STEP2 2001 -- Pure Mathematics

Exam: STEP2  |  Year: 2001  |  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	二项式展开 Binomial Expansion	Routine	二项式定理,级数截断,误差分析
2	取整函数与级数 Floor Function and Series	Standard	取整函数,分段分析,几何级数求和,极限
3	立体几何 Solid Geometry	Standard	向量点积,立体几何,不等式证明
4	三角方程 Trigonometric Equations	Challenging	三角恒等式,因式分解,判别式分析
5	微积分 Calculus	Standard	积分,驻点判断,曲线草图,指数函数
6	积分 Integration	Challenging	多项式长除法,分部积分,换元积分
7	复数 Complex Numbers	Challenging	复数几何表示,旋转,圆的方程求解
8	微分方程 Differential Equations	Challenging	周期函数性质,分离变量法,指数衰减

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Question 1

Topic: 二项式展开 Binomial Expansion  |  Difficulty: Routine  |  Marks: 20

Problem

1 Use the binomial expansion to obtain a polynomial of degree 2 which is a good approximation to $\sqrt{(1 - x)}$ when $x$ is small.

(i) By taking $x = 1/100$, show that $\sqrt{(11)} \approx 79599/24000$, and estimate, correct to 1 significant figure, the error in this approximation. (You may assume that the error is given approximately by the first neglected term in the binomial expansion.)

(ii) Find a rational number which approximates $\sqrt{(1111)}$ with an error of about $2 \times 10^{-12}$.

Model Solution

Binomial expansion. We expand $(1 - x)^{1/2}$ using the generalised binomial theorem:

$$(1 - x)^{1/2} = 1 + \binom{1/2}{1}(-x) + \binom{1/2}{2}(-x)^2 + \cdots$$

Computing the binomial coefficients:

$$\binom{1/2}{1} = \frac{1}{2}, \qquad \binom{1/2}{2} = \frac{\frac{1}{2} \cdot (-\frac{1}{2})}{2!} = -\frac{1}{8}$$

So the degree 2 approximation is:

$$\sqrt{1 - x} \approx 1 - \frac{x}{2} - \frac{x^2}{8} \qquad (*)$$

Part (i)

We write $\sqrt{11} = \frac{1}{3}\sqrt{99}$. Since $99 = 100(1 - 1/100)$:

$$\sqrt{99} = 10\sqrt{1 - \tfrac{1}{100}}$$

and so

$$\sqrt{11} = \frac{10}{3}\sqrt{1 - \tfrac{1}{100}}.$$

Setting $x = 1/100$ in the approximation:

$$\sqrt{1 - \tfrac{1}{100}} \approx 1 - \frac{1}{200} - \frac{1}{80000}.$$

Therefore:

$$\sqrt{11} \approx \frac{10}{3}\left(1 - \frac{1}{200} - \frac{1}{80000}\right) = \frac{10}{3} - \frac{10}{600} - \frac{10}{240000}.$$

Simplifying each term with common denominator $240000$:

$$\frac{10}{3} = \frac{800000}{240000}, \qquad \frac{10}{600} = \frac{4000}{240000}, \qquad \frac{10}{240000} = \frac{10}{240000}.$$ $$\sqrt{11} \approx \frac{800000 - 4000 - 10}{240000} = \frac{795990}{240000} = \frac{79599}{24000}.$$

Error estimate. The first neglected term in the binomial expansion is the $k = 3$ term:

$$\binom{1/2}{3}(-x)^3 = \frac{\frac{1}{2} \cdot (-\frac{1}{2}) \cdot (-\frac{3}{2})}{3!}(-x^3) = -\frac{x^3}{16}.$$

With $x = 1/100$:

$$\left|-\frac{1}{16 \times 10^6}\right| = \frac{1}{16\,000\,000} \approx 6 \times 10^{-8}.$$

Multiplying by $10/3$ (the factor relating $\sqrt{11}$ to $\sqrt{1 - 1/100}$):

$$\text{Error} \approx \frac{10}{3} \times 6 \times 10^{-8} \approx 2 \times 10^{-7}.$$

To 1 significant figure, the error is $\mathbf{2 \times 10^{-7}}$.

Part (ii)

We write $\sqrt{1111} = \frac{1}{3}\sqrt{9999}$. Since $9999 = 10000(1 - 10^{-4})$:

$$\sqrt{9999} = 100\sqrt{1 - 10^{-4}}, \qquad \sqrt{1111} = \frac{100}{3}\sqrt{1 - 10^{-4}}.$$

Setting $x = 10^{-4}$ in the degree 2 binomial approximation:

$$\sqrt{1 - 10^{-4}} \approx 1 - \frac{10^{-4}}{2} - \frac{10^{-8}}{8}.$$

Therefore:

$$\sqrt{1111} \approx \frac{100}{3} - \frac{100}{3} \cdot \frac{10^{-4}}{2} - \frac{100}{3} \cdot \frac{10^{-8}}{8} = \frac{100}{3} - \frac{1}{600} - \frac{1}{24\,000\,000}.$$

Combining over the common denominator $24\,000\,000$:

$$\frac{100}{3} = \frac{800\,000\,000}{24\,000\,000}, \qquad \frac{1}{600} = \frac{40\,000}{24\,000\,000}, \qquad \frac{1}{24\,000\,000} = \frac{1}{24\,000\,000}.$$ $$\sqrt{1111} \approx \frac{800\,000\,000 - 40\,000 - 1}{24\,000\,000} = \frac{799\,959\,999}{24\,000\,000}.$$

Error check. The first neglected ($k = 3$) term gives:

$$\text{Error} \approx \frac{100}{3} \times \frac{(10^{-4})^3}{16} = \frac{100}{3} \times \frac{10^{-12}}{16} \approx 2 \times 10^{-12}.$$

This confirms the approximation $\dfrac{799\,959\,999}{24\,000\,000}$ has an error of about $2 \times 10^{-12}$. $\blacksquare$

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Question 2

Topic: 取整函数与级数 Floor Function and Series  |  Difficulty: Standard  |  Marks: 20

Problem

2 Sketch the graph of the function $[x/N]$, for $0 < x < 2N$, where the notation $[y]$ means the integer part of $y$. (Thus $[2.9] = 2$, $[4] = 4$.)

