STEP3 2002 -- Pure Mathematics

Exam: STEP3  |  Year: 2002  |  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	换元积分,分部积分,旋转体体积公式,广义积分
2	级数与数列 Sequences and Series	Hard	数学归纳法,反正切加法公式,裂项相消,级数求和
3	函数与方程 Functions and Equations	Challenging	根式函数分析,函数图像描绘,不等式分析,代数方程求解
4	数论 Number Theory	Hard	不等式放缩,反证法,因式分解,整数分析
5	代数 Algebra	Challenging	配方法,待定系数法,多项式恒等式,因式分解
6	微分方程 Differential Equations	Challenging	可分离变量法,分情况讨论,解曲线作图
7	三角学与坐标几何 Trigonometry and Coordinate Geometry	Challenging	三角恒等式,坐标法,代数恒等变换,不等式放缩
8	复数 Complex Numbers	Hard	复数辐角计算,模的运算,几何解释,Ptolemy定理

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Question 1

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

Problem

1 Find the area of the region between the curve $y = \frac{\ln x}{x}$ and the $x$-axis, for $1 \leqslant x \leqslant a$. What happens to this area as $a$ tends to infinity?

Find the volume of the solid obtained when the region between the curve $y = \frac{\ln x}{x}$ and the $x$-axis, for $1 \leqslant x \leqslant a$, is rotated through $2\pi$ radians about the $x$-axis. What happens to this volume as $a$ tends to infinity?

Model Solution

Part (i): Area

The area is

$A = \int_1^a \frac{\ln x}{x}\, dx = \left[\frac{(\ln x)^2}{2}\right]_1^a = \frac{(\ln a)^2}{2}.$

As $a \to \infty$, $\ln a \to \infty$, so $A \to \infty$. The area diverges.

Part (ii): Volume of revolution

The volume obtained by rotating about the $x$-axis is

$V = \pi \int_1^a \left(\frac{\ln x}{x}\right)^2 dx = \pi \int_1^a \frac{(\ln x)^2}{x^2}\, dx.$

We evaluate this by parts. Let $u = (\ln x)^2$, $dv = x^{-2}\, dx$, so $du = \frac{2\ln x}{x}\, dx$, $v = -\frac{1}{x}$. Then

$\int_1^a \frac{(\ln x)^2}{x^2}\, dx = \left[-\frac{(\ln x)^2}{x}\right]_1^a + \int_1^a \frac{2\ln x}{x^2}\, dx = -\frac{(\ln a)^2}{a} + 2\int_1^a \frac{\ln x}{x^2}\, dx.$

For the remaining integral, let $u = \ln x$, $dv = x^{-2}\, dx$, so $du = \frac{1}{x}\, dx$, $v = -\frac{1}{x}$:

$\int_1^a \frac{\ln x}{x^2}\, dx = \left[-\frac{\ln x}{x}\right]_1^a + \int_1^a \frac{1}{x^2}\, dx = -\frac{\ln a}{a} + \left[-\frac{1}{x}\right]_1^a = -\frac{\ln a}{a} + 1 - \frac{1}{a}.$

Combining:

$\int_1^a \frac{(\ln x)^2}{x^2}\, dx = -\frac{(\ln a)^2}{a} - \frac{2\ln a}{a} + 2 - \frac{2}{a} = 2 - \frac{(\ln a)^2 + 2\ln a + 2}{a}.$

Therefore

$V = \pi\left[2 - \frac{(\ln a)^2 + 2\ln a + 2}{a}\right].$

As $a \to \infty$, the fraction $\frac{(\ln a)^2 + 2\ln a + 2}{a} \to 0$ (since polynomial growth of $\ln a$ is slower than linear growth of $a$), so $V \to 2\pi$.

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Question 2

Topic: 级数与数列 Sequences and Series  |  Difficulty: Hard  |  Marks: 20

Problem

2 Prove that $\arctan a + \arctan b = \arctan \left( \frac{a + b}{1 - ab} \right)$ when $0 < a < 1$ and $0 < b < 1$.

Prove by induction that, for $n \geqslant 1$,

$\sum_{r=1}^{n} \arctan \left( \frac{1}{r^2 + r + 1} \right) = \arctan \left( \frac{n}{n + 2} \right)$

and hence find

$\sum_{r=1}^{\infty} \arctan \left( \frac{1}{r^2 + r + 1} \right) .$

Hence prove that

$\sum_{r=1}^{\infty} \arctan \left( \frac{1}{r^2 - r + 1} \right) = \frac{\pi}{2} .$

Model Solution

Part (i): The addition formula

Let $\alpha = \arctan a$ and $\beta = \arctan b$, so $\tan\alpha = a$ and $\tan\beta = b$. Since $0 < a < 1$ and $0 < b < 1$, we have $0 < \alpha < \pi/4$ and $0 < \beta < \pi/4$, so $0 < \alpha + \beta < \pi/2$.

