1} \, \mathrm{d}x$

We simplify the integrand: $\frac{x + 2}{x + 1} = 1 + \frac{1}{x + 1}$, so

$\ln |u| = x + \ln|x + 1| + \text{const}$

$u = k(x + 1)e^x$

where $k$ is an arbitrary constant.

Part (ii)

Let $y = ze^{-x}$. Then:

$\frac{\mathrm{d}y}{\mathrm{d}x} = z'e^{-x} - ze^{-x} = (z' - z)e^{-x}$

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = (z'' - z')e^{-x} - (z' - z)e^{-x} = (z'' - 2z' + z)e^{-x}$

Substituting into $(x + 1)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x\frac{\mathrm{d}y}{\mathrm{d}x} - y = 0$:

$(x + 1)(z'' - 2z' + z)e^{-x} + x(z' - z)e^{-x} - ze^{-x} = 0$

Dividing through by $e^{-x}$:

$(x + 1)z'' - 2(x + 1)z' + (x + 1)z + xz' - xz - z = 0$

Collecting terms:

$(x + 1)z'' + [-2(x + 1) + x]z' + [(x + 1) - x - 1]z = 0$

$(x + 1)z'' - (x + 2)z' = 0$

This is a first-order equation in $z'$, which is exactly the equation from part (i) with $u = z'$. So:

$z' = k(x + 1)e^x$

Integrating (by parts, with $u = x + 1$, $\mathrm{d}v = e^x \mathrm{d}x$):

$z = k \int (x + 1)e^x \, \mathrm{d}x = k[xe^x] + c = kxe^x + c$

Therefore $y = ze^{-x} = kx + ce^{-x}$, i.e.,

$y = Ax + Be^{-x}$

where $A$ and $B$ are arbitrary constants.

Part (iii)

Using the same substitution $y = ze^{-x}$, the left-hand side of

$(x + 1)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x\frac{\mathrm{d}y}{\mathrm{d}x} - y = (x + 1)^2$

reduces (as in part (ii)) to $[(x + 1)z'' - (x + 2)z']e^{-x}$. So:

$[(x + 1)z'' - (x + 2)z']e^{-x} = (x + 1)^2$

$(x + 1)z'' - (x + 2)z' = (x + 1)^2 e^x$

Setting $u = z'$:

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

The integrating factor is $\frac{e^{-x}}{x + 1}$ (from part (i)). Multiplying through:

$\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{e^{-x}}{x + 1} u \right) = 1$

$\frac{e^{-x}}{x + 1} u = x + c_1$

$u = (x + c_1)(x + 1)e^x$

Since $u = z'$, integrating:

$z = \int (x + c_1)(x + 1)e^x \, \mathrm{d}x = \int (x^2 + (c_1 + 1)x + c_1)e^x \, \mathrm{d}x$

Using $\int xe^x \, \mathrm{d}x = xe^x - e^x$ and $\int x^2 e^x \, \mathrm{d}x = (x^2 - 2x + 2)e^x$:

$z = (x^2 - 2x + 2)e^x + (c_1 + 1)(xe^x - e^x) + c_1 e^x + c_2$

$= \left[ x^2 - 2x + 2 + (c_1 + 1)x - (c_1 + 1) + c_1 \right] e^x + c_2$

$= \left[ x^2 + (c_1 - 1)x + 1 \right] e^x + c_2$

Therefore $y = ze^{-x} = x^2 + (c_1 - 1)x + 1 + c_2 e^{-x}$.

Setting $A = c_1 - 1$ and $B = c_2$:

$y = (x^2 + 1) + Ax + Be^{-x}$

where $A$ and $B$ are arbitrary constants.

Examiner Notes

As might be expected, this was a very popular question, in fact the most popular being attempted by very nearly all the candidates. Fortunately, it was also generally well-attempted, with scores well above those for other questions. Apart from frequent algebraic errors and overlooking terms, especially when using results from a previous part that required adaptation, the main difficulties were in showing that (*) in part (ii) did indeed lead to a first order differential equation in $\frac{dz}{dx}$, and the consequent solution of that equation. Part (iii) was generally well done. At the other end of the scale, some candidates did leave their answers to part (i) in the form $\ln u =$.

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Question 2

Topic: 代数 Algebra  |  Difficulty: Standard  |  Marks: 20

Problem

2 The polynomial $f(x)$ is defined by $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_2x^2 + a_1x + a_0 ,$ where $n \geqslant 2$ and the coefficients $a_0, \dots, a_{n-1}$ are integers, with $a_0 \neq 0$. Suppose that the equation $f(x) = 0$ has a rational root $p/q$, where $p$ and $q$ are integers with no common factor greater than 1, and $q > 0$. By considering $q^{n-1}f(p/q)$, find the value of $q$ and deduce that any rational root of the equation $f(x) = 0$ must be an integer.

(i) Show that the $n$th root of 2 is irrational for $n \geqslant 2$.

(ii) Show that the cubic equation $x^3 - x + 1 = 0$ has no rational roots.

(iii) Show that the polynomial equation $x^n - 5x + 7 = 0$ has no rational roots for $n \geqslant 2$.

Hint

2. Parts (ii) and (iii) could be shown by applying the stem and then considering the left hand side of each equation for the cases $n$ even and $n$ odd.

Model Solution

Stem. Suppose $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ has a rational root $p/q$ where $\gcd(p, q) = 1$ and $q > 0$. Then $f(p/q) = 0$:

$\frac{p^n}{q^n} + a_{n-1}\frac{p^{n-1}}{q^{n-1}} + \dots + a_1 \frac{p}{q} + a_0 = 0$

Multiplying by $q^{n-1}$:

$\frac{p^n}{q} + a_{n-1}p^{n-1} + a_{n-2}p^{n-2}q + \dots + a_1 pq^{n-2} + a_0 q^{n-1} = 0$

All terms except the first are integers, so $\frac{p^n}{q}$ must be an integer. Since $\gcd(p, q) = 1$, we have $\gcd(p^n, q) = 1$, so $q = 1$. Therefore any rational root of $f(x) = 0$ must be an integer, and by the factor theorem it must divide $a_0$.

Part (i)

Suppose $\sqrt[n]{2}$ is rational for some $n \geqslant 2$. Then $x = \sqrt[n]{2}$ satisfies $x^n - 2 = 0$, which is a monic polynomial with integer coefficients and constant term $-2 \neq 0$. By the stem, any rational root must be an integer dividing $-2$, i.e., $x \in \{1, -1, 2, -2\}$.

• $x = 1$: $1 - 2 = -1 \neq 0$.
• $x = -1$: $(-1)^n - 2$. For $n$ even this is $-1$, for $n$ odd this is $-3$. Neither is zero.
• $x = 2$: $2^n - 2 \geqslant 2$ for $n \geqslant 2$. Not zero.
• $x = -2$: $(-2)^n - 2$. For $n = 2$ this is $2$; for $n \geqslant 3$, $|(-2)^n| \geqslant 8 > 2$. Not zero.

None of the candidates is a root, so $\sqrt[n]{2}$ is irrational for all $n \geqslant 2$.

Part (ii)

The equation $x^3 - x + 1 = 0$ is monic with integer coefficients and constant term $1$. By the stem, any rational root must be an integer dividing $1$, so $x = 1$ or $x = -1$.

