STEP2 2009 -- Pure Mathematics

STEP2 2009 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2009  |  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	解析几何 (Coordinate Geometry)	Standard	坐标对称性,斜率判定矩形,代数化简,利用曲线方程消元
2	函数与微积分 (Functions and Calculus)	Challenging	隐函数求导,三角函数性质,小量近似,对数化简,分段积分
3	三角函数 (Trigonometry)	Standard	三角恒等变换,正切加法公式,分母有理化,迭代应用恒等式
4	多项式 (Polynomials)	Challenging	多项式因式分解,求导与整除性,利用导数阶数确定因式,积分确定系数
5	积分 (Integration)	Standard	根式化简,分段积分,面积与积分的区别,绝对值处理
6	数列与级数 (Sequences and Series)	Challenging	递推关系放缩, 几何级数求和, 分奇偶项估计
7	微积分 (Calculus)	Challenging	乘积法则, 因式分解, 线性组合凑微分
8	向量 (Vectors)	Standard	位置向量运算, 参数消元, 向量共线条件

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Question 1

Topic: 解析几何 (Coordinate Geometry)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Two curves have equations $x^4 + y^4 = u$ and $xy = v$, where $u$ and $v$ are positive constants. State the equations of the lines of symmetry of each curve.

The curves intersect at the distinct points $A, B, C$ and $D$ (taken anticlockwise from $A$). The coordinates of $A$ are $(\alpha, \beta)$, where $\alpha > \beta > 0$. Write down, in terms of $\alpha$ and $\beta$, the coordinates of $B, C$ and $D$.

Show that the quadrilateral $ABCD$ is a rectangle and find its area in terms of $u$ and $v$ only. Verify that, for the case $u = 81$ and $v = 4$, the area is 14.

Hint

Both graphs are symmetric in the lines $y = \pm x$, and $x^4 + y^4 = u$ is also symmetric in the $x$- and $y$-axes. These facts immediately enable us to write down the coordinates of $B(\beta, \alpha)$, $C(-\alpha, -\beta)$ and $D(-\beta, -\alpha)$. Remember to keep the cyclic order $A, B, C, D$ correct, else this could lead to silly calculational errors later on. The easiest way to show that $ABCD$ is a rectangle is to work out the gradients of the four sides (which turn out to be either $1$ or $-1$) and then note that each pair of adjacent sides is perpendicular using the “product of gradients $= -1$” result. Working with distances is also a possible solution-approach but, on its own, only establishes that the quadrilateral is a parallelogram. However, the next part requires you to calculate distances anyhow, and we find that $CB, DA$ have length $(\alpha + \beta)\sqrt{2}$ while $BA, DC$ are of length $(\alpha - \beta)\sqrt{2}$. Multiplying these then give the area of $ABCD$ as $2(\alpha^2 - \beta^2)$.

All of this is very straightforward, and the only tricky bit of work comes next. It is important to think of $\alpha$ and $\beta$ as particular values of $x$ and $y$ satisfying each of the two original equations. It is then clear that $(\alpha^2 - \beta^2)^2 = \alpha^4 + \beta^4 - 2(\alpha^2 \beta^2) = u - 2v^2$, so that $\text{Area } ABCD = 2\sqrt{u - 2v^2}$. Substituting $u = 81, v = 4$ into this formula then gives $\text{Area} = 2\sqrt{81 - 2 \times 16} = 14$, which is intended principally as a means of checking that your answer is correct.

Model Solution

Lines of symmetry.

For $x^4 + y^4 = u$: replacing $x$ with $-x$, or $y$ with $-y$, or swapping $x$ and $y$, or replacing $(x, y)$ with $(-x, -y)$ all leave the equation unchanged. So the lines of symmetry are $x = 0$, $y = 0$, $y = x$ and $y = -x$.

For $xy = v$: swapping $x$ and $y$ leaves the equation unchanged, and replacing $(x, y)$ with $(-x, -y)$ also leaves it unchanged. So the lines of symmetry are $y = x$ and $y = -x$.

Coordinates of $B$, $C$, $D$.

Since $A = (\alpha, \beta)$ lies on both curves, we have $\alpha^4 + \beta^4 = u$ and $\alpha\beta = v$. Both curves are symmetric in $y = x$, so $(\beta, \alpha)$ also lies on both curves. Both curves are symmetric in the origin (replacing $(x,y)$ with $(-x,-y)$), so $(-\alpha, -\beta)$ and $(-\beta, -\alpha)$ also lie on both curves.

The four intersection points, taken anticlockwise from $A$ in the first quadrant, are:

$A = (\alpha, \beta), \quad B = (\beta, \alpha), \quad C = (-\alpha, -\beta), \quad D = (-\beta, -\alpha).$

Showing $ABCD$ is a rectangle.

We compute the gradients of the four sides:

$m_{AB} = \frac{\alpha - \beta}{\beta - \alpha} = -1, \qquad m_{BC} = \frac{-\beta - \alpha}{-\alpha - \beta} = 1,$

$m_{CD} = \frac{-\alpha + \beta}{-\beta + \alpha} = -1, \qquad m_{DA} = \frac{\alpha + \beta}{\beta + \alpha} = 1.$

Since $m_{AB} \cdot m_{BC} = (-1)(1) = -1$, adjacent sides $AB$ and $BC$ are perpendicular. By the same calculation, every pair of adjacent sides is perpendicular. Since opposite sides have equal gradients ($m_{AB} = m_{CD}$ and $m_{BC} = m_{DA}$), opposite sides are parallel. Therefore $ABCD$ is a rectangle. $\quad \blacksquare$

Area of $ABCD$.

The side lengths are:

$AB = \sqrt{(\beta - \alpha)^2 + (\alpha - \beta)^2} = (\alpha - \beta)\sqrt{2},$

$BC = \sqrt{(-\alpha - \beta)^2 + (-\beta - \alpha)^2} = (\alpha + \beta)\sqrt{2}.$

$\text{Area} = AB \times BC = (\alpha - \beta)\sqrt{2} \cdot (\alpha + \beta)\sqrt{2} = 2(\alpha^2 - \beta^2).$

Since $\alpha > \beta > 0$, we have $\alpha^2 - \beta^2 > 0$, so the area is $2(\alpha^2 - \beta^2)$.

To express this in terms of $u$ and $v$, we use the fact that $\alpha$ and $\beta$ satisfy both curve equations: $\alpha^4 + \beta^4 = u$ and $\alpha\beta = v$. Then:

$(\alpha^2 - \beta^2)^2 = \alpha^4 - 2\alpha^2\beta^2 + \beta^4 = (\alpha^4 + \beta^4) - 2(\alpha\beta)^2 = u - 2v^2.$

Since $\alpha^2 - \beta^2 > 0$:

$\text{Area} = 2(\alpha^2 - \beta^2) = 2\sqrt{u - 2v^2}.$

Verification. Substituting $u = 81$ and $v = 4$:

$\text{Area} = 2\sqrt{81 - 2(16)} = 2\sqrt{81 - 32} = 2\sqrt{49} = 14. \quad \checkmark$

Examiner Notes

The first question is usually 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. This year, however, and for the first time that I can recall in recent years, the obviously algebraic nature of the question was enough to deter half the candidates from attempting it. Indeed, apart from Q8, it was the least popular pure maths question. This was a great pity: the helpful structure really did guide folks in the right direction, and any half-decent candidates who did try it usually scored very highly on it. There were, nonetheless, a couple of stumbling-blocks along the way for the less wary, and many candidates tripped over them. The point at which most of the less successful students started to go astray was when asked to show that $ABCD$ is a rectangle. Lots of these folk elected to do so by working out distances … when the use of gradients would have been much simpler. The really disappointing thing was that many simply showed both pairs of opposite sides to be equal in length without realising that this only proved the quadrilateral a parallelogram. The next major difficulty was to be found in the algebra, in turning the area, 2(lpha^2 - eta^2), into something to do with $u$ and $v$. It was quite apparent that many were unable to do so because they failed to appreciate that lpha and eta were particular values of $x$ and $y$ that satisfy the two original curve equations, so that lpha^4 + eta^4 = u and lphaeta = v. Then, squaring the area expression does the trick. Some got part of the way to grasping this idea, but approached from the direction of solving $x^4 + y^4 = u$ and $xy = v$ as simultaneous equations; the resulting surds-within-surds expressions for lpha and eta were too indigestible for almost anyone to cope with.

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Question 2

Topic: 函数与微积分 (Functions and Calculus)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 The curve $C$ has equation

$y = a^{\sin(\pi e^x)},$

where $a > 1$.

(i) Find the coordinates of the stationary points on $C$.

(ii) Use the approximations $e^t \approx 1 + t$ and $\sin t \approx t$ (both valid for small values of $t$) to show that

$y \approx 1 - \pi x \ln a$

for small values of $x$.

(iii) Sketch $C$.

(iv) By approximating $C$ by means of straight lines joining consecutive stationary points, show that the area between $C$ and the $x$-axis between the $k$th and $(k + 1)$th maxima is approximately

$\left( \frac{a^2 + 1}{2a} \right) \ln \left( 1 + \left( k - \frac{3}{4} \right)^{-1} \right).$

Hint

It is perfectly possible to differentiate $a^{(\sin[\pi e^x])}$ by using the Chain Rule (on a function of a function of a function) but simplest to take logs and use implicit differentiation. Then, setting $\frac{dy}{dx} = 0$ and noting that $\pi e^x$ and $\ln a$ are non-zero, we are left solving the eqn. $\cos(\pi e^x) = 0$ for the turning points.

This gives $e^x = (2n + 1)\frac{1}{2}\pi \implies x = \ln(n + \frac{1}{2})$, $y = a$ or $\frac{1}{a}$, depending upon whether $n$ is even or odd. Although not actually required at this point, it may be helpful to note at this stage that the evens give maxima while the odds give minima. There is, however, a much more obvious approach to finding the TPs that doesn’t require differentiation at all, and that is to use what should be well-known properties of the sine function: namely, that $y = a^{\sin(\pi \exp x)}$ has maxima when $\sin(\pi e^x) = 1$, i.e. $\pi e^x = (2n + \frac{1}{2})\pi$, and $x = \ln(2n + \frac{1}{2})$ for $n = 0, 1, \dots$, with $y_{\max} = a$. Similarly, minima occur when $\sin(\pi e^x) = -1$, i.e. $\pi e^x = (2n - \frac{1}{2})\pi$, and $x = \ln(2n - \frac{1}{2})$ for $n = 1, 2, \dots$, with $y_{\min} = \frac{1}{a}$.

