STEP3 2020 -- Pure Mathematics

STEP3 2020 — Section A (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	绾暟	Challenging	鍒嗛儴绉垎, 澶嶅悎瑙掑叕寮? 鏁板褰掔撼娉?
2	绾暟	Challenging	闅愬嚱鏁版眰瀵? 鍒ゅ埆寮忓垎鏋? 鍙屾洸鍑芥暟鎬ц川, 鏇茬嚎瀵圭О鎬?
3	绾暟	Hard	澶嶆暟鎸囨暟褰㈠紡, 鏃嬭浆鍏紡, 閲嶅績鍏紡, 妯￠暱璁＄畻
4	绾暟	Hard	鍙嶅皠鐭╅樀鍏紡, 鐭╅樀鏂圭▼姹傝В, 鏃嬭浆鐭╅樀, 鍙樻崲澶嶅悎
5	绾暟	Challenging	閮ㄥ垎鍒嗗紡鍒嗚В, 浠ｆ暟鎭掔瓑寮? 姹傚鎶€宸? 鍙嶅父绉垎鏋侀檺
6	绾暟	Hard	鏋佸潗鏍囨洸绾挎弿缁? 灏忛噺杩戜技, 鏋佸潗鏍囬潰绉叕寮? 涓夎鎹㈠厓
7	绾暟	Challenging	闄嶉樁娉? 绉垎鍥犲瓙娉? 鍙傛暟鍖归厤, 鍒濆鏉′欢浠ｅ叆
8	绾暟	Hard	鍙嶈瘉娉? 鏁板褰掔撼娉? 閫掓帹鍒嗘瀽, 鏋勯€犳硶, Euclidean绠楁硶

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Question 1

Topic: 绾暟  |  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

PLACEHOLDER_Q1

Examiner Notes

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Question 2

Topic: 绾暟  |  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

PLACEHOLDER_Q2

Examiner Notes

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Question 3

Topic: 绾暟  |  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 ABC 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 $60°$ is equilateral (by the cosine rule, the third side equals the other two, and all angles are $60°$). Therefore $T_2$ is always an equilateral triangle.

Examiner Notes

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Question 4

Topic: 绾暟  |  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 $Q - \overrightarrow{PQ} = \mathbf{q} + (\mathbf{x} \cdot \mathbf{n})\mathbf{n} = \mathbf{x} - 2(\mathbf{x} \cdot \mathbf{n})\mathbf{n}$.

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

涓嶅彈娆㈣繋浣嗗悇閮ㄥ垎灏濊瘯鏃跺熀鏈纭€?i)鍗充娇鏃犳硶搴旂敤棰樺共锛屽鏁拌兘鐢ㄥ绉版€ф眰j鍜宬鐨勫儚銆?ii)鍑犱箮閮借兘瑙ｆ柟绋嬩絾璁＄畻甯稿嚭閿欍€?iii)鍙湁鍏ㄥ鎴栧叏閿欙紝闇€绫绘瘮(i)鐨勬妧鏈€?iv)鎻忚堪鍙樻崲鏃跺父瑙侀敊璇細璇存垚鍏充簬鍘熺偣鐨勬棆杞€岄潪鍗婅浆銆?

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Question 5

Topic: 绾暟  |  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 $N \to \infty$, $\left( \frac{N-1}{N} \right) \to 1$, so $3 \ln \left( \frac{N-1}{N} \right) \to 0$, 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

鐑棬涓旀垚鍔熺巼杈冮珮銆?i)(a)涓昏闂锛氭眰A鏃跺嚭閿欍€佸拷鐣=k澶勬湭瀹氫箟銆?i)(b)寰堝鑰冪敓涓嶇煡閬撻渶涓よ竟瀵箈姹傚锛屾眰瀵奸敊璇拰娣锋穯鏄殰纰嶃€?ii)澶氭暟鑳界敤(i)(b)缁撴灉锛屼絾鏋侀檺涓鏁伴」澶勭悊甯歌閬楁紡銆?

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Question 6

Topic: 绾暟  |  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.

---

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, the curve is 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 a closed loop 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.

---

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

绾?0%灏濊瘯浣嗕粎寰楃害35%鍒嗘暟銆?i)澶氭暟鎴愬姛浣嗗拷鐣ョ鐐规枩鐜囥€?ii)浣滃浘杈冨樊锛岀鐐硅涓哄鐞嗕笉濂姐€?iii)灏忛噺杩戜技璁鸿瘉涓嶅畬鏁达紱澶氭暟鏈彂鐜癈鈧佹槸C鈧傜殑涓€鏉″垎鏀紱鏋佸潗鏍囬潰绉绠楁槸涓昏闅滅锛屾垚鍔熻€呮剰璇嗗埌闇€鐢ㄤ袱涓瀬鍧愭爣绉垎涔嬪樊銆?

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Question 7

Topic: 绾暟  |  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

绗簩鐑棬涓旀渶鎴愬姛(~65%)銆?i)鍑犱箮閮借兘瀹屾垚銆?ii)鎵句竴闃禣DE寰堝皯鍑洪棶棰橈紱澶氭暟鐚渘=-1鍐嶉獙璇乲=2銆傜Н鍒嗗洜瀛愬拰绉垎璁＄畻鏄富瑕佸け鍒嗙偣銆傞儴鍒嗚€冪敓鐢ㄧ壒瑙?榻愭瑙ｆ柟娉曚絾榻愭瑙ｅ父鍑洪敊銆?

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Question 8

Topic: 绾暟  |  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 depends only on the values $(u_j, u_{j+1})$, not on the position $j$. Since $(u_m, u_{m+1}) = (u_n, u_{n+1})$, their predecessors are equal: $(u_{m-1}, u_m) = (u_{n-1}, u_n)$.

Repeating this argument, we eventually get $(u_1, u_2) = (u_{n-m+1}, u_{n-m+2})$, i.e.\ $(1, 1) = (u_{n-m+1}, u_{n-m+2})$. But the only consecutive pair equal to $(1, 1)$ is $(u_1, u_2)$ itself. So $n - m + 1 = 1$, which means $n = m$, 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

60%灏濊瘯浣嗕粎~33%鍒嗘暟銆?i)甯歌閿欒锛氳璁や负鎵€鏈夊伓鏁颁笅鏍囬」涓?銆?ii)闇€鑰冭檻(u_{2k-1},u_{2k})鍜?u_{2k},u_{2k+1})涓ょ鎯呭喌銆?iii)‘consecutive’鍚箟鐞嗚В鏄叧閿紝寰堝浜洪敊璇０绉伴噸鐜板繀鍦?^n路k浣嶇疆銆?iv)(v)寰堝皯瀹屾垚锛岄渶灏?iv)鐨凟uclidean绠楁硶鎬濇兂涓?v)鐨勬弧灏勬€ц仈绯汇€?