STEP3 2005 -- Pure Mathematics

STEP3 2005 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2005  |  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	三角函数 Trigonometry	Standard	三角恒等式变换,辅助角公式,不等式估计,函数图像分析
2	微分方程与曲线 Differential Equations and Curves	Challenging	分离变量法,隐函数求导,二阶导数判别法,参数分类讨论
3	多项式与方程 Polynomials and Equations	Standard	多项式复合,待定系数法,配方法,四次方程因式分解
4	数列与递推关系 Sequences and Recurrence Relations	Challenging	数学归纳法,递推关系化简,等比数列条件,周期性分析
5	二次曲线与切线 Quadratic Curves and Tangents	Challenging	切线方程推导,判别式分析,必要充分条件论证,线性方程解的存在性
6	三次方程与双曲函数	Cubic Equations and Hyperbolic Functions	Hard
7	积分技巧	Integration Techniques	Challenging
8	复数与几何	Complex Numbers and Geometry	Challenging

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Question 1

Topic: 三角函数 Trigonometry  |  Difficulty: Standard  |  Marks: 20

Problem

1 Show that $\sin A = \cos B$ if and only if $A = (4n + 1) \frac{\pi}{2} \pm B$ for some integer $n$.

Show also that $|\sin x \pm \cos x| \leqslant \sqrt{2}$ for all values of $x$ and deduce that there are no solutions to the equation $\sin (\sin x) = \cos (\cos x)$.

Sketch, on the same axes, the graphs of $y = \sin (\sin x)$ and $y = \cos (\cos x)$. Sketch, not on the previous axes, the graph of $y = \sin (2 \sin x)$.

Hint

1 To prove the first part, use the results: $\cos B = \sin \left( \frac{\pi}{2} - B \right)$, whatever the value of $B$; and

$\sin A = \sin Y \Leftrightarrow A = Y + 2n\pi \text{ or } A = \pi - Y + 2n\pi;$

thus here, replacing $Y$ by $\frac{\pi}{2} - B$, $A = 2n\pi + \frac{\pi}{2} \pm B$.

For the next part, it is probably easiest to use the fact that $a \sin x \pm b \cos x$ can be written in the form $R \sin(x \pm \alpha)$; here,

$\sin x \pm \cos x = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} \pm \cos x \sin \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x \pm \frac{\pi}{4} \right)$

so $|\sin x \pm \cos x| \leqslant \sqrt{2}$.

Now, from the first part,

$\sin (\sin x) = \cos (\cos x) \Leftrightarrow \sin x = 2n\pi + \frac{\pi}{2} \pm \cos x$

so

$|\sin x \pm \cos x| \geqslant \left| 2n\pi + \frac{\pi}{2} \right| \geqslant \frac{\pi}{2} > \sqrt{2},$

which is a contradiction.

All the curves asked for have period $2\pi$, so they will be sketched for $x$ in this range only.

For $y = \sin (\sin x)$, $y = 0$ when $\sin x = 0$ only (since $|\sin x| < \pi$), so at $0$, $\pi$ and $2\pi$; the turning points are at $\cos x \cos (\sin x) = 0$, so when $\cos x = 0$, that is at $x = \frac{\pi}{2}, \frac{3\pi}{2}$, or when $\cos (\sin x) = 0$, which is impossible since $|\sin x| < \frac{\pi}{2}$; the turning points are a maximum at $\left( \frac{\pi}{2}, \sin(1) \right)$ and a minimum at $\left( \frac{3\pi}{2}, -\sin(1) \right)$, where $\sin(1) \approx 0.84$.

For $y = \cos (\cos x)$, $y > 0$ for all $x$, since $|\cos x| \leqslant 1 < \frac{\pi}{2}$; the turning points are at $\sin x \sin (\cos x) = 0$, so when either $\sin x = 0$ or $\cos x = 0$, that is at $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$; the turning points are maxima at $\left( \frac{\pi}{2}, 1 \right)$ and $\left( \frac{3\pi}{2}, 1 \right)$, and minima at $(0, \cos(1))$, $(\pi, \cos(1))$, $(2\pi, \cos(1))$, where $\cos(1) \approx 0.54$.

x	y = cos(cos x)	y = sin(sin x)
0	0.54	0
π/2	1	0.84
π	0.54	0
3π/2	1	-0.84
2π	0.54	0

For the curve $y = \sin(2 \sin x)$, $y = 0$ if $2 \sin x$ is a multiple of $\pi$, which is only possible if $\sin x = 0$ (since $|2 \sin x| < \pi$), so when $x$ is $0$, $\pi$ and $2\pi$; the turning points are at $2 \cos x \cos(2 \sin x) = 0$; so when $\cos x = 0$, that is at $x = \frac{\pi}{2}, \frac{3\pi}{2}$, or when $2 \sin x$ is an odd multiple of $\frac{\pi}{2}$, which is only possible if $2 \sin x = \pm \frac{\pi}{2}$, so when $\sin x = \frac{\pi}{4} \approx \pm 0.8$; the turning points are a minimum at $\left(\frac{\pi}{2}, \sin 2\right)$, where $\sin 2 \approx 0.91$ and maxima either side of this, with $y$-coordinates $1$ and a maximum at $\left(\frac{3\pi}{2}, -\sin 2\right)$ with minima either side with $y$-coordinates $-1$.

x	y
0	0
π	0
2π	0

Model Solution

Part 1: Show that $\sin A = \cos B$ if and only if $A = (4n+1)\frac{\pi}{2} \pm B$

We start with the identity $\cos B = \sin\left(\frac{\pi}{2} - B\right)$, which holds for all $B$.

Therefore $\sin A = \cos B$ if and only if $\sin A = \sin\left(\frac{\pi}{2} - B\right)$.

The general solution of $\sin A = \sin Y$ is $A = 2n\pi + Y$ or $A = 2n\pi + (\pi - Y)$ for integer $n$.

Substituting $Y = \frac{\pi}{2} - B$:

$A = 2n\pi + \frac{\pi}{2} - B \qquad \text{or} \qquad A = 2n\pi + \frac{\pi}{2} + B$

which can be combined as

$A = 2n\pi + \frac{\pi}{2} \pm B = (4n + 1)\frac{\pi}{2} \pm B. \qquad \blacksquare$

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Part 2: Show that $|\sin x \pm \cos x| \leqslant \sqrt{2}$

We write $\sin x \pm \cos x$ in the form $R\sin(x \pm \alpha)$.

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

Similarly:

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

Since $|\sin\theta| \leqslant 1$ for all $\theta$, we have

$|\sin x \pm \cos x| = \sqrt{2}\left|\sin\left(x \pm \frac{\pi}{4}\right)\right| \leqslant \sqrt{2}. \qquad \blacksquare$

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Part 3: Deduce that $\sin(\sin x) = \cos(\cos x)$ has no solutions

From Part 1, $\sin(\sin x) = \cos(\cos x)$ holds if and only if

$\sin x = 2n\pi + \frac{\pi}{2} \pm \cos x$

for some integer $n$. Rearranging:

$\sin x \mp \cos x = 2n\pi + \frac{\pi}{2}$

Taking absolute values and applying Part 2:

$\sqrt{2} \geqslant |\sin x \mp \cos x| = \left|2n\pi + \frac{\pi}{2}\right|$

If $n \geqslant 1$: $\left|2n\pi + \frac{\pi}{2}\right| \geqslant 2\pi + \frac{\pi}{2} = \frac{5\pi}{2} > \sqrt{2}$.

If $n \leqslant -1$: $\left|2n\pi + \frac{\pi}{2}\right| \geqslant 2\pi - \frac{\pi}{2} = \frac{3\pi}{2} > \sqrt{2}$.

If $n = 0$: $|\sin x \mp \cos x| = \frac{\pi}{2}$. But $\frac{\pi}{2} \approx 1.571 > \sqrt{2} \approx 1.414$, so $\frac{\pi}{2} > \sqrt{2}$, which contradicts $|\sin x \mp \cos x| \leqslant \sqrt{2}$.

In all cases we reach a contradiction, so $\sin(\sin x) = \cos(\cos x)$ has no solutions. $\qquad \blacksquare$

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Part 4: Sketches

Graph of $y = \sin(\sin x)$:

This has period $2\pi$.

• Zeros: $\sin(\sin x) = 0$ when $\sin x = 0$, i.e. $x = 0, \pi, 2\pi$ (within $[0, 2\pi]$).
• Turning points: From $\frac{dy}{dx} = \cos x \cdot \cos(\sin x) = 0$, either $\cos x = 0$ or $\cos(\sin x) = 0$. The second requires $|\sin x| = \frac{\pi}{2}$, which is impossible since $|\sin x| \leqslant 1 < \frac{\pi}{2}$. So turning points occur at $x = \frac{\pi}{2}, \frac{3\pi}{2}$.
  • At $x = \frac{\pi}{2}$: $y = \sin(1) \approx 0.84$ (maximum).
  • At $x = \frac{3\pi}{2}$: $y = \sin(-1) = -\sin(1) \approx -0.84$ (minimum).

The curve is a smooth wave oscillating between $\pm\sin(1)$, resembling a sinusoid but slightly flattened.

Graph of $y = \cos(\cos x)$:

This has period $2\pi$.

