STEP3 2001 -- Pure Mathematics

Exam: STEP3  |  Year: 2001  |  Questions: Q1—Q8  |  Total marks per question: 20

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

Overview

Q	Topic	Difficulty	Key Techniques
1	微积分 Calculus	Challenging	隐函数求导,数学归纳法,递推关系,Maclaurin级数展开
2	微积分 Calculus	Challenging	反双曲函数恒等式,积分求面积,区域边界分析
3	代数与方程 Algebra and Equations	Standard	判别式分析,韦达定理,不等式约束,区域作图
4	微积分 Calculus	Challenging	反三角函数恒等式,分段函数分析,函数图像描绘
5	微积分 Calculus	Hard	三次方程分析,参数方程法线,轨迹方程推导,分类讨论
6	立体几何 / 3D Geometry	Challenging	向量法,平面方程,三角形特殊点,几何证明
7	微分方程 / Differential Equations	Challenging	正交轨线法,可分离变量微分方程,曲线族与包络
8	复数 / Complex Numbers	Hard	复数模和辐角,轨迹方程,分式线性变换,复平面几何

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Question 1

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

Problem

1 Given that $y = \ln (x + \sqrt{(x^2 + 1)})$, show that $\frac{dy}{dx} = \frac{1}{\sqrt{(x^2 + 1)}}$.

Prove by induction that, for $n \geqslant 0$,

$(x^2 + 1) y^{(n+2)} + (2n + 1) xy^{(n+1)} + n^2y^{(n)} = 0 ,$

where $y^{(n)} = \frac{d^ny}{dx^n}$ and $y^{(0)} = y$.

Using this result in the case $x = 0$, or otherwise, show that the Maclaurin series for $y$ begins

$x - \frac{x^3}{6} + \frac{3x^5}{40}$

and find the next non-zero term.

Model Solution

Part (a): Show that $\frac{dy}{dx} = \frac{1}{\sqrt{x^2 + 1}}$

Let $y = \ln\!\left(x + \sqrt{x^2 + 1}\right)$.

Differentiating using the chain rule:

$\frac{dy}{dx} = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \frac{d}{dx}\!\left(x + \sqrt{x^2 + 1}\right)$

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

$= \frac{1}{x + \sqrt{x^2 + 1}} \cdot \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}$

$= \frac{1}{\sqrt{x^2 + 1}} \qquad \blacksquare$

Part (b): Prove by induction that $(x^2 + 1) y^{(n+2)} + (2n+1) x\, y^{(n+1)} + n^2 y^{(n)} = 0$

First, from part (a), we have $(x^2 + 1)^{1/2} \frac{dy}{dx} = 1$, so:

$(x^2 + 1)\!\left(\frac{dy}{dx}\right)^2 = 1 \qquad \text{(*)}$

Differentiating (*) with respect to $x$:

$2x\!\left(\frac{dy}{dx}\right)^2 + 2(x^2 + 1)\frac{dy}{dx}\frac{d^2y}{dx^2} = 0$

Since $\frac{dy}{dx} \neq 0$:

$x\frac{dy}{dx} + (x^2 + 1)\frac{d^2y}{dx^2} = 0 \qquad \text{(**)}$

This is the case $n = 0$ of the recurrence (since $y^{(0)} = y$ does not appear, the $n=0$ case reduces to $(x^2+1)y'' + xy' = 0$, which is (**)).

Base case $n = 0$. We have just shown $(x^2 + 1)y^{(2)} + x\,y^{(1)} = 0$. (The term $0^2 \cdot y^{(0)} = 0$ vanishes.)

Inductive step. Assume the result holds for some $n \geqslant 0$:

$(x^2 + 1) y^{(n+2)} + (2n+1) x\, y^{(n+1)} + n^2 y^{(n)} = 0 \qquad \text{(IH)}$

Differentiate (IH) with respect to $x$:

$2x\, y^{(n+2)} + (x^2 + 1) y^{(n+3)} + (2n+1) y^{(n+1)} + (2n+1) x\, y^{(n+2)} + n^2 y^{(n+1)} = 0$

Collecting terms:

$(x^2 + 1) y^{(n+3)} + [2x + (2n+1)x]\, y^{(n+2)} + [(2n+1) + n^2]\, y^{(n+1)} = 0$

$(x^2 + 1) y^{(n+3)} + (2n+3) x\, y^{(n+2)} + (n+1)^2 y^{(n+1)} = 0$

This is exactly the recurrence with $n$ replaced by $n + 1$. By induction, the result holds for all $n \geqslant 0$. $\blacksquare$

Part (c): Maclaurin series

Setting $x = 0$ in the recurrence:

$(0 + 1) y^{(n+2)}(0) + 0 + n^2 y^{(n)}(0) = 0$

$y^{(n+2)}(0) = -n^2\, y^{(n)}(0) \qquad \text{(***)}$

We compute the derivatives at $x = 0$:

• $y(0) = \ln(0 + 1) = 0$, so $y^{(0)}(0) = 0$.
• $y'(0) = \frac{1}{\sqrt{0+1}} = 1$.
• From (***): $y''(0) = -0^2 \cdot y(0) = 0$.
• $y^{(3)}(0) = -1^2 \cdot y'(0) = -1$.
• $y^{(4)}(0) = -2^2 \cdot y''(0) = 0$.
• $y^{(5)}(0) = -3^2 \cdot y^{(3)}(0) = -9 \cdot (-1) = 9$.
• $y^{(6)}(0) = -4^2 \cdot y^{(4)}(0) = 0$.
• $y^{(7)}(0) = -5^2 \cdot y^{(5)}(0) = -25 \cdot 9 = -225$.