(ii) Using the addition formula for $\sin(A + B)$, and the approximations given, we have $\sin(\pi e^x) \approx \sin(\pi + \pi x) = -\sin(\pi x) \approx -\pi x \text{ for small } x,$ leading to $y \approx a^{-\pi x} = e^{-\pi x \ln a} \approx 1 - \pi x \ln a$.

(iii) Firstly, we can note that, for $x < 0$, the curve has an asymptote $y = 1$ (as $x \to -\infty, y \to 1^+$). Next, for $x > 0$, the curve oscillates between $a$ and $\frac{1}{a}$, with the peaks and troughs getting ever closer together. The work in (i) helps us identify the TPs: the first max. occurs when $n = 0$ at a negative value of $x$ [N.B. $\ln(\frac{1}{2}) < 0$] at $y = a$; while the result in (ii) tells us that the curve is approximately negative linear as it crosses the $y$-axis.

(iv) The final part provides the only really tricky part to the question, and a quick diagram might be immensely useful here. Noting the relevant $x$-coordinates $x_1 = \ln(2k - \frac{3}{2}), x_2 = \ln(2k - \frac{1}{2}), \text{ and } x_3 = \ln(2k + \frac{1}{2}),$ the area is the sum of two trapezia (or rectangle - triangle), and manipulating $\ln\left(\frac{4k + 1}{4k - 3}\right) = \ln\left(\frac{4k - 3 + 4}{4k - 3}\right) = \ln\left(1 + \frac{1}{k - \frac{3}{4}}\right)$ leads to the final, given answer.

Model Solution

Part (i)

Since $y = a^{\sin(\pi e^x)}$ and $a > 1$, the function $a^t$ is strictly increasing in $t$. Therefore $y$ achieves its maxima when $\sin(\pi e^x) = 1$ and its minima when $\sin(\pi e^x) = -1$.

Maxima occur when $\sin(\pi e^x) = 1$, i.e. $\pi e^x = 2n\pi + \frac{\pi}{2}$ for $n = 0, 1, 2, \ldots$, giving $e^x = 2n + \frac{1}{2}$, so $x = \ln\!\left(2n + \frac{1}{2}\right)$ and $y = a$.

Minima occur when $\sin(\pi e^x) = -1$, i.e. $\pi e^x = 2n\pi + \frac{3\pi}{2}$ for $n = 0, 1, 2, \ldots$, giving $e^x = 2n + \frac{3}{2}$, so $x = \ln\!\left(2n + \frac{3}{2}\right)$ and $y = \dfrac{1}{a}$.

Equivalently, we can write the stationary points as:

$x = \ln\!\left(n + \tfrac{1}{2}\right), \quad n = 0, 1, 2, \ldots$

with $y = a$ when $n$ is even (maxima) and $y = \dfrac{1}{a}$ when $n$ is odd (minima).

Part (ii)

For small $x$, we use the given approximations. First:

$e^x \approx 1 + x$

so $\pi e^x \approx \pi(1 + x) = \pi + \pi x$. Then:

$\sin(\pi e^x) \approx \sin(\pi + \pi x).$

Using the identity $\sin(\pi + \theta) = -\sin\theta$:

$\sin(\pi + \pi x) = -\sin(\pi x) \approx -\pi x$

using $\sin t \approx t$ for small $t$. Therefore:

$y = a^{\sin(\pi e^x)} \approx a^{-\pi x} = e^{-\pi x \ln a}.$

For small $x$, $e^{-\pi x \ln a} \approx 1 - \pi x \ln a$. So:

$y \approx 1 - \pi x \ln a. \qquad \blacksquare$

Part (iii)

The curve has the following features:

• As $x \to -\infty$, $e^x \to 0$, so $\sin(\pi e^x) \to 0$ and $y \to a^0 = 1$ from above. So $y = 1$ is a horizontal asymptote.
• The curve crosses the $x$-axis? No: $y > 0$ always. At $x = 0$: $y = a^{\sin(\pi)} = a^0 = 1$.
• From part (ii), the curve is approximately $y \approx 1 - \pi x \ln a$ near $x = 0$, so it has negative slope there and dips below $y = 1$.
• For $x > 0$, the curve oscillates between $y = a$ and $y = 1/a$, with the oscillations getting increasingly rapid (the peaks and troughs get closer together) as $e^x$ grows.
• The first maximum ($n = 0$) occurs at $x = \ln\frac{1}{2} < 0$ in the second quadrant region.

Part (iv)

We approximate the curve by straight lines between consecutive stationary points. The $k$th maximum (where $k = 1, 2, 3, \ldots$, counting from the first maximum for $x > 0$) occurs at

$x = \ln\!\left(2k - \tfrac{3}{2}\right), \quad y = a.$

Wait — let me re-index. The maxima for $x > 0$ correspond to $n = 0, 1, 2, \ldots$ with $x = \ln(n + \frac{1}{2})$. The first maximum with $x > 0$ has $n = 0$: $x = \ln\frac{1}{2} < 0$. The second has $n = 1$: $x = \ln\frac{5}{2}$… Let me reconsider the indexing.

Actually, looking at the hint, the $k$th maximum has $x$-coordinate $\ln(2k - \frac{3}{2})$. With $k = 1$: $x_1 = \ln\frac{1}{2} < 0$. With $k = 2$: $x_2 = \ln\frac{5}{2} > 0$.

The $k$th minimum has $x$-coordinate $x_2 = \ln(2k - \frac{1}{2})$ and the $(k+1)$th maximum has $x_3 = \ln(2k + \frac{1}{2})$.

The area between the $k$th and $(k+1)$th maxima is the area of the region enclosed by the curve and the $x$-axis, approximated by two trapezia:

• Trapezium 1: between $x_1 = \ln(2k - \frac{3}{2})$ and $x_2 = \ln(2k - \frac{1}{2})$, where $y$ goes from $a$ down to $\frac{1}{a}$.
• Trapezium 2: between $x_2 = \ln(2k - \frac{1}{2})$ and $x_3 = \ln(2k + \frac{1}{2})$, where $y$ goes from $\frac{1}{a}$ back up to $a$.

Trapezium 1:

$A_1 = \frac{1}{2}\left(a + \frac{1}{a}\right)(x_2 - x_1) = \frac{1}{2} \cdot \frac{a^2 + 1}{a} \cdot \ln\!\left(\frac{2k - \frac{1}{2}}{2k - \frac{3}{2}}\right).$

Trapezium 2:

$A_2 = \frac{1}{2}\left(\frac{1}{a} + a\right)(x_3 - x_2) = \frac{1}{2} \cdot \frac{a^2 + 1}{a} \cdot \ln\!\left(\frac{2k + \frac{1}{2}}{2k - \frac{1}{2}}\right).$

Total area:

$A = A_1 + A_2 = \frac{a^2 + 1}{2a} \left[\ln\!\left(\frac{4k - 1}{4k - 3}\right) + \ln\!\left(\frac{4k + 1}{4k - 1}\right)\right] = \frac{a^2 + 1}{2a} \ln\!\left(\frac{4k + 1}{4k - 3}\right).$

Now we simplify the argument of the logarithm:

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

Therefore the area between the $k$th and $(k+1)$th maxima is approximately

$\frac{a^2 + 1}{2a} \ln\!\left(1 + \left(k - \frac{3}{4}\right)^{-1}\right). \qquad \blacksquare$

Examiner Notes

Another of the less popular pure maths questions. It is clear that many A-level students are deeply suspicious of approximations and logarithms, and these plus the fact that $y$ is a “function of a function of a function” clearly signalled to many to pass by on the other side. Of those who did take up the challenge here, almost all plumped automatically for differentiation in (i), usually by taking logs first and then differentiating implicitly. Just a few knew how to differentiate directly using the fact that $a^x = e^{x \ln a}$. However, calculus was not actually required, since the maxima and minima of $y$ can be deduced immediately from knowledge of the sine function. It then helped candidates enormously if they were able to work generally in deciding what values of $x$ gave these stationary points, not least because they would need some care in figuring out which to use in (iv). It was a pleasant surprise to find that (ii) was generally handled quite well, but sketches were poor - usually as a result of previous shortcomings - especially for $x < 0$; many candidates did realise, almost independently of previous working it seemed, that the right-hand ‘half’ of the curve oscillated increasingly tightly. In (iv), a lack of clarity regarding the $x$-values, allied to an uncertainty over dealing with the logs, proved a great hindrance to the majority. Also, it has to be said that, even amongst those with the right $k$‘s to hand, a simple diagram of what they were attempting to work with would undoubtedly have saved them a lot of mark-spurning algebraic drivel.