• Since $|\cos x| \leqslant 1 < \frac{\pi}{2}$, we have $\cos(\cos x) > 0$ for all $x$. The curve never crosses the $x$-axis.
• Turning points: From $\frac{dy}{dx} = \sin x \cdot \sin(\cos x) = 0$, either $\sin x = 0$ or $\sin(\cos x) = 0$.
  • $\sin x = 0$ at $x = 0, \pi, 2\pi$: $y = \cos(1) \approx 0.54$ (minima).
  • $\sin(\cos x) = 0$ when $\cos x = 0$, i.e. $x = \frac{\pi}{2}, \frac{3\pi}{2}$: $y = \cos(0) = 1$ (maxima).

The curve oscillates between $\cos(1) \approx 0.54$ and $1$, always positive, with a “double-hump” shape per period.

Graph of $y = \sin(2\sin x)$:

This has period $2\pi$.

• Zeros: $\sin(2\sin x) = 0$ when $2\sin x = k\pi$ for integer $k$. Since $|2\sin x| \leqslant 2 < \pi$, the only possibility is $k = 0$, i.e. $\sin x = 0$, so $x = 0, \pi, 2\pi$.
• Turning points: From $\frac{dy}{dx} = 2\cos x \cdot \cos(2\sin x) = 0$:
  • $\cos x = 0$ at $x = \frac{\pi}{2}, \frac{3\pi}{2}$:
    • $x = \frac{\pi}{2}$: $y = \sin 2 \approx 0.91$ (local minimum — the curve dips here from the surrounding peaks).
    • $x = \frac{3\pi}{2}$: $y = \sin(-2) = -\sin 2 \approx -0.91$ (local maximum).
  • $\cos(2\sin x) = 0$ when $2\sin x = \pm\frac{\pi}{2}$, i.e. $\sin x = \pm\frac{\pi}{4} \approx \pm 0.785$. This gives four $x$-values in $[0, 2\pi]$ where $y = \pm 1$ (the global maxima and minima).

The curve oscillates between $-1$ and $1$ with a characteristic shape: two “side peaks” reaching $\pm 1$ on either side of each main turning point at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$, where $y \approx \pm 0.91$.

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Question 2

Topic: 微分方程与曲线 Differential Equations and Curves  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Find the general solution of the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2 + a^2}$, where $a \neq 0$, and show that it can be written in the form $y^2(x^2 + a^2) = c^2$, where $c$ is an arbitrary constant. Sketch this curve.

Find an expression for $\frac{\mathrm{d}}{\mathrm{d}x}(x^2 + y^2)$ and show that

$\frac{\mathrm{d}^2}{\mathrm{d}x^2}(x^2 + y^2) = 2 \left( 1 - \frac{c^2}{(x^2 + a^2)^2} \right) + \frac{8c^2x^2}{(x^2 + a^2)^3} \text{ .}$

(i) Show that, if $0 < c < a^2$, the points on the curve whose distance from the origin is least are $(0, \pm \frac{c}{a})$.

(ii) If $c > a^2$, determine the points on the curve whose distance from the origin is least.

Hint

2 This equation can be solved by separating the variables:

$\int \frac{2 \, dy}{y} = - \int \frac{2x \, dx}{x^2 + a^2} \quad \text{so} \quad \ln(y^2) = -\ln(x^2 + a^2) + k \quad \text{or} \quad y^2(x^2 + a^2) = c^2.$

The curve has two branches: one has $y > 0$, reflection symmetry about the $y$-axis, a maximum at $(0, \frac{c}{a})$ and $y \to 0$ as $|x| \to \infty$; the other has $y < 0$ and is a reflection of the first branch in the $x$-axis.

$\frac{d}{dx}(x^2 + y^2) = 2x - 2y \frac{dy}{dx} = 2x - \frac{2xy^2}{x^2 + a^2} = 2x - \frac{2xc^2}{(x^2 + a^2)^2}$

$\frac{d^2}{dx^2}(x^2 + y^2) = 2 - \frac{2c^2}{(x^2 + a^2)^2} + \frac{4xc^2 \cdot 2x}{(x^2 + a^2)^3} = 2 \left( 1 - \frac{c^2}{(x^2 + a^2)^2} \right) + \frac{8c^2x^2}{(x^2 + a^2)^3}$

(i) $\frac{\text{d}}{\text{d}x}(x^2 + y^2) = 0$ when $x = 0$ and when $c^2 = (x^2 + a^2)^2$, but the latter is not possible if $0 < c < a^2$. If $x = 0$, $y = \pm \frac{c}{a}$ and $\frac{\text{d}^2}{\text{d}x^2}(x^2 + y^2) = 1 - \frac{c^2}{a^4}$ which is positive if $0 < c < a^2$, indicating a local minimum. Hence the points on the curve whose distance from the origin is least are $(0, \pm \frac{c}{a})$.

(ii) If $c > a^2$ then $\frac{\text{d}^2}{\text{d}x^2}(x^2 + y^2)$ is negative at $x = 0$, indicating a local maximum there; but in this case there are further stationary points at $x^2 = c - a^2$, $y = \pm \sqrt{c}$ and at these points $\frac{\text{d}^2}{\text{d}x^2}(x^2 + y^2) = \frac{8x^2}{c} > 0$. Hence the points on the curve whose distance from the origin is least are $(\pm \sqrt{c - a^2}, \pm \sqrt{c})$.

Model Solution

Part 1: Solve the differential equation and show $y^2(x^2 + a^2) = c^2$

We have $\frac{dy}{dx} = -\frac{xy}{x^2 + a^2}$. This is separable.

For $y \neq 0$:

$\frac{dy}{y} = -\frac{x}{x^2 + a^2}\, dx$

Integrating both sides:

$\ln|y| = -\frac{1}{2}\ln(x^2 + a^2) + k$

where $k$ is an arbitrary constant. Exponentiating:

$|y| = \frac{e^k}{\sqrt{x^2 + a^2}}$

so $y = \frac{A}{\sqrt{x^2 + a^2}}$ where $A$ is an arbitrary constant (which can be positive, negative, or zero — the zero case $y = 0$ also satisfies the original ODE). Squaring:

$y^2 = \frac{A^2}{x^2 + a^2}$

$y^2(x^2 + a^2) = c^2 \qquad \text{where } c^2 = A^2. \qquad \blacksquare$

Sketch of the curve:

The curve $y^2(x^2 + a^2) = c^2$ has two branches:

• Upper branch ($y > 0$): $y = \frac{c}{\sqrt{x^2 + a^2}}$, which is symmetric about the $y$-axis, has a maximum of $y = \frac{c}{a}$ at $x = 0$, and $y \to 0$ as $|x| \to \infty$.
• Lower branch ($y < 0$): $y = -\frac{c}{\sqrt{x^2 + a^2}}$, which is the reflection of the upper branch in the $x$-axis.

The curve looks like a symmetric pair of bell-shaped curves, one above and one below the $x$-axis, never touching the $x$-axis.

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Part 2: Find $\frac{d}{dx}(x^2 + y^2)$ and verify the second derivative

Since $y^2(x^2 + a^2) = c^2$, we have $y^2 = \frac{c^2}{x^2 + a^2}$.

First derivative. Using implicit differentiation of $x^2 + y^2$:

$\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}$

Substituting $\frac{dy}{dx} = -\frac{xy}{x^2 + a^2}$:

$\frac{d}{dx}(x^2 + y^2) = 2x - \frac{2xy^2}{x^2 + a^2} = 2x - \frac{2x \cdot \frac{c^2}{x^2 + a^2}}{x^2 + a^2} = 2x - \frac{2xc^2}{(x^2 + a^2)^2}$

$= 2x\left(1 - \frac{c^2}{(x^2 + a^2)^2}\right). \qquad \text{(*)}$

Second derivative. Differentiating $(*)$ with respect to $x$:

$\frac{d^2}{dx^2}(x^2 + y^2) = \frac{d}{dx}\left[2x - \frac{2xc^2}{(x^2 + a^2)^2}\right]$

$= 2 - c^2 \cdot \frac{d}{dx}\left[\frac{2x}{(x^2 + a^2)^2}\right]$

Computing the derivative of $\frac{2x}{(x^2 + a^2)^2}$ using the quotient rule:

$\frac{d}{dx}\left[\frac{2x}{(x^2 + a^2)^2}\right] = \frac{2(x^2 + a^2)^2 - 2x \cdot 2(x^2 + a^2) \cdot 2x}{(x^2 + a^2)^4} = \frac{2(x^2 + a^2) - 8x^2}{(x^2 + a^2)^3} = \frac{2a^2 - 6x^2}{(x^2 + a^2)^3}$

Therefore:

$\frac{d^2}{dx^2}(x^2 + y^2) = 2 - c^2 \cdot \frac{2a^2 - 6x^2}{(x^2 + a^2)^3} = 2 - \frac{2a^2c^2 - 6c^2x^2}{(x^2 + a^2)^3}$

$= 2 - \frac{2a^2c^2}{(x^2 + a^2)^3} + \frac{6c^2x^2}{(x^2 + a^2)^3}$

Now we show this equals the required form. Observe that

$\frac{2a^2c^2}{(x^2 + a^2)^3} = \frac{2c^2(x^2 + a^2) - 2c^2x^2}{(x^2 + a^2)^3} = \frac{2c^2}{(x^2 + a^2)^2} - \frac{2c^2x^2}{(x^2 + a^2)^3}$

Substituting:

$\frac{d^2}{dx^2}(x^2 + y^2) = 2 - \frac{2c^2}{(x^2 + a^2)^2} + \frac{2c^2x^2}{(x^2 + a^2)^3} + \frac{6c^2x^2}{(x^2 + a^2)^3}$

$= 2\left(1 - \frac{c^2}{(x^2 + a^2)^2}\right) + \frac{8c^2x^2}{(x^2 + a^2)^3}. \qquad \blacksquare$

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Part (i): $0 < c < a^2$ — points closest to the origin

The distance from the origin squared is $x^2 + y^2$. Stationary points satisfy $\frac{d}{dx}(x^2 + y^2) = 0$.

