STEP2 2014 -- Pure Mathematics

Exam: STEP2  |  Year: 2014  |  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	几何与三角 (Geometry & Trigonometry)	Standard	三角恒等式cos²θ+sin²θ=1,二次方程判别式,坐标几何距离分析,延长线上的几何推理
2	微积分 (Calculus)	Challenging	三角函数积分计算,反例构造方法,二次函数代入不等式,三角函数代入不等式,π的近似值比较
3	微积分 (Calculus)	Challenging	点到直线距离公式,隐函数微分,微分方程求解,y”=0的特殊情形分析,圆的方程推导
4	微积分 (Calculus)	Standard	u=1/x代换,对称区间上的积分性质,arctan恒等式,参数选择技巧,无穷积分计算
5	微分方程 (Differential Equations)	Standard	齐次微分方程的y=xu代换,分离变量法,坐标平移消常数项,初值条件代入
6	三角函数与级数 (Trigonometric Series)	Challenging	三角恒等式求和, 导数求驻点, 数学归纳法, 不等式证明
7	绝对值函数与方程 (Absolute Value Functions & Equations)	Standard	分段函数分析, 图像绘制, 绝对值方程求解, 分类讨论
8	二项式展开与组合数学 (Binomial Expansion & Combinatorics)	Hard	二项式系数公式, 比值法判断单调性, 不等式放缩, 分类讨论

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Question 1

Topic: 几何与三角 (Geometry & Trigonometry)  |  Difficulty: Standard  |  Marks: 20

Problem

1 In the triangle $ABC$, the base $AB$ is of length 1 unit and the angles at $A$ and $B$ are $\alpha$ and $\beta$ respectively, where $0 < \alpha \leqslant \beta$. The points $P$ and $Q$ lie on the sides $AC$ and $BC$ respectively, with $AP = PQ = QB = x$. The line $PQ$ makes an angle of $\theta$ with the line through $P$ parallel to $AB$.

(i) Show that $x \cos \theta = 1 - x \cos \alpha - x \cos \beta$, and obtain an expression for $x \sin \theta$ in terms of $x$, $\alpha$ and $\beta$. Hence show that

$(1 + 2 \cos(\alpha + \beta))x^2 - 2(\cos \alpha + \cos \beta)x + 1 = 0 . \qquad \text{(*)}$

Show that $(*)$ is also satisfied if $P$ and $Q$ lie on $AC$ produced and $BC$ produced, respectively. [By definition, $P$ lies on $AC$ produced if $P$ lies on the line through $A$ and $C$ and the points are in the order $A, C, P$.]

(ii) State the condition on $\alpha$ and $\beta$ for $(*)$ to be linear in $x$. If this condition does not hold (but the condition $0 < \alpha \leqslant \beta$ still holds), show that $(*)$ has distinct real roots.

(iii) Find the possible values of $x$ in the two cases (a) $\alpha = \beta = 45^\circ$ and (b) $\alpha = 30^\circ$, $\beta = 90^\circ$, and illustrate each case with a sketch.

Hint

Drawing a diagram and considering the horizontal and vertical distances will establish the relationships for $x \cos \theta$ and $x \sin \theta$ easily. The quadratic equation will then follow from use of the identity $\cos^2 \theta + \sin^2 \theta \equiv 1$. The same reasoning applied to a diagram showing the case where P and Q lie on AC produced and BC produced will show that the same equation is satisfied.

() will be linear if the coefficient of $x^2$ is 0, so therefore $\cos(\alpha + \beta)$ will need to equal $-\frac{1}{2}$, which gives a relationship between $\alpha$ and $\beta$. For () to have distinct roots the discriminant must be positive. Using some trigonometric identities it can be shown that the discriminant is equal to $4(1 - (\sin \alpha - \sin \beta)^2)$ and it should be easy to explain why this must be greater than 0.

The first case in part (iii) leads to $x = \sqrt{2} \pm 1$ and so there are two diagrams to be drawn. In each case the line joining P to Q will be horizontal.

The second case in part (iii) is an example where (*) is linear. This leads to $x = \frac{\sqrt{3}}{3}$. Therefore Q is at the same point as C and so the point P is the midpoint of AC.

Model Solution

Part (i)

Set up coordinates with $A$ at the origin and $B$ at $(1, 0)$. Then $AC$ makes angle $\alpha$ with $AB$, and $BC$ makes angle $\pi - \beta$ with the positive $x$-axis (since $\beta$ is the interior angle at $B$).

Since $AP = x$, the coordinates of $P$ are:

$P = (x\cos\alpha,\; x\sin\alpha)$

Since $QB = x$ and $Q$ lies on $BC$ with $B$ at $(1,0)$, moving from $B$ toward $C$ by distance $x$:

$Q = (1 - x\cos\beta,\; x\sin\beta)$

The line $PQ$ has length $x$ and makes angle $\theta$ with the horizontal. The horizontal component of $\vec{PQ}$ is:

$x\cos\theta = (1 - x\cos\beta) - x\cos\alpha = 1 - x\cos\alpha - x\cos\beta \qquad \text{(...)}$

The vertical component of $\vec{PQ}$ is:

$x\sin\theta = x\sin\beta - x\sin\alpha \qquad \text{(...)}$

Using $\cos^2\theta + \sin^2\theta = 1$, we square and add:

$(1 - x\cos\alpha - x\cos\beta)^2 + (x\sin\beta - x\sin\alpha)^2 = x^2$

Expanding the left side:

$1 - 2x\cos\alpha - 2x\cos\beta + x^2\cos^2\alpha + 2x^2\cos\alpha\cos\beta + x^2\cos^2\beta + x^2\sin^2\beta - 2x^2\sin\alpha\sin\beta + x^2\sin^2\alpha$

Collecting the $x^2$ terms using $\cos^2\alpha + \sin^2\alpha = 1$ and $\cos^2\beta + \sin^2\beta = 1$:

$x^2(2 + 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta) = x^2(2 + 2\cos(\alpha + \beta))$

So the equation becomes:

$1 - 2x(\cos\alpha + \cos\beta) + x^2(2 + 2\cos(\alpha + \beta)) = x^2$

$1 - 2x(\cos\alpha + \cos\beta) + x^2(1 + 2\cos(\alpha + \beta)) = 0$

$(1 + 2\cos(\alpha + \beta))x^2 - 2(\cos\alpha + \cos\beta)x + 1 = 0 \qquad (*)$

Extension to $AC$ produced and $BC$ produced: If $P$ lies on $AC$ produced (order $A, C, P$) and $Q$ lies on $BC$ produced (order $B, C, Q$), then $P$ and $Q$ are on the opposite sides of $C$ from $A$ and $B$. The horizontal and vertical displacements become:

$x\cos\theta = 1 + x\cos\alpha + x\cos\beta \quad \text{or} \quad x\cos\theta = -(1 + x\cos\alpha + x\cos\beta)$

depending on the direction of $\vec{PQ}$. In either case, squaring gives $(1 + x\cos\alpha + x\cos\beta)^2$, and the vertical displacement squared is $(x\sin\beta - x\sin\alpha)^2$ (or its negative, which squares to the same). Adding and using $\cos^2\theta + \sin^2\theta = 1$:

$(1 + x\cos\alpha + x\cos\beta)^2 + (x\sin\beta - x\sin\alpha)^2 = x^2$

Expanding:

$1 + 2x\cos\alpha + 2x\cos\beta + x^2\cos^2\alpha + 2x^2\cos\alpha\cos\beta + x^2\cos^2\beta + x^2\sin^2\beta - 2x^2\sin\alpha\sin\beta + x^2\sin^2\alpha = x^2$

This simplifies to:

$(1 + 2\cos(\alpha + \beta))x^2 + 2(\cos\alpha + \cos\beta)x + 1 = 0$

Since $x$ can be negative in this configuration (measuring the “signed distance” along the produced line), replacing $x$ by $-x$ recovers $(*)$. Alternatively, since the quadratic in $x$ from the original case already accounts for both signs of $x$, the same equation $(*)$ is satisfied.

Part (ii)

$(*)$ is linear in $x$ when the coefficient of $x^2$ is zero:

$1 + 2\cos(\alpha + \beta) = 0 \implies \cos(\alpha + \beta) = -\frac{1}{2}$

Since $0 < \alpha \leqslant \beta$ and $\alpha + \beta < \pi$ (angles of a triangle), we need $\alpha + \beta = \frac{2\pi}{3}$, i.e., $\alpha + \beta = 120°$.