The Maclaurin series is:

$y = \sum_{k=0}^{\infty} \frac{y^{(k)}(0)}{k!} x^k = x - \frac{x^3}{3!} + \frac{9 x^5}{5!} - \frac{225 x^7}{7!} + \cdots$

$= x - \frac{x^3}{6} + \frac{9x^5}{120} - \frac{225 x^7}{5040} + \cdots$

$= x - \frac{x^3}{6} + \frac{3x^5}{40} - \frac{5x^7}{112} + \cdots$

The first three terms are $x - \frac{x^3}{6} + \frac{3x^5}{40}$ as required, and the next non-zero term is $-\dfrac{5x^7}{112}$.

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Question 2

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

Problem

2 Show that $\cosh^{-1} x = \ln (x + \sqrt{(x^2 - 1)})$.

Show that the area of the region defined by the inequalities $y^2 \geqslant x^2 - 8$ and $x^2 \geqslant 25y^2 - 16$ is $(72/5) \ln 2$.

Model Solution

Part (a): Show that $\cosh^{-1} x = \ln\!\left(x + \sqrt{x^2 - 1}\right)$

Let $y = \cosh^{-1} x$, so $x = \cosh y$ with $y \geqslant 0$.

From $\cosh y = \frac{e^y + e^{-y}}{2} = x$, multiply by $2e^y$:

$e^{2y} - 2x\, e^y + 1 = 0$

This is quadratic in $e^y$:

$e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}$

Since $y \geqslant 0$, we need $e^y \geqslant 1$. Both roots are positive (product = 1), so $x + \sqrt{x^2 - 1} \geqslant 1$ and $x - \sqrt{x^2 - 1} \leqslant 1$ (since their product is 1). Therefore $e^y = x + \sqrt{x^2 - 1}$, giving:

$\cosh^{-1} x = \ln\!\left(x + \sqrt{x^2 - 1}\right) \qquad \blacksquare$

Part (b): Area of the region

The region is defined by $y^2 \geqslant x^2 - 8$ and $x^2 \geqslant 25y^2 - 16$.

Both inequalities are unchanged when $x$ or $y$ is replaced by its negative, so the total area is four times the area in the first quadrant.

In the first quadrant, the second inequality gives

$0 \leqslant y \leqslant \frac{1}{5}\sqrt{x^2+16}.$

The first inequality is automatic when $0 \leqslant x < \sqrt8$, and gives $y \geqslant \sqrt{x^2-8}$ when $x \geqslant \sqrt8$.

The two upper boundary curves meet when

$\sqrt{x^2-8}=\frac15\sqrt{x^2+16}.$

Squaring gives $25x^2-200=x^2+16$, so $x^2=9$. Thus in the first quadrant the bounded part runs from $x=0$ to $x=3$, with the lower boundary changing at $x=\sqrt8$.

The first-quadrant area is therefore

$\int_0^{\sqrt{8}} \frac{1}{5}\sqrt{x^2+16}\, dx + \int_{\sqrt{8}}^{3} \left[\frac{1}{5}\sqrt{x^2+16} - \sqrt{x^2-8}\right] dx$

$= \frac{1}{5}\int_0^{3} \sqrt{x^2+16}\, dx - \int_{\sqrt{8}}^{3} \sqrt{x^2-8}\, dx$

Integral 1: $\int \sqrt{x^2 + 16}\, dx$.

Using the standard formula $\int \sqrt{x^2 + a^2}\, dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln\!\left(x + \sqrt{x^2+a^2}\right) + C$:

$\int_0^3 \sqrt{x^2+16}\, dx = \left[\frac{x}{2}\sqrt{x^2+16} + 8\ln\!\left(x + \sqrt{x^2+16}\right)\right]_0^3$

$= \frac{3}{2}\sqrt{25} + 8\ln(3+5) - 0 - 8\ln 4$

$= \frac{15}{2} + 8\ln 8 - 8\ln 4 = \frac{15}{2} + 8\ln 2$

Integral 2: $\int \sqrt{x^2 - 8}\, dx$.

