STEP2 2008 -- Pure Mathematics

STEP2 2008 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2008  |  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	序列与递推关系 (Sequences and Recurrence Relations)	Standard	代入消元, 因式分解, 判别式判断实根, 解的验证
2	部分分式与级数展开 (Partial Fractions and Series Expansion)	Challenging	部分分式分解, 二项式定理展开, 分类讨论, 级数求值
3	不等式与反证法 (Inequalities and Proof by Contradiction)	Challenging	求导求极值, 函数图像分析, 反证法, 不等式证明
4	隐函数求导与几何应用 (Implicit Differentiation and Geometry)	Challenging	隐函数求导, tan(A-B)角度公式, 代数化简, 极值分析
5	积分与不等式 (Integration and Inequalities)	Standard	三角换元, 部分分式积分, 二项式展开, 对数不等式
6	三角函数 (Trigonometry)	Challenging	周期计算,余弦方程对称性,和积公式,同时取最大值条件
7	微分方程 (Differential Equations)	Standard	变量替换,分离变量,归纳猜想
8	向量 (Vectors)	Challenging	比例定理,角平分线定理,标量积运算

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Question 1

Topic: 序列与递推关系 (Sequences and Recurrence Relations)  |  Difficulty: Standard  |  Marks: 20

Problem

1 A sequence of points $(x_1, y_1), (x_2, y_2), \dots$ in the cartesian plane is generated by first choosing $(x_1, y_1)$ then applying the rule, for $n = 1, 2, \dots$,

$(x_{n+1}, y_{n+1}) = (x_n^2 - y_n^2 + a, 2x_ny_n + b + 2),$

where $a$ and $b$ are given real constants.

(i) In the case $a = 1$ and $b = -1$, find the values of $(x_1, y_1)$ for which the sequence is constant.

(ii) Given that $(x_1, y_1) = (-1, 1)$, find the values of $a$ and $b$ for which the sequence has period 2.

Hint

1 (i) Given $(x_{n+1}, y_{n+1}) = (x_n^2 - y_n^2 + 1, 2x_n y_n + 1)$, it is easier to remove the subscripts and set $x^2 - y^2 + 1 = x$ and $2xy + 1 = y$. Then, identifying the $y$‘s (or $x$‘s) in each case, gives $y^2 = x^2 - x + 1$ and $y = \frac{1}{1 - 2x}$. Eliminating the $y$‘s leads to a polynomial equation in $x$; namely, $4x^4 - 8x^3 + 9x^2 - 5x = 0$.

Noting the obvious factor of $x$, and then finding a second linear factor (e.g. by the factor theorem) leads to $x(x - 1)(4x^2 - 4x + 5) = 0$. Here, the quadratic factor has no real roots, since the discriminant, $\Delta = 4^2 - 4 \cdot 4 \cdot 5 = -64 < 0$. [Alternatively, one could note that $4x^2 - 4x + 5 \equiv (2x - 1)^2 + 4 > 0 \ \forall x$.]

The two values of $x$, and the corresponding values of $y$, gained by substituting these $x$‘s into $y = \frac{1}{1 - 2x}$, are then $(x, y) = (0, 1)$ and $(1, -1)$.

(ii) Now $(x_1, y_1) = (-1, 1) \Rightarrow (x_2, y_2) = (a, b)$ and $(x_3, y_3) = (a^2 - b^2 + a, 2ab + b + 2)$. Setting both $a^2 - b^2 + a = -1$ and $2ab + b + 2 = 1$, so that the third term is equal to the first, and identifying the $b$‘s in each case, gives $b^2 = a^2 + a + 1$ and $b = \frac{-1}{1 + 2a}$.

One could go about this the long way, as before. However, it can be noted that the algebra is the same as in (i), but with $a = -x$ and $b = -y$. Either way, we obtain the two possible solution-pairs: $(a, b) = (0, -1)$ and $(-1, 1)$.

However, upon checking, the solution $(-1, 1)$ actually gives rise to a constant sequence (and remember that the working only required the third term to be the same as the first, which doesn’t preclude the possibility that it is also the same as the second term!), so we find that there is in fact just the one solution: $(a, b) = (0, -1)$.

Model Solution

Part (i)

For the sequence to be constant, we need $(x_{n+1}, y_{n+1}) = (x_n, y_n)$ for all $n$. Dropping subscripts and setting $a = 1$, $b = -1$:

$x^2 - y^2 + 1 = x \qquad \text{...(1)}$ $2xy - 1 + 2 = y \implies 2xy + 1 = y \qquad \text{...(2)}$

From (2): $y - 2xy = 1$, so $y(1 - 2x) = 1$, giving

$y = \frac{1}{1 - 2x} \qquad \text{provided } x \neq \tfrac{1}{2}. \qquad \text{...(3)}$

From (1): $y^2 = x^2 - x + 1$. \qquad …(4)

Squaring (3) and substituting into (4):

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

$(x^2 - x + 1)(1 - 2x)^2 = 1$

Expanding $(1 - 2x)^2 = 1 - 4x + 4x^2$:

$(x^2 - x + 1)(1 - 4x + 4x^2) = 4x^4 - 8x^3 + 9x^2 - 5x + 1$

Setting this equal to 1:

$4x^4 - 8x^3 + 9x^2 - 5x = 0$

$x(4x^3 - 8x^2 + 9x - 5) = 0$

Testing $x = 1$: $4 - 8 + 9 - 5 = 0$, so $(x - 1)$ is a factor. Dividing:

$4x^3 - 8x^2 + 9x - 5 = (x - 1)(4x^2 - 4x + 5)$

So the equation becomes $x(x - 1)(4x^2 - 4x + 5) = 0$.

The discriminant of $4x^2 - 4x + 5$ is $\Delta = (-4)^2 - 4(4)(5) = 16 - 80 = -64 < 0$, so the quadratic has no real roots. (Alternatively, $4x^2 - 4x + 5 = (2x - 1)^2 + 4 > 0$ for all real $x$.)

Therefore $x = 0$ or $x = 1$.

• $x = 0$: from (3), $y = \frac{1}{1} = 1$, so $(x, y) = (0, 1)$.
• $x = 1$: from (3), $y = \frac{1}{1 - 2} = -1$, so $(x, y) = (1, -1)$.

Part (ii)

With $(x_1, y_1) = (-1, 1)$:

$(x_2, y_2) = ((-1)^2 - 1^2 + a,\ 2(-1)(1) + b + 2) = (a,\ b)$

$(x_3, y_3) = (a^2 - b^2 + a,\ 2ab + b + 2)$

For period 2, we require $(x_3, y_3) = (x_1, y_1)$ while $(x_2, y_2) \neq (x_1, y_1)$. Setting $(x_3, y_3) = (-1, 1)$:

$a^2 - b^2 + a = -1 \qquad \text{...(5)}$ $2ab + b + 2 = 1 \implies 2ab + b = -1 \qquad \text{...(6)}$

From (6): $b(2a + 1) = -1$, so $b = \frac{-1}{2a + 1}$ (provided $a \neq -\frac{1}{2}$).

From (5): $b^2 = a^2 + a + 1$.

These equations are identical in form to those in part (i), with the substitutions $a = -x$ and $b = -y$. So the solutions are $(a, b) = (0, -1)$ and $(a, b) = (-1, 1)$.

We must check which give genuine period 2:

• $(a, b) = (0, -1)$: $(x_1, y_1) = (-1, 1)$, $(x_2, y_2) = (0, -1)$, and $(x_3, y_3) = (0 - 1 + 0,\ 0 - 1 + 2) = (-1, 1)$. Since $(x_2, y_2) \neq (x_1, y_1)$, this is period 2.

• $(a, b) = (-1, 1)$: $(x_2, y_2) = (-1, 1) = (x_1, y_1)$, so the sequence is constant (period 1), not period 2.

Therefore $(a, b) = (0, -1)$.

Examiner Notes

Question 1

The first question is invariably intended to be a gentle introduction to the paper, and to allow all candidates to gain some marks without making great demands on either memory or technical skills. As such, most candidates traditionally tend to begin with question 1, and this proved to be the case here. Almost 700 candidates attempted this question, making it (marginally) the second most popular question on the paper; and it gained the highest mean score of about 14 marks.

There were still several places where marks were commonly lost. In (i), setting $(x_2, y_2) = (x_1, y_1)$ and eliminating $y$ (for instance) leads to a quartic equation in $x$. There were two straightforward linear factors easily found to the quartic expression, leaving a quadratic factor which could yield no real roots. Many candidates failed to explain why, or show that, this was so. In (ii), the algebra again leads to two solutions, gained by setting $(x_3, y_3) = (x_1, y_1)$. However, one of them corresponds to one of the solutions already found in (i), where the sequence is constant, and most candidates omitted either to notice this or

to discover it by checking. Another very common oversight – although far less important in the sense that candidates could still gain all the marks by going the long way round – was that the algebra in (ii) was exactly the same as that in (i), but with $a = -x$ and $b = -y$. For the very few who noticed this, the working for the second half of the question was remarkably swift.

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Question 2

Topic: 部分分式与级数展开 (Partial Fractions and Series Expansion)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Let $a_n$ be the coefficient of $x^n$ in the series expansion, in ascending powers of $x$, of

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

where $|x| < 1$. Show, using partial fractions, that either $a_n = n + 1$ or $a_n = n + 2$ according to the value of $n$.