Distinct real roots: If $1 + 2\cos(\alpha + \beta) \neq 0$, the discriminant is:

$\Delta = 4(\cos\alpha + \cos\beta)^2 - 4(1 + 2\cos(\alpha + \beta))$

Using $\cos\alpha + \cos\beta = 2\cos\!\tfrac{\alpha+\beta}{2}\cos\!\tfrac{\alpha-\beta}{2}$:

$\Delta = 16\cos^2\!\tfrac{\alpha+\beta}{2}\cos^2\!\tfrac{\alpha-\beta}{2} - 4 - 8\cos(\alpha+\beta)$

Setting $\phi = \frac{\alpha+\beta}{2}$ and using $\cos(\alpha+\beta) = 2\cos^2\phi - 1$:

$\Delta = 16\cos^2\phi\cos^2\!\tfrac{\alpha-\beta}{2} - 8(2\cos^2\phi - 1) - 4 = 16\cos^2\phi\cos^2\!\tfrac{\alpha-\beta}{2} - 16\cos^2\phi + 4$

$= 4 - 16\cos^2\phi\sin^2\!\tfrac{\alpha-\beta}{2} = 4\left(1 - 4\cos^2\phi\sin^2\!\tfrac{\alpha-\beta}{2}\right)$

Since $0 < \alpha \leqslant \beta$ and $\alpha + \beta < \pi$, we have $0 < \phi < \frac{\pi}{2}$ so $\cos\phi < 1$ (strict, since $\alpha + \beta < \pi$), and $0 \leqslant \frac{\alpha-\beta}{2} < \frac{\pi}{2}$ so $\sin\!\tfrac{\alpha-\beta}{2} < 1$. Therefore:

$4\cos^2\phi\sin^2\!\tfrac{\alpha-\beta}{2} < 1$

which gives $\Delta > 0$, so $(*)$ has distinct real roots. $\square$

Part (iii)

(a) $\alpha = \beta = 45°$: $\cos(\alpha+\beta) = \cos 90° = 0$, $\cos\alpha + \cos\beta = \sqrt{2}$.

$x^2 - 2\sqrt{2}\,x + 1 = 0$

$x = \frac{2\sqrt{2} \pm \sqrt{8 - 4}}{2} = \sqrt{2} \pm 1$

So $x = \sqrt{2} + 1$ (extended case, $P$ and $Q$ beyond $C$) or $x = \sqrt{2} - 1$ (internal case, $P$ and $Q$ on $AC$ and $BC$).

In both cases $PQ$ is horizontal ($\theta = 0$) since $\alpha = \beta$ gives $x\sin\theta = x(\sin\beta - \sin\alpha) = 0$.

(b) $\alpha = 30°$, $\beta = 90°$: $\alpha + \beta = 120°$, so $(*)$ is linear:

$-2(\cos 30° + \cos 90°)x + 1 = 0 \implies -\sqrt{3}\,x + 1 = 0 \implies x = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Since $BC$ is vertical (angle $90°$ at $B$) and $QB = \frac{\sqrt{3}}{3}$, while $BC$ has length $\frac{\sqrt{3}}{3}$ (by the sine rule), $Q$ coincides with $C$. Then $P$ is the midpoint of $AC$.

Examiner Notes

While the first part of the question was successfully completed by many of the candidates, there were quite a few diagrams drawn showing the point P further from the line AB than Q. Those who established the expression for $x \cos \theta$ were usually able to find an expression for $x \sin \theta$ and good justifications of the quadratic equation were given. The case where P and Q lie on the lines AC produced and BC produced caused a lot of difficulty for many of the candidates, many of whom tried unsuccessfully to create an argument based on similar triangles.

The condition for (*) to be linear in $x$ did not cause much difficulty, although a number of candidates did not give the value of $\cos^{-1} \left( -\frac{1}{2} \right)$. Many candidates realised that the justification that the roots were distinct would involve the discriminant, although some solutions included the case where the discriminant could be equal to 0 were produced. However, very few solutions were able to give a clear justification that the discriminant must be greater than 0.

In the final part some candidates sketched the graph of the quadratic rather than sketching the triangle in the two cases given. In the second case many candidates did not realise that Q was at the same point as C.

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Question 2

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

Problem

2 This question concerns the inequality

$\int_{0}^{\pi} (f(x))^2 \, \mathrm{d}x \leqslant \int_{0}^{\pi} (f'(x))^2 \, \mathrm{d}x . \qquad \text{(*)}$

(i) Show that $(*)$ is satisfied in the case $f(x) = \sin nx$, where $n$ is a positive integer.

Show by means of counterexamples that $(*)$ is not necessarily satisfied if either $f(0) \neq 0$ or $f(\pi) \neq 0$.

(ii) You may now assume that $(*)$ is satisfied for any (differentiable) function $f$ for which $f(0) = f(\pi) = 0$.

By setting $f(x) = ax^2 + bx + c$, where $a, b$ and $c$ are suitably chosen, show that $\pi^2 \leqslant 10$.

By setting $f(x) = p \sin \frac{1}{2}x + q \cos \frac{1}{2}x + r$, where $p, q$ and $r$ are suitably chosen, obtain another inequality for $\pi$.

Which of these inequalities leads to a better estimate for $\pi^2$?

Hint

By rewriting in terms of $\cos 2nx$ it can be shown that $\int_{0}^{\pi} \sin^2 nx \, dx = \frac{\pi}{2}$ and $\int_{0}^{\pi} n^2 \cos^2 nx \, dx = \frac{n^2 \pi}{2}$. Therefore () must be satisfied as $n$ is a positive integer. The function $f(x) = x$ does not satisfy () and $f(0) = 0$ but $f(\pi) \neq 0$. The function $g(x) = f(\pi - x)$ will therefore provide a counterexample where $g(\pi) = 0$, but $g(0) \neq 0$.

In part (ii), $f(x) = x^2 - \pi x$ will need to be selected to be able to use the assumption that () is satisfied. The two sides of () can then be evaluated:

$\int_{0}^{\pi} x^4 - 2\pi x^3 + \pi^2 x^2 \, dx = \frac{\pi^5}{30}$

$\int_{0}^{\pi} 4x^2 - 4\pi x + \pi^2 \, dx = \frac{\pi^3}{3}$

Substitution into (*) then leads to the inequality $\pi^2 \leq 10$.

To satisfy the conditions on $f(x)$ for the second type of function, the values of $p$, $q$ and $r$ must satisfy $q + r = 0$ and $p + r = 0$. Evaluating the integrals then leads to $\pi \leq \frac{22}{7}$.

Since $(\frac{22}{7})^2 < 10$, $\pi \leq \frac{22}{7}$ leads to a better estimate for $\pi^2$.

Model Solution

Part (i)

Let $f(x) = \sin nx$ where $n$ is a positive integer. Then $f'(x) = n\cos nx$.

Left side:

$\int_0^\pi \sin^2 nx \, dx = \int_0^\pi \frac{1 - \cos 2nx}{2} \, dx = \frac{\pi}{2} - \frac{1}{2}\left[\frac{\sin 2nx}{2n}\right]_0^\pi = \frac{\pi}{2}$

since $\sin 2n\pi = 0$.

Right side:

$\int_0^\pi n^2\cos^2 nx \, dx = n^2 \int_0^\pi \frac{1 + \cos 2nx}{2} \, dx = \frac{n^2\pi}{2}$

Since $n \geqslant 1$, we have $\frac{n^2\pi}{2} \geqslant \frac{\pi}{2}$, so $(*)$ is satisfied. $\square$

Counterexamples:

Case $f(\pi) \neq 0$: Let $f(x) = x$. Then $f(0) = 0$ but $f(\pi) = \pi \neq 0$. We have $f'(x) = 1$, so:

$\int_0^\pi x^2 \, dx = \frac{\pi^3}{3}, \qquad \int_0^\pi 1 \, dx = \pi$

Since $\frac{\pi^3}{3} > \pi$ (as $\pi^2 > 3$), $(*)$ is violated.

Case $f(0) \neq 0$: Let $g(x) = f(\pi - x) = \pi - x$. Then $g(\pi) = 0$ but $g(0) = \pi \neq 0$. We have $g'(x) = -1$, so:

$\int_0^\pi (\pi - x)^2 \, dx = \frac{\pi^3}{3}, \qquad \int_0^\pi 1 \, dx = \pi$

Again $\frac{\pi^3}{3} > \pi$, so $(*)$ is violated. $\square$

Part (ii)

Quadratic function: Choose $f(x) = x^2 - \pi x$ so that $f(0) = 0$ and $f(\pi) = \pi^2 - \pi^2 = 0$.

