STEP3 2010 -- Pure Mathematics

STEP3 2010 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2010  |  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	统计学 (Statistics)	Standard	求和运算变形, 代数化简, 不等式证明
2	积分 (Integration)	Challenging	部分分式分解, 双曲函数恒等式, 反常积分计算
3	复数与多项式 (Complex Numbers and Polynomials)	Challenging	单位根性质, 分圆多项式构造, 多项式因式分解, 反证法
4	多项式 (Polynomials)	Standard	公共根消元法, 结式理论, 代数变形
5	坐标几何 (Coordinate Geometry)	Standard	坐标法, 直线交点计算, 垂直条件验证, 直尺作图
6	三维旋转与坐标几何 (3D Rotations and Coordinate Geometry)	Hard	旋转矩阵,坐标变换,向量叉积,三角恒等式
7	微分方程与级数展开 (Differential Equations and Series Expansion)	Hard	隐函数求导,数学归纳法,幂级数展开,三角多项式
8	积分技巧与微分方程 (Integration Techniques and Differential Equations)	Challenging	商的求导法则逆用,待定系数法,积分因子法

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Question 1

Topic: 统计学 (Statistics)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Let $x_1, x_2, \dots, x_n$ and $x_{n+1}$ be any fixed real numbers. The numbers $A$ and $B$ are defined by

$A = \frac{1}{n} \sum_{k=1}^{n} x_k, \quad B = \frac{1}{n} \sum_{k=1}^{n} (x_k - A)^2,$

and the numbers $C$ and $D$ are defined by

$C = \frac{1}{n+1} \sum_{k=1}^{n+1} x_k, \quad D = \frac{1}{n+1} \sum_{k=1}^{n+1} (x_k - C)^2.$

(i) Express $C$ in terms of $A, x_{n+1}$ and $n$.

(ii) Show that $B = \frac{1}{n} \sum_{k=1}^{n} x_k^2 - A^2$.

(iii) Express $D$ in terms of $B, A, x_{n+1}$ and $n$.

Hence show that $(n+1)D \geqslant nB$ for all values of $x_{n+1}$, but that $D < B$ if and only if

$A - \sqrt{\frac{(n+1)B}{n}} < x_{n+1} < A + \sqrt{\frac{(n+1)B}{n}}.$

Hint

1. The first two parts are obtained by separating off the final term of the summation and expanding the brackets respectively giving $C = \frac{1}{n+1}(nA + x_{n+1})$, and

$B = \frac{1}{n} \sum_{k=1}^{n} {x_k}^2 - A^2$

(the latter given in the question) .

By comparison with the expression for B,

$D = \frac{1}{n+1} \sum_{k=1}^{n+1} {x_k}^2 - C^2$

which by substituting for

$\frac{1}{n} \sum_{k=1}^{n} {x_k}^2$

from the expression for B gives

$D = \frac{1}{n+1} [n(B + A^2) + {x_{n+1}}^2] - C^2$

Substituting for C from the initial result, the required expression can be obtained which can most neatly be written

$D = \frac{n}{(n+1)^2} [(n+1)B + (A - x_{n+1})^2]$

Thus $(n+1)D = nB + \frac{n}{n+1} (A - x_{n+1})^2$ yielding the first inequality.

Also, $D - B = \frac{n}{(n+1)^2} (A - x_{n+1})^2 - \frac{1}{n+1} B$ and this quadratic expression is only negative if and only if $(A - x_{n+1})^2 < \frac{n+1}{n} B$.

Rearranging the inequality to make $x_{n+1}$ the subject yields the required result.

Model Solution

Part (i)

We have

$C = \frac{1}{n+1} \sum_{k=1}^{n+1} x_k = \frac{1}{n+1} \left( \sum_{k=1}^{n} x_k + x_{n+1} \right).$

Since $A = \frac{1}{n} \sum_{k=1}^{n} x_k$, we get $\sum_{k=1}^{n} x_k = nA$. Therefore

$C = \frac{1}{n+1}(nA + x_{n+1}).$

Part (ii)

Expanding the definition of $B$:

$B = \frac{1}{n} \sum_{k=1}^{n} (x_k - A)^2 = \frac{1}{n} \sum_{k=1}^{n} (x_k^2 - 2Ax_k + A^2).$

Separating the terms:

$B = \frac{1}{n} \sum_{k=1}^{n} x_k^2 - \frac{2A}{n} \sum_{k=1}^{n} x_k + \frac{1}{n} \cdot nA^2 = \frac{1}{n} \sum_{k=1}^{n} x_k^2 - 2A \cdot A + A^2 = \frac{1}{n} \sum_{k=1}^{n} x_k^2 - A^2.$

Part (iii)

By the same method as part (ii),

$D = \frac{1}{n+1} \sum_{k=1}^{n+1} x_k^2 - C^2. \qquad \text{(*)}$

From part (ii), $\frac{1}{n} \sum_{k=1}^{n} x_k^2 = B + A^2$, so $\sum_{k=1}^{n} x_k^2 = n(B + A^2)$. Substituting into (*):

$D = \frac{1}{n+1}\left[n(B + A^2) + x_{n+1}^2\right] - C^2.$

Using $C = \frac{nA + x_{n+1}}{n+1}$ from part (i):

$D = \frac{nB + nA^2 + x_{n+1}^2}{n+1} - \frac{(nA + x_{n+1})^2}{(n+1)^2}.$

Expanding $(nA + x_{n+1})^2 = n^2 A^2 + 2nA x_{n+1} + x_{n+1}^2$ and putting over a common denominator of $(n+1)^2$:

$D = \frac{(n+1)(nB + nA^2 + x_{n+1}^2) - (n^2 A^2 + 2nA x_{n+1} + x_{n+1}^2)}{(n+1)^2}.$

The numerator expands to:

$n(n+1)B + n(n+1)A^2 + (n+1)x_{n+1}^2 - n^2 A^2 - 2nA x_{n+1} - x_{n+1}^2$

$= n(n+1)B + nA^2 + nx_{n+1}^2 - 2nA x_{n+1}$

$= n(n+1)B + n(A - x_{n+1})^2.$

Therefore

$D = \frac{n}{(n+1)^2}\left[(n+1)B + (A - x_{n+1})^2\right].$

Showing $(n+1)D \geqslant nB$

From the expression above:

$(n+1)D = \frac{n}{n+1}\left[(n+1)B + (A - x_{n+1})^2\right] = nB + \frac{n}{n+1}(A - x_{n+1})^2.$

Since $\frac{n}{n+1}(A - x_{n+1})^2 \geqslant 0$, we have $(n+1)D \geqslant nB$, with equality if and only if $x_{n+1} = A$.

Showing $D < B$ if and only if the stated inequality holds

$D - B = \frac{n}{(n+1)^2}\left[(n+1)B + (A - x_{n+1})^2\right] - B.$

Writing $B = \frac{(n+1)^2 B}{(n+1)^2}$:

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

So $D < B$ if and only if $n(A - x_{n+1})^2 - (n+1)B < 0$, i.e.,

$(A - x_{n+1})^2 < \frac{(n+1)B}{n}.$

Taking square roots (this is valid since $B \geqslant 0$):

$|A - x_{n+1}| < \sqrt{\frac{(n+1)B}{n}},$

which gives

$A - \sqrt{\frac{(n+1)B}{n}} < x_{n+1} < A + \sqrt{\frac{(n+1)B}{n}}.$

Examiner Notes

1. This was a very popular question, and the first two parts usually scored full marks. The expression of $D$ in part (iii) caused some problems with inaccurate algebra which then made the last two results unobtainable. Those that simplified $D$ most neatly were in a stronger position to finish the question, though “if and only if” was frequently ignored, or only lip-service was paid to it. Consequently, scores were well-spread.

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Question 2

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

Problem

2 In this question, $a$ is a positive constant.

(i) Express $\cosh a$ in terms of exponentials.