Hence find a decimal approximation, to nine significant figures, for the fraction $\frac{11\,000}{8181}$. [You are not required to justify the accuracy of your approximation.]

Hint

2 The correct partial fraction form for the given algebraic fraction is

$\frac{1 + x}{(1 - x)^2(1 + x^2)} \equiv \frac{A}{1 - x} + \frac{B}{(1 - x)^2} + \frac{Cx + D}{1 + x^2},$

although these can also be put together in other correct ways that don’t materially hinder the progress of the solution. The standard procedure now is to multiply throughout by the denominator of the LHS and compare coefficients or substitute in suitable values: which leads to $A = \frac{1}{2}$, $B = 1$, $C = \frac{1}{2}$ and $D = -\frac{1}{2}$.

In order to apply the binomial theorem to these separate fractions, we now use index notation to turn

$\frac{1 + x}{(1 - x)^2(1 + x^2)} \equiv A(1 - x)^{-1} + B(1 - x)^{-2} + Cx(1 + x^2)^{-1} + D(1 + x^2)^{-1}$

into the infinite series

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

It should be clear at this point that the last two of these series have odd/even powers only, with alternating signs playing an extra part. The consequence of all this is that we need to

examine cases for $n$ modulo 4; i.e. depending upon whether $n$ leaves a remainder of 0, 1, 2 or 3 when divided by 4.

For $n \equiv 0 \pmod 4$, the coefft. of $x^n$ is $\frac{1}{2} + n + 1 + 0 - \frac{1}{2} = n + 1$;

A1 for $n \equiv 1 \pmod 4$, coefft. of $x^n$ is $\frac{1}{2} + n + 1 + \frac{1}{2} - 0 = n + 2$;

A1 for $n \equiv 2 \pmod 4$, coefft. of $x^n$ is $\frac{1}{2} + n + 1 + 0 + \frac{1}{2} = n + 2$;

A1 for $n \equiv 3 \pmod 4$, coefft. of $x^n$ is $\frac{1}{2} + n + 1 - \frac{1}{2} + 0 = n + 1$.

For the very final part of the question, we note that $\frac{11000}{8181} = \frac{1.1}{0.9^2 \times 1.01}$, is a cancelled form of our original expression, with $x = 0.1$. (N.B. $|x| < 1$ assures the convergence of the infinite series forms). Substituting this value of $x$ into

$1 + 3x + 4x^2 + 4x^3 + 5x^4 + 7x^5 + 8x^6 + 8x^7 + 9x^8 + \dots$

then gives 1.344 578 90 to 8dp.

Model Solution

Partial fraction decomposition

We seek constants $A, B, C, D$ such that

$\frac{1 + x}{(1 - x)^2(1 + x^2)} \equiv \frac{A}{1 - x} + \frac{B}{(1 - x)^2} + \frac{Cx + D}{1 + x^2}.$

Multiplying both sides by $(1 - x)^2(1 + x^2)$:

$1 + x = A(1 - x)(1 + x^2) + B(1 + x^2) + (Cx + D)(1 - x)^2.$

Setting $x = 1$: $2 = 2B$, so $B = 1$.

Setting $x = -1$: $0 = 2A(2) + 2B + (-C + D)(4) = 4A + 2 + 4(D - C)$, so $4A + 4D - 4C = -2$, i.e. $2A + 2D - 2C = -1$.

Setting $x = 0$: $1 = A + B + D = A + 1 + D$, so $A + D = 0$, i.e. $D = -A$.

Setting $x = 2$: $3 = A(-1)(5) + 5B + (2C + D)(1) = -5A + 5 + 2C + D$. So $-5A + 2C + D = -2$.

From $D = -A$: $-5A + 2C - A = -2$, so $-6A + 2C = -2$, i.e. $C = 3A - 1$.

From $2A + 2D - 2C = -1$ with $D = -A$: $2A - 2A - 2C = -1$, so $C = \frac{1}{2}$.

Then $3A - 1 = \frac{1}{2}$, so $A = \frac{1}{2}$, and $D = -\frac{1}{2}$.

Therefore:

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

Series expansion

We expand each term for $|x| < 1$:

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

$(1 - x)^{-2} = \sum_{n=0}^{\infty} (n+1) x^n \qquad \text{(standard binomial series)}.$

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

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

So the coefficient $a_n$ of $x^n$ is:

$a_n = \frac{1}{2} + (n + 1) + [\text{contribution from } \tfrac{x}{2}(1+x^2)^{-1}] + [\text{contribution from } -\tfrac{1}{2}(1+x^2)^{-1}].$

The third series $\frac{1}{2}\sum_{m=0}^{\infty}(-1)^m x^{2m+1}$ contributes only to odd powers $x^{2m+1}$. For $n$ odd with $n = 2m + 1$, the contribution is $\frac{1}{2}(-1)^m$; for $n$ even, the contribution is $0$.

The fourth series $-\frac{1}{2}\sum_{m=0}^{\infty}(-1)^m x^{2m}$ contributes only to even powers $x^{2m}$. For $n$ even with $n = 2m$, the contribution is $-\frac{1}{2}(-1)^m$; for $n$ odd, the contribution is $0$.

We examine the four cases for $n$ modulo 4:

Case $n \equiv 0 \pmod{4}$: $n = 4k$ (even, $m = 2k$ even). Contribution from the fourth series: $-\frac{1}{2}(-1)^{2k} = -\frac{1}{2}$. Third series contributes $0$.

$a_n = \frac{1}{2} + (n + 1) + 0 - \frac{1}{2} = n + 1.$

Case $n \equiv 1 \pmod{4}$: $n = 4k + 1$ (odd, $m = 2k$ even). Contribution from the third series: $\frac{1}{2}(-1)^{2k} = \frac{1}{2}$. Fourth series contributes $0$.

$a_n = \frac{1}{2} + (n + 1) + \frac{1}{2} + 0 = n + 2.$

Case $n \equiv 2 \pmod{4}$: $n = 4k + 2$ (even, $m = 2k + 1$ odd). Contribution from the fourth series: $-\frac{1}{2}(-1)^{2k+1} = \frac{1}{2}$. Third series contributes $0$.

$a_n = \frac{1}{2} + (n + 1) + 0 + \frac{1}{2} = n + 2.$

Case $n \equiv 3 \pmod{4}$: $n = 4k + 3$ (odd, $m = 2k + 1$ odd). Contribution from the third series: $\frac{1}{2}(-1)^{2k+1} = -\frac{1}{2}$. Fourth series contributes $0$.

$a_n = \frac{1}{2} + (n + 1) - \frac{1}{2} + 0 = n + 1.$

Therefore $a_n = n + 1$ when $n \equiv 0$ or $3 \pmod{4}$, and $a_n = n + 2$ when $n \equiv 1$ or $2 \pmod{4}$.

Decimal approximation for $\frac{11\,000}{8181}$

We verify that $\frac{11\,000}{8181}$ equals the original expression with $x = 0.1$:

$\frac{1 + x}{(1 - x)^2(1 + x^2)}\bigg|_{x = 0.1} = \frac{1.1}{(0.9)^2(1.01)} = \frac{1.1}{0.81 \times 1.01} = \frac{1.1}{0.8181} = \frac{11\,000}{8181}.$

Since $|0.1| < 1$, the series converges. Using the pattern of coefficients $a_0 = 1, a_1 = 3, a_2 = 4, a_3 = 4, a_4 = 5, a_5 = 7, a_6 = 8, a_7 = 8, a_8 = 9, a_9 = 11, \ldots$:

$\frac{11\,000}{8181} = 1 + 3(0.1) + 4(0.01) + 4(0.001) + 5(0.0001) + 7(0.00001) + 8(0.000001) + 8(0.0000001) + 9(0.00000001) + 11(0.000000001) + \cdots$

$= 1 + 0.3 + 0.04 + 0.004 + 0.0005 + 0.00007 + 0.000008 + 0.0000008 + 0.00000009 + 0.000000011 + \cdots$

$= 1.34457890\ldots$

Therefore $\frac{11\,000}{8181} \approx 1.344\,578\,91$ to nine significant figures.

Examiner Notes

Question 2

Noticeably less popular than Q1 – with only around 500 “hits” – and with a very much poorer mean mark of about 8, it was rather obvious that many candidates were very unsure as to what constituted the best partial fraction form for the given algebraic fraction to begin with. Then, with very little direct guidance being given in the question, candidates’ confidence seemed to ebb visibly as they proceeded, being required to turn the resulting collection of single algebraic fractions into series, using the Binomial Theorem, and then into a consideration of general terms. There was much fudging of these general terms in order to get the given answers of either $n + 1$ or $n + 2$ for the general term’s coefficients; even amongst those who did spot which one occurred when, there was often little visible justification to support the conclusions. As a result of all the hurdles to be cleared, those who managed to get to the numerical ending successfully were very few in number.

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Question 3

Topic: 不等式与反证法 (Inequalities and Proof by Contradiction)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Find the coordinates of the turning points of the curve $y = 27x^3 - 27x^2 + 4$. Sketch the curve and deduce that $x^2(1 - x) \leqslant 4/27$ for all $x \geqslant 0$.

Given that each of the numbers $a, b$ and $c$ lies between 0 and 1, prove by contradiction that at least one of the numbers $bc(1 - a)$, $ca(1 - b)$ and $ab(1 - c)$ is less than or equal to $4/27$.