(i) Prove that $\sum_{k=1}^{2N} (-1)^{[k/N]}k = 2N - N^2.$

(ii) Let $S_N = \sum_{k=1}^{2N} (-1)^{[k/N]}2^{-k}.$ Find $S_N$ in terms of $N$ and determine the limit of $S_N$ as $N \to \infty$.

Model Solution

Graph of $[x/N]$. For $0 < x < 2N$:

• When $0 < x < N$: $0 < x/N < 1$, so $[x/N] = 0$.
• When $N \le x < 2N$: $1 \le x/N < 2$, so $[x/N] = 1$.

The graph is a step function: value $0$ on $(0, N)$, jumping to value $1$ on $[N, 2N)$.

Part (i)

We split the sum according to the value of $[k/N]$:

• $k = 1, \ldots, N - 1$: $[k/N] = 0$, so $(-1)^{[k/N]} = +1$.
• $k = N, \ldots, 2N - 1$: $[k/N] = 1$, so $(-1)^{[k/N]} = -1$.
• $k = 2N$: $[k/N] = 2$, so $(-1)^{[k/N]} = +1$.

Therefore:

$$\sum_{k=1}^{2N} (-1)^{[k/N]}k = \sum_{k=1}^{N-1} k \;-\; \sum_{k=N}^{2N-1} k \;+\; 2N.$$

Using $\displaystyle\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$:

$$\sum_{k=1}^{N-1} k = \frac{(N-1)N}{2} = \frac{N^2 - N}{2}.$$

For the middle sum:

$$\sum_{k=N}^{2N-1} k = \sum_{k=1}^{2N-1} k - \sum_{k=1}^{N-1} k = \frac{(2N-1)(2N)}{2} - \frac{(N-1)N}{2} = \frac{2N(2N-1) - N(N-1)}{2}.$$

Expanding:

$$2N(2N-1) = 4N^2 - 2N, \qquad N(N-1) = N^2 - N.$$ $$\sum_{k=N}^{2N-1} k = \frac{4N^2 - 2N - N^2 + N}{2} = \frac{3N^2 - N}{2}.$$

Substituting back:

$$\sum_{k=1}^{2N} (-1)^{[k/N]}k = \frac{N^2 - N}{2} - \frac{3N^2 - N}{2} + 2N = \frac{N^2 - N - 3N^2 + N}{2} + 2N$$ $$= \frac{-2N^2}{2} + 2N = -N^2 + 2N = 2N - N^2. \qquad \blacksquare$$

Part (ii)

We split $S_N = \sum_{k=1}^{2N} (-1)^{[k/N]}2^{-k}$ using the same three ranges:

$$S_N = \underbrace{\sum_{k=1}^{N-1} 2^{-k}}_{\text{positive}} \;-\; \underbrace{\sum_{k=N}^{2N-1} 2^{-k}}_{\text{negative}} \;+\; 2^{-2N}.$$

Positive sum. This is a geometric series with first term $1/2$, common ratio $1/2$, and $N - 1$ terms:

$$\sum_{k=1}^{N-1} 2^{-k} = \frac{\frac{1}{2}(1 - (\frac{1}{2})^{N-1})}{1 - \frac{1}{2}} = 1 - 2^{1-N}.$$

(When $N = 1$ this sum is empty and equals $0$, which the formula also gives since $2^{0} = 1$.)

Negative sum. This is a geometric series with first term $2^{-N}$, common ratio $1/2$, and $N$ terms:

$$\sum_{k=N}^{2N-1} 2^{-k} = \frac{2^{-N}(1 - 2^{-N})}{1 - \frac{1}{2}} = 2 \cdot 2^{-N}(1 - 2^{-N}) = 2^{1-N}(1 - 2^{-N}).$$

Combining. Let $u = 2^{-N}$. Then $2^{1-N} = 2u$ and $2^{-2N} = u^2$, so:

$$S_N = (1 - 2u) - 2u(1 - u) + u^2 = 1 - 2u - 2u + 2u^2 + u^2 = 1 - 4u + 3u^2.$$

Factorising:

$$S_N = (1 - u)(1 - 3u) = (1 - 2^{-N})(1 - 3 \cdot 2^{-N}).$$

Verification. For $N = 1$: $S_1 = (1 - \frac{1}{2})(1 - \frac{3}{2}) = \frac{1}{2} \cdot (-\frac{1}{2}) = -\frac{1}{4}$.

Directly: $(-1)^1 \cdot 2^{-1} + (-1)^2 \cdot 2^{-2} = -\frac{1}{2} + \frac{1}{4} = -\frac{1}{4}$. Correct.

