STEP2 2005 -- Pure Mathematics

Exam: STEP2  |  Year: 2005  |  Questions: Q1—Q8  |  Total marks per question: 20

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

Overview

Q	Topic	Difficulty	Key Techniques
1	微积分 Calculus（求导与多项式构造）	Standard	乘积法则求导,提取公因式,构造微分方程求多项式
2	数论 Number Theory（Euler totient 函数）	Standard	素因数分解,构造反例,代数恒等式
3	微积分 Calculus（定积分与不等式）	Challenging	分部积分,三角恒等式降幂,不等式放缩
4	三角函数 Trigonometry（反三角函数恒等式）	Challenging	反正切加法公式,代数化简,参数赋值验证
5	几何 Geometry（圆与三角形）	Challenging	切线长相等性质,配方法求最值,面积比计算
6	幂级数 Power Series	Challenging	二项式级数展开,代入法求和,逐项微分
7	向量与三维几何 Vectors and 3D Geometry	Challenging	位置向量几何描述,标量积求夹角,三角不等式求时间区间
8	微分方程与曲线描绘 Differential Equations and Curve Sketching	Challenging	分离变量法,换元积分,渐近展开,曲线渐近行为分析

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Question 1

Topic: 微积分 Calculus（求导与多项式构造）  |  Difficulty: Standard  |  Marks: 20

Problem

1 Find the three values of $x$ for which the derivative of $x^2e^{-x^2}$ is zero.

Given that $a$ and $b$ are distinct positive numbers, find a polynomial $P(x)$ such that the derivative of $P(x)e^{-x^2}$ is zero for $x = 0$, $x = \pm a$ and $x = \pm b$, but for no other values of $x$.

Hint

Differentiation leads to f’(x) = 2xe^{-x^2} - 2x^3e^{-x^2}. Since e^{-x^2} is not equal to 0 for any finite x then f’(x) = 0 implies x - x^3 = 0, so x = 0, 1, -1.

For the rest of the question, observe first that P’(x) - 2xP(x) = x(x^2 - a^2)(x^2 - b^2) () within a multiplicative non-zero constant. Thus P(x) can take the form -x^4/2 + px^2 + q and hence substitution into () plus equating the coefficient of x^2 and constant terms leads to a possible result for P(x).

A similar argument based on setting P(x) = sum c_i x^i is feasible, but it involves more working and so is correspondingly more error prone.

Alternatively, one can multiply (*) by e^{-x^2} and then integrate with respect to x to obtain P(x)e^{-x^2} = integral of x(x^2 - a^2)(x^2 - b^2)e^{-x^2} dx. From integral of xe^{-x^2} dx = -(1/2)e^{-x^2} the integrals of x^3e^{-x^2} dx and x^5e^{-x^2} dx can be evaluated by use of the integration by parts rule. It then only remains to cancel out the factor e^{-x^2} to obtain P(x) = -x^4/2 + (a^2/2 + b^2/2 - 1)x^2 - 1 + a^2/2 + b^2/2 - a^2b^2/2.

Model Solution

Part 1: Find the three values of $x$ for which the derivative of $x^2 e^{-x^2}$ is zero.

Let $f(x) = x^2 e^{-x^2}$. By the product rule:

$f'(x) = 2x \cdot e^{-x^2} + x^2 \cdot (-2x)e^{-x^2} = 2x\,e^{-x^2} - 2x^3 e^{-x^2} = 2x\,e^{-x^2}(1 - x^2) \text{.}$

Since $e^{-x^2} > 0$ for all real $x$, the derivative is zero if and only if:

$2x(1 - x^2) = 0 \text{,}$

which gives $x = 0$ or $x^2 = 1$. Hence the three values are:

$x = 0, \quad x = 1, \quad x = -1 \text{.} \qquad \square$

Part 2: Find a polynomial $P(x)$ such that $\frac{d}{dx}\!\left[P(x)e^{-x^2}\right] = 0$ at $x = 0, \pm a, \pm b$ only.

We compute:

$\frac{d}{dx}\!\left[P(x)e^{-x^2}\right] = \left[P'(x) - 2xP(x)\right]e^{-x^2} \text{.}$

Since $e^{-x^2} \neq 0$, this is zero if and only if $P'(x) - 2xP(x) = 0$. We need this to hold at exactly $x = 0, \pm a, \pm b$, so we require:

$P'(x) - 2xP(x) = kx(x^2 - a^2)(x^2 - b^2) \qquad \text{(*)}$

for some non-zero constant $k$. The right side is a degree $5$ polynomial, so $P$ must have degree $4$. Since the right side is an odd function, we try $P(x) = \alpha x^4 + \beta x^2 + \gamma$ (an even polynomial). Then:

$P'(x) - 2xP(x) = (4\alpha x^3 + 2\beta x) - 2x(\alpha x^4 + \beta x^2 + \gamma) = -2\alpha x^5 + (4\alpha - 2\beta)x^3 + (2\beta - 2\gamma)x \text{.}$

Expanding the right side of $(*)$:

$kx(x^2 - a^2)(x^2 - b^2) = kx^5 - k(a^2 + b^2)x^3 + ka^2b^2 x \text{.}$

Matching coefficients of each power of $x$:

$x^5$: $-2\alpha = k$, so $\alpha = -\dfrac{k}{2}$.

$x^3$: $4\alpha - 2\beta = -k(a^2 + b^2)$, so $\beta = 2\alpha + \dfrac{k(a^2 + b^2)}{2} = \dfrac{k(a^2 + b^2 - 2)}{2}$.

$x^1$: $2\beta - 2\gamma = ka^2b^2$, so $\gamma = \beta - \dfrac{ka^2b^2}{2} = \dfrac{k(a^2 + b^2 - a^2b^2 - 2)}{2}$.

Setting $k = -2$ for a clean leading coefficient:

$P(x) = x^4 - (a^2 + b^2 - 2)x^2 + (a^2 + b^2 - a^2b^2 - 2) \text{.}$

Verification. We check $(*)$ with $k = -2$:

$P'(x) - 2xP(x) = 4x^3 - 2(a^2 + b^2 - 2)x - 2x^5 + 2(a^2 + b^2 - 2)x^3 - 2(a^2 + b^2 - a^2b^2 - 2)x$

$= -2x^5 + 2(a^2 + b^2)x^3 - 2a^2b^2 x = -2x(x^2 - a^2)(x^2 - b^2) \text{.}$

Since $a$ and $b$ are distinct positive numbers, the five zeros $x = 0, \pm a, \pm b$ are all distinct, and a degree $5$ polynomial has at most $5$ real zeros. Hence $\frac{d}{dx}[P(x)e^{-x^2}] = 0$ at exactly $x = 0, \pm a, \pm b$. $\qquad \square$

Note: Any non-zero scalar multiple of $P(x)$ also satisfies the requirements (since the equation $P' - 2xP = kx(x^2-a^2)(x^2-b^2)$ holds for any non-zero $k$). The choice $k = -2$ gives the simplest integer coefficients.

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Question 2

Topic: 数论 Number Theory（Euler totient 函数）  |  Difficulty: Standard  |  Marks: 20

Problem

2 For any positive integer $N$, the function $f(N)$ is defined by

$f(N) = N \left( 1 - \frac{1}{p_1} \right) \left( 1 - \frac{1}{p_2} \right) \cdots \left( 1 - \frac{1}{p_k} \right)$

where $p_1, p_2, \dots, p_k$ are the only prime numbers that are factors of $N$. Thus $f(80) = 80(1 - \frac{1}{2})(1 - \frac{1}{5})$.

(a) (i) Evaluate $f(12)$ and $f(180)$.