(ii) Given that each of the numbers $p$ and $q$ lies between 0 and 1, prove that at least one of the numbers $p(1 - q)$ and $q(1 - p)$ is less than or equal to $1/4$.

Hint

3 (i) Setting $\frac{dy}{dx} = 81x^2 - 54x = 0$ for TPs gives $(0, 4)$ and $(\frac{2}{3}, 0)$. You really ought to know the shape of such a (“positive”) cubic, and it is customary to find the crossing-points on the axes: $x = 0$ gives $y = 4$, and $y = 0$ leads to $x = -1$ and $x = \frac{2}{3}$ (twice). [If you have been paying attention, this latter zero for $y$ should come as no surprise!] The graph now shows that, for all $x \ge 0$, $y \ge 0$; which leads to the required result — $x^2(1 - x) \le \frac{4}{27}$ — with just a little bit of re-arrangement.

In order to prove the result by contradiction (reduction ad absurdum), we first assume that all three numbers exceed $\frac{4}{27}$. Then their product

$bc(1 - a)ca(1 - b)ab(1 - c) > \left( \frac{4}{27} \right)^3.$

However, this product can be re-written in the form

$a^2(1 - a) \cdot b^2(1 - b) \cdot c^2(1 - c),$

and the previous result guarantees that $x^2(1 - x) \le \frac{4}{27}$ for each of $a, b, c$, from which it follows that

$a^2(1 - a) \cdot b^2(1 - b) \cdot c^2(1 - c) \le \left( \frac{4}{27} \right)^3,$

which is the required contradiction. Hence, at least one of the three numbers $bc(1 - a)$, $ca(1 - b)$, $ab(1 - c)$ is less than, or equal to, $\frac{4}{27}$.

(ii) Drawing the graph of $y = x - x^2$ (there are, of course, other suitable choices, such as $y = (2x - 1)^2$ for example) and showing that it has a maximum at $(\frac{1}{2}, \frac{1}{4})$ gives

$x(1 - x) \le \frac{1}{4} \text{ for all } x.$

The assumption that $p(1 - q), q(1 - p) > \frac{1}{4} \implies p(1 - p) \cdot q(1 - q) > \left( \frac{1}{4} \right)^2$.

However, we know that $x(1 - x) \le \frac{1}{4}$ for each of $p$ and $q$, and this gives us that

$p(1 - p) \cdot q(1 - q) \le \left( \frac{1}{4} \right)^2.$

Hence, by contradiction, at least one of $p(1 - q), q(1 - p) \le \frac{1}{4}$.

Model Solution

Part (i)

We have $y = 27x^3 - 27x^2 + 4$. Differentiating:

$\frac{dy}{dx} = 81x^2 - 54x = 27x(3x - 2).$

Setting $\frac{dy}{dx} = 0$ gives $x = 0$ or $x = \frac{2}{3}$.

• At $x = 0$: $y = 4$, so one turning point is $(0, 4)$.
• At $x = \frac{2}{3}$: $y = 27 \cdot \frac{8}{27} - 27 \cdot \frac{4}{9} + 4 = 8 - 12 + 4 = 0$, so the other turning point is $\left(\frac{2}{3}, 0\right)$.

To determine the nature of each turning point, we compute $\frac{d^2y}{dx^2} = 162x - 54$.

• At $x = 0$: $\frac{d^2y}{dx^2} = -54 < 0$, so $(0, 4)$ is a local maximum.
• At $x = \frac{2}{3}$: $\frac{d^2y}{dx^2} = 108 - 54 = 54 > 0$, so $\left(\frac{2}{3}, 0\right)$ is a local minimum.

The $y$-intercept is $(0, 4)$. Setting $y = 0$:

$27x^3 - 27x^2 + 4 = 0.$

We can check that $x = \frac{2}{3}$ is a root (already found above), so $(3x - 2)$ is a factor. Dividing:

$27x^3 - 27x^2 + 4 = (3x - 2)(9x^2 - 3x - 2) = (3x - 2)(3x + 1)(3x - 2) = (3x - 2)^2(3x + 1).$

So $y = 0$ at $x = \frac{2}{3}$ (double root) and $x = -\frac{1}{3}$.

The sketch shows a positive cubic with a local maximum at $(0, 4)$, touching the $x$-axis at $\left(\frac{2}{3}, 0\right)$ (since it is a double root), and crossing at $x = -\frac{1}{3}$.

For $x \geq 0$, the curve starts at $y = 4$ when $x = 0$, decreases to touch $y = 0$ at $x = \frac{2}{3}$, then increases without bound. Therefore $y \geq 0$ for all $x \geq 0$, i.e.

$27x^3 - 27x^2 + 4 \geq 0 \quad \text{for all } x \geq 0.$

Rearranging: $27x^2(1 - x) \leq 4$, so

$x^2(1 - x) \leq \frac{4}{27} \quad \text{for all } x \geq 0. \qquad (\star)$

Now we prove the main result by contradiction. Suppose, for contradiction, that all three numbers $bc(1 - a)$, $ca(1 - b)$, $ab(1 - c)$ are strictly greater than $\frac{4}{27}$. Then their product satisfies:

$bc(1 - a) \cdot ca(1 - b) \cdot ab(1 - c) > \left(\frac{4}{27}\right)^3.$

The left-hand side can be rearranged:

$a^2 b^2 c^2 (1 - a)(1 - b)(1 - c) = a^2(1 - a) \cdot b^2(1 - b) \cdot c^2(1 - c).$

Since $0 < a < 1$, we have $a \geq 0$, so by $(\star)$ with $x = a$: $a^2(1 - a) \leq \frac{4}{27}$. Similarly $b^2(1 - b) \leq \frac{4}{27}$ and $c^2(1 - c) \leq \frac{4}{27}$.

Therefore:

$a^2(1 - a) \cdot b^2(1 - b) \cdot c^2(1 - c) \leq \left(\frac{4}{27}\right)^3.$

This contradicts our assumption. Hence at least one of $bc(1 - a)$, $ca(1 - b)$, $ab(1 - c)$ must be less than or equal to $\frac{4}{27}$.

Part (ii)

Consider the function $g(x) = x(1 - x) = x - x^2$. Then $g'(x) = 1 - 2x = 0$ gives $x = \frac{1}{2}$, and $g''(x) = -2 < 0$ confirms this is a maximum. The maximum value is $g\left(\frac{1}{2}\right) = \frac{1}{4}$.

Therefore $x(1 - x) \leq \frac{1}{4}$ for all $x$.

Now suppose, for contradiction, that both $p(1 - q) > \frac{1}{4}$ and $q(1 - p) > \frac{1}{4}$. Multiplying:

$p(1 - q) \cdot q(1 - p) > \frac{1}{16}.$

The left-hand side equals $p(1 - p) \cdot q(1 - q)$.

Since $0 < p < 1$: $p(1 - p) \leq \frac{1}{4}$. Since $0 < q < 1$: $q(1 - q) \leq \frac{1}{4}$.

Therefore $p(1 - p) \cdot q(1 - q) \leq \frac{1}{16}$, contradicting the assumption. Hence at least one of $p(1 - q)$, $q(1 - p)$ is less than or equal to $\frac{1}{4}$.

Examiner Notes

Question 3

This, the third most popular question on the paper, producing a mixed bag of responses. It strikes me that, although the A-level specifications require candidates to understand the process of proof by contradiction, this is never actually tested anywhere by any of the exam. boards. Nonetheless, it was very pleasing to see that so many candidates were able to grasp the basic idea of what to do, and many did so very successfully. The impartial observer might well note that the situation in (i) is very much tougher (in terms of degree) than that in (ii). However, candidates were very much more closely guided in (i) and then left to make their own way in (ii).

Apart from the standard, expected response to (i) – see the SOLUTIONS document for this – many other candidates produced a very pleasing alternative which they often dressed up as proof by contradiction but which was, in fact, a direct proof. It was, however, so mathematically sound and appealing an argument (and a legitimate imitation of a p by c) that we gave it all but one of the marks available in this part of the question. It ran like this:

Suppose w.l.o.g. that $0 < a \le b \le c < 1$. Then $ab(1 - c) \le b^2(1 - b) \le \frac{4}{27}$ by the previous result (namely $x^2(1 - x) \le \frac{4}{27}$ for all $x \ge 0$). QED.

[Note that we could have used $ab(1 - c) \le c^2(1 - c) \le \frac{4}{27}$ also.]

It has to be said that most other inequality arguments were rather poorly constructed and unconvincing, leaving the markers with little option but to put a line through (often) several pages of circular arguments, faulty assumptions, dubious conclusions, and occasionally correct statements with either no supporting reasoning or going nowhere useful.

There was one remarkable alternative which was produced by just a couple of candidates (that I know of) and is not included in the SOLUTIONS because it is such a rarity. However, for those who know of the AM – GM Inequality, it is sufficiently appealing to include it here for novelty value. It ran like this:

Assume that $bc(1 - a), ca(1 - b), ab(1 - c) > \frac{4}{27}$. Using the previous result, we have $a^2(1 - a), b^2(1 - b), c^2(1 - c) \le \frac{4}{27}$. Then, since all terms are positive, it follows that $a^2 \le bc, b^2 \le ca, c^2 \le ab$ so that $a^2 + b^2 + c^2 \le bc + ca + ab. (*)$ However, by the AM – GM Inequality (or directly by the Cauchy-Schwarz Inequality), $a^2 + b^2 \ge 2ab, b^2 + c^2 \ge 2bc \text{ and } c^2 + a^2 \ge 2ca.$ Adding and dividing by two then gives $a^2 + b^2 + c^2 \ge ab + bc + ca$, which contradicts the conclusion (*), etc., etc.

