STEP3 2023 -- Pure Mathematics

STEP3 2023 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2023  |  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	纯数	Challenging	坐标几何,切线方程,判别式,韦达定理,参数化方法
2	纯数	Challenging	极坐标面积公式,三角函数积分,极限计算,渐近分析
3	纯数	Hard	复数几何意义,韦达定理(三次方程),曲线草图,不等式证明
4	纯数	Hard	de Moivre定理,二项式展开,代入法求系数,多项式因式分解
5	纯数	Challenging	因数分解,素数性质,韦达定理,不等式推理,反证法
6	纯数	Challenging	Maclaurin级数展开,不等式证明,导数分析,驻点判定,函数图像
7	纯数	Challenging	积分换元,分部积分,完全平方式构造,非负函数性质
8	纯数	Hard	特征方程,分段求解,连续可微条件,函数对称性,递推构造

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Question 1

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

1 The distinct points $P(2ap, ap^2)$ and $Q(2aq, aq^2)$ lie on the curve $x^2 = 4ay$, where $a > 0$.

(i) Given that

$(p + q)^2 = p^2q^2 + 6pq + 5 , \qquad (\ast)$

show that the line through $P$ and $Q$ is a tangent to the circle with centre $(0, 3a)$ and radius $2a$.

(ii) Show that, for any given value of $p$ with $p^2 \neq 1$, there are two distinct real values of $q$ that satisfy equation $(\ast)$.

Let these values be $q_1$ and $q_2$. Find expressions, in terms of $p$, for $q_1 + q_2$ and $q_1q_2$.

(iii) Show that, for any given value of $p$ with $p^2 \neq 1$, there is a triangle with one vertex at $P$ such that all three vertices lie on the curve $x^2 = 4ay$ and all three sides are tangents to the circle with centre $(0, 3a)$ and radius $2a$.

Hint

(i) The line through $P$ and $Q$ is $\frac{y - ap^2}{x - 2ap} = \frac{y - aq^2}{x - 2aq}$

$\frac{y - ap^2}{x - 2ap} = \frac{ap^2 - aq^2}{2ap - 2aq} = \frac{p + q}{2}$

$\frac{y - aq^2}{x - 2aq} = \frac{ap^2 - aq^2}{2ap - 2aq} = \frac{p + q}{2}$

or multiplied to remove denominators.

M1

$(y - ap^2)(x - 2aq) = (y - aq^2)(x - 2ap)$

$(2ap - 2aq)y + 2a^2(p^2q - pq^2) = (ap^2 - aq^2)x$

$P$ and $Q$ are distinct thus $p \neq q$ and so $p - q \neq 0$

Therefore $2y + 2apq = (p + q)x$ that is $(p + q)x - 2y - 2apq = 0$

A1

The perpendicular distance of $(0, 3a)$ from the line $PQ$ is $2a$ which requires

$\left| \frac{-6a - 2apq}{\sqrt{((p + q)^2 + 4)}} \right| = 2a$

M1 A1 A1 A1

that is $(pq + 3)^2 = (p + q)^2 + 4$ M1 A1

i.e. $(p + q)^2 = p^2q^2 + 6pq + 9 - 4 = p^2q^2 + 6pq + 5$ $(\ast)$

A1* (9)

(I) $(p + q)x - 2y - 2apq = 0$ meets $x^2 + (y - 3a)^2 = 4a^2$ when

$4(y + apq)^2 + (p + q)^2(y^2 - 6ay + 5a^2) = 0$

M1 A1

$(4 + (p + q)^2)y^2 - (6a(p + q)^2 - 8apq)y + (5a^2(p + q)^2 + 4a^2p^2q^2) = 0$

A1

Thus using $(p + q)^2 = p^2q^2 + 6pq + 5$ , M1

$(pq + 3)^2y^2 - 2a(pq + 3)(3pq + 5)y + a^2(3pq + 5)^2 = 0$

which is a perfect square, A1

so $(pq + 3)y - a(3pq + 5) = 0$ which only has a single root so the line is a tangent. A1 A1*