Left side:

$\int_0^\pi (x^2 - \pi x)^2 \, dx = \int_0^\pi (x^4 - 2\pi x^3 + \pi^2 x^2) \, dx$

$= \left[\frac{x^5}{5} - \frac{2\pi x^4}{4} + \frac{\pi^2 x^3}{3}\right]_0^\pi = \frac{\pi^5}{5} - \frac{\pi^5}{2} + \frac{\pi^5}{3} = \pi^5\left(\frac{6 - 15 + 10}{30}\right) = \frac{\pi^5}{30}$

Right side: $f'(x) = 2x - \pi$, so:

$\int_0^\pi (2x - \pi)^2 \, dx = \int_0^\pi (4x^2 - 4\pi x + \pi^2) \, dx = \left[\frac{4x^3}{3} - 2\pi x^2 + \pi^2 x\right]_0^\pi$

$= \frac{4\pi^3}{3} - 2\pi^3 + \pi^3 = \pi^3\left(\frac{4}{3} - 1\right) = \frac{\pi^3}{3}$

By $(*)$: $\frac{\pi^5}{30} \leqslant \frac{\pi^3}{3}$, giving $\pi^2 \leqslant 10$. $\square$

Trigonometric function: Choose $f(x) = p\sin\frac{x}{2} + q\cos\frac{x}{2} + r$ with $f(0) = f(\pi) = 0$:

$f(0) = q + r = 0 \implies q = -r$ $f(\pi) = p + r = 0 \implies p = -r$

So $f(x) = r(-\sin\frac{x}{2} - \cos\frac{x}{2} + 1)$, and we may take $r = 1$ (the inequality is homogeneous in $r$), giving $f(x) = 1 - \sin\frac{x}{2} - \cos\frac{x}{2}$.

Left side: $f(x)^2 = 1 + \sin^2\frac{x}{2} + \cos^2\frac{x}{2} - 2\sin\frac{x}{2} - 2\cos\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}$

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

$\int_0^\pi f(x)^2 \, dx = 2\pi - 2\!\int_0^\pi\!\sin\frac{x}{2}\,dx - 2\!\int_0^\pi\!\cos\frac{x}{2}\,dx + \int_0^\pi\!\sin x\,dx$

$= 2\pi - 2\left[-2\cos\frac{x}{2}\right]_0^\pi - 2\left[2\sin\frac{x}{2}\right]_0^\pi + [-\cos x]_0^\pi$

$= 2\pi - 2(-2\cdot 0 + 2) - 2(2 - 0) + (1 + 1) = 2\pi - 4 - 4 + 2 = 2\pi - 6$

Right side: $f'(x) = -\frac{1}{2}\cos\frac{x}{2} + \frac{1}{2}\sin\frac{x}{2}$, so $f'(x)^2 = \frac{1}{4}\cos^2\frac{x}{2} + \frac{1}{4}\sin^2\frac{x}{2} - \frac{1}{2}\sin\frac{x}{2}\cos\frac{x}{2}$

$= \frac{1}{4} - \frac{1}{2}\sin\frac{x}{2}\cos\frac{x}{2} = \frac{1}{4} - \frac{1}{4}\sin x$

$\int_0^\pi f'(x)^2 \, dx = \frac{\pi}{4} - \frac{1}{4}\int_0^\pi \sin x \, dx = \frac{\pi}{4} - \frac{1}{4}\cdot 2 = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}$

By $(*)$: $2\pi - 6 \leqslant \frac{\pi - 2}{4}$, so $8\pi - 24 \leqslant \pi - 2$, giving $7\pi \leqslant 22$, i.e., $\pi \leqslant \frac{22}{7}$.

Comparison: Since $\left(\frac{22}{7}\right)^2 = \frac{484}{49} \approx 9.878 < 10$, the inequality $\pi \leqslant \frac{22}{7}$ gives a better estimate for $\pi^2$. $\square$

Examiner Notes

This was one of the more popular questions of the paper. Most candidates successfully showed that the first inequality was satisfied, but when producing counterexamples, some failed to show that either $f(x) \neq 0$ or $f(\pi) \neq 0$ for their chosen functions. In the second part many candidates did not attempt to choose values of $a$, $b$ and $c$, but substituted the general form of the quadratic function into the inequality instead. In the case where the function involved trigonometric functions, many of those who attempted it were able to deduce that $p = q = -r$, but several candidates made mistakes in the required integration. Those who established two inequalities were able to decide which gives the better estimate for $\pi$.

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Question 3

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

Problem

3 (i) Show, geometrically or otherwise, that the shortest distance between the origin and the line $y = mx + c$, where $c \geqslant 0$, is $c(m^2 + 1)^{-\frac{1}{2}}$.

(ii) The curve $C$ lies in the $x$-$y$ plane. Let the line $L$ be tangent to $C$ at a point $P$ on $C$, and let $a$ be the shortest distance between the origin and $L$. The curve $C$ has the property that the distance $a$ is the same for all points $P$ on $C$.

Let $P$ be the point on $C$ with coordinates $(x, y(x))$. Given that the tangent to $C$ at $P$ is not vertical, show that

$(y - xy')^2 = a^2(1 + (y')^2) . \qquad \text{(*)}$

By first differentiating $(*)$ with respect to $x$, show that either $y = mx \pm a(1 + m^2)^{\frac{1}{2}}$ for some $m$ or $x^2 + y^2 = a^2$.

(iii) Now suppose that $C$ (as defined above) is a continuous curve for $-\infty < x < \infty$, consisting of the arc of a circle and two straight lines. Sketch an example of such a curve which has a non-vertical tangent at each point.

Hint

By drawing a diagram and marking the shortest distance a pair of similar triangles can be used to show that $\frac{c/m}{c\sqrt{m^2+1}/m} = \frac{d}{c}$, which simplifies to $d = c/\sqrt{m^2 + 1}$.

For the second part, the tangent to the curve at the general point $(x, y)$ will have a gradient of $y'$ and so the $y$-intercept will be at the point $(0, y - xy')$. Therefore the result from part (i) can be applied using $m = y'$ and $c = y - xy'$ to give $a = (y - xy')/\sqrt{(y')^2 + 1}$, which rearranges to give the required result.

Differentiating the equation then gives $y''(a^2y' + x(y - xy')) = 0$ and so either $y'' = 0$ or

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

If $y'' = 0$ then the equation will be of a straight line and the $y$-intercept can be deduced in terms of $m$.

If $a^2y' + x(y - xy') = 0$, then the differential equation can be solved to give the equation of a circle.

Part (iii) then requires combining the two possible cases from part (ii) to construct a curve which satisfies the conditions given. This must be an arc of a circle with no vertical tangents, with straight lines at either end of the arc in the direction of the tangents to the circle at that point.

Model Solution

Part (i)

The line $y = mx + c$ can be written as $mx - y + c = 0$. The perpendicular from the origin to this line has gradient $-\frac{1}{m}$, so its equation is $y = -\frac{x}{m}$.

Substituting into $y = mx + c$:

$-\frac{x}{m} = mx + c \implies -x = m^2 x + mc \implies x = -\frac{mc}{m^2 + 1}$

$y = -\frac{x}{m} = \frac{c}{m^2 + 1}$

The distance from the origin to this point is:

$d = \sqrt{x^2 + y^2} = \sqrt{\frac{m^2 c^2}{(m^2+1)^2} + \frac{c^2}{(m^2+1)^2}} = \frac{c\sqrt{m^2 + 1}}{m^2 + 1} = \frac{c}{\sqrt{m^2 + 1}} = c(m^2+1)^{-1/2}$

since $c \geqslant 0$. $\square$

Part (ii)

The tangent to $C$ at $P = (x, y)$ has gradient $y'$, so its equation is:

$Y - y = y'(X - x) \implies Y = y'X + (y - xy')$

This has gradient $m = y'$ and $y$-intercept $c = y - xy'$. By part (i), the distance from the origin to this tangent is:

$a = \frac{y - xy'}{\sqrt{1 + (y')^2}}$

Squaring: $a^2(1 + (y')^2) = (y - xy')^2$. $\square$

Differentiating $(*)$: Differentiate both sides with respect to $x$:

$\text{LHS: } 2a^2 y' y''$

$\text{RHS: } 2(y - xy') \cdot \frac{d}{dx}(y - xy') = 2(y - xy')(y' - y' - xy'') = -2xy''(y - xy')$

So: $2a^2 y' y'' = -2xy''(y - xy')$

$y''\left(a^2 y' + x(y - xy')\right) = 0$

Case 1: $y'' = 0$. Then $y = mx + k$ for constants $m$ and $k$. Substituting into $(*)$:

$a^2(1 + m^2) = (y - xy')^2 = (mx + k - mx)^2 = k^2$

$k = \pm a\sqrt{1 + m^2}$

So $y = mx \pm a(1 + m^2)^{1/2}$. $\square$

Case 2: $a^2 y' + x(y - xy') = 0$. Rearranging:

$a^2 y' + xy - x^2 y' = 0 \implies y'(a^2 - x^2) = -xy \implies \frac{dy}{dx} = \frac{-xy}{a^2 - x^2}$

Separating variables: $\frac{dy}{y} = \frac{-x}{a^2 - x^2}\,dx = \frac{x}{x^2 - a^2}\,dx$

Integrating: $\ln|y| = \frac{1}{2}\ln|x^2 - a^2| + \text{const}$

$y^2 = K(x^2 - a^2)$

Substituting into $(*)$: $(y - xy')^2 = a^2(1 + (y')^2)$.