Using $\int \sqrt{x^2 - a^2}\, dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln\!\left|x + \sqrt{x^2-a^2}\right| + C$:

$\int_{\sqrt{8}}^{3} \sqrt{x^2-8}\, dx = \left[\frac{x}{2}\sqrt{x^2-8} - 4\ln\!\left(x + \sqrt{x^2-8}\right)\right]_{\sqrt{8}}^{3}$

$= \frac{3}{2}\sqrt{1} - 4\ln(3+1) - 0 + 4\ln(\sqrt{8})$

$= \frac{3}{2} - 4\ln 4 + 4\ln\sqrt{8} = \frac{3}{2} - 4\ln 4 + 4 \cdot \frac{3}{2}\ln 2$

$= \frac{3}{2} - 8\ln 2 + 6\ln 2 = \frac{3}{2} - 2\ln 2$

Thus the first-quadrant area is

$\frac{1}{5}\!\left(\frac{15}{2}+8\ln2\right)-\left(\frac{3}{2}-2\ln2\right)=\frac{18}{5}\ln2.$

Multiplying by $4$ for the four quadrants gives

$A=4\cdot\frac{18}{5}\ln2=\frac{72}{5}\ln2. \qquad \blacksquare$

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Question 3

Topic: 代数与方程 Algebra and Equations  |  Difficulty: Standard  |  Marks: 20

Problem

3 Consider the equation

$x^2 - bx + c = 0 ,$

where $b$ and $c$ are real numbers.

(i) Show that the roots of the equation are real and positive if and only if $b > 0$ and $b^2 \geqslant 4c > 0$, and sketch the region of the $b$-$c$ plane in which these conditions hold.

(ii) Sketch the region of the $b$-$c$ plane in which the roots of the equation are real and less than 1 in magnitude.

Model Solution

Part (i)

Let the roots be $\alpha$ and $\beta$. By Vieta’s formulas:

$\alpha + \beta = b \qquad \text{(1)}$

$\alpha \beta = c \qquad \text{(2)}$

We prove both directions of the “if and only if”.

Forward direction. Suppose both roots are real and positive.

Since $\alpha > 0$ and $\beta > 0$:

• From (1): $b = \alpha + \beta > 0$.
• From (2): $c = \alpha \beta > 0$.
• Real roots require discriminant $\Delta = b^2 - 4c \geqslant 0$, i.e., $b^2 \geqslant 4c$.

Combining: $b > 0$ and $b^2 \geqslant 4c > 0$.

Reverse direction. Suppose $b > 0$ and $b^2 \geqslant 4c > 0$.

• $b^2 \geqslant 4c$ means $\Delta \geqslant 0$, so the roots are real.
• $4c > 0$ gives $c > 0$, so $\alpha \beta = c > 0$: the roots have the same sign.
• $b > 0$ gives $\alpha + \beta = b > 0$: the roots are not both negative.

Since the roots have the same sign and their sum is positive, both roots are positive.

This completes the proof. $\blacksquare$

Sketch of the region. In the $b$-$c$ plane, the conditions are:

1. $b > 0$ (right of the $c$-axis, not including it),
2. $c > 0$ (above the $b$-axis, not including it),
3. $c \leqslant \frac{b^2}{4}$ (on or below the parabola $c = \frac{b^2}{4}$).

The region is bounded below by the positive $b$-axis, on the left by the $c$-axis (excluded), and above by the parabola $c = \frac{b^2}{4}$ (included). The parabola passes through the origin and opens upward. The boundary $c = 0$ (the $b$-axis) is excluded since $4c > 0$, while the parabola $c = \frac{b^2}{4}$ is included.

Part (ii)

We need the roots to satisfy $-1 < \alpha, \beta < 1$, i.e., both roots lie strictly between $-1$ and $1$.

Since the coefficient of $x^2$ is positive, the parabola $y = x^2 - bx + c$ opens upward. Both roots lie in $(-1, 1)$ if and only if:

(a) $f(1) > 0$: the value at $x = 1$ is positive (so $x = 1$ is to the right of both roots),

(b) $f(-1) > 0$: the value at $x = -1$ is positive (so $x = -1$ is to the left of both roots),

(c) $\Delta \geqslant 0$: the roots are real.

Evaluating:

$f(1) = 1 - b + c > 0 \implies c > b - 1 \qquad \text{(a)}$

$f(-1) = 1 + b + c > 0 \implies c > -b - 1 \qquad \text{(b)}$

$b^2 \geqslant 4c \qquad \text{(c)}$

Since $\alpha,\beta\in(-1,1)$, their sum satisfies $-2<\alpha+\beta<2$, so the region must also have $-2<b<2$. Conversely, the conditions

$-2<b<2,\qquad c>|b|-1,\qquad c\leqslant \frac{b^2}{4}$

give real roots, put the endpoints $x=-1$ and $x=1$ outside the interval between the roots, and keep the vertex between the endpoints. Hence both roots lie strictly in $(-1,1)$.

Sketch of the region. The boundary curves are:

• $c = b - 1$: a line with gradient 1 and $c$-intercept $-1$.
• $c = -b - 1$: a line with gradient $-1$ and $c$-intercept $-1$.
• $c = \frac{b^2}{4}$: the parabola from part (i).

The two lines meet at $(0, -1)$, forming a V-shape opening upward. The parabola intersects $c = b - 1$ at $b = 2$ (giving the point $(2, 1)$) and intersects $c = -b - 1$ at $b = -2$ (giving the point $(-2, 1)$).

The region is:

• Above both lines: $c > b - 1$ and $c > -b - 1$, equivalently $c > |b| - 1$.
• Below or on the parabola: $c \leqslant \frac{b^2}{4}$.
• Between the vertical lines $b=-2$ and $b=2$, with both vertical boundary lines excluded.

