STEP2 1998 -- Pure Mathematics

Exam: STEP2  |  Year: 1998  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Model solutions are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	代数与方程 (Algebra & Equations)	Standard	展开立方,奇偶性分析,整除性论证
2	级数展开 (Series Expansion)	Standard	二项式定理,分数指数展开,数值计算
3	级数求和 (Series & Summation)	Challenging	部分分式,裂项消去,递推数列
4	积分 (Integration)	Challenging	递推积分,三角恒等式,积差消去
5	复数 (Complex Numbers)	Challenging	复数模,三角不等式,数学归纳法,反证法
6	解析几何与微积分 (Coordinate Geometry & Calculus)	Challenging	参数方程求导,切线法线方程,三角恒等式
7	不等式与微积分 (Inequalities & Calculus)	Challenging	导数判号,不等式证明,对数求导法,三角函数性质
8	向量几何 (Vector Geometry)	Challenging	向量参数方程,中点公式,共点证明,重心坐标

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Question 1

Topic: 代数与方程 (Algebra & Equations)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Show that, if $n$ is an integer such that

$(n - 3)^3 + n^3 = (n + 3)^3, \qquad \text{(*)}$

then $n$ is even and $n^2$ is a factor of 54. Deduce that there is no integer $n$ which satisfies the equation $(*)$.

Show that, if $n$ is an integer such that

$(n - 6)^3 + n^3 = (n + 6)^3, \qquad \text{(**)}$

then $n$ is even. Deduce that there is no integer $n$ which satisfies the equation $(**)$.

Model Solution

Part (i): Show that if $(n-3)^3 + n^3 = (n+3)^3$ then $n$ is even and $n^2 \mid 54$, and deduce no integer solution exists.

Expand each cube using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(n-3)^3 = n^3 - 9n^2 + 27n - 27$

$(n+3)^3 = n^3 + 9n^2 + 27n + 27$

Substituting into the equation:

$(n^3 - 9n^2 + 27n - 27) + n^3 = n^3 + 9n^2 + 27n + 27$

$2n^3 - 9n^2 + 27n - 27 = n^3 + 9n^2 + 27n + 27$

Subtracting the right-hand side:

$n^3 - 18n^2 - 54 = 0$

Rearranging:

$n^2(n - 18) = 54 \qquad \text{($\star$)}$

$n$ is even. Suppose for contradiction that $n$ is odd. Then $n^2$ is odd and $n - 18$ is odd (odd minus even), so the product $n^2(n-18)$ is odd. But the right-hand side $54$ is even, a contradiction. Hence $n$ is even.

$n^2$ is a factor of 54. From ($\star$), since $n-18$ is an integer, $n^2$ divides $54$.

No integer solution exists. We have $54 = 2 \times 3^3$. The positive divisors of $54$ that are perfect squares are $1$ and $9$ only.

• $n^2 = 1 \implies n = \pm 1$: both are odd, contradicting $n$ even.
• $n^2 = 9 \implies n = \pm 3$: both are odd, contradicting $n$ even.

Since no even integer has a square dividing $54$, there is no integer $n$ satisfying ($\star$). $\quad\blacksquare$

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Part (ii): Show that if $(n-6)^3 + n^3 = (n+6)^3$ then $n$ is even, and deduce no integer solution exists.

Expand:

$(n-6)^3 = n^3 - 18n^2 + 108n - 216$

$(n+6)^3 = n^3 + 18n^2 + 108n + 216$

Substituting:

$(n^3 - 18n^2 + 108n - 216) + n^3 = n^3 + 18n^2 + 108n + 216$

$2n^3 - 18n^2 + 108n - 216 = n^3 + 18n^2 + 108n + 216$

$n^3 - 36n^2 - 432 = 0$

$n^2(n - 36) = 432 \qquad \text{($\star\star$)}$

$n$ is even. Suppose $n$ is odd. Then $n^2$ is odd and $n - 36$ is odd, so $n^2(n-36)$ is odd. But $432$ is even, a contradiction. Hence $n$ is even.

No integer solution exists. Since $n$ is even, write $n = 2m$. Substituting into ($\star\star$):

$4m^2(2m - 36) = 432$

$m^2(m - 18) = 54$

Now $54 = 2 \times 3^3$, and the only perfect squares dividing $54$ are $1$ and $9$.

• $m^2 = 1$: $m = \pm 1$, giving $m^2(m-18) = 1 \times (-17) = -17$ or $1 \times (-19) = -19$. Neither equals $54$.
• $m^2 = 9$: $m = \pm 3$, giving $m^2(m-18) = 9 \times (-15) = -135$ or $9 \times (-21) = -189$. Neither equals $54$.

No integer $m$ satisfies $m^2(m-18) = 54$, so no integer $n$ satisfies ($\star\star$). $\quad\blacksquare$

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Question 2

Topic: 级数展开 (Series Expansion)  |  Difficulty: Standard  |  Marks: 20

Problem

2 Use the first four terms of the binomial expansion of $(1 - 1/50)^{1/2}$, writing $1/50 = 2/100$ to simplify the calculation, to derive the approximation $\sqrt{2} \approx 1.414214$.

Calculate similarly an approximation to the cube root of 2 to six decimal places by considering $(1 + N/125)^a$, where $a$ and $N$ are suitable numbers.

[You need not justify the accuracy of your approximations.]