Using the tangent addition formula:

$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{a + b}{1 - ab}.$

Since $0 < a, b < 1$, we have $ab < 1$, so $1 - ab > 0$ and $\frac{a+b}{1-ab} > 0$. Also $\alpha + \beta \in (0, \pi/2)$, so

$\alpha + \beta = \arctan\left(\frac{a + b}{1 - ab}\right). \qquad \blacksquare$

Part (ii): Induction proof

Base case ($n = 1$): $\arctan\!\left(\frac{1}{1+1+1}\right) = \arctan\!\left(\frac{1}{3}\right) = \arctan\!\left(\frac{1}{1+2}\right)$, as required.

Inductive step: Assume $\sum_{r=1}^{n} \arctan\!\left(\frac{1}{r^2+r+1}\right) = \arctan\!\left(\frac{n}{n+2}\right)$. Then

$\sum_{r=1}^{n+1} \arctan\!\left(\frac{1}{r^2+r+1}\right) = \arctan\!\left(\frac{n}{n+2}\right) + \arctan\!\left(\frac{1}{(n+1)^2+(n+1)+1}\right).$

We have $(n+1)^2 + (n+1) + 1 = n^2 + 3n + 3$, so we need to compute

$\arctan\!\left(\frac{n}{n+2}\right) + \arctan\!\left(\frac{1}{n^2+3n+3}\right).$

Since $0 < \frac{n}{n+2} < 1$ and $0 < \frac{1}{n^2+3n+3} < 1$, we apply the addition formula:

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

Numerator: $n^3 + 3n^2 + 3n + n + 2 = n^3 + 3n^2 + 4n + 2$.

Denominator: $n^3 + 3n^2 + 3n + 2n^2 + 6n + 6 - n = n^3 + 5n^2 + 8n + 6$.

We can verify that $\frac{n^3 + 3n^2 + 4n + 2}{n^3 + 5n^2 + 8n + 6} = \frac{n+1}{n+3}$ by cross-multiplying:

$(n+1)(n^3 + 5n^2 + 8n + 6) = n^4 + 6n^3 + 13n^2 + 14n + 6$.

$(n+3)(n^3 + 3n^2 + 4n + 2) = n^4 + 6n^3 + 13n^2 + 14n + 6$, as required.

Therefore $\sum_{r=1}^{n+1} \arctan\!\left(\frac{1}{r^2+r+1}\right) = \arctan\!\left(\frac{n+1}{n+3}\right)$, completing the induction. $\qquad \blacksquare$

Part (iii): The infinite sum

$\sum_{r=1}^{\infty} \arctan\!\left(\frac{1}{r^2+r+1}\right) = \lim_{n\to\infty} \arctan\!\left(\frac{n}{n+2}\right) = \arctan(1) = \frac{\pi}{4}.$

Part (iv): The second identity

Note that $\frac{1}{r^2 - r + 1} = \frac{1}{(r-1)^2 + (r-1) + 1}$. Substituting $s = r - 1$:

$\sum_{r=1}^{\infty} \arctan\!\left(\frac{1}{r^2-r+1}\right) = \sum_{s=0}^{\infty} \arctan\!\left(\frac{1}{s^2+s+1}\right) = \arctan\!\left(\frac{1}{0+0+1}\right) + \sum_{s=1}^{\infty} \arctan\!\left(\frac{1}{s^2+s+1}\right)$

$= \arctan(1) + \frac{\pi}{4} = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}. \qquad \blacksquare$

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Question 3

Topic: 函数与方程 Functions and Equations  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Let

$f(x) = a\sqrt{x} - \sqrt{x - b} ,$

where $x \geqslant b > 0$ and $a > 1$. Sketch the graph of $f(x)$. Hence show that the equation $f(x) = c$, where $c > 0$, has no solution when $c^2 < b \left( a^2 - 1 \right)$. Find conditions on $c^2$ in terms of $a$ and $b$ for the equation to have exactly one or exactly two solutions.

Solve the equations (i) $3\sqrt{x} - \sqrt{x - 2} = 4$ and (ii) $3\sqrt{x} - \sqrt{x - 3} = 5$.

Model Solution

Part (i): Sketch and analysis

We have $f(x) = a\sqrt{x} - \sqrt{x-b}$ for $x \geqslant b > 0$, $a > 1$.

At $x = b$: $f(b) = a\sqrt{b}$.

As $x \to \infty$: $f(x) = \sqrt{x}(a - \sqrt{1 - b/x}) \to \infty$ since $a > 1$.

Derivative:

$f'(x) = \frac{a}{2\sqrt{x}} - \frac{1}{2\sqrt{x-b}} = \frac{a\sqrt{x-b} - \sqrt{x}}{2\sqrt{x}\sqrt{x-b}}.$

Setting $f'(x) = 0$: $a\sqrt{x-b} = \sqrt{x}$, so $a^2(x-b) = x$, giving $x = \frac{a^2 b}{a^2 - 1}$.

Since $a > 1$, we have $\frac{a^2}{a^2-1} > 1$, so $x_c = \frac{a^2 b}{a^2-1} > b$. This critical point lies in the domain.

For $b \leqslant x < x_c$: $f'(x) < 0$ (decreasing). For $x > x_c$: $f'(x) > 0$ (increasing).