• $x = 1$: $1 - 1 + 1 = 1 \neq 0$.
• $x = -1$: $-1 + 1 + 1 = 1 \neq 0$.

So $x^3 - x + 1 = 0$ has no rational roots.

Part (iii)

The equation $x^n - 5x + 7 = 0$ is monic with integer coefficients and constant term $7$. By the stem, any rational root must be an integer dividing $7$, so $x \in \{1, -1, 7, -7\}$.

• $x = 1$: $1 - 5 + 7 = 3 \neq 0$.
• $x = -1$: $(-1)^n + 5 + 7$. For $n$ even this is $13$; for $n$ odd this is $11$. Neither is zero.
• $x = 7$: $7^n - 35 + 7 = 7^n - 28$. For $n = 2$: $49 - 28 = 21 \neq 0$. For $n \geqslant 3$: $7^n \geqslant 343$, so $7^n - 28 > 0$. Not zero.
• $x = -7$: $(-7)^n + 35 + 7 = (-7)^n + 42$. For $n$ even: $7^n + 42 > 0$. For $n$ odd: $-7^n + 42$. For $n = 2$: $49 + 42 > 0$. For $n \geqslant 3$: $7^n \geqslant 343 > 42$, so $-7^n + 42 < 0$. Not zero.

None of the candidates is a root, so $x^n - 5x + 7 = 0$ has no rational roots for $n \geqslant 2$.

Examiner Notes

2. Of the attempts, about a third were close to completely correct, and nearly all the others were barely doing more than grasping at crumbs, reflecting the fact that candidates either did or did not know what they were doing. There was negligible middle ground.

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Question 3

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

Problem

3 Show that, provided $q^2 \neq 4p^3$, the polynomial

$x^3 - 3px + q \qquad (p \neq 0, \ q \neq 0)$

can be written in the form

$a(x - \alpha)^3 + b(x - \beta)^3 ,$

where $\alpha$ and $\beta$ are the roots of the quadratic equation $pt^2 - qt + p^2 = 0$, and $a$ and $b$ are constants which you should express in terms of $\alpha$ and $\beta$.

Hence show that one solution of the equation $x^3 - 24x + 48 = 0$ is

$x = \frac{2(2 - 2^{\frac{1}{3}})}{1 - 2^{\frac{1}{3}}}$

and obtain similar expressions for the other two solutions in terms of $\omega$, where $\omega = e^{2\pi i/3}$.

Find also the roots of $x^3 - 3px + q = 0$ when $p = r^2$ and $q = 2r^3$ for some non-zero constant $r$.

Hint

3. Considering the quadratic equation $pt^2 - qt + p^2 = 0$, the condition $q^2 \neq 4p^3$ shows, by considering the discriminant, that the roots are unequal. Supposing that $x^3 - 3px + q$ can be written as $a(x - \alpha)^3 + b(x - \beta)^3$, and equating coefficients generates the four equations $a + b = 1$ $-3a\alpha - 3b\beta = 0$ $3\alpha^2 a + 3\beta^2 b = -3p$ $-\alpha^3 a - \beta^3 b = q$

The first pair may be solved simultaneously to give $a = \frac{-\beta}{\alpha - \beta}$ and $b = \frac{\alpha}{\alpha - \beta}$. Substitution yields $p = \alpha\beta$ and $q = \alpha\beta(\alpha + \beta)$, or alternatively, $\alpha\beta = p$ and $\alpha + \beta = \frac{q}{p}$ and so $\alpha$ and $\beta$ satisfy $t^2 - \frac{q}{p}t + p = 0$ i.e. $pt^2 - qt + p^2 = 0$. For $p = 8$, $q = 48$, $q^2 - 4p^3 = 2^8 \neq 0$. Hence $\alpha$ and $\beta$ are the roots of $8t^2 - 48t + 64 = 0$, i.e. $t^2 - 6t + 8 = 0$ and wlog $\alpha = 2$, $\beta = 4$, $a = 2$, $b = -1$. So $x^3 - 24x + 48 = 0$ can be re-arranged as $\left(\frac{x-4}{x-2}\right)^3 = 2$

As $\omega^3 = 1$, $\frac{x-4}{x-2} = \sqrt[3]{2}$, $\omega\sqrt[3]{2}$, $\omega^2\sqrt[3]{2}$ and so $x = \frac{2(2-\sqrt[3]{2})}{1-\sqrt[3]{2}}, \frac{2(2-\omega\sqrt[3]{2})}{1-\omega\sqrt[3]{2}}, \frac{2(2-\omega^2\sqrt[3]{2})}{1-\omega^2\sqrt[3]{2}}$

If $q = 2r^3$ and $p = r^2$, $q^2 = 4p^3$ so the first part cannot be used. However, $x^3 - 3r^2x + 2r^3 = 0$ can be readily factorised as $(x - r)^2(x + 2r) = 0$ and so $x = r$ (repeated) or $-2r$

Model Solution

Part 1: Expressing the cubic in the required form

Suppose $x^3 - 3px + q = a(x - \alpha)^3 + b(x - \beta)^3$. Expanding the right-hand side:

$a(x - \alpha)^3 + b(x - \beta)^3 = a(x^3 - 3\alpha x^2 + 3\alpha^2 x - \alpha^3) + b(x^3 - 3\beta x^2 + 3\beta^2 x - \beta^3)$

$= (a + b)x^3 - 3(a\alpha + b\beta)x^2 + 3(a\alpha^2 + b\beta^2)x - (a\alpha^3 + b\beta^3).$

Equating coefficients with $x^3 - 3px + q$:

$a + b = 1 \qquad \cdots (1)$ $a\alpha + b\beta = 0 \qquad \cdots (2)$ $a\alpha^2 + b\beta^2 = -p \qquad \cdots (3)$ $-(a\alpha^3 + b\beta^3) = q \qquad \cdots (4)$

From equation (2), $a = -b\beta/\alpha$ (assuming $\alpha \neq 0$, which holds since $\alpha\beta = p \neq 0$). Substituting into (1):

$-\frac{b\beta}{\alpha} + b = 1 \implies b\left(\frac{\alpha - \beta}{\alpha}\right) = 1 \implies b = \frac{\alpha}{\alpha - \beta}.$

Then $a = 1 - b = \frac{-\beta}{\alpha - \beta}$.

We need $\alpha \neq \beta$ for these to be well-defined. The discriminant of $pt^2 - qt + p^2 = 0$ is $q^2 - 4p^3$, so $q^2 \neq 4p^3$ guarantees distinct roots.

Verifying equations (3) and (4) and finding the quadratic for $\alpha, \beta$:

From equation (2), $a\alpha + b\beta = 0$, so $a\alpha = -b\beta$.

For equation (3): $a\alpha^2 + b\beta^2 = \alpha(a\alpha) + \beta(b\beta) = \alpha(-b\beta) + \beta(b\beta) = b\beta(\beta - \alpha)$.

Substituting $b = \frac{\alpha}{\alpha - \beta}$:

$b\beta(\beta - \alpha) = \frac{\alpha\beta(\beta - \alpha)}{\alpha - \beta} = -\alpha\beta.$

So equation (3) gives $-\alpha\beta = -p$, hence $\alpha\beta = p$.