(ii) Show that $f(N)$ is an integer for all $N$.

(b) Prove, or disprove by means of a counterexample, each of the following:

(i) $f(m)f(n) = f(mn)$;

(ii) $f(p)f(q) = f(pq)$ if $p$ and $q$ are distinct prime numbers;

(iii) $f(p)f(q) = f(pq)$ only if $p$ and $q$ are distinct prime numbers.

(c) Find a positive integer $m$ and a prime number $p$ such that $f(p^m) = 146410$.

Hint

(a) (i) Following the definition of f(N), it is immediate that f(12) = 12(1 - 1/2)(1 - 1/3) = 4, and f(180) = 180(1 - 1/2)(1 - 1/3)(1 - 1/5) = 48.

(ii) The result may seem obvious but care must be taken in order to construct a complete proof. For example, N = p_1^{alpha_1} … p_k^{alpha_k} implies f(N) = p_1^{alpha_1-1} … p_k^{alpha_k-1}(p_1 - 1) … (p_k - 1). Thus as p_i is a positive integer and alpha_i - 1 is a non-negative integer for 1 <= i <= k, then f(N) is an integer.

(b) In each of (i), (ii), (iii), the conclusion must be made clear.

(i) As f(3)f(9) = 2 x 6 = 12 is not equal to f(27) = 18, then the statement is false.

(ii) For any two primes p and q, f(p)f(q) = p(1 - 1/p)q(1 - 1/q) = pq(1 - 1/p)(1 - 1/q) = f(pq). Hence the statement is true.

(iii) Consider f(5) = 4, f(6) = 2, f(30) = 8 = 2 x 4. Then as 6 is not a prime it is clear that the statement is false.

(c) Start with p^{m-1}(p - 1) = 146410, then without difficulty it will be found that p = 11 and m = 5 (not 4).

Model Solution

Part (a)(i) Evaluate $f(12)$ and $f(180)$.

For $N = 12 = 2^2 \times 3$, the prime factors are $2$ and $3$:

$f(12) = 12\left(1 - \frac{1}{2}\right)\!\left(1 - \frac{1}{3}\right) = 12 \cdot \frac{1}{2} \cdot \frac{2}{3} = 4 \text{.}$

For $N = 180 = 2^2 \times 3^2 \times 5$, the prime factors are $2$, $3$, and $5$:

$f(180) = 180\left(1 - \frac{1}{2}\right)\!\left(1 - \frac{1}{3}\right)\!\left(1 - \frac{1}{5}\right) = 180 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} = 48 \text{.}$

Part (a)(ii) Show that $f(N)$ is an integer for all $N$.

Write $N = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ where $p_1, p_2, \dots, p_k$ are distinct primes and $\alpha_i \geqslant 1$. Then:

$f(N) = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} \cdot \left(1 - \frac{1}{p_1}\right)\!\left(1 - \frac{1}{p_2}\right)\cdots\left(1 - \frac{1}{p_k}\right)$

$= p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} \cdot \frac{p_1 - 1}{p_1} \cdot \frac{p_2 - 1}{p_2} \cdots \frac{p_k - 1}{p_k}$

$= p_1^{\alpha_1 - 1} p_2^{\alpha_2 - 1} \cdots p_k^{\alpha_k - 1} \cdot (p_1 - 1)(p_2 - 1) \cdots (p_k - 1) \text{.}$

Since each $\alpha_i \geqslant 1$, every exponent $\alpha_i - 1 \geqslant 0$, and each $p_i - 1 \geqslant 1$. All factors are positive integers, so $f(N)$ is a positive integer. $\qquad \square$

Part (b)(i) Disprove: $f(m)f(n) = f(mn)$ for all $m, n$.

Counterexample. Take $m = 3$, $n = 9$. Then $mn = 27 = 3^3$.

$f(3) = 3\left(1 - \frac{1}{3}\right) = 2, \qquad f(9) = 9\left(1 - \frac{1}{3}\right) = 6, \qquad f(27) = 27\left(1 - \frac{1}{3}\right) = 18 \text{.}$

Since $f(3)f(9) = 2 \times 6 = 12 \neq 18 = f(27)$, the statement is false. $\qquad \square$

Part (b)(ii) Prove: $f(p)f(q) = f(pq)$ if $p$ and $q$ are distinct primes.

Since $p$ and $q$ are distinct primes, $pq$ has prime factors $p$ and $q$:

$f(pq) = pq\left(1 - \frac{1}{p}\right)\!\left(1 - \frac{1}{q}\right) = p\left(1 - \frac{1}{p}\right) \cdot q\left(1 - \frac{1}{q}\right) = f(p) \cdot f(q) \text{.}$

The statement is true. $\qquad \square$

Part (b)(iii) Disprove: $f(p)f(q) = f(pq)$ only if $p$ and $q$ are distinct primes.

We need a counterexample where $f(m)f(n) = f(mn)$ but at least one of $m, n$ is not prime.

Counterexample. Take $m = 5$ (prime), $n = 6 = 2 \times 3$ (not prime). Then $mn = 30 = 2 \times 3 \times 5$.

$f(5) = 5 \cdot \frac{4}{5} = 4, \qquad f(6) = 6 \cdot \frac{1}{2} \cdot \frac{2}{3} = 2, \qquad f(30) = 30 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} = 8 \text{.}$

Since $f(5)f(6) = 4 \times 2 = 8 = f(30)$, yet $6$ is not prime, the statement is false. $\qquad \square$

Part (c) Find $m$ and prime $p$ such that $f(p^m) = 146410$.

For a prime power $p^m$, the only prime factor is $p$, so:

$f(p^m) = p^m\left(1 - \frac{1}{p}\right) = p^{m-1}(p - 1) = 146410 \text{.}$

We factor $146410$: since $146410 = 10 \times 14641$ and $14641 = 11^4$, we get $146410 = 2 \times 5 \times 11^4$.

Since $p$ must be prime and $p - 1$ must divide $146410$, we try $p = 11$:

$p^{m-1}(p - 1) = 11^{m-1} \times 10 = 146410 \implies 11^{m-1} = 14641 = 11^4 \implies m = 5 \text{.}$

Verification: $f(11^5) = 11^4 \times (11 - 1) = 14641 \times 10 = 146410$. $\checkmark$

Therefore $p = 11$ and $m = 5$. $\qquad \square$

Note: No other solution exists. If $p = 2$, then $p - 1 = 1$ and $p^{m-1} = 146410$, but $146410$ is not a power of $2$. If $p = 5$, then $p - 1 = 4$ and $p^{m-1} = 36602.5$, which is not an integer. For $p > 11$, we would need $p^{m-1}(p-1) = 146410$, but $p - 1 \geqslant 12$ would require $p^{m-1} \leqslant 12200$, and checking small values of $m$ yields no prime solutions.

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Question 3

Topic: 微积分 Calculus（定积分与不等式）  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Give a sketch, for $0 \leqslant x \leqslant \pi/2$, of the curve

$y = (\sin x - x \cos x) \text{ ,}$

and show that $0 \leqslant y \leqslant 1$.

Show that:

(i) $\int_0^{\pi/2} y \, dx = 2 - \frac{\pi}{2}$ ;

(ii) $\int_0^{\pi/2} y^2 \, dx = \frac{\pi^3}{48} - \frac{\pi}{8}$ .

Deduce that $\pi^3 + 18\pi < 96$.

Hint

Here dy/dx = x sin x which is zero at x = 0 and is positive for 0 < x <= pi/2. A further differentiation will show that d^2y/dx^2 = 0 at x = 0 and positive for 0 < x <= pi/2. Also, as y(0) = 0 and y(pi/2) = 1, then 0 <= y <= 1 for 0 <= x <= pi/2, and the sketch can now be completed consistently with the above conclusions.