(II) Foot of perpendicular from $(0,3a)$ to $(p + q)x - 2y - 2apq = 0$ is at intersection with $(p + q)y + 2x = 3a(p + q)$ M1 A1

So solving $(p + q)^2y + 4y + 4apq = 3a(p + q)^2$ A1

$y = \frac{3a(p+q)^2-4apq}{(p+q)^2+4} \text{ and } x = \frac{(p+q)}{2} \left( 3a - \frac{3a(p+q)^2-4apq}{(p+q)^2+4} \right) = \frac{2a(p+q)(3+pq)}{(p+q)^2+4}$

A1

and so the square of the distance is M1 A1

$\left[ \frac{2a(p+q)(3+pq)}{(p+q)^2+4} \right]^2 + \left[ \frac{4a(3+pq)}{(p+q)^2+4} \right]^2 = \left[ \frac{2a(3+pq)}{(p+q)^2+4} \right]^2 (4 + (p+q)^2)$ $= \frac{4a^2(3+pq)^2}{4+(p+q)^2} = 4a^2$

using given condition. A1*

(III) Method is possible by differentiation of circle equation. Partial or incorrect solution by this method zero marks; completely correct solution full marks; completely correct solution except minor inaccuracy, withhold one accuracy mark and final accuracy mark.

(ii) $(\ast)$ can be re-written

$q^2(p^2 - 1) + 4pq + (5 - p^2) = 0$ M1

Considering this as a quadratic equation for $q$, to be two distinct roots, $p^2 - 1 \neq 0$ (it is given that $p^2 \neq 1$) E1 and the discriminant needs to be positive.

$16p^2 - 4(p^2 - 1)(5 - p^2) = 4(p^4 - 2p^2 + 5) = 4(p^2 - 1)^2 + 16 > 0$

as required. M1 A1

$q_1 + q_2 = \frac{-4p}{(p^2-1)}, \quad q_1q_2 = \frac{(5-p^2)}{(p^2-1)} \quad \text{A1 A1 (6)}$

(iii) Given $P$, with $p^2 \neq 1$, by (ii) points $Q_1$ and $Q_2$ can be defined with parameters $q_1$ and $q_2$ where $q_1$ and $q_2$ are the roots of $(\ast)$. So by (i), $PQ_1$ and $PQ_2$ are tangents to the circle centre $(0,3a)$ radius $2a$. E1

The perpendicular distance of $(0,3a)$ from the line $Q_1Q_2$ is

$\left| \frac{-6a - 2aq_1q_2}{\sqrt{((q_1 + q_2)^2 + 4)}} \right| = \left| \frac{-6a - 2a \left( \frac{5-p^2}{p^2-1} \right)}{\sqrt{\left( \left( \frac{-4p}{p^2-1} \right)^2 + 4 \right)}} \right| = 2a \left| \frac{3(p^2-1) + (5-p^2)}{\sqrt{16p^2 + 4(p^2-1)^2}} \right|$

M1 A1

$= 2a \left| \frac{2p^2 + 2}{\sqrt{4p^4 + 16p^2 + 4}} \right| = 2a$

A1

Alternative $Q_1Q_2$ is the third such line provided that $(q_1q_2 + 3)^2 = (q_1 + q_2)^2 + 4$

Diagrammatically, take $a$ as the midpoint of the base of the isosceles triangle, take $b$ as the vector from this midpoint to the remaining vertex, and take $s$ as the ratio of half the base length to the altitude length.

M1 A1 A1

Thus $PQ_1Q_2$ is the triangle required.