For equation (4): $a\alpha^3 + b\beta^3 = \alpha^2(a\alpha) + \beta^2(b\beta) = \alpha^2(-b\beta) + \beta^2(b\beta) = b\beta(\beta^2 - \alpha^2) = b\beta(\beta - \alpha)(\beta + \alpha)$.

$b\beta(\beta - \alpha)(\beta + \alpha) = \frac{\alpha\beta(\beta - \alpha)(\beta + \alpha)}{\alpha - \beta} = -\alpha\beta(\alpha + \beta) = -p(\alpha + \beta).$

So equation (4) gives $-(-p(\alpha + \beta)) = q$, hence $p(\alpha + \beta) = q$, giving $\alpha + \beta = q/p$.

Since $\alpha\beta = p$ and $\alpha + \beta = q/p$, the numbers $\alpha$ and $\beta$ are roots of:

$t^2 - \frac{q}{p}t + p = 0 \iff pt^2 - qt + p^2 = 0.$

This completes the proof that $x^3 - 3px + q = a(x - \alpha)^3 + b(x - \beta)^3$ where $\alpha, \beta$ are roots of $pt^2 - qt + p^2 = 0$, with $a = \frac{-\beta}{\alpha - \beta}$ and $b = \frac{\alpha}{\alpha - \beta}$.

Part 2: Solving $x^3 - 24x + 48 = 0$

Here $p = 8$ and $q = 48$. We check $q^2 - 4p^3 = 2304 - 2048 = 256 \neq 0$, so the result applies.

The quadratic for $\alpha, \beta$ is $8t^2 - 48t + 64 = 0$, i.e., $t^2 - 6t + 8 = 0$, giving $(t-2)(t-4) = 0$.

Taking $\alpha = 2$ and $\beta = 4$:

$a = \frac{-4}{2 - 4} = 2, \qquad b = \frac{2}{2 - 4} = -1.$

So the equation becomes:

$2(x - 2)^3 - (x - 4)^3 = 0 \implies (x - 4)^3 = 2(x - 2)^3.$

Provided $x \neq 2$, we divide:

$\left(\frac{x - 4}{x - 2}\right)^3 = 2 \implies \frac{x - 4}{x - 2} = 2^{1/3}\omega^k, \quad k = 0, 1, 2,$

where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity.

Solving $\frac{x - 4}{x - 2} = c$ for $x$: $x - 4 = c(x - 2)$, so $x(1 - c) = 4 - 2c$, giving $x = \frac{4 - 2c}{1 - c} = \frac{2(2 - c)}{1 - c}$.

For $k = 0$: $c = 2^{1/3}$, so

$x = \frac{2(2 - 2^{1/3})}{1 - 2^{1/3}}.$

For $k = 1$: $c = \omega \cdot 2^{1/3}$, so

$x = \frac{2(2 - \omega \cdot 2^{1/3})}{1 - \omega \cdot 2^{1/3}}.$

For $k = 2$: $c = \omega^2 \cdot 2^{1/3}$, so

$x = \frac{2(2 - \omega^2 \cdot 2^{1/3})}{1 - \omega^2 \cdot 2^{1/3}}.$

Part 3: Finding roots when $p = r^2$ and $q = 2r^3$

Here $q^2 - 4p^3 = 4r^6 - 4r^6 = 0$, so the method above does not apply (the quadratic has a repeated root). Instead, we factorise directly:

$x^3 - 3r^2 x + 2r^3 = 0.$

Testing $x = r$: $r^3 - 3r^3 + 2r^3 = 0$. So $(x - r)$ is a factor. Performing the division:

$x^3 - 3r^2 x + 2r^3 = (x - r)(x^2 + rx - 2r^2) = (x - r)(x - r)(x + 2r) = (x - r)^2(x + 2r).$

The roots are $x = r$ (repeated) and $x = -2r$.

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Question 4

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

Problem

4 The following result applies to any function $f$ which is continuous, has positive gradient and satisfies $f(0) = 0$:

$ab \leqslant \int_{0}^{a} f(x) \, dx + \int_{0}^{b} f^{-1}(y) \, dy , \qquad \text{(*)}$

where $f^{-1}$ denotes the inverse function of $f$, and $a \geqslant 0$ and $b \geqslant 0$.

(i) By considering the graph of $y = f(x)$, explain briefly why the inequality ($*$) holds. In the case $a > 0$ and $b > 0$, state a condition on $a$ and $b$ under which equality holds.

(ii) By taking $f(x) = x^{p-1}$ in ($*$), where $p > 1$, show that if $\frac{1}{p} + \frac{1}{q} = 1$ then

$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q} .$

Verify that equality holds under the condition you stated above.

(iii) Show that, for $0 \leqslant a \leqslant \frac{1}{2}\pi$ and $0 \leqslant b \leqslant 1$,

$ab \leqslant b \arcsin b + \sqrt{1 - b^2} - \cos a .$

Deduce that, for $t \geqslant 1$,

$\arcsin(t^{-1}) \geqslant t - \sqrt{t^2 - 1} .$

Hint

4. (i) $\int_0^a f(x)dx$ is the area between the curve $y = f(x)$, the $x$ axis, and the line $x = a$ $\int_0^b f^{-1}(y)dy$ is the area between the curve $y = f(x)$, the $y$ axis, and the line $y = b$.

The sum of these areas is greater than or equal to the area of the rectangle with corners $(0,0)$, $(a,0)$, $(a,b)$ and $(0,b)$, with equality holding if $b = f(a)$.

(ii) With $f(x) = x^{p-1}$, the sum of the two integrals is $\frac{1}{p}a^p + \frac{p-1}{p}b^{\frac{p}{p-1}}$

But as $\frac{1}{p} + \frac{1}{q} = 1$, $\frac{1}{q} = \frac{p-1}{p}$, and so the required result follows by applying the result of part (i). If $b = a^{p-1}$, simple algebra shows $a = b^{q-1}$, so $\frac{1}{p} a^p + \frac{1}{q} b^q = \frac{1}{p} ab + \frac{1}{q} ba = ab$ and equality is verified.

(iii) $f(x) = \sin x$ satisfies the conditions of part (i) So $\int_0^a f(x) dx = 1 - \cos a$, and, by parts, $\int_0^b f^{-1}(y) dy = b \sin^{-1} b + \sqrt{1 - b^2} - 1$ which together give the required result.

Choosing $a = 0$, and $b = t^{-1}$, part (i) gives $0 \le t^{-1} \sin^{-1}(t^{-1}) + \sqrt{1 - t^{-2}} - 1$ which can be re-arranged to give the required result.

Model Solution

Part (i)

The integral $\int_0^a f(x) \, \mathrm{d}x$ is the area between the curve $y = f(x)$, the $x$-axis, and the line $x = a$. The integral $\int_0^b f^{-1}(y) \, \mathrm{d}y$ is the area between the curve $y = f(x)$, the $y$-axis, and the line $y = b$.

Since $f$ is continuous with positive gradient and $f(0) = 0$, the function is strictly increasing. The sum of these two areas always covers the rectangle $[0, a] \times [0, b]$, so:

$ab \leqslant \int_0^a f(x) \, \mathrm{d}x + \int_0^b f^{-1}(y) \, \mathrm{d}y.$

Equality holds when $b = f(a)$ (equivalently $a = f^{-1}(b)$), since in this case the two integrals partition the rectangle exactly with no overlap or gap.

