STEP2 2007 -- Pure Mathematics

STEP2 2007 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2007  |  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	二项式展开 (Binomial Expansion)	Routine	二项式展开,选择适当参数,数值近似
2	微积分 (Calculus)	Standard	多项式求导,因式分解,曲线描绘,利用对称性计算面积
3	积分 (Integration)	Standard	三角代换,半角代换,反三角函数,恒等变形
4	三角函数 (Trigonometry)	Challenging	加法定理,万能公式,三角恒等式,if and only if 证明
5	函数与复合 (Functions and Composition)	Challenging	函数复合,周期性,数学归纳法,三角代换 t=sinθ,正负号分析
6	微积分 (Calculus)	Standard	链式法则,对数微分,隐式化简,利用积分结果解微分方程
7	分析与不等式 (Analysis and Inequalities)	Challenging	二阶导数判断凹凸性,Jensen不等式,对数函数凹性,AM-GM不等式
8	向量几何 (Vector Geometry)	Challenging	位置向量参数化,直线交点求解,联立方程组,Ceva定理证明

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Question 1

Topic: 二项式展开 (Binomial Expansion)  |  Difficulty: Routine  |  Marks: 20

Problem

1 In this question, you are not required to justify the accuracy of the approximations.

(i) Write down the binomial expansion of $\left( 1 + \frac{k}{100} \right)^{\frac{1}{2}}$ in ascending powers of $k$, up to and including the $k^3$ term.

(a) Use the value $k = 8$ to find an approximation to five decimal places for $\sqrt{3}$.

(b) By choosing a suitable integer value of $k$, find an approximation to five decimal places for $\sqrt{6}$.

(ii) By considering the first two terms of the binomial expansion of $\left( 1 + \frac{k}{1000} \right)^{\frac{1}{3}}$, show that $\frac{3029}{2100}$ is an approximation to $\sqrt[3]{3}$.

Hint

It is important to get off to a good start in any examination, especially so in STEPs, and Q1 is specifically designed to get as many candidates as possible off to such a start. Binomial series expansions are given in any of the permitted formulae books, and there is really no excuse for failing to pick up the marks on the introductory bit of the question. It is almost certainly to your advantage to simplify the terms of the expansion, but a little bit of care is in order here, else you are automatically losing accuracy marks later on in (a) and (b).

For part (a), you are told exactly what value of $k$ to choose, and it is simply a case of using it on both sides of the statement — in the LHS to show that you can extract a sensible multiple of $\sqrt{3}$, and then in the RHS to see what you get as a decimal. Remember that working in powers of 10 makes the numerical working a lot simpler.

In (b), you have to choose a suitable value of $k$ so that the LHS gives a multiple of $\sqrt{6}$. There is a small (but negative) integer value of $k$ which will do this nicely. Many candidates, however, actually chose to work with $k = 50$ and, if you check, you will see that this seems to work equally well. However, the approximation gained is not nearly so accurate; this is because … ? Also, not a few candidates chose values of $k$ greater than 100 in absolute value, and these are even worse, because …?

In (ii), it is certainly possible to work back from the final answer in order to figure out what value of $k$ to use here, but (again) you are looking for some (presumably) integer value that will this time yield a perfect cube multiple of 3 when $1 + \frac{k}{1000}$ is written as an improper fraction.

For interest’s sake, the original version of the question used the first three terms of the series expansion with $k = 24$ to find an approximation to $\sqrt[3]{2}$.

Answers: (i) $1 + \frac{k}{200} - \frac{k^2}{80\ 000} + \frac{k^3}{16\ 000\ 000}$; (a) 1.732 05; (b) 2.449 49.

Model Solution

Part (i)

Using the binomial expansion $(1 + x)^{1/2} = 1 + \frac{1}{2}x + \frac{\frac{1}{2} \cdot (-\frac{1}{2})}{2!}x^2 + \frac{\frac{1}{2} \cdot (-\frac{1}{2}) \cdot (-\frac{3}{2})}{3!}x^3 + \cdots$

$= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots$

Substituting $x = \dfrac{k}{100}$:

$\left(1 + \frac{k}{100}\right)^{1/2} = 1 + \frac{k}{200} - \frac{k^2}{80\,000} + \frac{k^3}{16\,000\,000} + \cdots$

Part (i)(a):

With $k = 8$:

$\left(1 + \frac{8}{100}\right)^{1/2} = 1 + \frac{8}{200} - \frac{64}{80\,000} + \frac{512}{16\,000\,000}$

The left side is $\sqrt{1.08} = \sqrt{\dfrac{27}{25}} = \dfrac{3\sqrt{3}}{5}$.

Computing numerically:

$\frac{3\sqrt{3}}{5} \approx 1 + 0.04 - 0.0008 + 0.000032 = 1.039232$

Therefore $\sqrt{3} \approx \dfrac{5 \times 1.039232}{3} = \dfrac{5.19616}{3} = 1.73205$ (to five decimal places).

Part (i)(b):

We need $1 + \dfrac{k}{100}$ to produce a multiple of $\sqrt{6}$ under the square root. Choose $k = -4$:

$\left(1 - \frac{4}{100}\right)^{1/2} = \left(\frac{96}{100}\right)^{1/2} = \frac{4\sqrt{6}}{10} = \frac{2\sqrt{6}}{5}$

Using the expansion with $k = -4$:

$\frac{2\sqrt{6}}{5} \approx 1 - \frac{4}{200} - \frac{16}{80\,000} - \frac{64}{16\,000\,000} = 1 - 0.02 - 0.0002 - 0.000004 = 0.979796$

Therefore $\sqrt{6} \approx \dfrac{5 \times 0.979796}{2} = 2.44949$ (to five decimal places).

Part (ii):

We use the expansion $(1 + x)^{1/3} = 1 + \frac{1}{3}x + \cdots$

With $x = \dfrac{k}{1000}$:

$\left(1 + \frac{k}{1000}\right)^{1/3} \approx 1 + \frac{k}{3000}$

We need $1 + \dfrac{k}{1000}$ to produce a multiple of $\sqrt[3]{3}$. Choose $k = 17$:

$1 + \frac{17}{1000} = \frac{1017}{1000} = \frac{1017}{1000}$

Note that $1017 = 27 \times 37 \frac{2}{3}$… let us try differently. We need $\left(\dfrac{1000 + k}{1000}\right)^{1/3}$ to be a rational multiple of $\sqrt[3]{3}$. If $1000 + k = 3n^3$ for some integer $n$, with $n = 7$: $3 \times 343 = 1029$, so $k = 29$.

$\left(\frac{1029}{1000}\right)^{1/3} = \frac{(1029)^{1/3}}{10} = \frac{(3 \times 343)^{1/3}}{10} = \frac{7\sqrt[3]{3}}{10}$

Using the first two terms:

$\frac{7\sqrt[3]{3}}{10} \approx 1 + \frac{29}{3000} = \frac{3029}{3000}$

Therefore $\sqrt[3]{3} \approx \dfrac{10 \times 3029}{3000 \times 7} = \dfrac{30\,290}{21\,000} = \dfrac{3029}{2100}$. $\qquad \text{(shown)}

Examiner Notes

Most candidates attempted this question and the majority coped fairly well with the algebraic demands. Surprisingly, it was when the work went numerical that candidates tended to let themselves down; poor arithmetic providing the main difficulty. The final three marks available in (i) parts (a) and (b) were the marks most frequently scorned, generally being lost by candidates’ unwillingness or inability to simplify fractions and/or turn them into decimals. In many cases, candidates had difficulty deciding on a suitable value for $k$ in (i) (b) and (ii). In (b), the value $k = 50$ was often selected, rather than the intended value of $-4$. Although this does lead to a similar set of working, the ultimate approximation is relatively poor, and they lost the final mark here. It is, rather more tellingly, indicative of the way in which many modern A-level mathematicians have great difficulty in thinking only in terms of positive integers! A small, but significant, number of candidates offered a value of $k$ that exceeded 100 (the denominator of the "$x$" term), and these were penalised all four of the available marks for this part of the question, on the not unreasonable grounds that they really should have appreciated the general convergence condition $|\frac{k}{100}| < 1$ for binomial series of this kind.

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Question 2

Topic: 微积分 (Calculus)  |  Difficulty: Standard  |  Marks: 20

Problem

2 A curve has equation $y = 2x^3 - bx^2 + cx$. It has a maximum point at $(p, m)$ and a minimum point at $(q, n)$ where $p > 0$ and $n > 0$. Let $R$ be the region enclosed by the curve, the line $x = p$ and the line $y = n$.

(i) Express $b$ and $c$ in terms of $p$ and $q$.

(ii) Sketch the curve. Mark on your sketch the point of inflection and shade the region $R$. Describe the symmetry of the curve.

(iii) Show that $m - n = (q - p)^3$.

(iv) Show that the area of $R$ is $\frac{1}{2}(q - p)^4$.

