STEP3 2020 -- Pure Mathematics

Exam: STEP3  |  Year: 2020  |  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	Pure Mathematics	Challenging	Integration by parts, compound-angle formulae, induction
2	Pure Mathematics	Challenging	Implicit differentiation, discriminants, hyperbolic functions, curve symmetry
3	纯数	Hard	复数指数形式, 旋转公式, 重心公式, 模长计算
4	纯数	Hard	反射矩阵公式, 矩阵方程求解, 旋转矩阵, 变换复合
5	Pure Mathematics	Challenging	Partial fractions, algebraic identities, differentiation, improper-integral limits
6	Pure Mathematics	Hard	Polar sketches, small-quantity approximations, polar area formulae, trigonometric substitutions
7	Pure Mathematics	Challenging	Reduction of order, integrating factors, parameter matching, initial conditions
8	Pure Mathematics	Hard	Contradiction, induction, recurrence analysis, construction, Euclidean algorithm

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Question 1

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

1 For non-negative integers $a$ and $b$, let $I(a, b) = \int_{0}^{\frac{\pi}{2}} \cos^{a} x \cos bx \ dx.$

(i) Show that for positive integers $a$ and $b$, $I(a, b) = \frac{a}{a + b} I(a - 1, b - 1).$

(ii) Prove by induction on $n$ that for non-negative integers $n$ and $m$, $\int_{0}^{\frac{\pi}{2}} \cos^{n} x \cos(n + 2m + 1)x \ dx = (-1)^{m} \frac{2^{n} n! (2m)! (n + m)!}{m! (2n + 2m + 1)!}.$

Hint

1. (i) Integrating by parts,

$u = \cos^a x \quad v' = \cos bx$ $u' = -a \cos^{a-1} x \sin x \quad v = \frac{1}{b} \sin bx$

$I(a, b) = \left[ \cos^a x \frac{1}{b} \sin bx \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} -a \cos^{a-1} x \sin x \frac{1}{b} \sin bx \ dx$ $= 0 + \int_0^{\frac{\pi}{2}} a \cos^{a-1} x \sin x \frac{1}{b} \sin bx \ dx = \frac{a}{b} \int_0^{\frac{\pi}{2}} \cos^{a-1} x \sin x \sin bx \ dx$

$\cos(b - 1)x = \cos bx \cos x + \sin bx \sin x$

So $I(a, b) = \frac{a}{b} \int_0^{\frac{\pi}{2}} \cos^{a-1} x \ (\cos(b - 1)x - \cos bx \cos x) \ dx$ $= \frac{a}{b} [I(a - 1, b - 1) - I(a, b)]$

Thus $I(a, b) = \frac{a}{a+b} \ I(a - 1, b - 1)$ as required.

(ii) Suppose

$I(k, k + 2m + 1) = (-1)^m \frac{2^k \ k! \ (2m)! \ (k + m)!}{m! \ (2k + 2m + 1)!}$

Then by (i),

$I(k + 1, k + 2m + 2) = \frac{k + 1}{2k + 2m + 3} I(k, k + 2m + 1)$

$= \frac{k + 1}{2k + 2m + 3} \ (-1)^m \frac{2^k \ k! \ (2m)! \ (k + m)!}{m! \ (2k + 2m + 1)!}$

$= \frac{k + 1}{2k + 2m + 3} \ (-1)^m \frac{2^k \ k! \ (2m)! \ (k + m)!}{m! \ (2k + 2m + 1)!} \times \frac{2k + 2m + 2}{2k + 2m + 2}$

$= (-1)^m \frac{2^{k+1} \ (k + 1)! \ (2m)! \ (k + m)!}{m! \ (2k + 2m + 3)!}$

$= (-1)^m \frac{2^{k+1} \ (k + 1)! \ (2m)! \ ((k + 1) + m)!}{m! \ (2(k + 1) + 2m + 1)!}$

which is the required result for $k + 1$

$I(0, 2m + 1) = \int\limits_{0}^{\frac{\pi}{2}} \cos(2m + 1)x \ dx = \frac{1}{2m + 1} \left[ \sin(2m + 1)x \right]_{0}^{\frac{\pi}{2}}$

$= \frac{1}{2m+1} \text{ if m is even or } = \frac{-1}{2m+1} \text{ if m is odd , or alternatively } (-1)^m \frac{1}{2m+1}$

If $n = 0$,

$(-1)^m \frac{2^n \ n! \ (2m)! \ (n + m)!}{m! \ (2n + 2m + 1)!} = (-1)^m \frac{(2m)! \ (m)!}{m! \ (2m + 1)!} = (-1)^m \frac{1}{2m + 1}$

so result is true for $n = 0$.

So by the principle of mathematical induction, the required result is true.

Alternative for (i)

$\cos^a x \cos bx = \cos^{a-1} x \left[ \cos x \cos bx \right]$ $\cos x \cos bx = \frac{1}{2} [\cos(b + 1)x + \cos(b - 1)x]$ $2I(a, b) = I(a - 1, b + 1) + I(a - 1, b - 1)$

Also

$\sin x \sin bx = \frac{1}{2} [\cos(b - 1)x - \cos(b + 1)x]$

so using integration by parts of main scheme,

$2I(a, b) = \frac{a}{b} [I(a - 1, b - 1) - I(a - 1, b + 1)]$

Eliminating $I(a - 1, b + 1)$ between these results gives required result.

Model Solution

Part (i). Integrate by parts, taking

$u=\cos^a x,\qquad dv=\cos bx\,dx.$

Then

$du=-a\cos^{a-1}x\sin x\,dx,\qquad v=\frac{\sin bx}{b}.$

The boundary term is zero, since $\sin bx=0$ at $x=0$ and $\cos^a x=0$ at $x=\frac{\pi}{2}$ for positive $a$. Hence

$$I(a,b)=\frac{a}{b}\int_0^{\pi/2}\cos^{a-1}x\sin x\sin bx\,dx.$$

Using

$\cos(b-1)x=\cos bx\cos x+\sin bx\sin x,$

we get

$\sin x\sin bx=\cos(b-1)x-\cos x\cos bx.$

Therefore

$$I(a,b)=\frac{a}{b}\left(I(a-1,b-1)-I(a,b)\right).$$

Thus

$$(a+b)I(a,b)=aI(a-1,b-1),$$

and so

$$I(a,b)=\frac{a}{a+b}I(a-1,b-1).$$

Part (ii). We prove the result by induction on $n$, with $m$ fixed.

For $n=0$,

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

The proposed formula gives

$$(-1)^m\frac{2^0\,0!\,(2m)!\,m!}{m!(2m+1)!} =\frac{(-1)^m}{2m+1},$$

so the base case holds.

Assume the formula holds for $n=k$, namely

$$I(k,k+2m+1) =(-1)^m\frac{2^k k!(2m)!(k+m)!}{m!(2k+2m+1)!}.$$

Applying part (i) with $a=k+1$ and $b=k+2m+2$ gives

$$I(k+1,k+2m+2) =\frac{k+1}{2k+2m+3}I(k,k+2m+1).$$

Substituting the induction hypothesis,

$$I(k+1,k+2m+2) =(-1)^m \frac{(k+1)2^k k!(2m)!(k+m)!}{m!(2k+2m+3)(2k+2m+1)!}.$$

Since $2k+2m+2=2(k+m+1)$, this is

$$(-1)^m \frac{2^{k+1}(k+1)!(2m)!(k+m+1)!}{m!(2k+2m+3)!},$$

which is the required formula with $n=k+1$. By induction, the formula holds for all non-negative integers $n$.

Examiner Notes

约70%的考生尝试本题，平均分约63%。第(i)部分常见问题是代数和符号错误，尤其是复合角公式的使用。第(ii)部分的归纳法表述常不够规范：遗漏基例、未明确归纳假设，或没有说明非负整数情形应从 $n=0$ 开始。

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Question 2

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

2 The curve $C$ has equation $\sinh x + \sinh y = 2k$, where $k$ is a positive constant.

(i) Show that the curve $C$ has no stationary points and that $\frac{d^{2}y}{dx^{2}} = 0$ at the point $(x, y)$ on the curve if and only if $1 + \sinh x \sinh y = 0.$ Find the co-ordinates of the points of inflection on the curve $C$, leaving your answers in terms of inverse hyperbolic functions.

(ii) Show that if $(x, y)$ lies on the curve $C$ and on the line $x + y = a$, then $e^{2x}(1 - e^{-a}) - 4ke^{x} + (e^{a} - 1) = 0$ and deduce that $1 < \cosh a \leqslant 2k^{2} + 1$.

(iii) Sketch the curve $C$.

Hint

2. (i) $\sinh x + \sinh y = 2k$

Differentiating with respect to $x$, $\cosh x + \cosh y \frac{dy}{dx} = 0$

$\frac{dy}{dx} = 0 \implies \cosh x = 0$ which is not possible as $\cosh x \geq 1 \forall x$, so there are no stationary points.

Differentiating again with respect to $x$, $\sinh x + \sinh y \left(\frac{dy}{dx}\right)^2 + \cosh y \frac{d^2y}{dx^2} = 0$

$\frac{dy}{dx} = \frac{-\cosh x}{\cosh y}$ and $\frac{d^2y}{dx^2} = 0$ implies $\sinh x + \sinh y \left(\frac{-\cosh x}{\cosh y}\right)^2 = 0$

$\cosh^2 y \sinh x + \cosh^2 x \sinh y = 0$

$(1 + \sinh^2 y) \sinh x + (1 + \sinh^2 x) \sinh y = 0$

$(\sinh x + \sinh y)(1 + \sinh x \sinh y) = 0$

But $\sinh x + \sinh y = 2k > 0$ so

$1 + \sinh x \sinh y = 0$

as required.

At a point of inflection, $\frac{d^2y}{dx^2} = 0$, so $\sinh x + \sinh y = 2k$ and $\sinh x \sinh y = -1$ and thus, $\sinh x$ (and $\sinh y$ as well) is a root of $\lambda^2 - 2k\lambda - 1 = 0$

$\lambda = \frac{2k \pm \sqrt{4k^2 + 4}}{2}$

$\sinh x = k + \sqrt{k^2 + 1}$, $\sinh y = \frac{-1}{k + \sqrt{k^2 + 1}} = \frac{-1}{k + \sqrt{k^2 + 1}} \times \frac{k - \sqrt{k^2 + 1}}{k - \sqrt{k^2 + 1}} = \frac{-(k - \sqrt{k^2 + 1})}{k^2 - (k^2 + 1)} = k - \sqrt{k^2 + 1}$ and vice versa.

So the points of inflection are

$(\sinh^{-1}(k + \sqrt{k^2 + 1}), \sinh^{-1}(k - \sqrt{k^2 + 1}))$ and $(\sinh^{-1}(k - \sqrt{k^2 + 1}), \sinh^{-1}(k + \sqrt{k^2 + 1}))$

(ii) $x + y = a \implies y = a - x$ so as $\sinh x + \sinh y = 2k$

$\frac{e^x - e^{-x}}{2} + \frac{e^{a-x} - e^{x-a}}{2} = 2k$

Multiplying by $2e^x$,

$e^{2x} - 1 + e^a - e^{2x}e^{-a} = 4ke^x$

$e^{2x}(1 - e^{-a}) - 4ke^x + (e^a - 1) = 0$

As $e^x$ is real, '$b^2 - 4ac \geq 0$', so $16k^2 - 4(1 - e^{-a})(e^a - 1) \geq 0$

$4k^2 - e^a - e^{-a} + 2 \geq 0$ $4k^2 - 2 \cosh a + 2 \geq 0$

So $\cosh a \leq 2k^2 + 1$

If $a = 0$, then $x = -y$ so $\sinh x = -\sinh y$ and thus $\sinh x + \sinh y = 2k = 0$ but $k > 0$.