By using partial fractions, prove that

$\int_{0}^{1} \frac{1}{x^2 + 2x \cosh a + 1} \, dx = \frac{a}{2 \sinh a}.$

(ii) Find, expressing your answers in terms of hyperbolic functions,

$\int_{1}^{\infty} \frac{1}{x^2 + 2x \sinh a - 1} \, dx$

and

$\int_{0}^{\infty} \frac{1}{x^4 + 2x^2 \cosh a + 1} \, dx.$

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Hint

2. The expression of $\cosh a$ in exponentials enables the integral to be written as

$\int_{0}^{1} \frac{1}{x^2 + x(e^a + e^{-a}) + 1} dx$

which can in turn can be expressed as

$\int_{0}^{1} \frac{1}{(x + e^a)(x + e^{-a})} dx$

and so employing partial fractions this is

$\frac{1}{(e^a - e^{-a})} \left[ \ln \left( \frac{x + e^{-a}}{x + e^a} \right) \right]_{0}^{1}$

The evaluation of this with simplification of logarithms yields $\frac{1}{2 \sinh a} \left( \ln \left( e^a \frac{1 + e^a}{1 + e^a} \right) \right)$ giving the required result. In part (ii), the same technique can be employed for both integrals giving, in the first case $\begin{aligned} \int_1^\infty \frac{1}{(x + e^a)(x - e^{-a})} dx &= \frac{1}{(e^a + e^{-a})} \left[ \ln \left( \frac{x - e^{-a}}{x + e^a} \right) \right]_1^\infty \\ &= \frac{1}{2 \cosh a} \left( a + \ln \left( \text{coth} \frac{a}{2} \right) \right) \end{aligned}$ and in the second $\begin{aligned} \int_0^\infty \frac{1}{(x^2 + e^a)(x^2 + e^{-a})} dx &= \frac{1}{(e^a - e^{-a})} \left[ \frac{1}{e^{\frac{a}{2}}} \tan^{-1} \left( \frac{x}{e^{\frac{a}{2}}} \right) - \frac{1}{e^{-\frac{a}{2}}} \tan^{-1} \left( \frac{x}{e^{-\frac{a}{2}}} \right) \right]_0^\infty \\ &= \frac{1}{2 \sinh a} \left( \frac{\pi}{2} \sinh \frac{a}{2} \right) \end{aligned}$ or alternatively $\frac{\pi}{4 \cosh \frac{a}{2}}$

Model Solution

Part (i)

By definition, $\cosh a = \frac{e^a + e^{-a}}{2}$.

The denominator of the integrand factors as:

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

Using partial fractions:

$\frac{1}{(x + e^a)(x + e^{-a})} = \frac{1}{e^a - e^{-a}}\left(\frac{1}{x + e^{-a}} - \frac{1}{x + e^a}\right) = \frac{1}{2\sinh a}\left(\frac{1}{x + e^{-a}} - \frac{1}{x + e^a}\right).$

Integrating:

$\int_0^1 \frac{dx}{(x + e^a)(x + e^{-a})} = \frac{1}{2\sinh a}\left[\ln\left(\frac{x + e^{-a}}{x + e^a}\right)\right]_0^1.$

Evaluating at the limits:

At $x = 1$: $\frac{1 + e^{-a}}{1 + e^a}$.

At $x = 0$: $\frac{e^{-a}}{e^a} = e^{-2a}$.

So the integral equals:

$\frac{1}{2\sinh a}\left[\ln\left(\frac{1 + e^{-a}}{1 + e^a}\right) - \ln(e^{-2a})\right] = \frac{1}{2\sinh a}\left[\ln\left(\frac{1 + e^{-a}}{1 + e^a}\right) + 2a\right].$

Simplifying the logarithm:

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

Therefore the integral equals:

$\frac{1}{2\sinh a}\left[\ln(e^{-a}) + 2a\right] = \frac{1}{2\sinh a}(-a + 2a) = \frac{a}{2\sinh a}. \qquad \text{(shown)}$

Part (ii) — First integral

We factor $x^2 + 2x\sinh a - 1$. Using $\sinh a = \frac{e^a - e^{-a}}{2}$:

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

We verify: $(x + e^a)(x - e^{-a}) = x^2 + x(e^a - e^{-a}) - e^a \cdot e^{-a} = x^2 + 2x\sinh a - 1$. Confirmed.

Using partial fractions:

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

Integrating from 1 to $\infty$:

$\int_1^{\infty} \frac{dx}{(x + e^a)(x - e^{-a})} = \frac{1}{2\cosh a}\left[\ln\left(\frac{x - e^{-a}}{x + e^a}\right)\right]_1^{\infty}.$

As $x \to \infty$: $\frac{x - e^{-a}}{x + e^a} \to 1$, so $\ln(1) = 0$.

At $x = 1$: $\frac{1 - e^{-a}}{1 + e^a}$.

Simplifying: $\frac{1 - e^{-a}}{1 + e^a} = \frac{e^{-a}(e^a - 1)}{e^a(e^{-a} + 1)} \cdot \frac{e^a}{e^{-a}} = \frac{e^a - 1}{e^a + 1} = \tanh\frac{a}{2}$.

Therefore the integral equals:

$\frac{1}{2\cosh a}\left[0 - \ln\left(\tanh\frac{a}{2}\right)\right] = \frac{1}{2\cosh a}\ln\left(\coth\frac{a}{2}\right).$

Part (ii) — Second integral

We factor $x^4 + 2x^2\cosh a + 1$ as a quadratic in $x^2$. The roots are:

$x^2 = \frac{-2\cosh a \pm \sqrt{4\cosh^2 a - 4}}{2} = -\cosh a \pm \sinh a = -e^{\pm a}.$

So $x^4 + 2x^2\cosh a + 1 = (x^2 + e^a)(x^2 + e^{-a})$.

Using partial fractions in $x^2$:

$\frac{1}{(x^2 + e^a)(x^2 + e^{-a})} = \frac{1}{e^a - e^{-a}}\left(\frac{1}{x^2 + e^{-a}} - \frac{1}{x^2 + e^a}\right) = \frac{1}{2\sinh a}\left(\frac{1}{x^2 + e^{-a}} - \frac{1}{x^2 + e^a}\right).$

Integrating from $0$ to $\infty$:

$\int_0^{\infty} \frac{dx}{x^2 + e^{-a}} = \frac{1}{\sqrt{e^{-a}}}\left[\arctan\frac{x}{\sqrt{e^{-a}}}\right]_0^{\infty} = \frac{1}{e^{-a/2}} \cdot \frac{\pi}{2} = \frac{\pi}{2}e^{a/2}.$

$\int_0^{\infty} \frac{dx}{x^2 + e^a} = \frac{1}{\sqrt{e^a}} \cdot \frac{\pi}{2} = \frac{\pi}{2}e^{-a/2}.$

Therefore the integral equals:

$\frac{1}{2\sinh a}\left(\frac{\pi}{2}e^{a/2} - \frac{\pi}{2}e^{-a/2}\right) = \frac{\pi}{4\sinh a}\left(e^{a/2} - e^{-a/2}\right) = \frac{\pi \cdot 2\sinh(a/2)}{4\sinh a}.$

Using $\sinh a = 2\sinh(a/2)\cosh(a/2)$:

$\frac{\pi \cdot 2\sinh(a/2)}{4 \cdot 2\sinh(a/2)\cosh(a/2)} = \frac{\pi}{4\cosh(a/2)}.$

Examiner Notes

2. The most popular question, the scoring rate was very similar to the first. Quite a few candidates did not take the hint provided in part (i) to express $\cosh a$ in terms of exponentials in order to perform the integration. However, apart from those that did not correctly substantiate the given result, many handled the partial fractions and exponentials well, and quite a number dealt with the infinite limit impressively. Problems arose later in the question with manipulating logarithms and the instruction to express answers in terms of hyperbolic functions was either overlooked or beyond their capacity.

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Question 3

Topic: 复数与多项式 (Complex Numbers and Polynomials)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 For any given positive integer $n$, a number $a$ (which may be complex) is said to be a primitive $n$th root of unity if $a^n = 1$ and there is no integer $m$ such that $0 < m < n$ and $a^m = 1$. Write down the two primitive 4th roots of unity.

Let $C_n(x)$ be the polynomial such that the roots of the equation $C_n(x) = 0$ are the primitive $n$th roots of unity, the coefficient of the highest power of $x$ is one and the equation has no repeated roots. Show that $C_4(x) = x^2 + 1$.

(i) Find $C_1(x)$, $C_2(x)$, $C_3(x)$, $C_5(x)$ and $C_6(x)$, giving your answers as unfactorised polynomials.

(ii) Find the value of $n$ for which $C_n(x) = x^4 + 1$.

(iii) Given that $p$ is prime, find an expression for $C_p(x)$, giving your answer as an unfactorised polynomial.

(iv) Prove that there are no positive integers $q$, $r$ and $s$ such that $C_q(x) \equiv C_r(x)C_s(x)$.