From $(*)$:

$2x\left(1 - \frac{c^2}{(x^2 + a^2)^2}\right) = 0$

Either $x = 0$, or $(x^2 + a^2)^2 = c^2$, i.e. $x^2 + a^2 = c$ (taking the positive root since $x^2 + a^2 > 0$), giving $x^2 = c - a^2$.

If $0 < c < a^2$, then $c - a^2 < 0$, so $x^2 = c - a^2$ has no real solutions. The only stationary point is at $x = 0$.

At $x = 0$: $y^2 = \frac{c^2}{a^2}$, so $y = \pm\frac{c}{a}$.

To confirm this is a minimum, we check the second derivative at $x = 0$:

$\frac{d^2}{dx^2}(x^2 + y^2)\bigg|_{x=0} = 2\left(1 - \frac{c^2}{a^4}\right) + 0 = 2\left(1 - \frac{c^2}{a^4}\right)$

Since $0 < c < a^2$, we have $c^2 < a^4$, so $1 - \frac{c^2}{a^4} > 0$. The second derivative is positive, confirming a local minimum.

Therefore, when $0 < c < a^2$, the points on the curve closest to the origin are $\left(0, \pm\frac{c}{a}\right)$. $\qquad \blacksquare$

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Part (ii): $c > a^2$ — points closest to the origin

When $c > a^2$, we have $c - a^2 > 0$, so $x^2 = c - a^2$ gives $x = \pm\sqrt{c - a^2}$.

Check $x = 0$: The second derivative is $2\left(1 - \frac{c^2}{a^4}\right)$. Since $c > a^2$, we have $c^2 > a^4$, so the second derivative is negative — this is a local maximum, not a minimum.

Check $x^2 = c - a^2$: At these points, $x^2 + a^2 = c$, so $(x^2 + a^2)^2 = c^2$. The second derivative becomes:

$\frac{d^2}{dx^2}(x^2 + y^2) = 2\left(1 - \frac{c^2}{c^2}\right) + \frac{8c^2x^2}{c^3} = 0 + \frac{8x^2}{c} = \frac{8(c - a^2)}{c} > 0$

since $c > a^2 > 0$. This confirms a local minimum.

At these points:

$y^2 = \frac{c^2}{x^2 + a^2} = \frac{c^2}{c} = c$

so $y = \pm\sqrt{c}$.

Therefore, when $c > a^2$, the points on the curve closest to the origin are $\left(\pm\sqrt{c - a^2}, \pm\sqrt{c}\right)$. $\qquad \blacksquare$

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Question 3

Topic: 多项式与方程 Polynomials and Equations  |  Difficulty: Standard  |  Marks: 20

Problem

3 Let $f(x) = x^2 + px + q$ and $g(x) = x^2 + rx + s$. Find an expression for $f(g(x))$ and hence find a necessary and sufficient condition on $a$, $b$ and $c$ for it to be possible to write the quartic expression $x^4 + ax^3 + bx^2 + cx + d$ in the form $f(g(x))$, for some choice of values of $p, q, r$ and $s$.

Show further that this condition holds if and only if it is possible to write the quartic expression $x^4 + ax^3 + bx^2 + cx + d$ in the form $(x^2 + vx + w)^2 - k$, for some choice of values of $v, w$ and $k$.

Find the roots of the quartic equation $x^4 - 4x^3 + 10x^2 - 12x + 4 = 0$.

ERRATUM NOTICE

OXFORD CAMBRIDGE AND RSA EXAMINATIONS

Sixth Term Examination Papers

MATHEMATICS III 9475

Wednesday 29 JUNE 2005 Afternoon 3 hours

Instructions to Invigilators

Before the start of the examination please give each candidate a copy of this erratum notice.

Page 4 Question 6.

There is an error in the sentence starting ‘Show that the equation …’ . This sentence should read:

‘Show that the equation $x^3 - 3a^2x = 2a^3 \cosh T$ is satisfied by $2a \cosh (\frac{1}{3}T)$ and hence that, if $c^2 \ge b^3 > 0$, one of the roots of the equation $x^3 - 3bx = 2c$ is $u + \frac{b}{u}$, where

$u = \left( c + \sqrt{c^2 - b^3} \right)^{\frac{1}{3}}.$

Page 5 Question 10(ii)

There is an error in this part question. The question should read:

‘Show that, if the discs collide at least once, their total kinetic energy just before the first collision is $\frac{4}{3}mgr(2 - 3\mu)$.’

Any enquiry about this notice should be referred to the Information Bureau on 01223 553 998 or helpdesk@ocr.org.uk

Jun05/erratum09

Hint

3 Direct substitution gives

$f(g(x)) = (x^2 + rx + s)^2 + p(x^2 + rx + s) + q$ $= x^4 + 2rx^3 + (r^2 + 2s + p)x^2 + (2rs + rp)x + s^2 + ps + q.$

If $x^4 + ax^3 + bx^2 + cx + d$ is to have this form then it is necessary to choose $r = \frac{a}{2}$ and to choose $s$ and $p$ to satisfy $2s + p = b - r^2 = b - \frac{a^2}{4}$ and $r(2s + p) = c$ or $a(2s + p) = 2c$. Thus $a\left(b - \frac{a^2}{4}\right) = 2c$ is a necessary condition for this to be possible.

It is also sufficient: in fact, pick $p = 0$; then $s = \frac{4b - a^2}{8}$ and $q = d - s^2$ will do.

Expanding the second form gives

$(x^2 + vx + w)^2 - k = x^4 + 2vx^3 + (v^2 + 2w)x^2 + 2vwx + w^2 - k,$

but this is identical to $x^4 + 2rx^3 + (r^2 + 2s + p)x^2 + (2rs + rp)x + s^2 + ps + q$ with $p = 0$, $v = r$, $w = s$ and $k = -q$, and so, since the sufficiency demonstrated above allowed the choice $p = 0$, the condition is the same.

To solve the final equation, write the quartic in the second form:

$x^4 - 4x^3 + 10x^2 - 12x + 4 = (x^2 - 2x + 3)^2 - 5 = 0$ so $x^2 - 2x + 3 - \sqrt{5} = 0 \text{ or } x^2 - 2x + 3 + \sqrt{5} = 0$ so $x = 1 \pm \sqrt{\sqrt{5} - 2} \text{ or } 1 \pm i\sqrt{\sqrt{5} + 2}.$

Model Solution

Part 1: Find $f(g(x))$ and the necessary and sufficient condition

We compute $f(g(x))$ directly:

$f(g(x)) = (g(x))^2 + p \cdot g(x) + q = (x^2 + rx + s)^2 + p(x^2 + rx + s) + q$

Expanding $(x^2 + rx + s)^2$:

$= x^4 + 2rx^3 + (r^2 + 2s)x^2 + 2rsx + s^2$

So:

$f(g(x)) = x^4 + 2rx^3 + (r^2 + 2s + p)x^2 + (2rs + rp)x + s^2 + ps + q$

For $x^4 + ax^3 + bx^2 + cx + d = f(g(x))$, we equate coefficients:

$2r = a \qquad \text{(coefficient of } x^3\text{)}$ $r^2 + 2s + p = b \qquad \text{(coefficient of } x^2\text{)}$ $2rs + rp = c \qquad \text{(coefficient of } x\text{)}$ $s^2 + ps + q = d \qquad \text{(constant term)}$

From the first equation: $r = \frac{a}{2}$.

From the third equation: $r(2s + p) = c$, so $\frac{a}{2}(2s + p) = c$, giving $2s + p = \frac{2c}{a}$ (provided $a \neq 0$).

From the second equation: $2s + p = b - r^2 = b - \frac{a^2}{4}$.

For consistency we need:

$b - \frac{a^2}{4} = \frac{2c}{a}$

$a\left(b - \frac{a^2}{4}\right) = 2c$

$ab - \frac{a^3}{4} = 2c$

$4ab - a^3 = 8c$

This is the necessary and sufficient condition. It is necessary because the coefficient equations force it. It is sufficient because, given this condition, we can choose $p = 0$, $r = \frac{a}{2}$, $s = \frac{4b - a^2}{8}$, and $q = d - s^2$, and all four coefficient equations are satisfied.

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Part 2: Equivalence with the form $(x^2 + vx + w)^2 - k$

Expanding $(x^2 + vx + w)^2 - k$:

$= x^4 + 2vx^3 + (v^2 + 2w)x^2 + 2vwx + w^2 - k$

This is exactly the same as $f(g(x))$ with $p = 0$, $v = r$, $w = s$, and $k = -q = s^2 - d$.