Limit. As $N \to \infty$, $u = 2^{-N} \to 0$, so:

$$\lim_{N \to \infty} S_N = (1 - 0)(1 - 0) = 1.$$

$\blacksquare$

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Question 3

Topic: 立体几何 Solid Geometry  |  Difficulty: Standard  |  Marks: 20

Problem

3 The cuboid $ABCDEFGH$ is such $AE, BF, CG, DH$ are perpendicular to the opposite faces $ABCD$ and $EFGH$, and $AB = 2, BC = 1, AE = \lambda$. Show that if $\alpha$ is the acute angle between the diagonals $AG$ and $BH$ then $\cos \alpha = \left| \frac{3 - \lambda^2}{5 + \lambda^2} \right|$

Let $R$ be the ratio of the volume of the cuboid to its surface area. Show that $R < 1/3$ for all possible values of $\lambda$.

Prove that, if $R \geqslant 1/4$, then $\alpha \leqslant \arccos(1/9)$.

Model Solution

Part (i): Finding $\cos\alpha$

Set up coordinates with $A$ at the origin. Let $\vec{AB}$ point along the $x$-axis, $\vec{AD}$ along the $y$-axis, and $\vec{AE}$ along the $z$-axis. Then:

$A = (0,0,0),\quad B = (2,0,0),\quad C = (2,1,0),\quad D = (0,1,0)$ $E = (0,0,\lambda),\quad F = (2,0,\lambda),\quad G = (2,1,\lambda),\quad H = (0,1,\lambda)$

The two diagonals are:

$\vec{AG} = (2,\, 1,\, \lambda), \qquad \vec{BH} = (-2,\, 1,\, \lambda)$

The dot product is:

$\vec{AG} \cdot \vec{BH} = (2)(-2) + (1)(1) + (\lambda)(\lambda) = -4 + 1 + \lambda^2 = \lambda^2 - 3$

The magnitudes are:

$|\vec{AG}| = \sqrt{4 + 1 + \lambda^2} = \sqrt{5 + \lambda^2}, \qquad |\vec{BH}| = \sqrt{4 + 1 + \lambda^2} = \sqrt{5 + \lambda^2}$

The angle $\theta$ between the two vectors satisfies:

$\cos\theta = \frac{\vec{AG} \cdot \vec{BH}}{|\vec{AG}||\vec{BH}|} = \frac{\lambda^2 - 3}{5 + \lambda^2}$

Since $\alpha$ is the acute angle between the diagonals, we require $\cos\alpha \geq 0$, so:

$\cos\alpha = \left| \frac{\lambda^2 - 3}{5 + \lambda^2} \right| = \left| \frac{3 - \lambda^2}{5 + \lambda^2} \right| \qquad \blacksquare$

Part (ii): Showing $R < 1/3$

The volume is:

$V = AB \times BC \times AE = 2 \times 1 \times \lambda = 2\lambda$

The surface area is:

$S = 2(AB \cdot BC) + 2(BC \cdot AE) + 2(AB \cdot AE) = 2(2) + 2(\lambda) + 2(2\lambda) = 4 + 6\lambda$

The ratio is:

$R = \frac{V}{S} = \frac{2\lambda}{4 + 6\lambda} = \frac{\lambda}{2 + 3\lambda}$

We show $R < 1/3$, i.e. $\dfrac{\lambda}{2 + 3\lambda} < \dfrac{1}{3}$.

Since $\lambda > 0$, the denominator $2 + 3\lambda > 0$, so cross-multiplying:

$3\lambda < 2 + 3\lambda \implies 0 < 2$

This holds for all $\lambda > 0$, so $R < 1/3$ for all possible values of $\lambda$. $\qquad \blacksquare$

Part (iii): If $R \geq 1/4$ then $\alpha \leq \arccos(1/9)$

First, find the constraint on $\lambda$ from $R \geq 1/4$:

$\frac{\lambda}{2 + 3\lambda} \geq \frac{1}{4}$

Since $2 + 3\lambda > 0$:

$4\lambda \geq 2 + 3\lambda \implies \lambda \geq 2$

Now, $\alpha \leq \arccos(1/9)$ is equivalent to $\cos\alpha \geq 1/9$ (since $\arccos$ is a decreasing function and $\alpha$ is acute so $\cos\alpha \geq 0$).

When $\lambda \geq 2$: $\lambda^2 \geq 4 > 3$, so $3 - \lambda^2 < 0$ and

$\cos\alpha = \frac{\lambda^2 - 3}{5 + \lambda^2}$

We show $\dfrac{\lambda^2 - 3}{5 + \lambda^2} \geq \dfrac{1}{9}$. Since $5 + \lambda^2 > 0$:

$9(\lambda^2 - 3) \geq 5 + \lambda^2$ $9\lambda^2 - 27 \geq 5 + \lambda^2$ $8\lambda^2 \geq 32$ $\lambda^2 \geq 4$ $\lambda \geq 2 \quad (\text{since } \lambda > 0)$

This is precisely the condition from $R \geq 1/4$. Therefore, if $R \geq 1/4$ then $\lambda \geq 2$, which gives $\cos\alpha \geq 1/9$, i.e. $\alpha \leq \arccos(1/9)$. $\qquad \blacksquare$

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Question 4

Topic: 三角方程 Trigonometric Equations  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Let $f(x) = P \sin x + Q \sin 2x + R \sin 3x .$ Show that if $Q^2 < 4R(P - R)$, then the only values of $x$ for which $f(x) = 0$ are given by $x = m\pi$, where $m$ is an integer. [You may assume that $\sin 3x = \sin x(4 \cos^2 x - 1)$.]

Now let $g(x) = \sin 2nx + \sin 4nx - \sin 6nx,$ where $n$ is a positive integer and $0 < x < \pi/2$. Find an expression for the largest root of the equation $g(x) = 0$, distinguishing between the cases where $n$ is even and $n$ is odd.