Part (ii)

Taking $f(x) = x^{p-1}$ for $p > 1$. This is continuous with positive gradient for $x > 0$ and $f(0) = 0$, so the conditions of ($*$) are satisfied.

The inverse function: $y = x^{p-1}$ gives $x = y^{1/(p-1)} = y^{q-1}$ since $\frac{1}{p} + \frac{1}{q} = 1$ implies $q - 1 = \frac{1}{p-1}$.

Computing the integrals:

$\int_0^a x^{p-1} \, \mathrm{d}x = \frac{a^p}{p}$

$\int_0^b y^{q-1} \, \mathrm{d}y = \frac{b^q}{q}$

By ($*$):

$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}$

Equality check. Equality in ($*$) holds when $b = f(a) = a^{p-1}$, i.e., $a = b^{1/(p-1)} = b^{q-1}$. Then $a^p = a \cdot a^{p-1} = ab$ and $b^q = b \cdot b^{q-1} = ba = ab$. So:

$\frac{a^p}{p} + \frac{b^q}{q} = \frac{ab}{p} + \frac{ab}{q} = ab\left(\frac{1}{p} + \frac{1}{q}\right) = ab.$

Equality is verified.

Part (iii)

Take $f(x) = \sin x$ for $0 \leqslant x \leqslant \frac{\pi}{2}$. This is continuous with positive gradient and $f(0) = 0$, so the conditions of ($*$) are satisfied.

The inverse function is $f^{-1}(y) = \arcsin y$ for $0 \leqslant y \leqslant 1$.

Computing the integrals:

$\int_0^a \sin x \, \mathrm{d}x = [-\cos x]_0^a = 1 - \cos a$

For $\int_0^b \arcsin y \, \mathrm{d}y$, integrate by parts with $u = \arcsin y$, $\mathrm{d}v = \mathrm{d}y$:

$\int_0^b \arcsin y \, \mathrm{d}y = [y \arcsin y]_0^b - \int_0^b \frac{y}{\sqrt{1 - y^2}} \, \mathrm{d}y$

$= b \arcsin b + [\sqrt{1 - y^2}]_0^b = b \arcsin b + \sqrt{1 - b^2} - 1$

By ($*$):

$ab \leqslant (1 - \cos a) + (b \arcsin b + \sqrt{1 - b^2} - 1)$

$ab \leqslant b \arcsin b + \sqrt{1 - b^2} - \cos a$

Deduction. In this inequality, set $a = 0$ (which gives $0 \leqslant$ something) and $b = t^{-1}$ for $t \geqslant 1$ (so $0 < b \leqslant 1$). Since $a = 0$, the left side is $0$:

$0 \leqslant t^{-1} \arcsin(t^{-1}) + \sqrt{1 - t^{-2}} - 1$

$1 \leqslant t^{-1} \arcsin(t^{-1}) + \sqrt{1 - t^{-2}}$

Multiply through by $t > 0$:

$t \leqslant \arcsin(t^{-1}) + t\sqrt{1 - t^{-2}}$

Since $t\sqrt{1 - t^{-2}} = \sqrt{t^2 - 1}$ for $t \geqslant 1$:

$t \leqslant \arcsin(t^{-1}) + \sqrt{t^2 - 1}$

$\arcsin(t^{-1}) \geqslant t - \sqrt{t^2 - 1}$

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Question 5

Topic: 几何与力学 Geometry and Mechanics  |  Difficulty: Hard  |  Marks: 20

Problem

5 A movable point $P$ has cartesian coordinates $(x, y)$, where $x$ and $y$ are functions of $t$. The polar coordinates of $P$ with respect to the origin $O$ are $r$ and $\theta$. Starting with the expression

$\frac{1}{2} \int r^2 \, \mathrm{d}\theta$

for the area swept out by $OP$, obtain the equivalent expression

$\frac{1}{2} \int \left( x \frac{\mathrm{d}y}{\mathrm{d}t} - y \frac{\mathrm{d}x}{\mathrm{d}t} \right) \mathrm{d}t . \qquad \text{(*)}$

The ends of a thin straight rod $AB$ lie on a closed convex curve $\mathcal{C}$. The point $P$ on the rod is a fixed distance $a$ from $A$ and a fixed distance $b$ from $B$. The angle between $AB$ and the positive $x$ direction is $t$. As $A$ and $B$ move anticlockwise round $\mathcal{C}$, the angle $t$ increases from $0$ to $2\pi$ and $P$ traces a closed convex curve $\mathcal{D}$ inside $\mathcal{C}$, with the origin $O$ lying inside $\mathcal{D}$, as shown in the diagram.

Let $(x, y)$ be the coordinates of $P$. Write down the coordinates of $A$ and $B$ in terms of $a, b, x, y$ and $t$.

The areas swept out by $OA, OB$ and $OP$ are denoted by $[A], [B]$ and $[P]$, respectively. Show, using $(*)$, that

$[A] = [P] + \pi a^2 - af$

where

$f = \frac{1}{2} \int_0^{2\pi} \left( \left( x + \frac{\mathrm{d}y}{\mathrm{d}t} \right) \cos t + \left( y - \frac{\mathrm{d}x}{\mathrm{d}t} \right) \sin t \right) \mathrm{d}t .$

Obtain a corresponding expression for $[B]$ involving $b$. Hence show that the area between the curves $\mathcal{C}$ and $\mathcal{D}$ is $\pi ab$.

Hint

$r^2 d\theta = (x^2 + y^2) \frac{d}{dt} \left( \tan^{-1} \left( \frac{y}{x} \right) \right) dt = (x^2 + y^2) \frac{1}{1 + \left( \frac{y}{x} \right)^2} \frac{\left( x \frac{dy}{dt} - y \frac{dx}{dt} \right)}{x^2} dt = \left( x \frac{dy}{dt} - y \frac{dx}{dt} \right) dt$

and hence integrating gives the result.

A is $(x - a \cos t, y - a \sin t)$ and B is $(x + b \cos t, y + b \sin t)$

$[A] = \frac{1}{2} \int_0^{2\pi} (x - a \cos t) \left( \frac{dy}{dt} - a \cos t \right) - (y - a \sin t) \left( \frac{dx}{dt} + a \sin t \right) dt$ using (*) which leads directly to $[A] = [P] - af + \pi a^2$.

Replacing $-a$ by $b$ gives $[B] = [P] + bf + \pi b^2$ As $[A] = [B]$, these expressions can be equated to give $f = \pi(a - b)$.

The area between curves $C$ and $D$ is $[A] - [P] = -af + \pi a^2$ which by substitution gives $\pi ab$ as required.

Model Solution

Deriving the Cartesian area formula.