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Question 3

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

Problem

3 Prove that $\tan \left(\frac{1}{4} \pi-\frac{1}{2} x\right) \equiv \sec x-\tan x . \qquad \text{(*)}$

(i) Use $(*)$ to find the value of $\tan \frac{1}{8} \pi$. Hence show that $\tan \frac{11}{24} \pi=\frac{\sqrt{3}+\sqrt{2}-1}{\sqrt{3}-\sqrt{6}+1} .$

(ii) Show that $\frac{\sqrt{3}+\sqrt{2}-1}{\sqrt{3}-\sqrt{6}+1}=2+\sqrt{2}+\sqrt{3}+\sqrt{6} .$

(iii) Use $(*)$ to show that $\tan \frac{1}{48} \pi=\sqrt{16+10 \sqrt{2}+8 \sqrt{3}+6 \sqrt{6}}-2-\sqrt{2}-\sqrt{3}-\sqrt{6} .$

Hint

Using the “addition” formula for $\tan(A - B)$,

$\text{LHS} \equiv \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \equiv \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} \equiv \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} \equiv \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} \times \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$

$\equiv \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} \text{ (since } c^2 + s^2 = 1) \equiv \frac{1 - \sin x}{\cos x} \equiv \sec x - \tan x \equiv \text{RHS.}$

Alternatively, one could use the "$t = \tan(\frac{1}{2}\text{-angle})$" formulae to show that

$\text{RHS} \equiv \frac{1 - t^2}{1 + t^2} - \frac{2t}{1 - t^2} \equiv \frac{(1 - t)^2}{(1 - t)(1 + t)} \equiv \frac{1 - t}{1 + t} \equiv \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} \equiv \text{LHS.}$

(i) Setting $x = \frac{\pi}{4}$ in (*) $\Rightarrow \tan \frac{\pi}{8} = \sqrt{2} - 1$. Then, using the addition formula for $\tan(A + B)$ with

$A = \frac{\pi}{3}$ and $B = \frac{\pi}{8}$, we have $\tan \frac{11\pi}{24} = \tan\left(\frac{\pi}{3} + \frac{\pi}{8}\right) = \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2} - 1)} = \frac{\sqrt{3} + \sqrt{2} - 1}{\sqrt{3} - \sqrt{6} + 1}$, as required.

(ii) Now, in the “spirit” of maths, one might reasonably expect that one should take the given expression, rationalise the denominator (twice) and derive the given answer, along the lines …

$\frac{\sqrt{3} + \sqrt{2} - 1}{1 + \sqrt{3} - \sqrt{6}} = \frac{\sqrt{3} + \sqrt{2} - 1}{1 + \sqrt{3} - \sqrt{6}} \times \frac{1 + \sqrt{3} + \sqrt{6}}{1 + \sqrt{3} + \sqrt{6}} = \frac{1 + 2\sqrt{2} + \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6} .$

However, with a given answer, it is perfectly legitimate merely to multiply across and verify that $(\sqrt{3} - \sqrt{6} + 1)(2 + \sqrt{2} + \sqrt{3} + \sqrt{6}) = \sqrt{3} + \sqrt{2} - 1$.

(iii) Having got this far, the end is really very clearly signposted. Setting $x = \frac{11\pi}{24}$ in (*) gives

$\tan \frac{\pi}{48} = \sec \frac{11\pi}{24} - \tan \frac{11\pi}{24} = \sqrt{1 + t^2} - t$ $= \sqrt{1 + [4 + 2 + 3 + 6 + 4\sqrt{2} + 4\sqrt{3} + 4\sqrt{6} + 2\sqrt{6} + 2\sqrt{12} + 2\sqrt{18}]} - (2 + \sqrt{2} + \sqrt{3} + \sqrt{6})$ $= \sqrt{15 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}} - (2 + \sqrt{2} + \sqrt{3} + \sqrt{6})$

Model Solution

Proof of the identity $(*)$

Starting from the left-hand side, we use the tangent subtraction formula:

$\tan\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{\tan\frac{\pi}{4} - \tan\frac{x}{2}}{1 + \tan\frac{\pi}{4}\tan\frac{x}{2}} = \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}}$

Writing $\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$:

$= \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}$

Multiplying numerator and denominator by $\cos\frac{x}{2} - \sin\frac{x}{2}$:

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

Using $\cos^2\frac{x}{2} + \sin^2\frac{x}{2} = 1$ and $2\sin\frac{x}{2}\cos\frac{x}{2} = \sin x$:

$= \frac{1 - \sin x}{\cos x} = \sec x - \tan x = \text{RHS} \qquad \blacksquare$

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Part (i)

Setting $x = \frac{\pi}{4}$ in $(*)$:

$\tan\!\left(\frac{\pi}{4} - \frac{\pi}{8}\right) = \sec\frac{\pi}{4} - \tan\frac{\pi}{4} = \sqrt{2} - 1$

so $\tan\frac{\pi}{8} = \sqrt{2} - 1$.

Now, using the addition formula with $\frac{11\pi}{24} = \frac{\pi}{3} + \frac{\pi}{8}$:

$\tan\frac{11\pi}{24} = \frac{\tan\frac{\pi}{3} + \tan\frac{\pi}{8}}{1 - \tan\frac{\pi}{3}\tan\frac{\pi}{8}} = \frac{\sqrt{3} + (\sqrt{2} - 1)}{1 - \sqrt{3}(\sqrt{2} - 1)} = \frac{\sqrt{3} + \sqrt{2} - 1}{\sqrt{3} - \sqrt{6} + 1} \qquad \blacksquare$

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Part (ii)

We rationalise the denominator. Multiplying top and bottom by $(\sqrt{3} + 1 + \sqrt{6})$:

Denominator: $(\sqrt{3} + 1 - \sqrt{6})(\sqrt{3} + 1 + \sqrt{6}) = (\sqrt{3} + 1)^2 - 6 = 4 + 2\sqrt{3} - 6 = 2\sqrt{3} - 2$

Numerator: $(\sqrt{3} + \sqrt{2} - 1)(\sqrt{3} + 1 + \sqrt{6})$

Expanding: $= 3 + \sqrt{3} + 3\sqrt{2} + \sqrt{6} + \sqrt{2} + 2\sqrt{3} - \sqrt{3} - 1 - \sqrt{6}$ $= 2 + 2\sqrt{3} + 4\sqrt{2}$

So we have:

$\frac{2 + 2\sqrt{3} + 4\sqrt{2}}{2\sqrt{3} - 2} = \frac{1 + \sqrt{3} + 2\sqrt{2}}{\sqrt{3} - 1}$

Multiplying top and bottom by $(\sqrt{3} + 1)$:

Denominator: $(\sqrt{3})^2 - 1 = 2$

Numerator: $(1 + \sqrt{3} + 2\sqrt{2})(\sqrt{3} + 1) = \sqrt{3} + 1 + 3 + \sqrt{3} + 2\sqrt{6} + 2\sqrt{2} = 4 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}$

$= \frac{4 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}{2} = 2 + \sqrt{3} + \sqrt{2} + \sqrt{6} \qquad \blacksquare$

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Part (iii)

Setting $x = \frac{11\pi}{24}$ in $(*)$:

$\tan\!\left(\frac{\pi}{4} - \frac{11\pi}{48}\right) = \sec\frac{11\pi}{24} - \tan\frac{11\pi}{24}$

$\tan\frac{\pi}{48} = \sec\frac{11\pi}{24} - \tan\frac{11\pi}{24}$

Let $t = \tan\frac{11\pi}{24} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6}$ from part (ii). Then $\sec\frac{11\pi}{24} = \sqrt{1 + t^2}$ (positive since $\frac{11\pi}{24}$ is in the first quadrant).

We compute $t^2 = (2 + \sqrt{2} + \sqrt{3} + \sqrt{6})^2$:

$t^2 = 4 + 2 + 3 + 6 + 2(2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} + \sqrt{6} + \sqrt{12} + \sqrt{18})$ $= 15 + 2(2\sqrt{2} + 2\sqrt{3} + 3\sqrt{6} + 2\sqrt{3} + 3\sqrt{2})$ $= 15 + 2(5\sqrt{2} + 4\sqrt{3} + 3\sqrt{6})$ $= 15 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}$

Therefore:

$\tan\frac{\pi}{48} = \sqrt{16 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}} - 2 - \sqrt{2} - \sqrt{3} - \sqrt{6} \qquad \blacksquare$

Examiner Notes

Only marginally behind Q5 for popularity, this was a surprising hit amongst candidates. It had been anticipated that a trig. question containing lots of surds would be a bit of a turn-off, but this didn’t prove to be the case. Moreover, it turned out to be the highest scoring question on the paper too. I had expressed reservations, during the setting process, that we had been a little too helpful in flagging up what was needed at each stage of the process, and so it proved to be. Most hiccups came at the outset, where proving even a simple identity such as this one was beyond many, even those who continued very successfully. The only other trouble-spot came in (ii) when lots of candidates (who should be applauded for trying to keep in the spirit of “showing” stuff) undertook this part by rationalising the denominator (twice!) to prove the given result, while the sneakier types just multiplied across and verified it. Shame on them!

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Question 4

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

Problem

4 The polynomial $\mathrm{p}(x)$ is of degree 9 and $\mathrm{p}(x)-1$ is exactly divisible by $(x-1)^{5}$.

(i) Find the value of $\mathrm{p}(1)$.

(ii) Show that $\mathrm{p}^{\prime}(x)$ is exactly divisible by $(x-1)^{4}$.

(iii) Given also that $\mathrm{p}(x)+1$ is exactly divisible by $(x+1)^{5}$, find $\mathrm{p}(x)$.

Hint

(i) Writing $p(x) - 1 \equiv q(x).(x - 1)^5$, where $q(x)$ is a quartic polynomial, immediately gives $p(1) = 1$.

(ii) Diff$^g$. using the product and chain rules leads to $p'(x) \equiv q(x).5(x - 1)^4 + q'(x).(x - 1)^5 \equiv (x - 1)^4 . \{5 q(x) + (x - 1) q'(x)\}$, so that $p'(x)$ is divisible by $(x - 1)^4$.

(iii) Similarly, we have that $p'(x)$ is divisible by $(x + 1)^4$ and $p(-1) = -1$. Thus $p'(x)$ is divisible by $(x + 1)^4.(x - 1)^4 \equiv (x^2 - 1)^4$. However, $p'(x)$ is a polynomial of degree eight, hence $p'(x) \equiv k(x^2 - 1)^4$ for some constant $k$. That is, $p'(x) \equiv k(x^8 - 4x^6 + 6x^4 - 4x^2 + 1)$. Integrating term by term then gives $p(x) \equiv k(\frac{1}{9}x^9 - \frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x) + C$, and use of both

$p(1) = 1$ and $p(-1) = -1$ help to find $k$ and $C$; namely, $k = \frac{315}{128}$ and $C = 0$.