So $\cosh a > 1$ as required.

(iii)

Alternative

(i) $\frac{dy}{dx} = \frac{-\cosh x}{\cosh y}$, $\frac{d^2y}{dx^2} = -\left\{ \frac{\cosh y \sinh x - \cosh x \sinh y \frac{dy}{dx}}{\cosh^2 y} \right\} = -\left\{ \frac{\cosh^2 y \sinh x + \cosh^2 x \sinh y}{\cosh^3 y} \right\}$

then as before.

(ii) Substituting $a = 0$ would imply $e^x = 0$ which is impossible.

Model Solution

Part (i). Differentiating

$\sinh x+\sinh y=2k$

with respect to $x$ gives

$$\cosh x+\cosh y\,\frac{dy}{dx}=0,$$

so

$$\frac{dy}{dx}=-\frac{\cosh x}{\cosh y}.$$

Since $\cosh x>0$ and $\cosh y>0$, the derivative is never zero. Hence $C$ has no stationary points.

Differentiating again,

$$\sinh x+\sinh y\left(\frac{dy}{dx}\right)^2+\cosh y\,\frac{d^2y}{dx^2}=0.$$

Thus $\frac{d^2y}{dx^2}=0$ if and only if

$$\sinh x+\sinh y\left(\frac{\cosh^2x}{\cosh^2y}\right)=0.$$

Multiplying by $\cosh^2y$ and using $\cosh^2t=1+\sinh^2t$,

$$(1+\sinh^2y)\sinh x+(1+\sinh^2x)\sinh y=0.$$

This factors as

$$(\sinh x+\sinh y)(1+\sinh x\sinh y)=0.$$

Since $\sinh x+\sinh y=2k>0$, we must have

$$1+\sinh x\sinh y=0.$$

At an inflection point, put $X=\sinh x$ and $Y=\sinh y$. Then

$$X+Y=2k,\qquad XY=-1.$$

So $X$ and $Y$ are the roots of

$$\lambda^2-2k\lambda-1=0,$$

namely

$$\lambda=k\pm\sqrt{k^2+1}.$$

The points of inflection are therefore

$$\left(\sinh^{-1}(k+\sqrt{k^2+1}),\ \sinh^{-1}(k-\sqrt{k^2+1})\right)$$

and

$$\left(\sinh^{-1}(k-\sqrt{k^2+1}),\ \sinh^{-1}(k+\sqrt{k^2+1})\right).$$

Part (ii). If the point also lies on $x+y=a$, then $y=a-x$. Substituting into the curve,

$$\frac{e^x-e^{-x}}{2}+\frac{e^{a-x}-e^{x-a}}{2}=2k.$$

Multiplying by $2e^x$ gives

$$e^{2x}-1+e^a-e^{2x-a}=4ke^x,$$

or

$$e^{2x}(1-e^{-a})-4ke^x+(e^a-1)=0.$$

This is a quadratic in $e^x$, so for a real point of intersection its discriminant must be non-negative:

$$16k^2-4(1-e^{-a})(e^a-1)\geq 0.$$

Now

$$(1-e^{-a})(e^a-1)=e^a+e^{-a}-2=2\cosh a-2.$$

Hence

$$16k^2-4(2\cosh a-2)\geq0,$$

so

$$\cosh a\leq 2k^2+1.$$

Also $a\neq0$, because $x+y=0$ would give $y=-x$ and hence $\sinh x+\sinh y=0$, contradicting $2k>0$. Therefore $\cosh a>1$, and

$$1<\cosh a\leq 2k^2+1.$$

Part (iii). The curve is symmetric in $x$ and $y$, hence symmetric about the line $y=x$. It has no stationary points and has two inflection points found in part (i). Part (ii) shows that a line $x+y=a$ can meet the curve only when

$$1<\cosh a\leq2k^2+1,$$

which gives the limiting range of such diagonal sections. The sketch should show a decreasing curve, symmetric about $y=x$, passing through the two inflection points and lying in the allowed strip determined by the bounds on $x+y$.

Examiner Notes

84%的考生尝试本题，平均分约45%。第(i)(ii)部分整体完成较好，但第(ii)部分中严格下界常被忽略。第(iii)部分作图时，很多答案没有充分利用第(ii)部分的边界条件，拐点和截距标注也不完整。曲线关于 $y=x$ 对称以及相关直线边界是作图的关键。

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Question 3

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

3 Given distinct points $A$ and $B$ in the complex plane, the point $G_{AB}$ is defined to be the centroid of the triangle $ABK$, where the point $K$ is the image of $B$ under rotation about $A$ through a clockwise angle of $\frac{1}{3}\pi$.

Note: if the points $P$, $Q$ and $R$ are represented in the complex plane by $p$, $q$ and $r$, the centroid of triangle $PQR$ is defined to be the point represented by $\frac{1}{3}(p + q + r)$.

(i) If $A$, $B$ and $G_{AB}$ are represented in the complex plane by $a$, $b$ and $g_{ab}$, show that

$g_{ab} = \frac{1}{\sqrt{3}}(\omega a + \omega^* b),$

where $\omega = e^{\frac{i\pi}{6}}$.

(ii) The quadrilateral $Q_1$ has vertices $A$, $B$, $C$ and $D$, in that order, and the quadrilateral $Q_2$ has vertices $G_{AB}$, $G_{BC}$, $G_{CD}$ and $G_{DA}$, in that order. Using the result in part (i), show that $Q_1$ is a parallelogram if and only if $Q_2$ is a parallelogram.

(iii) The triangle $T_1$ has vertices $A$, $B$ and $C$ and the triangle $T_2$ has vertices $G_{AB}$, $G_{BC}$ and $G_{CA}$. Using the result in part (i), show that $T_2$ is always an equilateral triangle.

Hint

3. (i)

$k - a = (b - a)e^{-\frac{i\pi}{3}}$

Therefore,

$g_{AB} = \frac{1}{3} \left[ a + b + \left( a + (b - a)e^{-\frac{i\pi}{3}} \right) \right]$

$= a \left( \frac{2 - e^{-\frac{i\pi}{3}}}{3} \right) + b \left( \frac{1 + e^{-\frac{i\pi}{3}}}{3} \right)$

$\omega = e^{\frac{i\pi}{6}} = \frac{\sqrt{3} + i}{2}$

and so

$\omega^* = \frac{\sqrt{3} - i}{2}$

$\frac{2 - e^{-\frac{i\pi}{3}}}{3} = \frac{2 - \left( \frac{1 - i\sqrt{3}}{2} \right)}{3} = \frac{3 + i\sqrt{3}}{6} = \frac{1}{\sqrt{3}} \frac{\sqrt{3} + i}{2} = \frac{1}{\sqrt{3}} \omega$

and

$\frac{1 + e^{-\frac{i\pi}{3}}}{3} = \frac{1 + \left( \frac{1 - i\sqrt{3}}{2} \right)}{3} = \frac{3 - i\sqrt{3}}{6} = \frac{1}{\sqrt{3}} \omega^*$

Thus $g_{AB} = \frac{1}{\sqrt{3}} (\omega a + \omega^* b)$ as required.

(ii) $g_{AB} = \frac{1}{\sqrt{3}} (\omega a + \omega^* b)$

$g_{BC} = \frac{1}{\sqrt{3}} (\omega b + \omega^* c)$

$g_{CD} = \frac{1}{\sqrt{3}} (\omega c + \omega^* d)$

$g_{DA} = \frac{1}{\sqrt{3}} (\omega d + \omega^* a)$

$Q_1$ parallelogram $\Rightarrow b - a = c - d \Leftrightarrow d - a = c - b$

$g_{BC} - g_{AB} = \frac{1}{\sqrt{3}} (\omega(b - a) + \omega^*(c - b)) = \frac{1}{\sqrt{3}} (\omega(c - d) + \omega^*(d - a)) = g_{CD} - g_{DA}$

$\Rightarrow Q_2$ parallelogram.

$Q_2$ parallelogram $\Rightarrow g_{BC} - g_{AB} = g_{CD} - g_{DA}$

$\frac{1}{\sqrt{3}} \{ \omega[(b - a) - (c - d)] + \omega^*[(c - b) - (d - a)] \} = 0$

$\frac{1}{\sqrt{3}} (\omega^* - \omega) [(a - b) - (d - c)] = 0$

As $\omega^* - \omega \neq 0$, $(a - b) - (d - c) = 0$ and so $Q_1$ is a parallelogram

(iii) $g_{BC} - g_{AB} = \frac{1}{\sqrt{3}}(\omega(b - a) + \omega^*(c - b))$ $g_{CA} - g_{AB} = \frac{1}{\sqrt{3}}(\omega(c - a) + \omega^*(a - b)) \quad (1)$ $\omega^2(g_{BC} - g_{AB}) = \frac{1}{\sqrt{3}}(\omega^3(b - a) + \omega(c - b)) \quad (2)$ $= \frac{1}{\sqrt{3}}(i(b - a) + \omega(c - b))$

The coefficient of $\frac{1}{\sqrt{3}}a$ in (1) is $\omega^* - \omega = \frac{\sqrt{3}-i}{2} - \frac{\sqrt{3}+i}{2} = -i$ The coefficient of $\frac{1}{\sqrt{3}}b$ in (1) is $-\omega^* = (i - \omega)$ The coefficient of $\frac{1}{\sqrt{3}}c$ in (1) is $\omega$

Thus $G_{AB}G_{BC}$ rotated through $\frac{\pi}{3}$ is $G_{AB}G_{CA}$ which means that $G_{AB}G_{BC}G_{CA}$ is an equilateral triangle.

(iii) Alternative

$x = g_{BC} - g_{AB} = \frac{1}{\sqrt{3}}(\omega(b - a) + \omega^*(c - b))$ $y = g_{CA} - g_{AB} = \frac{1}{\sqrt{3}}(\omega(c - a) + \omega^*(a - b))$ $x = \frac{1}{\sqrt{3}}\left(e^{\frac{i\pi}{6}}b - e^{\frac{i\pi}{6}}a + e^{\frac{-i\pi}{6}}c - e^{\frac{-i\pi}{6}}b\right)$ $= \frac{1}{\sqrt{3}}\left(e^{\frac{i\pi}{2}}b + e^{\frac{-i\pi}{6}}c + e^{\frac{i7\pi}{6}}a\right)$ $y = \frac{1}{\sqrt{3}}\left(e^{\frac{i\pi}{6}}c + e^{\frac{i3\pi}{2}}a + e^{\frac{i5\pi}{6}}b\right)$ $\frac{y}{x} = e^{\frac{i\pi}{3}}$

$e^{\frac{i\pi}{3}}$ means $y$ is $x$ rotated through $\frac{\pi}{3}$ and thus $G_{AB}G_{BC}G_{CA}$ is an equilateral triangle.

[or alternatively

$\left| \frac{y}{x} \right| = 1$ and similarly $\left| \frac{z}{y} \right| = 1$ and thus all three sides are equal length]

Model Solution

Part (i)

The point $K$ is obtained by rotating $B$ about $A$ through a clockwise angle of $\frac{\pi}{3}$. A clockwise rotation corresponds to multiplication by $e^{-i\pi/3}$ in the complex plane. To rotate $B$ about $A$:

1. Translate so that $A$ is at the origin: $b - a$.
2. Apply the rotation: $(b - a)e^{-i\pi/3}$.
3. Translate back: $k = a + (b - a)e^{-i\pi/3}$.