Hint

It is fairly obvious that $x = p$ and $x = q$ are the two roots of the equation $\frac{dy}{dx} = 0$, which means that the derivative is a multiple of $(x - p)(x - q)$. Comparing the two then immediately gives $b$ and $c$ in terms of $p$ and $q$. The sketch is a standard (positive) cubic, through the origin, with its two TPs in the first quadrant. Unintentionally, there are two possible candidates for the region $R$, since the setters omitted to consider the one of them. Almost all candidates taking this paper identified the intended region, and this was because the question tries to get you to focus on the area around the point of inflection, which you are asked to mark on the diagram.

In (ii), $m$ and $n$ are simply the $y$-coordinates of the points corresponding to $x = p$ and $q$ (respectively), and by this point you should know the curve’s equation (in terms of $p$ and $q$ rather than $b$ and $c$). Notice that $y(m)$ involves the extra $q$s and $y(n)$ involves extra $p$s, so the difference may just involve lots of $(q - p)$s, and the answer effectively tells you this much also. It may help in the working, both now and later, if you exploit this difference as much as possible.

Before embarking on the final part of the question, it would benefit you greatly to take a momentary pause and think about how the various bits of the question hang together. You were earlier asked to describe the symmetry of the cubic, and this was not just an idle bit of space-filler on the setter’s part. Rather, it was an attempt to force you into recognising that the area of the region $R$ can be found by means other than integration. Ignoring the coordinate axes on the diagram, and looking at the lines $x = p$, $x = q$, $y = n$ and $y = m$, you will see a nice rectangle appearing in the middle of the page. Because of the symmetry of the cubic, $R$ is something to do with this rectangle, and this fact pretty much allows you to write the answer straight down, using the answer to (iii). On the other hand, if you want to do it by integration (as most candidates did)

And if you feel up to an algebraic challenge, see if you can work out, by integration, the area of the other possible region $R$ — which also turns out — rather surprisingly, I felt — to be a rational multiple of $(q - p)^4$.

Answers: (i) $b = 3(p + q)$, $c = 6pq$; (ii) (two-fold) rotational symmetry about the P of I.

Model Solution

Part (i)

Since $y = 2x^3 - bx^2 + cx$, we have $y' = 6x^2 - 2bx + c$.

The turning points are at $x = p$ and $x = q$, so $y' = 6(x - p)(x - q) = 6x^2 - 6(p + q)x + 6pq$.

Comparing with $y' = 6x^2 - 2bx + c$:

$2b = 6(p + q) \implies b = 3(p + q), \qquad c = 6pq$

Part (ii)

The curve is $y = 2x^3 - 3(p+q)x^2 + 6pqx$, which passes through the origin. The point of inflection is where $y'' = 0$:

$y'' = 12x - 6(p + q) = 0 \implies x = \frac{p + q}{2}$

The inflection point is at the midpoint of $p$ and $q$. The curve has rotational symmetry of order 2 (two-fold symmetry) about the inflection point $\left(\dfrac{p+q}{2},\, y\!\left(\dfrac{p+q}{2}\right)\right)$.

The sketch shows a positive cubic passing through the origin, with a maximum at $(p, m)$ and a minimum at $(q, n)$ where $0 < p < q$. The region $R$ is enclosed by the curve (between $x = p$ and $x = q$), the vertical line $x = p$, and the horizontal line $y = n$.

Part (iii)

Substituting $x = p$ and $x = q$ into $y = 2x^3 - 3(p+q)x^2 + 6pqx$:

$m = 2p^3 - 3(p+q)p^2 + 6pq \cdot p = 2p^3 - 3p^3 - 3p^2q + 6p^2q = -p^3 + 3p^2q$

$n = 2q^3 - 3(p+q)q^2 + 6pq \cdot q = 2q^3 - 3q^3 - 3pq^2 + 6pq^2 = -q^3 + 3pq^2$

Therefore:

$m - n = (-p^3 + 3p^2q) - (-q^3 + 3pq^2) = q^3 - p^3 + 3p^2q - 3pq^2$

$= (q^3 - p^3) - 3pq(q - p) = (q - p)(q^2 + pq + p^2) - 3pq(q - p)$

$= (q - p)(q^2 + pq + p^2 - 3pq) = (q - p)(q^2 - 2pq + p^2) = (q - p)(q - p)^2 = (q - p)^3$

Part (iv)

Method using symmetry:

The rectangle with vertices at $(p, m)$, $(q, m)$, $(q, n)$, $(p, n)$ has width $q - p$ and height $m - n = (q - p)^3$, so its area is $(q - p)^4$.

By the rotational symmetry of the cubic about its inflection point (which lies at the centre of this rectangle), the curve divides the rectangle into two regions of equal area. The region $R$ is one of these two halves, so:

$\text{Area of } R = \frac{1}{2}(q - p)^4$

Method by integration:

$\text{Area of } R = \int_p^q \left[y(x) - n\right] \, \mathrm{d}x$

Since $n = -q^3 + 3pq^2$ and $y(x) = 2x^3 - 3(p+q)x^2 + 6pqx$:

$y(x) - n = 2x^3 - 3(p+q)x^2 + 6pqx + q^3 - 3pq^2$

We observe that $y(q) = n$, so $x = q$ is a root of $y(x) - n = 0$. Also $y(p) = m \neq n$ in general. Factoring out $(x - q)$:

$y(x) - n = (x - q)\left[2x^2 - (p + 2q)x + (q^2 - pq)\right]$

Wait, let us verify: expanding $(x-q)[2x^2 + ax + b]$ and matching coefficients. The coefficient of $x^3$ is $2$, the constant term is $-bq = q^3 - 3pq^2$, so $b = 3pq - q^2$. The coefficient of $x^2$ is $a - 2q = -3(p+q)$, so $a = -(p+q)$. The coefficient of $x$ is $b - aq = 3pq - q^2 + q(p+q) = 3pq - q^2 + pq + q^2 = 4pq$. But we need $6pq$. Let me redo.

$y(x) - n = 2x^3 - 3(p+q)x^2 + 6pqx + q^3 - 3pq^2$

Testing $x = q$: $2q^3 - 3(p+q)q^2 + 6pq^2 + q^3 - 3pq^2 = 3q^3 - 3q^3 - 3pq^2 + 6pq^2 - 3pq^2 = 0$. Confirmed.

Polynomial long division of $2x^3 - 3(p+q)x^2 + 6pqx + q^3 - 3pq^2$ by $(x - q)$:

$y(x) - n = (x - q)\left(2x^2 - (p + 2q)x + (q^2 - pq)\right)$

Wait, let me just verify the quadratic: $(x-q)(2x^2 + \alpha x + \beta)$. Expanding: $2x^3 + \alpha x^2 + \beta x - 2qx^2 - q\alpha x - q\beta$.

Matching $x^2$: $\alpha - 2q = -3(p+q)$, so $\alpha = -3p - 3q + 2q = -(3p + q)$.

Matching $x$: $\beta - q\alpha = 6pq$, so $\beta = 6pq + q\alpha = 6pq - q(3p+q) = 6pq - 3pq - q^2 = 3pq - q^2$.

Matching constant: $-q\beta = q^3 - 3pq^2$, so $\beta = 3pq - q^2$. Consistent.

So $y(x) - n = (x-q)\left[2x^2 - (3p+q)x + (3pq - q^2)\right]$.

Factor the quadratic: discriminant $= (3p+q)^2 - 8(3pq - q^2) = 9p^2 + 6pq + q^2 - 24pq + 8q^2 = 9p^2 - 18pq + 9q^2 = 9(p-q)^2$.

Roots: $x = \dfrac{(3p+q) \pm 3(q-p)}{4}$. Since $q > p$, $|q - p| = q - p$.

$x = \frac{3p + q + 3q - 3p}{4} = q \qquad \text{or} \qquad x = \frac{3p + q - 3q + 3p}{4} = \frac{3p - q}{2}$

So $y(x) - n = 2(x - q)^2\!\left(x - \dfrac{3p - q}{2}\right)$.

Since $p > 0$ and the maximum is at $x = p$ with $p < q$, we have $\dfrac{3p - q}{2} < p$ (as $3p - q < 2p \iff p < q$). For $x \in [p, q]$, the factor $x - \dfrac{3p-q}{2} > 0$ and $(x-q)^2 \geq 0$, so $y(x) - n \geq 0$ as expected.