Model Solution

Part (i)

The gradient of line $PQ$ is

$m = \frac{ap^2 - aq^2}{2ap - 2aq} = \frac{a(p - q)(p + q)}{2a(p - q)} = \frac{p + q}{2}$

since $p \neq q$ (the points are distinct). The equation of line $PQ$ through $P$ is

$y - ap^2 = \frac{p + q}{2}(x - 2ap)$

Rearranging:

$2y - 2ap^2 = (p + q)x - 2ap(p + q)$

$(p + q)x - 2y - 2ap(p + q) + 2ap^2 = 0$

$(p + q)x - 2y - 2apq = 0$

The perpendicular distance from the centre $(0, 3a)$ of the circle to this line is

$d = \frac{|(p + q)(0) - 2(3a) - 2apq|}{\sqrt{(p + q)^2 + 4}} = \frac{|-6a - 2apq|}{\sqrt{(p + q)^2 + 4}} = \frac{2a|pq + 3|}{\sqrt{(p + q)^2 + 4}}$

For the line to be tangent to the circle of radius $2a$, we require $d = 2a$:

$\frac{2a|pq + 3|}{\sqrt{(p + q)^2 + 4}} = 2a$

$|pq + 3| = \sqrt{(p + q)^2 + 4}$

Squaring both sides:

$(pq + 3)^2 = (p + q)^2 + 4$

Expanding the left side:

$p^2q^2 + 6pq + 9 = (p + q)^2 + 4$

$(p + q)^2 = p^2q^2 + 6pq + 5$

This is precisely the given condition. Therefore, the line through $P$ and $Q$ is tangent to the circle with centre $(0, 3a)$ and radius $2a$. $\qquad \blacksquare$

Part (ii)

Expanding the given condition $(p + q)^2 = p^2q^2 + 6pq + 5$:

$p^2 + 2pq + q^2 = p^2q^2 + 6pq + 5$

$q^2 - p^2q^2 + 2pq - 6pq + p^2 - 5 = 0$

$q^2(1 - p^2) - 4pq + (p^2 - 5) = 0$

$q^2(p^2 - 1) + 4pq + (5 - p^2) = 0$

This is a quadratic equation in $q$. Since $p^2 \neq 1$, the coefficient of $q^2$ is non-zero. For two distinct real roots, we need the discriminant to be strictly positive:

$\Delta = (4p)^2 - 4(p^2 - 1)(5 - p^2)$

$= 16p^2 - 4(5p^2 - p^4 - 5 + p^2)$

$= 16p^2 + 4p^4 - 24p^2 + 20$

$= 4p^4 - 8p^2 + 20$

$= 4(p^4 - 2p^2 + 1) + 16 = 4(p^2 - 1)^2 + 16$

Since $4(p^2 - 1)^2 \geq 0$, we have $\Delta \geq 16 > 0$ for all real $p$. Therefore, for any given $p$ with $p^2 \neq 1$, there are two distinct real values of $q$ satisfying the given equation.

By Vieta’s formulae applied to the quadratic in $q$ (with coefficients $A = p^2 - 1$, $B = 4p$, $C = 5 - p^2$):

$q_1 + q_2 = -\frac{4p}{p^2 - 1}, \qquad q_1 q_2 = \frac{5 - p^2}{p^2 - 1}$

Part (iii)

Given $P(2ap, ap^2)$ with $p^2 \neq 1$, part (ii) provides two distinct real values $q_1, q_2$ satisfying the given condition. Define $Q_1(2aq_1, aq_1^2)$ and $Q_2(2aq_2, aq_2^2)$ on the curve $x^2 = 4ay$.

By part (i), since each $q_i$ satisfies the condition with $p$, the lines $PQ_1$ and $PQ_2$ are both tangent to the circle. Moreover, $PQ_1$ and $PQ_2$ are distinct lines (since $q_1 \neq q_2$ gives different slopes $(p + q_1)/2 \neq (p + q_2)/2$).

It remains to show that $Q_1Q_2$ is also tangent to the circle. By the same criterion as part (i), the line through $Q_1$ and $Q_2$ is tangent if and only if

$(q_1 q_2 + 3)^2 = (q_1 + q_2)^2 + 4$

Using the expressions from part (ii):

$q_1 q_2 + 3 = \frac{5 - p^2}{p^2 - 1} + 3 = \frac{5 - p^2 + 3(p^2 - 1)}{p^2 - 1} = \frac{2(p^2 + 1)}{p^2 - 1}$

$(q_1 q_2 + 3)^2 = \frac{4(p^2 + 1)^2}{(p^2 - 1)^2}$

$(q_1 + q_2)^2 + 4 = \frac{16p^2}{(p^2 - 1)^2} + 4 = \frac{16p^2 + 4(p^2 - 1)^2}{(p^2 - 1)^2}$

$= \frac{4p^4 + 8p^2 + 4}{(p^2 - 1)^2} = \frac{4(p^2 + 1)^2}{(p^2 - 1)^2}$

These are equal, so $Q_1Q_2$ is tangent to the circle.