Model Solution

Part (i): Showing that $Q^2 < 4R(P-R)$ implies $f(x) = 0$ only when $x = m\pi$

We are given:

$f(x) = P\sin x + Q\sin 2x + R\sin 3x$

and the identity $\sin 3x = \sin x(4\cos^2 x - 1)$.

We also need $\sin 2x = 2\sin x\cos x$. Factor out $\sin x$:

$f(x) = \sin x \bigl[P + 2Q\cos x + R(4\cos^2 x - 1)\bigr]$

$= \sin x \bigl[4R\cos^2 x + 2Q\cos x + (P - R)\bigr]$

So $f(x) = 0$ when either:

• $\sin x = 0$, which gives $x = m\pi$ for integer $m$; or
• $4R\cos^2 x + 2Q\cos x + (P - R) = 0$.

Let $t = \cos x$. The second equation becomes the quadratic:

$4Rt^2 + 2Qt + (P - R) = 0$

We compute the discriminant:

$\Delta = (2Q)^2 - 4(4R)(P - R) = 4Q^2 - 16R(P - R) = 4\bigl[Q^2 - 4R(P - R)\bigr]$

The condition $Q^2 < 4R(P - R)$ is equivalent to $\Delta < 0$.

When $\Delta < 0$, the quadratic $4Rt^2 + 2Qt + (P - R) = 0$ has no real roots in $t$, and hence there is no real value of $\cos x$ satisfying it. (Note: we also need $4R > 0$, i.e. $R > 0$, for the quadratic to open upward and be strictly positive everywhere when $\Delta < 0$. If $R > 0$ and $\Delta < 0$, the quadratic is always positive, so it is never zero.)

Therefore, when $Q^2 < 4R(P - R)$ (and $R > 0$), the only solutions come from $\sin x = 0$, giving $x = m\pi$ for integer $m$. $\qquad \blacksquare$

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Part (ii): Largest root of $g(x) = 0$ for $0 < x < \pi/2$

We have:

$g(x) = \sin 2nx + \sin 4nx - \sin 6nx$

Using the sum-to-product formula $\sin A + \sin B = 2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)$ on the first two terms:

$\sin 2nx + \sin 4nx = 2\sin 3nx \cos nx$

Also, by the double angle formula:

$\sin 6nx = 2\sin 3nx \cos 3nx$

So:

$g(x) = 2\sin 3nx \cos nx - 2\sin 3nx \cos 3nx = 2\sin 3nx(\cos nx - \cos 3nx)$

Now apply the identity $\cos A - \cos B = -2\sin\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$:

$\cos nx - \cos 3nx = -2\sin 2nx \sin(-nx) = 2\sin 2nx \sin nx$

Therefore:

$g(x) = 2\sin 3nx \cdot 2\sin 2nx \sin nx = 4\sin nx \sin 2nx \sin 3nx$

So $g(x) = 0$ when $\sin nx = 0$, $\sin 2nx = 0$, or $\sin 3nx = 0$.

For $0 < x < \pi/2$:

• $\sin nx = 0 \implies nx = k\pi \implies x = \dfrac{k\pi}{n}$, for $k = 1, 2, \ldots, n-1$ (since $0 < k\pi/n < \pi/2$ requires $k < n/2$, i.e. $k = 1, \ldots, \lfloor(n-1)/2\rfloor$).

• $\sin 2nx = 0 \implies 2nx = k\pi \implies x = \dfrac{k\pi}{2n}$, for $k = 1, \ldots, n-1$.

• $\sin 3nx = 0 \implies 3nx = k\pi \implies x = \dfrac{k\pi}{3n}$, for $k = 1, \ldots, \lfloor(3n-1)/2\rfloor$.

The roots from $\sin nx = 0$ are a subset of the roots from $\sin 2nx = 0$ (since $k\pi/n = 2k\pi/2n$), which are in turn a subset of the roots from $\sin 3nx = 0$ (since $k\pi/2n = 3k\pi/6n$). So the complete set of roots is:

$x = \frac{k\pi}{3n}, \qquad k = 1, 2, \ldots, \left\lfloor \frac{3n-1}{2} \right\rfloor$

(The condition $x < \pi/2$ gives $k\pi/3n < \pi/2$, so $k < 3n/2$, i.e. $k \leq \lfloor(3n-1)/2\rfloor$.)

The largest root corresponds to $k = \lfloor(3n-1)/2\rfloor$.

Case 1: $n$ is even. Write $n = 2m$. Then $3n/2 = 3m$ is an integer, and $k < 3m$ gives $k_{\max} = 3m - 1 = \frac{3n}{2} - 1 = \frac{3n - 2}{2}$.

$x_{\max} = \frac{(3n - 2)\pi}{2 \cdot 3n} = \frac{(3n-2)\pi}{6n}$

Check: $\lfloor(3n-1)/2\rfloor = \lfloor 3m - 1/2\rfloor = 3m - 1 = (3n-2)/2$. Correct.

Case 2: $n$ is odd. Write $n = 2m + 1$. Then $3n/2 = 3m + 3/2$ is not an integer, and $k < 3m + 3/2$ gives $k_{\max} = 3m + 1 = \frac{3(n-1)}{2} + 1 = \frac{3n - 1}{2}$.

$x_{\max} = \frac{(3n-1)\pi}{2 \cdot 3n} = \frac{(3n-1)\pi}{6n}$

Check: $\lfloor(3n-1)/2\rfloor = \lfloor 3m + 1\rfloor = 3m + 1 = (3n-1)/2$. Correct.