This region lies between $b = -2$ and $b = 2$, bounded below by the V-shaped pair of lines (excluded) and above by the parabola (included). The region has a roughly lens-shaped appearance, widest near $b = 0$ where the gap between the parabola ($c = 0$) and the lines ($c = -1$) is greatest.

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Question 4

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

Problem

4 In this question, the function $\sin^{-1}$ is defined to have domain $-1 \leqslant x \leqslant 1$ and range $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$ and the function $\tan^{-1}$ is defined to have the real numbers as its domain and range $-\frac{\pi}{2} < x < \frac{\pi}{2}$.

(i) Let

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

Sketch the graph of $g(x)$ and state the range of $g$.

(ii) Let

$f(x) = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \text{ , } \quad -\infty < x < \infty \text{ .}$

Show that $f(x) = 2 \tan^{-1} x$ for $-1 \leqslant x \leqslant 1$ and $f(x) = \pi - 2 \tan^{-1} x$ for $x \geqslant 1$. Sketch the graph of $f(x)$.

Model Solution

Part (i)

We analyse $g(x) = \frac{2x}{1 + x^2}$.

Derivative. Using the quotient rule:

$g'(x) = \frac{2(1 + x^2) - 2x \cdot 2x}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$

Setting $g'(x) = 0$: $1 - x^2 = 0$, so $x = \pm 1$.

• $g'(x) > 0$ for $|x| < 1$: $g$ is increasing on $(-1, 1)$.
• $g'(x) < 0$ for $|x| > 1$: $g$ is decreasing on $(-\infty, -1)$ and $(1, \infty)$.

Key values. $g(-1) = \frac{-2}{2} = -1$ (local minimum), $g(1) = \frac{2}{2} = 1$ (local maximum), $g(0) = 0$.

Behaviour at infinity. $g(x) \to 0^+$ as $x \to +\infty$, and $g(x) \to 0^-$ as $x \to -\infty$.

Symmetry. $g(-x) = \frac{-2x}{1 + x^2} = -g(x)$, so $g$ is an odd function (symmetric about the origin).

Graph. The curve passes through the origin, rises to a maximum of $1$ at $x = 1$, then decays toward $0$ as $x \to \infty$. By odd symmetry, it falls to a minimum of $-1$ at $x = -1$ and approaches $0$ from below as $x \to -\infty$. The $x$-axis is a horizontal asymptote.

Range. The maximum value is $g(1) = 1$ and the minimum value is $g(-1) = -1$. Since $g$ is continuous and achieves all values between $-1$ and $1$:

$\text{range of } g = [-1, 1].$

Part (ii)

Let $\theta = \tan^{-1} x$, so $x = \tan\theta$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Then:

$\frac{2x}{1 + x^2} = \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta$

So $f(x) = \sin^{-1}(\sin 2\theta)$.

We evaluate $\sin^{-1}(\sin 2\theta)$ using the range of $\sin^{-1}$, which is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Case 1: $-1 \leqslant x \leqslant 1$.

Here $\theta = \tan^{-1} x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$, so $2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Since $2\theta$ lies in the range of $\sin^{-1}$:

$f(x) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x$

Case 2: $x \geqslant 1$.

Here $\theta = \tan^{-1} x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right)$, so $2\theta \in \left[\frac{\pi}{2}, \pi\right)$.

For $2\theta \in \left[\frac{\pi}{2}, \pi\right)$, we use the identity $\sin(\pi - 2\theta) = \sin 2\theta$ and note that $\pi - 2\theta \in \left(0, \frac{\pi}{2}\right] \subset \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$:

$f(x) = \sin^{-1}(\sin 2\theta) = \pi - 2\theta = \pi - 2\tan^{-1} x$

Case 3: $x \leqslant -1$ (by symmetry).

Since $g$ is odd, we have $f(-x) = \sin^{-1}(-g(x)) = -\sin^{-1}(g(x)) = -f(x)$, so $f$ is odd.

For $x \leqslant -1$, using $f(x) = -f(-x)$ and Case 2 (since $-x \geqslant 1$):

$f(x) = -[\pi - 2\tan^{-1}(-x)] = -\pi + 2\tan^{-1}(-x) = -\pi - 2\tan^{-1} x$

This completes the derivation. $\blacksquare$

Sketch of $f(x)$.

The piecewise definition is:

$f(x) = \begin{cases} -\pi - 2\tan^{-1} x & x \leqslant -1 \\ 2\tan^{-1} x & -1 \leqslant x \leqslant 1 \\ \pi - 2\tan^{-1} x & x \geqslant 1 \end{cases}$

Key features:

• $f$ is continuous everywhere (check at $x = \pm 1$: $f(1) = \frac{\pi}{2}$, $f(-1) = -\frac{\pi}{2}$).
• $f$ is an odd function: symmetric about the origin.
• On $[-1, 1]$, $f$ increases from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ with an inflection point at the origin (this portion is $2\tan^{-1} x$).
• For $x \geqslant 1$, $f$ decreases from $\frac{\pi}{2}$ toward $0$ as $x \to \infty$ (horizontal asymptote $y = 0$).
• For $x \leqslant -1$, $f$ increases from $-\frac{\pi}{2}$ toward $0$ as $x \to -\infty$ (horizontal asymptote $y = 0$).
• Not differentiable at $x = \pm 1$ (the left and right derivatives differ).