The line $Q_1Q_2$ cannot pass through $P$: if it did, $P$ would lie on a tangent to the circle through $Q_1$ and $Q_2$, meaning $PQ_1$ and $Q_1Q_2$ would be the same line, contradicting the fact that $PQ_1$ and $PQ_2$ are distinct. Therefore $P$, $Q_1$, $Q_2$ are non-collinear and form a triangle with all three sides tangent to the circle with centre $(0, 3a)$ and radius $2a$. $\qquad \blacksquare$

Examiner Notes

最受欢迎题目之一(93%作答),平均分9.5/20。常见错误:使用距离公式时漏掉绝对值符号;第(iii)部分很多考生重复前面的计算而非利用(i)(ii)的结果,解法不够简洁。

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Question 2

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

2 The polar curves $C_1$ and $C_2$ are defined for $0 \leqslant \theta \leqslant \pi$ by

$r = k(1 + \sin \theta)$

$r = k + \cos \theta$

respectively, where $k$ is a constant greater than 1.

(i) Sketch the curves on the same diagram. Show that if $\theta = \alpha$ at the point where the curves intersect, $\tan \alpha = \frac{1}{k}$.

(ii) The region A is defined by the inequalities

$0 \leqslant \theta \leqslant \alpha \quad \text{and} \quad r \leqslant k(1 + \sin \theta).$

Show that the area of A can be written as

$\frac{k^2}{4}(3\alpha - \sin \alpha \cos \alpha) + k^2(1 - \cos \alpha).$

(iii) The region B is defined by the inequalities

$\alpha \leqslant \theta \leqslant \pi \quad \text{and} \quad r \leqslant k + \cos \theta.$

Find an expression in terms of $k$ and $\alpha$ for the area of B.

(iv) The total area of regions A and B is denoted by $R$. The area of the region enclosed by $C_1$ and the lines $\theta = 0$ and $\theta = \pi$ is denoted by $S$. The area of the region enclosed by $C_2$ and the lines $\theta = 0$ and $\theta = \pi$ is denoted by $T$.

Show that as $k \to \infty$, $\frac{R}{T} \to 1$

and find the limit of $\frac{R}{S}$

as $k \to \infty$.

Hint

G1 G1 G1

At intersection, when $\theta = \alpha$ , $k(1 + \sin \theta) = k + \cos \theta$

Therefore, $k \sin \alpha = \cos \alpha$ , that is, $\tan \alpha = \frac{1}{k}$ B1* (5)

(ii) Area A is

$\frac{1}{2} \int_{0}^{\alpha} (k(1 + \sin \theta))^2 d\theta = \frac{k^2}{2} \int_{0}^{\alpha} 1 + 2 \sin \theta + \sin^2 \theta \ d\theta$

M1

$= \frac{k^2}{2} \int_{0}^{\alpha} 1 + 2 \sin \theta + \frac{1 - \cos 2\theta}{2} \ d\theta = \frac{k^2}{2} \left[ \frac{3}{2}\theta - 2 \cos \theta - \frac{1}{4} \sin 2\theta \right]_{0}^{\alpha}$

dM1 A1

$= \frac{k^2}{2} \left\{ \frac{3}{2}\alpha - 2 \cos \alpha - \frac{1}{4} \sin 2\alpha + 2 \right\} = \frac{k^2}{2} \left\{ \frac{3}{2}\alpha - 2 \cos \alpha - \frac{1}{2} \sin \alpha \cos \alpha + 2 \right\}$

$= \frac{k^2}{4} (3\alpha - \sin \alpha \cos \alpha) + k^2(1 - \cos \alpha)$

A1* (4)