Summary. The largest root of $g(x) = 0$ in $(0, \pi/2)$ is:

$x_{\max} = \begin{cases} \dfrac{(3n-1)\pi}{6n} & \text{if } n \text{ is odd} \\[8pt] \dfrac{(3n-2)\pi}{6n} & \text{if } n \text{ is even} \end{cases}$

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Question 5

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

Problem

5 The curve $C_1$ passes through the origin in the $x$–$y$ plane and its gradient is given by $\frac{dy}{dx} = x(1 - x^2)e^{-x^2}.$ Show that $C_1$ has a minimum point at the origin and a maximum point at $(1, \frac{1}{2}e^{-1})$. Find the coordinates of the other stationary point. Give a rough sketch of $C_1$.

The curve $C_2$ passes through the origin and its gradient is given by $\frac{dy}{dx} = x(1 - x^2)e^{-x^3}.$ Show that $C_2$ has a minimum point at the origin and a maximum point at $(1, k)$, where $k > \frac{1}{2}e^{-1}$. (You need not find $k$.)

Model Solution

Finding $y(x)$ for $C_1$.

We integrate $\dfrac{dy}{dx} = x(1 - x^2)e^{-x^2} = (x - x^3)e^{-x^2}$.

For $\int x\,e^{-x^2}\,dx$: set $u = x^2$, $du = 2x\,dx$, so $\int x\,e^{-x^2}\,dx = \tfrac{1}{2}\int e^{-u}\,du = -\tfrac{1}{2}e^{-x^2} + C.$

For $\int x^3 e^{-x^2}\,dx$: set $u = x^2$, $du = 2x\,dx$, so $\int x^3 e^{-x^2}\,dx = \tfrac{1}{2}\int u\,e^{-u}\,du.$ Integrate by parts with $a = u$, $db = e^{-u}\,du$: $\int u\,e^{-u}\,du = -u\,e^{-u} + \int e^{-u}\,du = -u\,e^{-u} - e^{-u} = -(u+1)e^{-u}.$ So $\int x^3 e^{-x^2}\,dx = -\tfrac{1}{2}(x^2 + 1)e^{-x^2} + C.$

Combining: $y = -\tfrac{1}{2}e^{-x^2} + \tfrac{1}{2}(x^2 + 1)e^{-x^2} + C = \tfrac{1}{2}x^2 e^{-x^2} + C.$

Since $C_1$ passes through the origin, $y(0) = 0$, giving $C = 0$. Therefore $y = \tfrac{1}{2}x^2 e^{-x^2}.$

Stationary points. Setting $\dfrac{dy}{dx} = 0$: since $e^{-x^2} \neq 0$, we need $x(1 - x^2) = 0$, giving $x = 0,\; x = 1,\; x = -1$.

At $x = 0$: $y = 0$, so the point is $(0, 0)$.

At $x = 1$: $y = \tfrac{1}{2}e^{-1}$, so the point is $(1, \tfrac{1}{2}e^{-1})$.

At $x = -1$: $y = \tfrac{1}{2}e^{-1}$, so the point is $(-1, \tfrac{1}{2}e^{-1})$.

Second derivative test. We differentiate the gradient: $\frac{d^2 y}{dx^2} = \frac{d}{dx}\bigl[(x - x^3)e^{-x^2}\bigr] = (1 - 3x^2)e^{-x^2} + (x - x^3)(-2x)e^{-x^2}$ $= e^{-x^2}\bigl[(1 - 3x^2) - 2x^2(1 - x^2)\bigr] = e^{-x^2}(1 - 5x^2 + 2x^4).$

At $x = 0$: $\dfrac{d^2y}{dx^2} = 1 > 0$, so $(0, 0)$ is a minimum point.

At $x = 1$: $\dfrac{d^2y}{dx^2} = e^{-1}(1 - 5 + 2) = -2e^{-1} < 0$, so $(1, \tfrac{1}{2}e^{-1})$ is a maximum point.

The other stationary point is $\mathbf{(-1,\;\tfrac{1}{2}e^{-1})}$.

Sketch of $C_1$. The curve passes through the origin (minimum), rises to a maximum at $(1, \tfrac{1}{2}e^{-1})$, and by symmetry about the $y$-axis (since $y = \tfrac{1}{2}x^2 e^{-x^2}$ is an even function) also has a maximum at $(-1, \tfrac{1}{2}e^{-1})$. As $x \to \pm\infty$, $y \to 0^+$. The curve has a characteristic bell-shaped hump centred on the $y$-axis.

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Curve $C_2$.

$C_2$ passes through the origin with $\dfrac{dy}{dx} = x(1 - x^2)e^{-x^3}$.

Stationary points. Setting $\dfrac{dy}{dx} = 0$: since $e^{-x^3} \neq 0$, we again need $x(1 - x^2) = 0$, giving $x = 0, \pm 1$.

Second derivative. Differentiating the gradient: $\frac{d^2y}{dx^2} = \frac{d}{dx}\bigl[(x - x^3)e^{-x^3}\bigr] = (1 - 3x^2)e^{-x^3} + (x - x^3)(-3x^2)e^{-x^3}$ $= e^{-x^3}\bigl[(1 - 3x^2) - 3x^2(1 - x^2)(x)\bigr] = e^{-x^3}(1 - 3x^2 - 3x^3 + 3x^5).$

At $x = 0$: $\dfrac{d^2y}{dx^2} = 1 > 0$, so $(0, 0)$ is a minimum point.