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Question 4

Topic: 隐函数求导与几何应用 (Implicit Differentiation and Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 A curve is given by

$x^2 + y^2 + 2axy = 1,$

where $a$ is a constant satisfying $0 < a < 1$. Show that the gradient of the curve at the point $P$ with coordinates $(x, y)$ is

$-\frac{x + ay}{ax + y},$

provided $ax + y \neq 0$. Show that $\theta$, the acute angle between $OP$ and the normal to the curve at $P$, satisfies

$\tan \theta = a|y^2 - x^2|.$

Show further that, if $\frac{d\theta}{dx} = 0$ at $P$, then:

(i) $a(x^2 + y^2) + 2xy = 0$;

(ii) $(1 + a)(x^2 + y^2 + 2xy) = 1$;

(iii) $\tan \theta = \frac{a}{\sqrt{1 - a^2}}$.

Hint

4 Differentiating implicitly gives $2\left( x + y\frac{dy}{dx} + ax\frac{dy}{dx} + ay \right) = 0$, from which it follows that

$\frac{dy}{dx} = -\frac{x + ay}{ax + y}$ and hence the gradient of the normal is $\frac{ax + y}{x + ay}$.

Using $\tan(A - B)$ on this and $\frac{y}{x}$ gives $\tan \theta = \left| \frac{\frac{y}{x} - \frac{ax + y}{x + ay}}{1 + \frac{y}{x} \times \frac{ax + y}{x + ay}} \right| = \left| \frac{xy + ay^2 - ax^2 - xy}{x^2 + axy + axy + y^2} \right|$.

However, we know that $x^2 + y^2 + 2axy = 1$ from the curve’s eqn., and so $\tan \theta = a|y^2 - x^2|.$

(i) Differentiating this w.r.t. $x$ then gives $\sec^2 \theta \frac{d\theta}{dx} = a\left( 2y\frac{dy}{dx} - 2x \right)$. Equating this to zero and using $\frac{dy}{dx} = -\frac{x + ay}{ax + y}$ from earlier then leads to $a(x^2 + y^2) + 2xy = 0$.

(ii) Adding $x^2 + y^2 + 2axy = 1$ and $a(x^2 + y^2) + 2xy = 0$ gives $(1 + a)(x + y)^2 = 1$.

(iii) However, subtracting these two eqns. instead gives $(1 - a)(y - x)^2 = 1$, and multiplying these two last results together yields $(1 - a^2)(y^2 - x^2)^2 = 1$.

Finally, using $\tan \theta = a|y^2 - x^2| \Rightarrow (y^2 - x^2)^2 = \frac{1}{a^2} \tan^2 \theta$, and substituting this

into the last result of (iii) then gives the required result: $\tan \theta = \frac{a}{\sqrt{1 - a^2}}$. All that

remains is to justify taking the positive square root, since $\tan \theta$ is | something |, which is necessarily non-negative.

Model Solution

Finding the gradient

Differentiating $x^2 + y^2 + 2axy = 1$ implicitly with respect to $x$:

$2x + 2y\frac{dy}{dx} + 2a\left(x\frac{dy}{dx} + y\right) = 0.$

Dividing by 2 and collecting terms in $\frac{dy}{dx}$:

$x + ay + (y + ax)\frac{dy}{dx} = 0.$

Therefore:

$\frac{dy}{dx} = -\frac{x + ay}{ax + y}, \qquad \text{provided } ax + y \neq 0.$

Finding $\tan\theta$

The gradient of $OP$ (from the origin to $P(x,y)$) is $\frac{y}{x}$, provided $x \neq 0$.

The gradient of the normal to the curve at $P$ is the negative reciprocal of $\frac{dy}{dx}$:

$m_{\text{normal}} = \frac{ax + y}{x + ay}.$

The angle $\theta$ between $OP$ and the normal is given by:

$\tan\theta = \left|\frac{\frac{y}{x} - \frac{ax + y}{x + ay}}{1 + \frac{y}{x} \cdot \frac{ax + y}{x + ay}}\right|.$

Computing the numerator:

$\frac{y}{x} - \frac{ax + y}{x + ay} = \frac{y(x + ay) - x(ax + y)}{x(x + ay)} = \frac{xy + ay^2 - ax^2 - xy}{x(x + ay)} = \frac{a(y^2 - x^2)}{x(x + ay)}.$

Computing the denominator:

$1 + \frac{y(ax + y)}{x(x + ay)} = \frac{x(x + ay) + y(ax + y)}{x(x + ay)} = \frac{x^2 + axy + axy + y^2}{x(x + ay)} = \frac{x^2 + y^2 + 2axy}{x(x + ay)}.$

Since $x^2 + y^2 + 2axy = 1$ (from the curve equation), the denominator simplifies to $\frac{1}{x(x + ay)}$.

Therefore:

$\tan\theta = \left|\frac{a(y^2 - x^2)}{x(x + ay)} \cdot x(x + ay)\right| = a|y^2 - x^2|.$

Part (i): Show that $\frac{d\theta}{dx} = 0$ implies $a(x^2 + y^2) + 2xy = 0$

We have $\tan\theta = a|y^2 - x^2|$. Without loss of generality, suppose $y^2 \geq x^2$ (the other case is analogous), so $\tan\theta = a(y^2 - x^2)$.

Differentiating both sides with respect to $x$:

$\sec^2\theta \cdot \frac{d\theta}{dx} = a\left(2y\frac{dy}{dx} - 2x\right).$

Setting $\frac{d\theta}{dx} = 0$ (and noting $\sec^2\theta \neq 0$):

$a\left(2y\frac{dy}{dx} - 2x\right) = 0.$

Since $a > 0$, we need $2y\frac{dy}{dx} - 2x = 0$, i.e. $y\frac{dy}{dx} = x$. Substituting $\frac{dy}{dx} = -\frac{x + ay}{ax + y}$:

$y \cdot \left(-\frac{x + ay}{ax + y}\right) = x.$

$-y(x + ay) = x(ax + y).$

$-xy - ay^2 = ax^2 + xy.$

$-ay^2 - ax^2 = 2xy.$

$a(x^2 + y^2) + 2xy = 0. \qquad \text{(i)}$

Part (ii): Show that $(1 + a)(x^2 + y^2 + 2xy) = 1$

We have two equations:

$x^2 + y^2 + 2axy = 1 \qquad \text{(curve equation)}$ $a(x^2 + y^2) + 2xy = 0 \qquad \text{(from part (i))}$

Adding these two equations:

$(1 + a)(x^2 + y^2) + 2axy + 2xy = 1.$

Factor the last two terms: $2axy + 2xy = 2xy(1 + a)$. Therefore:

$(1 + a)(x^2 + y^2) + 2xy(1 + a) = 1.$

$(1 + a)(x^2 + y^2 + 2xy) = 1. \qquad \text{(ii)}$

Part (iii): Show that $\tan\theta = \frac{a}{\sqrt{1 - a^2}}$

Subtracting the equation from part (i) from the curve equation:

$(x^2 + y^2 + 2axy) - a(x^2 + y^2) - 2xy = 1 - 0.$

$(1 - a)(x^2 + y^2) + 2xy(a - 1) = 1.$

$(1 - a)(x^2 + y^2 - 2xy) = 1.$

$(1 - a)(x - y)^2 = 1. \qquad (\star\star)$

From part (ii): $(1 + a)(x + y)^2 = 1$.

Multiplying $(\star\star)$ and the result of (ii):

$(1 - a)(1 + a)(x - y)^2(x + y)^2 = 1.$

$(1 - a^2)(y^2 - x^2)^2 = 1. \qquad (\star\star\star)$

Now recall $\tan\theta = a|y^2 - x^2|$, so $(y^2 - x^2)^2 = \frac{\tan^2\theta}{a^2}$.

Substituting into $(\star\star\star)$:

$(1 - a^2) \cdot \frac{\tan^2\theta}{a^2} = 1.$

$\tan^2\theta = \frac{a^2}{1 - a^2}.$

Since $0 < a < 1$, we have $1 - a^2 > 0$, and $\tan\theta \geq 0$ (as $\theta$ is acute). Taking the positive square root:

$\tan\theta = \frac{a}{\sqrt{1 - a^2}}. \qquad \text{(iii)}$

Examiner Notes

Q4

Another very popular question, poorly done (600 attempts, mean score below 7). Most efforts got little further than finding the gradient of the normal to the curve, and I strongly suspect that this question was frequently to be found amongst candidates’ non-contributing scorers. Using the $\tan(A - B)$ formula is a sufficiently common occurrence on past papers that there is little excuse for well-prepared candidates not to recognise when and how to apply it. Once that has been done, the question’s careful structuring guided able candidates over the hurdles one at a time, each result relying on the preceding result(s); yet most attempts had finished quite early on, and the majority of candidates failed to benefit from the setters’ kindness.