The centroid of triangle $ABK$ is:

$$g_{ab} = \frac{1}{3}(a + b + k) = \frac{1}{3}\left(a + b + a + (b - a)e^{-i\pi/3}\right)$$ $$= \frac{1}{3}\left(2a + b + (b - a)e^{-i\pi/3}\right)$$ $$= \frac{1}{3}\left(a(2 - e^{-i\pi/3}) + b(1 + e^{-i\pi/3})\right) \qquad (*)$$

Now evaluate the two coefficients. Since $e^{-i\pi/3} = \cos\frac{\pi}{3} - i\sin\frac{\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2}i$:

$$2 - e^{-i\pi/3} = 2 - \frac{1}{2} + \frac{\sqrt{3}}{2}i = \frac{3}{2} + \frac{\sqrt{3}}{2}i$$ $$1 + e^{-i\pi/3} = 1 + \frac{1}{2} - \frac{\sqrt{3}}{2}i = \frac{3}{2} - \frac{\sqrt{3}}{2}i$$

With $\omega = e^{i\pi/6} = \frac{\sqrt{3}+i}{2}$ and $\omega^* = \frac{\sqrt{3}-i}{2}$:

$$\sqrt{3}\,\omega = \sqrt{3}\cdot\frac{\sqrt{3}+i}{2} = \frac{3+i\sqrt{3}}{2} = \frac{3}{2} + \frac{\sqrt{3}}{2}i = 2 - e^{-i\pi/3}$$ $$\sqrt{3}\,\omega^* = \sqrt{3}\cdot\frac{\sqrt{3}-i}{2} = \frac{3-i\sqrt{3}}{2} = \frac{3}{2} - \frac{\sqrt{3}}{2}i = 1 + e^{-i\pi/3}$$

Substituting into $(*)$:

$$g_{ab} = \frac{1}{3}\left(\sqrt{3}\,\omega\cdot a + \sqrt{3}\,\omega^*\cdot b\right) = \frac{1}{\sqrt{3}}(\omega a + \omega^* b)$$

Part (ii)

From part (i), the four centroid points are:

$$g_{AB} = \frac{1}{\sqrt{3}}(\omega a + \omega^* b), \qquad g_{BC} = \frac{1}{\sqrt{3}}(\omega b + \omega^* c)$$ $$g_{CD} = \frac{1}{\sqrt{3}}(\omega c + \omega^* d), \qquad g_{DA} = \frac{1}{\sqrt{3}}(\omega d + \omega^* a)$$

$Q_1$ is a parallelogram if and only if its opposite sides are equal, i.e. $b - a = c - d$.

$Q_2$ is a parallelogram if and only if $g_{BC} - g_{AB} = g_{CD} - g_{DA}$.

Compute the two sides of $Q_2$:

$$g_{BC} - g_{AB} = \frac{1}{\sqrt{3}}\left(\omega(b - a) + \omega^*(c - b)\right)$$ $$g_{CD} - g_{DA} = \frac{1}{\sqrt{3}}\left(\omega(c - d) + \omega^*(d - a)\right)$$

So $Q_2$ is a parallelogram if and only if:

$$\omega(b - a) + \omega^*(c - b) = \omega(c - d) + \omega^*(d - a)$$

Rearranging:

$$\omega\left[(b - a) - (c - d)\right] + \omega^*\left[(c - b) - (d - a)\right] = 0$$

Since $(c - b) - (d - a) = (c - d) - (b - a) = -\left[(b - a) - (c - d)\right]$, this becomes:

$$\left[\omega - \omega^*\right]\left[(b - a) - (c - d)\right] = 0$$

Now $\omega - \omega^* = \frac{\sqrt{3}+i}{2} - \frac{\sqrt{3}-i}{2} = i \neq 0$, so:

$$Q_2 \text{ is a parallelogram} \iff (b - a) - (c - d) = 0 \iff b - a = c - d$$

This is exactly the condition for $Q_1$ to be a parallelogram. Therefore $Q_1$ is a parallelogram if and only if $Q_2$ is a parallelogram.

Part (iii)

The two side vectors of $T_2$ emanating from $G_{AB}$ are:

$$G_{BC} - G_{AB} = \frac{1}{\sqrt{3}}\left(\omega(b - a) + \omega^*(c - b)\right) \qquad (1)$$ $$G_{AB} - G_{CA} = \frac{1}{\sqrt{3}}\left(\omega(a - c) + \omega^*(b - a)\right) \qquad (2)$$

We show that multiplying (1) by $\omega^2$ gives $-(2)$, so that the two sides are related by a rotation of $\frac{\pi}{3}$.

Compute $\omega^2$ times (1):

$$\omega^2(G_{BC} - G_{AB}) = \frac{1}{\sqrt{3}}\left(\omega^3(b - a) + \omega^2\omega^*(c - b)\right)$$

Since $|\omega| = 1$ we have $\omega^* = \omega^{-1}$, so $\omega^2\omega^* = \omega$. Also $\omega^3 = e^{i\pi/2} = i$. Therefore:

$$\omega^2(G_{BC} - G_{AB}) = \frac{1}{\sqrt{3}}\left(i(b - a) + \omega(c - b)\right) \qquad (3)$$

Now expand (2):

$$G_{AB} - G_{CA} = \frac{1}{\sqrt{3}}\left(\omega a - \omega c + \omega^* b - \omega^* a\right) = \frac{1}{\sqrt{3}}\left((\omega - \omega^*)a + \omega^* b - \omega c\right)$$

Since $\omega - \omega^* = i$:

$$G_{AB} - G_{CA} = \frac{1}{\sqrt{3}}\left(ia + \omega^* b - \omega c\right) \qquad (4)$$

Expand (3):

$$\frac{1}{\sqrt{3}}\left(-ia + ib + \omega c - \omega b\right) = \frac{1}{\sqrt{3}}\left(-ia + (i - \omega)b + \omega c\right)$$

We check that $i - \omega = -\omega^*$:

$$i - \omega = i - \frac{\sqrt{3}+i}{2} = \frac{2i - \sqrt{3} - i}{2} = \frac{i - \sqrt{3}}{2} = -\frac{\sqrt{3}-i}{2} = -\omega^*$$

So (3) becomes:

$$\omega^2(G_{BC} - G_{AB}) = \frac{1}{\sqrt{3}}\left(-ia - \omega^* b + \omega c\right)$$

Comparing with (4), every coefficient in (3) is the negative of the corresponding coefficient in (4), so:

$$\omega^2(G_{BC} - G_{AB}) = -(G_{AB} - G_{CA})$$

Rearranging:

$$G_{CA} - G_{AB} = \omega^2(G_{BC} - G_{AB})$$

Since $|\omega^2| = 1$ and $\arg(\omega^2) = \frac{\pi}{3}$, the vector from $G_{AB}$ to $G_{CA}$ is obtained from the vector from $G_{AB}$ to $G_{BC}$ by a rotation through $\frac{\pi}{3}$. Therefore:

• $|G_{CA} - G_{AB}| = |G_{BC} - G_{AB}|$ (the two sides from vertex $G_{AB}$ have equal length);
• the angle at vertex $G_{AB}$ is $\frac{\pi}{3}$ (60 degrees).

A triangle with two equal sides and an included angle of $60^\circ$ is equilateral (by the cosine rule, the third side equals the other two, and all angles are $60^\circ$). Therefore $T_2$ is always an equilateral triangle.

Examiner Notes

这是较少人选择的纯数题。第(i)部分常见错误是把旋转方向搞反，或遗漏平移项。第(ii)部分中，许多考生试图同时证明两个方向；从 $Q_2$ 出发的证明通常更清楚。第(iii)部分需要用复数计算模长，把相关复数误当作实数是常见错误。

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Question 4

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

4 The plane $\Pi$ has equation $\mathbf{r} \cdot \mathbf{n} = 0$ where $\mathbf{n}$ is a unit vector. Let $P$ be a point with position vector $\mathbf{x}$ which does not lie on the plane $\Pi$. Show that the point $Q$ with position vector $\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\mathbf{n}$ lies on $\Pi$ and that $PQ$ is perpendicular to $\Pi$.

(i) Let transformation $T$ be a reflection in the plane $ax + by + cz = 0$, where $a^2 + b^2 + c^2 = 1$.

Show that the image of $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ under $T$ is $\begin{pmatrix} b^2 + c^2 - a^2 \\ -2ab \\ -2ac \end{pmatrix}$, and find the images of $\mathbf{j}$ and $\mathbf{k}$ under $T$.

Write down the matrix $\mathbf{M}$ which represents transformation $T$.

(ii) The matrix

$\begin{pmatrix} 0.64 & 0.48 & 0.6 \\ 0.48 & 0.36 & -0.8 \\ 0.6 & -0.8 & 0 \end{pmatrix}$

represents a reflection in a plane. Find the cartesian equation of the plane.

(iii) The matrix $\mathbf{N}$ represents a rotation through angle $\pi$ about the line through the origin parallel to $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$, where $a^2 + b^2 + c^2 = 1$. Find the matrix $\mathbf{N}$.

(iv) Identify the single transformation which is represented by the matrix $\mathbf{NM}$.

Hint

4. $\pi$ has equation $r.n = 0$ so $n$ is a vector perpendicular to this plane.

$Q$ lies on $\pi$ if $x - (x.n)n$ satisfies $r.n = 0$

$(x - (x.n)n).n = x.n - (x.n)n.n = x.n - x.n = 0$ so $Q$ lies on $\pi$ as required.

$PQ = (x - (x.n)n) - x = -(x.n)n$ which is parallel to $n$ and so is perpendicular to $\pi$.

(i) The image of a point with position vector $x$ under $T$ is $x - 2(x.n)n$, so as $n = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and

$i = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, the image of $i$ under $T$ is $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - 2 \left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} . \begin{pmatrix} a \\ b \\ c \end{pmatrix} \right) \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - 2a \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 1 - 2a^2 \\ -2ab \\ -2ac \end{pmatrix}$

But $a^2 + b^2 + c^2 = 1$ so $1 - 2a^2 = a^2 + b^2 + c^2 - 2a^2 = b^2 + c^2 - a^2$

Thus, the image of $i$ under $T$ is $\begin{pmatrix} b^2 + c^2 - a^2 \\ -2ab \\ -2ac \end{pmatrix}$ as required.

Similarly, the images of $j$ and $k$ are $\begin{pmatrix} -2ab \\ c^2 + a^2 - b^2 \\ -2bc \end{pmatrix}$ and $\begin{pmatrix} -2ac \\ -2bc \\ a^2 + b^2 - c^2 \end{pmatrix}$ respectively.

Thus $M = \begin{pmatrix} b^2 + c^2 - a^2 & -2ab & -2ac \\ -2ab & c^2 + a^2 - b^2 & -2bc \\ -2ac & -2bc & a^2 + b^2 - c^2 \end{pmatrix}$

(ii) $1 - 2a^2 = 0.64 \Rightarrow a = \pm 0.3\sqrt{2}$ and thus as $-2ab = 0.48$ and $-2ac = 0.6$,

$b = \mp 0.4\sqrt{2}$ and $c = \mp 0.5\sqrt{2}$ and the plane is $3x - 4y - 5z = 0$ (or $-3x + 4y + 5z = 0$)

(iii) Suppose the position vector of the point $Q$ on the given line such that $PQ$ is perpendicular to that line is $y$, then $y = \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ for some $\lambda$ and $(y - x) . \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 0$

So, $y . \begin{pmatrix} a \\ b \\ c \end{pmatrix} - x . \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 0$, i.e. $\lambda = x . \begin{pmatrix} a \\ b \\ c \end{pmatrix}$

So, the image of $P$ under the rotation, is $x + 2(y - x) = 2y - x = 2x . \begin{pmatrix} a \\ b \\ c \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} - x$

The image of $i$ under the rotation is thus $\begin{pmatrix} 2a^2 - 1 \\ 2ab \\ 2ac \end{pmatrix} = \begin{pmatrix} a^2 - b^2 - c^2 \\ 2ab \\ 2ac \end{pmatrix}$, and of $j$ and $k$ are $\begin{pmatrix} 2ab \\ b^2 - c^2 - a^2 \\ 2bc \end{pmatrix}$ and $\begin{pmatrix} 2ac \\ 2bc \\ c^2 - a^2 - b^2 \end{pmatrix}$ respectively.