The minimum value is

$f(x_c) = a\sqrt{\frac{a^2 b}{a^2-1}} - \sqrt{\frac{a^2 b}{a^2-1} - b} = \frac{a^2\sqrt{b}}{\sqrt{a^2-1}} - \frac{\sqrt{b}}{\sqrt{a^2-1}} = \frac{(a^2-1)\sqrt{b}}{\sqrt{a^2-1}} = \sqrt{b(a^2-1)}.$

The graph starts at $(b, a\sqrt{b})$, decreases to a minimum at $x_c$ with value $\sqrt{b(a^2-1)}$, then increases to infinity.

Part (ii): Conditions on $c$

The equation $f(x) = c$ with $c > 0$:

• No solution when $c < \sqrt{b(a^2-1)}$, i.e. $c^2 < b(a^2-1)$.
• Exactly one solution when $c = \sqrt{b(a^2-1)}$ (at the minimum).
• Exactly two solutions when $\sqrt{b(a^2-1)} < c \leq a\sqrt{b}$, i.e. $b(a^2-1) < c^2 \leq a^2 b$. At the endpoint $c=a\sqrt b$, the two solutions are $x=b$ and a second point on the increasing branch.
• Exactly one solution when $c > a\sqrt{b}$, i.e. $c^2 > a^2 b$.

Part (iii): Solving (i) $3\sqrt{x} - \sqrt{x-2} = 4$

Here $a = 3$, $b = 2$, $c = 4$. We have $f(2) = 3\sqrt{2} \approx 4.24$ and $\sqrt{b(a^2-1)} = \sqrt{16} = 4$.

Since $c = 4 = \sqrt{b(a^2-1)}$, the equation has exactly one solution (at the minimum).

Rearranging: $3\sqrt{x} = 4 + \sqrt{x-2}$. Squaring:

$9x = 16 + 8\sqrt{x-2} + x - 2 \implies 8x - 14 = 8\sqrt{x-2} \implies 4x - 7 = 4\sqrt{x-2}.$

Squaring again: $16x^2 - 56x + 49 = 16(x-2) = 16x - 32$, so $16x^2 - 72x + 81 = 0$.

Discriminant: $72^2 - 4(16)(81) = 5184 - 5184 = 0$, so $x = \frac{72}{32} = \frac{9}{4}$.

Substitution confirms $3\sqrt{9/4} - \sqrt{9/4 - 2} = 3 \cdot \frac{3}{2} - \sqrt{\frac{1}{4}} = \frac{9}{2} - \frac{1}{2} = 4$.

The unique solution is $x = \dfrac{9}{4}$.

Part (iv): Solving (ii) $3\sqrt{x} - \sqrt{x-3} = 5$

Here $a = 3$, $b = 3$, $c = 5$. We have $f(3) = 3\sqrt{3} \approx 5.20$ and $\sqrt{b(a^2-1)} = \sqrt{24} = 2\sqrt{6} \approx 4.90$.

Since $2\sqrt{6} < 5 < 3\sqrt{3}$, the equation has exactly two solutions.

Rearranging: $3\sqrt{x} = 5 + \sqrt{x-3}$. Squaring:

$9x = 25 + 10\sqrt{x-3} + x - 3 \implies 8x - 22 = 10\sqrt{x-3} \implies 4x - 11 = 5\sqrt{x-3}.$

Note: we need $4x - 11 \geqslant 0$, i.e. $x \geqslant 11/4$.

Squaring again: $16x^2 - 88x + 121 = 25(x-3) = 25x - 75$, so $16x^2 - 113x + 196 = 0$.

Discriminant: $113^2 - 4(16)(196) = 12769 - 12544 = 225 = 15^2$.

$x = \frac{113 \pm 15}{32} \implies x = 4 \quad \text{or} \quad x = \frac{49}{16}.$

Substitution gives $3\sqrt{4} - \sqrt{1} = 6 - 1 = 5$.

Similarly, $3\sqrt{49/16} - \sqrt{49/16 - 3} = \frac{21}{4} - \sqrt{\frac{1}{16}} = \frac{21}{4} - \frac{1}{4} = 5$.

The solutions are $x = \dfrac{49}{16}$ and $x = 4$.

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Question 4

Topic: 数论 Number Theory  |  Difficulty: Hard  |  Marks: 20

Problem

4 Show that if $x$ and $y$ are positive and $x^3 + x^2 = y^3 - y^2$ then $x < y$.

Show further that if $0 < x \leqslant y - 1$, then $x^3 + x^2 < y^3 - y^2$.

Prove that there does not exist a pair of positive integers such that the difference of their cubes is equal to the sum of their squares.

Find all the pairs of integers such that the difference of their cubes is equal to the sum of their squares.

Model Solution

Part (i): Show that $x < y$

Suppose for contradiction that $x \geqslant y$. If $x = y$, then $x^3 + x^2 = x^3 - x^2$, giving $2x^2 = 0$, so $x = 0$, contradicting $x > 0$.

If $x > y > 0$, then $x^3 > y^3$ and $x^2 > y^2$, so $x^3 + x^2 > y^3 + y^2 > y^3 - y^2$ (since $y^2 > 0$). This contradicts $x^3 + x^2 = y^3 - y^2$.