Since the sufficiency argument in Part 1 allowed us to choose $p = 0$, the condition $4ab - a^3 = 8c$ is both necessary and sufficient for writing the quartic in the form $(x^2 + vx + w)^2 - k$. $\qquad \blacksquare$

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Part 3: Find the roots of $x^4 - 4x^3 + 10x^2 - 12x + 4 = 0$

We check the condition: $a = -4$, $b = 10$, $c = -12$.

$4ab - a^3 = 4(-4)(10) - (-4)^3 = -160 + 64 = -96 = 8(-12) = 8c \qquad \checkmark$

So we can write this quartic as $(x^2 + vx + w)^2 - k$.

We need $v = \frac{a}{2} = -2$ and $w = \frac{4b - a^2}{8} = \frac{40 - 16}{8} = 3$.

Then $k = w^2 - d = 9 - 4 = 5$.

$x^4 - 4x^3 + 10x^2 - 12x + 4 = (x^2 - 2x + 3)^2 - 5$

Verification: $(x^2 - 2x + 3)^2 = x^4 - 4x^3 + 10x^2 - 12x + 9$, so $(x^2 - 2x + 3)^2 - 5 = x^4 - 4x^3 + 10x^2 - 12x + 4$. $\checkmark$

Setting $(x^2 - 2x + 3)^2 - 5 = 0$:

$(x^2 - 2x + 3)^2 = 5$

$x^2 - 2x + 3 = \sqrt{5} \qquad \text{or} \qquad x^2 - 2x + 3 = -\sqrt{5}$

Case 1: $x^2 - 2x + 3 = \sqrt{5}$

$x^2 - 2x + (3 - \sqrt{5}) = 0$

$x = \frac{2 \pm \sqrt{4 - 4(3 - \sqrt{5})}}{2} = 1 \pm \sqrt{1 - 3 + \sqrt{5}} = 1 \pm \sqrt{\sqrt{5} - 2}$

Since $\sqrt{5} \approx 2.236 > 2$, these are real roots.

Case 2: $x^2 - 2x + 3 = -\sqrt{5}$

$x^2 - 2x + (3 + \sqrt{5}) = 0$

$x = \frac{2 \pm \sqrt{4 - 4(3 + \sqrt{5})}}{2} = 1 \pm \sqrt{1 - 3 - \sqrt{5}} = 1 \pm \sqrt{-(2 + \sqrt{5})} = 1 \pm i\sqrt{2 + \sqrt{5}}$

The four roots are:

$x = 1 \pm \sqrt{\sqrt{5} - 2} \qquad \text{and} \qquad x = 1 \pm i\sqrt{\sqrt{5} + 2} \qquad \blacksquare$

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Question 4

Topic: 数列与递推关系 Sequences and Recurrence Relations  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The sequence $u_n$ ($n = 1, 2, \dots$) satisfies the recurrence relation

$u_{n+2} = \frac{u_{n+1}}{u_n}(ku_n - u_{n+1})$

where $k$ is a constant.

If $u_1 = a$ and $u_2 = b$, where $a$ and $b$ are non-zero and $b \neq ka$, prove by induction that

$u_{2n} = \left(\frac{b}{a}\right)u_{2n-1}$

$u_{2n+1} = cu_{2n}$

for $n \geq 1$, where $c$ is a constant to be found in terms of $k$, $a$ and $b$. Hence express $u_{2n}$ and $u_{2n-1}$ in terms of $a, b, c$ and $n$.

Find conditions on $a, b$ and $k$ in the three cases:

(i) the sequence $u_n$ is geometric;

(ii) the sequence $u_n$ has period 2;

(iii) the sequence $u_n$ has period 4.

Hint

4 For the base case you need to verify that $u_{2n} = \frac{b}{a}u_{2n-1}$ and $u_{2n+1} = cu_{2n}$ when $n = 1$:

$u_1 = a, u_2 = b \text{ so } u_{2n} = \frac{b}{a}u_{2n-1} \text{ when } n = 1;$

$u_3 = \frac{u_2}{u_1}(ku_1 - u_2) = u_2 \frac{ka - b}{a} \text{ so } u_{2n+1} = cu_{2n} \text{ when } n = 1, \text{ provided } c = k - \frac{b}{a}.$

For the induction step, assume that $u_{2n} = \frac{b}{a}u_{2n-1}$ and $u_{2n+1} = cu_{2n}$ when $n = N$ then

$u_{2N+2} = \frac{u_{2N+1}}{u_{2N}}(ku_{2N} - u_{2N+1})$ $= u_{2N+1}(k - c) \text{ (by the induction hypothesis)}$ $= \frac{b}{a}u_{2N+1} \text{ (by the definition of } c)$

and

$u_{2N+3} = \frac{u_{2N+2}}{u_{2N+1}}(ku_{2N+1} - u_{2N+2})$ $= u_{2N+2} \left( k - \frac{b}{a} \right) \text{ (by what has just been shown)}$ $= cu_{2N+2}.$

which completes the induction.

Hence $u_{2n} = \frac{bc}{a}u_{2n-2} = \dots = \left( \frac{bc}{a} \right)^{n-1} u_2 = b \left( \frac{bc}{a} \right)^{n-1}$ and $u_{2n-1} = \frac{bc}{a}u_{2n-3} = \dots = \left( \frac{bc}{a} \right)^{n-1} u_1 = a \left( \frac{bc}{a} \right)^{n-1}$

(i) For $u_n$ to be geometric requires $\frac{u_{2n}}{u_{2n-1}} = \frac{u_{2n+1}}{u_{2n}}$; that is, $\frac{b}{a} = c = k - \frac{b}{a}$ or $ak = 2b$;

(ii) For $u_n$ to have period 2 requires $u_{2n+1} = u_{2n-1}$, but $u_{2n+1} = cu_{2n} = \frac{cb}{a}u_{2n-1}$, so it is necessary that $\frac{cb}{a} = 1$ or $a^2 + b^2 = kab$;

(iii) For $u_n$ to have period 4 requires $u_{2n+3} = u_{2n-1}$ so, by the previous part, it is necessary that $\left( \frac{cb}{a} \right)^2 = 1$ but $\frac{cb}{a} \neq 1$ (to avoid period 2) so $\frac{cb}{a} = -1$ or $b^2 - a^2 = kab$.

Model Solution

Part 1: Prove by induction that $u_{2n} = \frac{b}{a}u_{2n-1}$ and $u_{2n+1} = cu_{2n}$

Finding $c$: From the recurrence with $n = 1$:

$u_3 = \frac{u_2}{u_1}(ku_1 - u_2) = \frac{b}{a}(ka - b) = b\left(k - \frac{b}{a}\right)$

So $u_3 = cu_2$ requires $c = k - \frac{b}{a}$.

Base case ($n = 1$):

$u_2 = b = \frac{b}{a} \cdot a = \frac{b}{a}u_1$ $\checkmark$

$u_3 = cu_2$ by definition of $c$. $\checkmark$

Induction step: Assume $u_{2N} = \frac{b}{a}u_{2N-1}$ and $u_{2N+1} = cu_{2N}$ for some $N \geq 1$.

We need to show $u_{2N+2} = \frac{b}{a}u_{2N+1}$ and $u_{2N+3} = cu_{2N+2}$.

From the recurrence:

$u_{2N+2} = \frac{u_{2N+1}}{u_{2N}}(ku_{2N} - u_{2N+1})$

Using the induction hypothesis $u_{2N+1} = cu_{2N}$:

$u_{2N+2} = \frac{cu_{2N}}{u_{2N}}(ku_{2N} - cu_{2N}) = c(k - c)u_{2N}$

Now $c = k - \frac{b}{a}$, so $k - c = \frac{b}{a}$. Therefore:

$u_{2N+2} = c \cdot \frac{b}{a} \cdot u_{2N} = \frac{b}{a} \cdot cu_{2N} = \frac{b}{a}u_{2N+1} \qquad \checkmark$

For the next term:

$u_{2N+3} = \frac{u_{2N+2}}{u_{2N+1}}(ku_{2N+1} - u_{2N+2})$

Using $u_{2N+2} = \frac{b}{a}u_{2N+1}$:

$u_{2N+3} = \frac{\frac{b}{a}u_{2N+1}}{u_{2N+1}}\left(ku_{2N+1} - \frac{b}{a}u_{2N+1}\right) = \frac{b}{a}\left(k - \frac{b}{a}\right)u_{2N+1} = \frac{b}{a} \cdot c \cdot u_{2N+1}$

But also $u_{2N+3} = cu_{2N+2}$, so we need $cu_{2N+2} = \frac{bc}{a}u_{2N+1}$. Since $u_{2N+2} = \frac{b}{a}u_{2N+1}$, this gives $c \cdot \frac{b}{a}u_{2N+1} = \frac{bc}{a}u_{2N+1}$. $\checkmark$

By induction, the result holds for all $n \geq 1$, with $c = k - \frac{b}{a}$. $\qquad \blacksquare$

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Part 2: Express $u_{2n}$ and $u_{2n-1}$ in terms of $a, b, c, n$

From the relations $u_{2n} = \frac{b}{a}u_{2n-1}$ and $u_{2n+1} = cu_{2n}$, we get:

$u_{2n+1} = cu_{2n} = \frac{bc}{a}u_{2n-1}$

So the odd-indexed terms satisfy $u_{2n+1} = \frac{bc}{a}u_{2n-1}$, which is a geometric progression with ratio $\frac{bc}{a}$.