Hint

3. The two primitive 4^th roots of unity are $\pm i$ so $C_4(x) = (x - i)(x + i) = x^2 + 1$ $C_1(x) = x - 1$, $x^2 - 1 = (x - 1)(x + 1)$ so $C_2(x) = x + 1$, $x^3 - 1 = (x - 1)(x^2 + x + 1)$ so $C_3(x) = x^2 + x + 1$ $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$ so $C_5(x) = x^4 + x^3 + x^2 + x + 1$ $x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x^3 - 1)(x + 1)(x^2 - x + 1)$ so $C_6(x) = x^2 - x + 1$ In part (ii), $C_n(x) = 0 \Rightarrow x^4 = -1 \Rightarrow x^8 = 1$ so n is a multiple of 8, and as there are 4 primitive 8^th roots of unity, n must be 8. $x^p = 1 \Rightarrow x^p - 1 = 0 \Rightarrow (x - 1)(x^{p-1} + x^{p-2} + x^{p-3} + \dots + 1)$ 1 is the only non-primitive root as no power of any other root less than the $p^{\text{th}}$ equals unity, because $p$ is prime, so $C_p(x) = x^{p-1} + x^{p-2} + x^{p-3} + \dots + 1$ No root of $C_n(x) = 0$ is a root of $C_t(x) = 0$ for any $t \neq n$. (For if $t < n$, by the definition of $C_n(x)$, there is no integer $t$ such that $a^t = 1$ when $a^n = 1$. Similarly, if $t > n$.) Thus if $C_q(x) \equiv C_r(x)C_s(x)$, and if $C_q(x) = 0$, then $C_r(x) = 0$ or $C_s(x) = 0$, so $q = r$ or $q = s$.

If $q = r$, then $C_q(x) \equiv C_r(x)$, and so $C_s(x) \equiv 1$ which is not possible for positive $s$, and likewise in the alternative case.

Model Solution

The primitive 4th roots of unity are $i$ and $-i$.

To see this, the 4th roots of unity satisfy $a^4 = 1$, giving $a \in \{1, -1, i, -i\}$. We check which have no smaller positive power equal to 1:

• $1^1 = 1$, so 1 is not primitive.
• $(-1)^2 = 1$, so $-1$ is not primitive.
• $i^2 = -1 \neq 1$ and $i^3 = -i \neq 1$, so $i$ is primitive.
• $(-i)^2 = -1 \neq 1$ and $(-i)^3 = i \neq 1$, so $-i$ is primitive.

Hence $C_4(x) = (x - i)(x + i) = x^2 + 1$.

Part (i)

$C_1(x)$: The only 1st root of unity is 1, which is trivially primitive. So $C_1(x) = x - 1$.

$C_2(x)$: The 2nd roots of unity are 1 and $-1$. Since $1$ is not primitive (it has order 1), the only primitive 2nd root is $-1$. So $C_2(x) = x + 1$.

$C_3(x)$: We have $x^3 - 1 = (x - 1)(x^2 + x + 1)$. The roots of $x^2 + x + 1$ are $\omega = e^{2\pi i/3}$ and $\omega^2 = e^{4\pi i/3}$. Since 3 is prime, neither $\omega$ nor $\omega^2$ satisfies $a^m = 1$ for any $0 < m < 3$. So $C_3(x) = x^2 + x + 1$.

$C_5(x)$: We have $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$. Since 5 is prime, every 5th root of unity other than 1 is primitive. So $C_5(x) = x^4 + x^3 + x^2 + x + 1$.

$C_6(x)$: We have $x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)$. The 6th roots of unity are $e^{2\pi i k/6}$ for $k = 0, 1, 2, 3, 4, 5$:

• $k = 0$: $1$ has order 1 (not primitive).
• $k = 1$: $e^{i\pi/3}$ has order 6 (primitive).
• $k = 2$: $e^{2i\pi/3}$ has order 3 (not primitive).
• $k = 3$: $-1$ has order 2 (not primitive).
• $k = 4$: $e^{4i\pi/3}$ has order 3 (not primitive).
• $k = 5$: $e^{5i\pi/3}$ has order 6 (primitive).

So the primitive 6th roots are $e^{i\pi/3}$ and $e^{5i\pi/3}$, which are the roots of $x^2 - x + 1$. Hence $C_6(x) = x^2 - x + 1$.

Part (ii)

We need $C_n(x) = x^4 + 1$. The roots of $x^4 + 1 = 0$ satisfy $x^4 = -1$, so $x^8 = 1$. Squaring gives $x^8 = 1$, so these roots are 8th roots of unity. We verify they are primitive: the four roots are $e^{i\pi/4}$, $e^{3i\pi/4}$, $e^{5i\pi/4}$, $e^{7i\pi/4}$, i.e., $e^{2\pi i k/8}$ for $k = 1, 3, 5, 7$. Each has order exactly 8 (since $\gcd(k, 8) = 1$ for $k \in \{1, 3, 5, 7\}$), confirming they are primitive 8th roots of unity.

Since $C_n(x)$ is monic with roots exactly the primitive $n$th roots of unity, and these four roots are precisely the primitive 8th roots, we conclude $n = 8$.

Part (iii)

For prime $p$, we have the factorisation $x^p - 1 = (x - 1)(x^{p-1} + x^{p-2} + \cdots + x + 1)$.

The $p$th roots of unity are $1, \zeta, \zeta^2, \dots, \zeta^{p-1}$ where $\zeta = e^{2\pi i/p}$. The root 1 is not primitive (it has order 1). For any other root $\zeta^k$ with $1 \leqslant k \leqslant p - 1$: if $(\zeta^k)^m = 1$ for some positive $m$, then $p \mid km$. Since $p$ is prime and $1 \leqslant k < p$, we have $\gcd(k, p) = 1$, so $p \mid m$. Thus $m \geqslant p$, meaning $\zeta^k$ has order exactly $p$ and is primitive.

Therefore all roots of $x^{p-1} + x^{p-2} + \cdots + x + 1 = 0$ are primitive $p$th roots, and this polynomial is monic with no repeated roots. Hence

$C_p(x) = x^{p-1} + x^{p-2} + \cdots + x + 1.$

Part (iv)

We prove this by contradiction. The key property of cyclotomic polynomials is that their roots are disjoint: no primitive $n$th root of unity is a primitive $m$th root for $m \neq n$.

Proof that roots are disjoint: Suppose $\alpha$ is both a primitive $n$th root and a primitive $m$th root with $n < m$. Since $\alpha$ is a primitive $n$th root, $\alpha^n = 1$. But $\alpha$ is also an $m$th root of unity, so $a^m = 1$. However, $m > n > 0$ and $\alpha^m = (\alpha^n)^{m/n}$… more precisely, since $\alpha^n = 1$, we have $\alpha^m = \alpha^{m - n} \cdot \alpha^n = \alpha^{m-n}$. If $m > n$ then $0 < m - n < m$ and $\alpha^{m-n} = 1$, contradicting $\alpha$ being a primitive $m$th root. A symmetric argument handles $n > m$.

Now suppose for contradiction that $C_q(x) \equiv C_r(x) C_s(x)$ for positive integers $q, r, s$. Let $\alpha$ be any root of $C_r(x) = 0$. Then $\alpha$ is a primitive $r$th root of unity, and since $C_q(\alpha) = C_r(\alpha) C_s(\alpha) = 0$, $\alpha$ is also a primitive $q$th root. By the disjointness property, $q = r$.

Similarly, let $\beta$ be any root of $C_s(x) = 0$. Then $C_q(\beta) = C_r(\beta) C_s(\beta) = 0$, so $\beta$ is a primitive $q$th root and also a primitive $s$th root, giving $q = s$.

Therefore $r = s = q$, and the equation becomes $C_q(x) \equiv (C_q(x))^2$. Since $C_q(x)$ has degree $\varphi(q) \geqslant 1$ for all positive $q$ (where $\varphi$ is Euler’s totient function), this is impossible: the left side has degree $\varphi(q)$ while the right side has degree $2\varphi(q)$.

Hence there are no positive integers $q, r, s$ such that $C_q(x) \equiv C_r(x) C_s(x)$.

Examiner Notes

3. Just over half the candidates attempted this question with most scores being quarter, half or three quarters in equal shares. Most candidates understood the idea of the question, the definition of a primitive root, and many wrote the roots of unity in (modulus)-argument form or exponential form. Failure to present a logical argument in parts (ii) and (iv) was a common problem and $C_6(x)$ tripped up quite a few.

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Question 4

Topic: 多项式 (Polynomials)  |  Difficulty: Standard  |  Marks: 20

Problem

4 (i) The number $\alpha$ is a common root of the equations $x^2 + ax + b = 0$ and $x^2 + cx + d = 0$ (that is, $\alpha$ satisfies both equations). Given that $a \neq c$, show that

$\alpha = -\frac{b - d}{a - c} .$

Hence, or otherwise, show that the equations have at least one common root if and only if

$(b - d)^2 - a(b - d)(a - c) + b(a - c)^2 = 0 .$

Does this result still hold if the condition $a \neq c$ is not imposed?

(ii) Show that the equations $x^2 + ax + b = 0$ and $x^3 + (a + 1)x^2 + qx + r = 0$ have at least one common root if and only if

$(b - r)^2 - a(b - r)(a + b - q) + b(a + b - q)^2 = 0 .$

Hence, or otherwise, find the values of $b$ for which the equations $2x^2 + 5x + 2b = 0$ and $2x^3 + 7x^2 + 5x + 1 = 0$ have at least one common root.