From the DE: $y' = \frac{-xy}{a^2 - x^2}$, so $y - xy' = y + \frac{x^2 y}{a^2 - x^2} = \frac{a^2 y}{a^2 - x^2}$.

$(*)$ gives: $\frac{a^4 y^2}{(a^2 - x^2)^2} = a^2\left(1 + \frac{x^2 y^2}{(a^2 - x^2)^2}\right) = a^2 \cdot \frac{(a^2 - x^2)^2 + x^2 y^2}{(a^2 - x^2)^2}$

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

$y^2(a^2 - x^2) = (a^2 - x^2)^2$

If $a^2 \neq x^2$: $y^2 = a^2 - x^2$, i.e., $x^2 + y^2 = a^2$. $\square$

Part (iii)

From part (ii), the curve $C$ consists of straight lines $y = mx \pm a\sqrt{1+m^2}$ and/or circular arcs $x^2 + y^2 = a^2$.

To construct a continuous curve for $-\infty < x < \infty$ with no vertical tangent, take the lower semicircle $x^2 + y^2 = a^2$ (i.e., $y = -\sqrt{a^2 - x^2}$) for $-a\cos\phi \leqslant x \leqslant a\cos\phi$ (where $0 < \phi < \frac{\pi}{2}$), and attach tangent lines at the endpoints.

At the point $(a\cos\phi, -a\sin\phi)$ on the circle, the tangent has gradient $-\cot\phi$ (from implicit differentiation $2x + 2yy' = 0$ gives $y' = -x/y = -\cot\phi$). The tangent line is:

$y + a\sin\phi = -\cot\phi(x - a\cos\phi)$

This line extends to $x \to +\infty$. Similarly, at $(-a\cos\phi, -a\sin\phi)$, the tangent has gradient $\cot\phi$ and extends to $x \to -\infty$.

The resulting curve is continuous, has non-vertical tangent at every point (choose $\phi$ so that $\cot\phi$ is finite, i.e., $\phi \neq \frac{\pi}{2}$), and consists of one circular arc and two straight lines. $\square$

Examiner Notes

Many candidates produced a correct solution to the first part of the question. There were a number of popular methods, such as the use of similar triangles, but an algebraic approach finding the intersection between the line and a perpendicular line through the origin was the most popular. Some candidates however, simply stated a formula for the shortest distance from a point to a line. Establishing the differential equation in the second part of the question was generally done well, but many candidates struggled with the solution of the differential equation. A common error was to ignore the case $y'' = 0$ and simply find the circle solution.

The final part of the question was attempted by only a few of the candidates, many of whom did not produce an example that satisfied all of the conditions stated in the question, in particular the condition that the tangents should not be vertical at any point was often missed.

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Question 4

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

Problem

4 (i) By using the substitution $u = 1/x$, show that for $b > 0$

$\int_{1/b}^{b} \frac{x \ln x}{(a^2 + x^2)(a^2x^2 + 1)} \, dx = 0 .$

(ii) By using the substitution $u = 1/x$, show that for $b > 0$,

$\int_{1/b}^{b} \frac{\arctan x}{x} \, dx = \frac{\pi \ln b}{2} .$

(iii) By using the result $\int_{0}^{\infty} \frac{1}{a^2 + x^2} \, dx = \frac{\pi}{2a}$ (where $a > 0$), and a substitution of the

form $u = k/x$, for suitable $k$, show that

$\int_{0}^{\infty} \frac{1}{(a^2 + x^2)^2} \, dx = \frac{\pi}{4a^3} \quad (a > 0) .$

Hint

In part (i), if the required integral is called $I$ then the given substitution leads to an integral which can be shown to be equal to $-I$. This means that $2I = 0$ and so $I = 0$.

In part (ii), once the substitution has been completed, the integral will simplify to $\int_{1/b}^{b} \frac{\arctan \frac{1}{u}}{u} du$. Since $\arctan x + \arctan \left(\frac{1}{x}\right) = \frac{\pi}{2}$ the integral can be shown to be equal to $\frac{1}{2} \int_{1/b}^{b} \frac{\pi}{2x} dx$, which then simplifies to the required result.

In part (iii), making with the substitution in terms of $k$ and simplifying will show that the integral is equivalent to

$\int_{0}^{\infty} \frac{ku^2}{(a^2u^2 + k^2)^2} du$

Therefore choosing $k = a^2$, the integral can be simplified further to

$\frac{1}{a^2} \int_{0}^{\infty} \frac{u^2}{(a^2 + u^2)^2} du = \frac{1}{a^2} \int_{0}^{\infty} \frac{1}{a^2 + u^2} du - \frac{1}{a^2} \int_{0}^{\infty} \frac{a^2}{(a^2 + u^2)^2} du$

The result then follows by using the given value for $\int_{0}^{\infty} \frac{1}{a^2+x^2} dx$.

Model Solution

Part (i)

Let $I = \int_{1/b}^{b} \frac{x\ln x}{(a^2 + x^2)(a^2 x^2 + 1)}\,dx$.

Substitute $u = 1/x$, so $x = 1/u$, $dx = -1/u^2\,du$. When $x = 1/b$, $u = b$; when $x = b$, $u = 1/b$.

$I = \int_{b}^{1/b} \frac{\frac{1}{u}\ln\frac{1}{u}}{\left(a^2 + \frac{1}{u^2}\right)\left(\frac{a^2}{u^2} + 1\right)} \cdot \frac{-1}{u^2}\,du$

Simplify the denominator:

$a^2 + \frac{1}{u^2} = \frac{a^2 u^2 + 1}{u^2}, \qquad \frac{a^2}{u^2} + 1 = \frac{a^2 + u^2}{u^2}$

So the denominator is $\frac{(a^2 u^2 + 1)(a^2 + u^2)}{u^4}$.

The integrand becomes:

$\frac{-\ln u / u}{(a^2 u^2 + 1)(a^2 + u^2)/u^4} \cdot \frac{-1}{u^2} = \frac{\ln u \cdot u}{(a^2 u^2 + 1)(a^2 + u^2)}$

Reversing the limits (which removes the negative sign):

$I = \int_{1/b}^{b} \frac{u\ln u}{(a^2 + u^2)(a^2 u^2 + 1)}\,du = I$

But we also had a sign change from $\ln(1/u) = -\ln u$, so actually $I = -I$, giving $2I = 0$, hence $I = 0$. $\square$

Part (ii)

Let $I = \int_{1/b}^{b} \frac{\arctan x}{x}\,dx$.

Substitute $u = 1/x$, $dx = -1/u^2\,du$. When $x = 1/b$, $u = b$; when $x = b$, $u = 1/b$.

$I = \int_{b}^{1/b} \frac{\arctan(1/u)}{1/u} \cdot \frac{-1}{u^2}\,du = \int_{1/b}^{b} \frac{\arctan(1/u)}{u}\,du$

Using the identity $\arctan(1/u) = \frac{\pi}{2} - \arctan u$ (for $u > 0$):

$I = \int_{1/b}^{b} \frac{\frac{\pi}{2} - \arctan u}{u}\,du = \frac{\pi}{2}\int_{1/b}^{b} \frac{du}{u} - \int_{1/b}^{b} \frac{\arctan u}{u}\,du$

Since $\int_{1/b}^{b} \frac{du}{u} = [\ln u]_{1/b}^{b} = \ln b - \ln(1/b) = 2\ln b$, and the second integral equals $I$:

$I = \frac{\pi}{2} \cdot 2\ln b - I = \pi\ln b - I$

$2I = \pi\ln b \implies I = \frac{\pi\ln b}{2} \qquad \square$

Part (iii)

Let $J = \int_0^\infty \frac{dx}{(a^2 + x^2)^2}$.