In polar coordinates, $r^2 = x^2 + y^2$ and $\theta = \arctan(y/x)$. Differentiating $\theta$ with respect to $t$:

$\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{1}{1 + (y/x)^2} \cdot \frac{x \frac{\mathrm{d}y}{\mathrm{d}t} - y \frac{\mathrm{d}x}{\mathrm{d}t}}{x^2} = \frac{x \frac{\mathrm{d}y}{\mathrm{d}t} - y \frac{\mathrm{d}x}{\mathrm{d}t}}{x^2 + y^2}$

So:

$r^2 \, \mathrm{d}\theta = (x^2 + y^2) \cdot \frac{x \frac{\mathrm{d}y}{\mathrm{d}t} - y \frac{\mathrm{d}x}{\mathrm{d}t}}{x^2 + y^2} \, \mathrm{d}t = \left( x \frac{\mathrm{d}y}{\mathrm{d}t} - y \frac{\mathrm{d}x}{\mathrm{d}t} \right) \mathrm{d}t$

Therefore:

$\frac{1}{2}\int r^2 \, \mathrm{d}\theta = \frac{1}{2}\int \left( x \frac{\mathrm{d}y}{\mathrm{d}t} - y \frac{\mathrm{d}x}{\mathrm{d}t} \right) \mathrm{d}t \qquad \text{(*)}$

Coordinates of $A$ and $B$.

Point $A$ is distance $a$ from $P$ along the rod in the direction opposite to $B$. Since the rod makes angle $t$ with the positive $x$-direction:

$A = (x - a\cos t, \quad y - a\sin t)$ $B = (x + b\cos t, \quad y + b\sin t)$

Computing $[A]$.

Using ($*$):

$[A] = \frac{1}{2}\int_0^{2\pi} \left( x_A \frac{\mathrm{d}y_A}{\mathrm{d}t} - y_A \frac{\mathrm{d}x_A}{\mathrm{d}t} \right) \mathrm{d}t$

where $x_A = x - a\cos t$ and $y_A = y - a\sin t$. Differentiating:

$\frac{\mathrm{d}x_A}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}t} + a\sin t, \qquad \frac{\mathrm{d}y_A}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}t} - a\cos t$

The integrand is:

$(x - a\cos t)\left(\frac{\mathrm{d}y}{\mathrm{d}t} - a\cos t\right) - (y - a\sin t)\left(\frac{\mathrm{d}x}{\mathrm{d}t} + a\sin t\right)$

Expanding and collecting terms gives:

$x_A \dot{y}_A = (x - a\cos t)(\dot{y} - a\cos t) = x\dot{y} - ax\cos t - a\dot{y}\cos t + a^2\cos^2 t.$

$y_A \dot{x}_A = (y - a\sin t)(\dot{x} + a\sin t) = y\dot{x} + ay\sin t - a\dot{x}\sin t - a^2\sin^2 t.$

$x_A\dot{y}_A - y_A\dot{x}_A = (x\dot{y} - y\dot{x}) - ax\cos t - a\dot{y}\cos t - ay\sin t + a\dot{x}\sin t + a^2\cos^2 t + a^2\sin^2 t$

$= (x\dot{y} - y\dot{x}) - a(x\cos t + y\sin t) - a(\dot{y}\cos t - \dot{x}\sin t) + a^2$

$= (x\dot{y} - y\dot{x}) - a[(x + \dot{y})\cos t + (y - \dot{x})\sin t] + a^2$

Thus the integrand for $[A]$ is:

$(x\dot{y} - y\dot{x}) - a[(x + \dot{y})\cos t + (y - \dot{x})\sin t] + a^2$

Now:

$[A] = \frac{1}{2}\int_0^{2\pi}\left[(x\dot{y} - y\dot{x}) - a\left(\left(x + \dot{y}\right)\cos t + (y - \dot{x})\sin t\right) + a^2\right]\mathrm{d}t$

$= [P] - a \cdot \frac{1}{2}\int_0^{2\pi}\left[\left(x + \dot{y}\right)\cos t + (y - \dot{x})\sin t\right]\mathrm{d}t + \frac{a^2}{2}\int_0^{2\pi}\mathrm{d}t$

$[A] = [P] - af + \pi a^2$

Computing $[B]$.

Replacing $a$ by $-b$ in the coordinates (since $B = (x + b\cos t, y + b\sin t) = (x - (-b)\cos t, y - (-b)\sin t)$):

$[B] = [P] - (-b)f + \pi(-b)^2 = [P] + bf + \pi b^2$

Finding the area between $\mathcal{C}$ and $\mathcal{D}$.

Since $A$ and $B$ trace the same closed curve $\mathcal{C}$, we have $[A] = [B]$. Equating:

$[P] - af + \pi a^2 = [P] + bf + \pi b^2$

$-af - bf = \pi b^2 - \pi a^2$

$-f(a + b) = -\pi(a^2 - b^2) = -\pi(a - b)(a + b)$

Since $a + b > 0$:

$f = \pi(a - b)$

The area enclosed by $\mathcal{C}$ is $[A]$ and by $\mathcal{D}$ is $[P]$. The area between the curves is:

$[A] - [P] = \pi a^2 - af = \pi a^2 - a\pi(a - b) = \pi a^2 - \pi a^2 + \pi ab = \pi ab$

Examiner Notes

Less than a third of the candidates attempted this. There were quite a few perfect scores, however the vast majority scored less than a quarter of the marks, which was the

mean mark. The general result at the start of the question was the key to success. Those that stumbled with handling four variables in terms of the fifth one, and the consequent calculus, did not attempt to make further progress into the rest of the question.

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Question 6

Topic: 积分 Integral Calculus  |  Difficulty: Challenging  |  Marks: 20

Problem

6 The definite integrals $T, U, V$ and $X$ are defined by

$T = \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\text{artanh } t}{t} \, dt \, , \qquad U = \int_{\ln 2}^{\ln 3} \frac{u}{2 \sinh u} \, du \, ,$

$V = -\int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\ln v}{1 - v^2} \, dv \, , \qquad X = \int_{\frac{1}{2} \ln 2}^{\frac{1}{2} \ln 3} \ln(\coth x) \, dx \, .$

Show, without evaluating any of them, that $T, U, V$ and $X$ are all equal.

Hint

Using the substitution $t = \tanh \left( \frac{u}{2} \right)$, then it can be shown that $T = U$, by making use of $2 \sinh \left( \frac{u}{2} \right) \cosh \left( \frac{u}{2} \right) = \sinh u$ to obtain the integrand, and $\tanh^{-1} t = \frac{1}{2} \ln \left( \frac{1+t}{1-t} \right)$ to obtain the limits.

If instead, integration by parts is used differentiating $\tanh^{-1} t$ and integrating $\frac{1}{t}$, and

employing $\tanh^{-1} t = \frac{1}{2} \ln \left( \frac{1+t}{1-t} \right)$ to demonstrate that $[\tanh^{-1} t \ln t]_{\frac{1}{3}}^1 = 0$, $T = V$.

The substitution $t = e^{-2x}$ can be used to demonstrate that $T = X$.

(Alternatively, starting from $U$, the substitution $u = 2 \tanh^{-1} t$ obtains $U = T$, the substitution $u = -\ln v$ obtains $U = V$, and the substitution $u = 2x$ followed by integration by parts yields $U = X$; starting from $V$, by parts it can be shown that $V = T$, using the substitution $v = e^{-u}$ that $V = U$, and the substitution $v = \tanh x$ that $V = X$; or starting from $X$, the substitution $x = -\frac{1}{2} \ln t$ gives $X = T$, integration by parts gives $X = U$, and the substitution $x = \tanh^{-1} v$ gives $X = V$.)

Model Solution

We show $T = U = V = X$ by establishing pairwise equalities through substitutions and integration by parts.