Hint

4. (i) As $\alpha$ satisfies both equations, $\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + c\alpha + d = 0$, so subtracting these the desired result is simply found.

If $(b - d)^2 - a(b - d)(a - c) + b(a - c)^2 = 0$, then we may divide by $(a - c)^2$, and find that $-\frac{(b-d)}{(a-c)}$ satisfies $x^2 + ax + b = 0$. But also,

$\left(\frac{(b-d)}{(a-c)}\right)^2 + c\left(-\frac{(b-d)}{(a-c)}\right) + d = \left(\frac{(b-d)}{(a-c)}\right)^2 + a\left(-\frac{(b-d)}{(a-c)}\right) + b + (c - a)\left(-\frac{(b-d)}{(a-c)}\right) + (d - b)$ and so $-\frac{(b-d)}{(a-c)}$ satisfies $x^2 + cx + d = 0$.

On the other hand if there is a common root, then it is found at the start of the question and as it satisfies $\alpha^2 + a\alpha + b = 0$, the required result is found.

If $(b - d)^2 - a(b - d)(a - c) + b(a - c)^2 = 0$ and $a = c$, then $b = d$ and so the two equations are one and trivially have a common root. Alternatively, if there is a common root and $a = c$, then the initial subtraction yields $b = d$, and so the result is trivially true.

(ii) If $(b - r)^2 - a(b - r)(a + b - q) + b(a + b - q)^2 = 0$, then $x^2 + ax + b = 0$ and $x^2 + (q - b)x + r = 0$ have a common root from (i), and so then do $x^2 + ax + b = 0$ and $x(x^2 + ax + b) + x^2 + (q - b)x + r = 0$ which is the required result. On the other hand, if the two equations have a common root $\alpha$, then $\alpha^2 + a\alpha + b = 0$ and $\alpha^3 + (a + 1)\alpha^2 + q\alpha + r = 0$, and thus so does

$\alpha^3 + (a + 1)\alpha^2 + q\alpha + r - \alpha(\alpha^2 + a\alpha + b) = 0$ which is a quadratic equation and we can use the result from (i) again.

Using $a = \frac{5}{2}$, $q = \frac{5}{2}$, $r = \frac{1}{2}$, in the given condition, we obtain a cubic equation in $b$, $b^3 - \frac{3}{2}b^2 + \frac{1}{4}b + \frac{1}{4} = 0$, which has a solution $b = 1$, meaning the other two can be simply obtained as $b = \frac{1 \pm \sqrt{5}}{4}$.

Model Solution

Part (i)

Since $\alpha$ is a common root, it satisfies both equations:

$\alpha^2 + a\alpha + b = 0 \qquad \text{and} \qquad \alpha^2 + c\alpha + d = 0.$

Subtracting the second from the first:

$(a - c)\alpha + (b - d) = 0.$

Since $a \neq c$:

$\alpha = -\frac{b - d}{a - c}. \qquad \text{(shown)}$

Showing the “if and only if” condition

$(\Rightarrow)$ Suppose there is a common root $\alpha$. Then $\alpha = -\frac{b-d}{a-c}$ and $\alpha^2 + a\alpha + b = 0$. Substituting:

$\frac{(b-d)^2}{(a-c)^2} - \frac{a(b-d)}{a-c} + b = 0.$

Multiplying by $(a-c)^2$:

$(b-d)^2 - a(b-d)(a-c) + b(a-c)^2 = 0. \qquad \text{(shown)}$

$(\Leftarrow)$ Suppose $(b-d)^2 - a(b-d)(a-c) + b(a-c)^2 = 0$. Dividing by $(a-c)^2$ (valid since $a \neq c$):

$\left(\frac{b-d}{a-c}\right)^2 - a\left(\frac{b-d}{a-c}\right) + b = 0.$

This shows $\alpha = -\frac{b-d}{a-c}$ satisfies $x^2 + ax + b = 0$. We verify it also satisfies $x^2 + cx + d = 0$:

$\alpha^2 + c\alpha + d = (\alpha^2 + a\alpha + b) + (c - a)\alpha + (d - b) = 0 + (c-a)\left(-\frac{b-d}{a-c}\right) + (d-b)$

$= (d-b) + (d-b) = 0.$

So $\alpha$ is a common root.

When $a = c$: If $a = c$, the condition becomes $(b-d)^2 + 0 + 0 = 0$, so $b = d$, meaning the two equations are identical and trivially have common roots. Conversely, if there is a common root and $a = c$, subtracting the equations gives $b - d = 0$. So the result holds regardless of whether $a \neq c$ is imposed.

Part (ii)

Let $\alpha$ be a common root of $x^2 + ax + b = 0$ and $x^3 + (a+1)x^2 + qx + r = 0$. From the cubic:

$\alpha^3 + (a+1)\alpha^2 + q\alpha + r = 0.$

Subtracting $\alpha \cdot (\alpha^2 + a\alpha + b) = 0$:

$\alpha^3 + (a+1)\alpha^2 + q\alpha + r - \alpha^3 - a\alpha^2 - b\alpha = 0$

$\alpha^2 + (q - b)\alpha + r = 0.$

So $\alpha$ is a common root of $x^2 + ax + b = 0$ and $x^2 + (q-b)x + r = 0$. By part (i) (the “if and only if” result with $a \to a$, $b \to b$, $c \to q-b$, $d \to r$):

$(b - r)^2 - a(b - r)(a - (q-b)) + b(a - (q-b))^2 = 0$

$(b - r)^2 - a(b - r)(a + b - q) + b(a + b - q)^2 = 0. \qquad \text{(shown)}$

This argument is reversible: if this condition holds, part (i) guarantees a common root of the two quadratics, which is also a root of the cubic (since the cubic = quadratic $\times x$ + the second quadratic).

Finding the values of $b$

The equations are $2x^2 + 5x + 2b = 0$ and $2x^3 + 7x^2 + 5x + 1 = 0$. Dividing the first by 2: $x^2 + \frac{5}{2}x + b = 0$, so $a = \frac{5}{2}$. Dividing the second by 2: $x^3 + \frac{7}{2}x^2 + \frac{5}{2}x + \frac{1}{2} = 0$, so $a + 1 = \frac{7}{2}$ (confirmed: $a = \frac{5}{2}$), $q = \frac{5}{2}$, $r = \frac{1}{2}$.

Now $a + b - q = \frac{5}{2} + b - \frac{5}{2} = b$ and $b - r = b - \frac{1}{2}$. Substituting into the condition:

$\left(b - \frac{1}{2}\right)^2 - \frac{5}{2}\left(b - \frac{1}{2}\right)b + b \cdot b^2 = 0.$

Expanding:

$b^2 - b + \frac{1}{4} - \frac{5}{2}b^2 + \frac{5}{4}b + b^3 = 0$

$b^3 - \frac{3}{2}b^2 + \frac{1}{4}b + \frac{1}{4} = 0.$

Multiplying by 4:

$4b^3 - 6b^2 + b + 1 = 0.$

Testing $b = 1$: $4 - 6 + 1 + 1 = 0$. So $(b - 1)$ is a factor. Dividing:

$4b^3 - 6b^2 + b + 1 = (b - 1)(4b^2 - 2b - 1).$

The quadratic $4b^2 - 2b - 1 = 0$ gives $b = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{1 \pm \sqrt{5}}{4}$.

The values of $b$ are $b = 1$, $b = \frac{1 + \sqrt{5}}{4}$, and $b = \frac{1 - \sqrt{5}}{4}$.

Examiner Notes

4. This was a popular question, though it was not generally well scored upon, with very few candidates earning full marks. Most began strongly, and finished by finding the values of $b$ correctly. However, basic sign errors did prevent some from achieving the numerical pay-off. Part (ii) was, as expected, found trickier than part (i). Overall, the non-triviality of “if and only if” was rarely addressed as an issue in either part.

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Question 5

Topic: 坐标几何 (Coordinate Geometry)  |  Difficulty: Standard  |  Marks: 20

Problem

5 The vertices $A, B, C$ and $D$ of a square have coordinates $(0, 0), (a, 0), (a, a)$ and $(0, a)$, respectively. The points $P$ and $Q$ have coordinates $(an, 0)$ and $(0, am)$ respectively, where $0 < m < n < 1$. The line $CP$ produced meets $DA$ produced at $R$ and the line $CQ$ produced meets $BA$ produced at $S$. The line $PQ$ produced meets the line $RS$ produced at $T$. Show that $TA$ is perpendicular to $AC$.

Explain how, given a square of area $a^2$, a square of area $2a^2$ may be constructed using only a straight-edge.

[Note: a straight-edge is a ruler with no markings on it; no measurements (and no use of compasses) are allowed in the construction.]