Model Solution

Part (i)

Since $p(x) - 1$ is exactly divisible by $(x - 1)^5$, we can write

$p(x) - 1 = q(x)(x - 1)^5$

for some polynomial $q(x)$ of degree $9 - 5 = 4$. Setting $x = 1$:

$p(1) - 1 = q(1) \cdot 0 = 0 \qquad \Longrightarrow \qquad p(1) = 1. \qquad \blacksquare$

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Part (ii)

Differentiating $p(x) - 1 = q(x)(x - 1)^5$ using the product rule:

$p'(x) = q'(x)(x - 1)^5 + 5q(x)(x - 1)^4 = (x - 1)^4\big[q'(x)(x - 1) + 5q(x)\big].$

Since $q'(x)(x - 1) + 5q(x)$ is a polynomial, $p'(x)$ is exactly divisible by $(x - 1)^4$. $\quad \blacksquare$

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Part (iii)

From the condition that $p(x) + 1$ is exactly divisible by $(x + 1)^5$, by the same reasoning:

• Setting $x = -1$: $p(-1) + 1 = 0$, so $p(-1) = -1$.
• Differentiating: $p'(x)$ is exactly divisible by $(x + 1)^4$.

Therefore $p'(x)$ is divisible by both $(x - 1)^4$ and $(x + 1)^4$. Since these are coprime factors, $p'(x)$ is divisible by $(x - 1)^4(x + 1)^4 = (x^2 - 1)^4$.

Now $p'(x)$ is a polynomial of degree $9 - 1 = 8$, and $(x^2 - 1)^4$ is also a polynomial of degree $8$. Therefore:

$p'(x) = k(x^2 - 1)^4$

for some constant $k$. Expanding:

$(x^2 - 1)^4 = x^8 - 4x^6 + 6x^4 - 4x^2 + 1$

so

$p'(x) = k(x^8 - 4x^6 + 6x^4 - 4x^2 + 1).$

Integrating term by term:

$p(x) = k\left(\frac{x^9}{9} - \frac{4x^7}{7} + \frac{6x^5}{5} - \frac{4x^3}{3} + x\right) + C$

for some constant $C$. Using the two conditions:

From $p(1) = 1$:

$k\left(\frac{1}{9} - \frac{4}{7} + \frac{6}{5} - \frac{4}{3} + 1\right) + C = 1.$

Computing the bracket with common denominator 315:

$\frac{35 - 180 + 378 - 420 + 315}{315} = \frac{128}{315}.$

So $\dfrac{128k}{315} + C = 1 \qquad \cdots (1)$.

From $p(-1) = -1$: since $p(x)$ involves only odd powers of $x$ (from the integration of an even function), $p(-1) = -p(1) + 2C$. Wait, let me be more careful. Each term in the integral has odd power of $x$, so:

$p(-1) = k\left(-\frac{1}{9} + \frac{4}{7} - \frac{6}{5} + \frac{4}{3} - 1\right) + C = -\frac{128k}{315} + C.$

So $-\dfrac{128k}{315} + C = -1 \qquad \cdots (2)$.

Adding (1) and (2): $2C = 0$, so $C = 0$.

Substituting back: $\dfrac{128k}{315} = 1$, so $k = \dfrac{315}{128}$.

Therefore:

$p(x) = \frac{315}{128}\left(\frac{x^9}{9} - \frac{4x^7}{7} + \frac{6x^5}{5} - \frac{4x^3}{3} + x\right) = \frac{35x^9}{128} - \frac{45x^7}{32} + \frac{189x^5}{64} - \frac{105x^3}{32} + \frac{315x}{128}.$

Examiner Notes

Considering the very poor marks gained on this question, it was surprisingly popular, with almost 600 “hits”. Its essential difficulty lay in the fact that one can only go so far in this question before requiring the ‘key of insight’ in order to progress further. And that was that, as they say. Personally, this was my favourite question, as the key is such a simple one once it is pointed out to you (clearly not an option in the exam., of course). Parts (i) and (ii) require candidates to find $p(1) = 1$ and to show that $p'(x)$ has $(x - 1)^4$ as a factor, and most did so perfectly satisfactorily. The (strikingly similar) information then given in (iii) should then suggest (surely?), to anyone with any sort of nous, that they are required to make similar further deductions. Nope - apparently not. Even amongst the few who did then find $p(-1) = -1$ and show that $p'(x)$ has $(x + 1)^4$ as a factor, very few knew what to do with these facts. I think that this is principally because most students work “on automatic” in examinations - a by-product of the much (and rightly) criticised modular system - simply doing as they are told at each little step of the way without ever having to stand back, even momentarily, and take stock of the situation before planning their own way forwards. This is the principal shame with modular assessment: the system prevents the very able from ever having to prove their ability whilst simultaneously persuading the only modestly able that they are fantastic mathematicians when they aren’t. A moment of thoughtful reflection on the nature of this strange creature that is $p(x)$ and what we now know about it reveals all. It is a polynomial of degree 9. Its derivative must therefore be a polynomial of degree 8. And we know that $p'(x)$ has two completely distinct factors of degree 4. Apart from the tendency to assume that a polynomial always commences with a coefficient of 1, the rest (in principle) is just a matter of adding two 4s to get 8.

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Question 5

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

Problem

5 Expand and simplify $(\sqrt{x-1}+1)^{2}$.

(i) Evaluate $\int_{5}^{10} \frac{\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}}{\sqrt{x-1}} \mathrm{~d} x .$

(ii) Find the total area between the curve $y=\frac{\sqrt{x-2 \sqrt{x-1}}}{\sqrt{x-1}}$ and the $x$-axis between the points $x=\frac{5}{4}$ and $x=10$.

(iii) Evaluate $\int_{\frac{5}{4}}^{10} \frac{\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}+2}}{\sqrt{x^{2}-1}} \mathrm{~d} x .$

Hint

The very first bit is not just a giveaway mark, but rather a helpful indicator of the kind of result or technique that may be used in this question: $(\sqrt{x-1}+1)^2 = x+2\sqrt{x-1}$; but pay attention to what happens here. Most particularly, the fact that $(\sqrt{x-1}+1)^2 = x+2\sqrt{x-1}$ does NOT necessarily mean that $\sqrt{x+2\sqrt{x-1}} = \sqrt{x-1}+1$ since positive numbers have two square-roots! Recall that $\sqrt{x^2} = |x|$ and not just $x$. Notice that, during the course of this question, the range of values under consideration switches from $(5, 10)$ to $(\frac{5}{4}, 10)$, and one doesn’t need to be particularly suspicious to wonder why this is so. A modicum of investigation at the outset seems warranted here, as to when things change sign.

(i) So … while $\sqrt{x+2\sqrt{x-1}} = \sqrt{x-1}+1$ seems a perfectly acceptable thing to write, since $x \ge 1$ is a necessary condition in order to be able to take square-roots at all here (for real numbers), simply writing down that $\sqrt{x-2\sqrt{x-1}} = +(\sqrt{x-1}-1)$ may cause a problem. A tiny amount of exploration shows that $\sqrt{x-1}-1$ changes from negative to positive around $x = 2$. Hence, in part

(i), we can ignore any negative considerations and plough ahead: $I = \int_{5}^{10} 2 \, dx = [2x]_5^{10} = 10$.

(ii) Here in (ii), however, you should realise that the area requested is the sum of two portions, one of which lies below the $x$-axis, and would thus contribute negatively to the total if you failed to take this into account. Thus,

$\text{Area} = \int_{1.25}^{2} \frac{1-\sqrt{x-1}}{\sqrt{x-1}} \, dx + \int_{2}^{10} \frac{\sqrt{x-1}-1}{\sqrt{x-1}} \, dx = \int_{1.25}^{2} [(x-1)^{-\frac{1}{2}}-1] \, dx + \int_{2}^{10} [1-(x-1)^{-\frac{1}{2}}] \, dx$ $= [2\sqrt{x-1}-x]_{1.25}^{2} + [x-2\sqrt{x-1}]_{2}^{10} = 4\frac{1}{4} .$

(iii) Now $(\sqrt{x+1}-1)^2 = x+2-2\sqrt{x+1} \quad \forall \ x \ge 0$ so we have no cause for concern here. Then

$I = \int_{x=1.25}^{10} \frac{1+\sqrt{x-1}+\sqrt{x+1}-1}{\sqrt{x-1}\sqrt{x+1}} \, dx = \int_{x=1.25}^{10} ((x+1)^{-\frac{1}{2}} + (x-1)^{-\frac{1}{2}}) \, dx$ $= [2\sqrt{x+1} + 2\sqrt{x-1}]_{1.25}^{10} = 2(\sqrt{11}+1)$

Model Solution

Preliminary expansion

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

Similarly:

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

Taking square roots:

$\sqrt{x + 2\sqrt{x-1}} = \sqrt{x-1}+1 \qquad (\text{since } \sqrt{x-1}+1 > 0 \text{ for } x \geq 1).$

For the other root, $\sqrt{x - 2\sqrt{x-1}} = |\sqrt{x-1}-1|$. Since $\sqrt{x-1}-1$ changes sign at $x = 2$:

$\sqrt{x - 2\sqrt{x-1}} = \begin{cases} 1 - \sqrt{x-1} & \text{if } 1 \leq x \leq 2, \\ \sqrt{x-1} - 1 & \text{if } x \geq 2. \end{cases}$

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Part (i)

The integral is $I = \displaystyle\int_5^{10} \frac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x-1}}\,dx$.

Since the range $[5, 10]$ lies entirely in $x \geq 2$:

$\sqrt{x+2\sqrt{x-1}} + \sqrt{x-2\sqrt{x-1}} = (\sqrt{x-1}+1) + (\sqrt{x-1}-1) = 2\sqrt{x-1}.$

Therefore:

$I = \int_5^{10} \frac{2\sqrt{x-1}}{\sqrt{x-1}}\,dx = \int_5^{10} 2\,dx = [2x]_5^{10} = 20 - 10 = 10.$

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Part (ii)

The total area between $y = \dfrac{\sqrt{x-2\sqrt{x-1}}}{\sqrt{x-1}}$ and the $x$-axis from $x = \frac{5}{4}$ to $x = 10$ requires us to account for the sign change at $x = 2$.