$u_{2n-1} = \left(\frac{bc}{a}\right)^{n-1} u_1 = a\left(\frac{bc}{a}\right)^{n-1}$

$u_{2n} = \frac{b}{a}u_{2n-1} = \frac{b}{a} \cdot a\left(\frac{bc}{a}\right)^{n-1} = b\left(\frac{bc}{a}\right)^{n-1}$

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Part (i): $u_n$ is geometric

For $u_n$ to be geometric, we need the common ratio $\frac{u_{n+1}}{u_n}$ to be constant for all $n$. In particular:

$\frac{u_2}{u_1} = \frac{u_3}{u_2}$

$\frac{b}{a} = \frac{u_3}{b} = \frac{cb}{b} = c$

So $c = \frac{b}{a}$, which means $k - \frac{b}{a} = \frac{b}{a}$, giving:

$k = \frac{2b}{a} \qquad \text{i.e.} \qquad ak = 2b \qquad \blacksquare$

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Part (ii): $u_n$ has period 2

Period 2 means $u_{n+2} = u_n$ for all $n$. In particular $u_3 = u_1$, i.e. $cu_2 = u_1$, so $cb = a$.

$cb = a \implies \left(k - \frac{b}{a}\right)b = a \implies kb - \frac{b^2}{a} = a \implies kab - b^2 = a^2$

$a^2 + b^2 = kab \qquad \blacksquare$

We can verify this gives period 2 for all terms: $u_{2n+1} = \frac{bc}{a}u_{2n-1} = \frac{bc}{a}u_{2n-1}$. With $cb = a$, this ratio is $\frac{bc}{a} = 1$, so $u_{2n+1} = u_{2n-1}$ for all $n$. Similarly $u_{2n} = u_{2n-2}$.

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Part (iii): $u_n$ has period 4

Period 4 means $u_{n+4} = u_n$ but $u_{n+2} \neq u_n$ (otherwise it would be period 2).

From Part 2, $u_{2n-1} = a\left(\frac{bc}{a}\right)^{n-1}$. For period 4, we need $u_{2n+3} = u_{2n-1}$, so $\left(\frac{bc}{a}\right)^2 = 1$, giving $\frac{bc}{a} = \pm 1$.

If $\frac{bc}{a} = 1$, this is the period 2 case (excluded). So we need $\frac{bc}{a} = -1$:

$bc = -a \implies \left(k - \frac{b}{a}\right)b = -a \implies kb - \frac{b^2}{a} = -a \implies kab - b^2 = -a^2$

$b^2 - a^2 = kab \qquad \blacksquare$

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Question 5

Topic: 二次曲线与切线 Quadratic Curves and Tangents  |  Difficulty: Challenging  |  Marks: 20

Problem

5 Let $P$ be the point on the curve $y = ax^2 + bx + c$ (where $a$ is non-zero) at which the gradient is $m$. Show that the equation of the tangent at $P$ is

$y - mx = c - \frac{(m - b)^2}{4a} .$

Show that the curves $y = a_1x^2 + b_1x + c_1$ and $y = a_2x^2 + b_2x + c_2$ (where $a_1$ and $a_2$ are non-zero) have a common tangent with gradient $m$ if and only if

$(a_2 - a_1)m^2 + 2(a_1b_2 - a_2b_1)m + 4a_1a_2(c_2 - c_1) + a_2b_1^2 - a_1b_2^2 = 0 .$

Show that, in the case $a_1 \neq a_2$, the two curves have exactly one common tangent if and only if they touch each other. In the case $a_1 = a_2$, find a necessary and sufficient condition for the two curves to have exactly one common tangent.

Hint

5 The point on the curve with the required gradient is given by

$\frac{dy}{dx} = 2ax + b = m \text{ or } x = \frac{m - b}{2a},$

with

$y = a \left( \frac{m - b}{2a} \right)^2 + b \left( \frac{m - b}{2a} \right) + c = \frac{m^2 - b^2}{4a} + c.$

The equation of the tangent is therefore:

$y - mx = a \left( \frac{m - b}{2a} \right)^2 + b \left( \frac{m - b}{2a} \right) + c - m \left( \frac{m - b}{2a} \right)$

$= c - \frac{(m - b)}{2a} \left( m - b - \frac{(m - b)}{2} \right) = c - \frac{(m - b)^2}{4a}.$

The curves have a common tangent with gradient $m$ if and only if the equations of the tangents to the two curves with gradient $m$ are identical; that is, have the same intercept, so if and only if

$c_1 - \frac{(m - b_1)^2}{4a_1} = c_2 - \frac{(m - b_2)^2}{4a_2}$

that is,

$4a_1a_2c_1 - a_2m^2 + 2a_2b_1m - a_2b_1^2 = 4a_1a_2c_2 - a_1m^2 + 2a_1b_2m - a_1b_2^2$

which gives the quoted result.

There is exactly one common tangent when $a_1 \neq a_2$ when the quadratic equation for $m$ has exactly one root, which occurs if and only if the discriminant of the equation is zero; that is

$4(a_1b_2 - a_2b_1)^2 = 4(a_2 - a_1)(4a_1a_2(c_2 - c_1) + a_2b_1^2 - a_1b_2^2)$

$\Leftrightarrow \quad 4a_1^2b_2^2 - 8a_1a_2b_1b_2 + 4a_2^2b_1^2 = 16a_1a_2(a_2 - a_1)(c_2 - c_1) + 4a_2^2b_1^2 + 4a_1^2b_2^2 - 4a_1a_2(b_1^2 + b_2^2)$

$\Leftrightarrow \quad 4a_1a_2(b_1^2 + b_2^2) - 8a_1a_2b_1b_2 = 16a_1a_2(a_2 - a_1)(c_2 - c_1)$

$\Leftrightarrow \quad b_1^2 + b_2^2 - 2b_1b_2 = 4(a_2 - a_1)(c_2 - c_1) \quad \text{(dividing by } 4a_1a_2 \text{ which is non-zero).}$

The curves touch if there is exactly one solution to the simultaneous equations

$y = a_1x^2 + b_1x + c_1 \text{ and } y = a_2x^2 + b_2x + c_2;$

that is, if the equation $(a_2 - a_1)x^2 + (b_2 - b_1)x + (c_2 - c_1) = 0$ has exactly one root so, again using the discriminant condition, if and only if $(b_2 - b_1)^2 = 4(a_2 - a_1)(c_2 - c_1)$, which is the same condition.

If $a_1 = a_2$ the curves have exactly one common tangent if there is exactly one solution to

$2m(b_2 - b_1) + 4a(c_2 - c_1) + (b_1^2 - b_2^2) = 0;$

since this is just a linear equation, the only condition is that $b_2 - b_1 \neq 0$.

Model Solution

Part 1: Equation of the tangent at $P$

The gradient of $y = ax^2 + bx + c$ is $\frac{dy}{dx} = 2ax + b$. Setting this equal to $m$:

$2ax + b = m \implies x = \frac{m - b}{2a}$

The $y$-coordinate at $P$ is:

$y = a\left(\frac{m - b}{2a}\right)^2 + b\left(\frac{m - b}{2a}\right) + c = \frac{(m - b)^2}{4a} + \frac{b(m - b)}{2a} + c$

$= \frac{(m - b)^2 + 2b(m - b)}{4a} + c = \frac{(m - b)(m - b + 2b)}{4a} + c = \frac{(m - b)(m + b)}{4a} + c = \frac{m^2 - b^2}{4a} + c$

The tangent at $P$ has equation $y - y_P = m(x - x_P)$, i.e. $y = mx + (y_P - mx_P)$:

$y - mx = \frac{m^2 - b^2}{4a} + c - m \cdot \frac{m - b}{2a} = c + \frac{m^2 - b^2 - 2m(m - b)}{4a}$

$= c + \frac{m^2 - b^2 - 2m^2 + 2mb}{4a} = c + \frac{-m^2 + 2mb - b^2}{4a} = c - \frac{(m - b)^2}{4a} \qquad \blacksquare$

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Part 2: Condition for a common tangent with gradient $m$

The tangent to the first curve with gradient $m$ is $y - mx = c_1 - \frac{(m - b_1)^2}{4a_1}$.

The tangent to the second curve with gradient $m$ is $y - mx = c_2 - \frac{(m - b_2)^2}{4a_2}$.

These are the same line if and only if their intercepts are equal:

$c_1 - \frac{(m - b_1)^2}{4a_1} = c_2 - \frac{(m - b_2)^2}{4a_2}$

Multiplying through by $4a_1a_2$:

$4a_1a_2c_1 - a_2(m - b_1)^2 = 4a_1a_2c_2 - a_1(m - b_2)^2$

$4a_1a_2c_1 - a_2(m^2 - 2b_1m + b_1^2) = 4a_1a_2c_2 - a_1(m^2 - 2b_2m + b_2^2)$

Rearranging:

$a_1m^2 - a_2m^2 + 2a_2b_1m - 2a_1b_2m + 4a_1a_2c_1 - 4a_1a_2c_2 - a_2b_1^2 + a_1b_2^2 = 0$

$(a_1 - a_2)m^2 + 2(a_2b_1 - a_1b_2)m + 4a_1a_2(c_1 - c_2) + a_1b_2^2 - a_2b_1^2 = 0$

Multiplying by $-1$:

$(a_2 - a_1)m^2 + 2(a_1b_2 - a_2b_1)m + 4a_1a_2(c_2 - c_1) + a_2b_1^2 - a_1b_2^2 = 0 \qquad \blacksquare$

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Part 3: Case $a_1 \neq a_2$ — exactly one common tangent iff the curves touch

When $a_1 \neq a_2$, the condition from Part 2 is a quadratic in $m$. There is exactly one common tangent if and only if this quadratic has exactly one root, i.e. its discriminant is zero.