Showing $T = U$:

Use the substitution $t = \tanh(u/2)$ in $T$.

Limits: $t = 1/3$ gives $\tanh(u/2) = 1/3$, so $u/2 = \text{artanh}(1/3) = \frac{1}{2}\ln\frac{1+1/3}{1-1/3} = \frac{1}{2}\ln 2$, hence $u = \ln 2$. Similarly $t = 1/2$ gives $u = \ln 3$.

Differential: $\mathrm{d}t = \frac{1}{2}\text{sech}^2(u/2) \, \mathrm{d}u$.

Integrand: $\text{artanh}(t) = u/2$ and $t = \tanh(u/2)$, so:

$\frac{\text{artanh}(t)}{t} \, \mathrm{d}t = \frac{u/2}{\tanh(u/2)} \cdot \frac{1}{2}\text{sech}^2(u/2) \, \mathrm{d}u = \frac{u \, \text{sech}^2(u/2)}{4\tanh(u/2)} \, \mathrm{d}u$

Since $\frac{\text{sech}^2(u/2)}{\tanh(u/2)} = \frac{\cosh^{-2}(u/2)}{\sinh(u/2)/\cosh(u/2)} = \frac{1}{\sinh(u/2)\cosh(u/2)} = \frac{2}{\sinh u}$:

$T = \int_{\ln 2}^{\ln 3} \frac{u}{4} \cdot \frac{2}{\sinh u} \, \mathrm{d}u = \int_{\ln 2}^{\ln 3} \frac{u}{2\sinh u} \, \mathrm{d}u = U$

Showing $T = V$:

Use the substitution $t = \tanh x$ in $T$.

Limits: $t = 1/3$ gives $x = \text{artanh}(1/3) = \frac{1}{2}\ln 2$. Similarly $t = 1/2$ gives $x = \frac{1}{2}\ln 3$.

Differential: $\mathrm{d}t = \text{sech}^2 x \, \mathrm{d}x$. Also $\text{artanh}(t) = x$ and $t = \tanh x$, so $1 - t^2 = 1 - \tanh^2 x = \text{sech}^2 x$.

$T = \int_{\frac{1}{2}\ln 2}^{\frac{1}{2}\ln 3} \frac{x}{\tanh x} \cdot \text{sech}^2 x \, \mathrm{d}x$

But $\frac{\text{sech}^2 x}{\tanh x} = \frac{1 - \tanh^2 x}{\tanh x} = \frac{1}{\tanh x} - \tanh x$, and we can also write:

$\frac{\text{sech}^2 x}{\tanh x} = \frac{1}{\sinh x \cosh x} = \frac{2}{\sinh 2x}$

To show $T=V$, integrate $T$ by parts:

$T = \int_{1/3}^{1/2} \text{artanh}(t) \cdot \frac{1}{t} \, \mathrm{d}t$

Let $u = \text{artanh}(t)$, $\mathrm{d}v = \frac{1}{t}\mathrm{d}t$, so $\mathrm{d}u = \frac{1}{1-t^2}\mathrm{d}t$ and $v = \ln t$.

$T = [\text{artanh}(t) \ln t]_{1/3}^{1/2} - \int_{1/3}^{1/2} \frac{\ln t}{1 - t^2} \, \mathrm{d}t$

Evaluating the boundary term: $\text{artanh}(t) = \frac{1}{2}\ln\frac{1+t}{1-t}$.

At $t = 1/2$: $\frac{1}{2}\ln 3 \cdot \ln(1/2) = -\frac{\ln 2 \ln 3}{2}$.

At $t = 1/3$: $\frac{1}{2}\ln 2 \cdot \ln(1/3) = -\frac{\ln 2 \ln 3}{2}$.

So $[\text{artanh}(t) \ln t]_{1/3}^{1/2} = -\frac{\ln 2 \ln 3}{2} - \left(-\frac{\ln 2 \ln 3}{2}\right) = 0$.

Therefore:

$T = -\int_{1/3}^{1/2} \frac{\ln t}{1 - t^2} \, \mathrm{d}t = V$

Showing $T = X$:

Use the substitution $t = e^{-2x}$ in $T$.

Limits: $t = 1/3$ gives $x = \frac{1}{2}\ln 3$. $t = 1/2$ gives $x = \frac{1}{2}\ln 2$.

Differential: $\mathrm{d}t = -2e^{-2x}\mathrm{d}x$.

For the integrand: $\text{artanh}(t) = \frac{1}{2}\ln\frac{1+e^{-2x}}{1-e^{-2x}} = \frac{1}{2}\ln\frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{1}{2}\ln(\coth x)$.

So:

$T = \int_{\frac{1}{2}\ln 3}^{\frac{1}{2}\ln 2} \frac{\frac{1}{2}\ln(\coth x)}{e^{-2x}} \cdot (-2e^{-2x}) \, \mathrm{d}x = \int_{\frac{1}{2}\ln 2}^{\frac{1}{2}\ln 3} \ln(\coth x) \, \mathrm{d}x = X$

Conclusion. Since $T = U$, $T = V$, and $T = X$, all four integrals are equal.

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Question 7

Topic: 数论与代数 Number Theory & Algebra  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Let

$T_n = \left( \sqrt{a+1} + \sqrt{a} \right)^n \, ,$

where $n$ is a positive integer and $a$ is any given positive integer.

(i) In the case when $n$ is even, show by induction that $T_n$ can be written in the form

$A_n + B_n \sqrt{a(a+1)} \, ,$

where $A_n$ and $B_n$ are integers (depending on $a$ and $n$) and $A_n^2 = a(a+1)B_n^2 + 1$.

(ii) In the case when $n$ is odd, show by considering $(\sqrt{a+1} + \sqrt{a})T_m$ where $m$ is even, or otherwise, that $T_n$ can be written in the form

$C_n \sqrt{a+1} + D_n \sqrt{a} \, ,$

where $C_n$ and $D_n$ are integers (depending on $a$ and $n$) and $(a+1)C_n^2 = aD_n^2 + 1$.

(iii) Deduce that, for each $n$, $T_n$ can be written as the sum of the square roots of two consecutive integers.

Hint

(i) The induction requires $T_{k+2} = A_{k+2} + B_{k+2} \sqrt{a(a+1)}$ and $A_{k+2}^2 - a(a+1) B_{k+2}^2 = 1$. $T_{k+2} = \left( A_k + B_k \sqrt{a(a+1)} \right) (\sqrt{a+1} + \sqrt{a})^2 = \left( A_k + B_k \sqrt{a(a+1)} \right) T_2$

$T_2 = \left( 2a + 1 + 2\sqrt{a(a + 1)} \right)$ and so $A_2 = 2a + 1$ and $B_2 = 2$, and $A_2^2 - a(a + 1)B_2^2 = (2a + 1)^2 - a(a + 1)2^2 = 1$ the result is true for $n = 2$. Evaluating $T_{k+2}$ using $\left( A_k + B_k\sqrt{a(a + 1)} \right) T_2$ then $A_{k+2} = (2a + 1)A_k + 2a(a + 1)B_k$ and $B_{k+2} = 2A_k + (2a + 1)B_k$, and so substituting and simplifying, $A_{k+2}^2 - a(a + 1)B_{k+2}^2 = A_k^2 - a(a + 1)B_k^2 = 1$ by the induction.