(i) integral from 0 to pi/2 of sin x dx = 1 and use of the integration by parts rule will show that integral from 0 to pi/2 of x cos x dx = pi/2 - 1. The displayed result for integral from 0 to pi/2 of y dx then follows immediately.

(ii) Start with integral from 0 to pi/2 of y^2 dx = integral of sin^2 x dx - integral of x sin 2x dx + integral of x^2 cos^2 x dx.

Next, express sin^2 x and cos^2 x in terms of cos 2x so that now there are only two essentially different integrals involving trigonometric terms, namely, integral of x sin 2x dx and integral of x^2 cos 2x dx. The second of these can be obtained from the first, again by application of the integration by parts rule. A correct application of this rule to the first integral and the careful collection of terms will lead to the displayed result.

For the final result, begin with the observation that y^2 < y for 0 < x < pi/2. From this, it is immediate that integral of y^2 dx < integral of y dx and hence that pi^3/48 - pi/8 < 2 - pi/2. The final displayed result can then be obtained without difficulty.

Model Solution

We have $y = \sin x - x\cos x$ for $0 \leqslant x \leqslant \pi/2$.

Derivative. By the product rule,

$\frac{dy}{dx} = \cos x - (\cos x - x\sin x) = x\sin x \text{.}$

For $0 < x \leqslant \pi/2$, we have $x > 0$ and $\sin x > 0$, so $\frac{dy}{dx} > 0$. At $x = 0$, $\frac{dy}{dx} = 0$. Thus $y$ is non-decreasing on $[0, \pi/2]$ with $y(0) = \sin 0 - 0\cdot\cos 0 = 0$ and $y(\pi/2) = \sin(\pi/2) - \frac{\pi}{2}\cos(\pi/2) = 1$. Since $y$ increases from $0$ to $1$, we have $0 \leqslant y \leqslant 1$.

The second derivative is $\frac{d^2y}{dx^2} = \sin x + x\cos x$, which is positive for $0 < x \leqslant \pi/2$, so the curve is concave up throughout. The sketch shows a curve starting at the origin with zero gradient, increasing monotonically with increasing slope, reaching $y = 1$ at $x = \pi/2$.

Part (i)

$\int_0^{\pi/2} y\,dx = \int_0^{\pi/2} \sin x\,dx - \int_0^{\pi/2} x\cos x\,dx \text{.}$

The first integral is $\int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = 0-(-1) = 1$.

For the second integral, integrate by parts with $u = x$, $dv = \cos x\,dx$ (so $du = dx$, $v = \sin x$):

$\int_0^{\pi/2} x\cos x\,dx = [x\sin x]_0^{\pi/2} - \int_0^{\pi/2} \sin x\,dx = \frac{\pi}{2} - 1 \text{.}$

Therefore

$\int_0^{\pi/2} y\,dx = 1 - \left(\frac{\pi}{2} - 1\right) = 2 - \frac{\pi}{2} \text{.} \qquad \square$

Part (ii) We expand $y^2$:

$y^2 = \sin^2 x - 2x\sin x\cos x + x^2\cos^2 x = \sin^2 x - x\sin 2x + x^2\cos^2 x \text{.}$

Using $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$:

$\int_0^{\pi/2} y^2\,dx = \underbrace{\int_0^{\pi/2} \frac{1-\cos 2x}{2}\,dx}_{I_1} - \underbrace{\int_0^{\pi/2} x\sin 2x\,dx}_{I_2} + \underbrace{\int_0^{\pi/2} \frac{x^2(1+\cos 2x)}{2}\,dx}_{I_3} \text{.}$

Evaluating $I_1$:

$I_1 = \frac{1}{2}\!\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4} \text{.}$

Evaluating $I_2$: Integrate by parts with $u = x$, $dv = \sin 2x\,dx$ (so $du = dx$, $v = -\frac{\cos 2x}{2}$):

$I_2 = \left[-\frac{x\cos 2x}{2}\right]_0^{\pi/2} + \int_0^{\pi/2} \frac{\cos 2x}{2}\,dx = -\frac{\frac{\pi}{2}\cdot(-1)}{2} + \left[\frac{\sin 2x}{4}\right]_0^{\pi/2} = \frac{\pi}{4} + 0 = \frac{\pi}{4} \text{.}$

Evaluating $I_3$:

$I_3 = \frac{1}{2}\int_0^{\pi/2} x^2\,dx + \frac{1}{2}\int_0^{\pi/2} x^2\cos 2x\,dx = \frac{\pi^3}{48} + \frac{1}{2}\int_0^{\pi/2} x^2\cos 2x\,dx \text{.}$

For $\int_0^{\pi/2} x^2\cos 2x\,dx$, integrate by parts with $u = x^2$, $dv = \cos 2x\,dx$ (so $du = 2x\,dx$, $v = \frac{\sin 2x}{2}$):

$\int_0^{\pi/2} x^2\cos 2x\,dx = \left[\frac{x^2\sin 2x}{2}\right]_0^{\pi/2} - \int_0^{\pi/2} x\sin 2x\,dx = 0 - I_2 = -\frac{\pi}{4} \text{.}$

Therefore $I_3 = \frac{\pi^3}{48} + \frac{1}{2}\!\left(-\frac{\pi}{4}\right) = \frac{\pi^3}{48} - \frac{\pi}{8}$.

Combining:

$\int_0^{\pi/2} y^2\,dx = I_1 - I_2 + I_3 = \frac{\pi}{4} - \frac{\pi}{4} + \frac{\pi^3}{48} - \frac{\pi}{8} = \frac{\pi^3}{48} - \frac{\pi}{8} \text{.} \qquad \square$

Deduction. For $0 < x < \pi/2$ we have $0 < y < 1$, so $y^2 < y$. Integrating over $[0, \pi/2]$:

$\int_0^{\pi/2} y^2\,dx < \int_0^{\pi/2} y\,dx$

$\frac{\pi^3}{48} - \frac{\pi}{8} < 2 - \frac{\pi}{2}$

Multiplying both sides by $48$:

$\pi^3 - 6\pi < 96 - 24\pi$

$\pi^3 + 18\pi < 96 \text{.} \qquad \square$

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Question 4

Topic: 三角函数 Trigonometry（反三角函数恒等式）  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The positive numbers $a$, $b$ and $c$ satisfy $bc = a^2 + 1$. Prove that

$\arctan \left( \frac{1}{a+b} \right) + \arctan \left( \frac{1}{a+c} \right) = \arctan \left( \frac{1}{a} \right) .$

The positive numbers $p, q, r, s, t, u$ and $v$ satisfy

$st = (p + q)^2 + 1 , \quad uv = (p + r)^2 + 1 , \quad qr = p^2 + 1 .$

Prove that

$\arctan \left( \frac{1}{p + q + s} \right) + \arctan \left( \frac{1}{p + q + t} \right) + \arctan \left( \frac{1}{p + r + u} \right) + \arctan \left( \frac{1}{p + r + v} \right) = \arctan \left( \frac{1}{p} \right) .$

Hence show that

$\arctan \left( \frac{1}{13} \right) + \arctan \left( \frac{1}{21} \right) + \arctan \left( \frac{1}{82} \right) + \arctan \left( \frac{1}{187} \right) = \arctan \left( \frac{1}{7} \right) .$

[Note that $\arctan x$ is another notation for $\tan^{-1} x$.]