Therefore $x < y$. $\qquad \blacksquare$

Part (ii): Show that $0 < x \leqslant y-1$ implies $x^3 + x^2 < y^3 - y^2$

Since $x \leqslant y - 1$, we have $x + 1 \leqslant y$, so

$x^3 + x^2 = x^2(x+1) \leqslant x^2 y.$

Also $y^3 - y^2 = y^2(y-1) \geqslant y^2 x$ (since $y - 1 \geqslant x$).

So $x^3 + x^2 \leqslant x^2 y$ and $y^3 - y^2 \geqslant xy^2$. We need $x^2 y < xy^2$, i.e. $xy(y - x) > 0$, which holds since $x, y > 0$ and $y > x$.

But we need strict inequality throughout. Since $x \leqslant y-1$ and $x > 0$, we have $y > 1$, so $y - 1 > 0$. Also $x < y$ (from Part (i) or directly from $x \leqslant y-1$).

More precisely: $x^2(x+1) \leqslant x^2 y$ with equality iff $x + 1 = y$. And $y^2(y-1) \geqslant xy^2$ with equality iff $x = y - 1$. Even if $x = y - 1$, we get $x^2 y$ vs $xy^2$, and $x^2 y < xy^2$ since $x < y$. Therefore $x^3 + x^2 < y^3 - y^2$. $\qquad \blacksquare$

Part (iii): No pair of positive integers

Suppose $m, n$ are positive integers with $m^3 - n^3 = m^2 + n^2$. Since $m^3 - n^3 = (m-n)(m^2 + mn + n^2) > 0$ and $m^2 + n^2 > 0$, we need $m > n$, so $m \geqslant n + 1$.

Setting $x = n$ and $y = m$ in Part (ii): since $0 < n \leqslant m - 1$, we have $n^3 + n^2 < m^3 - m^2$.

But we need $m^3 - n^3 = m^2 + n^2$, i.e. $n^3 + n^2 = m^3 - m^2$. This contradicts $n^3 + n^2 < m^3 - m^2$.

Therefore no such pair exists. $\qquad \blacksquare$

Part (iv): Find all integer pairs

The equation is $y^3 - x^3 = y^2 + x^2$, i.e. $y^3 - y^2 = x^3 + x^2$, which gives

$(y - x)(y^2 + xy + x^2) = x^2 + y^2. \qquad (\star)$

Case 1: $x = 0$. Then $y^3 = y^2$, so $y = 0$ or $y = 1$. Solutions: $(0, 0)$ and $(0, 1)$.

Case 2: $y = 0$. Then $-x^3 = x^2$, so $x = 0$ or $x = -1$. Solution: $(-1, 0)$.

Case 3: $x > 0$. If $y<0$, then $y^3-x^3<0$ while $x^2+y^2>0$, which is impossible. If $y>0$, Part (iii) shows that no positive integer pair exists.

Case 4: $x < 0$, $y > 0$. Let $x = -p$ with $p \geqslant 1$. Then $(\star)$ becomes

$(y + p)(y^2 - py + p^2) = p^2 + y^2.$

The left side equals $y^3+p^3$. For $p\geq2$ and $y\geq2$, we have $p^3>p^2$ and $y^3>y^2$, so $p^3+y^3>p^2+y^2$. Thus only the cases with $p=1$ or $y=1$ can remain.

$p = 1, y = 1$: $(1+1)(1-1+1) = 2$ and $1 + 1 = 2$. This gives the solution $(-1,1)$.

$p = 1, y \geqslant 2$: $(y+1)(y^2 - y + 1) = y^3 + 1 > y^2 + 1$. No solution.

$p = 2, y = 1$: $(3)(4-2+1) = 9$ and $4 + 1 = 5$. No solution.

$p \geqslant 2, y = 1$: $(1+p)(1-p+p^2) = 1 + p^3 \geqslant 9 > p^2 + 1$. No solution.

Case 5: $x < 0$, $y < 0$. Let $x = -p, y = -q$ with $p, q \geqslant 1$. Then

$(-q+p)(q^2 - pq + p^2) = p^2 + q^2, \quad \text{i.e.} \quad (p-q)(p^2 - pq + q^2) = p^2 + q^2.$

This is $p^3 - q^3 = p^2 + q^2$. If $p = q$: $0 = 2p^2$, impossible. If $p > q$: by Part (iii) applied to positive integers $q, p$ with $p^3 - q^3 = p^2 + q^2$, no solution exists (same argument). If $p < q$: left side is negative, right side is positive. No solution.

Case 6: $x < 0$, $y = 0$. Already covered in Case 2.

Therefore the complete set of integer solutions is

$(x, y) \in \{(0, 0),\ (0, 1),\ (-1, 0),\ (-1, 1)\}.$

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Question 5

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

Problem

5 Give a condition that must be satisfied by $p$, $q$ and $r$ for it to be possible to write the quadratic polynomial $px^2 + qx + r$ in the form $p(x + h)^2$, for some $h$.