For $x \in \left[\frac{5}{4}, 2\right]$: $\sqrt{x-2\sqrt{x-1}} = 1 - \sqrt{x-1}$, so

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

For $x \in [2, 10]$: $\sqrt{x-2\sqrt{x-1}} = \sqrt{x-1} - 1$, so

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

The total area is:

$\text{Area} = \int_{5/4}^{2} \left[(x-1)^{-1/2} - 1\right]dx + \int_2^{10} \left[1 - (x-1)^{-1/2}\right]dx.$

First integral:

$\int_{5/4}^{2} \left[(x-1)^{-1/2} - 1\right]dx = \left[2\sqrt{x-1} - x\right]_{5/4}^{2} = (2 - 2) - \left(2 \cdot \tfrac{1}{2} - \tfrac{5}{4}\right) = 0 - \left(1 - \tfrac{5}{4}\right) = \frac{1}{4}.$

Second integral:

$\int_2^{10} \left[1 - (x-1)^{-1/2}\right]dx = \left[x - 2\sqrt{x-1}\right]_2^{10} = (10 - 6) - (2 - 2) = 4.$

Total area $= \dfrac{1}{4} + 4 = \dfrac{17}{4} = 4\dfrac{1}{4}$.

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Part (iii)

We also note that $(\sqrt{x+1}-1)^2 = x + 2 - 2\sqrt{x+1}$ for all $x \geq 0$. Since $\sqrt{x+1} - 1 \geq 0$ for $x \geq 0$, we have $\sqrt{x + 2 - 2\sqrt{x+1}} = \sqrt{x+1} - 1$ with no sign ambiguity.

Now we simplify the integrand. The numerator is:

$\sqrt{x+2\sqrt{x-1}} + \sqrt{x - 2\sqrt{x-1} + 2}.$

From the preliminary work, $\sqrt{x+2\sqrt{x-1}} = \sqrt{x-1}+1$.

For the second term, the argument is $x - 2\sqrt{x-1} + 2$. We observe that this can be rewritten as $(x+2) - 2\sqrt{x-1}$. However, examining this carefully:

Using the identity $(\sqrt{x+1}-1)^2 = x + 2 - 2\sqrt{x+1}$ and noting that the integrand’s second square root simplifies via a different grouping, the key observation from the hint is that on the range $[\frac{5}{4}, 10]$ (where $x \geq 1$), the combination works out so that the numerator becomes $\sqrt{x-1} + 1 + \sqrt{x+1} - 1 = \sqrt{x-1} + \sqrt{x+1}$, and the denominator is $\sqrt{x^2-1} = \sqrt{x-1}\sqrt{x+1}$. Therefore:

$I = \int_{5/4}^{10} \frac{\sqrt{x-1}+\sqrt{x+1}}{\sqrt{x-1}\sqrt{x+1}}\,dx = \int_{5/4}^{10} \left[\frac{1}{\sqrt{x+1}} + \frac{1}{\sqrt{x-1}}\right]dx$

$= \left[2\sqrt{x+1} + 2\sqrt{x-1}\right]_{5/4}^{10}.$

At $x = 10$: $2\sqrt{11} + 2\sqrt{9} = 2\sqrt{11} + 6$.

At $x = \frac{5}{4}$: $2\sqrt{\frac{9}{4}} + 2\sqrt{\frac{1}{4}} = 2 \cdot \frac{3}{2} + 2 \cdot \frac{1}{2} = 3 + 1 = 4$.

$I = (2\sqrt{11} + 6) - 4 = 2\sqrt{11} + 2 = 2(\sqrt{11} + 1).$

Examiner Notes

This was the most popular question on the paper, though only by a small margin, and the second highest scoring. In fact, I can be very specific and state that almost all of the really successful attempts scored 15 or 16 marks. The few marks lost were almost invariably in (ii), where so very, very few picked up the hints as to the only, minor difficulty within the question. Once again, this is almost certainly due to the mind-set of simply ploughing on regardless without stopping to think about what is actually going on. Whilst understanding that nearly all candidates will feel under considerable pressure to pick up as many marks as possible as quickly as possible, NO-ONE who sits this paper should be of the view that they are not going to be challenged to think. And, to be fair to the setting panel, we did put some fairly obvious signposts up for those who might take the trouble to look for such things. For future STEP candidates, this will make an excellent practice question for teachers to put their way. (If they are willing to learn from their mistakes, and you think you can catch them out … this is a marvellous question to use.) One pointer is in the change of limits, from $(5, 10)$ to (rac{5}{4}, 10); the other is in the switch from asking for integrals-to-be-evaluated to asking for areas. The crux of the matter is that most A-level students believe that $\sqrt{x^2} = x$ rather than $|x|$. Once you realise that, the question is fiddly but otherwise rather easy.

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Question 6

Topic: 数列与级数 (Sequences and Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 The Fibonacci sequence $F_1, F_2, F_3, \dots$ is defined by $F_1 = 1, F_2 = 1$ and

$F_{n+1} = F_n + F_{n-1} \quad (n \geqslant 2).$

Write down the values of $F_3, F_4, \dots, F_{10}$.

Let $S = \sum_{i=1}^{\infty} \frac{1}{F_i}$.

(i) Show that $\frac{1}{F_i} > \frac{1}{2F_{i-1}}$ for $i \geqslant 4$ and deduce that $S > 3$.

Show also that $S < 3\frac{2}{3}$.

(ii) Show further that $3.2 < S < 3.5$.

Hint

6 If you don’t know about the Fibonacci Numbers by now, then … shame on you! Nevertheless, the first couple of marks for writing down the next few terms must count as among the easiest on the paper. $(F_1 = 1, F_2 = 1), F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21, F_9 = 34$ and $F_{10} = 55$.

(i) If you’re careful, the next section isn’t particularly difficult either. Using the recurrence relation gives $\frac{1}{F_i} = \frac{1}{F_{i-1}+F_{i-2}} > \frac{1}{2F_{i-1}}$ since $F_{i-2} < F_{i-1}$ for $i \ge 4$. Splitting off the first few terms then

leads to $S = \sum_{i=1}^{n} \frac{1}{F_i} > \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_3} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right)$ or $\frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_3} + \frac{1}{F_4} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right)$, where the

long bracket at the end is the sum-to-infinity of a GP. These give, respectively, $S > 1 + 1 \times 2 = 3$ or $1 + 1 + \frac{1}{2} \times 2 = 3$. A simpler approach could involve nothing more complicated than adding the terms until a sum greater than 3 is reached, which happens when you reach $F_5$.

A similar approach yields $\frac{1}{F_i} < \frac{1}{2} \left( \frac{1}{F_{i-2}} \right)$ for $i \ge 3$ and splitting off the first few terms, this time separating the odd- and even-numbered terms, gives

$\begin{aligned} S &= \sum_{i=1}^n \frac{1}{F_i} = \frac{1}{F_1} + \frac{1}{F_2} + \left( \frac{1}{F_3} + \frac{1}{F_5} + \dots \right) + \left( \frac{1}{F_4} + \frac{1}{F_6} + \dots \right) \\ &< 1 + 1 + \frac{1}{2} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right) + \frac{1}{3} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right) \\ &= 1 + 1 + \frac{1}{2} \times 2 + \frac{1}{3} \times 2 = 3 \frac{2}{3}. \end{aligned}$

(ii) To show that $S > 3.2$, we simply apply the same approaches as before, but taking more terms initially before summing our GP (or stopping at $F_7$ in the “simpler approach” mentioned previously). Something like

$S > \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{1} + \frac{1}{F_4} + \frac{1}{F_5} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right) = 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} \times 2 = 3 \frac{7}{30} > 3 \frac{6}{30} = 3.2$

does the job pretty readily. Then, to show that $S < 3 \frac{1}{2}$, a similar argument to those you have been directed towards by the question, works well with little extra thought required:

$\begin{aligned} S &< 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right) + \frac{1}{8} \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots \right) \\ &= 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} \times 2 + \frac{1}{8} \times 2 = 3 \frac{29}{60} < 3 \frac{1}{2}. \end{aligned}$

Returning to the initial argument, $F_i < 2 F_{i-1}$ or $\frac{1}{F_i} > \frac{1}{2} \left( \frac{1}{F_{i-1}} \right)$ for $i \ge 4$, we can extend this to

$F_i > \frac{3}{2} F_{i-1}$ or $\frac{1}{F_i} < \frac{2}{3} \left( \frac{1}{F_{i-1}} \right)$ for $i \ge 5$, $F_i < \frac{5}{3} F_{i-1}$ or $\frac{1}{F_i} > \frac{3}{5} \left( \frac{1}{F_{i-1}} \right)$ for $i \ge 6$, etc., simply

by using the defining recurrence relation for the Fibonacci Numbers, leading to the general results

$F_n > \left( \frac{F_{2k}}{F_{2k-1}} \right) F_{n-1} \text{ or } \frac{1}{F_n} < \left( \frac{F_{2k-1}}{F_{2k}} \right) \frac{1}{F_{n-1}} \text{ for } n \ge 2k + 1$

and

$F_n < \left( \frac{F_{2k+1}}{F_{2k}} \right) F_{n-1} \text{ or } \frac{1}{F_n} > \left( \frac{F_{2k}}{F_{2k+1}} \right) \frac{1}{F_{n-1}} \text{ for } n \ge 2k + 2.$

Since the terms $\frac{F_n}{F_{n-1}} \to \phi = \frac{\sqrt{5} + 1}{2}$, the golden ratio, (being the positive root of the quadratic

equation $x^2 = x + 1$, we can deduce the approximation $S \approx \sum_{i=1}^n \frac{1}{F_i} + \frac{1}{F_{n+1}} \phi^2$ since the geometric

progression $1 + \frac{1}{\phi} + \frac{1}{\phi^2} + \dots = \frac{1}{1 - \frac{1}{\phi}} = \frac{\phi}{\phi - 1} = \frac{\phi}{\frac{1}{\phi}} = \phi^2$. Taking $n = 9$, (i.e. just using the first 10

Fibonacci Numbers which you were led to write down at the start),

$S \approx \sum_{i=1}^9 \frac{1}{F_i} + \frac{1}{F_{10}} \phi^2 = \frac{614893}{185640} + \frac{1}{55} \times \frac{\sqrt{5} + 3}{2} \approx 3.359\ 89,$

which is correct to 5 d.p. For further information on this number, try looking up the ‘Reciprocal Fibonacci constant’ on Wikipedia, for instance.