The discriminant is:

$\Delta = 4(a_1b_2 - a_2b_1)^2 - 4(a_2 - a_1)\left[4a_1a_2(c_2 - c_1) + a_2b_1^2 - a_1b_2^2\right]$

Setting $\Delta = 0$ and dividing by 4:

$(a_1b_2 - a_2b_1)^2 = (a_2 - a_1)\left[4a_1a_2(c_2 - c_1) + a_2b_1^2 - a_1b_2^2\right]$

Expanding the left side: $a_1^2b_2^2 - 2a_1a_2b_1b_2 + a_2^2b_1^2$.

Expanding the right side: $4a_1a_2(a_2 - a_1)(c_2 - c_1) + a_2^2b_1^2 - a_1a_2b_2^2 - a_1a_2b_1^2 + a_1^2b_2^2$.

Subtracting the common terms $a_1^2b_2^2 + a_2^2b_1^2$ from both sides:

$-2a_1a_2b_1b_2 = 4a_1a_2(a_2 - a_1)(c_2 - c_1) - a_1a_2(b_1^2 + b_2^2)$

Dividing by $a_1a_2$ (both non-zero):

$-2b_1b_2 = 4(a_2 - a_1)(c_2 - c_1) - b_1^2 - b_2^2$

$b_1^2 + b_2^2 - 2b_1b_2 = 4(a_2 - a_1)(c_2 - c_1)$

$(b_2 - b_1)^2 = 4(a_2 - a_1)(c_2 - c_1) \qquad \text{(*)}$

Now the curves touch when the equation $a_1x^2 + b_1x + c_1 = a_2x^2 + b_2x + c_2$ has exactly one solution, i.e.

$(a_2 - a_1)x^2 + (b_2 - b_1)x + (c_2 - c_1) = 0$

has discriminant zero:

$(b_2 - b_1)^2 = 4(a_2 - a_1)(c_2 - c_1)$

This is exactly condition $(*)$. Therefore, when $a_1 \neq a_2$, the curves have exactly one common tangent if and only if they touch each other. $\qquad \blacksquare$

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Part 4: Case $a_1 = a_2$ — condition for exactly one common tangent

When $a_1 = a_2 = a$, the quadratic from Part 2 becomes linear (the $m^2$ term vanishes):

$2(ab_2 - ab_1)m + 4a^2(c_2 - c_1) + ab_1^2 - ab_2^2 = 0$

$2a(b_2 - b_1)m + 4a^2(c_2 - c_1) + a(b_1^2 - b_2^2) = 0$

Dividing by $a$ (non-zero):

$2(b_2 - b_1)m + 4a(c_2 - c_1) + (b_1^2 - b_2^2) = 0$

This is a linear equation in $m$. It has exactly one solution if and only if the coefficient of $m$ is non-zero, i.e.

$b_2 \neq b_1 \qquad \blacksquare$

(If $b_2 = b_1$, the equation becomes $4a(c_2 - c_1) = 0$. If also $c_2 = c_1$, the curves are identical and every tangent is common; if $c_2 \neq c_1$, there are no common tangents.)

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Question 6

Topic: 三次方程与双曲函数 | Cubic Equations and Hyperbolic Functions  |  Difficulty: Hard  |  Marks: 20

Problem

6 In this question, you may use without proof the results

$4 \cosh^3 y - 3 \cosh y = \cosh(3y) \quad \text{and} \quad \text{arcosh } y = \ln (y + \sqrt{y^2 - 1}) .$

[ Note: $\text{arcosh } y$ is another notation for $\cosh^{-1} y$ ]

Show that the equation $x^3 - 3a^2x = 2a^3 \cosh T$ is satisfied by $2a \cosh (\frac{1}{3}T)$ and hence that, if $c^2 \geqslant b^3$, one of the roots of the equation $x^3 - 3bx = 2c$ is $u + \frac{b}{u}$, where $u = \left( c + \sqrt{c^2 - b^3} \right)^{\frac{1}{3}}$.

Show that the other two roots of the equation $x^3 - 3bx = 2c$ are the roots of the quadratic equation $x^2 + (u + \frac{b}{u})x + u^2 + \frac{b^2}{u^2} - b = 0$, and find these roots in terms of $u$, $b$ and $\omega$, where $\omega = \frac{1}{2}(-1 + i\sqrt{3})$.

Solve completely the equation $x^3 - 6x = 6$.

Hint

6 Direct substitution of x = 2a cosh(T/3) into the left hand side of the equation gives

(2a cosh(T/3))^3 - 6a^3 cosh(T/3) = 2a^3(4(cosh(T/3))^3 - 3 cosh(T/3)) = 2a^3 cosh T

(by the first result given at the start of the question).

Let a^2 = b, which is possible since b > 0, and cosh T = c/a^3, which requires c/a^3 >= 1; but this holds if you choose a to have the same sign as c, since then c/a^3 > 0 and c^2 > b^3 = a^6.

Then, by the second result given at the start of the question,

T = ln(c/a^3 + sqrt(c^2/a^6 - 1)) = ln((c + sqrt(c^2 - b^3))/a^3) = 3 ln(u/a),

so one of the roots of the equation x^3 - 3bx = 2c is

2a cosh(ln(u/a)) = 2a * (u/a + a/u)/2 = u + b/u.

Note that, since u + b/u is a root of the equation x^3 - 3bx = 2c,

2c = (u + b/u)^3 - 3b(u + b/u) = (u + b/u)(u^2 + b^2/u^2 - b)

and that

u^2 + b^2/u^2 - b - (u + b/u)^2 = -3b,

so

x^3 - 3bx - 2c = (x - (u + b/u))(x^2 + (u + b/u)x + u^2 + b^2/u^2 - b),

so the other two roots of x^3 - 3bx = 2c are the roots of x^2 + (u + b/u)x + u^2 + b^2/u^2 - b = 0, which are

1/2 * (-(u + b/u) +/- sqrt((u + b/u)^2 - 4(u^2 + b^2/u^2 - b))) = 1/2 * (-(u + b/u) +/- sqrt(-3(u^2 + b^2/u^2 - 2b))) = 1/2 * (-(u + b/u) +/- sqrt(3)j(u - b/u)),

that is omegau + omega^2(b/u) and omega^2u + omega(b/u).

In x^3 - 6x = 6, b = 2, c = 3, so a = sqrt(2) and so u = cbrt(3 + 1) = 2^(2/3) and b/u = 2^(1/3), so the solutions are 2^(1/3) + 2^(2/3), omega2^(1/3) + omega^22^(2/3) and omega^22^(1/3) + omega2^(2/3).

Model Solution

Part 1: Show that $x = 2a\cosh\left(\frac{T}{3}\right)$ satisfies $x^3 - 3a^2x = 2a^3\cosh T$

Substituting $x = 2a\cosh\left(\frac{T}{3}\right)$:

$x^3 - 3a^2x = 8a^3\cosh^3\left(\frac{T}{3}\right) - 6a^3\cosh\left(\frac{T}{3}\right)$

$= 2a^3\left(4\cosh^3\left(\frac{T}{3}\right) - 3\cosh\left(\frac{T}{3}\right)\right)$

Using the identity $4\cosh^3 y - 3\cosh y = \cosh(3y)$ with $y = \frac{T}{3}$:

$= 2a^3\cosh T \qquad \blacksquare$

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Part 2: One root of $x^3 - 3bx = 2c$ is $u + \frac{b}{u}$

We need to reduce $x^3 - 3bx = 2c$ to the form $x^3 - 3a^2x = 2a^3\cosh T$.

Setting $a^2 = b$ (so $a = \sqrt{b}$, which is valid since $c^2 \geqslant b^3 > 0$ implies $b > 0$) and $a^3\cosh T = c$, i.e. $\cosh T = \frac{c}{a^3} = \frac{c}{b^{3/2}}$.

This requires $\frac{c}{b^{3/2}} \geqslant 1$, i.e. $c \geqslant b^{3/2}$, which holds since $c^2 \geqslant b^3$ and we can choose $a$ to have the same sign as $c$ (so $c/a^3 > 0$).

From Part 1, one root is $x = 2a\cosh\left(\frac{T}{3}\right)$.