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Question 5

Topic: 积分与不等式 (Integration and Inequalities)  |  Difficulty: Standard  |  Marks: 20

Problem

5 Evaluate the integrals

$\int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x} dx \quad \text{and} \quad \int_0^{\pi/2} \frac{\sin x}{1 + \sin^2 x} dx.$

Show, using the binomial expansion, that $(1 + \sqrt{2})^5 < 99$. Show also that $\sqrt{2} > 1.4$. Deduce that $2^{\sqrt{2}} > 1 + \sqrt{2}$. Use this result to determine which of the above integrals is greater.

Hint

5 Using a well-known double-angle formula gives $\int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x} dx = \int_0^{\pi/2} \frac{2 \sin x \cos x}{1 + \sin^2 x} dx$, and

this should suggest an obvious substitution: letting $s = \sin x$ turns this into the integral $\int_0^1 \frac{2s}{1 + s^2} ds.$

This is just a standard log. integral (the numerator being the derivative of the denominator), leading to the answer $\ln 2$.

Alternatively, one could use the identity $\sin^2 x \equiv \frac{1}{2} - \frac{1}{2} \cos 2x$ to end up with $\int_0^{\pi/2} \frac{2 \sin 2x}{3 - \cos 2x} dx.$

This, again, gives a log. integral, but without the substitution.

A suitable substitution for the second integral is $c = \cos x$, which leads to $\int_0^1 \frac{1}{2 - c^2} dc$.

Now you can either attack this using partial fractions, or you could look up what is a fairly standard result in your formula booklet. In each case, you get (after a bit of careful log and

surd work) $\frac{1}{\sqrt{2}} \ln(1 + \sqrt{2})$.

Now $(1 + \sqrt{2})^5 = 1 + 5\sqrt{2} + 20 + 20\sqrt{2} + 20 + 4\sqrt{2} = 41 + 29\sqrt{2}$ (using the binomial theorem, for instance), and

$41 + 29\sqrt{2} < 99 \Leftrightarrow 29\sqrt{2} < 58 \Leftrightarrow \sqrt{2} < 2,$

which is obviously the case. Also, $1.96 < 2 \Rightarrow 1.4 < \sqrt{2}$. Thereafter, an argument such as

$2^{1.4} > 1 + \sqrt{2} \Leftrightarrow 2^7 > (1 + \sqrt{2})^5 \Leftrightarrow 128 > 41 + 29\sqrt{2}$

$\Leftrightarrow 87 > 29\sqrt{2} \Leftrightarrow 3 > \sqrt{2}$

from which it follows that $2^{\sqrt{2}} > 2^{7/5} > 1 + \sqrt{2}$.

Taking logs in this result then gives $\sqrt{2} \ln 2 > \ln(1 + \sqrt{2}) \Rightarrow \ln 2 > \frac{1}{\sqrt{2}} \ln(1 + \sqrt{2})$; and

$\int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x} dx > \int_0^{\pi/2} \frac{\sin x}{1 + \sin^2 x} dx.$

Model Solution

First integral

Using $\sin 2x = 2\sin x \cos x$:

$\int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x}\, dx = \int_0^{\pi/2} \frac{2\sin x \cos x}{1 + \sin^2 x}\, dx.$

Let $s = \sin x$, so $ds = \cos x\, dx$. When $x = 0$, $s = 0$; when $x = \pi/2$, $s = 1$.

$= \int_0^1 \frac{2s}{1 + s^2}\, ds = \left[\ln(1 + s^2)\right]_0^1 = \ln 2 - \ln 1 = \ln 2.$

Second integral

$\int_0^{\pi/2} \frac{\sin x}{1 + \sin^2 x}\, dx = \int_0^{\pi/2} \frac{\sin x}{2 - \cos^2 x}\, dx \qquad \text{using } \sin^2 x = 1 - \cos^2 x.$

Let $c = \cos x$, so $dc = -\sin x\, dx$. When $x = 0$, $c = 1$; when $x = \pi/2$, $c = 0$.

$= \int_1^0 \frac{-dc}{2 - c^2} = \int_0^1 \frac{dc}{2 - c^2}.$

Using partial fractions: $\frac{1}{2 - c^2} = \frac{1}{(\sqrt{2} - c)(\sqrt{2} + c)} = \frac{1}{2\sqrt{2}}\left(\frac{1}{\sqrt{2} - c} + \frac{1}{\sqrt{2} + c}\right)$.

$= \frac{1}{2\sqrt{2}}\left[-\ln|\sqrt{2} - c| + \ln|\sqrt{2} + c|\right]_0^1 = \frac{1}{2\sqrt{2}}\left[\ln\frac{\sqrt{2} + c}{\sqrt{2} - c}\right]_0^1$

$= \frac{1}{2\sqrt{2}}\left(\ln\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \ln 1\right) = \frac{1}{2\sqrt{2}}\ln\frac{\sqrt{2} + 1}{\sqrt{2} - 1}.$

Rationalising: $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(\sqrt{2} + 1)^2}{1} = (\sqrt{2} + 1)^2$.

So the integral equals $\frac{1}{2\sqrt{2}} \cdot 2\ln(\sqrt{2} + 1) = \frac{1}{\sqrt{2}}\ln(1 + \sqrt{2})$.

Showing $(1 + \sqrt{2})^5 < 99$

By the binomial theorem:

$(1 + \sqrt{2})^5 = \binom{5}{0} + \binom{5}{1}\sqrt{2} + \binom{5}{2}(\sqrt{2})^2 + \binom{5}{3}(\sqrt{2})^3 + \binom{5}{4}(\sqrt{2})^4 + \binom{5}{5}(\sqrt{2})^5$

$= 1 + 5\sqrt{2} + 10 \cdot 2 + 10 \cdot 2\sqrt{2} + 5 \cdot 4 + 4\sqrt{2}$

$= 1 + 5\sqrt{2} + 20 + 20\sqrt{2} + 20 + 4\sqrt{2} = 41 + 29\sqrt{2}.$

Now $41 + 29\sqrt{2} < 99 \iff 29\sqrt{2} < 58 \iff \sqrt{2} < 2$, which is true since $2^2 = 4 > 2$.

Showing $\sqrt{2} > 1.4$

$(1.4)^2 = 1.96 < 2$, so $\sqrt{2} > 1.4$.

Deducing $2^{\sqrt{2}} > 1 + \sqrt{2}$

Since $\sqrt{2} > \frac{7}{5}$ (equivalently $5\sqrt{2} > 7$, i.e. $50 > 49$, which is true), and the function $2^t$ is increasing:

$2^{\sqrt{2}} > 2^{7/5}.$

It suffices to show $2^{7/5} > 1 + \sqrt{2}$, i.e. $2^7 > (1 + \sqrt{2})^5$. We have $2^7 = 128$ and $(1 + \sqrt{2})^5 = 41 + 29\sqrt{2}$. Since $\sqrt{2} < 3$:

$41 + 29\sqrt{2} < 41 + 87 = 128.$

Therefore $2^{\sqrt{2}} > 2^{7/5} > 1 + \sqrt{2}$.

Comparing the integrals

From $2^{\sqrt{2}} > 1 + \sqrt{2}$, taking natural logarithms:

$\sqrt{2}\ln 2 > \ln(1 + \sqrt{2}).$

Dividing by $\sqrt{2}$:

$\ln 2 > \frac{1}{\sqrt{2}}\ln(1 + \sqrt{2}).$

Therefore $\int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x}\, dx = \ln 2 > \frac{1}{\sqrt{2}}\ln(1 + \sqrt{2}) = \int_0^{\pi/2} \frac{\sin x}{1 + \sin^2 x}\, dx$.

The first integral is greater.

Examiner Notes

Q5

This was the most popular question on the paper (by a small margin) and with the second highest mean mark (12) of all the pure questions. Those who were able to spot the two standard trig. substitutions $s = \sin x$ and $c = \cos x$ for the first two parts generally made excellent progress, although the log. and surd work required to tidy up the second integral’s answer left many with a correct answer that wasn’t easy to do anything much useful with at the very end, when deciding which was numerically the greater. The binomial expansion of $(a + b)^5$ was handled very comfortably, as was much of the following inequality work. However, the very final conclusion was very seldom successfully handled as any little mistakes, unhelpful forms of answers, etc., prevented candidates’ final thoughts from being sufficiently relevant.

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Question 6

Topic: 三角函数 (Trigonometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 A curve has the equation $y = f(x)$, where

$f(x) = \cos \left( 2x + \frac{\pi}{3} \right) + \sin \left( \frac{3x}{2} - \frac{\pi}{4} \right).$

(i) Find the period of $f(x)$.

(ii) Determine all values of $x$ in the interval $-\pi \leqslant x \leqslant \pi$ for which $f(x) = 0$. Find a value of $x$ in this interval at which the curve touches the $x$-axis without crossing it.

(iii) Find the value or values of $x$ in the interval $0 \leqslant x \leqslant 2\pi$ for which $f(x) = 2$.

Hint

(i) Firstly, $\cos x$ has period $2\pi \Rightarrow \cos(2x)$ has period $\pi$; and $\sin x$ has period $2\pi \Rightarrow \sin\left(\frac{3x}{2}\right)$ has period $\frac{4}{3}\pi$.