Thus $N = \begin{pmatrix} a^2 - b^2 - c^2 & 2ab & 2ac \\ 2ab & b^2 - c^2 - a^2 & 2bc \\ 2ac & 2bc & c^2 - a^2 - b^2 \end{pmatrix}$, which, incidentally $= -M$.

(iv) $NM = -MM = -I$ as $M$ is self-inverse.

Thus the single transformation is an enlargement, scale factor -1, with centre of enlargement the origin.

alternative for (iii)

$x = u + v$ where $u \in \Pi$ and $v \perp \Pi$

$Mv = -v \qquad Mu = u \qquad Mx = u - v$

$Nx = v - u = -Mx$

$N = -M$

alternative for (ii) the matrix represents a reflection, an invariant point under the reflection lies on the plane of reflection.

Therefore,

$\begin{pmatrix} 0.64 & 0.48 & 0.6 \\ 0.48 & 0.36 & -0.8 \\ 0.6 & -0.8 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

Taking the simplest component equation

$0.6x - 0.8y = z$

(although the other two give equivalent equations).

This simplifies to

$3x - 4y - 5z = 0$

Model Solution

Preliminary

$Q$ has position vector $\mathbf{q} = \mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\mathbf{n}$.

To show $Q$ lies on $\Pi$, we verify $\mathbf{q} \cdot \mathbf{n} = 0$:

$$\mathbf{q} \cdot \mathbf{n} = \mathbf{x} \cdot \mathbf{n} - (\mathbf{x} \cdot \mathbf{n})(\mathbf{n} \cdot \mathbf{n}) = \mathbf{x} \cdot \mathbf{n} - \mathbf{x} \cdot \mathbf{n} = 0$$

using $\mathbf{n} \cdot \mathbf{n} = 1$ (since $\mathbf{n}$ is a unit vector). So $Q$ lies on $\Pi$.

The vector $\overrightarrow{PQ} = \mathbf{q} - \mathbf{x} = -(\mathbf{x} \cdot \mathbf{n})\mathbf{n}$, which is a scalar multiple of $\mathbf{n}$. Since $\mathbf{n}$ is perpendicular to $\Pi$, the line $PQ$ is perpendicular to $\Pi$.

Part (i)

The reflection $T$ in the plane $ax + by + cz = 0$ (with $\mathbf{n} = (a, b, c)^T$ a unit vector) maps a point with position vector $\mathbf{r}$ to $\mathbf{r} - 2(\mathbf{r} \cdot \mathbf{n})\mathbf{n}$. This follows from the preliminary result: the reflection sends $P$ to the point on the other side of the plane at the same distance, which is $Q+\overrightarrow{PQ}=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{n})\mathbf{n}$.

Image of $\mathbf{i}$:

$$T(\mathbf{i}) = \begin{pmatrix}1\\0\\0\end{pmatrix} - 2\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\cdot\begin{pmatrix}a\\b\\c\end{pmatrix}\right)\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}1\\0\\0\end{pmatrix} - 2a\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}1-2a^2\\-2ab\\-2ac\end{pmatrix}$$

Since $a^2 + b^2 + c^2 = 1$: $1 - 2a^2 = (a^2 + b^2 + c^2) - 2a^2 = b^2 + c^2 - a^2$.

$$T(\mathbf{i}) = \begin{pmatrix}b^2+c^2-a^2\\-2ab\\-2ac\end{pmatrix}$$

Image of $\mathbf{j}$:

$$T(\mathbf{j}) = \begin{pmatrix}0\\1\\0\end{pmatrix} - 2b\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}-2ab\\1-2b^2\\-2bc\end{pmatrix} = \begin{pmatrix}-2ab\\a^2+c^2-b^2\\-2bc\end{pmatrix}$$

Image of $\mathbf{k}$:

$$T(\mathbf{k}) = \begin{pmatrix}0\\0\\1\end{pmatrix} - 2c\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}-2ac\\-2bc\\1-2c^2\end{pmatrix} = \begin{pmatrix}-2ac\\-2bc\\a^2+b^2-c^2\end{pmatrix}$$

The columns of $\mathbf{M}$ are the images of $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$:

$$\mathbf{M} = \begin{pmatrix}b^2+c^2-a^2 & -2ab & -2ac \\ -2ab & a^2+c^2-b^2 & -2bc \\ -2ac & -2bc & a^2+b^2-c^2\end{pmatrix}$$

Part (ii)

We match the given matrix entries against $\mathbf{M}$.

From the $(1,1)$ entry: $b^2 + c^2 - a^2 = 0.64$. Since $a^2 + b^2 + c^2 = 1$, we get $1 - 2a^2 = 0.64$, so $a^2 = 0.18$.

From the $(1,2)$ entry: $-2ab = 0.48$, so $ab = -0.24$.

From the $(1,3)$ entry: $-2ac = 0.6$, so $ac = -0.3$.

Taking $a^2 = 0.18$, so $a = \pm\frac{3}{10}\sqrt{2} = \pm 0.3\sqrt{2}$.

Taking $a = 0.3\sqrt{2}$:

$$b = \frac{-0.24}{0.3\sqrt{2}} = \frac{-0.8}{\sqrt{2}} = -0.4\sqrt{2}$$ $$c = \frac{-0.3}{0.3\sqrt{2}} = \frac{-1}{\sqrt{2}} = -0.5\sqrt{2}$$

Verification: $a^2 + b^2 + c^2 = 0.18 + 0.32 + 0.5 = 1$. ✓

Check remaining entries:

• $(2,2)$: $a^2 + c^2 - b^2 = 0.18 + 0.5 - 0.32 = 0.36$ ✓
• $(2,3)$: $-2bc = -2(-0.4\sqrt{2})(-0.5\sqrt{2}) = -2(0.4)(0.5)(2) = -0.8$ ✓
• $(3,3)$: $a^2 + b^2 - c^2 = 0.18 + 0.32 - 0.5 = 0$ ✓

The plane equation is $ax + by + cz = 0$:

$$0.3\sqrt{2}\,x - 0.4\sqrt{2}\,y - 0.5\sqrt{2}\,z = 0$$

Dividing by $0.1\sqrt{2}$:

$$3x - 4y - 5z = 0$$

(The alternative sign choice $a = -0.3\sqrt{2}$ gives $-3x + 4y + 5z = 0$, which is the same plane.)

Part (iii)

A rotation through angle $\pi$ about a line through the origin maps each point to its reflection in that line. For a unit direction vector $\mathbf{n} = (a, b, c)^T$, the foot of the perpendicular from $\mathbf{x}$ to the line is $(\mathbf{x} \cdot \mathbf{n})\mathbf{n}$. The image of $\mathbf{x}$ is the point diametrically opposite across the foot:

$$T(\mathbf{x}) = 2(\mathbf{x} \cdot \mathbf{n})\mathbf{n} - \mathbf{x}$$

Image of $\mathbf{i}$:

$$\mathbf{N}(\mathbf{i}) = 2a\begin{pmatrix}a\\b\\c\end{pmatrix} - \begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}2a^2-1\\2ab\\2ac\end{pmatrix}$$

Since $2a^2 - 1 = 2a^2 - (a^2+b^2+c^2) = a^2 - b^2 - c^2$:

$$\mathbf{N}(\mathbf{i}) = \begin{pmatrix}a^2-b^2-c^2\\2ab\\2ac\end{pmatrix}$$

Similarly:

$$\mathbf{N}(\mathbf{j}) = \begin{pmatrix}2ab\\b^2-a^2-c^2\\2bc\end{pmatrix}, \qquad \mathbf{N}(\mathbf{k}) = \begin{pmatrix}2ac\\2bc\\c^2-a^2-b^2\end{pmatrix}$$ $$\mathbf{N} = \begin{pmatrix}a^2-b^2-c^2 & 2ab & 2ac \\ 2ab & b^2-a^2-c^2 & 2bc \\ 2ac & 2bc & c^2-a^2-b^2\end{pmatrix}$$

Comparing with $\mathbf{M}$, every entry of $\mathbf{N}$ is the negative of the corresponding entry of $\mathbf{M}$, so $\mathbf{N} = -\mathbf{M}$.

Part (iv)

$$\mathbf{NM} = (-\mathbf{M})\mathbf{M} = -\mathbf{M}^2$$

Since $\mathbf{M}$ represents a reflection, it is self-inverse: $\mathbf{M}^{-1} = \mathbf{M}$, and since $\mathbf{M}^2 = \mathbf{M}\mathbf{M}^{-1} = \mathbf{I}$:

$$\mathbf{NM} = -\mathbf{I}$$

The matrix $-\mathbf{I}$ maps every vector $\mathbf{x}$ to $-\mathbf{x}$. This is an enlargement with scale factor $-1$ and centre the origin (equivalently, a point reflection or inversion through the origin).

Examiner Notes

本题不太受欢迎，但尝试者在各小问中的基本思路通常正确。第(i)部分即使没有完全使用题干，多数考生也能用对称性求出 $\mathbf{j}$ 和 $\mathbf{k}$ 的像。第(ii)部分方程求解较直接，但计算错误较常见。第(iii)部分需要类比第(i)部分的技术。第(iv)部分描述变换时常见错误是说成“关于原点的旋转”，而没有明确这是半转。

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Question 5

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

5 Show that for positive integer $n$, $x^n - y^n = (x - y) \sum_{r=1}^{n} x^{n-r} y^{r-1}$.

(i) Let F be defined by

$F(x) = \frac{1}{x^n(x - k)} \quad \text{for } x \neq 0, k$

where $n$ is a positive integer and $k \neq 0$.

(a) Given that

$F(x) = \frac{A}{x - k} + \frac{f(x)}{x^n},$

where $A$ is a constant and $f(x)$ is a polynomial, show that

$f(x) = \frac{1}{x - k} \left( 1 - \left( \frac{x}{k} \right)^n \right).$

Deduce that

$F(x) = \frac{1}{k^n(x - k)} - \frac{1}{k} \sum_{r=1}^{n} \frac{1}{k^{n-r} x^r}.$

(b) By differentiating $x^n F(x)$, prove that

$\frac{1}{x^n(x - k)^2} = \frac{1}{k^n(x - k)^2} - \frac{n}{x k^n(x - k)} + \sum_{r=1}^{n} \frac{n - r}{k^{n+1-r} x^{r+1}}.$

(ii) Hence evaluate the limit of

$\int_{2}^{N} \frac{1}{x^3(x - 1)^2} \text{ d}x$

as $N \to \infty$, justifying your answer.

Hint

5. $(x - y)(x^{n-1} + x^{n-2}y + \dots + y^{n-1}) = x^n - x^{n-1}y + x^{n-1}y - x^{n-2}y + \dots + xy^{n-1} - y^n$

$= x^n - y^n$

as each even numbered term cancels with its subsequent term.

(i) If

$F(x) = \frac{1}{x^n(x - k)} = \frac{A}{x - k} + \frac{f(x)}{x^n}$

then multiplying by $x^n(x - k)$

$1 = Ax^n + (x - k)f(x)$

$x = k \Rightarrow A = \frac{1}{k^n}$

so

$1 = \frac{x^n}{k^n} + (x - k)f(x)$

and

$f(x) = \frac{1}{x - k} \left( 1 - \left( \frac{x}{k} \right)^n \right)$

as required.