Hint

The first two parts of this question depend on the identity tan^{-1} A + tan^{-1} B = tan^{-1}[(A + B)/(1 - AB)] which is simply another way of writing tan(alpha + beta) = [tan alpha + tan beta]/[1 - tan alpha tan beta].

In the second part, it follows from the data that

tan^{-1}[1/(p + q + s)] + tan^{-1}[1/(p + q + t)] = tan^{-1}[1/(p + q)]

and

tan^{-1}[1/(p + r + u)] + tan^{-1}[1/(p + r + v)] = tan^{-1}[1/(p + r)].

As also from the data,

tan^{-1}[1/(p + q)] + tan^{-1}[1/(p + r)] = tan^{-1}(1/p),

then the proof is complete.

For the final part, it is clear that $p = 7$ and this leads to $st = (7 + q)^2 + 1$, $uv = (7 + r)^2 + 1$, $qr = 50$. Pair the smaller targets with $r$: from $p+r+u = 13$ and $p+r+v = 21$, we have $r+u = 6$ and $r+v = 14$, so $(6-r)(14-r) = (7+r)^2 + 1$, which gives $r = 1$, and hence $u = 5$, $v = 13$. Similarly, pairing the larger targets with $q$ gives $q+s = 75$, $q+t = 180$, so $(75-q)(180-q) = (7+q)^2 + 1$, giving $q = 50$, $s = 25$, $t = 130$.

The solution given above is not unique. Moreover, other plausible strategies may lead to incorrect solutions. It is important, therefore, to check that the solution obtained not only satisfies the displayed identity, but also the given conditions.

Model Solution

We use the identity $\arctan A + \arctan B = \arctan\!\dfrac{A+B}{1-AB}$, valid when $AB < 1$.

Part 1. Let $A = \dfrac{1}{a+b}$ and $B = \dfrac{1}{a+c}$. Then:

$A + B = \frac{a+c+a+b}{(a+b)(a+c)} = \frac{2a+b+c}{(a+b)(a+c)}$

$1 - AB = 1 - \frac{1}{(a+b)(a+c)} = \frac{(a+b)(a+c)-1}{(a+b)(a+c)} = \frac{a^2+ab+ac+bc-1}{(a+b)(a+c)}$

Using $bc = a^2+1$:

$a^2 + ab + ac + bc - 1 = a^2 + ab + ac + a^2 + 1 - 1 = 2a^2 + ab + ac = a(2a+b+c)$

Therefore:

$\frac{A+B}{1-AB} = \frac{2a+b+c}{a(2a+b+c)} = \frac{1}{a}$

Since $a, b, c > 0$, we have $AB = \frac{1}{(a+b)(a+c)} < 1$ (as $(a+b)(a+c) > 1$), so:

$\arctan\frac{1}{a+b} + \arctan\frac{1}{a+c} = \arctan\frac{1}{a} \text{.} \qquad \square$

Part 2. We group the four terms into two pairs and apply the result from Part 1.

First pair: $\arctan\dfrac{1}{p+q+s} + \arctan\dfrac{1}{p+q+t}$.

This has the form $\arctan\dfrac{1}{a'+b'} + \arctan\dfrac{1}{a'+c'}$ with $a' = p+q$, $b' = s$, $c' = t$. The condition $b'c' = a'^2+1$ becomes $st = (p+q)^2+1$, which holds by hypothesis. By Part 1:

$\arctan\frac{1}{p+q+s} + \arctan\frac{1}{p+q+t} = \arctan\frac{1}{p+q} \text{.} \qquad \text{(*)}$

Second pair: $\arctan\dfrac{1}{p+r+u} + \arctan\dfrac{1}{p+r+v}$.

Similarly, with $a' = p+r$, $b' = u$, $c' = v$ and $uv = (p+r)^2+1$:

$\arctan\frac{1}{p+r+u} + \arctan\frac{1}{p+r+v} = \arctan\frac{1}{p+r} \text{.} \qquad \text{(**)}$

Combining: Adding $(*)$ and $(**)$, we need $\arctan\dfrac{1}{p+q} + \arctan\dfrac{1}{p+r}$. Applying Part 1 once more with $a = p$, $b = q$, $c = r$ and the condition $qr = p^2+1$:

$\arctan\frac{1}{p+q} + \arctan\frac{1}{p+r} = \arctan\frac{1}{p} \text{.} \qquad \square$

Numerical verification. We seek $p, q, r, s, t, u, v$ satisfying the three constraints and matching the four target arguments. Setting $p = 7$ and assigning:

$p+r+u = 13, \quad p+r+v = 21, \quad p+q+s = 82, \quad p+q+t = 187.$

First pair (with $a' = p+r$): $r+u = 6$ and $r+v = 14$, so $u = 6-r$ and $v = 14-r$. From $uv = (7+r)^2+1$:

$(6-r)(14-r) = (7+r)^2+1$

$r^2 - 20r + 84 = r^2 + 14r + 50$

$34 = 34r \implies r = 1$

So $u = 5$ and $v = 13$.

Second pair (with $a' = p+q$): $q+s = 75$ and $q+t = 180$, so $s = 75-q$ and $t = 180-q$. From $st = (7+q)^2+1$:

$(75-q)(180-q) = (7+q)^2+1$

$q^2 - 255q + 13500 = q^2 + 14q + 50$

$13450 = 269q \implies q = 50$

So $s = 25$ and $t = 130$.

Verification of all three constraints:

• $qr = 50 \times 1 = 50 = 7^2 + 1$ $\checkmark$
• $st = 25 \times 130 = 3250 = (7+50)^2 + 1 = 3250$ $\checkmark$
• $uv = 5 \times 13 = 65 = (7+1)^2 + 1 = 65$ $\checkmark$

Verification of all four arguments:

• $p+r+u = 7+1+5 = 13$ $\checkmark$
• $p+r+v = 7+1+13 = 21$ $\checkmark$
• $p+q+s = 7+50+25 = 82$ $\checkmark$
• $p+q+t = 7+50+130 = 187$ $\checkmark$

Since addition is commutative, the four-term sum equals $\arctan\frac{1}{13}+\arctan\frac{1}{21}+\arctan\frac{1}{82}+\arctan\frac{1}{187}$, which by the proved identity equals $\arctan\frac{1}{7}$. $\qquad \square$

Note: The correct assignment pairs the smaller targets ($13$, $21$) with $r = 1$ and the larger targets ($82$, $187$) with $q = 50$. All seven values $p=7$, $q=50$, $r=1$, $s=25$, $t=130$, $u=5$, $v=13$ are positive, as required.

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Question 5

Topic: 几何 Geometry（圆与三角形）  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The angle $A$ of triangle $ABC$ is a right angle and the sides $BC, CA$ and $AB$ are of lengths $a$, $b$ and $c$, respectively. Each side of the triangle is tangent to the circle $S_1$ which is of radius $r$. Show that $2r = b + c - a$.

Each vertex of the triangle lies on the circle $S_2$. The ratio of the area of the region between $S_1$ and the triangle to the area of $S_2$ is denoted by $R$. Show that

$\pi R = -(\pi - 1)q^2 + 2\pi q - (\pi + 1) ,$

where $q = \frac{b+c}{a}$. Deduce that

$R \le \frac{1}{\pi(\pi - 1)} .$

Hint

At the outset it should be emphasised that a large, well annotated diagram will enable insight into this question.

The first result may be obtained expeditiously by observing that if S_1 touches the sides BC, CA, AB at P, Q, R, respectively, then AQ = AR = r implies BR = c - r, CQ = b - r. Thus b - r + c - r = a implies 2r = b + c - a.