Substitute $u = a^2/x$ (so $k = a^2$), $x = a^2/u$, $dx = -a^2/u^2\,du$. When $x = 0$, $u = \infty$; when $x = \infty$, $u = 0$.

$J = \int_{\infty}^{0} \frac{1}{\left(a^2 + \frac{a^4}{u^2}\right)^2} \cdot \frac{-a^2}{u^2}\,du = \int_0^\infty \frac{a^2/u^2}{\left(\frac{a^2 u^2 + a^4}{u^2}\right)^2}\,du$

$= \int_0^\infty \frac{a^2 u^2}{a^4(u^2 + a^2)^2}\,du = \frac{1}{a^2}\int_0^\infty \frac{u^2}{(u^2 + a^2)^2}\,du$

Write $\frac{u^2}{(u^2 + a^2)^2} = \frac{(u^2 + a^2) - a^2}{(u^2 + a^2)^2} = \frac{1}{u^2 + a^2} - \frac{a^2}{(u^2 + a^2)^2}$

So: $J = \frac{1}{a^2}\int_0^\infty \frac{du}{u^2 + a^2} - \frac{1}{a^2}\int_0^\infty \frac{a^2\,du}{(u^2 + a^2)^2} = \frac{1}{a^2} \cdot \frac{\pi}{2a} - J$

$2J = \frac{\pi}{2a^3} \implies J = \frac{\pi}{4a^3} \qquad \square$

Examiner Notes

Many candidates were able to perform the given substitution correctly and then correctly explain how this demonstrates that the integral is equal to 1. The second part caused more difficulty, particularly with candidates not able to state the relationship between $\arctan x$ and $\arctan (\frac{1}{x})$. Attempts to integrate with the substitution $v = \arctan (\frac{1}{u})$ often resulted in an incorrect application of the chain rule when finding $\frac{dv}{du}$.

In the final part of the question many candidates attempted to use integration by parts to reach the given answer.

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Question 5

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

Problem

5 Given that $y = xu$, where $u$ is a function of $x$, write down an expression for $\frac{dy}{dx}$.

(i) Use the substitution $y = xu$ to solve

$\frac{dy}{dx} = \frac{2y + x}{y - 2x}$

given that the solution curve passes through the point $(1, 1)$.

Give your answer in the form of a quadratic in $x$ and $y$.

(ii) Using the substitutions $x = X + a$ and $y = Y + b$ for appropriate values of $a$ and $b$, or otherwise, solve

$\frac{dy}{dx} = \frac{x - 2y - 4}{2x + y - 3},$

given that the solution curve passes through the point $(1, 1)$.

Hint

Using the substitution $y = xu$, the differential equation can be simplified to

$x \frac{du}{dx} = \frac{1 + 4u - u^2}{u - 2}$

This can be solved by separating the variables after which making the substitution $u = \frac{y}{x}$ and substituting the point on the curve gives the required quadratic in $x$ and $y$.

In part (ii), $\frac{dY}{dX}$ can be shown to be equal to $\frac{dy}{dx}$. The values of $a$ and $b$ need to be chosen so that the right hand side of the differential equation has no constant terms in the numerator or denominator. This leads to the simultaneous equations:

$a - 2b - 4 = 0$

$2a + b - 3 = 0$

Solving these and substituting the values into the differential equation gives $\frac{dY}{dX} = \frac{X - 2Y}{2X + Y}$, and so

$\frac{dX}{dY} = \frac{2X + Y}{X - 2Y}$

This is the same differential equation as in part (i), with $x = Y$ and $y = X$. Most of the solution in part (i) can therefore be applied, but the point on the curve is different, so the constant in the final solution will need to be calculated for this case.

Model Solution

If $y = xu$, then $\frac{dy}{dx} = u + x\frac{du}{dx}$.

Part (i)

Substituting into $\frac{dy}{dx} = \frac{2y + x}{y - 2x}$:

$u + x\frac{du}{dx} = \frac{2xu + x}{xu - 2x} = \frac{2u + 1}{u - 2}$

$x\frac{du}{dx} = \frac{2u + 1}{u - 2} - u = \frac{2u + 1 - u(u - 2)}{u - 2} = \frac{-u^2 + 4u + 1}{u - 2}$

Separating variables:

$\frac{u - 2}{-u^2 + 4u + 1}\,du = \frac{dx}{x}$

Factor the denominator: $-u^2 + 4u + 1 = -(u^2 - 4u - 1) = -(u - 2 - \sqrt{3})(u - 2 + \sqrt{3})$.

Let $v = u - 2$. Then the integrand becomes $\frac{-v}{(v - \sqrt{3})(v + \sqrt{3})}$. Partial fractions:

$\frac{-v}{(v - \sqrt{3})(v + \sqrt{3})} = \frac{A}{v - \sqrt{3}} + \frac{B}{v + \sqrt{3}}$

$A = \frac{-\sqrt{3}}{2\sqrt{3}} = -\frac{1}{2}$, $\quad B = \frac{\sqrt{3}}{-2\sqrt{3}} = -\frac{1}{2}$

So: $\int \left(\frac{-1/2}{v - \sqrt{3}} + \frac{-1/2}{v + \sqrt{3}}\right)dv = \ln|x| + C$

$-\frac{1}{2}\ln|v - \sqrt{3}| - \frac{1}{2}\ln|v + \sqrt{3}| = \ln|x| + C$

$-\frac{1}{2}\ln|v^2 - 3| = \ln|x| + C$

$\ln|v^2 - 3| = -2\ln|x| - 2C = \ln(x^{-2}) + C'$

$|v^2 - 3| = \frac{K}{x^2}$

Substituting $v = u - 2 = \frac{y}{x} - 2 = \frac{y - 2x}{x}$:

$\left(\frac{y - 2x}{x}\right)^2 - 3 = \frac{K}{x^2}$

$(y - 2x)^2 - 3x^2 = K$

$y^2 - 4xy + 4x^2 - 3x^2 = K$

$x^2 - 4xy + y^2 = K$

Applying the initial condition $(1, 1)$: $1 - 4 + 1 = K$, so $K = -2$.

$x^2 - 4xy + y^2 = -2$

or equivalently: $y^2 - 4xy + x^2 + 2 = 0$. $\square$

Part (ii)

With $x = X + a$, $y = Y + b$: $\frac{dy}{dx} = \frac{dY}{dX}$. The DE becomes:

$\frac{dY}{dX} = \frac{(X + a) - 2(Y + b) - 4}{2(X + a) + (Y + b) - 3} = \frac{X - 2Y + (a - 2b - 4)}{2X + Y + (2a + b - 3)}$

Set constant terms to zero:

$a - 2b = 4 \qquad 2a + b = 3$

From the first: $a = 4 + 2b$. Substituting: $8 + 4b + b = 3 \implies b = -1$, $a = 2$.

The DE becomes $\frac{dY}{dX} = \frac{X - 2Y}{2X + Y}$, which is the same form as part (i) (with $X$ and $Y$ in place of $x$ and $y$).

From part (i), the solution is $Y^2 - 4XY + X^2 = K$.

The point $(x, y) = (1, 1)$ corresponds to $(X, Y) = (1 - 2, 1 - (-1)) = (-1, 2)$.

$4 - 4(-1)(2) + 1 = K \implies K = 13$

Substituting back $X = x - 2$, $Y = y + 1$:

$(y + 1)^2 - 4(x - 2)(y + 1) + (x - 2)^2 = 13$

Expanding:

$y^2 + 2y + 1 - 4(xy + x - 2y - 2) + x^2 - 4x + 4 = 13$

$y^2 + 2y + 1 - 4xy - 4x + 8y + 8 + x^2 - 4x + 4 = 13$

$x^2 - 4xy + y^2 - 8x + 10y = 0$

$y^2 - 4xy + x^2 - 8x + 10y = 0 \qquad \square$

Examiner Notes

This was the most popular question on the paper and the question which had the highest average score. Most candidates correctly solved the differential equation in the first part of the question, but many then calculated the constant term incorrectly. In the second part of the question most candidates were able to find the appropriate values of a and b, but then did not see how to apply the result from part (i) and so did the integration again or just copied the answer from the first part. Some candidates again struggled to obtain the correct constant for the integration and others did not substitute the correct values for the point on the curve (taking $(X, Y)$ as $(1, 1)$ rather than $(x, y)$).