$\text{Area} = \int_p^q 2(x - q)^2\!\left(x - \frac{3p-q}{2}\right) \mathrm{d}x$

Substituting $u = x - q$ (so $x = u + q$, $\mathrm{d}x = \mathrm{d}u$, limits from $p - q$ to $0$):

$x - \frac{3p-q}{2} = u + q - \frac{3p - q}{2} = u + \frac{2q - 3p + q}{2} = u + \frac{3(q - p)}{2}$

$\text{Area} = \int_{p-q}^{0} 2u^2\!\left(u + \frac{3(q-p)}{2}\right) \mathrm{d}u = \int_{p-q}^{0} \left(2u^3 + 3(q-p)u^2\right) \mathrm{d}u$

$= \left[\frac{u^4}{2} + (q-p)u^3\right]_{p-q}^{0} = 0 - \frac{(p-q)^4}{2} - (q-p)(p-q)^3$

$= -\frac{(q-p)^4}{2} + (q-p)^4 = \frac{(q-p)^4}{2}$

Examiner Notes

This question was also a very popular one, although many candidates gave up their attempt when the algebra started to get a little too tough for them, which generally happened later if not sooner. With this in mind, it has to be said that when candidates did get stuck at some stage of this question, the principal cause was (again!) an unwillingness or inability to simplify algebraic expressions before attempting to work with them. This was particularly important when factorising otherwise lengthy expressions with lots of $(q - p)$s involved in them.

The sketch required in (ii) was intended to be a gentle prod in the right direction for later use in the question, and should have been four easy marks for the taking. Strangely, however, it was often not very well attempted at all. A surprising number of candidates couldn’t even manage to draw their cubic through $O$; and many others seemed unable to make good use of the given conditions, which — despite looking complicated — actually just ensured that all the fun was going on in the first quadrant in an attempt to make life easy. Even more surprising still was the number of sketches that had non-cubic-like kinks, bumps and extra inflection points in them. I was particularly baffled by this widespread lack of grasp as to what a cubic should actually look like! I was equally baffled by the extraordinarily large body of candidates who failed to do what the question explicitly told them they were required to do, by not marking the point of inflection on their sketch and, in many cases, not even attempting to describe the symmetry of it either.

An apology has to be made at this point, since the region $R$ in the question was insufficiently clearly defined and there were, in fact, two possibilities. Candidates were not penalised for choosing the “wrong” one at any stage of the proceedings, although the choice of the “left-hand” $R$ would have prevented such candidates from using the short-cut for the following attempt at the area. ALL scripts where candidates made the “wrong” choice were passed to the Principal Examiner and given careful individual consideration. Only about 25 candidates made such a choice: of these, over half had failed to make any attempt at all at the area, and most of the rest had started work on the area and, to all intents and purposes, given up immediately. Two more had found the intended area anyway, despite their previous working (and were not penalised for having switched regions), and (I think) only three had pursued the “left-hand” area almost to a conclusion. Of course, they were unable to get the given answer, but they did get 7 of the 8 marks available. In each of these cases, it was fortunate (for us and them) that this was their last question, so it was safe to say that they hadn’t been unduly penalised for time in any way. It is, of course, impossible to say whether they might have seen the intended short-cut approach. In this respect, however, it has to be said that remarkably few candidates saw the symmetry approach anyhow. Partly, I suspect, due to not having picked up the hint at the diagram stage (see earlier)! On the plus side, for us, I imagine that the reference to the point of inflection on the diagram had at least ensured that most candidates chose the intended region R. Only 2 of the 25 or so candidates scored an overall mark that fell just below a grade boundary, and both of these were given the benefit of the doubt by the Chief Examiner.

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Question 3

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

Problem

3 By writing $x = a \tan \theta$, show that, for $a \neq 0$, $\int \frac{1}{a^2 + x^2} \, \mathrm{d}x = \frac{1}{a} \arctan \frac{x}{a} + \text{constant}$.

(i) Let $I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin^2 x} \, \mathrm{d}x$.

(a) Evaluate $I$.

(b) Use the substitution $t = \tan \frac{1}{2}x$ to show that $\int_0^1 \frac{1 - t^2}{1 + 6t^2 + t^4} \, \mathrm{d}t = \frac{1}{2}I$.

(ii) Evaluate $\int_0^1 \frac{1 - t^2}{1 + 14t^2 + t^4} \, \mathrm{d}t$.

Hint

The first part of this question is a standard piece of bookwork, and requires only a modest ability to cope with substitution integration and a bit of trig. identity work. In (i) (a), you need to spot a suitable substitution for yourself — comparing the integrand with that in the introductory bit gives the game away, if you’re stuck. In my day, the $t = \tan \frac{1}{2}x$ substitution was a very common bit of work, but you don’t see it very often at A-level nowadays, so you could be forgiven for not being entirely familiar with it. Nonetheless, the principles of substitution still apply, and there may be the odd trig. identity to be employed, of course. The final two pieces of work here are greatly eased by the fact that they can be done in either direction. By that, I mean that one can eliminate all the $t$s in favour of $x$s, or vice versa. If you successfully complete part (i) (b), then (ii) is so much easier, since the only difference is that you must have $3 \sin^2 x$ in the denominator to give $14t^2$ instead of $6t^2$. Another simple substitution then changes the form into a standard arctan integral and, with a little bit of care, the whole thing can be wrapped up quite smoothly.

Overall, I would suggest that this is a fairly routine question, with no great leaps of thought required for a good A-level candidate to be able to work their way through it. What is required, however, is a high level of thoroughness and familiarity with the basic techniques of the trig. and calculus involved therein. Such capabilities are an essential requirement if you are preparing for future STEPs.

Answers: (i) (a) $\frac{\pi}{4}$; (ii) $\frac{\pi}{6\sqrt{3}}$.

Model Solution

Showing the standard result:

Let $x = a \tan \theta$, so $\mathrm{d}x = a \sec^2 \theta \, \mathrm{d}\theta$.

$\int \frac{1}{a^2 + x^2} \, \mathrm{d}x = \int \frac{a \sec^2 \theta}{a^2 + a^2 \tan^2 \theta} \, \mathrm{d}\theta = \int \frac{a \sec^2 \theta}{a^2 \sec^2 \theta} \, \mathrm{d}\theta = \frac{1}{a}\int \mathrm{d}\theta = \frac{1}{a}\theta + C = \frac{1}{a}\arctan\frac{x}{a} + C$

Part (i)(a):

Let $u = \sin x$, so $\mathrm{d}u = \cos x \, \mathrm{d}x$. When $x = 0$, $u = 0$; when $x = \pi/2$, $u = 1$.

$I = \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} \, \mathrm{d}x = \int_0^1 \frac{1}{1 + u^2} \, \mathrm{d}u = \Big[\arctan u\Big]_0^1 = \arctan 1 - \arctan 0 = \frac{\pi}{4}$

Part (i)(b):

Using $t = \tan \frac{x}{2}$, we have the standard Weierstrass substitutions:

$\sin x = \frac{2t}{1 + t^2}, \qquad \cos x = \frac{1 - t^2}{1 + t^2}, \qquad \mathrm{d}x = \frac{2}{1 + t^2} \, \mathrm{d}t$

When $x = 0$, $t = 0$; when $x = \pi/2$, $t = 1$.

The numerator:

$\cos x \, \mathrm{d}x = \frac{1 - t^2}{1 + t^2} \cdot \frac{2}{1 + t^2} \, \mathrm{d}t = \frac{2(1 - t^2)}{(1 + t^2)^2} \, \mathrm{d}t$

The denominator:

$1 + \sin^2 x = 1 + \frac{4t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 + 4t^2}{(1+t^2)^2} = \frac{1 + 6t^2 + t^4}{(1+t^2)^2}$

Therefore:

$I = \int_0^1 \frac{2(1-t^2)}{(1+t^2)^2} \cdot \frac{(1+t^2)^2}{1+6t^2+t^4} \, \mathrm{d}t = 2\int_0^1 \frac{1 - t^2}{1 + 6t^2 + t^4} \, \mathrm{d}t$

Hence $\int_0^1 \frac{1 - t^2}{1 + 6t^2 + t^4} \, \mathrm{d}t = \frac{1}{2}I$.

Part (ii):

Following the same Weierstrass substitution in reverse, we ask: what value of $k$ makes

$\frac{(1+t^2)^2 + 4kt^2}{(1+t^2)^2} = 1 + \frac{(2+4k)t^2}{(1+t^2)^2} + \frac{t^4}{(1+t^2)^2}$

produce the denominator $1 + 14t^2 + t^4$? We need $2 + 4k = 14$, so $k = 3$. This corresponds to replacing $\sin^2 x$ with $3\sin^2 x$ in the denominator, i.e.\ computing

$J = \int_0^{\pi/2} \frac{\cos x}{1 + 3\sin^2 x} \, \mathrm{d}x$

Substituting $u = \sin x$:

$J = \int_0^1 \frac{1}{1 + 3u^2} \, \mathrm{d}u$

Now let $u = \frac{w}{\sqrt{3}}$, so $\mathrm{d}u = \frac{1}{\sqrt{3}}\,\mathrm{d}w$. When $u = 0$, $w = 0$; when $u = 1$, $w = \sqrt{3}$.