The graph looks like a stretched S-curve on $[-1, 1]$ that flattens out to the $x$-axis on both sides.

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Question 5

Topic: 微积分 Calculus  |  Difficulty: Hard  |  Marks: 20

Problem

5 Show that the equation $x^3 + px + q = 0$ has exactly one real solution if $p \geqslant 0$.

A parabola $C$ is given parametrically by

$x = at^2, \quad y = 2at \quad (a > 0) \text{ .}$

Find an equation which must be satisfied by $t$ at points on $C$ at which the normal passes through the point $(h, k)$. Hence show that, if $h \leqslant 2a$, exactly one normal to $C$ will pass through $(h, k)$.

Find, in Cartesian form, the equation of the locus of the points from which exactly two normals can be drawn to $C$. Sketch the locus.

Model Solution

Part (a): Show that $x^3 + px + q = 0$ has exactly one real solution if $p \geqslant 0$

Let $f(x) = x^3 + px + q$.

$f'(x) = 3x^2 + p$

Since $p \geqslant 0$ and $3x^2 \geqslant 0$ for all $x$, we have $f'(x) \geqslant 0$ for all $x$. Moreover, $f'(x) = 0$ only at $x = 0$ (when $p = 0$), so $f$ is strictly increasing.

A strictly increasing continuous function can cross the $x$-axis at most once. Since $f(x) \to +\infty$ as $x \to +\infty$ and $f(x) \to -\infty$ as $x \to -\infty$, by the intermediate value theorem, $f$ has exactly one real root. $\blacksquare$

Part (b): Find the equation for $t$ at normal points through $(h, k)$

For the parabola $x = at^2$, $y = 2at$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$

The slope of the normal at parameter $t$ is $-t$. The normal at $(at^2, 2at)$ is:

$y - 2at = -t(x - at^2)$

For this to pass through $(h, k)$:

$k - 2at = -t(h - at^2)$

$k - 2at = -th + at^3$

$at^3 + (2a - h)t - k = 0 \qquad \text{(*)}$

This is the required equation. $\blacksquare$

Part (c): If $h \leqslant 2a$, exactly one normal passes through $(h, k)$

Equation (*) can be written as $t^3 + \frac{2a - h}{a}\,t - \frac{k}{a} = 0$.

Since $h \leqslant 2a$, we have $2a - h \geqslant 0$, so the coefficient $\frac{2a - h}{a} \geqslant 0$ (using $a > 0$).

By part (a), this cubic in $t$ has exactly one real solution. Each real value of $t$ gives a distinct point on the parabola, so exactly one normal to $C$ passes through $(h, k)$. $\blacksquare$

Part (d): Locus of points from which exactly two normals can be drawn

Exactly two normals means the cubic (*) has exactly two distinct real roots, i.e., one root is repeated. Let the roots be $s, s, t_3$ with $s \neq t_3$.

From Vieta’s formulas for $t^3 + \frac{2a-h}{a}\,t - \frac{k}{a} = 0$:

$2s + t_3 = 0 \implies t_3 = -2s \qquad \text{(1)}$

$s^2 + 2st_3 = \frac{2a - h}{a} \qquad \text{(2)}$

$s^2 t_3 = \frac{k}{a} \qquad \text{(3)}$

From (1) and (3):

$s^2(-2s) = \frac{k}{a} \implies k = -2as^3$

From (1) and (2):

$s^2 + 2s(-2s) = \frac{2a - h}{a} \implies -3s^2 = \frac{2a - h}{a}$

$h = 2a + 3as^2 = a(2 + 3s^2)$

To eliminate $s$: from $k = -2as^3$, we get $s^3 = -\frac{k}{2a}$, so $s^6 = \frac{k^2}{4a^2}$.

From $h = a(2 + 3s^2)$: $s^2 = \frac{h - 2a}{3a}$, so $s^6 = \frac{(h-2a)^3}{27a^3}$.

Equating:

$\frac{k^2}{4a^2} = \frac{(h - 2a)^3}{27a^3}$

$27ak^2 = 4(h - 2a)^3$

This is a semicubical parabola (Neile’s parabola) with a cusp at $(2a, 0)$. It exists only for $h \geqslant 2a$ (since $s^2 \geqslant 0$ requires $h \geqslant 2a$).

The locus is the curve:

$\boxed{27ak^2 = 4(h - 2a)^3}$

This is the evolute of the parabola $C$. The cusp is at $(2a, 0)$, which is the focus of the parabola. The curve extends to the right (for $h > 2a$), and the region inside this evolute is where three normals can be drawn to $C$.

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Question 6

Topic: 立体几何 / 3D Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

6 The plane

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

meets the co-ordinate axes at the points $A, B$ and $C$. The point $M$ has coordinates $(\frac{1}{2}a, \frac{1}{2}b, \frac{1}{2}c)$ and $O$ is the origin.