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Question 6

Topic: 三角函数与级数 (Trigonometric Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 By simplifying $\sin(r + \frac{1}{2})x - \sin(r - \frac{1}{2})x$ or otherwise show that, for $\sin \frac{1}{2}x \neq 0$,

$\cos x + \cos 2x + \dots + \cos nx = \frac{\sin(n + \frac{1}{2})x - \sin \frac{1}{2}x}{2 \sin \frac{1}{2}x}.$

The functions $S_n$, for $n = 1, 2, \dots$, are defined by

$S_n(x) = \sum_{r=1}^{n} \frac{1}{r} \sin rx \quad (0 \leqslant x \leqslant \pi).$

(i) Find the stationary points of $S_2(x)$ for $0 \leqslant x \leqslant \pi$, and sketch this function.

(ii) Show that if $S_n(x)$ has a stationary point at $x = x_0$, where $0 < x_0 < \pi$, then

$\sin nx_0 = (1 - \cos nx_0) \tan \frac{1}{2}x_0$

and hence that $S_n(x_0) \geqslant S_{n-1}(x_0)$. Deduce that if $S_{n-1}(x) > 0$ for all $x$ in the interval $0 < x < \pi$, then $S_n(x) > 0$ for all $x$ in this interval.

(iii) Prove that $S_n(x) \geqslant 0$ for $n \geqslant 1$ and $0 \leqslant x \leqslant \pi$.

Hint

One of the standard trigonometric formulas can be used to show that

$\sin \left( r + \frac{1}{2} \right) x - \sin \left( r - \frac{1}{2} \right) x = 2 \cos rx \sin \frac{1}{2} x.$

Summing these from $r = 1$ to $r = n$ will then give the required result.

In part (i), the definition can be rewritten as $S_2(x) = \sin x + \frac{1}{2} \sin 2x$. The stationary points can then be evaluated by differentiating the function. The sketch is then easy to complete.

For part (ii), differentiating the function gives $S'_n(x) = \cos x + \cos 2x + \dots + \cos nx$. Applying the result from the start of the question, this can be written as

$S'_n(x) = \frac{\sin(n + \frac{1}{2}x) - \sin \frac{1}{2}x}{2 \sin \frac{1}{2}x}$

Since $\sin \frac{1}{2}x \neq 0$ in the given range, the stationary points are where $\sin (n + \frac{1}{2})x - \sin \frac{1}{2}x = 0$. This can then be simplified to the required form by splitting $(n + \frac{1}{2})x$ into functions of $nx$ and $\frac{1}{2}x$ and noting that $\sin \frac{1}{2}x \neq 0$ and $\cos \frac{1}{2}x \neq 0$ in the given range, so both can be divided by. By noting that the difference between $S_{n-1}(x)$ and $S_n(x)$ is $\frac{1}{n} \sin nx$ the result just shown can be used to show the final result of part (ii). Part (iii) then follows by induction.

Model Solution

Preliminary result: Using the identity $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$:

$\sin\!\left(r + \tfrac{1}{2}\right)x - \sin\!\left(r - \tfrac{1}{2}\right)x = 2\cos(rx)\sin\!\tfrac{x}{2}$

Summing from $r = 1$ to $r = n$:

$\sum_{r=1}^{n} 2\cos(rx)\sin\!\tfrac{x}{2} = \sum_{r=1}^{n}\left[\sin\!\left(r + \tfrac{1}{2}\right)x - \sin\!\left(r - \tfrac{1}{2}\right)x\right]$

The right side telescopes:

$= \sin\!\left(n + \tfrac{1}{2}\right)x - \sin\!\tfrac{x}{2}$

Dividing by $2\sin\frac{x}{2}$ (which is nonzero for $\sin\frac{x}{2} \neq 0$):

$\cos x + \cos 2x + \cdots + \cos nx = \frac{\sin(n + \frac{1}{2})x - \sin\frac{x}{2}}{2\sin\frac{x}{2}} \qquad \square$

Part (i)

$S_2(x) = \sin x + \frac{1}{2}\sin 2x$.

$S_2'(x) = \cos x + \cos 2x$

Using $\cos 2x = 2\cos^2 x - 1$:

$S_2'(x) = 2\cos^2 x + \cos x - 1 = (2\cos x - 1)(\cos x + 1)$

Setting $S_2'(x) = 0$: $\cos x = \frac{1}{2}$ (giving $x = \frac{\pi}{3}$) or $\cos x = -1$ (giving $x = \pi$).

At $x = 0$: $S_2(0) = 0$. At $x = \frac{\pi}{3}$: $S_2 = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$. At $x = \pi$: $S_2(\pi) = 0$.

The stationary points are $x = \frac{\pi}{3}$ (maximum) and $x = \pi$ (endpoint minimum).

Sketch: $S_2$ starts at $0$, rises to $\frac{3\sqrt{3}}{4} \approx 1.30$ at $x = \frac{\pi}{3}$, then falls back to $0$ at $x = \pi$. The curve is smooth and positive on $(0, \pi)$.

Part (ii)

$S_n'(x) = \cos x + \cos 2x + \cdots + \cos nx = \frac{\sin(n + \frac{1}{2})x - \sin\frac{x}{2}}{2\sin\frac{x}{2}}$

At a stationary point $x_0 \in (0, \pi)$, we have $S_n'(x_0) = 0$, so:

$\sin\!\left(n + \tfrac{1}{2}\right)x_0 = \sin\!\tfrac{x_0}{2}$

Since $\sin\frac{x_0}{2} \neq 0$ and $\cos\frac{x_0}{2} \neq 0$ for $x_0 \in (0, \pi)$, we expand $\sin(n + \frac{1}{2})x_0 = \sin(nx_0)\cos\frac{x_0}{2} + \cos(nx_0)\sin\frac{x_0}{2}$:

$\sin(nx_0)\cos\!\tfrac{x_0}{2} + \cos(nx_0)\sin\!\tfrac{x_0}{2} = \sin\!\tfrac{x_0}{2}$

$\sin(nx_0)\cos\!\tfrac{x_0}{2} = \sin\!\tfrac{x_0}{2}(1 - \cos(nx_0))$

Dividing by $\cos\frac{x_0}{2}$:

$\sin(nx_0) = (1 - \cos(nx_0))\tan\!\tfrac{x_0}{2} \qquad \square$

Since $\tan\frac{x_0}{2} > 0$ for $x_0 \in (0, \pi)$ and $1 - \cos(nx_0) \geqslant 0$, we conclude $\sin(nx_0) \geqslant 0$.

Now $S_n(x_0) = S_{n-1}(x_0) + \frac{1}{n}\sin(nx_0)$, so:

$S_n(x_0) \geqslant S_{n-1}(x_0) \qquad \square$

Deduction: If $S_{n-1}(x) > 0$ for all $x \in (0, \pi)$, then at every interior stationary point of $S_n$ we have $S_n(x_0) \geqslant S_{n-1}(x_0) > 0$. Since $S_n(0) = S_n(\pi) = 0$, the minimum of $S_n$ on $[0, \pi]$ is either $0$ (at the endpoints) or occurs at an interior stationary point where $S_n > 0$. In either case, $S_n(x) \geqslant 0$ for $0 \leqslant x \leqslant \pi$, and since $S_n$ is not identically zero, $S_n(x) > 0$ for $0 < x < \pi$. $\square$

Part (iii)

We prove $S_n(x) \geqslant 0$ for all $n \geqslant 1$ and $0 \leqslant x \leqslant \pi$ by induction.

Base case: $S_1(x) = \sin x \geqslant 0$ for $x \in [0, \pi]$. $\checkmark$

Inductive step: Assume $S_{n-1}(x) \geqslant 0$ for $x \in [0, \pi]$.

$S_n$ is continuous on $[0, \pi]$ with $S_n(0) = 0$ and $S_n(\pi) = 0$. If $S_n$ achieves a negative value, its minimum on $[0, \pi]$ must be negative and occur at an interior point $x_0 \in (0, \pi)$ (since $S_n$ is non-negative at the endpoints). At such a minimum, $S_n'(x_0) = 0$.

By part (ii), $S_n(x_0) \geqslant S_{n-1}(x_0) \geqslant 0$ (by the inductive hypothesis). This contradicts $S_n(x_0) < 0$.