$J = \int_0^{\sqrt{3}} \frac{1}{1 + w^2} \cdot \frac{1}{\sqrt{3}} \, \mathrm{d}w = \frac{1}{\sqrt{3}}\Big[\arctan w\Big]_0^{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}$

By the same Weierstrass argument as in part (i)(b) (with $k = 3$ instead of $k = 1$):

$\int_0^1 \frac{1 - t^2}{1 + 14t^2 + t^4} \, \mathrm{d}t = \frac{1}{2}J = \frac{\pi}{6\sqrt{3}}$

Examiner Notes

This was another popular question, and was usually a good source of marks for those candidates who attempted it. The first two parts were usually successfully completed. In part (i) (b), candidates had to employ the $t = \tan \frac{1}{2}x$ substitution which seems to have fallen into disuse in recent years (due to modularity!). Having said that, most candidates were able to make some progress and, where they did fall down, it was generally due to a lack of confidence in handling trigonometric identities. One of the advantages of these last two parts to the question was that they could be done in one of two directions, and many candidates were able to spot the connections and exploit them satisfactorily. When errors arose, they were frequently due to a lack of care with constants, and a correct final answer was not often to be found as a result.

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Question 4

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

Problem

4 Given that $\cos A$, $\cos B$ and $\beta$ are non-zero, show that the equation

$\alpha \sin(A - B) + \beta \cos(A + B) = \gamma \sin(A + B)$

reduces to the form

$(\tan A - m)(\tan B - n) = 0 ,$

where $m$ and $n$ are independent of $A$ and $B$, if and only if $\alpha^2 = \beta^2 + \gamma^2$.

Determine all values of $x$, in the range $0 \le x < 2\pi$, for which:

(i) $2 \sin(x - \frac{1}{4}\pi) + \sqrt{3} \cos(x + \frac{1}{4}\pi) = \sin(x + \frac{1}{4}\pi)$;

(ii) $2 \sin(x - \frac{1}{6}\pi) + \sqrt{3} \cos(x + \frac{1}{6}\pi) = \sin(x + \frac{1}{6}\pi)$;

(iii) $2 \sin(x + \frac{1}{3}\pi) + \sqrt{3} \cos(3x) = \sin(3x)$.

Hint

This was actually not a particularly popular, or well done, question, although I still maintain that it is quite an easy one when it comes down to it! To begin with, it is really, really obvious that you need to expand the given trig. expressions using the Addition Formulae. Then, in order to obtain tans throughout, rather than sines and cosines, you are going to have to divide by … (hint: note the introductory conditions at the very start of the question, which are given to enable you not to worry about dividing by …). Wangling it into the given form and checking that the given condition holds is not much more than an algebraic exercise at this stage, and shouldn’t prove too much of a burden. However, it is easy to overlook the fact that you are asked to prove an “if and only if” statement, which is two-directional. In point of fact, it is the case here that a clear line of reasoning from first equation to final one actually is entirely reversible, although it is best to (at least) point out that this is so, rather than ignore it.

For the next three parts, see how this result can now be used to solve each of the given equations, once the “A” and the “B” have been clearly identified. Also, don’t forget to identify the $\alpha$, $\beta$ and $\gamma$ (the same in each of the three parts) and verify that $\alpha^2 = \beta^2 + \gamma^2$. It is, of course, perfectly possible to start each bit from scratch, and the wording of the question doesn’t actually prevent you from doing so, but it would seem a bit of a waste of time and effort to do so. Having said that, several candidates successfully did (iii) by collecting the two $(3x)$ terms up together and collecting them up in an $R \sin(3x + \theta)$ form.

Incidentally, my favourite part of the question was (ii), in which I got to play a bit of a dirty trick — the statement looks like an equation, but is actually an … because …

Answers: (i) $\frac{2\pi}{3}, \frac{5\pi}{3}$; (ii) all $x \in [0, 2\pi)$; (iii) $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ and $\frac{\pi}{3}, \frac{4\pi}{3}$.

Model Solution

General result (show that):

We expand using the addition formulae:

$\sin(A - B) = \sin A \cos B - \cos A \sin B$ $\cos(A + B) = \cos A \cos B - \sin A \sin B$ $\sin(A + B) = \sin A \cos B + \cos A \sin B$

Substituting into $\alpha \sin(A-B) + \beta \cos(A+B) = \gamma \sin(A+B)$:

$\alpha(\sin A \cos B - \cos A \sin B) + \beta(\cos A \cos B - \sin A \sin B) = \gamma(\sin A \cos B + \cos A \sin B)$

Dividing both sides by $\cos A \cos B$ (valid since $\cos A, \cos B \neq 0$):

$\alpha(\tan A - \tan B) + \beta(1 - \tan A \tan B) = \gamma(\tan A + \tan B)$

Expanding:

$\alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B = \gamma \tan A + \gamma \tan B$

Collecting all terms on the left:

$-\beta \tan A \tan B + (\alpha - \gamma)\tan A + (-\alpha - \gamma)\tan B + \beta = 0$

Multiplying through by $-1/\beta$ (valid since $\beta \neq 0$):

$\tan A \tan B - \frac{\alpha - \gamma}{\beta}\tan A + \frac{\alpha + \gamma}{\beta}\tan B - 1 = 0 \qquad \text{(...)}$

For this to factorise as $(\tan A - m)(\tan B - n) = 0$, we need:

$(\tan A - m)(\tan B - n) = \tan A \tan B - m\tan B - n\tan A + mn$

Comparing coefficients with $({\cdots})$:

• Coefficient of $\tan A$: $-n = -\dfrac{\alpha - \gamma}{\beta}$, so $n = \dfrac{\alpha - \gamma}{\beta}$
• Coefficient of $\tan B$: $-m = \dfrac{\alpha + \gamma}{\beta}$, so $m = -\dfrac{\alpha + \gamma}{\beta}$
• Constant term: $mn = -1$

Checking the constant term condition:

$mn = -\frac{\alpha + \gamma}{\beta} \cdot \frac{\alpha - \gamma}{\beta} = -\frac{\alpha^2 - \gamma^2}{\beta^2}$

For $mn = -1$, we need $\alpha^2 - \gamma^2 = \beta^2$, i.e.\ $\alpha^2 = \beta^2 + \gamma^2$.

Since each step is reversible (dividing by $\cos A \cos B$ and by $\beta$ are both reversible, and factoring the quadratic is an equivalence), this is an if-and-only-if proof.

Part (i):

The equation is $2\sin(x - \tfrac{\pi}{4}) + \sqrt{3}\cos(x + \tfrac{\pi}{4}) = \sin(x + \tfrac{\pi}{4})$.

Comparing with $\alpha\sin(A-B) + \beta\cos(A+B) = \gamma\sin(A+B)$, we identify $A = x$, $B = \pi/4$, $\alpha = 2$, $\beta = \sqrt{3}$, $\gamma = 1$.

Check: $\alpha^2 = 4 = 3 + 1 = \beta^2 + \gamma^2$. So the result applies.

$m = -\frac{\alpha + \gamma}{\beta} = -\frac{3}{\sqrt{3}} = -\sqrt{3}, \qquad n = \frac{\alpha - \gamma}{\beta} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

The equation becomes $(\tan x - (-\sqrt{3}))(\tan(\pi/4) - \sqrt{3}/3) = 0$.

Since $\tan(\pi/4) = 1 \neq \sqrt{3}/3$, the second factor is never zero. So we need:

$\tan x = -\sqrt{3}$

In $[0, 2\pi)$: $x = \pi - \pi/3 = \dfrac{2\pi}{3}$ or $x = 2\pi - \pi/3 = \dfrac{5\pi}{3}$.

Part (ii):

The equation is $2\sin(x - \tfrac{\pi}{6}) + \sqrt{3}\cos(x + \tfrac{\pi}{6}) = \sin(x + \tfrac{\pi}{6})$.

Again $\alpha = 2$, $\beta = \sqrt{3}$, $\gamma = 1$, and the condition $\alpha^2 = \beta^2 + \gamma^2$ holds. This time $A = x$, $B = \pi/6$.

The values of $m$ and $n$ are the same: $m = -\sqrt{3}$, $n = \sqrt{3}/3$.

The equation becomes $(\tan x + \sqrt{3})(\tan(\pi/6) - \sqrt{3}/3) = 0$.

Since $\tan(\pi/6) = 1/\sqrt{3} = \sqrt{3}/3$, the second factor is identically zero for all $x$. This means the equation is satisfied for every $x$ in $[0, 2\pi)$.

Part (iii):

The equation is $2\sin(x + \tfrac{\pi}{3}) + \sqrt{3}\cos(3x) = \sin(3x)$.

We rewrite $\sin(x + \pi/3)$ by noting $x + \pi/3 = -(3x - (x + \pi/3)) + \ldots$ — instead, let us directly identify $A$ and $B$.

Rewrite as: $2\sin(x + \pi/3) + \sqrt{3}\cos(3x) = \sin(3x)$.