Model Solution

Part (i): Approximating $\sqrt{2}$

Write $1/50 = 2/100$ and expand $(1 - 2/100)^{1/2}$ using the binomial theorem with $\alpha = 1/2$, $x = -2/100$:

$(1+x)^{1/2} = 1 + \frac{1}{2}x + \frac{\tfrac{1}{2} \cdot (-\tfrac{1}{2})}{2!}x^2 + \frac{\tfrac{1}{2} \cdot (-\tfrac{1}{2}) \cdot (-\tfrac{3}{2})}{3!}x^3 + \cdots$

Computing each coefficient:

$\frac{\tfrac{1}{2} \cdot (-\tfrac{1}{2})}{2!} = -\frac{1}{8}, \qquad \frac{\tfrac{1}{2} \cdot (-\tfrac{1}{2}) \cdot (-\tfrac{3}{2})}{3!} = \frac{1}{16}.$

With $x = -2/100$:

Term 1: $1$

Term 2: $\dfrac{1}{2} \times \left(-\dfrac{2}{100}\right) = -\dfrac{1}{100} = -0.01$

Term 3: $-\dfrac{1}{8} \times \left(-\dfrac{2}{100}\right)^2 = -\dfrac{1}{8} \times \dfrac{4}{10000} = -\dfrac{1}{20000} = -0.00005$

Term 4: $\dfrac{1}{16} \times \left(-\dfrac{2}{100}\right)^3 = \dfrac{1}{16} \times \left(-\dfrac{8}{10^6}\right) = -\dfrac{1}{2000000} = -0.0000005$

Sum:

$1 - \frac{1}{100} - \frac{1}{20000} - \frac{1}{2000000} = \frac{2000000 - 20000 - 100 - 1}{2000000} = \frac{1979899}{2000000} = 0.9899495$

Now $(1 - 1/50)^{1/2} = (49/50)^{1/2} = 7/\sqrt{50} = 7\sqrt{2}/10$, so:

$\frac{7\sqrt{2}}{10} \approx 0.9899495 \implies \sqrt{2} \approx \frac{10}{7} \times 0.9899495 = \frac{1979899}{1400000} \approx 1.414213571...$

Rounding to six decimal places: $\sqrt{2} \approx 1.414214$. $\quad\blacksquare$

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Part (ii): Approximating $2^{1/3}$

We seek $a$ and $N$ such that $(1 + N/125)^{a}$ relates to $2^{1/3}$. Since $5^3 = 125$ and $4^3 = 64$, observe that

$\left(\frac{5}{4}\right)^3 \times \frac{128}{125} = \frac{125}{64} \times \frac{128}{125} = 2.$

Taking cube roots:

$2^{1/3} = \frac{5}{4}\left(\frac{128}{125}\right)^{1/3} = \frac{5}{4}\left(1 + \frac{3}{125}\right)^{1/3}.$

So we take $a = 1/3$ and $N = 3$. Expanding $(1 + x)^{1/3}$ with $x = 3/125$:

$(1+x)^{1/3} = 1 + \frac{1}{3}x + \frac{\tfrac{1}{3}\cdot(-\tfrac{2}{3})}{2!}x^2 + \frac{\tfrac{1}{3}\cdot(-\tfrac{2}{3})\cdot(-\tfrac{5}{3})}{3!}x^3 + \cdots$

Computing the binomial coefficients:

$\frac{\tfrac{1}{3} \cdot (-\tfrac{2}{3})}{2} = -\frac{1}{9}, \qquad \frac{\tfrac{1}{3} \cdot (-\tfrac{2}{3}) \cdot (-\tfrac{5}{3})}{6} = \frac{5}{81}.$

With $x = 3/125$:

Term 1: $1$

Term 2: $\dfrac{1}{3} \times \dfrac{3}{125} = \dfrac{1}{125}$

Term 3: $-\dfrac{1}{9} \times \left(\dfrac{3}{125}\right)^2 = -\dfrac{1}{9} \times \dfrac{9}{15625} = -\dfrac{1}{15625}$

Term 4: $\dfrac{5}{81} \times \left(\dfrac{3}{125}\right)^3 = \dfrac{5}{81} \times \dfrac{27}{1953125} = \dfrac{135}{158203125} = \dfrac{1}{1171875}$

Sum with a common denominator of $1171875$:

$1 + \frac{1}{125} - \frac{1}{15625} + \frac{1}{1171875} = \frac{1171875 + 9375 - 75 + 1}{1171875} = \frac{1181176}{1171875}$

Therefore:

$2^{1/3} \approx \frac{5}{4} \times \frac{1181176}{1171875} = \frac{5905880}{4687500} = \frac{295294}{234375} \approx 1.259921067...$

To six decimal places: $2^{1/3} \approx 1.259921$. $\quad\blacksquare$

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Question 3

Topic: 级数求和 (Series & Summation)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Show that the sum $S_N$ of the first $N$ terms of the series

$\frac{1}{1.2.3} + \frac{3}{2.3.4} + \frac{5}{3.4.5} + \dots + \frac{2n - 1}{n(n + 1)(n + 2)} + \dots$

is

$\frac{1}{2} \left( \frac{3}{2} + \frac{1}{N + 1} - \frac{5}{N + 2} \right).$

What is the limit of $S_N$ as $N \to \infty$?

The numbers $a_n$ are such that

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

Find an expression for $a_n/a_1$ and hence, or otherwise, evaluate $\sum_{n=1}^{\infty} a_n$ when $a_1 = \frac{2}{9}$.