Therefore $S_n(x) \geqslant 0$ for all $x \in [0, \pi]$. By induction, the result holds for all $n \geqslant 1$. $\square$

Examiner Notes

This was one of the less popular of the pure maths questions, but the average mark achieved on this paper was one of the highest for the paper. The first section did not present too much difficulty for the majority of candidates, with a variety of methods being used to show the first result such as proof by induction or use of $e^x = \cos x + i \sin x$. In the second part of the question many of the candidates struggled to explain the reasoning clearly to show the required result. Most candidates who reached the final part of the question realised that the previous part provides the basis for a proof by induction.

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Question 7

Topic: 绝对值函数与方程 (Absolute Value Functions & Equations)  |  Difficulty: Standard  |  Marks: 20

Problem

7 (i) The function $f$ is defined by $f(x) = |x - a| + |x - b|$, where $a < b$. Sketch the graph of $f(x)$, giving the gradient in each of the regions $x < a$, $a < x < b$ and $x > b$. Sketch on the same diagram the graph of $g(x)$, where $g(x) = |2x - a - b|$.

What shape is the quadrilateral with vertices $(a, 0)$, $(b, 0)$, $(b, f(b))$ and $(a, f(a))$?

(ii) Show graphically that the equation

$|x - a| + |x - b| = |x - c|,$

where $a < b$, has 0, 1 or 2 solutions, stating the relationship of $c$ to $a$ and $b$ in each case.

(iii) For the equation

$|x - a| + |x - b| = |x - c| + |x - d|,$

where $a < b$, $c < d$ and $d - c < b - a$, determine the number of solutions in the various cases that arise, stating the relationship between $a, b, c$ and $d$ in each case.

Hint

By considering the regions $x \le a$, $a < x < b$ and $x \ge b$, $f(x)$ can be written as

$f(x) = \begin{matrix} a + b - 2x & x \le a \\ b - a & a < x < b \\ 2x - a - b & x \ge b \end{matrix}$

Therefore the graph of $y = f(x)$ will be made up of two sloping sections (with gradients 2 and -2 and a horizontal section). The graph of $y = g(x)$ will have the same definition in the regions $x \le a$ and $x \ge b$, with the sloping edges extending to a point of intersection on the $x$-axis. The quadrilateral with therefore have sides of equal length and right angles at each vertex, so it is a square.

In part (ii), sketches of the cases where $c = a$ and $c = b$ show that these cases give just one solution. If $a < c < b$ there will be no solutions and in the other regions there will be two solutions.

In part (iii) the graphs for the two sides of the equation can be related to graphs of the form of $g(x)$ (apart from the section which is replaced by a horizontal line) in the first part of the question. Since $d - c < b - a$, the horizontal sections of the two graphs must be at different heights so the number of solutions can be seen to be the same as the number of intersections of the graphs of the form of $g(x)$.

Model Solution

Part (i)

For $f(x) = |x - a| + |x - b|$ with $a < b$, consider three regions:

• $x \leqslant a$: $f(x) = (a - x) + (b - x) = a + b - 2x$ (gradient $-2$)
• $a < x < b$: $f(x) = (x - a) + (b - x) = b - a$ (gradient $0$)
• $x \geqslant b$: $f(x) = (x - a) + (x - b) = 2x - a - b$ (gradient $+2$)

The graph is a “tent” shape: two sloped lines meeting a flat section at height $b - a$ over $[a, b]$.

For $g(x) = |2x - a - b| = 2|x - \frac{a+b}{2}|$:

• $x \leqslant \frac{a+b}{2}$: $g(x) = a + b - 2x$ (gradient $-2$)
• $x \geqslant \frac{a+b}{2}$: $g(x) = 2x - a - b$ (gradient $+2$)

The graph of $g$ is a V-shape with vertex at $(\frac{a+b}{2}, 0)$. Note that $g$ and $f$ share the same sloped edges for $x \leqslant a$ and $x \geqslant b$.

The quadrilateral: The vertices are $(a, 0)$, $(b, 0)$, $(b, b-a)$, $(a, b-a)$. This has:

• Bottom edge from $(a,0)$ to $(b,0)$: length $b - a$
• Right edge from $(b,0)$ to $(b,b-a)$: length $b - a$
• Top edge from $(b,b-a)$ to $(a,b-a)$: length $b - a$
• Left edge from $(a,b-a)$ to $(a,0)$: length $b - a$

All sides equal and all angles $90°$, so the quadrilateral is a square. $\square$

Part (ii)

The equation $|x - a| + |x - b| = |x - c|$ asks for intersections of $y = f(x)$ (the tent) with $y = |x - c|$ (a V-shape with vertex at $(c, 0)$).

Since the vertex of $|x - c|$ lies on the $x$-axis (the bottom edge of the square), the V starts at the base of the tent.

• $a < c < b$: The V-vertex is at $(c, 0)$, below the flat section of $f$ (which is at height $b - a > 0$). The V’s slopes ($\pm 1$) are shallower than $f$‘s slopes ($\pm 2$), so the V stays below $f$ everywhere. 0 solutions.

• $c < a$ or $c > b$: The V-vertex is outside $[a, b]$. On the side nearest to $[a, b]$, the V rises with slope $1$ while $f$ rises with slope $2$, so $f$ starts above the V and the V cannot catch up. On the opposite side, the V rises with slope $1$ and $f$ rises with slope $2$, but $f$ starts above the V. However, on the far side of $[a, b]$, $f$ and the V have the same slope directions but $f$ starts at height $b - a > 0$ while the V starts from $0$. The V intersects each sloped edge of $f$ exactly once. 2 solutions.

• $c = a$ or $c = b$: The V-vertex coincides with a corner of the square. One entire sloped edge of $f$ overlaps with one ray of the V. Infinitely many solutions (all $x$ on the overlapping edge).

In summary: 0 solutions if $a < c < b$; 2 solutions if $c < a$ or $c > b$; infinitely many if $c = a$ or $c = b$.

Part (iii)

Let $f(x) = |x - a| + |x - b|$ and $h(x) = |x - c| + |x - d|$. Both are tent-shaped with flat sections of heights $b - a$ and $d - c$ respectively. Since $d - c < b - a$, $f$‘s flat section is higher.

For $x$ outside both intervals, $f$ and $h$ have the same gradient ($\pm 2$), so $f - h$ is constant in those regions. The key is to compare the flat sections and the transitions between regions.

Case 1: $[c, d] \cap [a, b] = \emptyset$ (i.e., $d \leqslant a$ or $b \leqslant c$): In the region between the two intervals, one function is flat and the other is sloped, and they cross. In the outer regions, both have gradient $\pm 2$ with $f$ above $h$. 2 solutions.

Case 2: $a \leqslant c < d \leqslant b$ ($[c,d] \subseteq [a,b]$): The flat section of $h$ (height $d - c$) sits below the flat section of $f$ (height $b - a$). At $x = c$, $f$ is still decreasing (gradient $-2$) while $h$ becomes flat; they cross once on the left. By symmetry, they cross once on the right. 2 solutions.

Case 3: $c < a < b < d$ ($[a,b] \subseteq [c,d]$): The flat section of $h$ (height $d - c$) is lower than $f$‘s flat section (height $b - a$). In $[a, b]$, $f = b - a > d - c = h$. For $x \leqslant c$, both decrease with gradient $-2$ and $h$ is below $f$. For $x \geqslant d$, both increase with gradient $2$ and $h$ is below $f$. 0 solutions.

Case 4: $c < a < d < b$ (partial overlap, $[c,d]$ extends left of $[a,b]$): The flat sections are at different heights ($d - c < b - a$). Depending on the relative positions:

• If $c + d < a + b$: one crossing on the left transition and one below $f$‘s flat section that does not occur (the flat section of $h$ is too low). 1 solution.
• If $c + d = a + b$: the right-hand outer edges coincide for $x \geqslant d$. Infinitely many solutions.
• If $c + d > a + b$: the right-hand outer edges are separated with $f$ below $h$, giving an additional crossing. 2 solutions.

Case 5: $a < c < b < d$ (partial overlap, $[c,d]$ extends right of $[a,b]$): By symmetry with Case 4:

• If $c + d < a + b$: 2 solutions.
• If $c + d = a + b$: Infinitely many solutions.
• If $c + d > a + b$: 1 solution.