Show that $OM$ meets the plane at the centroid $(\frac{1}{3}a, \frac{1}{3}b, \frac{1}{3}c)$ of triangle $ABC$. Show also that the perpendiculars to the plane from $O$ and from $M$ meet the plane at the orthocentre and at the circumcentre of triangle $ABC$ respectively.

Hence prove that the centroid of a triangle lies on the line segment joining its orthocentre and circumcentre, and that it divides this line segment in the ratio $2 : 1$.

[The orthocentre of a triangle is the point at which the three altitudes intersect; the circumcentre of a triangle is the point equidistant from the three vertices.]

Model Solution

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ meets the coordinate axes at $A = (a, 0, 0)$, $B = (0, b, 0)$, $C = (0, 0, c)$.

Part (a): $OM$ meets the plane at the centroid of $\triangle ABC$

The line from $O = (0,0,0)$ through $M = (\frac{1}{2}a, \frac{1}{2}b, \frac{1}{2}c)$ is:

$(x, y, z) = t\!\left(\tfrac{1}{2}a, \tfrac{1}{2}b, \tfrac{1}{2}c\right) = \left(\tfrac{ta}{2}, \tfrac{tb}{2}, \tfrac{tc}{2}\right)$

Substituting into the plane equation:

$\frac{ta/2}{a} + \frac{tb/2}{b} + \frac{tc/2}{c} = 1 \implies \frac{3t}{2} = 1 \implies t = \frac{2}{3}$

The intersection point is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.

The centroid of $\triangle ABC$ is $G = \frac{A + B + C}{3} = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.

These are the same point. $\blacksquare$

Part (b): The perpendicular from $O$ meets the plane at the orthocentre

The normal to the plane is $\mathbf{n} = \left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right)$. The perpendicular from $O$ to the plane is:

$(x, y, z) = t\!\left(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\right)$

Substituting into the plane equation:

$\frac{t}{a^2} + \frac{t}{b^2} + \frac{t}{c^2} = 1 \implies t = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$

Let $S = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. The foot of the perpendicular is:

$H = \left(\frac{1}{aS}, \frac{1}{bS}, \frac{1}{cS}\right)$

We now verify this is the orthocentre of $\triangle ABC$. It is enough to show that $H$ lies on the three altitudes. For example,

$\overrightarrow{AH} = \left(\frac{1}{aS} - a, \frac{1}{bS}, \frac{1}{cS}\right), \quad \overrightarrow{BC} = (0, -b, c)$

so

$\overrightarrow{AH} \cdot \overrightarrow{BC} = -\frac{b}{bS} + \frac{c}{cS} = -\frac{1}{S} + \frac{1}{S} = 0.$

By symmetry (cyclic permutation of $a, b, c$):

$\overrightarrow{BH} \cdot \overrightarrow{CA} = 0,\qquad \overrightarrow{CH} \cdot \overrightarrow{AB} = 0.$

Thus $H$ lies on all three altitudes, confirming it is the orthocentre. $\blacksquare$

Part (c): The perpendicular from $M$ meets the plane at the circumcentre

The perpendicular from $M = (\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$ in direction $\mathbf{n} = (\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$:

$(x, y, z) = \left(\frac{a}{2} + \frac{t}{a}, \frac{b}{2} + \frac{t}{b}, \frac{c}{2} + \frac{t}{c}\right)$

Substituting into the plane:

$\frac{1}{2} + \frac{t}{a^2} + \frac{1}{2} + \frac{t}{b^2} + \frac{1}{2} + \frac{t}{c^2} = 1$

$\frac{3}{2} + tS = 1 \implies t = -\frac{1}{2S}$

The foot of the perpendicular is:

$O_c = \left(\frac{a}{2} - \frac{1}{2aS}, \frac{b}{2} - \frac{1}{2bS}, \frac{c}{2} - \frac{1}{2cS}\right)$

We verify this is equidistant from $A$, $B$, $C$. Let $\alpha = \frac{1}{a^2 S}$, $\beta = \frac{1}{b^2 S}$, $\gamma = \frac{1}{c^2 S}$, so $\alpha + \beta + \gamma = 1$.

Then $O_c = \left(\frac{a(1-\alpha)}{2}, \frac{b(1-\beta)}{2}, \frac{c(1-\gamma)}{2}\right)$.

Distance to $A = (a, 0, 0)$:

$|O_c A|^2 = \frac{a^2(1+\alpha)^2}{4} + \frac{b^2(1-\beta)^2}{4} + \frac{c^2(1-\gamma)^2}{4}$

Distance to $B = (0, b, 0)$:

$|O_c B|^2 = \frac{a^2(1-\alpha)^2}{4} + \frac{b^2(1+\beta)^2}{4} + \frac{c^2(1-\gamma)^2}{4}$

Distance to $C = (0, 0, c)$:

$|O_c C|^2 = \frac{a^2(1-\alpha)^2}{4} + \frac{b^2(1-\beta)^2}{4} + \frac{c^2(1+\gamma)^2}{4}$

Computing $|O_c A|^2 - |O_c B|^2$:

$= \frac{a^2[(1+\alpha)^2 - (1-\alpha)^2] - b^2[(1+\beta)^2 - (1-\beta)^2]}{4}$

$= \frac{4a^2\alpha - 4b^2\beta}{4} = a^2\alpha - b^2\beta = \frac{1}{S} - \frac{1}{S} = 0$

Similarly $|O_c B|^2 - |O_c C|^2 = 0$. So $O_c$ is equidistant from all three vertices, confirming it is the circumcentre. $\blacksquare$

Part (d): The centroid divides the orthocentre—circumcentre segment in the ratio $2:1$

The centroid, orthocentre, and circumcentre all lie in the plane of $\triangle ABC$. We show $G = \frac{2O_c + H}{3}$.