Model Solution

Part 1: Evaluating $S_N$

The $n$-th term is

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

Partial fractions. Write

$\frac{2n - 1}{n(n + 1)(n + 2)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2}.$

Multiplying through by $n(n+1)(n+2)$:

$2n - 1 = A(n + 1)(n + 2) + Bn(n + 2) + Cn(n + 1).$

• $n = 0$: $-1 = 2A \implies A = -\tfrac{1}{2}$
• $n = -1$: $-3 = -B \implies B = 3$
• $n = -2$: $-5 = 2C \implies C = -\tfrac{5}{2}$

Therefore

$u_n = -\frac{1}{2n} + \frac{3}{n + 1} - \frac{5}{2(n + 2)}.$

Verification. $n = 1$: $-\frac{1}{2} + \frac{3}{2} - \frac{5}{6} = \frac{1}{6} = \frac{1}{1 \cdot 2 \cdot 3}$. $\checkmark$

$n = 2$: $-\frac{1}{4} + 1 - \frac{5}{8} = \frac{1}{8} = \frac{3}{24}$. $\checkmark$

Summing. Let $H_N = \sum_{n=1}^{N} \frac{1}{n}$. Then

$S_N = -\frac{1}{2}H_N + 3\sum_{n=1}^{N}\frac{1}{n+1} - \frac{5}{2}\sum_{n=1}^{N}\frac{1}{n+2}.$

Re-indexing the second sum ($m = n + 1$):

$\sum_{n=1}^{N}\frac{1}{n+1} = \sum_{m=2}^{N+1}\frac{1}{m} = H_N + \frac{1}{N+1} - 1.$

Re-indexing the third sum ($m = n + 2$):

$\sum_{n=1}^{N}\frac{1}{n+2} = \sum_{m=3}^{N+2}\frac{1}{m} = H_N + \frac{1}{N+1} + \frac{1}{N+2} - \frac{3}{2}.$

Substituting:

$S_N = -\frac{1}{2}H_N + 3\!\left(H_N + \frac{1}{N+1} - 1\right) - \frac{5}{2}\!\left(H_N + \frac{1}{N+1} + \frac{1}{N+2} - \frac{3}{2}\right).$

The coefficient of $H_N$ is $-\frac{1}{2} + 3 - \frac{5}{2} = 0$, so all harmonic terms cancel. Collecting the rest:

$S_N = \frac{3}{N+1} - 3 - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} + \frac{15}{4}$

$= \frac{1}{2(N+1)} - \frac{5}{2(N+2)} + \frac{3}{4}.$

Rewriting with a common factor of $\frac{1}{2}$:

$S_N = \frac{1}{2}\!\left(\frac{3}{2} + \frac{1}{N+1} - \frac{5}{N+2}\right). \qquad \text{(as required)}$

Check. $N = 1$: $\frac{1}{2}(\frac{3}{2} + \frac{1}{2} - \frac{5}{3}) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} = \frac{1}{1 \cdot 2 \cdot 3}$. $\checkmark$

$N = 2$: $\frac{1}{2}(\frac{3}{2} + \frac{1}{3} - \frac{5}{4}) = \frac{1}{2} \cdot \frac{18 + 4 - 15}{12} = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24}$. And $S_2 = \frac{1}{6} + \frac{3}{24} = \frac{7}{24}$. $\checkmark$

Limit. As $N \to \infty$:

$\lim_{N\to\infty} S_N = \frac{1}{2}\!\left(\frac{3}{2} + 0 - 0\right) = \boxed{\frac{3}{4}}.$

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Part 2: Using the recurrence for $a_n$

For $n \geq 2$,

$\frac{a_n}{a_1} = \prod_{k=2}^{n} \frac{(k - 1)(2k - 1)}{(k + 2)(2k - 3)}.$

Separate the product into two telescoping parts:

$\prod_{k=2}^{n} \frac{k - 1}{k + 2} = \frac{1 \cdot 2 \cdot 3 \cdots (n - 1)}{4 \cdot 5 \cdot 6 \cdots (n + 2)} = \frac{6}{n(n + 1)(n + 2)},$

and

$\prod_{k=2}^{n} \frac{2k - 1}{2k - 3} = \frac{3}{1}\cdot\frac{5}{3}\cdot\frac{7}{5}\cdots\frac{2n - 1}{2n - 3} = 2n - 1.$

Hence

$\frac{a_n}{a_1} = \frac{6(2n - 1)}{n(n + 1)(n + 2)}.$

This also gives $\frac{a_1}{a_1}=1$ when $n=1$, so the formula is valid for all $n \geq 1$.

If $a_1 = \frac{2}{9}$, then

$a_n = \frac{2}{9}\cdot\frac{6(2n - 1)}{n(n + 1)(n + 2)} = \frac{4(2n - 1)}{3n(n + 1)(n + 2)}.$

Using the first part of the question,

$\sum_{n=1}^{\infty} \frac{2n - 1}{n(n + 1)(n + 2)} = \lim_{N\to\infty} S_N = \frac{3}{4}.$

Therefore

$\sum_{n=1}^{\infty} a_n = \frac{4}{3}\cdot\frac{3}{4} = \boxed{1}.$

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Question 4

Topic: 积分 (Integration)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The integral $I_n$ is defined by

$I_n = \int_{0}^{\pi} (\pi/2 - x) \sin(nx + x/2) \operatorname{cosec}(x/2) \, dx,$

where $n$ is a positive integer. Evaluate $I_n - I_{n-1}$, and hence evaluate $I_n$ leaving your answer in the form of a sum.