Now using $\text{arcosh } y = \ln(y + \sqrt{y^2 - 1})$:

$T = \text{arcosh}\left(\frac{c}{b^{3/2}}\right) = \ln\left(\frac{c}{b^{3/2}} + \sqrt{\frac{c^2}{b^3} - 1}\right) = \ln\left(\frac{c + \sqrt{c^2 - b^3}}{b^{3/2}}\right)$

Let $u = \left(c + \sqrt{c^2 - b^3}\right)^{1/3}$. Then:

$\frac{T}{3} = \frac{1}{3}\ln\left(\frac{c + \sqrt{c^2 - b^3}}{b^{3/2}}\right) = \ln\left(\frac{u}{b^{1/2}}\right) = \ln\left(\frac{u}{a}\right)$

Therefore:

$\cosh\left(\frac{T}{3}\right) = \cosh\left(\ln\frac{u}{a}\right) = \frac{1}{2}\left(\frac{u}{a} + \frac{a}{u}\right)$

and the root is:

$x = 2a \cdot \frac{1}{2}\left(\frac{u}{a} + \frac{a}{u}\right) = u + \frac{a^2}{u} = u + \frac{b}{u} \qquad \blacksquare$

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Part 3: The other two roots are roots of $x^2 + \left(u + \frac{b}{u}\right)x + u^2 + \frac{b^2}{u^2} - b = 0$

Since $u + \frac{b}{u}$ is a root of $x^3 - 3bx - 2c = 0$, we can factor:

$x^3 - 3bx - 2c = \left(x - u - \frac{b}{u}\right)(x^2 + px + q)$

for some $p, q$. Expanding and comparing with $x^3 - 3bx - 2c$:

The coefficient of $x^2$: $p - u - \frac{b}{u} = 0$, so $p = u + \frac{b}{u}$.

The coefficient of $x$: $q - p\left(u + \frac{b}{u}\right) = -3b$, so $q = p^2 - 3b = \left(u + \frac{b}{u}\right)^2 - 3b$.

$q = u^2 + 2b + \frac{b^2}{u^2} - 3b = u^2 + \frac{b^2}{u^2} - b$

The constant term: $-q\left(u + \frac{b}{u}\right) = -2c$. We verify: since $u^3 = c + \sqrt{c^2 - b^3}$ and $\frac{b^3}{u^3} = c - \sqrt{c^2 - b^3}$ (as $\frac{b^3}{u^3} = \frac{b^3}{c + \sqrt{c^2 - b^3}} = \frac{b^3(c - \sqrt{c^2 - b^3})}{b^3} = c - \sqrt{c^2 - b^3}$), we have $u^3 + \frac{b^3}{u^3} = 2c$.

Also $u^3 + \frac{b^3}{u^3} = \left(u + \frac{b}{u}\right)\left(u^2 - b + \frac{b^2}{u^2}\right) = \left(u + \frac{b}{u}\right)q$, confirming $q\left(u + \frac{b}{u}\right) = 2c$. $\checkmark$

Therefore the other two roots satisfy:

$x^2 + \left(u + \frac{b}{u}\right)x + u^2 + \frac{b^2}{u^2} - b = 0 \qquad \blacksquare$

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Part 4: Find these roots in terms of $u, b, \omega$

Using the quadratic formula:

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

The discriminant:

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

$= -3u^2 + 6b - \frac{3b^2}{u^2} = -3\left(u^2 - 2b + \frac{b^2}{u^2}\right) = -3\left(u - \frac{b}{u}\right)^2$

So:

$x = \frac{-\left(u + \frac{b}{u}\right) \pm \sqrt{-3\left(u - \frac{b}{u}\right)^2}}{2} = \frac{-\left(u + \frac{b}{u}\right) \pm i\sqrt{3}\left(u - \frac{b}{u}\right)}{2}$

$= -\frac{1}{2}\left(u + \frac{b}{u}\right) \pm \frac{i\sqrt{3}}{2}\left(u - \frac{b}{u}\right)$

With $\omega = \frac{1}{2}(-1 + i\sqrt{3})$ and $\omega^2 = \frac{1}{2}(-1 - i\sqrt{3})$, we can write:

Taking the $+$ sign:

$x = -\frac{1}{2}u - \frac{b}{2u} + \frac{i\sqrt{3}}{2}u - \frac{i\sqrt{3}b}{2u} = u\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right) + \frac{b}{u}\left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right) = \omega u + \omega^2 \cdot \frac{b}{u}$

Wait, let me recheck. $\omega = \frac{1}{2}(-1 + i\sqrt{3})$, so $-\frac{1}{2} + \frac{i\sqrt{3}}{2} = \omega$ and $-\frac{1}{2} - \frac{i\sqrt{3}}{2} = \omega^2$. So:

$x_+ = \omega u + \omega^2 \cdot \frac{b}{u}$

Taking the $-$ sign:

$x_- = \omega^2 u + \omega \cdot \frac{b}{u}$

The other two roots are:

$\omega u + \frac{\omega^2 b}{u} \qquad \text{and} \qquad \omega^2 u + \frac{\omega b}{u} \qquad \blacksquare$

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Part 5: Solve $x^3 - 6x = 6$ completely

Here $b = 2$ and $c = 3$. Check: $c^2 = 9 \geqslant b^3 = 8$. $\checkmark$

$u = \left(3 + \sqrt{9 - 8}\right)^{1/3} = (3 + 1)^{1/3} = 4^{1/3} = 2^{2/3}$

$\frac{b}{u} = \frac{2}{2^{2/3}} = 2^{1/3}$

The three roots are:

$x_1 = u + \frac{b}{u} = 2^{2/3} + 2^{1/3}$

$x_2 = \omega \cdot 2^{2/3} + \omega^2 \cdot 2^{1/3}$

$x_3 = \omega^2 \cdot 2^{2/3} + \omega \cdot 2^{1/3}$

where $\omega = \frac{1}{2}(-1 + i\sqrt{3})$ and $\omega^2 = \frac{1}{2}(-1 - i\sqrt{3})$.

Numerically: $x_1 \approx 1.587 + 1.260 = 2.847$.

The complex roots can also be written as:

$x_{2,3} = -\frac{1}{2}(2^{2/3} + 2^{1/3}) \pm \frac{i\sqrt{3}}{2}(2^{2/3} - 2^{1/3})$

$\qquad \blacksquare$

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Question 7

Topic: 积分技巧 | Integration Techniques  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Show that if $\int \frac{1}{u f(u)} \, du = F(u) + c$, then $\int \frac{m}{x f(x^m)} \, dx = F(x^m) + c$, where $m \neq 0$.

Find:

(i) $\int \frac{1}{x^n - x} \, dx$;

(ii) $\int \frac{1}{\sqrt{x^n + x^2}} \, dx$.

Hint

7 Substituting u = x^m gives

integral m dx/(x f(x^m)) = integral m x^(m-1) dx/(x^m f(x^m)) = integral du/(u f(u)) = F(u) + c = F(x^m) + c

(i) integral dx/(x^n - x) = integral dx/(x(x^(n-1) - 1)), so letting u = x^(n-1) and f(u) = u - 1, integral (n-1) dx/(x^n - x) = integral du/(u(u-1)) = integral (1/(u-1) - 1/u) du = ln|(u-1)/u| so integral dx/(x^n - x) = (1/(n-1)) ln|(x^(n-1) - 1)/x^(n-1)| + c.

(ii) integral dx/sqrt(x^n + x^2) = integral dx/(x sqrt(x^(n-2) + 1)) (for x > 0) so letting u = x^(n-2) and f(u) = sqrt(u+1) (and assuming n != 2) integral (n-2) dx/sqrt(x^n + x^2) = integral du/(u sqrt(u+1)). Substituting u = v^2 - 1 with v > 0, integral du/(u sqrt(u+1)) = integral 2v dv/((v^2 - 1)v) = integral (1/(v-1) - 1/(v+1)) dv = ln|(v-1)/(v+1)| so integral dx/sqrt(x^n + x^2) = (1/(n-2)) ln|(sqrt(x^(n-2) + 1) - 1)/(sqrt(x^(n-2) + 1) + 1)| + c.

Model Solution

Part 1: Show the general result

Given that $\int \frac{1}{uf(u)}\,du = F(u) + c$, we substitute $u = x^m$, so $du = mx^{m-1}\,dx$.

$\int \frac{m}{xf(x^m)}\,dx = \int \frac{mx^{m-1}}{x^m f(x^m)}\,dx = \int \frac{1}{uf(u)}\,du = F(u) + c = F(x^m) + c \qquad \blacksquare$

(Here we used the substitution $u = x^m$ to convert the integral, noting that $\frac{mx^{m-1}}{x^m} = \frac{m}{x}$.)

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Part (i): Find $\int \frac{1}{x^n - x}\,dx$

$\int \frac{1}{x^n - x}\,dx = \int \frac{1}{x(x^{n-1} - 1)}\,dx$

We want to use the general result with $m = n - 1$ and $f(u) = u - 1$. Then $f(x^m) = f(x^{n-1}) = x^{n-1} - 1$, and $xf(x^m) = x(x^{n-1} - 1) = x^n - x$.

We need $\int \frac{1}{uf(u)}\,du = \int \frac{1}{u(u - 1)}\,du$.

Using partial fractions:

$\frac{1}{u(u - 1)} = \frac{1}{u - 1} - \frac{1}{u}$

$\int \frac{1}{u(u - 1)}\,du = \ln|u - 1| - \ln|u| + c = \ln\left|\frac{u - 1}{u}\right| + c$

So $F(u) = \ln\left|\frac{u - 1}{u}\right|$. By the general result with $m = n - 1$:

$\int \frac{n - 1}{x(x^{n-1} - 1)}\,dx = F(x^{n-1}) + c$

Therefore:

$\int \frac{1}{x^n - x}\,dx = \frac{1}{n - 1}\ln\left|\frac{x^{n-1} - 1}{x^{n-1}}\right| + c \qquad \blacksquare$

(valid for $n \neq 1$.)