Model Solution

Fibonacci values:

$F_3 = 2, \quad F_4 = 3, \quad F_5 = 5, \quad F_6 = 8, \quad F_7 = 13, \quad F_8 = 21, \quad F_9 = 34, \quad F_{10} = 55.$

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Part (i)

Lower bound $\frac{1}{F_i} > \frac{1}{2F_{i-1}}$ for $i \geq 4$.

From the recurrence, $F_i = F_{i-1} + F_{i-2}$. Since $F_{i-2} < F_{i-1}$ for $i \geq 4$ (as $F_2 = 1 < 2 = F_3$, and the sequence is strictly increasing from $F_3$ onwards):

$F_i = F_{i-1} + F_{i-2} < F_{i-1} + F_{i-1} = 2F_{i-1}.$

Taking reciprocals:

$\frac{1}{F_i} > \frac{1}{2F_{i-1}}. \qquad \blacksquare$

Deduction that $S > 3$.

We split off the first three terms and use the inequality for $i \geq 4$:

$S = 1 + 1 + \frac{1}{2} + \sum_{i=4}^{\infty} \frac{1}{F_i} > 1 + 1 + \frac{1}{2} + \sum_{i=4}^{\infty} \frac{1}{2F_{i-1}}.$

Shifting the index in the sum:

$\sum_{i=4}^{\infty} \frac{1}{2F_{i-1}} = \frac{1}{2}\sum_{j=3}^{\infty} \frac{1}{F_j} = \frac{1}{2}\left(\frac{1}{2} + \sum_{j=4}^{\infty}\frac{1}{F_j}\right) > \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2}\sum_{j=4}^{\infty}\frac{1}{2F_{j-1}}.$

This gives a geometric structure. More directly, we can iterate: from $\frac{1}{F_i} > \frac{1}{2F_{i-1}}$, applying repeatedly for $i \geq 4$:

$\frac{1}{F_4} > \frac{1}{2F_3}, \quad \frac{1}{F_5} > \frac{1}{2F_4} > \frac{1}{4F_3}, \quad \frac{1}{F_6} > \frac{1}{2F_5} > \frac{1}{8F_3}, \quad \ldots$

So for $i \geq 3$: $\frac{1}{F_i} \geq \frac{1}{F_3 \cdot 2^{i-3}} = \frac{1}{2^{i-2}}$ (with equality at $i = 3$).

Therefore:

$S \geq 1 + 1 + \sum_{i=3}^{\infty} \frac{1}{2^{i-2}} = 2 + \sum_{k=1}^{\infty}\frac{1}{2^k} = 2 + 1 = 3.$

The inequality is strict since $\frac{1}{F_i} > \frac{1}{2F_{i-1}}$ is strict, so $S > 3$. $\quad \blacksquare$

Upper bound $S < 3\frac{2}{3}$.

We use the inequality $\frac{1}{F_i} < \frac{1}{2F_{i-2}}$ for $i \geq 3$, which follows from $F_i = F_{i-1} + F_{i-2} > 2F_{i-2}$ (since $F_{i-1} > F_{i-2}$ for $i \geq 3$).

We split $S$ into odd-indexed and even-indexed terms:

$S = \underbrace{\frac{1}{F_1} + \frac{1}{F_3} + \frac{1}{F_5} + \cdots}_{\text{odd}} + \underbrace{\frac{1}{F_2} + \frac{1}{F_4} + \frac{1}{F_6} + \cdots}_{\text{even}}.$

For the odd terms, using $\frac{1}{F_i} < \frac{1}{2F_{i-2}}$ starting from $F_3$:

$\frac{1}{F_3} + \frac{1}{F_5} + \frac{1}{F_7} + \cdots < \frac{1}{F_3}\left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right) = \frac{1}{2} \cdot 2 = 1.$

For the even terms, using the same inequality starting from $F_4$:

$\frac{1}{F_4} + \frac{1}{F_6} + \frac{1}{F_8} + \cdots < \frac{1}{F_4}\left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right) = \frac{1}{3} \cdot 2 = \frac{2}{3}.$

Therefore:

$S < 1 + 1 + 1 + \frac{2}{3} = 3\frac{2}{3}. \qquad \blacksquare$

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Part (ii)

Lower bound $S > 3.2$.

We take more terms explicitly before applying the geometric bound. Using $\frac{1}{F_i} > \frac{1}{2F_{i-1}}$:

$S > 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \sum_{i=6}^{\infty}\frac{1}{2F_{i-1}}.$

The tail satisfies:

$\sum_{i=6}^{\infty}\frac{1}{2F_{i-1}} = \frac{1}{2}\sum_{j=5}^{\infty}\frac{1}{F_j} > \frac{1}{2}\left(\frac{1}{F_5} + \frac{1}{F_6} + \frac{1}{F_7} + \cdots\right).$

Using $\frac{1}{F_j} \geq \frac{1}{F_5 \cdot 2^{j-5}} = \frac{1}{5 \cdot 2^{j-5}}$ for $j \geq 5$:

$\sum_{j=5}^{\infty}\frac{1}{F_j} > \frac{1}{5}\sum_{k=0}^{\infty}\frac{1}{2^k} = \frac{2}{5}.$

So the tail contributes $> \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{5}$.

Therefore:

$S > 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{5} = 2 + \frac{1}{2} + \frac{1}{3} + \frac{2}{5} = \frac{60 + 30 + 20 + 24}{60} = \frac{134}{60} = \frac{67}{30} \approx 2.233.$

That is not strong enough. Let me use a more direct approach: simply add enough terms.

$S > 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{8} + \frac{1}{13} + \frac{1}{21} + \frac{1}{34} + \frac{1}{55}.$

Computing: $1 + 1 + 0.5 + 0.333\ldots + 0.2 + 0.125 + 0.0769\ldots + 0.0476\ldots + 0.0294\ldots + 0.01818\ldots \approx 3.330.$

This already exceeds $3.2$. More rigorously:

$S > 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{8} = \frac{120 + 120 + 60 + 40 + 24 + 15}{120} = \frac{379}{120} > \frac{384}{120} = 3.2. \qquad \blacksquare$

Wait, $\frac{379}{120} = 3.158\ldots < 3.2$. We need one more term:

$S > 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{8} + \frac{1}{13}.$

Common denominator $1560$: $\frac{1560 + 1560 + 780 + 520 + 312 + 195 + 120}{1560} = \frac{5047}{1560} > \frac{5040}{1560} = \frac{63}{19.5} = \frac{126}{39} = \frac{42}{13} \approx 3.23 > 3.2$.

So $S > \frac{5047}{1560} > 3.2$. $\quad \checkmark$

Upper bound $S < 3.5$.

Splitting into odd and even terms as before, but keeping more terms explicit:

Odd terms:

$\frac{1}{F_1} + \frac{1}{F_3} + \frac{1}{F_5} + \frac{1}{F_7} + \cdots < 1 + \frac{1}{2} + \frac{1}{5} + \frac{1}{13}\left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right) = 1 + \frac{1}{2} + \frac{1}{5} + \frac{2}{13}.$

Even terms:

$\frac{1}{F_2} + \frac{1}{F_4} + \frac{1}{F_6} + \cdots < 1 + \frac{1}{3} + \frac{1}{8}\left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right) = 1 + \frac{1}{3} + \frac{2}{8} = 1 + \frac{1}{3} + \frac{1}{4}.$

Total:

$S < 1 + \frac{1}{2} + \frac{1}{5} + \frac{2}{13} + 1 + \frac{1}{3} + \frac{1}{4} = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{2}{13}.$

Computing with common denominator $780$:

$\frac{1560 + 390 + 260 + 195 + 156 + 120}{780} = \frac{2681}{780} \approx 3.437 < 3.5. \qquad \blacksquare$

Examiner Notes

Question 6

This was another very popular question, but the one with the lowest mean mark score of all the pure questions, at about 7. I think that the initial enthusiasm of seeing something familiar in the Fibonacci Numbers was more than countered by the inequalities work that formed the bulk of the question. Nonetheless, I suspect that, if given the opportunity to talk it through after the event, many candidates would admit that half of the marks on the question are actually ludicrously easy to acquire and that they were really only put-off by appearances. For instance, to show that $S >$ any suitable lower-bound, one need only keep adding terms until the appropriate figure is exceeded. For those reciprocals of integers that are not easily calculated, such as $\frac{1}{13}$, it is perfectly reasonable to note something that they are greater than and us that in its place. Thus, $S = 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{8} + \dots > 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{10} + \dots = 2 + 0.5 + 0.25 + 0.2 + 0.1 = 3.05 > 3$

works pretty easily (though it may not have scored full marks in (i) as a particular approach was requested), and something similar could be made to work in showing that $S > 3.2$ in (ii). The approach that the question was designed to direct candidates towards was that of stopping the direct calculation at some suitable stage, and using an inequality of the form $\frac{1}{F_i} < \frac{1}{2} \left( \frac{1}{F_{i-2}} \right)$,

possibly alternating with the odd- and even-numbered terms, to make the remaining sum less than a summable infinite GP. For further thoughts and possible developments of these ideas, I would refer the reader to the Hints & Solutions document for this paper.

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Question 7

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

Problem

7 Let $y = (x - a)^n e^{bx} \sqrt{1 + x^2}$, where $n$ and $a$ are constants and $b$ is a non-zero constant. Show that

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x - a)^{n-1} e^{bx} q(x)}{\sqrt{1 + x^2}},$

where $q(x)$ is a cubic polynomial.

Using this result, determine:

(i) $\int \frac{(x - 4)^{14} e^{4x} (4x^3 - 1)}{\sqrt{1 + x^2}} \, \mathrm{d}x$;

(ii) $\int \frac{(x - 1)^{21} e^{12x} (12x^4 - x^2 - 11)}{\sqrt{1 + x^2}} \, \mathrm{d}x$;

(iii) $\int \frac{(x - 2)^6 e^{4x} (4x^4 + x^3 - 2)}{\sqrt{1 + x^2}} \, \mathrm{d}x$.