At $x = 1$: $\dfrac{d^2y}{dx^2} = e^{-1}(1 - 3 - 3 + 3) = -2e^{-1} < 0$, so $(1, k)$ is a maximum point.

Showing $k > \frac{1}{2}e^{-1}$. Since $C_2$ passes through the origin, the $y$-value at the maximum is $k = \int_0^1 x(1 - x^2)e^{-x^3}\,dx.$

For $C_1$, the $y$-value at its maximum is $\tfrac{1}{2}e^{-1} = \int_0^1 x(1 - x^2)e^{-x^2}\,dx$ which we can verify: $\int_0^1 (x - x^3)e^{-x^2}\,dx = \bigl[\tfrac{1}{2}x^2 e^{-x^2}\bigr]_0^1 = \tfrac{1}{2}e^{-1}$.

For $0 < x < 1$: $x^3 < x^2$, so $-x^3 > -x^2$, giving $e^{-x^3} > e^{-x^2}$. Also $x(1 - x^2) > 0$ for $0 < x < 1$.

Therefore $x(1 - x^2)e^{-x^3} > x(1 - x^2)e^{-x^2}$ for all $x \in (0, 1)$, and since the integrands are continuous and agree at the endpoints, we conclude $k = \int_0^1 x(1 - x^2)e^{-x^3}\,dx > \int_0^1 x(1 - x^2)e^{-x^2}\,dx = \tfrac{1}{2}e^{-1}.$

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Question 6

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

Problem

6 Show that $\int_0^1 \frac{x^4}{1 + x^2} \text{ d}x = \frac{\pi}{4} - \frac{2}{3} .$ Determine the values of (i) $\int_0^1 x^3 \tan^{-1} \left( \frac{1 - x}{1 + x} \right) \text{ d}x$ , (ii) $\int_0^1 \frac{(1 - y)^3}{(1 + y)^5} \tan^{-1} y \text{ d}y$ .

Model Solution

Preliminary result.

Perform polynomial long division: $\frac{x^4}{1 + x^2} = x^2 - 1 + \frac{1}{1 + x^2}$ since $x^4 = (x^2 - 1)(x^2 + 1) + 1$.

Therefore $\int_0^1 \frac{x^4}{1 + x^2}\,dx = \int_0^1 (x^2 - 1)\,dx + \int_0^1 \frac{1}{1 + x^2}\,dx = \left[\frac{x^3}{3} - x\right]_0^1 + \left[\tan^{-1}x\right]_0^1$ $= \left(\frac{1}{3} - 1\right) + \frac{\pi}{4} = \frac{\pi}{4} - \frac{2}{3}. \qquad \checkmark$

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Part (i).

We use integration by parts. First we compute the derivative of $\tan^{-1}\!\left(\dfrac{1-x}{1+x}\right)$.

Set $u = \dfrac{1-x}{1+x}$, so $\dfrac{du}{dx} = \dfrac{-(1+x) - (1-x)}{(1+x)^2} = \dfrac{-2}{(1+x)^2}$.

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

By the chain rule: $\frac{d}{dx}\tan^{-1}\!\left(\frac{1-x}{1+x}\right) = \frac{1}{1 + u^2}\cdot\frac{du}{dx} = \frac{(1+x)^2}{2(1+x^2)}\cdot\frac{-2}{(1+x)^2} = \frac{-1}{1+x^2}.$

Now integrate by parts with $U = \tan^{-1}\!\left(\dfrac{1-x}{1+x}\right)$ and $dV = x^3\,dx$: $dU = \frac{-1}{1+x^2}\,dx, \qquad V = \frac{x^4}{4}.$

$\int_0^1 x^3 \tan^{-1}\!\left(\frac{1-x}{1+x}\right)dx = \left[\frac{x^4}{4}\tan^{-1}\!\left(\frac{1-x}{1+x}\right)\right]_0^1 - \int_0^1 \frac{x^4}{4}\cdot\frac{-1}{1+x^2}\,dx$ $= \left[\frac{x^4}{4}\tan^{-1}\!\left(\frac{1-x}{1+x}\right)\right]_0^1 + \frac{1}{4}\int_0^1 \frac{x^4}{1+x^2}\,dx.$

Evaluating the boundary term:

• At $x = 1$: $\dfrac{1}{4}\tan^{-1}(0) = 0$.
• At $x = 0$: $0 \cdot \tan^{-1}(1) = 0$.

So the boundary term vanishes, and $\int_0^1 x^3 \tan^{-1}\!\left(\frac{1-x}{1+x}\right)dx = \frac{1}{4}\int_0^1 \frac{x^4}{1+x^2}\,dx = \frac{1}{4}\left(\frac{\pi}{4} - \frac{2}{3}\right) = \frac{3\pi - 8}{48}.$

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Part (ii).

Use the substitution $y = \dfrac{1-t}{1+t}$, so $t = \dfrac{1-y}{1+y}$.

When $y = 0$: $t = 1$. When $y = 1$: $t = 0$.

$\frac{dy}{dt} = \frac{-(1+t) - (1-t)}{(1+t)^2} = \frac{-2}{(1+t)^2}.$

We compute the key building blocks: $1 + y = 1 + \frac{1-t}{1+t} = \frac{(1+t) + (1-t)}{1+t} = \frac{2}{1+t},$ $1 - y = 1 - \frac{1-t}{1+t} = \frac{(1+t) - (1-t)}{1+t} = \frac{2t}{1+t}.$

Therefore $\frac{(1-y)^3}{(1+y)^5} = \frac{[2t/(1+t)]^3}{[2/(1+t)]^5} = \frac{8t^3/(1+t)^3}{32/(1+t)^5} = \frac{t^3(1+t)^2}{4}.$

Also $\tan^{-1}y = \tan^{-1}\!\left(\dfrac{1-t}{1+t}\right)$.