Hint

5. The line CP can be shown to have equation $(1 - n)y = x - an$ and so R is $\left(0, \frac{an}{n-1}\right)$

So, similarly, S must be $\left(\frac{am}{m-1}, 0\right)$. Thus RS has equation $n(m - 1)x + m(n - 1)y = amn$ and PQ has equation $mx + ny = amn$. As the coordinates of T satisfy both equations, they satisfy their difference which is $(mn - n - m)(x + y) = 0$. As RS and PQ intersect, $\frac{n}{m} \neq \frac{m(n-1)}{n(m-1)}$ which yields $(m - n)(mn - m - n) \neq 0$ and hence $(mn - m - n) \neq 0$ implying that T’s coordinates satisfy $x + y = 0$ giving the desired result. (Alternatively, $mn - m - n = 0 \Leftrightarrow n = \frac{m}{m-1} < 0$, which is a contradiction.)

The construction can be achieved more than one way, but one is to label the given square ABCD anti-clockwise, choose points on AB and AD different distances from A, label them P and Q, construct CP and CQ, and find their intersections with AD and AB, R and S, respectively, and find the intersection of PQ and RS, label it T, then TA is perpendicular to AC. Rotating the labelling through a right angle and repeating three more times achieves the desired square.

Model Solution

Setting up coordinates

The vertices of the square are $A(0,0)$, $B(a,0)$, $C(a,a)$, $D(0,a)$. The points are $P(an, 0)$ and $Q(0, am)$ with $0 < m < n < 1$.

Finding $R$

The line $CP$ passes through $C(a,a)$ and $P(an, 0)$. Its parametric form:

$\frac{x - a}{an - a} = \frac{y - a}{0 - a}, \quad \text{i.e.,} \quad \frac{x - a}{a(n-1)} = \frac{y - a}{-a}.$

Simplifying: $-(x - a) = (n-1)(y - a)$, giving $x - a = -(n-1)(y - a)$, so $x = a - (n-1)(y - a) = a + (n-1)(a - y) = a(1 + n - 1) - (n-1)y = an - (n-1)y$.

Wait, let me redo this more carefully. The line through $C(a,a)$ and $P(an, 0)$:

Direction: $(an - a, 0 - a) = (a(n-1), -a)$.

Parametric: $(x, y) = (a, a) + t(a(n-1), -a)$.

So $x = a + ta(n-1)$, $y = a - ta$.

The line $DA$ produced is the line $x = 0$ extended beyond $A$, i.e., $x = 0$ with $y < 0$.

Setting $x = 0$: $a + ta(n-1) = 0$, so $t = \frac{-1}{n-1} = \frac{1}{1-n}$.

Then $y = a - \frac{a}{1-n} = a\left(1 - \frac{1}{1-n}\right) = a \cdot \frac{1-n-1}{1-n} = \frac{-an}{1-n} = \frac{an}{n-1}$.

Since $0 < n < 1$, we have $n - 1 < 0$, so $y < 0$. Thus $R = \left(0, \frac{an}{n-1}\right)$.

Finding $S$

The line $CQ$ passes through $C(a,a)$ and $Q(0, am)$.

Direction: $(0 - a, am - a) = (-a, a(m-1))$.

Parametric: $(x, y) = (a, a) + t(-a, a(m-1))$.

So $x = a - ta$, $y = a + ta(m-1)$.

The line $BA$ produced is the line $y = 0$ extended beyond $A$, i.e., $y = 0$ with $x < 0$.

Setting $y = 0$: $a + ta(m-1) = 0$, so $t = \frac{-1}{m-1} = \frac{1}{1-m}$.

Then $x = a - \frac{a}{1-m} = a\left(1 - \frac{1}{1-m}\right) = a \cdot \frac{-m}{1-m} = \frac{-am}{1-m} = \frac{am}{m-1}$.

Since $0 < m < 1$, we have $m - 1 < 0$, so $x < 0$. Thus $S = \left(\frac{am}{m-1}, 0\right)$.

Equations of lines $RS$ and $PQ$

Line $RS$ through $R\left(0, \frac{an}{n-1}\right)$ and $S\left(\frac{am}{m-1}, 0\right)$:

$\frac{x}{\frac{am}{m-1}} + \frac{y}{\frac{an}{n-1}} = 1.$

Multiplying through: $\frac{(m-1)x}{am} + \frac{(n-1)y}{an} = 1$, so $n(m-1)x + m(n-1)y = amn$. $\qquad \text{(RS)}$

Line $PQ$ through $P(an, 0)$ and $Q(0, am)$:

$\frac{x}{an} + \frac{y}{am} = 1, \quad \text{i.e.,} \quad mx + ny = amn. \qquad \text{(PQ)}$

Finding $T$

$T$ lies on both lines. Subtracting equation (PQ) from equation (RS):

$n(m-1)x + m(n-1)y - mx - ny = 0$

$(mn - n - m)x + (mn - m - n)y = 0$

$(mn - m - n)(x + y) = 0.$

We need $mn - m - n \neq 0$. If $mn - m - n = 0$, then $n(m-1) = m$, so $n = \frac{m}{m-1}$. But $m < 1$ means $m - 1 < 0$, so $n = \frac{m}{m-1} < 0$, contradicting $n > 0$.

Therefore $x + y = 0$ for the point $T$.

Showing $TA \perp AC$

Since $T = (t, -t)$ for some value $t$, and $A = (0,0)$, the direction of $TA$ is $(t, -t) = t(1, -1)$.

The diagonal $AC$ has direction $(a, a) = a(1, 1)$.

The dot product: $(1)(1) + (-1)(1) = 0$, so $TA \perp AC$. $\qquad \text{(shown)}$

Construction with a straight-edge

Given a square $ABCD$ of area $a^2$:

1. Label the vertices $A, B, C, D$ anti-clockwise.
2. Choose points $P$ on $AB$ and $Q$ on $AD$, at different distances from $A$.
3. Draw line $CP$ and extend it to meet line $DA$ produced at $R$.
4. Draw line $CQ$ and extend it to meet line $BA$ produced at $S$.
5. Draw line $PQ$ and line $RS$; let them meet at $T$.
6. By the result above, $TA \perp AC$. Since $A$ is the centre of the square with diagonal $AC$, the line $TA$ is the perpendicular bisector of $AC$, meaning $T$ lies on the line through $A$ perpendicular to the diagonal.
7. Now rotate the labels $A \to B \to C \to D \to A$ and repeat the construction. This gives a point $T'$ such that $T'B \perp BD$. Repeating for all four vertices gives four points, each on the perpendicular to a diagonal through a vertex.
8. These four points form a square whose diagonal equals the side of the original square times $\sqrt{2}$, so its area is $2a^2$.

More precisely: the original square has diagonal $a\sqrt{2}$. A square with this diagonal as its side has area $(a\sqrt{2})^2 = 2a^2$. The construction achieves this by locating points that define a square of the required dimensions using only straight lines.

Examiner Notes

5. This question resembled question 3 in popularity and success. Most were able to derive line equations reliably, and address the intersection problem. (Those that used an equally valid vector formulism had a low success rate for no apparent reason.) Very few addressed whether or not factors that were being divided by were non-zero. Mistaking $m$ for $n$ and vice versa, careless algebraic errors, and overlooking which equation represented which line caused problems in trying to find $T$. The idea of explaining the construction verbally in the last part exposed that many candidates are not used to expressing a formal argument in words. The nicety of this question is that whilst all candidates will have encountered geometrical constructions involving straight edge and compass, few will have previously met one that only requires a straight edge.

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Question 6

Topic: 三维旋转与坐标几何 (3D Rotations and Coordinate Geometry)  |  Difficulty: Hard  |  Marks: 20

Problem

6 The points $P, Q$ and $R$ lie on a sphere of unit radius centred at the origin, $O$, which is fixed. Initially, $P$ is at $P_0(1, 0, 0), Q$ is at $Q_0(0, 1, 0)$ and $R$ is at $R_0(0, 0, 1)$.

(i) The sphere is then rotated about the $z$-axis, so that the line $OP$ turns directly towards the positive $y$-axis through an angle $\phi$. The position of $P$ after this rotation is denoted by $P_1$. Write down the coordinates of $P_1$.

(ii) The sphere is now rotated about the line in the $x$-$y$ plane perpendicular to $OP_1$, so that the line $OP$ turns directly towards the positive $z$-axis through an angle $\lambda$. The position of $P$ after this rotation is denoted by $P_2$. Find the coordinates of $P_2$. Find also the coordinates of the points $Q_2$ and $R_2$, which are the positions of $Q$ and $R$ after the two rotations.