Obtain an equation, which you need not simplify, that must be satisfied by $t$ if it is possible to write $\left(x^2 + \frac{1}{2}bx + t\right)^2 - \left(x^4 + bx^3 + cx^2 + dx + e\right)$ in the form $k(x + h)^2$, for some $k$ and $h$.

Hence, or otherwise, write $x^4 + 6x^3 + 9x^2 - 2x - 7$ as a product of two quadratic factors.

Model Solution

Part (i): Condition for $px^2 + qx + r = p(x+h)^2$

Expanding: $p(x+h)^2 = px^2 + 2phx + ph^2$. Comparing coefficients with $px^2 + qx + r$:

$q = 2ph, \qquad r = ph^2.$

From the first equation $h = q/(2p)$, so $r = p \cdot q^2/(4p^2) = q^2/(4p)$.

The necessary and sufficient condition is $q^2 = 4pr$, i.e. the discriminant is zero. $\qquad \blacksquare$

Part (ii): Equation for $t$

We want

$\left(x^2 + \tfrac{1}{2}bx + t\right)^2 - \left(x^4 + bx^3 + cx^2 + dx + e\right) = k(x+h)^2.$

The left side must be a perfect square (as a quadratic in $x$). Expanding the square:

$\left(x^2 + \tfrac{1}{2}bx + t\right)^2 = x^4 + bx^3 + \left(\tfrac{1}{4}b^2 + 2t\right)x^2 + btx + t^2.$

Subtracting:

$\left(\tfrac{1}{4}b^2 + 2t - c\right)x^2 + (bt - d)x + (t^2 - e) = k(x+h)^2.$

From Part (i), the condition for this quadratic $Ax^2 + Bx + C$ to be a perfect square is $B^2 = 4AC$:

$(bt - d)^2 = 4\left(\tfrac{1}{4}b^2 + 2t - c\right)(t^2 - e). \qquad \blacksquare$

Part (iii): Factorizing $x^4 + 6x^3 + 9x^2 - 2x - 7$

Here $b = 6$, $c = 9$, $d = -2$, $e = -7$. The condition becomes

$(6t + 2)^2 = 4(9 + 2t - 9)(t^2 + 7) = 4 \cdot 2t \cdot (t^2 + 7).$

Simplifying: $(3t+1)^2 = 2t(t^2 + 7)$, i.e. $9t^2 + 6t + 1 = 2t^3 + 14t$.

$2t^3 - 9t^2 + 8t - 1 = 0.$

Testing $t = 1$ gives $2 - 9 + 8 - 1 = 0$, so we may take $t = 1$. Then

$\left(x^2 + 3x + 1\right)^2 - \left(x^4 + 6x^3 + 9x^2 - 2x - 7\right) = 2x^2 + 8x + 8 = 2(x+2)^2.$

Therefore

$x^4 + 6x^3 + 9x^2 - 2x - 7 = \left(x^2 + 3x + 1\right)^2 - \left(\sqrt{2}(x+2)\right)^2.$

Using the difference of two squares:

$x^4 + 6x^3 + 9x^2 - 2x - 7 = \left(x^2 + (3-\sqrt{2})x + 1 - 2\sqrt{2}\right)\left(x^2 + (3+\sqrt{2})x + 1 + 2\sqrt{2}\right).$

Verification: The $x$ coefficient is $(3-\sqrt{2})(1+2\sqrt{2}) + (3+\sqrt{2})(1-2\sqrt{2})$.

$(3-\sqrt{2})(1+2\sqrt{2}) = 3 + 6\sqrt{2} - \sqrt{2} - 4 = -1 + 5\sqrt{2}$.

$(3+\sqrt{2})(1-2\sqrt{2}) = 3 - 6\sqrt{2} + \sqrt{2} - 4 = -1 - 5\sqrt{2}$.

Sum $= -2$, as required.

The constant term is $(1-2\sqrt{2})(1+2\sqrt{2}) = 1 - 8 = -7$.

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Question 6

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

Problem

6 Find all the solution curves of the differential equation $y^4 \left(\frac{dy}{dx}\right)^4 = \left(y^2 - 1\right)^2$ that pass through either of the points (i) $\left(0, \frac{\sqrt{3}}{2}\right)$, (ii) $\left(0, \frac{\sqrt{5}}{2}\right)$.

Show also that $y = 1$ and $y = -1$ are solutions of the differential equation. Sketch all these solution curves on a single set of axes.

Model Solution

Part (i): Solving the differential equation

The equation $y^4\left(\frac{dy}{dx}\right)^4 = (y^2-1)^2$ can be written as $\left(y\frac{dy}{dx}\right)^4 = (y^2-1)^2$.

Taking square roots (both sides are non-negative): $\left(y\frac{dy}{dx}\right)^2 = |y^2 - 1|$.

So $y\frac{dy}{dx} = \pm\sqrt{|y^2-1|}$, which gives four ODEs depending on the signs.

Case A: $|y| > 1$ (so $y^2 - 1 > 0$)

$y\frac{dy}{dx} = \pm\sqrt{y^2-1}$, i.e. $\frac{y}{\sqrt{y^2-1}}\, dy = \pm\, dx$.