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Part (ii): Find $\int \frac{1}{\sqrt{x^n + x^2}}\,dx$

$\int \frac{1}{\sqrt{x^n + x^2}}\,dx = \int \frac{1}{x\sqrt{x^{n-2} + 1}}\,dx \qquad \text{(for } x > 0\text{)}$

We use the general result with $m = n - 2$ and $f(u) = \sqrt{u + 1}$. Then $f(x^m) = \sqrt{x^{n-2} + 1}$, and $xf(x^m) = x\sqrt{x^{n-2} + 1} = \sqrt{x^n + x^2}$.

We need $\int \frac{1}{uf(u)}\,du = \int \frac{1}{u\sqrt{u + 1}}\,du$.

Substitute $u = v^2 - 1$ (with $v > 0$), so $du = 2v\,dv$ and $\sqrt{u + 1} = v$:

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

Using partial fractions:

$\frac{2}{v^2 - 1} = \frac{2}{(v - 1)(v + 1)} = \frac{1}{v - 1} - \frac{1}{v + 1}$

$\int \frac{2}{v^2 - 1}\,dv = \ln|v - 1| - \ln|v + 1| + c = \ln\left|\frac{v - 1}{v + 1}\right| + c$

So $F(u) = \ln\left|\frac{\sqrt{u + 1} - 1}{\sqrt{u + 1} + 1}\right|$. By the general result with $m = n - 2$:

$\int \frac{n - 2}{x\sqrt{x^{n-2} + 1}}\,dx = F(x^{n-2}) + c$

Therefore:

$\int \frac{1}{\sqrt{x^n + x^2}}\,dx = \frac{1}{n - 2}\ln\left|\frac{\sqrt{x^{n-2} + 1} - 1}{\sqrt{x^{n-2} + 1} + 1}\right| + c \qquad \blacksquare$

(valid for $n \neq 2$.)

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Question 8

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

Problem

8 In this question, $a$ and $c$ are distinct non-zero complex numbers. The complex conjugate of any complex number $z$ is denoted by $z^*$.

Show that

$|a - c|^2 = aa^* + cc^* - ac^* - ca^*$

and hence prove that the triangle $OAC$ in the Argand diagram, whose vertices are represented by $0$, $a$ and $c$ respectively, is right angled at $A$ if and only if $2aa^* = ac^* + ca^*$.

Points $P$ and $P'$ in the Argand diagram are represented by the complex numbers $ab$ and $\frac{a}{b^*}$, where $b$ is a non-zero complex number. A circle in the Argand diagram has centre $C$ and passes through the point $A$, and is such that $OA$ is tangent to the circle. Show that the point $P$ lies on the circle if and only if point $P'$ lies on the circle.

Conversely, show that if the points represented by the complex numbers $ab$ and $\frac{a}{b^*}$, for some non-zero complex number $b$ with $bb^* \neq 1$, both lie on a circle centre $C$ in the Argand diagram which passes through $A$, then $OA$ is a tangent to the circle.

Hint

8 Direct use of the important result |z|^2 = zz* gives

|a - c|^2 = (a - c)(a* - c*) = aa* + cc* - ac* - ca*.

OAC is a right angle if and only if |AC|^2 + |OA|^2 = |OC|^2; that is, |a - c|^2 + |a|^2 = |c|^2 or, using the result above, 2aa* - ac* - ca* = 0.

The circle has centre C and radius AC, so complex numbers representing points on the circle satisfy |z - c|^2 = |a - c|^2 or zz* - zc* - cz* = aa* - ac* - ca*.

Because OA is a tangent to the circle, angle OAC is a right angle and so 2aa* - ac* - ca* = 0 as above; thus the condition for points to lie on the circle becomes zz* - zc* - cz* + aa* = 0.

P lies on this circle if and only if

abab - abc* - cab + aa* = 0

and P’ lies on the circle if and only if

aa*/(bb*) - a/b* * c* - c * a*/b + aa* = 0

but multiplying this by bb* (which is not equal to zero) gives the same condition.

Conversely, if the points lie on the circle represented by |z - c|^2 = |a - c|^2,

abab - abc* - cab + aa* = 2aa* - ac* - ca*

and

aa*/(bb*) - a/b* * c* - c * a*/b = 2aa* - ac* - ca*,

so that

abab - abc* - cab + aa* = bb*(2aa* - ac* - ca*)

and so, provided bb* != 1, it must be the case that 2aa* - ac* - ca* = 0, and this shows that OAC is a right angle and hence that OA is a tangent to the circle.

Model Solution

Part 1: Show that $|a - c|^2 = aa^* + cc^* - ac^* - ca^*$

Using the result $|z|^2 = zz^*$:

$|a - c|^2 = (a - c)(a - c)^* = (a - c)(a^* - c^*)$

$= aa^* - ac^* - ca^* + cc^* = aa^* + cc^* - ac^* - ca^* \qquad \blacksquare$

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Part 2: Triangle $OAC$ is right-angled at $A$ iff $2aa^* = ac^* + ca^*$

Triangle $OAC$ is right-angled at $A$ if and only if $|OA|^2 + |AC|^2 = |OC|^2$ (Pythagoras).

$|OA|^2 = |a|^2 = aa^*$ $|AC|^2 = |a - c|^2 = aa^* + cc^* - ac^* - ca^*$ $|OC|^2 = |c|^2 = cc^*$

The condition $|OA|^2 + |AC|^2 = |OC|^2$ becomes:

$aa^* + aa^* + cc^* - ac^* - ca^* = cc^*$

$2aa^* - ac^* - ca^* = 0$

$2aa^* = ac^* + ca^* \qquad \blacksquare$

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Part 3: $P$ lies on the circle iff $P'$ lies on the circle

The circle has centre $C$ and passes through $A$, so a point $z$ lies on the circle if and only if:

$|z - c|^2 = |a - c|^2$

Expanding using $|z|^2 = zz^*$:

$zz^* - zc^* - cz^* + cc^* = aa^* - ac^* - ca^* + cc^*$

$zz^* - zc^* - cz^* = aa^* - ac^* - ca^* \qquad \text{(*)}$

Since $OA$ is tangent to the circle, angle $OAC$ is a right angle, so by Part 2: $2aa^* = ac^* + ca^*$, i.e. $ac^* + ca^* = 2aa^*$.

Substituting into $(*)$:

$zz^* - zc^* - cz^* = aa^* - 2aa* = -aa*$

$zz^* - zc^* - cz^* + aa^* = 0 \qquad \text{(**)}$

$P$ lies on the circle: $z = ab$, so $z^* = a^*b^*$.

$(ab)(a^*b^*) - (ab)c^* - c(a^*b^*) + aa^* = 0$

$aa^*bb^* - abc^* - a^*b^*c + aa^* = 0 \qquad \text{(P)}$

$P'$ lies on the circle: $z = \frac{a}{b^*}$, so $z^* = \frac{a^*}{b}$.

$\frac{a}{b^*} \cdot \frac{a^*}{b} - \frac{a}{b^*}c^* - c\frac{a^*}{b} + aa^* = 0$

$\frac{aa^*}{bb^*} - \frac{ac^*}{b^*} - \frac{ca^*}{b} + aa^* = 0$

Multiplying through by $bb^*$ (which is non-zero):

$aa^* - abc^* - a^*b^*c + aa^*bb^* = 0 \qquad \text{(P')}$

Conditions (P) and (P’) are identical. Therefore $P$ lies on the circle if and only if $P'$ lies on the circle. $\qquad \blacksquare$

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Part 4: Converse — if both $ab$ and $\frac{a}{b^*}$ lie on a circle through $A$ with centre $C$, then $OA$ is tangent

Suppose both $z = ab$ and $z = \frac{a}{b^*}$ lie on the circle $|z - c|^2 = |a - c|^2$.

From the general circle condition $(*)$:

For $z = ab$: $\quad aa^*bb^* - abc^* - a^*b^*c = aa^* - ac^* - ca^* \qquad \text{(I)}$

For $z = \frac{a}{b^*}$: $\quad \frac{aa^*}{bb^*} - \frac{ac^*}{b^*} - \frac{ca^*}{b} = aa^* - ac^* - ca^* \qquad \text{(II)}$

Let $K = aa^* - ac^* - ca^*$ (the right-hand side of both equations).

From (II), multiplying by $bb^*$:

$aa^* - abc^* - a^*b^*c = bb^* K \qquad \text{(II')}$

From (I): the left-hand side equals $K$, so:

$K = aa^*bb^* - abc^* - a^*b^*c$

But from (II’): $abc^* + a^*b^*c = aa^* - bb^*K$, substituting into (I):

$K = aa^*bb^* - (aa^* - bb^*K) = aa^*bb^* - aa^* + bb^*K$

$K - bb^*K = aa^*(bb^* - 1)$

$K(1 - bb^*) = aa^*(bb^* - 1)$

$K = -aa^*$

(since $bb^* \neq 1$, we can divide by $(1 - bb^*) = -(bb^* - 1)$.)

So $K = aa^* - ac^* - ca^* = -aa^*$, which gives:

$2aa^* = ac^* + ca^*$

By Part 2, this means triangle $OAC$ is right-angled at $A$, so $OA$ is perpendicular to $AC$. Since $AC$ is a radius of the circle and $OA$ passes through $A$, this means $OA$ is tangent to the circle. $\qquad \blacksquare$