Thus

$F(x) = \frac{\frac{1}{k^n}}{x - k} + \frac{\frac{1}{x - k} \left( 1 - \left( \frac{x}{k} \right)^n \right)}{x^n}$

$= \frac{1}{k^n(x - k)} - \frac{x^n - k^n}{k^n x^n (x - k)}$

and so, by the result of the stem,

$F(x) = \frac{1}{k^n(x - k)} - \frac{1}{k^n x^n} \sum_{r=1}^{n} x^{n-r} k^{r-1}$

$= \frac{1}{k^n(x - k)} - \frac{1}{k} \sum_{r=1}^{n} \frac{1}{k^{n-r} x^r}$

(ii) $x^n F(x) = \frac{1}{x - k} = \frac{x^n}{k^n(x - k)} - \frac{1}{k} \sum_{r=1}^{n} \frac{x^{n-r}}{k^{n-r}}$

Differentiating with respect to $x$,

$\frac{-1}{(x - k)^2} = \frac{nx^{n-1}}{k^n(x - k)} - \frac{x^n}{k^n(x - k)^2} - \frac{1}{k} \sum_{r=1}^{n} \frac{(n - r)x^{n-r-1}}{k^{n-r}}$

Multiplying by $\frac{-1}{x^n}$

$\frac{1}{x^n(x - k)^2} = \frac{-n}{xk^n(x - k)} + \frac{1}{k^n(x - k)^2} + \sum_{r=1}^{n} \frac{n - r}{k^{n+1-r}x^{r+1}}$

(iii) $\int_{2}^{N} \frac{1}{x^3(x - 1)^2} dx = \int_{2}^{N} \frac{-3}{x(x - 1)} + \frac{1}{(x - 1)^2} + \sum_{r=1}^{3} \frac{3 - r}{x^{r+1}} dx$

$= \int_{2}^{N} \frac{3}{x} - \frac{3}{(x - 1)} + \frac{1}{(x - 1)^2} + \sum_{r=1}^{3} \frac{3 - r}{x^{r+1}} dx$

$= \left[ 3 \ln x - 3 \ln(x - 1) - \frac{1}{x - 1} - \sum_{r=1}^{3} \frac{3 - r}{rx^r} \right]_{2}^{N}$ $= \left[ 3 \ln \left( \frac{x}{x - 1} \right) - \frac{1}{x - 1} - \sum_{r=1}^{3} \frac{3 - r}{rx^r} \right]_{2}^{N}$

$= 3 \ln \left( \frac{N}{N - 1} \right) - \frac{1}{N - 1} - \sum_{r=1}^{3} \frac{3 - r}{rN^r} - 3 \ln(2) + 1 + \sum_{r=1}^{3} \frac{3 - r}{r2^r}$

As $N \to \infty$, $\left( \frac{N}{N-1} \right) \to 1$, so $3 \ln \left( \frac{N}{N-1} \right) \to 0$, and $\frac{1}{N-1} \to 0$, $\frac{1}{N^r} \to 0$

So the limit of the integral is,

$-3 \ln 2 + 1 + \frac{2}{2} + \frac{1}{8} = -3 \ln 2 + \frac{17}{8}$

Alternatives for stem using sum of GP or proof by induction

Model Solution

Preliminary result. We show $x^n - y^n = (x - y)\sum_{r=1}^{n} x^{n-r} y^{r-1}$.

Expanding the right side:

$$(x - y)\sum_{r=1}^{n} x^{n-r} y^{r-1} = \sum_{r=1}^{n} x^{n-r+1} y^{r-1} - \sum_{r=1}^{n} x^{n-r} y^{r}$$

In the first sum, let $s = r - 1$: $\sum_{s=0}^{n-1} x^{n-s} y^{s} = x^n + \sum_{s=1}^{n-1} x^{n-s} y^{s}$.

In the second sum, let $s = r$: $\sum_{s=1}^{n} x^{n-s} y^{s} = \sum_{s=1}^{n-1} x^{n-s} y^{s} + y^n$.

Subtracting, all intermediate terms cancel, leaving $x^n - y^n$. $\qquad \square$

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Part (i)(a). From $F(x) = \frac{A}{x - k} + \frac{f(x)}{x^n}$, multiply through by $x^n(x - k)$:

$$1 = Ax^n + (x - k)f(x)$$

Setting $x = k$: $1 = Ak^n$, so $A = \dfrac{1}{k^n}$.

Rearranging: $(x - k)f(x) = 1 - \dfrac{x^n}{k^n}$, hence

$$f(x) = \frac{1}{x - k}\left(1 - \left(\frac{x}{k}\right)^n\right) \qquad \checkmark$$

Deduction. We simplify $\dfrac{f(x)}{x^n}$. Write

$$f(x) = \frac{1}{x - k} \cdot \frac{k^n - x^n}{k^n}$$

Applying the preliminary result with $k$ in place of $x$ and $x$ in place of $y$:

$$k^n - x^n = (k - x)\sum_{r=1}^{n} k^{n-r} x^{r-1} = -(x - k)\sum_{r=1}^{n} k^{n-r} x^{r-1}$$

So

$$f(x) = \frac{-(x - k)\sum_{r=1}^{n} k^{n-r} x^{r-1}}{k^n(x - k)} = -\sum_{r=1}^{n} \frac{x^{r-1}}{k^r}$$

Dividing by $x^n$:

$$\frac{f(x)}{x^n} = -\sum_{r=1}^{n} \frac{1}{k^r \, x^{n-r+1}}$$

Reindex with $s = n + 1 - r$ (so $r = n + 1 - s$, and $r: 1 \to n$ gives $s: n \to 1$):

$$\sum_{r=1}^{n} \frac{1}{k^r \, x^{n+1-r}} = \sum_{s=1}^{n} \frac{1}{k^{n+1-s} x^s} = \frac{1}{k}\sum_{s=1}^{n} \frac{1}{k^{n-s} x^s}$$

Therefore

$$F(x) = \frac{1}{k^n(x - k)} + \frac{f(x)}{x^n} = \frac{1}{k^n(x - k)} - \frac{1}{k}\sum_{r=1}^{n} \frac{1}{k^{n-r} x^r} \qquad \square$$

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Part (i)(b). From $F(x) = \dfrac{1}{x^n(x - k)}$, we have $x^n F(x) = \dfrac{1}{x - k}$.

Using the result of (a), multiply through by $x^n$:

$$x^n F(x) = \frac{x^n}{k^n(x - k)} - \frac{1}{k}\sum_{r=1}^{n} \frac{x^{n-r}}{k^{n-r}} = \frac{x^n}{k^n(x - k)} - \sum_{r=1}^{n} \frac{x^{n-r}}{k^{n-r+1}}$$

Differentiate both sides with respect to $x$. The left side:

$$\frac{d}{dx}\left[\frac{1}{x - k}\right] = \frac{-1}{(x - k)^2}$$

For the right side, use the product rule on $\dfrac{x^n}{k^n} \cdot \dfrac{1}{x - k}$ and the power rule on each sum term:

$$\frac{d}{dx}\left[\frac{x^n}{k^n(x - k)}\right] = \frac{nx^{n-1}}{k^n(x - k)} - \frac{x^n}{k^n(x - k)^2}$$ $$\frac{d}{dx}\left[\sum_{r=1}^{n} \frac{x^{n-r}}{k^{n-r+1}}\right] = \sum_{r=1}^{n} \frac{(n - r)\,x^{n-r-1}}{k^{n-r+1}}$$

Equating the derivatives:

$$\frac{-1}{(x - k)^2} = \frac{nx^{n-1}}{k^n(x - k)} - \frac{x^n}{k^n(x - k)^2} - \sum_{r=1}^{n} \frac{(n - r)\,x^{n-r-1}}{k^{n-r+1}}$$

Multiply both sides by $\dfrac{-1}{x^n}$:

$$\frac{1}{x^n(x - k)^2} = \frac{1}{k^n(x - k)^2} - \frac{n}{x\,k^n(x - k)} + \sum_{r=1}^{n} \frac{(n - r)\,x^{n-r-1}}{k^{n-r+1}\,x^n}$$

Since $\dfrac{x^{n-r-1}}{x^n} = \dfrac{1}{x^{r+1}}$:

$$\frac{1}{x^n(x - k)^2} = \frac{1}{k^n(x - k)^2} - \frac{n}{x\,k^n(x - k)} + \sum_{r=1}^{n} \frac{n - r}{k^{n+1-r}\,x^{r+1}} \qquad \square$$

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Part (ii). Set $n = 3$ and $k = 1$ in the result of (b):

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

For the $\dfrac{3}{x(x - 1)}$ term, use partial fractions:

$$\frac{3}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1} \implies 3 = A(x - 1) + Bx$$

$x = 0$: $A = -3$. $\quad$ $x = 1$: $B = 3$. $\quad$ So $\dfrac{3}{x(x - 1)} = \dfrac{3}{x - 1} - \dfrac{3}{x}$.

Therefore

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

Integrating term by term:

$$\int \frac{dx}{x^3(x - 1)^2} = -\frac{1}{x - 1} - 3\ln(x - 1) + 3\ln x - \frac{2}{x} - \frac{1}{2x^2} + C$$ $$= 3\ln\frac{x}{x - 1} - \frac{1}{x - 1} - \frac{2}{x} - \frac{1}{2x^2} + C$$

Evaluating from $2$ to $N$:

$$\int_2^N = \left[3\ln\frac{x}{x - 1} - \frac{1}{x - 1} - \frac{2}{x} - \frac{1}{2x^2}\right]_2^N$$

At $x = N$:

$$3\ln\frac{N}{N - 1} - \frac{1}{N - 1} - \frac{2}{N} - \frac{1}{2N^2}$$

At $x = 2$:

$$3\ln 2 - 1 - 1 - \frac{1}{8} = 3\ln 2 - \frac{17}{8}$$

So

$$\int_2^N \frac{dx}{x^3(x - 1)^2} = 3\ln\frac{N}{N - 1} - \frac{1}{N - 1} - \frac{2}{N} - \frac{1}{2N^2} - 3\ln 2 + \frac{17}{8}$$

As $N \to \infty$:

• $\ln\dfrac{N}{N - 1} = -\ln\!\left(1 - \dfrac{1}{N}\right) \to 0$
• $\dfrac{1}{N - 1} \to 0$, $\quad \dfrac{2}{N} \to 0$, $\quad \dfrac{1}{2N^2} \to 0$

Therefore the limit is

$$\frac{17}{8} - 3\ln 2$$

Examiner Notes

本题较热门，成功率也较高。第(i)(a)部分主要问题是求 $A$ 时出错，或忽略 $x=k$ 处未定义。第(i)(b)部分不少考生没有意识到需要两边对 $x$ 求导，求导错误和变量混淆是主要障碍。第(ii)部分多数考生能使用第(i)(b)部分的结果，但极限中的对数项处理常被遗漏。

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Question 6

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

6 (i) Sketch the curve $y = \cos x + \sqrt{\cos 2x}$ for $-\frac{1}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi$.

(ii) The equation of curve $C_1$ in polar co-ordinates is

$r = \cos \theta + \sqrt{\cos 2\theta} \quad -\frac{1}{4}\pi \leqslant \theta \leqslant \frac{1}{4}\pi.$

Sketch the curve $C_1$.

(iii) The equation of curve $C_2$ in polar co-ordinates is

$r^2 - 2r \cos \theta + \sin^2 \theta = 0 \quad -\frac{1}{4}\pi \leqslant \theta \leqslant \frac{1}{4}\pi.$

Find the value of $r$ when $\theta = \pm \frac{1}{4}\pi$.