Then $f(x) = \cos\left(2x + \frac{\pi}{3}\right) + \sin\left(\frac{3x}{2} - \frac{\pi}{4}\right)$ has period $4\pi = \text{lcm}(\pi, \frac{4}{3}\pi)$.

(ii) Any approach here is going to require the use of some trig. identity work. The most straightforward is to note that $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin \theta$ so that $f(x) = 0$ reduces to

$\cos\left(2x + \frac{\pi}{3}\right) = \cos\left(\frac{3x}{2} + \frac{\pi}{4}\right), \text{ from which it follows that } 2x + \frac{\pi}{3} = 2n\pi \pm \left(\frac{3x}{2} + \frac{\pi}{4}\right)$

where $n$ is an integer, using the symmetric and periodic properties of the cosine curve. Taking suitable values of $n$, so that $x$ is in the required interval, leads to the answers

$x = -\frac{31\pi}{42}$ (from $n = -1$, with the $-$ sign), $x = -\frac{\pi}{6}$ ($n = 0$, with both $+$ and $-$ signs),

$x = \frac{17\pi}{42}$ ($n = 1$, $-$ sign) and $x = \frac{41\pi}{42}$ ($n = 2$, $-$ sign).

Since $x = -\frac{\pi}{6}$ is a repeated root (occurring twice in the above list), the curve of $y = f(x)$ touches the $x$-axis at this point.

For those who are aware of the results that appear in all the formula books, but which seem to be on the edge of the various syllabuses, that I know by the title of the Sum-and-Product Formulae, such as

$\cos A + \cos B \equiv 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, there is a

second straightforward approach available here. For example, noting that

$\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$ gives $\cos\left(2x + \frac{\pi}{3}\right) + \cos\left(\frac{3\pi}{4} - \frac{3x}{2}\right) = 0$ which (from the above

identity) then gives $2 \cos \left( \frac{x}{4} + \frac{13\pi}{24} \right) \cos \left( \frac{7x}{4} - \frac{5\pi}{24} \right) = 0$, and setting each of these two

cosine terms equal to zero, in turn, yields the same values of $x$ as before, including the repeat.

(iii) The key observation here is that $y = 2$ if and only if both $\cos \left( 2x + \frac{\pi}{3} \right) = 1$ and $\sin \left( \frac{3x}{2} - \frac{\pi}{4} \right) = 1$, simultaneously. So we must solve $\cos \left( 2x + \frac{\pi}{3} \right) = 1 \Rightarrow 2x + \frac{\pi}{3} = 0, 2\pi, 4\pi, \dots$, giving $x = \frac{5\pi}{6}, \frac{11\pi}{6}, \dots$; and $\sin \left( \frac{3x}{2} - \frac{\pi}{4} \right) = 1 \Rightarrow \frac{3x}{2} - \frac{\pi}{4} = \frac{\pi}{2}, \frac{5\pi}{2}, \dots$, giving $x = \frac{\pi}{2}, \frac{11\pi}{6}, \dots$. Both equations are satisfied when $x = \frac{11\pi}{6}$, and this is the required answer.

Model Solution

Part (i): Period of $f(x)$

The term $\cos(2x + \pi/3)$ has period $\frac{2\pi}{2} = \pi$.

The term $\sin\!\left(\frac{3x}{2} - \frac{\pi}{4}\right)$ has period $\frac{2\pi}{3/2} = \frac{4\pi}{3}$.

The period of $f(x)$ is $\text{lcm}\!\left(\pi,\, \frac{4\pi}{3}\right)$. Writing $\pi = \frac{3\pi}{3}$, we need the smallest positive $T$ such that $T = m\pi = n \cdot \frac{4\pi}{3}$ for positive integers $m, n$. This gives $3m = 4n$, so the smallest solution is $m = 4, n = 3$, giving $T = 4\pi$.

The period of $f(x)$ is $4\pi$.

Part (ii): Solving $f(x) = 0$ for $-\pi \leq x \leq \pi$

We rewrite the sine term using $\sin\theta = -\cos\!\left(\frac{\pi}{2} + \theta\right)$:

$\sin\!\left(\frac{3x}{2} - \frac{\pi}{4}\right) = -\cos\!\left(\frac{\pi}{2} + \frac{3x}{2} - \frac{\pi}{4}\right) = -\cos\!\left(\frac{3x}{2} + \frac{\pi}{4}\right).$

So $f(x) = 0$ becomes:

$\cos\!\left(2x + \frac{\pi}{3}\right) = \cos\!\left(\frac{3x}{2} + \frac{\pi}{4}\right).$

Using $\cos A = \cos B \iff A = 2n\pi \pm B$:

$2x + \frac{\pi}{3} = 2n\pi \pm \left(\frac{3x}{2} + \frac{\pi}{4}\right).$

Case 1 (+ sign):

$2x + \frac{\pi}{3} = 2n\pi + \frac{3x}{2} + \frac{\pi}{4}$

$\frac{x}{2} = 2n\pi + \frac{\pi}{4} - \frac{\pi}{3} = 2n\pi - \frac{\pi}{12}$

$x = 4n\pi - \frac{\pi}{6}.$

For $x \in [-\pi, \pi]$: only $n = 0$ gives $x = -\frac{\pi}{6}$.

Case 2 ($-$ sign):

$2x + \frac{\pi}{3} = 2n\pi - \frac{3x}{2} - \frac{\pi}{4}$

$\frac{7x}{2} = 2n\pi - \frac{\pi}{4} - \frac{\pi}{3} = 2n\pi - \frac{7\pi}{12}$

$x = \frac{4n\pi}{7} - \frac{\pi}{6} = \frac{24n\pi - 7\pi}{42} = \frac{(24n - 7)\pi}{42}.$

For $x \in [-\pi, \pi]$, we need $-42 \leq 24n - 7 \leq 42$, i.e. $-35 \leq 24n \leq 49$, so $n = -1, 0, 1, 2$:

• $n = -1$: $x = \frac{-31\pi}{42}$
• $n = 0$: $x = \frac{-7\pi}{42} = -\frac{\pi}{6}$
• $n = 1$: $x = \frac{17\pi}{42}$
• $n = 2$: $x = \frac{41\pi}{42}$

The complete set of solutions is:

$x = -\frac{31\pi}{42},\quad x = -\frac{\pi}{6},\quad x = \frac{17\pi}{42},\quad x = \frac{41\pi}{42}.$

Since $x = -\frac{\pi}{6}$ arises from both the ”+” case (Case 1) and the "$-$" case (Case 2 with $n = 0$), it is a repeated root. This means the curve touches the $x$-axis at $x = -\frac{\pi}{6}$ without crossing it.

Part (iii): Solving $f(x) = 2$ for $0 \leq x \leq 2\pi$

Since $-1 \leq \cos\!\left(2x + \frac{\pi}{3}\right) \leq 1$ and $-1 \leq \sin\!\left(\frac{3x}{2} - \frac{\pi}{4}\right) \leq 1$, and their sum equals 2, both terms must simultaneously equal 1:

$\cos\!\left(2x + \frac{\pi}{3}\right) = 1 \qquad \text{and} \qquad \sin\!\left(\frac{3x}{2} - \frac{\pi}{4}\right) = 1.$

From the first equation: $2x + \frac{\pi}{3} = 2k\pi$ for integer $k$, so $x = k\pi - \frac{\pi}{6}$.

For $x \in [0, 2\pi]$: $k = 1$ gives $x = \frac{5\pi}{6}$; $k = 2$ gives $x = \frac{11\pi}{6}$.

From the second equation: $\frac{3x}{2} - \frac{\pi}{4} = \frac{\pi}{2} + 2m\pi$ for integer $m$, so $\frac{3x}{2} = \frac{3\pi}{4} + 2m\pi$, giving $x = \frac{\pi}{2} + \frac{4m\pi}{3}$.

For $x \in [0, 2\pi]$: $m = 0$ gives $x = \frac{\pi}{2}$; $m = 1$ gives $x = \frac{\pi}{2} + \frac{4\pi}{3} = \frac{11\pi}{6}$.

The only value satisfying both equations is $x = \frac{11\pi}{6}$.

Examiner Notes

This was the least popular of the pure maths questions. Although there were 300 starts to the question, most of these barely got into the very opening part before the attempt was abandoned in favour of another question. Most attempts failed to show that $f(x)$ has a period of $4\pi$. As mentioned, few proceeded further. Of those who did, efforts were generally very poor indeed – as testified to by the very low mean mark of 4 – with the necessary comfort in handling even the most basic of trig. identities being very conspicuous by its absence. Part (iii) was my personal favourite amongst the pure questions, as it contained a very uncommon – yet remarkably simple – idea in order to get started on the road to a solution. The idea is simply this: $f(x)$, being the sum of a cosine term and sine term, is equal to 2 if and only if each of these separate terms is simultaneously at its maximum of 1. That is, the question is actually two very easy trig. equations disguised as one very complicated-looking one. Once realised, the whole thing becomes very straightforward indeed, but only a few candidates had persevered this far.

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Question 7

Topic: 微分方程 (Differential Equations)  |  Difficulty: Standard  |  Marks: 20

Problem

7 (i) By writing $y = u(1 + x^2)^{\frac{1}{2}}$, where $u$ is a function of $x$, find the solution of the equation $\frac{1}{y} \frac{dy}{dx} = xy + \frac{x}{1 + x^2}$ for which $y = 1$ when $x = 0$.