Substituting into the integral (and reversing limits to absorb the negative sign from $dy/dt$): $\int_0^1 \frac{(1-y)^3}{(1+y)^5}\tan^{-1}y\,dy = \int_1^0 \frac{t^3(1+t)^2}{4}\tan^{-1}\!\left(\frac{1-t}{1+t}\right)\cdot\frac{-2}{(1+t)^2}\,dt$ $= \frac{1}{2}\int_0^1 t^3\tan^{-1}\!\left(\frac{1-t}{1+t}\right)dt.$

This is exactly $\dfrac{1}{2}$ times the integral evaluated in part (i). Therefore

$\int_0^1 \frac{(1-y)^3}{(1+y)^5}\tan^{-1}y\,dy = \frac{1}{2}\cdot\frac{3\pi - 8}{48} = \frac{3\pi - 8}{96}.$

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Question 7

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

Problem

7 In an Argand diagram, $O$ is the origin and $P$ is the point $2 + 0i$. The points $Q$, $R$ and $S$ are such that the lengths $OP$, $PQ$, $QR$ and $RS$ are all equal, and the angles $OPQ$, $PQR$ and $QRS$ are all equal to $5\pi/6$, so that the points $O$, $P$, $Q$, $R$ and $S$ are five vertices of a regular 12-sided polygon lying in the upper half of the Argand diagram. Show that $Q$ is the point $2 + \sqrt{3} + i$ and find $S$.

The point $C$ is the centre of the circle that passes through the points $O$, $P$ and $Q$. Show that, if the polygon is rotated anticlockwise about $O$ until $C$ first lies on the real axis, the new position of $S$ is $-\frac{1}{2}(3\sqrt{2} + \sqrt{6})(\sqrt{3} - i) .$

Model Solution

Setting up the geometry.

The five points $O, P, Q, R, S$ are consecutive vertices of a regular 12-gon in the upper half-plane. The interior angle of a regular 12-gon is $5\pi/6$ (since each exterior angle is $\pi/6$). Each side has length $|OP| = 2$.

Finding $Q$.

Starting from $P = 2$, the vector $\vec{PQ}$ has length 2 and makes an angle of $\pi - 5\pi/6 = \pi/6$ with the positive real axis (the interior angle $OPQ = 5\pi/6$ means $PQ$ turns by $\pi/6$ from the direction $PO$).

So $\vec{PQ} = 2(\cos(\pi/6) + i\sin(\pi/6)) = 2\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \sqrt{3} + i$.

$Q = P + \vec{PQ} = 2 + \sqrt{3} + i \qquad \blacksquare$

Finding $S$.

Each step rotates by $-\pi/6$ (turning left by the exterior angle $\pi/6$) in the complex plane. We can express each successive vertex by multiplying by $e^{i\pi/6}$ (rotating by $\pi/6$) and scaling.

Alternatively, in the regular 12-gon centred at some point, each vertex is obtained from the previous by rotation through $2\pi/12 = \pi/6$. Let us use the multiplicative structure directly.

The vector from $O$ to $P$ is $2$. The vector from $P$ to $Q$ is $\sqrt{3} + i = 2e^{i\pi/6}$. Each successive side is obtained by rotating the previous side by $e^{i\pi/6}$:

• $\vec{OP} = 2$
• $\vec{PQ} = 2e^{i\pi/6}$
• $\vec{QR} = 2e^{i\pi/3}$
• $\vec{RS} = 2e^{i\pi/2} = 2i$

So: $Q = 2 + 2e^{i\pi/6} = 2 + \sqrt{3} + i$ $R = Q + 2e^{i\pi/3} = (2 + \sqrt{3} + i) + (1 + i\sqrt{3}) = 3 + \sqrt{3} + i(1 + \sqrt{3})$ $S = R + 2i = 3 + \sqrt{3} + i(3 + \sqrt{3})$

Therefore:

$S = (3 + \sqrt{3})(1 + i)$

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Finding the centre $C$ of the circle through $O$, $P$, $Q$.

The circumcircle of triangle $OPQ$ has centre $C$ equidistant from $O$, $P$, and $Q$.

Let $C = a + bi$. Then:

From $|CO| = |CP|$:

$a^2 + b^2 = (a - 2)^2 + b^2 \implies a^2 = a^2 - 4a + 4 \implies 4a = 4 \implies a = 1$

From $|CO| = |CQ|$ with $Q = 2 + \sqrt{3} + i$:

$1 + b^2 = (1 - 2 - \sqrt{3})^2 + (b - 1)^2$ $1 + b^2 = (1 + \sqrt{3})^2 + b^2 - 2b + 1$ $1 = 1 + 2\sqrt{3} + 3 - 2b + 1$ $1 = 5 + 2\sqrt{3} - 2b$ $2b = 4 + 2\sqrt{3}$ $b = 2 + \sqrt{3}$

So $C = 1 + (2 + \sqrt{3})i$.

Rotating until $C$ lies on the real axis.

When we rotate anticlockwise about $O$ by angle $\phi$, the new position of $C$ is $Ce^{i\phi}$. We need $\text{Im}(Ce^{i\phi}) = 0$ and $\phi > 0$ minimal.

Write $C = 1 + (2 + \sqrt{3})i$. In polar form:

$|C| = \sqrt{1 + (2 + \sqrt{3})^2} = \sqrt{1 + 4 + 4\sqrt{3} + 3} = \sqrt{8 + 4\sqrt{3}}$

The argument of $C$ is $\theta_C = \tan^{-1}(2 + \sqrt{3})$.