Show that, when $r$ is small, $r \approx \frac{1}{2}\theta^2$.

Sketch the curve $C_2$, indicating clearly the behaviour of the curve near $r = 0$ and near $\theta = \pm \frac{1}{4}\pi$.

Show that the area enclosed by curve $C_2$ and above the line $\theta = 0$ is $\frac{\pi}{2\sqrt{2}}$.

Hint

6. (i)

$y = \cos x + \sqrt{\cos 2x}$

$x = 0, y = 2$ There is symmetry in $x = 0$ $x = \pm \frac{\pi}{4}, y = \frac{1}{\sqrt{2}}$

$\frac{dy}{dx} = -\sin x - \frac{\sin 2x}{\sqrt{\cos 2x}} \quad \text{so} \quad x = 0, \frac{dy}{dx} = 0 \quad x > 0, \frac{dy}{dx} < 0 \text{ and vice versa}$

as $x \to \frac{\pi}{4}, \frac{dy}{dx} \to -\infty$

(ii)

(iii) $\theta = \pm \frac{\pi}{4}, r = \frac{1}{\sqrt{2}}$

$r^2 - 2r \cos \theta + \sin^2 \theta = 0$

$(r - \cos \theta)^2 = \cos^2 \theta - \sin^2 \theta = \cos 2\theta$

Therefore, $r - \cos \theta = \pm \sqrt{\cos 2\theta}$, i.e. $r = \cos \theta \pm \sqrt{\cos 2\theta}$

From (i), $r$ is only small on the branch, $r = \cos \theta - \sqrt{\cos 2\theta}$. For $\theta = 0, r = 0$

Otherwise, $\cos \theta - \sqrt{\cos 2\theta} = 0, \cos 2\theta = \cos^2 \theta, 2\cos^2 \theta - 1 = \cos^2 \theta, \cos \theta = \pm 1$ so for $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}, \theta = 0$ is the only value for which $r = 0$

So $r$ small implies $r = \cos \theta - \sqrt{\cos 2\theta}$ and $\theta$ is small

Thus $r \approx 1 - \frac{\theta^2}{2} - \left(1 - \frac{(2\theta)^2}{2}\right)^{\frac{1}{2}} \approx 1 - \frac{\theta^2}{2} - 1 + \theta^2 = \frac{\theta^2}{2}$ as required.

Area required is

$\frac{1}{2} \int_{0}^{\frac{\pi}{4}} (\cos \theta + \sqrt{\cos 2\theta})^2 \, d\theta - \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (\cos \theta - \sqrt{\cos 2\theta})^2 \, d\theta$

$= 2 \int_{0}^{\frac{\pi}{4}} \cos \theta \sqrt{\cos 2\theta} \, d\theta$

$= 2 \int_{0}^{\frac{\pi}{4}} \cos \theta \sqrt{1 - 2 \sin^2 \theta} \, d\theta$

Let $\sqrt{2} \sin \theta = \sin u$, then $\sqrt{2} \cos \theta \frac{d\theta}{du} = \cos u$,

So the integral becomes

$2 \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2} u}{\sqrt{2}} \ du = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos 2u + 1}{2} \ du = \sqrt{2} \left[ \frac{\sin 2u}{4} + \frac{u}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2\sqrt{2}}$

(iii) alternative

$r \ll 1 \Rightarrow -2r \cos \theta + \sin^{2} \theta \approx 0$ $r \approx \frac{\sin^{2} \theta}{2 \cos \theta} = \frac{1}{2} \sin \theta \tan \theta \approx \frac{\theta^{2}}{2}$

Model Solution

Part (i). Let $y = \cos x + \sqrt{\cos 2x}$ for $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$.

Key values:

• $x = 0$: $y = 1 + 1 = 2$
• $x = \pm\dfrac{\pi}{4}$: $y = \dfrac{1}{\sqrt{2}} + 0 = \dfrac{1}{\sqrt{2}}$

Symmetry: Both $\cos x$ and $\cos 2x$ are even functions, so $y(-x) = y(x)$. The curve is symmetric about $x = 0$.

Derivative:

$$y' = -\sin x - \frac{\sin 2x}{\sqrt{\cos 2x}} = -\sin x - \frac{2\sin x \cos x}{\sqrt{\cos 2x}} = -\sin x\left(1 + \frac{2\cos x}{\sqrt{\cos 2x}}\right)$$

At $x = 0$: $y'(0) = 0$ (horizontal tangent at the maximum).

For $0 < x < \frac{\pi}{4}$: $\sin x > 0$, $\cos x > 0$, $\cos 2x > 0$, so $y' < 0$ (decreasing).

As $x \to \frac{\pi}{4}^-$: $\cos 2x \to 0^+$, so $\dfrac{\sin 2x}{\sqrt{\cos 2x}} \to +\infty$, hence $y' \to -\infty$ (vertical tangent at both endpoints).

The curve has a maximum of $y = 2$ at $x = 0$, decreases to $y = \dfrac{1}{\sqrt{2}}$ at both endpoints with vertical tangents, and is symmetric about the $y$-axis.

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Part (ii). The polar curve $C_1$ is $r = \cos\theta + \sqrt{\cos 2\theta}$ for $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$.

This is the same function as in part (i), now plotted as $r$ against $\theta$ in polar coordinates. Since $r > 0$ throughout the domain and the endpoint rays are distinct, this is an open polar arc rather than a closed loop:

• At $\theta = 0$: $r = 2$, so the curve reaches the point $(2, 0)$ in Cartesian coordinates.
• At $\theta = \pm\frac{\pi}{4}$: $r = \frac{1}{\sqrt{2}}$, so the curve meets the lines $\theta = \pm\frac{\pi}{4}$ at distance $\frac{1}{\sqrt{2}}$ from the origin (Cartesian: $\left(\frac{1}{2}, \pm\frac{1}{2}\right)$).
• The curve is symmetric about the initial line $\theta = 0$.
• It forms an arc bulging outward to $r = 2$ along the positive $x$-axis.

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Part (iii).

Finding $r$ when $\theta = \pm\frac{\pi}{4}$:

$$r^2 - 2r\cos\frac{\pi}{4} + \sin^2\frac{\pi}{4} = 0$$ $$r^2 - \sqrt{2}\,r + \frac{1}{2} = 0$$ $$\left(r - \frac{1}{\sqrt{2}}\right)^2 = 0$$

So $r = \dfrac{1}{\sqrt{2}}$. $\qquad \checkmark$

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Showing $r \approx \frac{1}{2}\theta^2$ for small $r$:

Complete the square in the equation $r^2 - 2r\cos\theta + \sin^2\theta = 0$:

$$(r - \cos\theta)^2 = \cos^2\theta - \sin^2\theta = \cos 2\theta$$

So $r = \cos\theta \pm \sqrt{\cos 2\theta}$.

For small $r$, we need the $-$ branch: $r = \cos\theta - \sqrt{\cos 2\theta}$, since at $\theta = 0$ this gives $r = 1 - 1 = 0$. (The $+$ branch gives $r = 2$ at $\theta = 0$.)

For small $\theta$, using Taylor expansions:

$$\cos\theta \approx 1 - \frac{\theta^2}{2}, \qquad \cos 2\theta \approx 1 - 2\theta^2$$ $$\sqrt{\cos 2\theta} = (1 - 2\theta^2)^{1/2} \approx 1 - \theta^2$$

Therefore

$$r \approx \left(1 - \frac{\theta^2}{2}\right) - (1 - \theta^2) = \frac{\theta^2}{2} \qquad \checkmark$$

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Sketch of $C_2$:

The curve $C_2$ consists of two branches meeting at $\theta = \pm\frac{\pi}{4}$:

• Outer branch: $r = \cos\theta + \sqrt{\cos 2\theta}$ (the same as $C_1$; maximum $r = 2$ at $\theta = 0$).
• Inner branch: $r = \cos\theta - \sqrt{\cos 2\theta}$ (passes through the origin at $\theta = 0$; reaches $r = \frac{1}{\sqrt{2}}$ at $\theta = \pm\frac{\pi}{4}$).

Near $r = 0$: the inner branch has $r \approx \frac{\theta^2}{2}$, so the curve approaches the origin tangentially to the initial line $\theta = 0$ (since $r/\theta \to 0$ as $\theta \to 0$).

Near $\theta = \pm\frac{\pi}{4}$: both branches give $r = \frac{1}{\sqrt{2}}$, and $\frac{dr}{d\theta} \to \pm\infty$ on each branch, so the curve has cusps at these junction points.

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Area enclosed by $C_2$ above $\theta = 0$:

The required area is the region between the two branches for $0 \le \theta \le \frac{\pi}{4}$:

$$A = \frac{1}{2}\int_0^{\pi/4} r_+^2 \, d\theta - \frac{1}{2}\int_0^{\pi/4} r_-^2 \, d\theta$$

where $r_{\pm} = \cos\theta \pm \sqrt{\cos 2\theta}$.

Using the difference of squares:

$$r_+^2 - r_-^2 = (r_+ + r_-)(r_+ - r_-) = (2\cos\theta)(2\sqrt{\cos 2\theta}) = 4\cos\theta\sqrt{\cos 2\theta}$$

So

$$A = \frac{1}{2}\int_0^{\pi/4} 4\cos\theta\sqrt{\cos 2\theta} \, d\theta = 2\int_0^{\pi/4} \cos\theta\sqrt{1 - 2\sin^2\theta} \, d\theta$$

Substitute $\sqrt{2}\sin\theta = \sin\phi$. Then $\sqrt{2}\cos\theta \, d\theta = \cos\phi \, d\phi$, so

$$\cos\theta \, d\theta = \frac{\cos\phi}{\sqrt{2}} \, d\phi$$

When $\theta = 0$: $\phi = 0$. When $\theta = \frac{\pi}{4}$: $\sin\phi = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$, so $\phi = \frac{\pi}{2}$.

$$\sqrt{1 - 2\sin^2\theta} = \sqrt{1 - \sin^2\phi} = \cos\phi \qquad (\text{since } \phi \in [0, \tfrac{\pi}{2}])$$

Therefore

$$A = 2\int_0^{\pi/2} \cos\phi \cdot \frac{\cos\phi}{\sqrt{2}} \, d\phi = \frac{2}{\sqrt{2}}\int_0^{\pi/2} \cos^2\phi \, d\phi$$

Using $\cos^2\phi = \frac{1}{2}(1 + \cos 2\phi)$:

$$\int_0^{\pi/2} \cos^2\phi \, d\phi = \frac{1}{2}\left[\phi + \frac{\sin 2\phi}{2}\right]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$

So

$$A = \frac{2}{\sqrt{2}} \cdot \frac{\pi}{4} = \sqrt{2} \cdot \frac{\pi}{4} = \frac{\pi}{2\sqrt{2}} \qquad \square$$

Examiner Notes

约70%的考生尝试本题，但平均得分仅约35%。第(i)部分多数考生能完成，但常忽略端点斜率。第(ii)部分作图质量较弱，端点行为处理不好。第(iii)部分的小量近似论证常不完整；很多考生没有发现 $C_1$ 是 $C_2$ 的一条分支。极坐标面积计算是主要障碍，成功者通常意识到需要取两个极坐标积分之差。

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Question 7

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

7 (i) Given that the variables $x$, $y$ and $u$ are connected by the differential equations

$\frac{\mathrm{d}u}{\mathrm{d}x} + \mathrm{f}(x)u = \mathrm{h}(x) \quad \text{and} \quad \frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}(x)y = u,$

show that

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (\mathrm{g}(x) + \mathrm{f}(x))\frac{\mathrm{d}y}{\mathrm{d}x} + (\mathrm{g}'(x) + \mathrm{f}(x)\mathrm{g}(x))y = \mathrm{h}(x). \qquad \text{(1)}$

(ii) Given that the differential equation

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(1 + \frac{4}{x}\right)\frac{\mathrm{d}y}{\mathrm{d}x} + \left(\frac{2}{x} + \frac{2}{x^2}\right)y = 4x + 12 \qquad \text{(2)}$

can be written in the same form as (1), find a first order differential equation which is satisfied by $\mathrm{g}(x)$.