Model Solution

Step 1: Evaluate $I_n - I_{n-1}$

Write the difference as a single integral:

$I_n - I_{n-1} = \int_0^\pi (\pi/2 - x)\bigl[\sin\!\bigl((n + \tfrac{1}{2})x\bigr) - \sin\!\bigl((n - \tfrac{1}{2})x\bigr)\bigr]\csc(x/2)\,dx.$

Using $\sin A - \sin B = 2\cos\!\bigl(\frac{A+B}{2}\bigr)\sin\!\bigl(\frac{A-B}{2}\bigr)$ with $A = (n + \frac{1}{2})x$ and $B = (n - \frac{1}{2})x$:

$\frac{A+B}{2} = nx, \qquad \frac{A-B}{2} = x/2,$

$\sin\!\bigl((n + \tfrac{1}{2})x\bigr) - \sin\!\bigl((n - \tfrac{1}{2})x\bigr) = 2\cos(nx)\sin(x/2).$

Since $\sin(x/2)\csc(x/2) = 1$:

$I_n - I_{n-1} = \int_0^\pi 2(\pi/2 - x)\cos(nx)\,dx. \qquad \text{($\star$)}$

Evaluate by parts with $u = \pi/2 - x$ and $dv = \cos(nx)\,dx$:

$I_n - I_{n-1} = \left[\frac{2(\pi/2 - x)\sin(nx)}{n}\right]_0^\pi + \frac{2}{n}\int_0^\pi \sin(nx)\,dx.$

The boundary term vanishes at both limits: at $x = 0$, $\sin 0 = 0$; at $x = \pi$, $\sin(n\pi) = 0$ for integer $n$.

$I_n - I_{n-1} = \frac{2}{n}\left[-\frac{\cos(nx)}{n}\right]_0^\pi = \frac{2}{n^2}\bigl(1 - \cos(n\pi)\bigr) = \frac{2\bigl(1 - (-1)^n\bigr)}{n^2}. \qquad \text{($\star\star$)}$

This equals $\frac{4}{n^2}$ when $n$ is odd, and $0$ when $n$ is even.

Check. For $n = 1$: $\frac{2(1-(-1))}{1} = 4$. For $n = 2$: $\frac{2(1-1)}{4} = 0$. For $n = 3$: $\frac{4}{9}$. $\checkmark$

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Step 2: Evaluate $I_1$

We use the identity

$\csc(x/2)\sin\!\bigl((n + \tfrac{1}{2})x\bigr) = 1 + 2\sum_{k=1}^{n}\cos(kx), \qquad \text{($\star\star\star$)}$

which follows by induction: the base case $n = 0$ gives $1$ on both sides, and the inductive step uses $2\cos(nx)\sin(x/2) = \sin((n+\frac{1}{2})x) - \sin((n-\frac{1}{2})x)$.

For $n = 1$: $\csc(x/2)\sin(3x/2) = 1 + 2\cos x$, so

$I_1 = \int_0^\pi (\pi/2 - x)(1 + 2\cos x)\,dx = \underbrace{\int_0^\pi (\pi/2 - x)\,dx}_{= \, 0} + 2\int_0^\pi (\pi/2 - x)\cos x\,dx.$

The first integral vanishes since $\pi/2 - x$ is antisymmetric about $x = \pi/2$. For the second, integrate by parts:

$2\Bigl[(\pi/2 - x)\sin x\Bigr]_0^\pi + 2\int_0^\pi \sin x\,dx = 0 + 2[-\cos x]_0^\pi = 2(1 + 1) = 4.$

Therefore $I_1 = 4$.

Check. Also, from ($\star\star$) with $n = 1$: $I_1 - I_0 = 4$, where $I_0 = \int_0^\pi (\pi/2 - x)\csc(x/2)\sin(x/2)\,dx = \int_0^\pi (\pi/2 - x)\,dx = 0$. So $I_1 = 4$. $\checkmark$

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Step 3: Express $I_n$ as a sum

By telescoping from $I_0 = 0$:

$I_n = \sum_{k=1}^{n}(I_k - I_{k-1}) = \sum_{k=1}^{n}\frac{2(1 - (-1)^k)}{k^2}.$

Since $1 - (-1)^k = 0$ for even $k$ and $2$ for odd $k$:

$I_n = 4\sum_{\substack{k = 1 \\ k \text{ odd}}}^{n}\frac{1}{k^2} = 4\sum_{m=1}^{\lceil n/2 \rceil}\frac{1}{(2m-1)^2}. \qquad \text{($\dagger$)}$

Equivalently:

$I_n = 4\left(1 + \frac{1}{3^2} + \frac{1}{5^2} + \cdots + \frac{1}{(2\lceil n/2\rceil - 1)^2}\right).$

Verification. $n = 1$: $I_1 = 4/1^2 = 4$. $\checkmark$

$n = 2$: $I_2 = 4/1^2 = 4$ (only odd $k \leq 2$ is $k = 1$). $\checkmark$

$n = 3$: $I_3 = 4(1 + 1/9) = 40/9$. $\checkmark$

$n = 4$: $I_4 = 4(1 + 1/9) = 40/9$. $\checkmark$

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Question 5

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

Problem

5 Define the modulus of a complex number $z$ and give the geometric interpretation of $|z_1 - z_2|$ for two complex numbers $z_1$ and $z_2$. On the basis of this interpretation establish the inequality

$|z_1 + z_2| \leqslant |z_1| + |z_2|.$

Use this result to prove, by induction, the corresponding inequality for $|z_1 + \dots + z_n|$.

The complex numbers $a_1, a_2, \dots, a_n$ satisfy $|a_i| \leqslant 3$ ($i = 1, 2, \dots, n$). Prove that the equation

$a_1 z + a_2 z^2 + \dots + a_n z^n = 1$

has no solution $z$ with $|z| \leqslant 1/4$.

Model Solution

Definition and geometric interpretation

Let $z = x + iy$ where $x, y \in \mathbb{R}$. The modulus of $z$ is defined as

$|z| = \sqrt{x^2 + y^2}.$

Geometrically, $|z|$ is the distance from the origin to the point $(x, y)$ in the Argand diagram. For two complex numbers $z_1$ and $z_2$, the quantity $|z_1 - z_2|$ is the distance between the points representing $z_1$ and $z_2$ in the complex plane.