$\frac{2O_c + H}{3} = \frac{1}{3}\left(a(1-\alpha) + \frac{1}{aS}, \; b(1-\beta) + \frac{1}{bS}, \; c(1-\gamma) + \frac{1}{cS}\right)$

Since $\frac{1}{aS} = a\alpha$:

$a(1-\alpha) + a\alpha = a$

So the first component is $\frac{a}{3}$. By symmetry, the second and third components are $\frac{b}{3}$ and $\frac{c}{3}$.

Therefore $\frac{2O_c + H}{3} = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) = G$.

This means $G$ lies on the segment $HO_c$ and divides it in the ratio $HG : GO_c = 2 : 1$ (with $G$ closer to $O_c$). $\blacksquare$

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Question 7

Topic: 微分方程 / Differential Equations  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Sketch the graph of the function $\ln x - \frac{1}{2}x^2$.

Show that the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2xy}{x^2 - 1}$ describes a family of parabolas each of which passes through the points $(1, 0)$ and $(-1, 0)$ and has its vertex on the $y$-axis.

Hence find the equation of the curve that passes through the point $(1, 1)$ and intersects each of the above parabolas orthogonally. Sketch this curve.

[Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]

Model Solution

Part (a): Sketch of $f(x) = \ln x - \frac{1}{2}x^2$

Domain: $x > 0$.

Derivative. $f'(x) = \frac{1}{x} - x = \frac{1 - x^2}{x}$.

Setting $f'(x) = 0$: $x = 1$ (since $x > 0$).

• $f'(x) > 0$ for $0 < x < 1$: $f$ is increasing.
• $f'(x) < 0$ for $x > 1$: $f$ is decreasing.

So $x = 1$ is a maximum. $f(1) = 0 - \frac{1}{2} = -\frac{1}{2}$.

Behaviour at boundaries. $f(x) \to -\infty$ as $x \to 0^+$ (since $\ln x \to -\infty$) and $f(x) \to -\infty$ as $x \to +\infty$ (since $-\frac{1}{2}x^2$ dominates).

Root. Since $f(1) = -\frac{1}{2} < 0$ and $f(x) < 0$ at both endpoints, $f$ is negative everywhere. The graph has a single maximum at $(1, -\frac{1}{2})$ and lies entirely below the $x$-axis.

The curve is an inverted bell shape, rising from $-\infty$ near $x = 0$, reaching a peak of $-\frac{1}{2}$ at $x = 1$, then falling back to $-\infty$.

Part (b): The differential equation describes a family of parabolas

$\frac{dy}{dx} = \frac{2xy}{x^2 - 1}$

Separating variables:

$\frac{dy}{y} = \frac{2x}{x^2 - 1}\,dx$

Integrating:

$\ln|y| = \ln|x^2 - 1| + C$

$|y| = e^C |x^2 - 1|$

$y = A(x^2 - 1)$

where $A$ is an arbitrary constant. This is a family of parabolas $y = A(x^2 - 1)$, each with vertex at $(0, -A)$ (on the $y$-axis) and each passing through $(1, 0)$ and $(-1, 0)$. $\blacksquare$

Part (c): Orthogonal trajectory through $(1, 1)$

For the family $y = A(x^2 - 1)$, the slope is $\frac{dy}{dx} = 2Ax$. The orthogonal trajectory has slope $-\frac{1}{2Ax} = -\frac{x^2 - 1}{2xy}$ (using $A = \frac{y}{x^2 - 1}$).

So the orthogonal trajectory satisfies:

$\frac{dy}{dx} = -\frac{x^2 - 1}{2xy}$

Separating variables:

$2y\,dy = -\frac{x^2 - 1}{x}\,dx = \left(-x + \frac{1}{x}\right) dx$

Integrating:

$y^2 = -\frac{x^2}{2} + \ln|x| + C$

Since we need the curve to pass through $(1, 1)$:

$1 = -\frac{1}{2} + \ln 1 + C \implies C = \frac{3}{2}$

The orthogonal trajectory is:

$y^2 + \frac{x^2}{2} - \ln x = \frac{3}{2} \qquad (x > 0)$

or equivalently:

$y = \pm\sqrt{\frac{3}{2} - \frac{x^2}{2} + \ln x}$

Substitution confirms that $(1,1)$ lies on the curve. The slope of the orthogonal trajectory at $(1,1)$ is $0$, as given by $\frac{dy}{dx}=-\frac{x^2-1}{2xy}$.