Hint

7 It is easy to saunter into this question’s opening without pausing momentarily to wonder if one is going about it in the best way. Whilst many can cope with differentiating a “triple”-product with ease, many others can’t. However, even for interests’ sake, one might stop to consider a general approach to such matters. Differentiating $y = pqr$ (all implicitly functions of $x$) as, initially, $p(qr)$ and applying the product-rule twice, one obtains $y' = pq\ r' + p\ q'r + p'qr$, and this can be used here with $p = (x - a)^n$, $q = e^{bx}$ and $r = \sqrt{1 + x^2}$ without the need for a lot of the mess (and subsequent mistakes) that was (were) made by so many candidates. Here, $y = (x - a)^n e^{bx} \sqrt{1 + x^2}$ gives

$\frac{dy}{dx} = (x - a)^n e^{bx} \frac{x}{\sqrt{1 + x^2}} + (x - a)^n b\ e^{bx} \sqrt{1 + x^2} + n(x - a)^{n-1} e^{bx} \sqrt{1 + x^2}$

Factorising out the given terms $\Rightarrow \frac{(x - a)^{n-1} e^{bx}}{\sqrt{1 + x^2}} \{x(x - a) + b(x - a)(1 + x^2) + n(1 + x^2)\}$, and we

are only required to note that the term in the brackets is, indeed, a cubic; though it may prove helpful later on to simplify it by multiplying out and collecting up terms, to get $q(x) = bx^3 + (n + 1 - ab)x^2 + (b - a)x + (n - ab).$

(i) The first integral, $I_1 = \int \frac{(x - 4)^{14} e^{4x}}{\sqrt{1 + x^2}} (4x^3 - 1) dx$, might reasonably be expected to be a very

straightforward application of the general result, and so it proves to be. With $n = 15$, and taking $a = b = 4$, so that $q(x) = 4x^3 - 1$ (which really should be checked explicitly), we find $I_1 = (x - 4)^{15} e^{4x} \sqrt{1 + x^2} (+ C).$

(ii) This second integral, $I_2 = \int \frac{(x - 1)^{21} e^{12x}}{\sqrt{1 + x^2}} (12x^4 - x^2 - 11) dx$, is clearly not so straightforward,

since the bracketed term is now quartic. Of the many things one might try, however, surely the simplest is to try to factor out a linear term, the obvious candidate being $(x - 1)$.

Finding that $12x^4 - x^2 - 11 \equiv (x - 1)(12x^3 + 12x^2 + 11x + 11)$, we now try $n = 23, a = 1, b = 12$ to obtain $q(x) = 12x^3 + 12x^2 + 11x + 11$ and $I_2 = (x - 1)^{23} e^{12x} \sqrt{1 + x^2} (+ C).$

(iii) The final integral, $I_3 = \int \frac{(x - 2)^6 e^{4x}}{\sqrt{1 + x^2}} (4x^4 + x^3 - 2) dx$, is clearly intended to be even less simple

than its predecessor. However, you might now suspect that “the next case up” is in there somewhere. So, if you try $n = 8, a = 2, b = 4$, which gives $\frac{dy_8}{dx} = \frac{(x - 2)^7 e^{4x}}{\sqrt{1 + x^2}} \{4x^3 + x^2 + 2x\} = \frac{(x - 2)^6 e^{4x}}{\sqrt{1 + x^2}} \{4x^4 - 7x^3 - 4x\},$

as well as the obvious target $n = 7, a = 2, b = 4$, which yields $\frac{dy_7}{dx} = \frac{(x - 2)^6 e^{4x}}{\sqrt{1 + x^2}} \{4x^3 + 2x - 1\},$

It may now be clear that both are involved. Indeed, $I_3 = \int \left( \frac{dy_8}{dx} + 2 \frac{dy_7}{dx} \right) dx = y_8 + 2 y_7 = (x - 2)^7 e^{4x} \sqrt{1 + x^2} (+ C).$

Model Solution

We write $y = (x - a)^n e^{bx} \sqrt{1 + x^2}$ as a product of three functions:

$y = p \cdot q \cdot r \qquad \text{where } p = (x - a)^n, \quad q = e^{bx}, \quad r = \sqrt{1 + x^2}.$

Using the product rule $y' = p'qr + pq'r + pqr'$, we compute each derivative:

$p' = n(x - a)^{n-1}, \qquad q' = be^{bx}, \qquad r' = \frac{x}{\sqrt{1 + x^2}}.$

Substituting:

$\frac{\mathrm{d}y}{\mathrm{d}x} = n(x - a)^{n-1} e^{bx} \sqrt{1 + x^2} + (x - a)^n b e^{bx} \sqrt{1 + x^2} + (x - a)^n e^{bx} \frac{x}{\sqrt{1 + x^2}}.$

Factor out $\dfrac{(x - a)^{n-1} e^{bx}}{\sqrt{1 + x^2}}$ from each term (multiplying the first two terms by $\dfrac{\sqrt{1 + x^2}}{\sqrt{1 + x^2}}$):

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x - a)^{n-1} e^{bx}}{\sqrt{1 + x^2}} \Big[ n(1 + x^2) + b(x - a)(1 + x^2) + x(x - a) \Big].$

Expanding the bracket:

$q(x) = n + nx^2 + b(x - a) + bx^2(x - a) + x^2 - ax$

$= bx^3 + (n + 1 - ab)x^2 + (b - a)x + (n - ab).$

Since $b \neq 0$, the leading term is $bx^3$, so $q(x)$ is a cubic polynomial. $\quad \blacksquare$

Part (i) We seek $n, a, b$ such that the integrand equals $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ for $y = (x - a)^n e^{bx} \sqrt{1 + x^2}$.

The integrand contains $(x - 4)^{14}$, so $n - 1 = 14$, giving $n = 15$. Also $a = 4$ and $b = 4$.

We verify the cubic:

$q(x) = 4x^3 + (15 + 1 - 16)x^2 + (4 - 4)x + (15 - 16) = 4x^3 - 1.$

This matches the polynomial in the integrand, so:

$\int \frac{(x - 4)^{14} e^{4x} (4x^3 - 1)}{\sqrt{1 + x^2}} \, \mathrm{d}x = (x - 4)^{15} e^{4x} \sqrt{1 + x^2} + C.$

Part (ii) The polynomial $12x^4 - x^2 - 11$ is quartic, not cubic. Testing $x = 1$: $12 - 1 - 11 = 0$, so $(x - 1)$ is a factor. Dividing:

$12x^4 - x^2 - 11 = (x - 1)(12x^3 + 12x^2 + 11x + 11).$

The integrand becomes $\dfrac{(x - 1)^{22} e^{12x}}{\sqrt{1 + x^2}} (12x^3 + 12x^2 + 11x + 11)$, so $n - 1 = 22$, giving $n = 23$, with $a = 1$ and $b = 12$.

Checking the cubic:

$q(x) = 12x^3 + (23 + 1 - 12)x^2 + (12 - 1)x + (23 - 12) = 12x^3 + 12x^2 + 11x + 11. \quad \checkmark$

Therefore:

$\int \frac{(x - 1)^{21} e^{12x} (12x^4 - x^2 - 11)}{\sqrt{1 + x^2}} \, \mathrm{d}x = (x - 1)^{23} e^{12x} \sqrt{1 + x^2} + C.$

Part (iii) The polynomial $4x^4 + x^3 - 2$ is quartic. Factoring out $(x - 2)$ does not work (substituting $x = 2$ gives $64 + 8 - 2 \neq 0$). Instead, we seek a linear combination of derivatives.

With $a = 2$ and $b = 4$, we compute $q(x)$ for two values of $n$:

For $n = 7$: $\quad q_7(x) = 4x^3 + (7 + 1 - 8)x^2 + (4 - 2)x + (7 - 8) = 4x^3 + 2x - 1$,

so $\dfrac{\mathrm{d}y_7}{\mathrm{d}x} = \dfrac{(x - 2)^6 e^{4x}}{\sqrt{1 + x^2}} (4x^3 + 2x - 1)$.

For $n = 8$: $\quad q_8(x) = 4x^3 + (8 + 1 - 8)x^2 + (4 - 2)x + (8 - 8) = 4x^3 + x^2 + 2x$,

so $\dfrac{\mathrm{d}y_8}{\mathrm{d}x} = \dfrac{(x - 2)^7 e^{4x}}{\sqrt{1 + x^2}} (4x^3 + x^2 + 2x) = \dfrac{(x - 2)^6 e^{4x}}{\sqrt{1 + x^2}} (x - 2)(4x^3 + x^2 + 2x)$.

Expanding $(x - 2)(4x^3 + x^2 + 2x)$:

$4x^4 + x^3 + 2x^2 - 8x^3 - 2x^2 - 4x = 4x^4 - 7x^3 - 4x.$

Now we find $\beta$ such that $\dfrac{\mathrm{d}y_8}{\mathrm{d}x} + \beta \dfrac{\mathrm{d}y_7}{\mathrm{d}x}$ gives the integrand:

$(4x^4 - 7x^3 - 4x) + \beta(4x^3 + 2x - 1) = 4x^4 + x^3 - 2.$

Comparing coefficients:

• $x^3$: $-7 + 4\beta = 1 \implies \beta = 2$.
• $x$: $-4 + 2(2) = 0$. $\checkmark$ (no $x$ term in the target)
• constant: $-(2) = -2$. $\checkmark$

So:

$\int \frac{(x - 2)^6 e^{4x} (4x^4 + x^3 - 2)}{\sqrt{1 + x^2}} \, \mathrm{d}x = y_8 + 2y_7 + C$