Establishing the triangle inequality

Consider the triangle with vertices at $0$, $z_1$, and $z_1 + z_2$ in the Argand diagram. The three sides of this triangle have lengths:

• $|z_1 - 0| = |z_1|$ (distance from origin to $z_1$),
• $|(z_1 + z_2) - z_1| = |z_2|$ (distance from $z_1$ to $z_1 + z_2$),
• $|(z_1 + z_2) - 0| = |z_1 + z_2|$ (distance from origin to $z_1 + z_2$).

By the triangle inequality for Euclidean distance (the length of one side of a triangle cannot exceed the sum of the lengths of the other two sides):

$|z_1 + z_2| \leqslant |z_1| + |z_2|. \qquad \text{($\triangle$ inequality)}$

Equality holds if and only if $z_1$ and $z_2$ lie on the same ray from the origin, i.e., $z_2 = \lambda z_1$ for some $\lambda \geqslant 0$.

Proof by induction for the general case

We prove that for any $n \geqslant 2$ and complex numbers $z_1, z_2, \dots, z_n$:

$|z_1 + z_2 + \dots + z_n| \leqslant |z_1| + |z_2| + \dots + |z_n|.$

Base case ($n = 2$): This is the inequality established above.

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

$|z_1 + z_2 + \dots + z_k| \leqslant |z_1| + |z_2| + \dots + |z_k|.$

For $n = k + 1$, let $w = z_1 + z_2 + \dots + z_k$. Then:

$|z_1 + z_2 + \dots + z_k + z_{k+1}| = |w + z_{k+1}| \leqslant |w| + |z_{k+1}|$

by the base case. Applying the inductive hypothesis to $|w|$:

$|w| + |z_{k+1}| = |z_1 + z_2 + \dots + z_k| + |z_{k+1}| \leqslant |z_1| + |z_2| + \dots + |z_k| + |z_{k+1}|.$

Combining: $|z_1 + z_2 + \dots + z_{k+1}| \leqslant |z_1| + |z_2| + \dots + |z_{k+1}|$.

By induction, the inequality holds for all $n \geqslant 2$.

Proving the equation has no solution with $|z| \leqslant 1/4$

Suppose for contradiction that there exists $z$ with $|z| \leqslant 1/4$ satisfying

$a_1 z + a_2 z^2 + \dots + a_n z^n = 1.$

Taking the modulus of both sides and applying the generalised triangle inequality:

$1 = |a_1 z + a_2 z^2 + \dots + a_n z^n| \leqslant |a_1 z| + |a_2 z^2| + \dots + |a_n z^n|.$

Since $|a_i| \leqslant 3$ and $|z^k| = |z|^k$:

$1 \leqslant 3|z| + 3|z|^2 + \dots + 3|z|^n = 3|z|\bigl(1 + |z| + |z|^2 + \dots + |z|^{n-1}\bigr).$

Since $|z| \leqslant 1/4 < 1$, the geometric series satisfies:

$1 + |z| + |z|^2 + \dots + |z|^{n-1} = \frac{1 - |z|^n}{1 - |z|} \leqslant \frac{1}{1 - 1/4} = \frac{4}{3}.$

Therefore:

$1 \leqslant 3|z| \cdot \frac{4}{3} = 4|z| \leqslant 4 \cdot \frac{1}{4} = 1.$

This forces every inequality to be an equality, which requires both $|z| = 1/4$ and $|z|^n = 0$. But $|z| = 1/4$ gives $|z|^n = (1/4)^n > 0$, so $\frac{1 - (1/4)^n}{1 - 1/4} < \frac{4}{3}$. Hence:

$1 \leqslant 4 \cdot \frac{1}{4} \cdot \left(1 - \frac{1}{4^n}\right) = 1 - \frac{1}{4^n} < 1,$

which is a contradiction. Therefore the equation has no solution with $|z| \leqslant 1/4$. $\quad\blacksquare$

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Question 6

Topic: 解析几何与微积分 (Coordinate Geometry & Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Two curves are given parametrically by

(1) $x_1 = (\theta + \sin \theta)$, $y_1 = (1 + \cos \theta)$,

and

(2) $x_2 = (\theta - \sin \theta)$, $y_2 = -(1 + \cos \theta)$.

Find the gradients of the tangents to the curves at the points where $\theta = \pi/2$ and $\theta = 3\pi/2$. Sketch, using the same axes, the curves for $0 \leqslant \theta \leqslant 2\pi$.

Find the equation of the normal to the curve (1) at the point with parameter $\theta$. Show that this normal is a tangent to the curve (2).

Model Solution

Gradients of the tangents

Curve (1): $x_1 = \theta + \sin\theta$, $\;y_1 = 1 + \cos\theta$.

$\frac{dx_1}{d\theta} = 1 + \cos\theta, \qquad \frac{dy_1}{d\theta} = -\sin\theta.$

$\frac{dy_1}{dx_1} = \frac{-\sin\theta}{1 + \cos\theta}.$

At $\theta = \pi/2$: $\quad dx_1/d\theta = 1 + 0 = 1$, $\quad dy_1/d\theta = -1$, so $\dfrac{dy_1}{dx_1} = -1$.

At $\theta = 3\pi/2$: $\quad dx_1/d\theta = 1 + 0 = 1$, $\quad dy_1/d\theta = -(-1) = 1$, so $\dfrac{dy_1}{dx_1} = 1$.

Curve (2): $x_2 = \theta - \sin\theta$, $\;y_2 = -(1 + \cos\theta)$.