(iii) The sphere is now rotated for a third time, so that $P$ returns from $P_2$ to its original position $P_0$. During the rotation, $P$ remains in the plane containing $P_0, P_2$ and $O$. Show that the angle of this rotation, $\theta$, satisfies

$\cos \theta = \cos \phi \cos \lambda ,$

and find a vector in the direction of the axis about which this rotation takes place.

Hint

6. $P_1$ is $(\cos \varphi, \sin \varphi, 0)$, $P_2$ is $(\cos \varphi \cos \lambda, \sin \varphi \cos \lambda, \sin \lambda)$, $Q_1$ is $(-\sin \varphi, \cos \varphi, 0)$, $Q_2$ is $(-\sin \varphi, \cos \varphi, 0)$, $R_1$ is $(0,0,1)$ and $R_2$ is $(-\cos \varphi \sin \lambda, -\sin \varphi \sin \lambda, \cos \lambda)$.

The scalar product $OP_2 \cdot OP_0$ gives the quoted result immediately. The direction of the axis can

be found from the vector product $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} \cos \varphi \cos \lambda \\ \sin \varphi \cos \lambda \\ \sin \lambda \end{pmatrix}$ giving the direction of the axis as

$\begin{pmatrix} 0 \\ -\sin \lambda \\ \sin \varphi \cos \lambda \end{pmatrix}$.

Model Solution

Part (i)

Rotating about the $z$-axis by angle $\phi$ (turning $OP$ towards the positive $y$-axis), the rotation matrix is:

$R_z(\phi) = \begin{pmatrix} \cos\phi & -\sin\phi & 0 \\ \sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix}.$

Applying to $P_0 = (1, 0, 0)$:

$P_1 = (\cos\phi, \sin\phi, 0).$

Part (ii)

The second rotation is about the line in the $xy$-plane perpendicular to $OP_1$, turning $OP$ towards the positive $z$-axis through angle $\lambda$.

The axis of rotation is perpendicular to both $OP_1 = (\cos\phi, \sin\phi, 0)$ and $\hat{k} = (0, 0, 1)$. Since $OP_1$ lies in the $xy$-plane, the perpendicular direction in the $xy$-plane is $\hat{u} = (-\sin\phi, \cos\phi, 0)$.

This rotation takes $P_1 = (\cos\phi, \sin\phi, 0)$ towards the $z$-axis. The component of $P_1$ along $\hat{u}$ is zero (since $OP_1 \perp \hat{u}$), so $P_1$ lies entirely in the plane spanned by $\hat{u}^{\perp}$ (i.e., $OP_1$ direction) and $\hat{k}$. The rotation by $\lambda$ about $\hat{u}$ gives:

$P_2 = \cos\lambda \cdot (\cos\phi, \sin\phi, 0) + \sin\lambda \cdot (0, 0, 1) = (\cos\phi\cos\lambda, \sin\phi\cos\lambda, \sin\lambda).$

Finding $Q_2$

After the first rotation ($z$-axis by $\phi$):

$Q_1 = R_z(\phi) \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = (-\sin\phi, \cos\phi, 0).$

Note that $Q_1$ is exactly the direction $\hat{u}$, the axis of the second rotation. A point on the rotation axis is unchanged, so:

$Q_2 = Q_1 = (-\sin\phi, \cos\phi, 0).$

Finding $R_2$

After the first rotation, $R_1 = (0, 0, 1)$ (the $z$-axis rotation does not affect points on the $z$-axis).

For the second rotation about $\hat{u} = (-\sin\phi, \cos\phi, 0)$ by angle $\lambda$: $R_1 = (0, 0, 1)$ has component along $\hat{u}$ equal to $0$ (perpendicular to the axis), and its component perpendicular to both $\hat{u}$ and $\hat{k}$ is $(-\cos\phi, -\sin\phi, 0)$ (which is the direction $-OP_1$).

So $R_1 = 0 \cdot \hat{u} + 0 \cdot (-\cos\phi, -\sin\phi, 0) + 1 \cdot \hat{k}$. Under rotation by $\lambda$ about $\hat{u}$, the $\hat{k}$ component and the $(-\cos\phi, -\sin\phi, 0)$ component mix:

$R_2 = \cos\lambda \cdot (0, 0, 1) + \sin\lambda \cdot (-\cos\phi, -\sin\phi, 0) = (-\cos\phi\sin\lambda, -\sin\phi\sin\lambda, \cos\lambda).$

Let me verify with the hint: $R_2 = (-\cos\phi\sin\lambda, -\sin\phi\sin\lambda, \cos\lambda)$. Confirmed.

Part (iii)

The third rotation takes $P_2 = (\cos\phi\cos\lambda, \sin\phi\cos\lambda, \sin\lambda)$ back to $P_0 = (1, 0, 0)$.

Finding $\theta$:

Since $P_0$ and $P_2$ are both on the unit sphere, the angle $\theta$ between them satisfies:

$\cos\theta = OP_0 \cdot OP_2 = 1 \cdot \cos\phi\cos\lambda + 0 + 0 = \cos\phi\cos\lambda. \qquad \text{(shown)}$

Finding the axis direction:

The axis of rotation must be perpendicular to the plane containing $O$, $P_0$, and $P_2$. This direction is given by the cross product:

$P_0 \times P_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} \cos\phi\cos\lambda \\ \sin\phi\cos\lambda \\ \sin\lambda \end{pmatrix} = \begin{pmatrix} 0 \cdot \sin\lambda - 0 \cdot \sin\phi\cos\lambda \\ 0 \cdot \cos\phi\cos\lambda - 1 \cdot \sin\lambda \\ 1 \cdot \sin\phi\cos\lambda - 0 \cdot \cos\phi\cos\lambda \end{pmatrix} = \begin{pmatrix} 0 \\ -\sin\lambda \\ \sin\phi\cos\lambda \end{pmatrix}.$

The axis of rotation is in the direction $(0, -\sin\lambda, \sin\phi\cos\lambda)$.

Examiner Notes

Question 6

About a tenth of the candidates attempted this, with less success than nearly all other questions on the paper. Part (i) caused few problems, but at some point in part (ii), errors were frequently made or lack of attention to which of the two angles in parts (i) and (ii) was being employed in which rotation, and so even those few that knew how to attempt part (iii) were thwarted.

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Question 7

Topic: 微分方程与级数展开 (Differential Equations and Series Expansion)  |  Difficulty: Hard  |  Marks: 20

Problem

7 Given that $y = \cos(m \arcsin x)$, for $|x| < 1$, prove that

$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + m^2y = 0 \, .$

Obtain a similar equation relating $\frac{d^3y}{dx^3}$, $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$, and a similar equation relating $\frac{d^4y}{dx^4}$, $\frac{d^3y}{dx^3}$ and $\frac{d^2y}{dx^2}$.

Conjecture and prove a relation between $\frac{d^{n+2}y}{dx^{n+2}}$, $\frac{d^{n+1}y}{dx^{n+1}}$ and $\frac{d^ny}{dx^n}$.

Obtain the first three non-zero terms of the Maclaurin series for $y$. Show that, if $m$ is an even integer, $\cos m\theta$ may be written as a polynomial in $\sin \theta$ beginning

$1 - \frac{m^2 \sin^2 \theta}{2!} + \frac{m^2(m^2 - 2^2) \sin^4 \theta}{4!} - \cdots \, . \qquad (| \theta | < \frac{1}{2}\pi)$

State the degree of the polynomial.

Hint

7. The initial result can be obtained by differentiating y directly twice obtaining

$\frac{dy}{dx} = -\sin(m \sin^{-1} x) \frac{m}{\sqrt{1-x^2}}$ $\frac{d^2y}{dx^2} = -\cos(m \sin^{-1} x) \frac{m^2}{1-x^2} - \sin(m \sin^{-1} x) \frac{mx}{(1-x^2)^{\frac{3}{2}}}$ and substituting into the LHS.

(Slightly more elegant is to rearrange as $\cos^{-1} y = m \sin^{-1} x$, differentiate and then square to

obtain $(1-x^2) \left(\frac{dy}{dx}\right)^2 = m^2(1-y^2)$ and then differentiate a second time.)

The two similar results are $(1-x^2) \frac{d^3y}{dx^3} - 3x \frac{d^2y}{dx^2} + (m^2 - 1) \frac{dy}{dx} = 0$ and

$(1-x^2) \frac{d^4y}{dx^4} - 5x \frac{d^3y}{dx^3} + (m^2 - 4) \frac{d^2y}{dx^2} = 0$, which lead to the conjecture

$(1-x^2) \frac{d^{n+2}y}{dx^{n+2}} - (2n+1)x \frac{d^{n+1}y}{dx^{n+1}} + (m^2 - n^2) \frac{d^ny}{dx^n} = 0$ which is proved simply by induction.