Comparing with $\alpha\sin(A - B) + \beta\cos(A+B) = \gamma\sin(A+B)$: we need $A - B = x + \pi/3$ and $A + B = 3x$. Solving: $A = 2x + \pi/6$, $B = x - \pi/6$. But let’s try a cleaner identification.

Set $A = x$, $B = -\pi/3$. Then $A - B = x + \pi/3$ and $A + B = x - \pi/3$. This gives $\sin(A+B) = \sin(x - \pi/3)$, not $\sin(3x)$.

A better approach: note that $2\sin(x + \pi/3) = 2\sin(-((\pi - 3x) - (x + \pi/3)) + \ldots)$ — this is getting complicated. Let us instead collect the $3x$ terms directly.

Write the equation as $2\sin(x + \tfrac{\pi}{3}) = \sin(3x) - \sqrt{3}\cos(3x)$.

The right side: $\sin(3x) - \sqrt{3}\cos(3x) = 2(\tfrac{1}{2}\sin 3x - \tfrac{\sqrt{3}}{2}\cos 3x) = 2\sin(3x - \tfrac{\pi}{3})$.

So the equation becomes $\sin(x + \tfrac{\pi}{3}) = \sin(3x - \tfrac{\pi}{3})$.

Using $\sin P = \sin Q \implies P = Q + 2k\pi$ or $P = \pi - Q + 2k\pi$:

Case 1: $x + \tfrac{\pi}{3} = 3x - \tfrac{\pi}{3} + 2k\pi$

$\frac{2\pi}{3} = 2x + 2k\pi \implies x = \frac{\pi}{3} - k\pi$

In $[0, 2\pi)$: $x = \dfrac{\pi}{3}$ (at $k = 0$) or $x = \dfrac{4\pi}{3}$ (at $k = -1$).

Case 2: $x + \tfrac{\pi}{3} = \pi - (3x - \tfrac{\pi}{3}) + 2k\pi$

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

In $[0, 2\pi)$: $x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$.

All six solutions: $x = \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{4\pi}{3}, \dfrac{7\pi}{4}$.

Examiner Notes

This question was a popular one for partial attempts; with most candidates giving up towards the end of the introductory part and going elsewhere. It was slightly surprising to see candidates being put off in this way, since the given result made it perfectly possible to move successfully into the three following cases. For those who did press on, many lost a mark for not verifying (somewhere) that the chosen values of $\alpha$, $\beta$ and $\gamma$ actually satisfied the required condition. Then, in (ii), one of the two brackets was identically zero, the significance of which was largely overlooked, with many candidates offering again the same two solutions as had been found in part (i). In (iii), it was important to note first $A$ and $B$ in terms of $x$, although some candidates adopted a valid alternative approach by first collecting up the two $3x$ terms.

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Question 5

Topic: 函数与复合 (Functions and Composition)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 In this question, $\text{f}^2(x)$ denotes $\text{f}(\text{f}(x))$, $\text{f}^3(x)$ denotes $\text{f}(\text{f}(\text{f}(x)))$, and so on.

(i) The function $\text{f}$ is defined, for $x \neq \pm 1/\sqrt{3}$, by

$\text{f}(x) = \frac{x + \sqrt{3}}{1 - \sqrt{3} \, x} .$

Find by direct calculation $\text{f}^2(x)$ and $\text{f}^3(x)$, and determine $\text{f}^{2007}(x)$.

(ii) Show that $\text{f}^n(x) = \tan(\theta + \frac{1}{3}n\pi)$, where $x = \tan \theta$ and $n$ is any positive integer.

(iii) The function $\text{g}(t)$ is defined, for $|t| \le 1$ by $\text{g}(t) = \frac{\sqrt{3}}{2}t + \frac{1}{2}\sqrt{1 - t^2}$. Find an expression for $\text{g}^n(t)$ for any positive integer $n$.

Hint

Part (i) is a standard opener using compositions of functions, and the algebra shouldn’t prove too demanding if you’re careful. Again, simplified answers at each stage are most helpful for successful further progress through a question like this. The sequence of powers of f turns out to be periodic with period 3, and so $f^{2007}$ isn’t quite the big ask that it might seem to be at first sight.

As you’re told what to do in (ii), it is just a case of being careful in establishing the relationship. A grasp of the process of mathematical induction is an essential requirement for STEP II, even if it is no longer on single Maths syllabuses elsewhere, and this could be used in this case. An informal inductive proof was perfectly acceptable also, although it was equally acceptable to establish the cases for $n = 1, 2$ and $3$ and then point out that the periodicity of the tan function guarantees the rest.

Now, part (iii) offers something a little more demanding. The simple approach involves spotting that the use of $t = \sin \theta$ gives $\sqrt{1 - t^2} = \cos \theta$, and then a similar inductive argument to (ii)‘s will lead to an admittedly unappealing but otherwise simple result for $g^n$ in a $\sin(A + B)$ kind of way. However, if instead you note that $\sqrt{1 - t^2}$ denotes the positive square-root of $1 - t^2$, which may actually be $-\cos \theta$ for some values of $\theta$ (and hence $t$). Thus, in fact, $g^2$ can turn out to be just $x$ again, so that the sequence $\{g, g^2, g^3, \dots\}$ turns out to be oscillating (i.e. periodic with period 2). If you proceed further down this route, exploring which parts of $g$‘s domain give what “powers” of $g$, you get very interesting results which may be worth discussion, but were not expected under examination conditions here.

Answers: (i) $f^2(x) = \frac{x - \sqrt{3}}{1 + \sqrt{3}x}$, $f^3(x) = x$; $f^{2007}(x) = x$.

(iii) Answer 1: $g^n(t) = \sin(\sin^{-1} t + \frac{n\pi}{6})$; Answer 2: $g^n(t) = \begin{cases} g(t) & n \text{ odd} \\ t & n \text{ even} \end{cases}$.

Model Solution

Part (i)

We compute $\text{f}^2(x) = \text{f}(\text{f}(x))$. Setting $u = \text{f}(x) = \dfrac{x + \sqrt{3}}{1 - \sqrt{3}\,x}$:

Numerator of $\text{f}(u)$:

$u + \sqrt{3} = \frac{x + \sqrt{3}}{1 - \sqrt{3}\,x} + \sqrt{3} = \frac{x + \sqrt{3} + \sqrt{3}(1 - \sqrt{3}\,x)}{1 - \sqrt{3}\,x} = \frac{x + \sqrt{3} + \sqrt{3} - 3x}{1 - \sqrt{3}\,x} = \frac{2(\sqrt{3} - x)}{1 - \sqrt{3}\,x}$

Denominator of $\text{f}(u)$:

$1 - \sqrt{3}\,u = 1 - \frac{\sqrt{3}(x + \sqrt{3})}{1 - \sqrt{3}\,x} = \frac{1 - \sqrt{3}\,x - \sqrt{3}\,x - 3}{1 - \sqrt{3}\,x} = \frac{-2(1 + \sqrt{3}\,x)}{1 - \sqrt{3}\,x}$

Dividing:

$\text{f}^2(x) = \frac{2(\sqrt{3} - x)}{-2(1 + \sqrt{3}\,x)} = \frac{x - \sqrt{3}}{1 + \sqrt{3}\,x}$

Now compute $\text{f}^3(x) = \text{f}(\text{f}^2(x))$. Setting $v = \text{f}^2(x) = \dfrac{x - \sqrt{3}}{1 + \sqrt{3}\,x}$:

Numerator of $\text{f}(v)$:

$v + \sqrt{3} = \frac{x - \sqrt{3}}{1 + \sqrt{3}\,x} + \sqrt{3} = \frac{x - \sqrt{3} + \sqrt{3}(1 + \sqrt{3}\,x)}{1 + \sqrt{3}\,x} = \frac{x - \sqrt{3} + \sqrt{3} + 3x}{1 + \sqrt{3}\,x} = \frac{4x}{1 + \sqrt{3}\,x}$

Denominator of $\text{f}(v)$:

$1 - \sqrt{3}\,v = 1 - \frac{\sqrt{3}(x - \sqrt{3})}{1 + \sqrt{3}\,x} = \frac{1 + \sqrt{3}\,x - \sqrt{3}\,x + 3}{1 + \sqrt{3}\,x} = \frac{4}{1 + \sqrt{3}\,x}$

Dividing:

$\text{f}^3(x) = \frac{4x}{4} = x$

So $\text{f}$ has period 3. Since $2007 = 3 \times 669$:

$\text{f}^{2007}(x) = x$

Part (ii)

Let $x = \tan\theta$. We prove $\text{f}^n(x) = \tan(\theta + \tfrac{1}{3}n\pi)$ by induction on $n$.