(ii) $T_n = (\sqrt{a + 1} + \sqrt{a})T_m = (\sqrt{a + 1} + \sqrt{a})\left( A_m + B_m\sqrt{a(a + 1)} \right)$

$= (A_m + aB_m)\sqrt{a + 1} + (A_m + (a + 1)B_m)\sqrt{a}$ which is of required form because

$C_n = A_m + aB_m$ and $D_n = A_m + (a + 1)B_m$ are integers and

$(a + 1)C_n^2 - aD_n^2 = (a + 1)(A_m + aB_m)^2 - a(A_m + (a + 1)B_m)^2$

$= A_m^2 - a(a + 1)B_m^2 = 1$ as required.

Trivially the case $n = 1$ is true.

(iii) In the case that $n$ is even,

$T_n = A_n + B_n\sqrt{a(a + 1)} = \sqrt{A_n^2} + \sqrt{a(a + 1)B_n^2} = \sqrt{a(a + 1)B_n^2 + 1} + \sqrt{a(a + 1)B_n^2}$ as required,

and in the case that $n$ is odd, $T_n = C_n\sqrt{a + 1} + D_n\sqrt{a} = \sqrt{(a + 1)C_n^2} + \sqrt{aD_n^2} =$ $\sqrt{aD_n^2 + 1} + \sqrt{aD_n^2}$ as required.

Model Solution

Part (i): $n$ even.

We prove by induction on even $n$ that $T_n = A_n + B_n\sqrt{a(a+1)}$ where $A_n, B_n$ are integers and $A_n^2 = a(a+1)B_n^2 + 1$.

Base case $n = 2$:

$T_2 = (\sqrt{a+1} + \sqrt{a})^2 = (a+1) + 2\sqrt{a(a+1)} + a = (2a+1) + 2\sqrt{a(a+1)}$

So $A_2 = 2a+1$ and $B_2 = 2$, which are integers. Checking:

$A_2^2 - a(a+1)B_2^2 = (2a+1)^2 - a(a+1) \cdot 4 = 4a^2 + 4a + 1 - 4a^2 - 4a = 1$

Inductive step. Suppose $T_k = A_k + B_k\sqrt{a(a+1)}$ with $A_k^2 - a(a+1)B_k^2 = 1$ for some even $k$. Then:

$T_{k+2} = T_k \cdot T_2 = (A_k + B_k\sqrt{a(a+1)})((2a+1) + 2\sqrt{a(a+1)})$

Expanding:

$A_{k+2} = (2a+1)A_k + 2a(a+1)B_k$

$B_{k+2} = 2A_k + (2a+1)B_k$

Both are integers (since $A_k, B_k, a$ are integers). Now verify $A_{k+2}^2 - a(a+1)B_{k+2}^2 = 1$:

$A_{k+2}^2 = [(2a+1)A_k + 2a(a+1)B_k]^2 = (2a+1)^2 A_k^2 + 4a(a+1)(2a+1)A_k B_k + 4a^2(a+1)^2 B_k^2$

$a(a+1)B_{k+2}^2 = a(a+1)[2A_k + (2a+1)B_k]^2 = 4a(a+1)A_k^2 + 4a(a+1)(2a+1)A_k B_k + a(a+1)(2a+1)^2 B_k^2$

Subtracting (the cross terms cancel):

$A_{k+2}^2 - a(a+1)B_{k+2}^2 = [(2a+1)^2 - 4a(a+1)]A_k^2 + [4a^2(a+1)^2 - a(a+1)(2a+1)^2]B_k^2$

Computing the coefficients:

$(2a+1)^2 - 4a(a+1) = 4a^2 + 4a + 1 - 4a^2 - 4a = 1$

$4a^2(a+1)^2 - a(a+1)(2a+1)^2 = a(a+1)[4a(a+1) - (2a+1)^2] = a(a+1)[4a^2+4a - 4a^2-4a-1] = -a(a+1)$

So:

$A_{k+2}^2 - a(a+1)B_{k+2}^2 = A_k^2 - a(a+1)B_k^2 = 1$

by the induction hypothesis. This completes the induction.

Part (ii): $n$ odd.

For odd $n$, write $n = m + 1$ where $m$ is even. Then:

$T_n = (\sqrt{a+1} + \sqrt{a}) \cdot T_m = (\sqrt{a+1} + \sqrt{a})(A_m + B_m\sqrt{a(a+1)})$

Expanding:

$= A_m\sqrt{a+1} + B_m(a+1)\sqrt{a} + A_m\sqrt{a} + B_m a\sqrt{a+1}$

$= (A_m + aB_m)\sqrt{a+1} + (A_m + (a+1)B_m)\sqrt{a}$

So $C_n = A_m + aB_m$ and $D_n = A_m + (a+1)B_m$, which are integers.

Checking the identity:

$(a+1)C_n^2 - aD_n^2 = (a+1)(A_m + aB_m)^2 - a(A_m + (a+1)B_m)^2$

Expanding:

$= (a+1)(A_m^2 + 2aA_m B_m + a^2 B_m^2) - a(A_m^2 + 2(a+1)A_m B_m + (a+1)^2 B_m^2)$

$= (a+1)A_m^2 + 2a(a+1)A_m B_m + a^2(a+1)B_m^2 - aA_m^2 - 2a(a+1)A_m B_m - a(a+1)^2 B_m^2$

The cross terms cancel:

$= [(a+1) - a]A_m^2 + [a^2(a+1) - a(a+1)^2]B_m^2 = A_m^2 + a(a+1)[a - (a+1)]B_m^2$

$= A_m^2 - a(a+1)B_m^2 = 1$

by part (i). (For $n = 1$: $T_1 = \sqrt{a+1} + \sqrt{a}$, so $C_1 = D_1 = 1$ and $(a+1) - a = 1$.)

Part (iii): $T_n$ as a sum of square roots of consecutive integers.

Case 1: $n$ even. We have $T_n = A_n + B_n\sqrt{a(a+1)}$ with $A_n^2 = a(a+1)B_n^2 + 1$.

$T_n = \sqrt{A_n^2} + \sqrt{a(a+1)B_n^2} = \sqrt{a(a+1)B_n^2 + 1} + \sqrt{a(a+1)B_n^2}$

The quantities under the square roots, $a(a+1)B_n^2 + 1$ and $a(a+1)B_n^2$, are consecutive integers.

Case 2: $n$ odd. We have $T_n = C_n\sqrt{a+1} + D_n\sqrt{a}$ with $(a+1)C_n^2 = aD_n^2 + 1$.

$T_n = \sqrt{(a+1)C_n^2} + \sqrt{aD_n^2} = \sqrt{aD_n^2 + 1} + \sqrt{aD_n^2}$

Again, $aD_n^2 + 1$ and $aD_n^2$ are consecutive integers.

In both cases, $T_n$ is the sum of the square roots of two consecutive integers.

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Question 8

Topic: 复数 Complex Numbers  |  Difficulty: Challenging  |  Marks: 20

Problem

8 The complex numbers $z$ and $w$ are related by

$w = \frac{1 + iz}{i + z} \, .$

Let $z = x + iy$ and $w = u + iv$, where $x, y, u$ and $v$ are real. Express $u$ and $v$ in terms of $x$ and $y$.