If $\mathrm{g}(x) = kx^n$, find a possible value of $n$ and the corresponding value of $k$.

Hence find a solution of (2) with $y = 5$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = -3$ at $x = 1$.

Hint

7. (i) $u = \frac{dy}{dx} + g(x)y$

Thus

$\frac{du}{dx} = \frac{d^2y}{dx^2} + g(x)\frac{dy}{dx} + g'(x)y$

As $\frac{du}{dx} + f(x)u = h(x)$

$\frac{d^2y}{dx^2} + g(x)\frac{dy}{dx} + g'(x)y + f(x)\left(\frac{dy}{dx} + g(x)y\right) = h(x)$

that is

$\frac{d^2y}{dx^2} + (g(x) + f(x))\frac{dy}{dx} + (g'(x) + f(x)g(x))y = h(x)$

as required.

(ii)

$g(x) + f(x) = 1 + \frac{4}{x}$

and so $f(x) = 1 + \frac{4}{x} - g(x)$

$g'(x) + f(x)g(x) = \frac{2}{x} + \frac{2}{x^2}$

so

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

as requested.

If $g(x) = kx^n$, $g'(x) = knx^{n-1}$

$knx^{n-1} + \left(1 + \frac{4}{x} - kx^n\right)kx^n = \frac{2}{x} + \frac{2}{x^2}$

$-k^2x^{2n+2} + kx^{n+2} + k(n + 4)x^{n+1} - 2x - 2 = 0$

Considering the $x^{2n+2}$ term,

either it is eliminated by the $x^{n+2}$ term, in which case, $2n + 2 = n + 2$ and $-k^2 + k = 0$

which would imply $n = 0$ and $k = 0$ or $k = 1$

$k = 0$ is not possible $(-2x - 2 = 0)$; $n = 0, k = 1$ would give $4x - 2x - 2 = 0$ so not possible

Or it is eliminated by the $x^{n+1}$ term, in which case, $2n + 2 = n + 1$ which implies $n = -1$ and thus $-k^2 + k(n + 4) - 2 = 0$ and considering the other two terms $k - 2 = 0$

$k = 2$ and $n = -1$ satisfy $-k^2 + k(n + 4) - 2 = 0$ so these are possible values.

So $g(x) = \frac{2}{x}$ and as $f(x) = 1 + \frac{4}{x} - g(x)$, $f(x) = 1 + \frac{2}{x}$ $h(x) = 4x + 12$

$\frac{du}{dx} + f(x)u = h(x)$ is thus $\frac{du}{dx} + \left(1 + \frac{2}{x}\right)u = 4x + 12$

The integrating factor is

$e^{\int \left(1 + \frac{2}{x}\right) dx} = e^{x + 2 \ln x} = x^2 e^x$

Thus

$x^2 e^x \frac{du}{dx} + (x^2 + 2x)e^x u = (4x + 12)x^2 e^x = (4x^3 + 12x^2)e^x$

Integrating with respect to $x$

$x^2 e^x u = \int (4x^3 + 12x^2)e^x dx = 4x^3 e^x + c$

As $u = \frac{dy}{dx} + g(x)y$, and $g(x) = \frac{2}{x}$, when $x = 1$, $y = 5$, $\frac{dy}{dx} = -3$, we have $u = -3 + 2 \times 5$

That is $u = 7$, so $7e = 4e + c$, which means $c = 3e$

So $u = 4x + 3e \frac{e^{-x}}{x^2}$

$\frac{dy}{dx} + \frac{2}{x}y = 4x + 3e \frac{e^{-x}}{x^2}$

This has integrating factor

$e^{\int \frac{2}{x} dx} = e^{2 \ln x} = x^2$

So

$x^2 \frac{dy}{dx} + 2xy = 4x^3 + 3e e^{-x}$

Integrating with respect to $x$

$x^2 y = \int 4x^3 + 3e e^{-x} dx = x^4 - 3ee^{-x} + c'$

when $x = 1$, $y = 5$ so $5 = 1 - 3 + c'$ which means $c' = 7$

Therefore,

$y = x^2 + \frac{7}{x^2} - \frac{3e^{-x+1}}{x^2}$

Model Solution

Part (i)

From the second equation, $u = \dfrac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}(x)y$.

Differentiating both sides with respect to $x$ using the product rule:

$\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \mathrm{g}(x)\frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}'(x)y.$

Substituting into the first equation $\dfrac{\mathrm{d}u}{\mathrm{d}x} + \mathrm{f}(x)u = \mathrm{h}(x)$:

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \mathrm{g}(x)\frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}'(x)y + \mathrm{f}(x)\!\left(\frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}(x)y\right) = \mathrm{h}(x).$

Collecting terms in $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and $y$:

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \bigl(\mathrm{g}(x) + \mathrm{f}(x)\bigr)\frac{\mathrm{d}y}{\mathrm{d}x} + \bigl(\mathrm{g}'(x) + \mathrm{f}(x)\mathrm{g}(x)\bigr)y = \mathrm{h}(x). \qquad \text{(1)}$

Part (ii)

Comparing equation (2) with (1), we identify:

$\mathrm{g}(x) + \mathrm{f}(x) = 1 + \frac{4}{x}, \qquad \mathrm{g}'(x) + \mathrm{f}(x)\mathrm{g}(x) = \frac{2}{x} + \frac{2}{x^2}.$

From the first relation, $\mathrm{f}(x) = 1 + \dfrac{4}{x} - \mathrm{g}(x)$. Substituting into the second:

$\mathrm{g}'(x) + \left(1 + \frac{4}{x} - \mathrm{g}(x)\right)\mathrm{g}(x) = \frac{2}{x} + \frac{2}{x^2}.$

This is the required first order ODE for $\mathrm{g}(x)$.

Now suppose $\mathrm{g}(x) = kx^n$, so $\mathrm{g}'(x) = knx^{n-1}$. Substituting:

$knx^{n-1} + \left(1 + \frac{4}{x} - kx^n\right)kx^n = \frac{2}{x} + \frac{2}{x^2}.$

Expanding the left side:

$knx^{n-1} + kx^n + 4kx^{n-1} - k^2x^{2n} = \frac{2}{x} + \frac{2}{x^2}.$

$-k^2x^{2n} + kx^n + k(n+4)x^{n-1} = \frac{2}{x} + \frac{2}{x^2}.$

Multiplying through by $x^2$:

$-k^2x^{2n+2} + kx^{n+2} + k(n+4)x^{n+1} = 2x + 2. \qquad (\ast)$

The right side has terms in $x^1$ and $x^0$ only. We try $n = -1$, which gives:

$-k^2 + kx + 3k = 2x + 2.$

Comparing coefficients of $x$: $k = 2$.

Comparing constant terms: $-k^2 + 3k = -4 + 6 = 2$. This matches.

So $n = -1$ and $k = 2$, giving $\mathrm{g}(x) = \dfrac{2}{x}$.

It follows that $\mathrm{f}(x) = 1 + \dfrac{4}{x} - \dfrac{2}{x} = 1 + \dfrac{2}{x}$ and $\mathrm{h}(x) = 4x + 12$.

Solving for $y$:

The first equation for $u$ is:

$\frac{\mathrm{d}u}{\mathrm{d}x} + \left(1 + \frac{2}{x}\right)u = 4x + 12.$

The integrating factor is $e^{\int(1 + 2/x)\,\mathrm{d}x} = e^{x + 2\ln x} = x^2e^x$. Multiplying through:

$\frac{\mathrm{d}}{\mathrm{d}x}\!\left(x^2 e^x u\right) = (4x + 12)x^2 e^x = (4x^3 + 12x^2)e^x.$

We evaluate $\displaystyle\int(4x^3 + 12x^2)e^x\,\mathrm{d}x$ by repeated integration by parts:

$\int 4x^3 e^x\,\mathrm{d}x = 4x^3 e^x - 12\int x^2 e^x\,\mathrm{d}x = 4x^3 e^x - 12x^2 e^x + 24\int x e^x\,\mathrm{d}x$ $= 4x^3 e^x - 12x^2 e^x + 24xe^x - 24e^x.$

$\int 12x^2 e^x\,\mathrm{d}x = 12x^2 e^x - 24\int xe^x\,\mathrm{d}x = 12x^2 e^x - 24xe^x + 24e^x.$

Adding:

$\int(4x^3 + 12x^2)e^x\,\mathrm{d}x = 4x^3 e^x + C.$

So $x^2 e^x u = 4x^3 e^x + C$, i.e.\ $u = 4x + Ce^{-x}/x^2$.

From the second equation, $u = \dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{2}{x}y$. At $x = 1$: $y = 5$, $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3$, so $u = -3 + 10 = 7$.

Setting $x = 1$, $u = 7$: $7 = 4 + Ce^{-1}$, so $C = 3e$.

Thus $u = 4x + \dfrac{3e^{1-x}}{x^2}$.

Now solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{2}{x}y = 4x + \dfrac{3e^{1-x}}{x^2}$.

The integrating factor is $e^{\int 2/x\,\mathrm{d}x} = x^2$. Multiplying through:

$\frac{\mathrm{d}}{\mathrm{d}x}(x^2 y) = 4x^3 + 3e^{1-x}.$

Integrating:

$x^2 y = x^4 - 3e^{1-x} + C'.$

At $x = 1$, $y = 5$: $5 = 1 - 3 + C'$, so $C' = 7$.

$y = x^2 + \frac{7}{x^2} - \frac{3e^{1-x}}{x^2}.$

Examiner Notes

这是第二热门且最成功的题目，平均得分约65%。第(i)部分几乎所有考生都能完成。第(ii)部分中，找到一阶微分方程通常不是主要问题；许多考生会先猜 $n=-1$，再验证 $k=2$。积分因子和积分计算是主要失分点。部分考生使用特解加齐次解的方法，但齐次解常出错。

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Question 8

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

8 A sequence $u_k$, for integer $k \geqslant 1$, is defined as follows.

$u_1 = 1$ $u_{2k} = u_k \text{ for } k \geqslant 1$ $u_{2k+1} = u_k + u_{k+1} \text{ for } k \geqslant 1$

(i) Show that, for every pair of consecutive terms of this sequence, except the first pair, the term with odd subscript is larger than the term with even subscript.

(ii) Suppose that two consecutive terms in this sequence have a common factor greater than one. Show that there are then two consecutive terms earlier in the sequence which have the same common factor. Deduce that any two consecutive terms in this sequence are co-prime (do not have a common factor greater than one).

(iii) Prove that it is not possible for two positive integers to appear consecutively in the same order in two different places in the sequence.

(iv) Suppose that $a$ and $b$ are two co-prime positive integers which do not occur consecutively in the sequence with $b$ following $a$. If $a > b$, show that $a - b$ and $b$ are two co-prime positive integers which do not occur consecutively in the sequence with $b$ following $a - b$, and whose sum is smaller than $a + b$. Find a similar result for $a < b$.

(v) For each integer $n \geqslant 1$, define the function f from the positive integers to the positive rational numbers by $f(n) = \frac{u_n}{u_{n+1}}$. Show that the range of f is all the positive rational numbers, and that f has an inverse.

Hint

8. (i) All terms of the sequence are positive integers because they are all either equal to a previous term or the sum of two previous terms which are positive integers.