Integrating: $\sqrt{y^2-1} = \pm x + C$, so

$y^2 = (x - C)^2 + 1 \qquad \text{(choosing appropriate sign)}. \qquad (\star)$

Case B: $|y| < 1$ (so $1 - y^2 > 0$)

$y\frac{dy}{dx} = \pm\sqrt{1-y^2}$, i.e. $\frac{y}{\sqrt{1-y^2}}\, dy = \pm\, dx$.

Integrating: $-\sqrt{1-y^2} = \pm x + C$, so $\sqrt{1-y^2} = \mp x + C'$, giving

$y^2 = 1 - (x - C)^2. \qquad (\star\star)$

Solution curves through $(0, \sqrt{3}/2)$

Since $y = \sqrt{3}/2 < 1$, we use Case B. From $(\star\star)$: $\frac{3}{4} = 1 - C^2$, so $C = \pm\frac{1}{2}$.

This gives two circles:

$\left(x - \tfrac{1}{2}\right)^2 + y^2 = 1 \qquad \text{and} \qquad \left(x + \tfrac{1}{2}\right)^2 + y^2 = 1.$

Both are valid solution curves (the upper semicircles satisfy the ODE with $y > 0$).

Solution curves through $(0, \sqrt{5}/2)$

Since $y = \sqrt{5}/2 > 1$, we use Case A. From $(\star)$: $\frac{5}{4} = C^2 + 1$, so $C = \pm\frac{1}{2}$.

This gives two hyperbolas:

$y^2 - \left(x - \tfrac{1}{2}\right)^2 = 1 \qquad \text{and} \qquad y^2 - \left(x + \tfrac{1}{2}\right)^2 = 1.$

Part (ii): Constant solutions

If $y = 1$, then the left side is $1 \cdot 0 = 0$ and the right side is $0$.

If $y = -1$, then the left side is $1 \cdot 0 = 0$ and the right side is $0$.

Part (iii): Sketch

The solution curves are:

• Two circles: $\left(x \pm \frac{1}{2}\right)^2 + y^2 = 1$ (valid for $|y| < 1$)
• Two hyperbolas: $y^2 - \left(x \pm \frac{1}{2}\right)^2 = 1$ (valid for $|y| > 1$)
• Two horizontal lines: $y = 1$ and $y = -1$

The circles are centred at $(\pm 1/2, 0)$ with radius 1, fitting within the strip $|y| < 1$. The hyperbolas have vertices at $(\pm 1/2, \pm 1)$ and open vertically, with slant asymptotes $y = \pm(x-\frac12)$ and $y = \pm(x+\frac12)$ as appropriate. The horizontal lines $y = \pm1$ are constant solution curves and boundary levels touched at the vertices, not asymptotes.

The curve through $(0, \sqrt{3}/2)$ consists of the upper arcs of both circles. The curve through $(0, \sqrt{5}/2)$ consists of the upper branches of both hyperbolas.

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Question 7

Topic: 三角学与坐标几何 Trigonometry and Coordinate Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Given that $\alpha$ and $\beta$ are acute angles, show that $\alpha + \beta = \pi/2$ if and only if $\cos^2 \alpha + \cos^2 \beta = 1$.

In the $x$–$y$ plane, the point $A$ has coordinates $(0, s)$ and the point $C$ has coordinates $(s, 0)$, where $s > 0$. The point $B$ lies in the first quadrant ($x > 0, y > 0$). The lengths of $AB$, $OB$ and $CB$ are respectively $a, b$ and $c$.

Show that

$(s^2 + b^2 - a^2)^2 + (s^2 + b^2 - c^2)^2 = 4s^2b^2$

and hence that

$(2s^2 - a^2 - c^2)^2 + (2b^2 - a^2 - c^2)^2 = 4a^2c^2 .$

Deduce that

$(a - c)^2 \leqslant 2b^2 \leqslant (a + c)^2 .$

Model Solution

Part (i): $\alpha + \beta = \pi/2$ if and only if $\cos^2\alpha + \cos^2\beta = 1$

Forward direction. Suppose $\alpha + \beta = \pi/2$. Then $\beta = \pi/2 - \alpha$, so

$\cos\beta = \cos\!\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha.$

Since $\alpha$ is acute, both $\sin\alpha$ and $\cos\alpha$ are non-negative, so $\cos^2\beta = \sin^2\alpha$. Therefore

$\cos^2\alpha + \cos^2\beta = \cos^2\alpha + \sin^2\alpha = 1.$

Reverse direction. Suppose $\cos^2\alpha + \cos^2\beta = 1$. Using $\cos^2\alpha = 1 - \sin^2\alpha$, we get $\sin^2\alpha + \cos^2\beta = 1$, so $\cos^2\beta = \sin^2\alpha$.

Since $\alpha$ and $\beta$ are acute, $\cos\beta > 0$ and $\sin\alpha > 0$, so $\cos\beta = \sin\alpha = \cos(\pi/2 - \alpha)$. Since $\cos$ is strictly decreasing on $[0, \pi]$ and both $\beta$ and $\pi/2 - \alpha$ lie in $(0, \pi/2)$, we conclude $\beta = \pi/2 - \alpha$, i.e. $\alpha + \beta = \pi/2$. $\qquad \blacksquare$

Part (ii): The first identity

Let $B = (x, y)$ with $x > 0, y > 0$. Let $\theta = \angle BOx$, so $x = b\cos\theta$ and $y = b\sin\theta$.