Examiner Notes

This was another of the less popular pure maths questions. The nature of this question meant that many solutions involved a series of sketches of graphs with very little written explanation. Most candidates were able to identify that the sloping edges of $y = f(x)$ would have the same gradient as the sloping edges of $y = g(x)$, but many did not have both sloping edges overlapping for the two graphs. In some cases only one sloping edge of $y = g(x)$ was drawn. A large number of candidates who correctly sketched the graphs identified the quadrilateral as a rectangle, rather than a square. In the second part of the question, sketches of the case with one solution often did not have the graph of $y = |x - c|$ meeting the $x$-axis at one corner of the square identified in part (i), although many candidates were able to identify the different cases that could occur. Unfortunately in the final part of the question very few candidates used the result from the first part of the question and so considered a number of possibilities that do not exist for any values of a, b, c and d.

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Question 8

Topic: 二项式展开与组合数学 (Binomial Expansion & Combinatorics)  |  Difficulty: Hard  |  Marks: 20

Problem

8 For positive integers $n, a$ and $b$, the integer $c_r$ ($0 \leqslant r \leqslant n$) is defined to be the coefficient of $x^r$ in the expansion in powers of $x$ of $(a + bx)^n$. Write down an expression for $c_r$ in terms of $r, n, a$ and $b$.

For given $n, a$ and $b$, let $m$ denote a value of $r$ for which $c_r$ is greatest (that is, $c_m \geqslant c_r$ for $0 \leqslant r \leqslant n$).

Show that

$\frac{b(n + 1)}{a + b} - 1 \leqslant m \leqslant \frac{b(n + 1)}{a + b}.$

Deduce that $m$ is either a unique integer or one of two consecutive integers.

Let $G(n, a, b)$ denote the unique value of $m$ (if there is one) or the larger of the two possible values of $m$.

(i) Evaluate $G(9, 1, 3)$ and $G(9, 2, 3)$.

(ii) For any positive integer $k$, find $G(2k, a, a)$ and $G(2k - 1, a, a)$ in terms of $k$.

(iii) For fixed $n$ and $b$, determine a value of $a$ for which $G(n, a, b)$ is greatest.

(iv) For fixed $n$, find the greatest possible value of $G(n, 1, b)$. For which values of $b$ is this greatest value achieved?

Hint

The coefficients from the binomial expansion should be easily written down. It can then be shown that

$\frac{c_{r+1}}{c_r} = \frac{b(n - r)}{a(r + 1)}$

This will be greater than 1 (indicating that the value of $c_r$ is increasing) while $b(n - r) > a(r + 1)$, which simplifies to $r < \frac{nb - a}{a + b}$. Similarly, $\frac{c_{r+1}}{c_r} = 1$ if $r = \frac{nb - a}{a + b}$ and $\frac{c_{r+1}}{c_r} < 1$ if $r > \frac{nb - a}{a + b}$. Therefore the maximum value of $c_r$ will be the first integer after $\frac{nb - a}{a + b}$ (and there will be two maximum values for $c_r$ if $\frac{nb - a}{a + b}$ is an integer. The required inequality summarises this information.

In parts (i) and (ii) the values need to be substituted into the inequality. Where there are two possible values, it needs to be checked that they are equal before taking the higher if this has not been justified in the first case.

In part (iii) the greatest value will be achieved when the denominator takes the smallest possible value, therefore $a = 1$, and then in part (iv) the greatest value will be achieved by maximising the numerator. Since the maximum possible value of $G(n, a, b)$ is $n$, $b \ge n$ will achieve this maximum.

Model Solution

General setup: By the binomial theorem, $(a + bx)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r x^r$, so:

$c_r = \binom{n}{r} a^{n-r} b^r$

Bounding $m$: Consider the ratio:

$\frac{c_{r+1}}{c_r} = \frac{\binom{n}{r+1} a^{n-r-1} b^{r+1}}{\binom{n}{r} a^{n-r} b^r} = \frac{n - r}{r + 1} \cdot \frac{b}{a}$

$c_{r+1} > c_r$ when $\frac{b(n - r)}{a(r + 1)} > 1$, i.e., $bn - br > ar + a$, i.e., $r(a + b) < nb - a$:

$r < \frac{nb - a}{a + b}$

Similarly, $c_{r+1} = c_r$ when $r = \frac{nb - a}{a + b}$ and $c_{r+1} < c_r$ when $r > \frac{nb - a}{a + b}$.

So $c_r$ increases while $r < \frac{nb - a}{a + b}$ and decreases after. The maximum $c_m$ occurs at:

$m = \left\lceil \frac{nb - a}{a + b} \right\rceil$

If $\frac{nb - a}{a + b}$ is an integer, then $c_m = c_{m-1}$ and both $m$ and $m - 1$ are maximizers.

Writing $\frac{nb - a}{a + b} = \frac{nb}{a + b} - \frac{a}{a + b}$, and noting $0 < \frac{a}{a+b} < 1$:

$\frac{nb}{a + b} - 1 < m \leqslant \frac{nb}{a + b} \qquad \square$

$m$ is unique or two consecutive integers: The interval $\left(\frac{nb}{a+b} - 1, \frac{nb}{a+b}\right]$ has length $1$. If $\frac{nb}{a+b}$ is not an integer, it contains exactly one integer (namely $\lceil \frac{nb}{a+b} \rceil$). If $\frac{nb}{a+b}$ is an integer, it contains exactly two integers: $\frac{nb}{a+b}$ and $\frac{nb}{a+b} - 1$. $\square$

Part (i)

$G(9, 1, 3)$: $\frac{nb}{a+b} = \frac{9 \times 3}{1 + 3} = \frac{27}{4} = 6.75$.

The interval is $(5.75, 6.75]$, so $m = 6$. Since $6.75$ is not an integer, $m$ is unique.

$G(9, 2, 3)$: $\frac{nb}{a+b} = \frac{9 \times 3}{2 + 3} = \frac{27}{5} = 5.4$.

The interval is $(4.4, 5.4]$, so $m = 5$. Since $5.4$ is not an integer, $m$ is unique. $\square$

Part (ii)

$G(2k, a, a)$: $\frac{nb - a}{a+b} = \frac{2ka - a}{2a} = k - \frac{1}{2}$.

This is not an integer, so $m = \lceil k - \frac{1}{2} \rceil = k$ is unique. $G(2k, a, a) = k$. $\square$

$G(2k - 1, a, a)$: We use the condition on $\frac{nb - a}{a+b}$:

$\frac{nb - a}{a+b} = \frac{(2k-1)a - a}{2a} = k - 1$

This is an integer, so $c_{k-1} = c_k$ (both are maximizers). $G$ takes the larger: $G(2k - 1, a, a) = k$. $\square$

Part (iii)

$G(n, a, b) = \left\lceil \frac{nb}{a + b} \right\rceil$ (taking the larger value when there are two maximizers).

For fixed $n$ and $b$, to maximize $G$ we need to maximize $\frac{nb}{a + b}$. Since $nb$ is fixed, we minimize $a + b$, which means minimizing $a$. Since $a$ is a positive integer, $a = 1$.

$G(n, 1, b) = \left\lceil \frac{nb}{1 + b} \right\rceil \qquad \square$

Part (iv)

$G(n, 1, b) = \left\lceil \frac{nb}{b + 1} \right\rceil$. Note that $\frac{nb}{b+1} = n - \frac{n}{b+1}$.

This is maximized when $\frac{n}{b+1}$ is minimized, i.e., when $b + 1$ divides $n$ and $b + 1$ is as large as possible. But even without divisibility, $\frac{nb}{b+1} \to n$ as $b \to \infty$.

For $G(n, 1, b) = n$, we need $\frac{nb}{b+1} > n - 1$, i.e., $nb > (n-1)(b+1) = nb + n - b - 1$, i.e., $b > n - 1$, i.e., $b \geqslant n$.

When $b \geqslant n$: $\frac{nb}{b+1} \geqslant \frac{n^2}{n+1} = n - \frac{n}{n+1} > n - 1$, so $G(n, 1, b) = n$.

When $b < n$: $\frac{nb}{b+1} < \frac{n(n-1)}{n} = n - 1$, so $G(n, 1, b) \leqslant n - 1$.

Therefore the greatest possible value of $G(n, 1, b)$ is $n$, achieved for all $b \geqslant n$. $\square$

Examiner Notes

This was the least popular of the pure maths questions and also the one with the lowest average score. Many of the candidates were able to show the required result at the start of the question, although very few candidates explained that $m$ could be either of the two integers when the range included two integers. Parts (i) and (ii) were then quite straightforward for most candidates, although many calculated the range of values but did not justify their choice in the case where there were two possibilities. In the final two parts of the question some candidates mistakenly chose the value 0 when asked for a positive integer.