Using $x=0$, we find that $y=1, \frac{dy}{dx}=0, \frac{d^2y}{dx^2}=-m^2, \frac{d^3y}{dx^3}=0, \frac{d^4y}{dx^4}=m^2(m^2-4)$ and so the Maclaurin series commences $y = 1 - \frac{m^2}{2!}x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 + \dots$

Now replacing x by $\sin \theta$,

$\cos m\theta = 1 - \frac{m^2}{2!}x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 + \dots = 1 - \frac{m^2}{2!}\sin^2 \theta + \frac{m^2(m^2-2^2)}{4!}\sin^4 \theta + \dots$

All the odd differentials are zero, and the even ones are $(-1)^{k+1}m^2(m^2-2^2) \dots (m^2-(2k)^2)$, so if $m$ is even all the terms are zero from a certain point (when $m=2k$) and thus the series terminates and is a polynomial in $\sin \theta$, of degree $m$.

Model Solution

Proving the differential equation

Let $u = \arcsin x$, so $y = \cos(mu)$ and $x = \sin u$, giving $\frac{dx}{du} = \cos u = \sqrt{1 - x^2}$.

First derivative:

$\frac{dy}{dx} = \frac{dy/du}{dx/du} = \frac{-m\sin(mu)}{\cos u} = \frac{-m\sin(mu)}{\sqrt{1-x^2}}.$

So $\sqrt{1-x^2}\frac{dy}{dx} = -m\sin(mu)$. $\qquad \text{(*)}$

Differentiating $\sqrt{1-x^2}\frac{dy}{dx} = -m\sin(mu)$ with respect to $x$:

$\frac{-x}{\sqrt{1-x^2}}\frac{dy}{dx} + \sqrt{1-x^2}\frac{d^2y}{dx^2} = -m\cos(mu) \cdot m \cdot \frac{du}{dx} = \frac{-m^2\cos(mu)}{\sqrt{1-x^2}}.$

Multiplying by $\sqrt{1-x^2}$:

$-x\frac{dy}{dx} + (1-x^2)\frac{d^2y}{dx^2} = -m^2\cos(mu) = -m^2 y.$

Therefore:

$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2 y = 0. \qquad \text{(shown)}$

Deriving the equation for the third derivative

Differentiating $(1-x^2)y'' - xy' + m^2 y = 0$ with respect to $x$:

$-2xy'' + (1-x^2)y''' - y' - xy'' + m^2 y' = 0$

$(1-x^2)y''' - 3xy'' + (m^2 - 1)y' = 0.$

Deriving the equation for the fourth derivative

Differentiating again:

$-2xy''' + (1-x^2)y'''' - 3y'' - 3xy''' + (m^2 - 1)y'' = 0$

$(1-x^2)y'''' - 5xy''' + (m^2 - 4)y'' = 0.$

Conjecture and proof

The pattern suggests:

$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+1)x\frac{d^{n+1}y}{dx^{n+1}} + (m^2 - n^2)\frac{d^ny}{dx^n} = 0. \qquad \text{(**)}$

Proof by induction. The base case $n = 0$ is the equation we proved. Assume (**) holds for some $n \geqslant 0$. Differentiating with respect to $x$:

$-2x\frac{d^{n+2}y}{dx^{n+2}} + (1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+1)\frac{d^{n+1}y}{dx^{n+1}} - (2n+1)x\frac{d^{n+2}y}{dx^{n+2}} + (m^2 - n^2)\frac{d^{n+1}y}{dx^{n+1}} = 0.$

Collecting terms:

$(1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+1+2)x\frac{d^{n+2}y}{dx^{n+2}} + (m^2 - n^2 - 2n - 1)\frac{d^{n+1}y}{dx^{n+1}} = 0$

$(1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+3)x\frac{d^{n+2}y}{dx^{n+2}} + (m^2 - (n+1)^2)\frac{d^{n+1}y}{dx^{n+1}} = 0.$

This is (**) with $n$ replaced by $n+1$, completing the induction.

Maclaurin series

Setting $x = 0$ in the relations:

• From the base equation ($n = 0$): $y''(0) + m^2 y(0) = 0$. Since $y(0) = \cos(0) = 1$, we get $y''(0) = -m^2$.
• From $n = 1$: $y'''(0) + (m^2 - 1)y'(0) = 0$. Since $y'(0) = 0$, we get $y'''(0) = 0$.
• From $n = 2$: $y^{(4)}(0) + (m^2 - 4)y''(0) = 0$, so $y^{(4)}(0) = -(m^2-4)(-m^2) = m^2(m^2 - 4)$.
• All odd derivatives at $x = 0$ are zero (since $y'(0) = 0$ and the recurrence preserves this).

The Maclaurin series is:

$y = 1 - \frac{m^2}{2!}x^2 + \frac{m^2(m^2 - 4)}{4!}x^4 - \cdots$

More generally, the even-order derivatives satisfy $y^{(2k)}(0) = (-1)^k m^2(m^2 - 2^2)(m^2 - 4^2)\cdots(m^2 - (2k-2)^2)$ for $k \geqslant 1$.

Polynomial expression for $\cos m\theta$

Setting $x = \sin\theta$ (valid for $|\theta| < \frac{\pi}{2}$), so $y = \cos(m\arcsin(\sin\theta)) = \cos(m\theta)$:

$\cos m\theta = 1 - \frac{m^2}{2!}\sin^2\theta + \frac{m^2(m^2 - 2^2)}{4!}\sin^4\theta - \cdots. \qquad \text{(shown)}$

When $m$ is an even integer, say $m = 2k$, the factor $m^2 - (2k)^2 = m^2 - m^2 = 0$ appears in the coefficient of $\sin^{2k+2}\theta$ and all subsequent terms. Therefore the series terminates and $\cos m\theta$ is a polynomial in $\sin\theta$.

The highest surviving term has $\sin^{2k}\theta = \sin^m\theta$, so the polynomial has degree $m$.

Examiner Notes

Question 7

Just over 60% attempted this question, achieving moderate success. The opening result was well done, but the two similar equations foundered frequently on incorrect differentiation. If these two were correctly obtained, then the conjecture and induction were usually correct. Appreciating that the final expression was actually a polynomial, and what this entails, passed most by.

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Question 8

Topic: 积分技巧与微分方程 (Integration Techniques and Differential Equations)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Given that $P(x) = Q(x)R'(x) - Q'(x)R(x)$, write down an expression for

$\int \frac{P(x)}{(Q(x))^2} \, dx \, .$

(i) By choosing the function $R(x)$ to be of the form $a + bx + cx^2$, find

$\int \frac{5x^2 - 4x - 3}{(1 + 2x + 3x^2)^2} \, dx \, .$

Show that the choice of $R(x)$ is not unique and, by comparing the two functions $R(x)$ corresponding to two different values of $a$, explain how the different choices are related.

(ii) Find the general solution of

$(1 + \cos x + 2 \sin x) \frac{dy}{dx} + (\sin x - 2 \cos x)y = 5 - 3 \cos x + 4 \sin x \, .$

Hint

8. Substituting for $P(x)$, the desired integral is seen to be the reverse of the quotient rule, i.e.

$\frac{R(x)}{Q(x)} (+k)$

To choose a suitable function $R(x)$ in part (i), substitution of $R(x) = a + bx + cx^2$ and $Q(x) = 1 + 2x + 3x^2$ in the given expression yields a quadratic equation, and equating the coefficients of the powers of $x$ gives $5 = -3b + 2c$, $-2 = -3a + c$, $-3 = -2a + b$. These three equations are linearly dependent and so their solution is not unique.

Choosing, for example $a=0, b=-3, c=-2$ and then $a=1, b=-1, c=1$ gives solutions which are related by $\frac{1-x+x^2}{1+2x+3x^2} = \frac{1+2x+3x^2-3x-2x^2}{1+2x+3x^2} = 1 + \frac{-3x-2x^2}{1+2x+3x^2}$ i.e. the same bar the

arbitrary constant.