Sketch. The curve is symmetric about the $x$-axis. It exists for $x > 0$ where $\frac{3}{2} - \frac{x^2}{2} + \ln x \geqslant 0$. This expression tends to $-\infty$ as $x\to0^+$ and also as $x\to\infty$, and has its maximum at $x=1$. Hence the curve is a closed loop passing through $(1,\pm1)$, with two positive $x$-intercepts determined by $\frac{3}{2} - \frac{x^2}{2} + \ln x = 0$.

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Question 8

Topic: 复数 / Complex Numbers  |  Difficulty: Hard  |  Marks: 20

Problem

8 (i) Prove that the equations $|z - (1 + i)|^2 = 2 \qquad \text{(*)}$ and $|z - (1 - i)|^2 = 2|z - 1|^2$ describe the same locus in the complex $z$-plane. Sketch this locus.

(ii) Prove that the equation $\arg\left(\frac{z - 2}{z}\right) = \frac{\pi}{4} \qquad \text{(**)}$ describes part of this same locus, and show on your sketch which part.

(iii) The complex number $w$ is related to $z$ by $w = \frac{2}{z} .$ Determine the locus produced in the complex $w$-plane if $z$ satisfies (*). Sketch this locus and indicate the part of this locus that corresponds to (**).

Model Solution

Part (i): The two equations describe the same locus

Let $z = x + iy$.

Equation (*): $|z - (1+i)|^2 = 2$

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

$x^2 + y^2 - 2x - 2y = 0 \qquad \text{(I)}$

Second equation: $|z - (1-i)|^2 = 2|z - 1|^2$

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

$x^2 - 2x + 1 + y^2 + 2y + 1 = 2x^2 - 4x + 2 + 2y^2$

$0 = x^2 + y^2 - 2x - 2y$

This is identical to (I). Therefore both equations describe the same locus: a circle centred at $(1, 1)$ with radius $\sqrt{2}$. $\blacksquare$

Sketch. The circle passes through the origin and through $(2, 0)$, $(0, 2)$, and $(2, 2)$.

Part (ii): The arg equation describes part of this locus

Let $z = x + iy$. We compute:

$\frac{z - 2}{z} = \frac{(x - 2) + iy}{x + iy}$

Multiplying numerator and denominator by the conjugate of the denominator:

$= \frac{[(x-2) + iy](x - iy)}{x^2 + y^2}$

Numerator: $(x-2)x + y^2 + i[yx - (x-2)y] = (x^2 - 2x + y^2) + i(2y)$

$\frac{z - 2}{z} = \frac{x^2 + y^2 - 2x}{x^2 + y^2} + i\frac{2y}{x^2 + y^2}$

The condition $\arg\left(\frac{z-2}{z}\right) = \frac{\pi}{4}$ requires:

$\text{Im} = \text{Re} > 0$

Setting Im $=$ Re:

$\frac{2y}{x^2 + y^2} = \frac{x^2 + y^2 - 2x}{x^2 + y^2}$

$2y = x^2 + y^2 - 2x$

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

This is exactly the circle from part (i). $\blacksquare$

The additional constraint Re $> 0$ gives $x^2+y^2-2x>0$. On the circle from part (i), this is equivalent to $2y>0$, so $y>0$. Thus (**) describes the upper arc of the circle from $z=0$ to $z=2$, with both endpoints excluded because the argument is undefined there.

Part (iii): Locus in the $w$-plane

Given $w = \frac{2}{z}$, so $z = \frac{2}{w} = \frac{2\bar{w}}{|w|^2}$. With $w = u + iv$:

$z = \frac{2(u - iv)}{u^2 + v^2} = \frac{2u}{u^2 + v^2} - i\frac{2v}{u^2 + v^2}$

Substituting into equation (*): $|z - (1+i)|^2 = 2$, i.e., $(x-1)^2 + (y-1)^2 = 2$ where $x = \frac{2u}{u^2+v^2}$, $y = \frac{-2v}{u^2+v^2}$:

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

Let $r^2 = u^2 + v^2$. Expanding:

$\frac{(2u - r^2)^2 + (2v + r^2)^2}{r^4} = 2$

$(2u - r^2)^2 + (2v + r^2)^2 = 2r^4$

Expanding the left side:

$4u^2 - 4ur^2 + r^4 + 4v^2 + 4vr^2 + r^4 = 2r^4$

$4(u^2 + v^2) - 4r^2(u - v) + 2r^4 = 2r^4$

$4r^2 - 4r^2(u - v) = 0$

Since $r^2 \neq 0$ (as $z \neq 0$):

$1 - (u - v) = 0 \implies u - v = 1$

The locus is the line $u - v = 1$, i.e., $\operatorname{Re}(w) - \operatorname{Im}(w) = 1$. $\blacksquare$

Sketch. This is a straight line in the $w$-plane passing through $(1, 0)$ and $(0, -1)$ with gradient $1$.

Which part corresponds to () :**

For the arc in part (ii), $z=x+iy$ lies on the circle with $y>0$. Since $w=2/z$, we have

$v=\operatorname{Im}(w)=-\frac{2y}{x^2+y^2}<0.$

Thus the corresponding part of the line is

$u-v=1,\qquad v<0,$

with endpoint $(1,0)$ excluded and the other endpoint approached at $(0,-1)$ as $z\to0$ along the arc.