Using $A = (0, s)$, $C = (s, 0)$, $O = (0, 0)$:

$a^2 = |AB|^2 = x^2 + (y - s)^2 = x^2 + y^2 - 2sy + s^2 = b^2 + s^2 - 2sb\sin\theta$

$c^2 = |CB|^2 = (x - s)^2 + y^2 = x^2 - 2sx + s^2 + y^2 = b^2 + s^2 - 2sb\cos\theta$

Therefore

$s^2 + b^2 - a^2 = 2sb\sin\theta, \qquad s^2 + b^2 - c^2 = 2sb\cos\theta.$

Squaring and adding:

$(s^2 + b^2 - a^2)^2 + (s^2 + b^2 - c^2)^2 = 4s^2b^2\sin^2\theta + 4s^2b^2\cos^2\theta = 4s^2b^2. \qquad \blacksquare$

Part (iii): The second identity

From Part (ii), $a^2 = s^2 + b^2 - 2sb\sin\theta$ and $c^2 = s^2 + b^2 - 2sb\cos\theta$, so

$a^2 + c^2 = 2(s^2 + b^2) - 2sb(\sin\theta + \cos\theta).$

Hence

$2s^2 - a^2 - c^2 = 2s^2 - 2(s^2 + b^2) + 2sb(\sin\theta + \cos\theta) = 2sb(\sin\theta + \cos\theta) - 2b^2$

$2b^2 - a^2 - c^2 = 2b^2 - 2(s^2 + b^2) + 2sb(\sin\theta + \cos\theta) = 2sb(\sin\theta + \cos\theta) - 2s^2.$

Let $u = a^2 + c^2$, $m = s^2 + b^2$ and $w = \sin\theta + \cos\theta$. Then

$(2s^2 - u)^2 + (2b^2 - u)^2 = (2s^2 - a^2 - c^2)^2 + (2b^2 - a^2 - c^2)^2$

$= [2s^2 - 2m + 2sbw]^2 + [2b^2 - 2m + 2sbw]^2$

$= [-2b^2 + 2sbw]^2 + [-2s^2 + 2sbw]^2$

$= 4b^2(sw - b)^2 + 4s^2(bw - s)^2$

$= 4[b^2(s^2w^2 - 2bsw + b^2) + s^2(b^2w^2 - 2bsw + s^2)]$

$= 4[s^2b^2w^2 - 2b^3sw + b^4 + s^2b^2w^2 - 2s^3bw + s^4]$

$= 4[2s^2b^2w^2 - 2sbw(b^2 + s^2) + s^4 + b^4]$

$= 4[2s^2b^2w^2 - 2sbwm + m^2 - 2s^2b^2]$

Now $4a^2c^2 = 4[m^2 - 2sbmw + 4s^2b^2\sin\theta\cos\theta]$.

But $w^2 - 1 = (\sin\theta + \cos\theta)^2 - 1 = 2\sin\theta\cos\theta$, so

$2s^2b^2w^2 - 2s^2b^2 = 4s^2b^2\sin\theta\cos\theta.$

Thus

$(2s^2 - a^2 - c^2)^2 + (2b^2 - a^2 - c^2)^2 = 4a^2c^2. \qquad \blacksquare$

Part (iv): The deduction

From Part (iii), rearranging: $(2b^2 - a^2 - c^2)^2 = 4a^2c^2 - (2s^2 - a^2 - c^2)^2$.

Since $(2s^2 - a^2 - c^2)^2 \geqslant 0$, we have

$(2b^2 - a^2 - c^2)^2 \leqslant 4a^2c^2,$

so $|2b^2 - a^2 - c^2| \leqslant 2ac$. This gives

$-2ac \leqslant 2b^2 - a^2 - c^2 \leqslant 2ac.$

Adding $a^2 + c^2$ throughout: $a^2 + c^2 - 2ac \leqslant 2b^2 \leqslant a^2 + c^2 + 2ac$, i.e.

$(a - c)^2 \leqslant 2b^2 \leqslant (a + c)^2. \qquad \blacksquare$

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Question 8

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

Problem

8 Four complex numbers $u_1, u_2, u_3$ and $u_4$ have unit modulus, and arguments $\theta_1, \theta_2, \theta_3$ and $\theta_4$, respectively, with $-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi$.

Show that

$\arg (u_1 - u_2) = \frac{1}{2} (\theta_1 + \theta_2 - \pi) + 2n\pi$

where $n = 0$ or $1$. Deduce that

$\arg ((u_1 - u_2) (u_4 - u_3)) = \arg ((u_1 - u_4) (u_3 - u_2)) + 2n\pi$

for some integer $n$.