(ii) Rearranging the equation to be solved as $\frac{dy}{dx} + \frac{(\sin x - 2 \cos x)}{(1 + \cos x + 2 \sin x)} y = \frac{(5 - 3 \cos x + 4 \sin x)}{(1 + \cos x + 2 \sin x)}$, the integrating factor is $e^{\int \frac{(\sin x - 2 \cos x)}{(1 + \cos x + 2 \sin x)} dx} = e^{-\ln(1 + \cos x + 2 \sin x)} = \frac{1}{1 + \cos x + 2 \sin x}$ As a result, the RHS we require to integrate is $\frac{(5 - 3 \cos x + 4 \sin x)}{(1 + \cos x + 2 \sin x)^2}$ Repeating similar working to part (i), except with $Q(x) = 1 + \cos x + 2 \sin x$ and $R(x) = a + b \sin x + c \cos x$, gives three linearly dependent equations, $5 = b - 2c$, $-3 = b - 2a$, $4 = a - c$ Choosing e.g. $a = 4$, $b = 5$, $c = 0$, the solution is $y = 4 + 5 \sin x + k(1 + \cos x + 2 \sin x)$

Section B: Mechanics

Model Solution

The integral

Given $P(x) = Q(x)R'(x) - Q'(x)R(x)$:

$\int \frac{P(x)}{(Q(x))^2} \, dx = \int \frac{Q(x)R'(x) - Q'(x)R(x)}{(Q(x))^2} \, dx = \int \frac{d}{dx}\left(\frac{R(x)}{Q(x)}\right) dx = \frac{R(x)}{Q(x)} + k.$

Part (i)

We need to find $R(x) = a + bx + cx^2$ such that $Q(x)R'(x) - Q'(x)R(x) = 5x^2 - 4x - 3$, where $Q(x) = 1 + 2x + 3x^2$.

We have $Q'(x) = 2 + 6x$ and $R'(x) = b + 2cx$.

$Q(x)R'(x) = (1 + 2x + 3x^2)(b + 2cx) = b + 2cx + 2bx + 4cx^2 + 3bx^2 + 6cx^3$

$= b + (2c + 2b)x + (4c + 3b)x^2 + 6cx^3.$

$Q'(x)R(x) = (2 + 6x)(a + bx + cx^2) = 2a + 2bx + 2cx^2 + 6ax + 6bx^2 + 6cx^3$

$= 2a + (2b + 6a)x + (2c + 6b)x^2 + 6cx^3.$

Subtracting:

$P(x) = (b - 2a) + (2c + 2b - 2b - 6a)x + (4c + 3b - 2c - 6b)x^2 + (6c - 6c)x^3$

$= (b - 2a) + (2c - 6a)x + (2c - 3b)x^2.$

Setting equal to $5x^2 - 4x - 3$ and comparing coefficients:

$x^0: \quad b - 2a = -3 \qquad \text{(I)}$ $x^1: \quad 2c - 6a = -4 \qquad \text{(II)}$ $x^2: \quad 2c - 3b = 5 \qquad \text{(III)}$

From (I): $b = 2a - 3$. From (II): $c = 3a - 2$. Substituting into (III):

$2(3a - 2) - 3(2a - 3) = 5$ $6a - 4 - 6a + 9 = 5$ $5 = 5. \qquad \text{(always true)}$

So the equations are linearly dependent and the solution is a one-parameter family:

$b = 2a - 3, \quad c = 3a - 2, \quad \text{for any } a.$

Therefore:

$\int \frac{5x^2 - 4x - 3}{(1 + 2x + 3x^2)^2} \, dx = \frac{a + (2a-3)x + (3a-2)x^2}{1 + 2x + 3x^2} + k.$

Non-uniqueness: Different choices of $a$ give different functions $R(x)$ but the same integral (up to the constant of integration). For example, $a = 0$ gives $R(x) = -3x - 2x^2$, while $a = 1$ gives $R(x) = 1 - x + x^2$. The difference is:

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

In general, changing $a$ by $\delta$ adds $\frac{\delta(1 + 2x + 3x^2)}{1 + 2x + 3x^2} = \delta$ to the result, which is absorbed into the constant of integration. Different choices of $R(x)$ correspond to the same antiderivative up to an additive constant.

Part (ii)

Dividing by $(1 + \cos x + 2\sin x)$:

$\frac{dy}{dx} + \frac{\sin x - 2\cos x}{1 + \cos x + 2\sin x} y = \frac{5 - 3\cos x + 4\sin x}{1 + \cos x + 2\sin x}.$

The integrating factor is:

$\mu = e^{\int \frac{\sin x - 2\cos x}{1 + \cos x + 2\sin x} dx}.$

Let $u = 1 + \cos x + 2\sin x$, so $du = (-\sin x + 2\cos x)dx = -(\sin x - 2\cos x)dx$. Therefore:

$\int \frac{\sin x - 2\cos x}{1 + \cos x + 2\sin x} dx = -\int \frac{du}{u} = -\ln|u|.$

So $\mu = e^{-\ln(1 + \cos x + 2\sin x)} = \frac{1}{1 + \cos x + 2\sin x}$.

Multiplying through by $\mu$:

$\frac{d}{dx}\left(\frac{y}{1 + \cos x + 2\sin x}\right) = \frac{5 - 3\cos x + 4\sin x}{(1 + \cos x + 2\sin x)^2}.$

We need to integrate the right side. Using the technique from part (i), we seek $R(x) = a + b\sin x + c\cos x$ such that $Q(x)R'(x) - Q'(x)R(x) = 5 - 3\cos x + 4\sin x$, where $Q(x) = 1 + \cos x + 2\sin x$ and $Q'(x) = -\sin x + 2\cos x$.

$R'(x) = b\cos x - c\sin x.$

$Q(x)R'(x) = (1 + \cos x + 2\sin x)(b\cos x - c\sin x)$

$= b\cos x - c\sin x + b\cos^2 x - c\sin x\cos x + 2b\sin x\cos x - 2c\sin^2 x.$

Using $\cos^2 x = \frac{1+\cos 2x}{2}$, $\sin^2 x = \frac{1-\cos 2x}{2}$, $\sin x\cos x = \frac{\sin 2x}{2}$ gets complicated. Instead, let us directly compare coefficients of $1$, $\sin x$, $\cos x$.

Working through the products and collecting terms of constant, $\sin x$, $\cos x$:

$Q(x)R'(x)$:

• Constant terms: from $b\cos^2 x \to b/2$ (from $\cos^2 x$ average) and $-2c\sin^2 x \to -c$ (from average). Actually, let me just expand and collect using $\cos^2 x = 1 - \sin^2 x$ etc. This gets messy. Let me use a different approach.

Since we know the method works (from the hint), let me substitute $R(x) = a + b\sin x + c\cos x$ and $Q(x) = 1 + \cos x + 2\sin x$ directly into $P = QR' - Q'R$ and match coefficients of $1$, $\sin x$, $\cos x$.

$QR' = (1 + \cos x + 2\sin x)(b\cos x - c\sin x)$ $= b\cos x - c\sin x + b\cos^2 x - c\sin x\cos x + 2b\sin x\cos x - 2c\sin^2 x$

$Q'R = (-\sin x + 2\cos x)(a + b\sin x + c\cos x)$ $= -a\sin x - b\sin^2 x - c\sin x\cos x + 2a\cos x + 2b\sin x\cos x + 2c\cos^2 x$

$P = QR' - Q'R$: $= b\cos x - c\sin x + b\cos^2 x - c\sin x\cos x + 2b\sin x\cos x - 2c\sin^2 x$ $\quad + a\sin x + b\sin^2 x + c\sin x\cos x - 2a\cos x - 2b\sin x\cos x - 2c\cos^2 x$

Collecting:

• $\cos x$: $b - 2a$
• $\sin x$: $-c + a = a - c$
• $\cos^2 x$: $b - 2c$
• $\sin^2 x$: $-2c + b = b - 2c$
• $\sin x\cos x$: $-c + 2b + c - 2b = 0$

So $P = (b - 2a)\cos x + (a - c)\sin x + (b - 2c)(\cos^2 x + \sin^2 x) = (b - 2c) + (a - c)\sin x + (b - 2a)\cos x$.

Setting equal to $5 - 3\cos x + 4\sin x$:

$b - 2c = 5 \qquad \text{(I)}$ $a - c = 4 \qquad \text{(II)}$ $b - 2a = -3 \qquad \text{(III)}$

From (II): $a = c + 4$. From (I): $b = 2c + 5$. Substituting into (III):

$2c + 5 - 2(c + 4) = 2c + 5 - 2c - 8 = -3. \qquad \text{(always true)}$

Again linearly dependent. Choose $c = 0$: $a = 4$, $b = 5$.

So $R(x) = 4 + 5\sin x$ and:

$\frac{y}{1 + \cos x + 2\sin x} = \frac{4 + 5\sin x}{1 + \cos x + 2\sin x} + k.$

Therefore the general solution is:

$y = 4 + 5\sin x + k(1 + \cos x + 2\sin x),$

where $k$ is an arbitrary constant.

Examiner Notes

Question 8

Three quarters of the candidates had a go at this, with moderate success. Most understood the method intended for part (i) and were aware of the method of using an integrating factor. Algebraic slips led to incorrect simultaneous equations in part (i), and few dealt with the non-uniqueness of $R(x)$ satisfactorily. Having found the integrating factor for part (ii), most did not proceed further. Some candidates introduced a sign error into part (ii) which trivialized the left hand side to a differential of a product. A small number of candidates produced elegant solutions to part (ii) using the tan half angle substitution.

Section B: Mechanics