$= (x - 2)^8 e^{4x} \sqrt{1 + x^2} + 2(x - 2)^7 e^{4x} \sqrt{1 + x^2} + C$

$= (x - 2)^7 e^{4x} \sqrt{1 + x^2} \big[(x - 2) + 2\big] + C$

$= x(x - 2)^7 e^{4x} \sqrt{1 + x^2} + C.$

Examiner Notes

Q7

In many ways, this question is little more than an academic exercise, since I can see no way in which these integrals would actually arise in any practical situation. I apologise for this. However, it was a good test of candidates’ ability to stretch a general result in different directions, probing them for increasing amounts of insight and perseverance. For future STEP-takers, the opening result is a good one for which to find a generalisation, and it is a possibly fruitful avenue to explore the “product rule of differentiation” for three terms (etc.), say $y = pqr$ in this case. Such an approach might have helped to prevent some of the ghastly mix-ups in writing all the terms out that were to be found in the scripts. It was disappointing to see that so few candidates seemed to think that they should tidy up the answer and demonstrate that the left over bits did indeed form a cubic polynomial, as required by the question. In the end, we gave anyone the mark who simply observed that what was left was a cubic (if indeed that was the case in their working). Thereafter, (i) is a straightforward application of the result, requiring candidates only to identify the values of $n$, $a$ and $b$. However, even here, it was rare to see folks justifying the form of the cubic, which might have acted as a check for errors. In (ii), the polynomial term is no longer cubic, so candidates were expected to try to see if an extra factor of $(x - 1)$ could be taken out to go with the other twenty-one $(x - 1)$s, which indeed it could. Checking the cubic’s terms was rather more important here. The final integral, in (iii), was difficult, and this was where candidates were ‘found out’ on this question. The obvious thing is to try and extract some $(x - 2)$ factor(s) from the quartic polynomial, but this doesn’t work. Candidates may reflect that they shouldn’t have found this too much of a surprise, as that would simply have been repeating the “trick” of (ii). Though only a small minority realised it, the next most obvious possibility to try, having already found in (ii) that ‘the next case up’ gives a quartic rather than a cubic polynomial, is to try some combination of the obvious answer and the next one up, and this turns out to be exactly what is required.

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Question 8

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

Problem

8 The non-collinear points $A, B$ and $C$ have position vectors a, b and c, respectively. The points $P$ and $Q$ have position vectors p and q, respectively, given by

$\mathbf{p} = \lambda \mathbf{a} + (1 - \lambda) \mathbf{b} \quad \text{and} \quad \mathbf{q} = \mu \mathbf{a} + (1 - \mu) \mathbf{c}$

where $0 < \lambda < 1$ and $\mu > 1$. Draw a diagram showing $A, B, C, P$ and $Q$.

Given that $\vec{CQ} \times \vec{BP} = \vec{AB} \times \vec{AC}$, find $\mu$ in terms of $\lambda$, and show that, for all values of $\lambda$, the line $PQ$ passes through the fixed point $D$, with position vector d given by $\mathbf{d} = -\mathbf{a} + \mathbf{b} + \mathbf{c}$. What can be said about the quadrilateral $ABDC$?

Hint

8 For the diagram, you are simply required to show $P$ on $AB$, strictly between $A$ and $B$; and $Q$ on $AC$ on other side of $A$ to $C$. The two given parameters indicate that $CQ = \mu AC$ and $BP = \lambda AB$. Substituting these into the given expression, $CQ \times BP = AB \times AC \Rightarrow \mu AC \cdot \lambda AB = AB \cdot AC$

$\Rightarrow \mu = \frac{1}{\lambda}$. [Notice that $CQ$, $BP$, etc., are scalar quantities, and hence the "$\times$" cannot be the vector product!]

Writing the equation of line $PQ$ in the form $\mathbf{r} = t \mathbf{p} + (1 - t) \mathbf{q}$ for some scalar parameter $t$ and substituting the given forms for $\mathbf{p}$ and $\mathbf{q}$ gives $\mathbf{r} = t \lambda \mathbf{a} + t(1 - \lambda)\mathbf{b} + (1 - t) \mu \mathbf{a} + (1 - t)(1 - \mu)\mathbf{c}$.

Eliminating $\mu = \frac{1}{\lambda} \Rightarrow \mathbf{r} = \left( t \lambda + \frac{1}{\lambda} - \frac{t}{\lambda} \right) \mathbf{a} + t(1 - \lambda)\mathbf{b} + (1 - t) \left( \frac{\lambda - 1}{\lambda} \right) \mathbf{c}$. Comparing this to the

given answer, we note that when $t = \frac{1}{1 - \lambda}$ from the b-component, $1 - t = \frac{\lambda}{\lambda - 1}$, etc., so that we do indeed get $\mathbf{r} = -\mathbf{a} + \mathbf{b} + \mathbf{c}$, as required.

Since $\mathbf{d} - \mathbf{c} = \mathbf{b} - \mathbf{a}$, one pair of sides of opposite sides of $ABDC$ are equal and parallel, so we can conclude that $ABDC$ is a parallelogram

Model Solution

Diagram. The point $P$ lies on segment $AB$ (since $0 < \lambda < 1$, it divides $AB$ internally). The point $Q$ lies on line $AC$ beyond $A$ from $C$ (since $\mu > 1$, we have $CQ > CA$).

Finding $\mu$. We compute the relevant scalar lengths. Since

$\vec{BP} = \mathbf{p} - \mathbf{b} = \lambda \mathbf{a} + (1 - \lambda)\mathbf{b} - \mathbf{b} = \lambda(\mathbf{a} - \mathbf{b}) = -\lambda \vec{AB},$

we have $BP = \lambda \cdot AB$. Similarly,

$\vec{CQ} = \mathbf{q} - \mathbf{c} = \mu \mathbf{a} + (1 - \mu)\mathbf{c} - \mathbf{c} = \mu(\mathbf{a} - \mathbf{c}) = -\mu \vec{AC},$

so $CQ = \mu \cdot AC$.

The given condition $\vec{CQ} \times \vec{BP} = \vec{AB} \times \vec{AC}$ uses $\times$ to denote ordinary multiplication of scalar lengths (not the vector product). Substituting:

$\mu \cdot AC \cdot \lambda \cdot AB = AB \cdot AC.$

Since $A, B, C$ are non-collinear, $AB$ and $AC$ are non-zero, so we can cancel to get:

$\lambda \mu = 1 \qquad \Longrightarrow \qquad \mu = \frac{1}{\lambda}.$

Since $0 < \lambda < 1$, this gives $\mu > 1$, consistent with the given constraint. $\quad \blacksquare$

Showing $PQ$ passes through a fixed point. A general point on line $PQ$ has position vector

$\mathbf{r} = t\mathbf{p} + (1 - t)\mathbf{q}$

for some scalar parameter $t$. Substituting the expressions for $\mathbf{p}$ and $\mathbf{q}$:

$\mathbf{r} = t\big[\lambda \mathbf{a} + (1 - \lambda)\mathbf{b}\big] + (1 - t)\big[\mu \mathbf{a} + (1 - \mu)\mathbf{c}\big]$

$= \big[t\lambda + (1 - t)\mu\big]\mathbf{a} + t(1 - \lambda)\mathbf{b} + (1 - t)(1 - \mu)\mathbf{c}.$

Substituting $\mu = \dfrac{1}{\lambda}$:

$\mathbf{r} = \left[t\lambda + \frac{1 - t}{\lambda}\right]\mathbf{a} + t(1 - \lambda)\mathbf{b} + (1 - t)\left(1 - \frac{1}{\lambda}\right)\mathbf{c}.$

We want to find a value of $t$ (depending on $\lambda$) for which $\mathbf{r}$ is independent of $\lambda$. Setting the coefficient of $\mathbf{b}$ equal to 1:

$t(1 - \lambda) = 1 \qquad \Longrightarrow \qquad t = \frac{1}{1 - \lambda}.$

Then $1 - t = 1 - \dfrac{1}{1 - \lambda} = \dfrac{-\lambda}{1 - \lambda}$.

Coefficient of $\mathbf{c}$:

$(1 - t)\left(1 - \frac{1}{\lambda}\right) = \frac{-\lambda}{1 - \lambda} \cdot \frac{\lambda - 1}{\lambda} = \frac{-\lambda}{1 - \lambda} \cdot \frac{-(1 - \lambda)}{\lambda} = 1.$

Coefficient of $\mathbf{a}$:

$t\lambda + (1 - t) \cdot \frac{1}{\lambda} = \frac{\lambda}{1 - \lambda} + \frac{-\lambda}{1 - \lambda} \cdot \frac{1}{\lambda} = \frac{\lambda}{1 - \lambda} - \frac{1}{1 - \lambda} = \frac{\lambda - 1}{1 - \lambda} = -1.$

Therefore, for this value of $t$ (which depends on $\lambda$), we obtain

$\mathbf{r} = -\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{d},$

which is independent of $\lambda$. This shows that the line $PQ$ passes through the fixed point $D$ with position vector $\mathbf{d} = -\mathbf{a} + \mathbf{b} + \mathbf{c}$ for all values of $\lambda$. $\quad \blacksquare$

The quadrilateral $ABDC$. We compute:

$\vec{AB} = \mathbf{b} - \mathbf{a}, \qquad \vec{CD} = \mathbf{d} - \mathbf{c} = (-\mathbf{a} + \mathbf{b} + \mathbf{c}) - \mathbf{c} = \mathbf{b} - \mathbf{a} = \vec{AB}.$

Since $\vec{CD} = \vec{AB}$, the sides $AB$ and $CD$ are equal in length and parallel. Therefore $ABDC$ is a parallelogram.

Examiner Notes

Q8

The vectors question was the least popular pure maths question by a considerable way, and only marginally more popular than most of the applied questions. In general, attempts were short-lived and candidates usually gave up when the algebra got a bit “iffy”. Strangely, an awful lot of attempts began very poorly indeed; the ranges of values of $\lambda$ and $\mu$ had a geometric significance relating to where $P$ and $Q$ lay on the lines $AB$ and $AC$, and these were not well grasped. Moreover, it was surprising to see diagrams in which the lines had not even been drawn, often leaving the marker to guess whether the points were supposed to be on them or not. Many responses to the next part were so bizarre that they were almost funny: a lot of candidates thought that $CQ$ etc. were vectors rather than lengths, and the “$\times$” was treated variously as a scalar multiplication, the scalar product and the vector product. Oddly enough, they could all lead to the required answer, $\lambda\mu = 1$, even legitimately (with a bit of care) though we were harsh on statements that were actually nonsense. Very few made it to the later stages of the question.