(ii) Find the solution of the equation $\frac{1}{y} \frac{dy}{dx} = x^2y + \frac{x^2}{1 + x^3}$ for which $y = 1$ when $x = 0$.

(iii) Give, without proof, a conjecture for the solution of the equation $\frac{1}{y} \frac{dy}{dx} = x^{n-1}y + \frac{x^{n-1}}{1 + x^n}$ for which $y = 1$ when $x = 0$, where $n$ is an integer greater than 1.

Hint

(i) Differentiating $y = u \sqrt{1 + x^2}$ gives $\frac{\text{d}y}{\text{d}x} = u \cdot \frac{x}{\sqrt{1 + x^2}} + \sqrt{1 + x^2} \cdot \frac{\text{d}u}{\text{d}x}$; so that $\frac{1}{y} \frac{\text{d}y}{\text{d}x} = xy + \frac{x}{1 + x^2}$ becomes $\frac{1}{u \sqrt{1 + x^2}} \left\{ \frac{ux}{\sqrt{1 + x^2}} + \sqrt{1 + x^2} \cdot \frac{\text{d}u}{\text{d}x} \right\} = xu \sqrt{1 + x^2} + \frac{x}{1 + x^2}$. Simplifying and cancelling the common term on both sides leads to $\frac{1}{u} \cdot \frac{\text{d}u}{\text{d}x} = xu \sqrt{1 + x^2}$. This is a standard form for a first-order differential equation, involving the separation of variables and integration: $\int \frac{1}{u^2} \cdot \text{d}u = \int x \sqrt{1 + x^2} \text{ d}x \Rightarrow -\frac{1}{u} = \frac{1}{3} (1 + x^2)^{3/2} (+ C)$. Using $x = 0, y = 1$ ($u = 1$) to find $C$ leads to the final answer, $y = \frac{3 \sqrt{1 + x^2}}{4 - (1 + x^2)^{3/2}}$.

(ii) The key here is to choose the appropriate function of $x$. If you have really got a feel for what has happened in the previous bit of the question, then this isn’t too demanding. If you haven’t really grasped fully what’s going on then you may well need to try one or two possibilities first. The product that needs to be identified here is $y = u(1 + x^3)^{1/3}$. Once you have found this, the process of (i) pretty much repeats itself. $\frac{\text{d}y}{\text{d}x} = u \cdot x^2 (1 + x^3)^{-2/3} + (1 + x^3)^{1/3} \frac{\text{d}u}{\text{d}x}$ means that $\frac{1}{y} \frac{\text{d}y}{\text{d}x} = x^2 y + \frac{x^2}{1 + x^3}$ becomes $\frac{1}{u} \cdot \frac{\text{d}u}{\text{d}x} = x^2 u (1 + x^3)^{1/3}$. Separating variables and integrating: $\int \frac{1}{u^2} \cdot \text{d}u = \int x^2 (1 + x^3)^{1/3} \text{ d}x = -\frac{1}{u} = \frac{1}{4} (1 + x^3)^{4/3} (+ C)$;

and $x = 0, y = 1$ ($u = 1$) gives $C$ and the answer $y = \frac{4(1 + x^3)^{1/3}}{5 - (1 + x^3)^{4/3}}$.

(iii) Note that the question didn’t actually require you to simplify the two answers in (i) and (ii), but doing so certainly enables you to have a better idea as to how to generalise the results:

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

Model Solution

Part (i)

Let $y = u(1 + x^2)^{1/2}$. Differentiating using the product rule:

$\frac{dy}{dx} = u \cdot \frac{x}{(1 + x^2)^{1/2}} + (1 + x^2)^{1/2}\frac{du}{dx}.$

Dividing by $y = u(1 + x^2)^{1/2}$:

$\frac{1}{y}\frac{dy}{dx} = \frac{x}{1 + x^2} + \frac{1}{u}\frac{du}{dx}.$

The original equation is $\frac{1}{y}\frac{dy}{dx} = xy + \frac{x}{1 + x^2}$. Substituting the expression above:

$\frac{x}{1 + x^2} + \frac{1}{u}\frac{du}{dx} = xu\sqrt{1 + x^2} + \frac{x}{1 + x^2}.$

Cancelling $\frac{x}{1 + x^2}$ from both sides:

$\frac{1}{u}\frac{du}{dx} = xu\sqrt{1 + x^2}.$

$\frac{du}{dx} = xu^2\sqrt{1 + x^2}.$

Separating variables:

$\int \frac{du}{u^2} = \int x\sqrt{1 + x^2}\, dx.$

For the right-hand side, let $w = 1 + x^2$, $dw = 2x\, dx$:

$\int x\sqrt{1 + x^2}\, dx = \frac{1}{2}\int w^{1/2}\, dw = \frac{1}{3}w^{3/2} = \frac{1}{3}(1 + x^2)^{3/2}.$

So:

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

Using the condition $y = 1$ when $x = 0$: $u = \frac{y}{(1 + x^2)^{1/2}}\big|_{x=0} = 1$.

$-1 = \frac{1}{3} + C \implies C = -\frac{4}{3}.$

$-\frac{1}{u} = \frac{1}{3}(1 + x^2)^{3/2} - \frac{4}{3} = \frac{(1 + x^2)^{3/2} - 4}{3}.$

$u = \frac{3}{4 - (1 + x^2)^{3/2}}.$

Therefore:

$y = u(1 + x^2)^{1/2} = \frac{3\sqrt{1 + x^2}}{4 - (1 + x^2)^{3/2}}.$

Part (ii)

The structure of part (i) suggests the substitution $y = u(1 + x^3)^{1/3}$. Differentiating:

$\frac{dy}{dx} = u \cdot x^2(1 + x^3)^{-2/3} + (1 + x^3)^{1/3}\frac{du}{dx}.$

Dividing by $y = u(1 + x^3)^{1/3}$:

$\frac{1}{y}\frac{dy}{dx} = \frac{x^2}{1 + x^3} + \frac{1}{u}\frac{du}{dx}.$

The original equation is $\frac{1}{y}\frac{dy}{dx} = x^2 y + \frac{x^2}{1 + x^3}$. Substituting:

$\frac{x^2}{1 + x^3} + \frac{1}{u}\frac{du}{dx} = x^2 u(1 + x^3)^{1/3} + \frac{x^2}{1 + x^3}.$

Cancelling $\frac{x^2}{1 + x^3}$:

$\frac{1}{u}\frac{du}{dx} = x^2 u(1 + x^3)^{1/3}.$

$\frac{du}{dx} = x^2 u^2(1 + x^3)^{1/3}.$

Separating variables:

$\int \frac{du}{u^2} = \int x^2(1 + x^3)^{1/3}\, dx.$

Let $w = 1 + x^3$, $dw = 3x^2\, dx$:

$\int x^2(1 + x^3)^{1/3}\, dx = \frac{1}{3}\int w^{1/3}\, dw = \frac{1}{3} \cdot \frac{3}{4}w^{4/3} = \frac{1}{4}(1 + x^3)^{4/3}.$

So:

$-\frac{1}{u} = \frac{1}{4}(1 + x^3)^{4/3} + C.$

Using $y = 1$ when $x = 0$: $u = 1$.

$-1 = \frac{1}{4} + C \implies C = -\frac{5}{4}.$

$-\frac{1}{u} = \frac{(1 + x^3)^{4/3} - 5}{4}, \qquad u = \frac{4}{5 - (1 + x^3)^{4/3}}.$

Therefore:

$y = \frac{4(1 + x^3)^{1/3}}{5 - (1 + x^3)^{4/3}}.$

Part (iii): Conjecture

The pattern from parts (i) and (ii) generalises. For the equation

$\frac{1}{y}\frac{dy}{dx} = x^{n-1}y + \frac{x^{n-1}}{1 + x^n},$

the substitution $y = u(1 + x^n)^{1/n}$ leads to the solution

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

Examiner Notes

In many ways, part (i) of the question was very routine, requiring little more than technical competence to see the differential equation, using the given substitution, through to a correct solution. Part (ii) then required candidates to spot a slightly different substitution on the basis of having gained a feel for what had gone on previously. I had thought that many more candidates would try something involving the square root of $1 + x^3$ or the cube root of $1 + x^2$, rather than cube root of $1 + x^3$, but many solutions that I saw went straight for the right thing. Once this had been successfully pushed through – with the working mimicking that of (i) very closely indeed – it was not difficult to spot the general answer required, unproven, in (iii). Overall, however, it seems that a lot of candidates failed to spot the right thing for part (ii) and their solutions stopped at this point. With almost 600 attempts, the mean score on this question was 10.

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Question 8

Topic: 向量 (Vectors)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 The points $A$ and $B$ have position vectors a and b, respectively, relative to the origin $O$. The points $A$, $B$ and $O$ are not collinear. The point $P$ lies on $AB$ between $A$ and $B$ such that $AP : PB = (1 - \lambda) : \lambda .$ Write down the position vector of $P$ in terms of a, b and $\lambda$. Given that $OP$ bisects $\angle AOB$, determine $\lambda$ in terms of $a$ and $b$, where $a = |\mathbf{a}|$ and $b = |\mathbf{b}|$.