Prove that

$| (u_1 - u_2) (u_4 - u_3) | + | (u_1 - u_4) (u_3 - u_2) | = | (u_1 - u_3) (u_4 - u_2) | .$

Model Solution

Part (i): Argument of $u_1 - u_2$

Write $u_k = e^{i\theta_k}$. Then

$u_1 - u_2 = e^{i\theta_1} - e^{i\theta_2} = e^{i(\theta_1+\theta_2)/2}\left(e^{i(\theta_1-\theta_2)/2} - e^{-i(\theta_1-\theta_2)/2}\right) = 2i\sin\!\left(\frac{\theta_1-\theta_2}{2}\right)e^{i(\theta_1+\theta_2)/2}.$

Let $\alpha = \frac{\theta_1-\theta_2}{2}$. Since $\theta_1 < \theta_2$, we have $\alpha < 0$ and therefore $\sin\alpha < 0$. Hence $2i\sin\alpha$ has argument $-\pi/2$ modulo $2\pi$. Thus

$\arg(u_1 - u_2) = \frac{\theta_1+\theta_2}{2} + \frac{3\pi}{2} = \frac{\theta_1+\theta_2}{2} - \frac{\pi}{2} \pmod{2\pi}.$

Since $\frac{\theta_1+\theta_2-\pi}{2}$ is equivalent to $\frac{\theta_1+\theta_2}{2} - \frac{\pi}{2}$, we have

$\arg(u_1 - u_2) = \frac{1}{2}(\theta_1 + \theta_2 - \pi) + 2n\pi, \qquad n = 0 \text{ or } 1. \qquad \blacksquare$

Part (ii): Argument of products

Using the result from Part (i):

$\arg(u_1 - u_2) = \frac{\theta_1+\theta_2-\pi}{2}, \qquad \arg(u_4 - u_3) = \frac{\theta_3+\theta_4-\pi}{2},$

$\arg(u_1 - u_4) = \frac{\theta_1+\theta_4-\pi}{2}, \qquad \arg(u_3 - u_2) = \frac{\theta_2+\theta_3-\pi}{2}.$

Therefore

$\arg((u_1-u_2)(u_4-u_3)) = \frac{\theta_1+\theta_2+\theta_3+\theta_4 - 2\pi}{2} \pmod{2\pi},$

$\arg((u_1-u_4)(u_3-u_2)) = \frac{\theta_1+\theta_2+\theta_3+\theta_4 - 2\pi}{2} \pmod{2\pi}.$

These are equal modulo $2\pi$, so

$\arg((u_1-u_2)(u_4-u_3)) = \arg((u_1-u_4)(u_3-u_2)) + 2n\pi. \qquad \blacksquare$

Part (iii): The magnitude identity

Since all $u_k$ lie on the unit circle, $|u_i - u_j| = 2\left|\sin\frac{\theta_i-\theta_j}{2}\right|$. With the ordering $-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi$, all differences $\theta_j - \theta_i$ for $i < j$ are positive and less than $2\pi$, so $|u_i - u_j| = 2\sin\frac{\theta_j-\theta_i}{2}$.

Let $\alpha = \theta_2 - \theta_1 > 0$, $\beta = \theta_3 - \theta_2 > 0$, $\gamma = \theta_4 - \theta_3 > 0$. Then

$|(u_1-u_2)(u_4-u_3)| = 4\sin\frac{\alpha}{2}\sin\frac{\gamma}{2},$

$|(u_1-u_4)(u_3-u_2)| = 4\sin\frac{\alpha+\beta+\gamma}{2}\sin\frac{\beta}{2},$

$|(u_1-u_3)(u_4-u_2)| = 4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}.$

We need to show:

$\sin\frac{\alpha}{2}\sin\frac{\gamma}{2} + \sin\frac{\alpha+\beta+\gamma}{2}\sin\frac{\beta}{2} = \sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}. \qquad (\dagger)$

Let $A = \alpha/2$, $B = \beta/2$, $C = \gamma/2$. Then $(\dagger)$ becomes

$\sin A\sin C + \sin(A+B+C)\sin B = \sin(A+B)\sin(B+C).$

Using the product-to-sum formula $\sin X \sin Y = \frac{1}{2}[\cos(X-Y) - \cos(X+Y)]$:

RHS: $\sin(A+B)\sin(B+C) = \frac{1}{2}[\cos(A-C) - \cos(A+2B+C)]$.

LHS: $\sin A\sin C = \frac{1}{2}[\cos(A-C) - \cos(A+C)]$ and

$\sin(A+B+C)\sin B = \frac{1}{2}[\cos(A+C) - \cos(A+2B+C)]$.

Adding: $\text{LHS} = \frac{1}{2}[\cos(A-C) - \cos(A+2B+C)] = \text{RHS}$. $\qquad \blacksquare$

Geometric interpretation: The four points $u_1, u_2, u_3, u_4$ lie in order on the unit circle, forming a cyclic quadrilateral. The identity $|(u_1-u_2)(u_4-u_3)| + |(u_1-u_4)(u_3-u_2)| = |(u_1-u_3)(u_4-u_2)|$ is precisely Ptolemy’s theorem: for a cyclic quadrilateral, the sum of the products of opposite sides equals the product of the diagonals.