This result leads to r = a(q - 1)/2 which is the key to the remainder of the question. In fact R = [2bc - pi a^2(q - 1)^2]/(pi a^2) (*).

From the data a^2 = b^2 + c^2 implies 2bc = (a + b + c)(b + c - a) implies bc/a^2 = (q^2 - 1)/2 which together with (*) leads to the second displayed result.

The obtaining of the turning value of a quadratic function is routine. In this context the method of completion of the square is to be preferred to the use of the calculus. Where the critical value of q is obtained from dR/dq = 0, it is important to give a reason as to why this defines an upper bound for R.

Model Solution

Show that $2r = b + c - a$.

Let $S_1$ touch $BC$, $CA$, $AB$ at $P$, $Q$, $R$ respectively. Since the two tangents from an external point to a circle are equal in length:

$AQ = AR = r, \quad BQ = BP, \quad CP = CQ.$

From these:

$BR = AB - AR = c - r, \quad CQ = CA - AQ = b - r.$

Since $BP = BR$ and $CP = CQ$:

$a = BC = BP + CP = BR + CQ = (c - r) + (b - r) = b + c - 2r.$

Therefore $2r = b + c - a$. $\qquad \square$

Show that $\pi R = -(\pi - 1)q^2 + 2\pi q - (\pi + 1)$.

Since $A$ is a right angle, by Pythagoras $a^2 = b^2 + c^2$, so $BC$ is the diameter of the circumcircle $S_2$. The circumradius is $a/2$, giving:

$\text{Area}(S_2) = \pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{4}.$

The area of triangle $ABC$ is $\frac{1}{2}bc$, and the area of $S_1$ is $\pi r^2$. Thus:

$R = \frac{\frac{1}{2}bc - \pi r^2}{\frac{\pi a^2}{4}} = \frac{2bc - \pi a^2(q-1)^2}{\pi a^2}$

where we substituted $r = \frac{a(q-1)}{2}$ from $2r = b + c - a = a(q - 1)$.

We now express $bc/a^2$ in terms of $q$. From $a^2 = b^2 + c^2$:

$(b+c)^2 = b^2 + 2bc + c^2 = a^2 + 2bc$

$\Longrightarrow \quad 2bc = (b+c)^2 - a^2 = a^2q^2 - a^2 = a^2(q^2 - 1)$

$\Longrightarrow \quad \frac{bc}{a^2} = \frac{q^2 - 1}{2}.$

Substituting into the expression for $\pi R$:

$\pi R = \frac{2bc}{a^2} - \pi(q-1)^2 = (q^2 - 1) - \pi(q-1)^2.$

Expanding $(q-1)^2 = q^2 - 2q + 1$:

$\pi R = q^2 - 1 - \pi q^2 + 2\pi q - \pi = -(\pi - 1)q^2 + 2\pi q - (\pi + 1). \qquad \square$

Deduce that $R \leqslant \frac{1}{\pi(\pi - 1)}$.

We have $\pi R = -(\pi - 1)q^2 + 2\pi q - (\pi + 1)$. Since $\pi - 1 > 0$, this is a downward-opening quadratic in $q$ with a maximum. Completing the square:

$\pi R = -(\pi - 1)\!\left[q^2 - \frac{2\pi}{\pi - 1}\,q\right] - (\pi + 1)$

$= -(\pi - 1)\!\left[q - \frac{\pi}{\pi - 1}\right]^2 + \frac{\pi^2}{\pi - 1} - (\pi + 1).$

The constant term simplifies:

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

So:

$\pi R = -(\pi - 1)\!\left[q - \frac{\pi}{\pi - 1}\right]^2 + \frac{1}{\pi - 1}.$

Since the squared term is non-negative, $\pi R \leqslant \frac{1}{\pi - 1}$, and therefore:

$R \leqslant \frac{1}{\pi(\pi - 1)}. \qquad \square$

The completed square gives the required upper bound; the unconstrained vertex $q = \frac{\pi}{\pi - 1}$ lies outside the feasible range $1 < q \leqslant \sqrt{2}$ for a right triangle with legs $b, c$.

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Question 6

Topic: 幂级数 Power Series  |  Difficulty: Challenging  |  Marks: 20

Problem

6 (i) Write down the general term in the expansion in powers of $x$ of $(1 - x)^{-1}$, $(1 - x)^{-2}$ and $(1 - x)^{-3}$, where $|x| < 1$.

Evaluate $\displaystyle \sum_{n=1}^{\infty} n 2^{-n}$ and $\displaystyle \sum_{n=1}^{\infty} n^2 2^{-n} .$

(ii) Show that $\displaystyle (1 - x)^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^2} \frac{x^n}{2^{2n}}$, for $|x| < 1$.

Evaluate $\displaystyle \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^2 2^{2n} 3^n}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{n(2n)!}{(n!)^2 2^{2n} 3^n} .$

Hint

(i) The power series representations of $(1 + x)^{-k}$ for $k = 1, 2, 3$ are standard and should be well known by any candidate for this examination. In fact the general terms of these three series are $x^n$, $(n + 1)x^n$, $(1/2)(n + 1)(n + 2)x^n$, respectively.

The displayed series may be summed by using the general terms obtained. Thus, $\sum_{n=1}^{\infty} n 2^{-n} = (1/2)(1 - 1/2)^{-2} = 2$, and as $\sum_{n=1}^{\infty} n(n + 1)2^{-n} = 8$, then $\sum_{n=1}^{\infty} n^2 2^{-n} = 8 - 2 = 6$.

(ii) The obtaining of the general term of the power series for $(1 - x)^{-1/2}$ is a straightforward application of the binomial series for a general exponent.

To sum the penultimate series, put $x = 1/3$ and to sum the final series, first differentiate with respect to $x$ and then put $x = 1/3$. The sums will be found to be $\sqrt{3/2}$ and $(1/4)\sqrt{3/2}$, respectively.

Model Solution

Part (i)

The general terms in the expansions of $(1-x)^{-k}$ for $|x| < 1$:

For $(1 - x)^{-1}$: the $n$th term (starting at $n = 0$) is $x^n$.

For $(1 - x)^{-2}$: using the identity $(1-x)^{-2} = \frac{d}{dx}(1-x)^{-1}$, the $n$th term is $(n+1)x^n$.

For $(1 - x)^{-3}$: using $(1-x)^{-3} = \frac{1}{2}\frac{d^2}{dx^2}(1-x)^{-1}$, the $n$th term is $\frac{1}{2}(n+1)(n+2)x^n$.

Evaluate $\displaystyle\sum_{n=1}^{\infty} n\,2^{-n}$.

From the expansion $(1-x)^{-2} = \sum_{n=0}^{\infty}(n+1)x^n$, we have:

$\sum_{n=0}^{\infty}(n+1)x^n = \frac{1}{(1-x)^2}$

$\Longrightarrow \quad \sum_{n=0}^{\infty} nx^n + \sum_{n=0}^{\infty} x^n = \frac{1}{(1-x)^2}$

$\Longrightarrow \quad \sum_{n=1}^{\infty} nx^n = \frac{1}{(1-x)^2} - \frac{1}{1-x} = \frac{x}{(1-x)^2}.$

Setting $x = \frac{1}{2}$:

$\sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2. \qquad \text{(a)}$

Evaluate $\displaystyle\sum_{n=1}^{\infty} n^2\,2^{-n}$.