(i) By setting $x = \tan(\theta/2)$, or otherwise, show that if the locus of $z$ is the real axis $y = 0$, $-\infty < x < \infty$, then the locus of $w$ is the circle $u^2 + v^2 = 1$ with one point omitted.

(ii) Find the locus of $w$ when the locus of $z$ is the line segment $y = 0$, $-1 < x < 1$.

(iii) Find the locus of $w$ when the locus of $z$ is the line segment $x = 0$, $-1 < y < 1$.

(iv) Find the locus of $w$ when the locus of $z$ is the line $y = 1$, $-\infty < x < \infty$.

Hint

8. $w = u + iv = \frac{1+i(x+iy)}{i+(x+iy)} = \frac{2x}{x^2+(1+y)^2} + i\frac{x^2-(1-y^2)}{x^2+(1+y)^2}$ using the complex conjugate, so $u = \frac{2x}{x^2+(1+y)^2}$ and $v = \frac{x^2-(1-y^2)}{x^2+(1+y)^2}$

(i) If $x = \tan \frac{\theta}{2}$, $y = 0$, then $u = \sin \theta$, and $v = -\cos \theta$, using the general result and so $u^2 + v^2 = 1$ but the point $\theta = \pi$ i.e. (0,1) is not included.

(ii) If $-1 < x < 1$, and $y = 0$, then it is the same locus as (i) except $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, and so it is the semi-circle that is the part of $u^2 + v^2 = 1$ below the $u$ axis.

(iii) $x = 0$, then $u = 0$ and $v = \frac{y-1}{y+1}$, and as $-1 < y < 1$ which is that part of the $v$ axis below the $u$ axis, i.e. $-\infty < v < 0$.

(iv) Let $x = 2 \tan \frac{\theta}{2}$ and $y = 1$, so as $-\infty < x < \infty$, $-\pi < \theta < \pi$, then $u = \frac{1}{2} \sin \theta$ and $v = \frac{1}{2}(1 - \cos \theta)$, so the locus is the circle $u^2 + \left( v - \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^2$ excluding the point $\theta = \pi$, which is (0,1).

Model Solution

Expressing $u$ and $v$.

With $z = x + iy$:

$w = \frac{1 + i(x + iy)}{i + (x + iy)} = \frac{1 - y + ix}{x + i(1 + y)}$

Multiplying numerator and denominator by the conjugate of the denominator:

$w = \frac{(1 - y + ix)(x - i(1 + y))}{x^2 + (1 + y)^2}$

Numerator:

$(1-y)x + (1-y)(-i)(1+y) + ix \cdot x + ix(-i)(1+y)$

$= x(1-y) + x^2 i - i(1-y^2) + x(1+y)$

$= [x(1-y) + x(1+y)] + i[x^2 - (1-y^2)]$

$= 2x + i(x^2 + y^2 - 1)$

So:

$u = \frac{2x}{x^2 + (1+y)^2}, \qquad v = \frac{x^2 + y^2 - 1}{x^2 + (1+y)^2}$

Part (i): Locus when $y = 0$, $-\infty < x < \infty$.

Setting $y = 0$:

$u = \frac{2x}{x^2 + 1}, \qquad v = \frac{x^2 - 1}{x^2 + 1}$

Set $x = \tan(\theta/2)$ where $-\pi < \theta < \pi$. Using the half-angle identities:

$u = \frac{2\tan(\theta/2)}{1 + \tan^2(\theta/2)} = \sin\theta$

$v = \frac{\tan^2(\theta/2) - 1}{\tan^2(\theta/2) + 1} = -\cos\theta$

So $u^2 + v^2 = \sin^2\theta + \cos^2\theta = 1$, i.e., the locus lies on the unit circle.

As $x$ ranges over $(-\infty, \infty)$, $\theta$ ranges over $(-\pi, \pi)$. The point $\theta = \pi$ (i.e., $u = 0$, $v = 1$) corresponds to $x = \tan(\pi/2)$ which is undefined. So the locus is the circle $u^2 + v^2 = 1$ with the point $(0, 1)$ omitted.

Part (ii): Locus when $y = 0$, $-1 < x < 1$.

With $y = 0$ and $x = \tan(\theta/2)$, the condition $-1 < x < 1$ gives $-1 < \tan(\theta/2) < 1$, so $-\pi/4 < \theta/2 < \pi/4$, i.e., $-\pi/2 < \theta < \pi/2$.

For $\theta \in (-\pi/2, \pi/2)$: $\sin\theta$ ranges over $(-1, 1)$ and $\cos\theta > 0$, so $v = -\cos\theta < 0$.

The locus is the semicircle $u^2 + v^2 = 1$ with $v < 0$ (the lower half of the unit circle, excluding the endpoints $(\pm 1, 0)$).

Part (iii): Locus when $x = 0$, $-1 < y < 1$.

Setting $x = 0$:

$u = 0, \qquad v = \frac{y^2 - 1}{(1 + y)^2} = \frac{(y-1)(y+1)}{(y+1)^2} = \frac{y - 1}{y + 1}$

For $-1 < y < 1$: $y - 1 < 0$ and $y + 1 > 0$, so $v < 0$. As $y \to -1^+$, $v \to -\infty$; as $y \to 1^-$, $v \to 0^-$.

The locus is the negative $v$-axis: the set of points $(0, v)$ with $v < 0$.

Part (iv): Locus when $y = 1$, $-\infty < x < \infty$.

Setting $y = 1$:

$u = \frac{2x}{x^2 + 4}, \qquad v = \frac{x^2}{x^2 + 4}$

Note that $v = \frac{x^2}{x^2 + 4}$ and $u = \frac{2x}{x^2 + 4}$, so $v = \frac{x}{2} \cdot u$ (when $x \neq 0$), giving $x = 2v/u$.

Alternatively, observe that $v = 1 - \frac{4}{x^2 + 4}$. We compute:

$u^2 + \left(v - \frac{1}{2}\right)^2 = \frac{4x^2}{(x^2+4)^2} + \left(\frac{x^2}{x^2+4} - \frac{1}{2}\right)^2$

$= \frac{4x^2}{(x^2+4)^2} + \left(\frac{x^2 - 4}{2(x^2+4)}\right)^2 = \frac{4x^2}{(x^2+4)^2} + \frac{(x^2-4)^2}{4(x^2+4)^2}$

$= \frac{16x^2 + x^4 - 8x^2 + 16}{4(x^2+4)^2} = \frac{x^4 + 8x^2 + 16}{4(x^2+4)^2} = \frac{(x^2+4)^2}{4(x^2+4)^2} = \frac{1}{4}$

So the locus lies on the circle $u^2 + (v - \frac{1}{2})^2 = \frac{1}{4}$, centered at $(0, \frac{1}{2})$ with radius $\frac{1}{2}$.

The point $(0, 1)$ on this circle requires $u = 0$ and $v = 1$, which gives $x = 0$ and $\frac{0}{4} = 0 \neq 1$. Checking: $u = 0$ requires $x = 0$, then $v = 0$. So $(0, 1)$ is not attained. As $x \to \pm\infty$, $u \to 0$ and $v \to 1$, approaching $(0,1)$ but never reaching it.

The locus is the circle $u^2 + (v - \frac{1}{2})^2 = \frac{1}{4}$ with the point $(0, 1)$ omitted.