$\frac{dx_2}{d\theta} = 1 - \cos\theta, \qquad \frac{dy_2}{d\theta} = \sin\theta.$

$\frac{dy_2}{dx_2} = \frac{\sin\theta}{1 - \cos\theta}.$

At $\theta = \pi/2$: $\quad dx_2/d\theta = 1 - 0 = 1$, $\quad dy_2/d\theta = 1$, so $\dfrac{dy_2}{dx_2} = 1$.

At $\theta = 3\pi/2$: $\quad dx_2/d\theta = 1 - 0 = 1$, $\quad dy_2/d\theta = -1$, so $\dfrac{dy_2}{dx_2} = -1$.

Summary of gradients:

	Curve (1)	Curve (2)
$\theta = \pi/2$	$-1$	$1$
$\theta = 3\pi/2$	$1$	$-1$

Sketch

Both curves are cycloids (the path traced by a point on the rim of a circle of radius 1 rolling along a line).

Curve (1) is an upward-opening cycloid arch. Key points at $\theta = 0, \pi/2, \pi, 3\pi/2, 2\pi$:

$(0,\, 2),\quad (\pi/2,\, 1),\quad (\pi,\, 0),\quad (3\pi/2,\, 1),\quad (2\pi,\, 2).$

It starts at $(0, 2)$, descends to a cusp at $(\pi, 0)$ where it touches the $x$-axis, then returns to $(2\pi, 2)$. The curve is symmetric about $x = \pi$.

Curve (2) is an inverted cycloid. Key points:

$(0,\, -2),\quad (\pi/2,\, -1),\quad (\pi,\, 0),\quad (3\pi/2,\, -1),\quad (2\pi,\, -2).$

It starts at $(0, -2)$, rises to a cusp at $(\pi, 0)$ where it touches the $x$-axis, then returns to $(2\pi, -2)$. The two curves are reflections of each other in the line $y = 0$.

Normal to curve (1) at parameter $\theta$

The tangent to curve (1) at parameter $\theta$ has slope

$\frac{dy_1}{dx_1} = \frac{-\sin\theta}{1 + \cos\theta}.$

The normal at this point has slope equal to the negative reciprocal:

$m_{\text{normal}} = \frac{1 + \cos\theta}{\sin\theta}.$

(When $\sin\theta = 0$, the tangent is horizontal and the normal is vertical.)

The point on curve (1) is $(\theta + \sin\theta,\; 1 + \cos\theta)$, so the normal equation is:

$y - (1 + \cos\theta) = \frac{1 + \cos\theta}{\sin\theta}\,\bigl(x - (\theta + \sin\theta)\bigr).$

Multiplying both sides by $\sin\theta$:

$\bigl(y - 1 - \cos\theta\bigr)\sin\theta = (1 + \cos\theta)\bigl(x - \theta - \sin\theta\bigr).$

Expanding both sides:

$y\sin\theta - \sin\theta - \sin\theta\cos\theta = x(1 + \cos\theta) - \theta(1 + \cos\theta) - \sin\theta(1 + \cos\theta).$

Expanding the last term on the right: $\sin\theta(1 + \cos\theta) = \sin\theta + \sin\theta\cos\theta$, so:

$y\sin\theta - \sin\theta - \sin\theta\cos\theta = x(1 + \cos\theta) - \theta(1 + \cos\theta) - \sin\theta - \sin\theta\cos\theta.$

The terms $-\sin\theta - \sin\theta\cos\theta$ cancel from both sides, leaving:

$y\sin\theta = x(1 + \cos\theta) - \theta(1 + \cos\theta).$

Rearranging, the normal equation is:

$x(1 + \cos\theta) - y\sin\theta = \theta(1 + \cos\theta). \qquad \text{($\star$)}$

Showing the normal is tangent to curve (2)

Take the point $P_2$ on curve (2) at the same parameter value $\theta$:

$P_2 = \bigl(\theta - \sin\theta,\; -(1 + \cos\theta)\bigr).$

Step 1: $P_2$ lies on the normal ($\star$). Substitute into the left-hand side:

$(\theta - \sin\theta)(1 + \cos\theta) - \bigl(-(1 + \cos\theta)\bigr)\sin\theta.$

Expanding:

$= \theta(1 + \cos\theta) - \sin\theta(1 + \cos\theta) + \sin\theta(1 + \cos\theta) = \theta(1 + \cos\theta).$

This equals the right-hand side of ($\star$), so $P_2$ lies on the normal to curve (1) at parameter $\theta$.

Step 2: The normal has the same slope as the tangent to curve (2) at $P_2$.

The tangent to curve (2) at parameter $\theta$ has slope

$\frac{dy_2}{dx_2} = \frac{\sin\theta}{1 - \cos\theta}.$

We verify this equals the normal slope $\frac{1 + \cos\theta}{\sin\theta}$:

$\frac{1 + \cos\theta}{\sin\theta} = \frac{(1 + \cos\theta)(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} = \frac{1 - \cos^2\theta}{\sin\theta(1 - \cos\theta)} = \frac{\sin^2\theta}{\sin\theta(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}.$

So the slopes are equal. (When $\sin\theta = 0$, both the normal and the tangent to curve (2) are vertical, corresponding to the cusps at $\theta = 0$ and $\theta = \pi$.)

Conclusion: The normal to curve (1) at parameter $\theta$ passes through $P_2$ on curve (2) and has the same slope as the tangent to curve (2) at $P_2$. Therefore the normal to curve (1) is a tangent to curve (2). $\quad\blacksquare$

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Question 7

Topic: 不等式与微积分 (Inequalities & Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Let $\begin{aligned} f(x) &= \tan x - x, \\ g(x) &= 2 - 2 \cos x - x \sin x, \\ h(x) &= 2x + x \cos 2x - \frac{3}{2} \sin 2x, \\ F(x) &= \frac{x(\cos x)^{1/3}}{\sin x}. \end{aligned}$

(i) By considering $f(0)$ and $f'(x)$, show that $f(x) > 0$ for $0 < x < \pi/2$.