From the expansion $(1-x)^{-3} = \sum_{n=0}^{\infty}\frac{1}{2}(n+1)(n+2)x^n$:

$\sum_{n=0}^{\infty}\frac{1}{2}(n+1)(n+2)x^n = \frac{1}{(1-x)^3}$

$\Longrightarrow \quad \sum_{n=0}^{\infty}(n+1)(n+2)x^n = \frac{2}{(1-x)^3}.$

Expanding $(n+1)(n+2) = n^2 + 3n + 2$:

$\sum_{n=1}^{\infty} n^2 x^n + 3\sum_{n=1}^{\infty} nx^n + 2\sum_{n=0}^{\infty} x^n = \frac{2}{(1-x)^3}$

$\Longrightarrow \quad \sum_{n=1}^{\infty} n^2 x^n = \frac{2}{(1-x)^3} - \frac{3x}{(1-x)^2} - \frac{2}{1-x}.$

Setting $x = \frac{1}{2}$:

$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = \frac{2}{\left(\frac{1}{2}\right)^3} - \frac{3 \cdot \frac{1}{2}}{\left(\frac{1}{2}\right)^2} - \frac{2}{\frac{1}{2}} = 16 - 6 - 4 = 6. \qquad \text{(b)}$

Part (ii)

Show that $\displaystyle (1-x)^{-1/2} = \sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}\frac{x^n}{2^{2n}}$.

Using the general binomial expansion for $(1-x)^{\alpha}$ with $\alpha = -\frac{1}{2}$:

$(1-x)^{-1/2} = \sum_{n=0}^{\infty}\binom{-1/2}{n}(-x)^n = \sum_{n=0}^{\infty}\binom{-1/2}{n}(-1)^n x^n.$

We compute:

$\binom{-1/2}{n} = \frac{(-\tfrac{1}{2})(-\tfrac{1}{2} - 1)(-\tfrac{1}{2} - 2)\cdots(-\tfrac{1}{2} - n + 1)}{n!}$

$= \frac{(-1)^n \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdots \frac{2n-1}{2}}{n!} = \frac{(-1)^n}{2^n \cdot n!}\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1).$

Therefore:

$\binom{-1/2}{n}(-1)^n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^n \cdot n!}.$

The product $1 \cdot 3 \cdot 5 \cdots (2n-1)$ equals $\frac{(2n)!}{2^n \cdot n!}$, since $(2n)! = [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots 2n] = [1 \cdot 3 \cdots (2n-1)] \cdot 2^n \cdot n!$.

So:

$\binom{-1/2}{n}(-1)^n = \frac{(2n)!}{2^n \cdot n! \cdot 2^n \cdot n!} = \frac{(2n)!}{4^n(n!)^2} = \frac{(2n)!}{(n!)^2 \cdot 2^{2n}}.$

Thus:

$(1-x)^{-1/2} = \sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}\frac{x^n}{2^{2n}} \qquad \text{for } |x| < 1. \qquad \square$

Evaluate $\displaystyle\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2\,2^{2n}\,3^n}$.

Setting $x = \frac{1}{3}$ in the expansion:

$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2\,2^{2n}} \cdot \frac{1}{3^n} = \left(1 - \frac{1}{3}\right)^{-1/2} = \left(\frac{2}{3}\right)^{-1/2} = \left(\frac{3}{2}\right)^{1/2}.$

$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2\,2^{2n}\,3^n} = \sqrt{\frac{3}{2}}. \qquad \text{(c)}$

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{n(2n)!}{(n!)^2\,2^{2n}\,3^n}$.

Differentiating the expansion term by term with respect to $x$:

$\frac{d}{dx}(1-x)^{-1/2} = \frac{1}{2}(1-x)^{-3/2} = \sum_{n=1}^{\infty}\frac{n(2n)!}{(n!)^2\,2^{2n}}\,x^{n-1}.$

Multiplying both sides by $x$:

$\frac{x}{2}(1-x)^{-3/2} = \sum_{n=1}^{\infty}\frac{n(2n)!}{(n!)^2\,2^{2n}}\,x^{n}.$

Setting $x = \frac{1}{3}$:

$\sum_{n=1}^{\infty}\frac{n(2n)!}{(n!)^2\,2^{2n}\,3^n} = \frac{1}{2}\cdot\frac{1}{3}\cdot\left(\frac{2}{3}\right)^{-3/2} = \frac{1}{6}\left(\frac{3}{2}\right)^{3/2}.$

Since $\left(\frac{3}{2}\right)^{3/2} = \frac{3}{2}\cdot\sqrt{\frac{3}{2}}$:

$\sum_{n=1}^{\infty}\frac{n(2n)!}{(n!)^2\,2^{2n}\,3^n} = \frac{1}{6}\cdot\frac{3}{2}\cdot\sqrt{\frac{3}{2}} = \frac{1}{4}\sqrt{\frac{3}{2}}. \qquad \text{(d)}$

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Question 7

Topic: 向量与三维几何 Vectors and 3D Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

7 The position vectors, relative to an origin $O$, at time $t$ of the particles $P$ and $Q$ are

$\cos t \mathbf{i} + \sin t \mathbf{j} + 0 \mathbf{k} \quad \text{and} \quad \cos(t + \frac{1}{4}\pi) \left[ \frac{3}{2}\mathbf{i} + \frac{3\sqrt{3}}{2}\mathbf{k} \right] + 3 \sin(t + \frac{1}{4}\pi) \mathbf{j} \text{ ,}$

respectively, where $0 \leqslant t \leqslant 2\pi$.

(i) Give a geometrical description of the motion of $P$ and $Q$.

(ii) Let $\theta$ be the angle $POQ$ at time $t$ that satisfies $0 \leqslant \theta \leqslant 2\pi$. Show that

$\cos \theta = \frac{3\sqrt{2}}{8} - \frac{1}{4} \cos(2t + \frac{1}{4}\pi) \text{.}$

(iii) Show that the total time for which $\theta \geqslant \frac{1}{4}\pi$ is $\frac{3}{2}\pi$.

Hint

(i) The absence of a k component in the specification of the locus of P, shows immediately that its motion takes place in the plane $z = 0$, i.e. in the $x - y$ plane. Also, it is obvious that $x^2 + y^2 = 1$. Hence P describes a circle centre O and radius 1 in the $x - y$ plane.

For the locus of Q, it is helpful to write $x = (3/2) \cos(t + \pi/4)$, $y = 3 \sin(t + \pi/4)$, $z = (3\sqrt{3}/2) \cos(t + \pi/4)$. It is then evident that $\sqrt{3}x - z = 0$ and so this defines the plane in which the motion of Q takes place. Furthermore, it is clear that $x^2 + y^2 + z^2 = 9$ which shows that the distance of Q from O is constant and equal to 3. Hence Q describes the circle centre O and radius 3 in the plane $\sqrt{3}x - z = 0$.

(ii) Use of the scalar product leads to $\cos \theta = |(1/2) \cos t \cos(t + \pi/4) + \sin t \sin(t + \pi/4)| = ... = |3/(4\sqrt{2}) - (1/4) \cos(2t + \pi/4)|$.

(iii) From the result just obtained, it is immediate that $\theta \geq \pi/4 \Rightarrow -1/\sqrt{2} \leq 3/(4\sqrt{2}) - c/4 \leq 1/\sqrt{2}$ where $c \equiv \cos(2t + \pi/4)$, so that $c$ is restricted to the interval $-1/\sqrt{2} \leq c \leq 1$, and as $c \le 1$, then $t \notin [\pi/4, \pi/2]$ and $t \notin [5\pi/4, 3\pi/2]$ are required. Hence $T = 3\pi/2$.