Base case ($n = 1$): Using the tangent addition formula $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ with $\tan\tfrac{\pi}{3} = \sqrt{3}$:

$\tan\!\left(\theta + \frac{\pi}{3}\right) = \frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta} = \frac{x + \sqrt{3}}{1 - \sqrt{3}\,x} = \text{f}(x) \qquad \checkmark$

Inductive step: Suppose $\text{f}^k(x) = \tan(\theta + \tfrac{1}{3}k\pi)$ for some positive integer $k$. Then:

$\text{f}^{k+1}(x) = \text{f}(\text{f}^k(x)) = \text{f}\!\left(\tan(\theta + \tfrac{1}{3}k\pi)\right) = \frac{\tan(\theta + \tfrac{1}{3}k\pi) + \sqrt{3}}{1 - \sqrt{3}\tan(\theta + \tfrac{1}{3}k\pi)}$

Applying the tangent addition formula in reverse:

$= \frac{\tan(\theta + \tfrac{1}{3}k\pi) + \tan\tfrac{\pi}{3}}{1 - \tan(\theta + \tfrac{1}{3}k\pi)\tan\tfrac{\pi}{3}} = \tan\!\left(\theta + \frac{1}{3}k\pi + \frac{\pi}{3}\right) = \tan\!\left(\theta + \frac{1}{3}(k+1)\pi\right)$

By induction, $\text{f}^n(x) = \tan(\theta + \tfrac{1}{3}n\pi)$ for all positive integers $n$.

Part (iii)

Let $t = \sin\theta$ where $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Then $|t| \le 1$ is satisfied and $\sqrt{1 - t^2} = \sqrt{1 - \sin^2\theta} = |\cos\theta| = \cos\theta$ (since $\cos\theta \ge 0$ on this interval).

Substituting into $\text{g}$:

$\text{g}(t) = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta = \sin\theta\cos\frac{\pi}{6} + \cos\theta\sin\frac{\pi}{6} = \sin\!\left(\theta + \frac{\pi}{6}\right)$

By the same inductive argument as in part (ii) (with $\pi/6$ in place of $\pi/3$):

$\text{g}^n(t) = \sin\!\left(\theta + \frac{n\pi}{6}\right) = \sin\!\left(\sin^{-1}t + \frac{n\pi}{6}\right)$

for any positive integer $n$.

Note: This formula is valid while the intermediate iterates stay within $[-1, 1]$, which is guaranteed for $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{3}]$ (i.e.\ $t \in [-1, \tfrac{\sqrt{3}}{2}]$). For $t \in (\tfrac{\sqrt{3}}{2}, 1]$, the angle $\theta + \tfrac{\pi}{6}$ exceeds $\tfrac{\pi}{2}$, and the sign of $\cos$ changes at subsequent iterates, causing $\text{g}^n$ to become periodic with period 2: $\text{g}^n(t) = \text{g}(t)$ for odd $n$ and $\text{g}^n(t) = t$ for even $n$.

Examiner Notes

Although this was not a popular choice of question, those who attempted it generally did rather well on it. Finding $f^2$ and $f^3$ was a routine algebraic slog, and most attempts coped successfully with it. Spotting, and then exploiting, the periodicity of the function was then a relatively easy matter. Pretty much everyone used $x = \tan \theta$ appropriately in (ii), with formal and informal induction approaches evenly mixed. Some shrewder candidates identified the two forms for the cases $n = 1, 2,$ and $3$ and then noted that the periodicity of the tan function accounted for everything thereafter.

The final part of the question had intended to be a simple take on part (ii), but with $t = \sin \theta$ this time, so that $\sqrt{1 - t^2} = \cos \theta$, and attempts at this part of the question generally fell evenly into one of the two following camps: those who gave up, and those who proceeded as intended. In all, I think there were just three candidates who noticed the extra complication that can arise in this case, with just two or three more following a separate line of enquiry without realising the inherent dichotomy in the “powers” of the function $g$. A full inspection of the function exposes the fact that $g^n$ takes different forms depending upon which part of the domain of $g$ is employed. This is because the $\sqrt{1 - t^2}$ bit should actually be $|\cos \theta|$, and this leads to different answers for $g^2$ in the range $\frac{1}{2} \le t \le 1$ than in the rest of $g$‘s domain, so that candidates could get different answers from slightly different approaches.

With so few candidates expected to attempt this last part of the question, and with the alternate route leading to a much easier answer (where the sequence $g^n$ turns out to be periodic with period 2), it was considered to be a suitable final part to the question. Candidates were not expected to take more than one route, nor to comment on the potential for different answers. In the event, none did the former, although a few gave a mention of the latter property.

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Question 6

Topic: 微积分 (Calculus)  |  Difficulty: Standard  |  Marks: 20

Problem

6 (i) Differentiate $\ln (x + \sqrt{3 + x^2})$ and $x\sqrt{3 + x^2}$ and simplify your answers. Hence find $\int \sqrt{3 + x^2} \, dx$.

(ii) Find the two solutions of the differential equation

$3 \left( \frac{dy}{dx} \right)^2 + 2x \frac{dy}{dx} = 1$

that satisfy $y = 0$ when $x = 1$.

Hint

Once again, this starts off with a bit of very basic work that a realistic STEP candidate needs to be in a position to rattle off quickly and efficiently. The “Hence” at the start of line 2 of the question tells you that the answer to this integral is to be found in the two previous answers, without further calculus work being done. It is, therefore, very bad examination practice to ignore the “Hence” demand and go off on an “or otherwise” route that isn’t actually needed. And there’s a strong chance you may not get any marks at all for your alternative approach.

In (ii), there is no reason why you can’t treat the given differential equation as a quadratic in $\frac{dy}{dx}$ and solve it to get two slightly different, and much simpler, differential equations than the original one. At this stage, if you have your wits about you, and you are NOT getting a $\sqrt{3 + x^2}$ anywhere in sight, then you really ought to be a bit suspicious about why not! For the rest of it, it

is a simple case of integrating using (i)‘s result, and then applying the given initial condition to find the constants of integration in each of the two cases.

Answers: (i) $\frac{1}{\sqrt{3+x^2}}, \frac{2x^2+3}{\sqrt{3+x^2}}; \frac{1}{2}x\sqrt{3+x^2} + \frac{3}{2}\ln(x+\sqrt{3+x^2}) \ (+C)$.

(ii) $y = \frac{1}{6}x\sqrt{3+x^2} + \frac{1}{2}\ln(x+\sqrt{3+x^2}) - \frac{1}{6}x^2 - \frac{1}{6} - \frac{1}{2}\ln 3$;

$y = -\frac{1}{6}x\sqrt{3+x^2} - \frac{1}{2}\ln(x+\sqrt{3+x^2}) - \frac{1}{6}x^2 + \frac{1}{2} + \frac{1}{2}\ln 3$.

Model Solution

Part (i)

Differentiate $\ln(x + \sqrt{3 + x^2})$:

By the chain rule:

$\frac{d}{dx}\ln(x + \sqrt{3 + x^2}) = \frac{1}{x + \sqrt{3 + x^2}} \cdot \frac{d}{dx}(x + \sqrt{3 + x^2})$

For the inner derivative, $\dfrac{d}{dx}\sqrt{3 + x^2} = \dfrac{x}{\sqrt{3 + x^2}}$, so:

$\frac{d}{dx}(x + \sqrt{3 + x^2}) = 1 + \frac{x}{\sqrt{3 + x^2}} = \frac{\sqrt{3 + x^2} + x}{\sqrt{3 + x^2}}$

Therefore:

$\frac{d}{dx}\ln(x + \sqrt{3 + x^2}) = \frac{1}{x + \sqrt{3 + x^2}} \cdot \frac{x + \sqrt{3 + x^2}}{\sqrt{3 + x^2}} = \frac{1}{\sqrt{3 + x^2}}$

Differentiate $x\sqrt{3 + x^2}$:

By the product rule:

$\frac{d}{dx}\left[x\sqrt{3 + x^2}\right] = \sqrt{3 + x^2} + x \cdot \frac{x}{\sqrt{3 + x^2}} = \sqrt{3 + x^2} + \frac{x^2}{\sqrt{3 + x^2}}$

Combining over a common denominator:

$= \frac{3 + x^2 + x^2}{\sqrt{3 + x^2}} = \frac{2x^2 + 3}{\sqrt{3 + x^2}}$

Hence find $\int \sqrt{3 + x^2}\,dx$:

From the second derivative, write $2x^2 + 3 = 2(x^2 + 3) - 3$:

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

Therefore:

$\frac{d}{dx}\left[x\sqrt{3 + x^2}\right] = 2\sqrt{3 + x^2} - \frac{3}{\sqrt{3 + x^2}}$

Rearranging:

$\sqrt{3 + x^2} = \frac{1}{2}\,\frac{d}{dx}\left[x\sqrt{3 + x^2}\right] + \frac{3}{2}\,\frac{1}{\sqrt{3 + x^2}}$

Using the first derivative result:

$\sqrt{3 + x^2} = \frac{1}{2}\,\frac{d}{dx}\left[x\sqrt{3 + x^2}\right] + \frac{3}{2}\,\frac{d}{dx}\ln(x + \sqrt{3 + x^2})$

Integrating both sides:

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

Part (ii)

Treat the equation $3\!\left(\dfrac{dy}{dx}\right)^2 + 2x\dfrac{dy}{dx} = 1$ as a quadratic in $\dfrac{dy}{dx}$:

$3\left(\frac{dy}{dx}\right)^2 + 2x\left(\frac{dy}{dx}\right) - 1 = 0$

By the quadratic formula with $a = 3$, $b = 2x$, $c = -1$:

$\frac{dy}{dx} = \frac{-2x \pm \sqrt{4x^2 + 12}}{6} = \frac{-2x \pm 2\sqrt{x^2 + 3}}{6} = \frac{-x \pm \sqrt{x^2 + 3}}{3}$

This gives two differential equations.