(ii) Show similarly that $g(x) > 0$ for $0 < x < \pi/2$.

(iii) Show that $h(x) > 0$ for $0 < x < \pi/4$, and hence that

$x(\sin^2 x + 3 \cos^2 x) - 3 \sin x \cos x > 0$

for $0 < x < \pi/4$.

(iv) By considering $\frac{F'(x)}{F(x)}$, show that $F'(x) < 0$ for $0 < x < \pi/4$.

Model Solution

Part (i)

We have $f(x) = \tan x - x$, so $f(0) = \tan 0 - 0 = 0$.

Differentiating:

$f'(x) = \sec^2 x - 1 = \tan^2 x.$

For $0 < x < \pi/2$, we have $\tan x > 0$, so $f'(x) = \tan^2 x > 0$.

Since $f(0) = 0$ and $f'(x) > 0$ for $0 < x < \pi/2$, the function $f$ is strictly increasing on $[0, \pi/2)$. Therefore $f(x) > f(0) = 0$ for $0 < x < \pi/2$. $\blacksquare$

Part (ii)

We have $g(x) = 2 - 2\cos x - x\sin x$, so $g(0) = 2 - 2 - 0 = 0$.

Differentiating:

$g'(x) = 2\sin x - \sin x - x\cos x = \sin x - x\cos x.$

At $x = 0$: $g'(0) = 0 - 0 = 0$.

Differentiating again:

$g''(x) = \cos x - (\cos x - x\sin x) = x\sin x.$

For $0 < x < \pi/2$, we have $x > 0$ and $\sin x > 0$, so $g''(x) = x\sin x > 0$.

Since $g''(x) > 0$ for $0 < x < \pi/2$ and $g'(0) = 0$, the function $g'$ is strictly increasing on $[0, \pi/2)$, so $g'(x) > g'(0) = 0$ for $0 < x < \pi/2$.

Since $g'(x) > 0$ for $0 < x < \pi/2$ and $g(0) = 0$, the function $g$ is strictly increasing on $[0, \pi/2)$, so $g(x) > g(0) = 0$ for $0 < x < \pi/2$. $\blacksquare$

Part (iii)

We have $h(x) = 2x + x\cos 2x - \tfrac{3}{2}\sin 2x$, so $h(0) = 0 + 0 - 0 = 0$.

Differentiating:

$h'(x) = 2 + \cos 2x - 2x\sin 2x - 3\cos 2x = 2 - 2\cos 2x - 2x\sin 2x.$

At $x = 0$: $h'(0) = 2 - 2 - 0 = 0$.

Note that $h'(x) = g(2x)$ where $g$ is the function from part (ii). From part (ii), $g(t) > 0$ for $0 < t < \pi/2$, so for $0 < 2x < \pi/2$, i.e.\ $0 < x < \pi/4$, we have $h'(x) = g(2x) > 0$.

Since $h'(x) > 0$ for $0 < x < \pi/4$ and $h(0) = 0$, the function $h$ is strictly increasing on $[0, \pi/4)$, so $h(x) > h(0) = 0$ for $0 < x < \pi/4$. $\blacksquare$

For the deduction, we need to show

$x(\sin^2 x + 3\cos^2 x) - 3\sin x\cos x > 0 \qquad \text{for } 0 < x < \pi/4.$

Using the double-angle identities $\sin x\cos x = \tfrac{1}{2}\sin 2x$ and $\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$:

$\sin^2 x + 3\cos^2 x = 1 + 2\cos^2 x = 1 + (1 + \cos 2x) = 2 + \cos 2x.$

So the left-hand side becomes:

$x(2 + \cos 2x) - \tfrac{3}{2}\sin 2x = 2x + x\cos 2x - \tfrac{3}{2}\sin 2x = h(x).$

Since we have shown $h(x) > 0$ for $0 < x < \pi/4$, the inequality follows immediately. $\blacksquare$

Part (iv)

We have $F(x) = \dfrac{x(\cos x)^{1/3}}{\sin x}$, which is positive for $0 < x < \pi/4$.

Taking logarithms:

$\ln F(x) = \ln x + \tfrac{1}{3}\ln(\cos x) - \ln(\sin x).$

Differentiating:

$\frac{F'(x)}{F(x)} = \frac{1}{x} + \frac{1}{3} \cdot \frac{-\sin x}{\cos x} - \frac{\cos x}{\sin x} = \frac{1}{x} - \frac{1}{3}\tan x - \cot x.$

To show $F'(x) < 0$, since $F(x) > 0$ it suffices to show $\dfrac{F'(x)}{F(x)} < 0$, i.e.

$\frac{1}{x} < \frac{1}{3}\tan x + \cot x.$

Multiplying both sides by $3x > 0$:

$3 < x\tan x + 3x\cot x = \frac{x\sin x}{\cos x} + \frac{3x\cos x}{\sin x} = \frac{x\sin^2 x + 3x\cos^2 x}{\sin x\cos x} = \frac{x(\sin^2 x + 3\cos^2 x)}{\sin x\cos x}.$

Multiplying both sides by $\sin x\cos x > 0$:

$3\sin x\cos x < x(\sin^2 x + 3\cos^2 x),$

which is exactly the inequality established in part (iii). Therefore $\dfrac{F'(x)}{F(x)} < 0$, and since $F(x) > 0$ for $0 < x < \pi/4$, we conclude $F'(x) < 0$ for $0 < x < \pi/4$. $\blacksquare$

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Question 8

Topic: 向量几何 (Vector Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Points $A, B, C$ in three dimensions have coordinate vectors a, b, c, respectively. Show that the lines joining the vertices of the triangle $ABC$ to the mid-points of the opposite sides meet at a point $R$.