Model Solution

Part (i)

The position vector of $P$ is $\mathbf{p} = \cos t\,\mathbf{i} + \sin t\,\mathbf{j}$, which has no $\mathbf{k}$ component. Thus $P$ moves in the plane $z = 0$ (the $xy$-plane). Since $x^2 + y^2 = \cos^2 t + \sin^2 t = 1$, the particle $P$ describes a circle of centre $O$ and radius $1$ in the $xy$-plane.

The position vector of $Q$ is $\mathbf{q} = \tfrac{3}{2}\cos(t + \tfrac{\pi}{4})\,\mathbf{i} + 3\sin(t + \tfrac{\pi}{4})\,\mathbf{j} + \tfrac{3\sqrt{3}}{2}\cos(t + \tfrac{\pi}{4})\,\mathbf{k}$. We compute:

$x^2 + y^2 + z^2 = \tfrac{9}{4}\cos^2(t+\tfrac{\pi}{4}) + 9\sin^2(t+\tfrac{\pi}{4}) + \tfrac{27}{4}\cos^2(t+\tfrac{\pi}{4}) = 9\cos^2(t+\tfrac{\pi}{4}) + 9\sin^2(t+\tfrac{\pi}{4}) = 9 \text{,}$

so $Q$ remains at distance $3$ from $O$. Also, the ratio $z/x = \tfrac{3\sqrt{3}/2}{3/2} = \sqrt{3}$ (when $\cos(t+\tfrac{\pi}{4}) \neq 0$), so $z = \sqrt{3}\,x$. Therefore $Q$ describes a circle of centre $O$ and radius $3$ in the plane $\sqrt{3}\,x - z = 0$.

Part (ii)

We compute $\cos\theta$ using $\cos\theta = \dfrac{\mathbf{p}\cdot\mathbf{q}}{|\mathbf{p}||\mathbf{q}|}$.

Since $|\mathbf{p}| = 1$ and $|\mathbf{q}| = 3$:

$\mathbf{p}\cdot\mathbf{q} = \tfrac{3}{2}\cos t\cos(t+\tfrac{\pi}{4}) + 3\sin t\sin(t+\tfrac{\pi}{4}) + 0 \text{.}$

Using the product-to-sum identities $\cos A\cos B = \tfrac{1}{2}[\cos(A-B)+\cos(A+B)]$ and $\sin A\sin B = \tfrac{1}{2}[\cos(A-B)-\cos(A+B)]$:

$\cos t\cos(t+\tfrac{\pi}{4}) = \tfrac{1}{2}[\cos(-\tfrac{\pi}{4}) + \cos(2t+\tfrac{\pi}{4})] = \tfrac{1}{2}[\cos\tfrac{\pi}{4} + \cos(2t+\tfrac{\pi}{4})]$

$\sin t\sin(t+\tfrac{\pi}{4}) = \tfrac{1}{2}[\cos(-\tfrac{\pi}{4}) - \cos(2t+\tfrac{\pi}{4})] = \tfrac{1}{2}[\cos\tfrac{\pi}{4} - \cos(2t+\tfrac{\pi}{4})]$

Substituting:

$\mathbf{p}\cdot\mathbf{q} = \tfrac{3}{2}\cdot\tfrac{1}{2}[\cos\tfrac{\pi}{4} + \cos(2t+\tfrac{\pi}{4})] + 3\cdot\tfrac{1}{2}[\cos\tfrac{\pi}{4} - \cos(2t+\tfrac{\pi}{4})]$

$= \tfrac{3}{4}\cos\tfrac{\pi}{4} + \tfrac{3}{4}\cos(2t+\tfrac{\pi}{4}) + \tfrac{3}{2}\cos\tfrac{\pi}{4} - \tfrac{3}{2}\cos(2t+\tfrac{\pi}{4})$

$= \tfrac{9}{4}\cos\tfrac{\pi}{4} - \tfrac{3}{4}\cos(2t+\tfrac{\pi}{4}) = \tfrac{9\sqrt{2}}{8} - \tfrac{3}{4}\cos(2t+\tfrac{\pi}{4}) \text{.}$

Therefore:

$\cos\theta = \frac{\mathbf{p}\cdot\mathbf{q}}{3} = \frac{3\sqrt{2}}{8} - \frac{1}{4}\cos(2t+\tfrac{\pi}{4}) \text{.} \qquad \square$

Part (iii)

Since $\theta \in [0, \pi]$, the condition $\theta \geqslant \tfrac{\pi}{4}$ is equivalent to $\cos\theta \leqslant \cos\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}$. Substituting:

$\frac{3\sqrt{2}}{8} - \frac{1}{4}\cos(2t+\tfrac{\pi}{4}) \leqslant \frac{\sqrt{2}}{2}$

$-\frac{1}{4}\cos(2t+\tfrac{\pi}{4}) \leqslant \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{8} = \frac{\sqrt{2}}{8}$

$\cos(2t+\tfrac{\pi}{4}) \geqslant -\frac{\sqrt{2}}{2} \text{.}$

We now find the values of $t \in [0, 2\pi]$ for which this fails, i.e., where $\cos(2t+\tfrac{\pi}{4}) < -\tfrac{\sqrt{2}}{2}$. Setting $\varphi = 2t + \tfrac{\pi}{4}$, the condition $\cos\varphi < -\tfrac{\sqrt{2}}{2}$ holds when $\varphi \in (\tfrac{3\pi}{4} + 2k\pi, \tfrac{5\pi}{4} + 2k\pi)$ for integer $k$.

For $t \in [0, 2\pi]$, we have $\varphi \in [\tfrac{\pi}{4}, \tfrac{17\pi}{4}]$. The excluded intervals are:

• $k = 0$: $\varphi \in (\tfrac{3\pi}{4}, \tfrac{5\pi}{4})$, giving $t \in (\tfrac{\pi}{4}, \tfrac{\pi}{2})$.
• $k = 1$: $\varphi \in (\tfrac{11\pi}{4}, \tfrac{13\pi}{4})$, giving $t \in (\tfrac{5\pi}{4}, \tfrac{3\pi}{2})$.

The total excluded time is:

$(\tfrac{\pi}{2} - \tfrac{\pi}{4}) + (\tfrac{3\pi}{2} - \tfrac{5\pi}{4}) = \tfrac{\pi}{4} + \tfrac{\pi}{4} = \tfrac{\pi}{2} \text{.}$

Hence the total time for which $\theta \geqslant \tfrac{\pi}{4}$ is:

$2\pi - \frac{\pi}{2} = \frac{3}{2}\pi \text{.} \qquad \square$

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Question 8

Topic: 微分方程与曲线描绘 Differential Equations and Curve Sketching  |  Difficulty: Challenging  |  Marks: 20

Problem

8 For $x \geqslant 0$ the curve $C$ is defined by

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3 y^2}{(1 + x^2)^{5/2}}$

with $y = 1$ when $x = 0$. Show that

$\frac{1}{y} = \frac{2 + 3x^2}{3(1 + x^2)^{3/2}} + \frac{1}{3}$

and hence that for large positive $x$

$y \approx 3 - \frac{9}{x} \text{.}$

Draw a sketch of $C$.

On a separate diagram draw a sketch of the two curves defined for $x \geqslant 0$ by

$\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{x^3 z^3}{2(1 + x^2)^{5/2}}$

with $z = 1$ at $x = 0$ on one curve, and $z = -1$ at $x = 0$ on the other.