Thus, for $k \geq 1$, as $u_{2k} = u_k$ and $u_{2k+1} = u_k + u_{k+1}$, $u_{2k+1} - u_{2k} = u_{k+1} \geq 1$

Also, $u_{2k+1} - u_{2k+2} = u_k + u_{k+1} - u_{k+1} = u_k \geq 1$. Thus, the required result is proved for terms from the third onwards. (The only terms not included in this proof are the first two, which are in case both equal to 1).

(ii) Suppose that $u_{2k} = c$, and that $u_{2k+1} = d$, for $k \geq 1$, where d and c share a common factor greater than one, then $u_k = c$, as $u_{2k} = u_k$, and $u_{k+1} = d - c \geq 1$ as $u_{2k+1} = u_k + u_{k+1}$ and using (i). Then as d and c share a common factor greater than one, d-c and c share a common factor greater than one. So, two earlier terms in the sequence do share the same common factor.

Likewise, suppose that $u_{2k+2} = c$, and that $u_{2k+1} = d$, for $k \geq 1$, where d and c share a common factor greater than one, then $u_{k+1} = c$ and $u_k = d - c$ giving the same result.

This is true for pairs of consecutive terms from the second term (and third) onwards. Repeating this argument, we find that it would imply that the first two terms would share a common factor greater than one, which is a contradiction. Hence any two consecutive terms are co-prime.

(iii) For $k \geq 1$, and $m \geq 1$ suppose that $u_{2k} = c$ and $u_{2k+1} = d$, and that $u_{2k+m} = c$ and $u_{2k+m+1} = d$, then as $d > c$, $2k + m$ is even, so $m$ is even, say $2n$. Thus, $u_k = c$ and $u_{k+1} = d - c$, and $u_{k+n} = c$ and $u_{k+n+1} = d - c$. That is, an earlier pair of terms would appear consecutively.

Likewise, if $u_{2k+2} = c$ and $u_{2k+1} = d$, and that $u_{2k+m+2} = c$ and $u_{2k+m+1} = d$, the same argument applies.

So the argument can be repeated down to the first two terms, which are of course equal, and it would imply a later pair are likewise which contradicts (i).

(iv) If $(a, b)$ does not occur, where $a$ and $b$ are coprime and $a > b$, then there does not exist $k$ such that $u_{2k+1} = a$ and $u_{2k+2} = b$. Therefore there cannot exist a $k$ such that $u_{k+1} = b$ and $u_k = a - b$, the sum of which is $a$, which is smaller than $a + b$.

If $(a, b)$ does not occur, where $a$ and $b$ are coprime and $a < b$, then there does not exist $k$ such that $u_{2k} = a$ and $u_{2k+1} = b$. Therefore there cannot exist a $k$ such that $u_k = a$ and $u_{k+1} = b - a$, the sum of which is $b$, which is smaller than $a + b$.

(v) Suppose that there exists an ordered pair of coprime integers (a,b) which does not occur consecutively in the sequence. Then by part (iv) the pair (a-b, b) [if a>b] or (a, b-a) [if b>a] (which has a smaller sum) does not occur. Repeating this means that a coprime pair with sum <3 does not occur. The only coprime pair of integers with sum <3 is (1, 1) which are the first two terms. Contradiction and so every ordered pair of coprime integers occurs in the sequence and by (iii) only occurs once. Therefore, there exists an $n$, and that $n$ is unique such that

$q = \frac{u_n}{u_{n+1}}$, for any positive rational $q$ (which is expressed in lowest form). So the inverse of f exists.

Model Solution

First, let us list the first several terms to build intuition:

$u_1 = 1,\; u_2 = 1,\; u_3 = 2,\; u_4 = 1,\; u_5 = 3,\; u_6 = 2,\; u_7 = 3,\; u_8 = 1,\; u_9 = 4,\; u_{10} = 3,\; u_{11} = 5,\; u_{12} = 2, \ldots$

All terms are positive integers, since every term is either equal to a previous term ($u_{2k} = u_k$) or a sum of two previous terms ($u_{2k+1} = u_k + u_{k+1}$), with $u_1 = 1 > 0$.

Part (i)

For $k \geqslant 1$, we compare consecutive pairs $(u_{2k}, u_{2k+1})$ and $(u_{2k+1}, u_{2k+2})$.

$u_{2k+1} - u_{2k} = (u_k + u_{k+1}) - u_k = u_{k+1} \geqslant 1.$

$u_{2k+1} - u_{2k+2} = (u_k + u_{k+1}) - u_{k+1} = u_k \geqslant 1.$

So for every pair of consecutive terms from the third term onwards, the odd-subscripted term exceeds the even-subscripted term. The only exception is the first pair $(u_1, u_2) = (1, 1)$, where both are equal.

Part (ii)

We show that if two consecutive terms share a common factor $d > 1$, then an earlier consecutive pair also shares the factor $d$.

Case 1: Suppose $u_{2k}$ and $u_{2k+1}$ share a common factor $d > 1$ (for $k \geqslant 1$). Then $u_k = u_{2k}$ is divisible by $d$, and $u_{k+1} = u_{2k+1} - u_k$ is also divisible by $d$. So the earlier consecutive pair $(u_k, u_{k+1})$ shares the factor $d$.

Case 2: Suppose $u_{2k+1}$ and $u_{2k+2}$ share a common factor $d > 1$ (for $k \geqslant 1$). Then $u_{k+1} = u_{2k+2}$ is divisible by $d$, and $u_k = u_{2k+1} - u_{k+1}$ is also divisible by $d$. So the earlier consecutive pair $(u_k, u_{k+1})$ shares the factor $d$.

In either case, we can step back to an earlier consecutive pair sharing the same factor $d > 1$. Repeating this process, we eventually reach the pair $(u_1, u_2) = (1, 1)$. But $\gcd(1, 1) = 1$, which contradicts the existence of a common factor $d > 1$.

Therefore, any two consecutive terms in the sequence are coprime.

Part (iii)

Suppose for contradiction that two positive integers $a, b$ appear consecutively in the same order at two different positions: $(u_m, u_{m+1}) = (u_n, u_{n+1}) = (a, b)$ with $m < n$.

We first establish that the predecessor of any consecutive pair is uniquely determined. Every consecutive pair $(u_j, u_{j+1})$ for $j \geqslant 2$ arises from an earlier pair $(u_k, u_{k+1})$ via one of two maps:

$(u_{2k}, u_{2k+1}) = (u_k,\; u_k + u_{k+1}) \qquad \text{and} \qquad (u_{2k+1}, u_{2k+2}) = (u_k + u_{k+1},\; u_{k+1}).$

By part (i), if $j$ is even then $u_{j+1} > u_j$, and if $j$ is odd then $u_j > u_{j+1}$. So:

• If $j$ is even, $j = 2k$: $(u_j, u_{j+1}) = (u_k, u_k + u_{k+1})$, so the predecessor is $(u_k, u_{k+1}) = (u_j,\; u_{j+1} - u_j)$.

• If $j$ is odd, $j = 2k+1$: $(u_j, u_{j+1}) = (u_k + u_{k+1}, u_{k+1})$, so the predecessor is $(u_k, u_{k+1}) = (u_j - u_{j+1},\; u_{j+1})$.

In both cases, the predecessor pair and its parent index are determined uniquely by the values $(u_j,u_{j+1})$: if the first entry is smaller, the pair came from an even index $2k$ and has predecessor $(u_j,u_{j+1}-u_j)$; if the first entry is larger, it came from an odd index $2k+1$ and has predecessor $(u_j-u_{j+1},u_{j+1})$. Since $(u_m,u_{m+1})=(u_n,u_{n+1})$, repeatedly applying this unique reverse step gives equal earlier ancestor pairs at every level.

Continuing the reverse process must eventually reach the initial pair $(u_1,u_2)=(1,1)$ for both occurrences. The same sequence of parity choices is forced by the pair values at each reverse step, so the binary path back from the root is the same for both positions. Hence the two positions are identical, contradicting $m<n$.

Therefore, no two positive integers can appear consecutively in the same order at two different positions.

Part (iv)

Case $a > b$: Suppose $(a, b)$ does not occur as a consecutive pair in the sequence. We claim $(a - b, b)$ also does not occur as a consecutive pair with $b$ following $a - b$.

Suppose for contradiction that $(a - b, b)$ does occur, say $(u_k, u_{k+1}) = (a - b, b)$. Then $(u_{2k+1}, u_{2k+2}) = (u_k + u_{k+1}, u_{k+1}) = (a, b)$, contradicting our assumption. So $(a - b, b)$ does not occur as a consecutive pair with $b$ following $a - b$.

Furthermore, $\gcd(a - b, b) = \gcd(a, b) = 1$, so $a - b$ and $b$ are coprime positive integers (positive since $a > b$). And $(a - b) + b = a < a + b$.

Case $a < b$: By an analogous argument, if $(a, b)$ does not occur, then $(a, b - a)$ does not occur as a consecutive pair with $b - a$ following $a$. These are coprime positive integers with sum $b < a + b$.

Explicitly: if $(a, b - a) = (u_k, u_{k+1})$, then $(u_{2k}, u_{2k+1}) = (u_k, u_k + u_{k+1}) = (a, b)$, a contradiction.

Part (v)

Surjectivity: Let $\dfrac{p}{q}$ be any positive rational number, where $p, q$ are coprime positive integers. We prove by strong induction on $p + q$ that the pair $(p, q)$ occurs as consecutive terms $(u_n, u_{n+1})$ for some $n$.

Base case: $p + q = 2$ gives $p = q = 1$. We have $(u_1, u_2) = (1, 1)$.

Inductive step: Suppose the result holds for all coprime pairs with sum less than $p + q$, where $p + q > 2$. Since $\gcd(p, q) = 1$ and $p + q > 2$, we cannot have $p = q$, so either $p > q$ or $p < q$.

If $p > q$: consider $(p - q, q)$, which is a coprime pair with sum $p < p + q$. By the inductive hypothesis, $(p - q, q) = (u_k, u_{k+1})$ for some $k$. Then $(u_{2k+1}, u_{2k+2}) = (u_k + u_{k+1}, u_{k+1}) = (p, q)$.

If $p < q$: consider $(p, q - p)$, which is a coprime pair with sum $q < p + q$. By the inductive hypothesis, $(p, q - p) = (u_k, u_{k+1})$ for some $k$. Then $(u_{2k}, u_{2k+1}) = (u_k, u_k + u_{k+1}) = (p, q)$.

In both cases, $(p, q)$ occurs as consecutive terms. So every positive rational $\dfrac{p}{q}$ is in the range of $f$.

Injectivity: Suppose $f(m) = f(n)$, i.e., $\dfrac{u_m}{u_{m+1}} = \dfrac{u_n}{u_{n+1}}$. Writing both fractions in lowest terms, we get the same coprime pair $(p, q)$. By part (iii), no ordered pair of positive integers appears consecutively at two different positions, so $m = n$. Thus $f$ is injective.

Since $f$ is both surjective and injective, it is a bijection from the positive integers to the positive rationals, and therefore has an inverse.

Examiner Notes

约60%的考生尝试本题，但平均得分仅约33%。第(i)部分常见错误是误认为所有偶数下标项都满足所需性质。第(ii)部分需要分别考虑 $(u_{2k-1},u_{2k})$ 和 $(u_{2k},u_{2k+1})$ 两种相邻情形。第(iii)部分对 “consecutive” 的理解是关键，很多考生错误地声称重现必定发生在 $2^n k$ 一类位置。第(iv)(v)部分很少有人完成，需要把第(iv)部分的 Euclidean algorithm 思想和第(v)部分的满射性联系起来。