$P$ is a point which is not in the plane $ABC$. Lines are drawn through the mid-points of $BC$, $CA$ and $AB$ parallel to $PA$, $PB$ and $PC$ respectively. Write down the vector equations of the lines and show by inspection that these lines meet at a common point $Q$.

Prove further that the line $PQ$ meets the plane $ABC$ at $R$.

Model Solution

Medians meeting at a point $R$

The mid-point of $BC$ is $\frac{\mathbf{b}+\mathbf{c}}{2}$, so the line from $A$ to this mid-point has equation

$\mathbf{r} = \mathbf{a} + \lambda\!\left(\frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a}\right) = (1-\lambda)\,\mathbf{a} + \frac{\lambda}{2}\,\mathbf{b} + \frac{\lambda}{2}\,\mathbf{c}. \qquad \text{(I)}$

The mid-point of $AC$ is $\frac{\mathbf{a}+\mathbf{c}}{2}$, so the line from $B$ to this mid-point has equation

$\mathbf{r} = \mathbf{b} + \mu\!\left(\frac{\mathbf{a}+\mathbf{c}}{2} - \mathbf{b}\right) = \frac{\mu}{2}\,\mathbf{a} + (1-\mu)\,\mathbf{b} + \frac{\mu}{2}\,\mathbf{c}. \qquad \text{(II)}$

At an intersection, equating coefficients of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ (since $A$, $B$, $C$ are not collinear):

$1-\lambda = \frac{\mu}{2}, \qquad \frac{\lambda}{2} = 1-\mu, \qquad \frac{\lambda}{2} = \frac{\mu}{2}.$

From the third equation, $\lambda = \mu$. Substituting into the first: $1 - \lambda = \lambda/2$, giving $\lambda = 2/3$. We verify the second equation: $\frac{1}{3} = 1 - \frac{2}{3} = \frac{1}{3}$. $\checkmark$

Substituting $\lambda = 2/3$ into (I):

$\mathbf{r} = \frac{1}{3}\,\mathbf{a} + \frac{1}{3}\,\mathbf{b} + \frac{1}{3}\,\mathbf{c} = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}.$

The third median (from $C$ to the mid-point of $AB$) has equation $\mathbf{r} = (1-\nu)\,\mathbf{c} + \frac{\nu}{2}\,\mathbf{a} + \frac{\nu}{2}\,\mathbf{b}$. Setting $\nu = 2/3$ gives $\mathbf{r} = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$, confirming all three medians meet at

$R = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}. \qquad \blacksquare$

Lines through mid-points parallel to $PA$, $PB$, $PC$

The line through the mid-point of $BC$ parallel to $PA$ has direction $\mathbf{a} - \mathbf{p}$:

$\ell_1:\quad \mathbf{r} = \frac{\mathbf{b}+\mathbf{c}}{2} + \lambda(\mathbf{a} - \mathbf{p}). \qquad \text{(III)}$

The line through the mid-point of $CA$ parallel to $PB$ has direction $\mathbf{b} - \mathbf{p}$:

$\ell_2:\quad \mathbf{r} = \frac{\mathbf{a}+\mathbf{c}}{2} + \mu(\mathbf{b} - \mathbf{p}). \qquad \text{(IV)}$

The line through the mid-point of $AB$ parallel to $PC$ has direction $\mathbf{c} - \mathbf{p}$:

$\ell_3:\quad \mathbf{r} = \frac{\mathbf{a}+\mathbf{b}}{2} + \nu(\mathbf{c} - \mathbf{p}). \qquad \text{(V)}$

We can find a common point by inspection. Setting $\lambda = 1/2$ in (III):

$\mathbf{r} = \frac{\mathbf{b}+\mathbf{c}}{2} + \frac{1}{2}(\mathbf{a}-\mathbf{p}) = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}-\mathbf{p}}{2}.$

Setting $\mu = 1/2$ in (IV):

$\mathbf{r} = \frac{\mathbf{a}+\mathbf{c}}{2} + \frac{1}{2}(\mathbf{b}-\mathbf{p}) = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}-\mathbf{p}}{2}.$

Setting $\nu = 1/2$ in (V):

$\mathbf{r} = \frac{\mathbf{a}+\mathbf{b}}{2} + \frac{1}{2}(\mathbf{c}-\mathbf{p}) = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}-\mathbf{p}}{2}.$

All three give the same result, so all three lines meet at

$Q = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}-\mathbf{p}}{2}. \qquad \blacksquare$

Line $PQ$ meets plane $ABC$ at $R$

The line $PQ$ has equation

$\mathbf{r} = \mathbf{p} + t\!\left(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}-\mathbf{p}}{2} - \mathbf{p}\right) = \mathbf{p} + t\cdot\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}-3\mathbf{p}}{2}.$

A point on this line is

$\mathbf{r} = \left(1 - \frac{3t}{2}\right)\mathbf{p} + \frac{t}{2}\,(\mathbf{a}+\mathbf{b}+\mathbf{c}).$

For $PQ$ to meet the plane $ABC$, the coefficient of $\mathbf{p}$ must be zero (since $P$ is not in the plane, and any point in the plane is a combination of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ only). Setting $1 - \frac{3t}{2} = 0$ gives $t = \frac{2}{3}$.

Substituting:

$\mathbf{r} = \frac{2/3}{2}\,(\mathbf{a}+\mathbf{b}+\mathbf{c}) = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3} = R.$

Therefore the line $PQ$ meets the plane $ABC$ at the point $R = \dfrac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$. $\blacksquare$