Hint

Separation of variables will lead to $A - 1/y = \int x^3(1 + x^2)^{-5/2} \, dx,$ where A is a constant. There are several possible strategies for the evaluation of the integral on the right; by parts, or by any of such substitutions as $w = 1 + x^2$, $x = \tan t$, $x = \sinh v$. One way or another, the result of this integration correctly carried out will lead to the equivalent of $A - 1/y = -(1/3)x^2(1 + x^2)^{-3/2} - (2/3)(1 + x^2)^{-1/2}.$

Use of the initial condition $y(0) = 1$ will then show that $A = 1/3$ and the required result follows at once: $1/y = 1/3 + (2 + 3x^2)/[3(1 + x^2)^{3/2}].$

To obtain the required approximation for $y$ for large positive $x$, first write $1/y \approx 1/3 + (2 + 3x^2)(1 - 3/2x^2)/3x^3,$ from which it follows that $1/y = 1/3 + 1/x + O(1/x^3)$ and this can easily be worked to the displayed approximation for $y$.

For the sketch, the main features are that it has a zero gradient at $x = 0$ and that for $x > 0$, it is monotonically increasing and has exactly one point of inflexion and that it is asymptotic to the line $y = 3$.

The two given differential equations are related by $y = z^2$. Thus it is unnecessary to solve the second differential equation independently of the first. In any case, the question does not require a formal solution for $z$. Nevertheless, it is helpful to obtain the approximation $z \approx \pm\sqrt{3} \mp 3\sqrt{3}/(2x)$ from $y \approx 3 - 9/x$, for large positive $x$.

All the main features of the $x - z$ sketch may be derived from those listed above for the $x - y$ sketch. The two curves which make up the sketch of $z$ are reflections of each other in the $x$-axis. They start at $(0, \pm 1)$ and are asymptotic to the lines $z = \pm\sqrt{3}$.

Model Solution

Show that $\dfrac{1}{y} = \dfrac{2 + 3x^2}{3(1 + x^2)^{3/2}} + \dfrac{1}{3}$.

We separate variables in $\dfrac{dy}{dx} = \dfrac{x^3 y^2}{(1+x^2)^{5/2}}$:

$\frac{dy}{y^2} = \frac{x^3}{(1+x^2)^{5/2}}\,dx$

Integrating both sides:

$-\frac{1}{y} = \int \frac{x^3}{(1+x^2)^{5/2}}\,dx + C \text{.}$

For the integral on the right, substitute $w = 1 + x^2$, so $dw = 2x\,dx$ and $x^2 = w - 1$:

$\int \frac{x^3}{(1+x^2)^{5/2}}\,dx = \int \frac{x^2 \cdot x}{(1+x^2)^{5/2}}\,dx = \int \frac{w - 1}{2w^{5/2}}\,dw = \frac{1}{2}\int \left(w^{-3/2} - w^{-5/2}\right) dw$

$= \frac{1}{2}\left[-2w^{-1/2} + \frac{2}{3}w^{-3/2}\right] = -w^{-1/2} + \frac{1}{3}w^{-3/2} = -(1+x^2)^{-1/2} + \frac{1}{3}(1+x^2)^{-3/2} \text{.}$

So:

$-\frac{1}{y} = -(1+x^2)^{-1/2} + \frac{1}{3}(1+x^2)^{-3/2} + C \text{.}$

Using the initial condition $y = 1$ when $x = 0$:

$-1 = -1 + \frac{1}{3} + C \implies C = -\frac{1}{3} \text{.}$

Therefore:

$\frac{1}{y} = (1+x^2)^{-1/2} - \frac{1}{3}(1+x^2)^{-3/2} + \frac{1}{3} \text{.}$

Writing $(1+x^2)^{-1/2} = \dfrac{1+x^2}{(1+x^2)^{3/2}}$ to combine the first two terms:

$\frac{1}{y} = \frac{1+x^2}{(1+x^2)^{3/2}} - \frac{1}{3(1+x^2)^{3/2}} + \frac{1}{3} = \frac{3(1+x^2) - 1}{3(1+x^2)^{3/2}} + \frac{1}{3}$

$= \frac{2 + 3x^2}{3(1+x^2)^{3/2}} + \frac{1}{3} \text{.} \qquad \square$

Show that $y \approx 3 - \dfrac{9}{x}$ for large positive $x$.

For large $x$:

$(1+x^2)^{3/2} = x^3\left(1 + \frac{1}{x^2}\right)^{3/2} = x^3\left(1 + \frac{3}{2x^2} + O(x^{-4})\right) \text{,}$

so:

$\frac{2 + 3x^2}{3(1+x^2)^{3/2}} = \frac{3x^2 + 2}{3x^3\left(1 + \frac{3}{2x^2} + \cdots\right)} = \left(\frac{1}{x} + \frac{2}{3x^3}\right)\left(1 - \frac{3}{2x^2} + \cdots\right) = \frac{1}{x} + O(x^{-3}) \text{.}$

Therefore:

$\frac{1}{y} = \frac{1}{3} + \frac{1}{x} + O(x^{-3}) \text{.}$

Inverting:

$y = \frac{1}{\frac{1}{3} + \frac{1}{x} + O(x^{-3})} = \frac{3x}{x + 3 + O(x^{-2})} = \frac{3}{1 + \frac{3}{x} + O(x^{-3})}$

$= 3\left(1 - \frac{3}{x} + O(x^{-2})\right) = 3 - \frac{9}{x} + O(x^{-2}) \text{.}$

Hence $y \approx 3 - \dfrac{9}{x}$ for large positive $x$. $\qquad \square$

Sketch of $C$.

Key features:

• At $x = 0$: $y = 1$, and $\dfrac{dy}{dx} = 0$ (zero gradient at the origin).
• For $x > 0$: $\dfrac{dy}{dx} = \dfrac{x^3 y^2}{(1+x^2)^{5/2}} > 0$, so $y$ is strictly increasing.
• As $x \to \infty$: $y \to 3$ asymptotically, with $y \approx 3 - 9/x$.
• The curve has one point of inflection (the gradient increases from $0$, reaches a maximum, then decreases toward $0$ as $y \to 3$).

The sketch shows a curve starting at $(0, 1)$ with zero slope, rising monotonically, bending to approach the horizontal asymptote $y = 3$ from below.

The $z$-equation and its two curves.

Observe that if $y = z^2$, then $\dfrac{dy}{dx} = 2z\dfrac{dz}{dx}$. The given $z$-equation is:

$\frac{dz}{dx} = \frac{x^3 z^3}{2(1+x^2)^{5/2}} \text{,}$

so $2z\dfrac{dz}{dx} = \dfrac{x^3 z^4}{(1+x^2)^{5/2}} = \dfrac{x^3 y^2}{(1+x^2)^{5/2}} = \dfrac{dy}{dx}$.

Therefore $y = z^2$ satisfies the $y$-equation. Both initial conditions $z = 1$ and $z = -1$ at $x = 0$ give $y = 1$ at $x = 0$, the same initial condition as before.

The two curves are therefore $z = +\sqrt{y}$ and $z = -\sqrt{y}$, where $y$ is the solution found above.

For large positive $x$, using $y \approx 3 - 9/x$:

$z = \pm\sqrt{y} \approx \pm\sqrt{3 - \frac{9}{x}} = \pm\sqrt{3}\sqrt{1 - \frac{3}{x}} \approx \pm\sqrt{3}\left(1 - \frac{3}{2x}\right) = \pm\sqrt{3} \mp \frac{3\sqrt{3}}{2x} \text{.}$

Key features of the $z$-sketch:

• The two curves are reflections of each other in the $x$-axis.
• The upper curve starts at $(0, 1)$ with zero gradient, increases monotonically, and is asymptotic to $z = \sqrt{3}$.
• The lower curve starts at $(0, -1)$ with zero gradient, decreases monotonically, and is asymptotic to $z = -\sqrt{3}$.
• Each curve has one point of inflection, with the same structure as $C$ but reflected/scaled appropriately.