The point $Q$ also lies on $AB$ between $A$ and $B$, and is such that $AP = BQ$. Prove that $OQ^2 - OP^2 = (b - a)^2 .$

Hint

The first result is an example of what is known as the Ratio Theorem:

$AP : PB = 1 - \lambda : \lambda \Rightarrow \mathbf{p} = \lambda \mathbf{a} + (1 - \lambda) \mathbf{b}.$

Alternatively, it can be deduced from the standard approach to the vector equation of a straight line, via $\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a})$.

Using the scalar product twice then gives

$\mathbf{a} \bullet \mathbf{p} = \lambda a^2 + (1 - \lambda)(\mathbf{a} \bullet \mathbf{b}) \text{ and } \mathbf{b} \bullet \mathbf{p} = \lambda(\mathbf{a} \bullet \mathbf{b}) + (1 - \lambda) b^2.$

Equating these two expressions for $\cos \theta$, $\frac{\mathbf{a} \bullet \mathbf{p}}{ap} = \frac{\mathbf{b} \bullet \mathbf{p}}{bp}$, re-arranging and collecting up

like terms, then gives $ab\{\lambda(a + b) - b\} = \mathbf{a} \bullet \mathbf{b} \{\lambda(a + b) - b\}$. There are two possible consequences to this statement, and both of them should be considered. Either $ab = \mathbf{a} \bullet \mathbf{b}$, which gives $\cos 2\theta = 1$, $\theta = 0$, $A = B$ and violates the non-collinearity of $O, A$ & $B$; or the bracketed factor on each side is zero, which gives

$\lambda = \frac{b}{a + b}.$

However, if you know the Angle Bisector Theorem, the working is short-circuited quite dramatically:

$\frac{AP}{PB} = \frac{OA}{OB} \Rightarrow \frac{(1 - \lambda)(AB)}{\lambda(AB)} = \frac{a}{b} \Rightarrow b - b\lambda = a\lambda \Rightarrow \lambda = \frac{b}{a + b}.$

Next, $AQ : QB = \lambda : 1 - \lambda \Rightarrow \mathbf{q} = (1 - \lambda)\mathbf{a} + \lambda\mathbf{b}$. Then

$OQ^2 = \mathbf{q} \bullet \mathbf{q} = (1 - \lambda)^2 a^2 + \lambda^2 b^2 + 2\lambda(1 - \lambda) \mathbf{a} \bullet \mathbf{b}$

and $OP^2 = \mathbf{p} \bullet \mathbf{p} = (1 - \lambda)^2 b^2 + \lambda^2 a^2 + 2\lambda(1 - \lambda) \mathbf{a} \bullet \mathbf{b}$.

[N.B. This working can also be done by the Cosine Rule.] Subtracting:

$OQ^2 - OP^2 = (b^2 - a^2) [\lambda^2 - (1 - \lambda)^2] = (b^2 - a^2) (2\lambda - 1)$

and, substituting $\lambda$ in terms of $a$ and $b$ into this expression, gives the required answer

$= (b - a)(b + a) \times \frac{b - a}{b + a} = (b - a)^2.$

Model Solution

Position vector of $P$

Since $P$ lies on $AB$ with $AP : PB = (1 - \lambda) : \lambda$, by the section formula (ratio theorem):

$\mathbf{p} = \lambda\mathbf{a} + (1 - \lambda)\mathbf{b}.$

This can be seen from the vector equation of the line $AB$: $\mathbf{r} = \mathbf{a} + t(\mathbf{b} - \mathbf{a})$, where $P$ corresponds to $t = \lambda$ (since $AP = \lambda \cdot AB$ and $PB = (1 - \lambda) \cdot AB$).

Finding $\lambda$ when $OP$ bisects $\angle AOB$

Let $\angle AOB = 2\theta$. If $OP$ bisects $\angle AOB$, then $\angle AOP = \angle BOP = \theta$. Using the scalar product:

$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{a \cdot p} = \frac{\mathbf{b} \cdot \mathbf{p}}{b \cdot p}$

so $b(\mathbf{a} \cdot \mathbf{p}) = a(\mathbf{b} \cdot \mathbf{p})$. We compute each dot product:

$\mathbf{a} \cdot \mathbf{p} = \mathbf{a} \cdot (\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}) = \lambda a^2 + (1 - \lambda)(\mathbf{a} \cdot \mathbf{b})$

$\mathbf{b} \cdot \mathbf{p} = \mathbf{b} \cdot (\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}) = \lambda(\mathbf{a} \cdot \mathbf{b}) + (1 - \lambda)b^2$

Substituting into $b(\mathbf{a} \cdot \mathbf{p}) = a(\mathbf{b} \cdot \mathbf{p})$:

$b\left[\lambda a^2 + (1 - \lambda)(\mathbf{a} \cdot \mathbf{b})\right] = a\left[\lambda(\mathbf{a} \cdot \mathbf{b}) + (1 - \lambda)b^2\right]$

Expanding both sides:

$\lambda a^2 b + (1 - \lambda)b(\mathbf{a} \cdot \mathbf{b}) = \lambda a(\mathbf{a} \cdot \mathbf{b}) + (1 - \lambda)ab^2$

Collecting terms with $\lambda$ on the left and terms with $(1 - \lambda)$ on the right:

$\lambda\left[a^2 b - a(\mathbf{a} \cdot \mathbf{b})\right] = (1 - \lambda)\left[ab^2 - b(\mathbf{a} \cdot \mathbf{b})\right]$

$\lambda \cdot a\left[ab - \mathbf{a} \cdot \mathbf{b}\right] = (1 - \lambda) \cdot b\left[ab - \mathbf{a} \cdot \mathbf{b}\right]$

Since $O$, $A$, $B$ are not collinear, $\mathbf{a} \cdot \mathbf{b} \neq ab$ (the angle between $\mathbf{a}$ and $\mathbf{b}$ is not zero), so we can divide both sides by $ab - \mathbf{a} \cdot \mathbf{b}$:

$\lambda a = (1 - \lambda)b$

$\lambda a + \lambda b = b$

$\lambda = \frac{b}{a + b}$

Alternatively, this follows immediately from the Angle Bisector Theorem: $\frac{AP}{PB} = \frac{OA}{OB} = \frac{a}{b}$, so $\frac{1 - \lambda}{\lambda} = \frac{a}{b}$, giving $\lambda = \frac{b}{a + b}$.

Proving $OQ^2 - OP^2 = (b - a)^2$

Since $AP = BQ$, and $AP = (1 - \lambda) \cdot AB$, we have $BQ = (1 - \lambda) \cdot AB$. Since $Q$ lies on $AB$ between $A$ and $B$:

$AQ = AB - BQ = AB - (1 - \lambda) \cdot AB = \lambda \cdot AB.$

So $AQ : QB = \lambda : (1 - \lambda)$.

By the section formula:

$\mathbf{q} = (1 - \lambda)\mathbf{a} + \lambda\mathbf{b}.$

Now we compute $OP^2$ and $OQ^2$:

$OP^2 = \mathbf{p} \cdot \mathbf{p} = (\lambda\mathbf{a} + (1 - \lambda)\mathbf{b}) \cdot (\lambda\mathbf{a} + (1 - \lambda)\mathbf{b})$

$= \lambda^2 a^2 + 2\lambda(1 - \lambda)(\mathbf{a} \cdot \mathbf{b}) + (1 - \lambda)^2 b^2.$

$OQ^2 = \mathbf{q} \cdot \mathbf{q} = ((1 - \lambda)\mathbf{a} + \lambda\mathbf{b}) \cdot ((1 - \lambda)\mathbf{a} + \lambda\mathbf{b})$

$= (1 - \lambda)^2 a^2 + 2\lambda(1 - \lambda)(\mathbf{a} \cdot \mathbf{b}) + \lambda^2 b^2.$

Subtracting:

$OQ^2 - OP^2 = \left[(1 - \lambda)^2 a^2 + \lambda^2 b^2\right] - \left[\lambda^2 a^2 + (1 - \lambda)^2 b^2\right]$

$= a^2\left[(1 - \lambda)^2 - \lambda^2\right] + b^2\left[\lambda^2 - (1 - \lambda)^2\right]$

$= (b^2 - a^2)\left[\lambda^2 - (1 - \lambda)^2\right].$

Expanding $\lambda^2 - (1 - \lambda)^2 = \lambda^2 - 1 + 2\lambda - \lambda^2 = 2\lambda - 1$.

So:

$OQ^2 - OP^2 = (b^2 - a^2)(2\lambda - 1).$

Substituting $\lambda = \frac{b}{a + b}$:

$2\lambda - 1 = \frac{2b}{a + b} - 1 = \frac{2b - a - b}{a + b} = \frac{b - a}{a + b}.$

Therefore:

$OQ^2 - OP^2 = (b^2 - a^2) \cdot \frac{b - a}{a + b} = (b - a)(b + a) \cdot \frac{b - a}{a + b} = (b - a)^2.$

Examiner Notes

As with Q6, this was both an unpopular question and poorly done. Those candidates who did do well generally did so after spotting that they could use the Angle Bisector Theorem to polish off the first half of the question, expressing $\lambda$ in terms of $a$ and $b$ almost immediately. Predominantly, the whole thing relied almost exclusively upon the use of the scalar product (or, alternatively, the Cosine Rule) and a bit of manipulation. The fact that the mean mark on this question was below 7 is simply indicative of the general lack of confidence amongst candidates where vectors are concerned.