We note that $\tan(5\pi/12) = \tan 75^\circ = 2 + \sqrt{3}$, so $\theta_C = 5\pi/12$.

To bring $C$ onto the real axis, we rotate by $\phi = -\theta_C$ (clockwise, which is negative anticlockwise) to land on the positive real axis, or by $\phi = \pi - \theta_C$ to land on the negative real axis. The question asks for the first time $C$ lands on the real axis as we rotate anticlockwise (positive $\phi$), so:

• $\phi = 2\pi - 5\pi/12 = 19\pi/12$ lands on the positive real axis (large rotation).
• $\phi = \pi - 5\pi/12 = 7\pi/12$ lands on the negative real axis (smaller rotation).

The first (smallest positive) rotation is $\phi = 7\pi/12$.

New position of $S$.

The rotated position of $S$ is:

$S' = S \cdot e^{i \cdot 7\pi/12}$

We have $S = (3 + \sqrt{3})(1 + i)$.

Write $1 + i = \sqrt{2}\,e^{i\pi/4}$, so:

$S' = (3 + \sqrt{3})\sqrt{2}\,e^{i(\pi/4 + 7\pi/12)} = (3 + \sqrt{3})\sqrt{2}\,e^{i \cdot 10\pi/12} = (3 + \sqrt{3})\sqrt{2}\,e^{i \cdot 5\pi/6}$

Now $e^{i \cdot 5\pi/6} = \cos(5\pi/6) + i\sin(5\pi/6) = -\frac{\sqrt{3}}{2} + \frac{i}{2} = \frac{1}{2}(-\sqrt{3} + i)$.

So:

$S' = (3 + \sqrt{3})\sqrt{2} \cdot \frac{1}{2}(-\sqrt{3} + i) = \frac{(3 + \sqrt{3})\sqrt{2}}{2}(-\sqrt{3} + i)$

We simplify $(3 + \sqrt{3})\sqrt{2}$:

$(3 + \sqrt{3})\sqrt{2} = 3\sqrt{2} + \sqrt{6}$

Therefore:

$S' = \frac{3\sqrt{2} + \sqrt{6}}{2}(-\sqrt{3} + i) = -\frac{1}{2}(3\sqrt{2} + \sqrt{6})(\sqrt{3} - i) \qquad \blacksquare$

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Question 8

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

Problem

8 The function $f$ satisfies $f(x + 1) = f(x)$ and $f(x) > 0$ for all $x$.

(i) Give an example of such a function.

(ii) The function $F$ satisfies $\frac{\mathrm{d}F}{\mathrm{d}x} = f(x)$ and $F(0) = 0$. Show that $F(n) = nF(1)$, for any positive integer $n$.

(iii) Let $y$ be the solution of the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} + f(x)y = 0$ that satisfies $y = 1$ when $x = 0$. Show that $y(n) \to 0$ as $n \to \infty$, where $n = 1, 2, 3, \dots$

Model Solution

Part (i): Example

$f(x) = 2 + \sin 2\pi x$ satisfies $f(x+1) = 2 + \sin 2\pi(x+1) = 2 + \sin(2\pi x + 2\pi) = 2 + \sin 2\pi x = f(x)$, and $f(x) \geq 2 - 1 = 1 > 0$ for all $x$. $\qquad \blacksquare$

(Any positive periodic function with period 1 works, e.g. $f(x) = 2 + \cos 2\pi x$.)

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Part (ii): Showing $F(n) = nF(1)$

Since $F'(x) = f(x)$ and $F(0) = 0$:

$F(n) = \int_0^n f(x)\,dx$

Split the integral into unit intervals:

$F(n) = \int_0^1 f(x)\,dx + \int_1^2 f(x)\,dx + \cdots + \int_{n-1}^n f(x)\,dx = \sum_{k=0}^{n-1} \int_k^{k+1} f(x)\,dx$

In each integral $\int_k^{k+1} f(x)\,dx$, substitute $x = t + k$:

$\int_k^{k+1} f(x)\,dx = \int_0^1 f(t + k)\,dt = \int_0^1 f(t)\,dt = F(1) - F(0) = F(1)$

where we used $f(t + k) = f(t)$ (since $f$ has period 1 and $k$ is an integer).

Therefore:

$F(n) = \sum_{k=0}^{n-1} F(1) = nF(1) \qquad \blacksquare$

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Part (iii): Showing $y(n) \to 0$ as $n \to \infty$

The differential equation is:

$\frac{dy}{dx} + f(x)y = 0 \implies \frac{dy}{dx} = -f(x)y$

This is separable:

$\frac{dy}{y} = -f(x)\,dx$

Integrating from $0$ to $x$:

$\ln|y(x)| - \ln|y(0)| = -\int_0^x f(t)\,dt = -F(x)$

Since $y(0) = 1$:

$\ln y(x) = -F(x) \implies y(x) = e^{-F(x)}$

(We can drop the absolute value since $y(0) = 1 > 0$ and $y$ is continuous, so $y > 0$ everywhere.)

At $x = n$:

$y(n) = e^{-F(n)} = e^{-nF(1)}$

From part (ii). Since $f(x) > 0$ for all $x$, we have $F(1) = \int_0^1 f(x)\,dx > 0$.

Therefore $nF(1) \to \infty$ as $n \to \infty$, and:

$y(n) = e^{-nF(1)} \to 0 \quad \text{as } n \to \infty \qquad \blacksquare$