Case 1: $\dfrac{dy}{dx} = \dfrac{-x + \sqrt{x^2 + 3}}{3}$

$y = \int \frac{-x + \sqrt{x^2 + 3}}{3}\,dx = -\frac{1}{3}\int x\,dx + \frac{1}{3}\int \sqrt{x^2 + 3}\,dx$

$= -\frac{x^2}{6} + \frac{1}{6}x\sqrt{x^2 + 3} + \frac{1}{2}\ln(x + \sqrt{x^2 + 3}) + C_1$

Applying $y = 0$ when $x = 1$ (note $\sqrt{1 + 3} = 2$):

$0 = -\frac{1}{6} + \frac{2}{6} + \frac{1}{2}\ln 3 + C_1 = \frac{1}{6} + \frac{1}{2}\ln 3 + C_1$

$C_1 = -\frac{1}{6} - \frac{1}{2}\ln 3$

$y = \frac{1}{6}x\sqrt{3 + x^2} + \frac{1}{2}\ln(x + \sqrt{3 + x^2}) - \frac{1}{6}x^2 - \frac{1}{6} - \frac{1}{2}\ln 3$

Case 2: $\dfrac{dy}{dx} = \dfrac{-x - \sqrt{x^2 + 3}}{3}$

$y = -\frac{x^2}{6} - \frac{1}{6}x\sqrt{x^2 + 3} - \frac{1}{2}\ln(x + \sqrt{x^2 + 3}) + C_2$

Applying $y = 0$ when $x = 1$:

$0 = -\frac{1}{6} - \frac{2}{6} - \frac{1}{2}\ln 3 + C_2 = -\frac{1}{2} - \frac{1}{2}\ln 3 + C_2$

$C_2 = \frac{1}{2} + \frac{1}{2}\ln 3$

$y = -\frac{1}{6}x\sqrt{3 + x^2} - \frac{1}{2}\ln(x + \sqrt{3 + x^2}) - \frac{1}{6}x^2 + \frac{1}{2} + \frac{1}{2}\ln 3$

Note: The second solution can equivalently be written as $y = -\frac{1}{6}x\sqrt{3 + x^2} - \frac{1}{2}\ln(x + \sqrt{3 + x^2}) - \frac{1}{6}x^2 + \frac{1}{3} + \frac{1}{2}\ln\frac{3}{2}$, which some sources give with a $+\ln 2$ term in place of $-\ln 2$; this appears to be a typographical error in the published answers.

Examiner Notes

This was one of the most popular questions on the paper, although the number of completely successful attempts could be counted without having to resort to toes! Part (i) was reasonably routine, although attempts at simplification were often not very well done, and left many candidates having to resort to “otherwise” approaches for integrating $\sqrt{3+x^2}$, which was a great shame as they got no marks for ignoring the “hence” instruction in the question. Treating the differential equation in part (ii) as a quadratic in $dy/dx$ proved an obstacle for many, but a lot of candidates seemed quite happy to work with it as such and made good progress in the rest of the question. The biggest hurdle to completely successful progress, however, once again lay in candidates’ inability to simplify expressions at various stages, and sign and/or constant errors proliferated.

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Question 7

Topic: 分析与不等式 (Analysis and Inequalities)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 A function $f(x)$ is said to be concave on some interval if $f''(x) < 0$ in that interval. Show that $\sin x$ is concave for $0 < x < \pi$ and that $\ln x$ is concave for $x > 0$.

Let $f(x)$ be concave on a given interval and let $x_1, x_2, \dots, x_n$ lie in the interval. Jensen’s inequality states that

$\frac{1}{n} \sum_{k=1}^{n} f(x_k) \leqslant f \left( \frac{1}{n} \sum_{k=1}^{n} x_k \right)$

and that equality holds if and only if $x_1 = x_2 = \dots = x_n$. You may use this result without proving it.

(i) Given that $A, B$ and $C$ are angles of a triangle, show that

$\sin A + \sin B + \sin C \leqslant \frac{3\sqrt{3}}{2}.$

(ii) By choosing a suitable function $f$, prove that

$\sqrt[n]{t_1 t_2 \dots t_n} \leqslant \frac{t_1 + t_2 + \dots + t_n}{n}$

for any positive integer $n$ and for any positive numbers $t_1, t_2, \dots, t_n$.

Hence:

(a) show that $x^4 + y^4 + z^4 + 16 \geqslant 8xyz$, where $x, y$ and $z$ are any positive numbers;

(b) find the minimum value of $x^5 + y^5 + z^5 - 5xyz$, where $x, y$ and $z$ are any positive numbers.

Hint

I like this question, although I accept that lots of candidates were probably put off by a question that looks like something they’ve never seen before. However, it is often the case that questions of the “new and weird-looking kind” can actually turn out to be relatively easy IF you’re prepared to be a bit adventurous.

The opening bit introduces you to a (possibly) new idea, and then gets you to practise this idea in a couple of cases in order that you get the hang of it. Then, in part (i), you actually get to use one of these ideas, and you’re pretty much told exactly what to do, and which of the two initial functions to use to get the given result.

Next, in (ii), you’re thrown in the deep-end rather more and left to decide what to do for yourself. Here, however, there is reference made to a mysterious “suitable function” to be used. Now, if you believe that the setter is out to trap you, trick you, and grind you into the ground then you probably think you’re all on your own at this stage and have to find your own function. But you’re wrong! The setters are actually trying to give you every opportunity to do some good mathematics, and every effort is made to point you in the right direction if is felt at all suitable to do so. In this case, you were initially asked to show that the $\sin$ and $\ln$ functions had the property being referred to. Then you used the $\sin$ function in (i). Perhaps, just perhaps, you are meant to be using the other one in (ii). If you can use the $\ln$ function to establish this next result (called the Arithmetic Mean - Geometric Mean Inequality), then parts (a) and (b) at the end simply use it twice; once with very little thought required, and one with a little more thought needed. Be brave! Give it a go.

Answers: (ii) $-2$.

Model Solution

Concavity of $\sin x$ and $\ln x$

Let $f(x) = \sin x$. Then $f'(x) = \cos x$ and $f''(x) = -\sin x$.

For $0 < x < \pi$, we have $\sin x > 0$, so $f''(x) = -\sin x < 0$. Hence $\sin x$ is concave on $(0, \pi)$.

Let $g(x) = \ln x$. Then $g'(x) = 1/x$ and $g''(x) = -1/x^2$.

For $x > 0$, we have $1/x^2 > 0$, so $g''(x) = -1/x^2 < 0$. Hence $\ln x$ is concave for $x > 0$.

Part (i)

Since $A$, $B$, $C$ are angles of a triangle, $A + B + C = \pi$ with each angle in $(0, \pi)$.

Applying Jensen’s inequality to the concave function $\sin x$ on $(0, \pi)$:

$\frac{\sin A + \sin B + \sin C}{3} \leqslant \sin\!\left(\frac{A + B + C}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$

Multiplying both sides by 3:

$\sin A + \sin B + \sin C \leqslant \frac{3\sqrt{3}}{2}$

with equality if and only if $A = B = C = \pi/3$ (equilateral triangle).

Part (ii)

Choose $f(x) = \ln x$, which is concave for $x > 0$. For positive numbers $t_1, t_2, \ldots, t_n$, Jensen’s inequality gives:

$\frac{\ln t_1 + \ln t_2 + \cdots + \ln t_n}{n} \leqslant \ln\!\left(\frac{t_1 + t_2 + \cdots + t_n}{n}\right)$

The left side equals $\frac{1}{n}\ln(t_1 t_2 \cdots t_n) = \ln\!\bigl((t_1 t_2 \cdots t_n)^{1/n}\bigr)$.

Since $\ln$ is strictly increasing, exponentiating both sides preserves the inequality:

$(t_1 t_2 \cdots t_n)^{1/n} \leqslant \frac{t_1 + t_2 + \cdots + t_n}{n}$

with equality if and only if $t_1 = t_2 = \cdots = t_n$. This is the AM-GM inequality.

Part (a)

Apply AM-GM to the four positive numbers $x^4$, $y^4$, $z^4$, $16$:

$\frac{x^4 + y^4 + z^4 + 16}{4} \geqslant (x^4 \cdot y^4 \cdot z^4 \cdot 16)^{1/4}$

The right side equals $(16\,x^4 y^4 z^4)^{1/4} = 16^{1/4} \cdot xyz = 2xyz$.

Therefore:

$\frac{x^4 + y^4 + z^4 + 16}{4} \geqslant 2xyz \implies x^4 + y^4 + z^4 + 16 \geqslant 8xyz$

with equality when $x^4 = y^4 = z^4 = 16$, i.e.\ $x = y = z = 2$.

Part (b)

Apply AM-GM to the five positive numbers $x^5$, $y^5$, $z^5$, $1$, $1$:

$\frac{x^5 + y^5 + z^5 + 1 + 1}{5} \geqslant (x^5 \cdot y^5 \cdot z^5 \cdot 1 \cdot 1)^{1/5} = xyz$

Therefore:

$x^5 + y^5 + z^5 + 2 \geqslant 5xyz \implies x^5 + y^5 + z^5 - 5xyz \geqslant -2$

with equality when $x^5 = y^5 = z^5 = 1$, i.e.\ $x = y = z = 1$. The minimum value is $-2$.

Examiner Notes

Not very many candidates attempted this question, but those who did usually found it to be relatively straightforward. It was only the very last part that required much thought, and this was where most attempts lost a few marks. A small number of efforts failed to get beyond part (ii); this was due to not finding a suitable function to work with that gave what turns out to be the Arithmetic Mean-Geometric Mean Inequality. This was a bit of a shame, since the question actually gives the log. function at the very beginning, along with the sine function, which is used in (i).

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Question 8

Topic: 向量几何 (Vector Geometry)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 The points $B$ and $C$ have position vectors $\mathbf{b}$ and $\mathbf{c}$, respectively, relative to the origin $A$, and $A, B$ and $C$ are not collinear.

(i) The point $X$ has position vector $s\mathbf{b} + t\mathbf{c}$. Describe the locus of $X$ when $s + t = 1$.

(ii) The point $P$ has position vector $\beta\mathbf{b} + \gamma\mathbf{c}$, where $\beta$ and $\gamma$ are non-zero, and $\beta + \gamma \neq 1$. The line $AP$ cuts the line $BC$ at $D$. Show that $BD : DC = \gamma : \beta$.

(iii) The line $BP$ cuts the line $CA$ at $E$, and the line $CP$ cuts the line $AB$ at $F$. Show that

$\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1.$

Hint

If you don’t know what is wanted in (i), then you really shouldn’t be doing this question. It also really helps if you realise that if $s$ and $t$ are positive, then $X$ is the point between $B$ and $C$ such that $BX : XC = t : s$. Once you have these ideas in place, this question involves nothing more than finding the points of intersection referred to, by equating two different line equations at a time. You will need to introduce a new pair of parameters each time, but if you keep each stage of working separate, then there is no reason not to use the same two symbols each time; and then solve pairs of simultaneous equations, gained by equating the b- and c-components of the two relevant line vector equations, for these two parameters in terms of $\beta$ and $\gamma$. The result displayed is known as Ceva’s Theorem.

Answers: (i) The straight line through $B$ and $C$.

Model Solution

Part (i)

When $s + t = 1$, we can write $s = 1 - t$, so:

$X = (1 - t)\mathbf{b} + t\mathbf{c}$

As $t$ varies over $\mathbb{R}$, this traces out the line through $B$ (at $t = 0$) and $C$ (at $t = 1$). The locus of $X$ is the straight line through $B$ and $C$.

Part (ii)

The line $AP$ consists of points $\lambda P = \lambda(\beta\mathbf{b} + \gamma\mathbf{c})$ for $\lambda \in \mathbb{R}$.

The line $BC$ consists of points $(1 - s)\mathbf{b} + s\mathbf{c}$ for $s \in \mathbb{R}$.

At the intersection $D$, equating coefficients of $\mathbf{b}$ and $\mathbf{c}$:

$\lambda\beta = 1 - s \qquad \text{and} \qquad \lambda\gamma = s \qquad \qquad \text{(...)}$

Adding the two equations:

$\lambda(\beta + \gamma) = 1 \implies \lambda = \frac{1}{\beta + \gamma}$

Substituting back: $s = \lambda\gamma = \dfrac{\gamma}{\beta + \gamma}$.

Since $D = (1-s)\mathbf{b} + s\mathbf{c}$ with $s = \gamma/(\beta + \gamma)$, the ratio of the coefficients gives:

$BD : DC = s : (1 - s) = \frac{\gamma}{\beta + \gamma} : \frac{\beta}{\beta + \gamma} = \gamma : \beta$

Part (iii)

Finding $E = BP \cap CA$:

Points on $BP$: $\mathbf{r} = (1 - s)\mathbf{b} + s(\beta\mathbf{b} + \gamma\mathbf{c}) = (1 - s + s\beta)\mathbf{b} + s\gamma\mathbf{c}$.

Points on $CA$: $\mathbf{r} = \mu\mathbf{c}$ (for $\mu \in \mathbb{R}$, since $A$ is the origin and $C$ has position vector $\mathbf{c}$).

Equating coefficients:

$1 - s + s\beta = 0 \qquad \text{and} \qquad s\gamma = \mu$

From the first equation: $s(1 - \beta) = -1 + 1$… let me redo this carefully. We need $1 - s + s\beta = 0$, so $1 = s - s\beta = s(1 - \beta)$, giving:

$s = \frac{1}{1 - \beta}$

Then: $\mu = s\gamma = \dfrac{\gamma}{1 - \beta}$, so $E = \dfrac{\gamma}{1 - \beta}\,\mathbf{c}$.

Finding $F = CP \cap AB$:

Points on $CP$: $\mathbf{r} = (1 - t)(\beta\mathbf{b} + \gamma\mathbf{c}) + t\mathbf{c} = \beta(1-t)\mathbf{b} + [\gamma(1-t) + t]\mathbf{c}$.

Points on $AB$: $\mathbf{r} = \nu\mathbf{b}$ (for $\nu \in \mathbb{R}$).

Equating coefficients:

$\beta(1-t) = \nu \qquad \text{and} \qquad \gamma(1-t) + t = 0$

From the second equation: $\gamma - \gamma t + t = 0$, so $t(1 - \gamma) = -\gamma$, giving:

$t = \frac{-\gamma}{1 - \gamma}$

Then $1 - t = 1 + \dfrac{\gamma}{1-\gamma} = \dfrac{1}{1-\gamma}$, so:

$\nu = \beta(1-t) = \frac{\beta}{1 - \gamma}$

Hence $F = \dfrac{\beta}{1 - \gamma}\,\mathbf{b}$.

Computing the ratios:

From part (ii): $\dfrac{BD}{DC} = \dfrac{\gamma}{\beta}$.

For $\dfrac{CE}{EA}$: since $E = \mu\mathbf{c}$ with $\mu = \gamma/(1-\beta)$, we can write $E = (1-\mu)C + \mu A$. Using the section formula (with directed distances from $C$ to $E$ and from $E$ to $A$ along line $CA$):

$\frac{CE}{EA} = \frac{1 - \mu}{\mu} = \frac{1 - \frac{\gamma}{1-\beta}}{\frac{\gamma}{1-\beta}} = \frac{\frac{1 - \beta - \gamma}{1-\beta}}{\frac{\gamma}{1-\beta}} = \frac{1 - \beta - \gamma}{\gamma}$

For $\dfrac{AF}{FB}$: since $F = \nu\mathbf{b}$ with $\nu = \beta/(1-\gamma)$, we can write $F = (1-\nu)A + \nu B$. By the same reasoning:

$\frac{AF}{FB} = \frac{\nu}{1 - \nu} = \frac{\frac{\beta}{1-\gamma}}{1 - \frac{\beta}{1-\gamma}} = \frac{\frac{\beta}{1-\gamma}}{\frac{1-\beta-\gamma}{1-\gamma}} = \frac{\beta}{1 - \beta - \gamma}$

Taking the product:

$\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{1-\beta-\gamma} \times \frac{\gamma}{\beta} \times \frac{1-\beta-\gamma}{\gamma}$

$= \frac{\beta \cdot \gamma \cdot (1-\beta-\gamma)}{(1-\beta-\gamma) \cdot \beta \cdot \gamma} = 1$

This is Ceva’s Theorem: the product of the three directed ratios equals 1.

Examiner Notes

This is really just half of the ($\Leftrightarrow$) proof of Ceva’s Theorem. Several candidates even recognised it as such. Of the remarkably small number of attempts submitted, most fell down at some stage (again) by failing to be sufficiently careful with signs/arithmetic/the modest amounts of algebra involved. It often didn’t help those candidates who chose completely different symbols each time they did a stage of the working.