(iii) Area B is $\frac{1}{2} \int_{\alpha}^{\pi} (k + \cos \theta)^2 d\theta = \frac{1}{2} \int_{\alpha}^{\pi} k^2 + 2k \cos \theta + \cos^2 \theta \ d\theta = \frac{1}{2} \int_{\alpha}^{\pi} k^2 + 2k \cos \theta + \frac{1 + \cos 2\theta}{2} \ d\theta$

M1 M1

$= \frac{1}{2} \left[ k^2\theta + 2k \sin \theta + \frac{1}{2}\theta + \frac{1}{4} \sin 2\theta \right]_{\alpha}^{\pi} = \frac{1}{2} \left\{ k^2\pi + \frac{\pi}{2} - k^2\alpha - 2k \sin \alpha - \frac{\alpha}{2} - \frac{1}{4} \sin 2\alpha \right\}$

A1

$= \frac{1}{2} \left\{ k^2 \pi + \frac{\pi}{2} - k^2 \alpha - 2k \sin \alpha - \frac{\alpha}{2} - \frac{1}{2} \sin \alpha \cos \alpha \right\}$

$= \frac{1}{4} \left\{ 2k^2 \pi + \pi - 2k^2 \alpha - 4k \sin \alpha - \alpha - \sin \alpha \cos \alpha \right\}$

A1 (4)

(iv) As $k \to \infty$, $\alpha$ is small as $\tan \alpha = \frac{1}{k}$ so $\alpha \approx \sin \alpha \approx \tan \alpha = \frac{1}{k}$ and $\cos \alpha \approx 1 - \frac{1}{2k^2}$ M1

Area A is $\frac{k}{2} +$ terms of lower order in $k$ A1

Area B is $\frac{k^2 \pi}{2} +$ terms of lower order in $k$ A1

So, area R is $\frac{k^2 \pi}{2} +$ terms of lower order in $k$

Area T is

$\frac{1}{2} \int_{0}^{\pi} (k + \cos \theta)^2 d\theta = \frac{1}{4} (2k^2 \pi + \pi)$

or alternatively, use of result from (iii) with $\alpha = 0$

which is $\frac{k^2 \pi}{2} +$ terms of lower order in $k$ B1

Thus, as required,

$\frac{area\ of\ R}{area\ of\ T} = \frac{\frac{k^2 \pi}{2} + \text{terms of lower order in } k}{\frac{k^2 \pi}{2} + \text{terms of lower order in } k} \to 1$

E1

Area S is

$\frac{1}{2} \int_{0}^{\pi} (k(1 + \sin \theta))^2 d\theta = \frac{k^2}{4} \times 3\pi + 2k^2 = k^2 \left( \frac{3\pi}{4} + 2 \right)$

or alternatively, use of result from (ii) with $\alpha = \pi$

B1

Thus

$\frac{area\ of\ R}{area\ of\ S} \to \frac{\frac{\pi}{2}}{\left( \frac{3\pi}{4} + 2 \right)} = \frac{2\pi}{3\pi + 8}$

A1 (7)

Model Solution

Part (i)

$C_1: r = k(1 + \sin\theta)$ is a cardioid-like curve. At $\theta = 0$, $r = k$; at $\theta = \pi/2$, $r = 2k$ (maximum); at $\theta = \pi$, $r = k$. It is symmetric about $\theta = \pi/2$ and always positive since $k > 1$.

$C_2: r = k + \cos\theta$ is roughly circular. At $\theta = 0$, $r = k + 1$ (maximum); at $\theta = \pi/2$, $r = k$; at $\theta = \pi$, $r = k - 1 > 0$.

At $\theta = 0$: $r_1 = k < k + 1 = r_2$, so $C_1$ is inside $C_2$. At $\theta = \pi/2$: $r_1 = 2k > k = r_2$, so $C_1$ is outside $C_2$. Therefore the curves cross once in $(0, \pi/2)$ at $\theta = \alpha$.

At the intersection, $k(1 + \sin\alpha) = k + \cos\alpha$:

$k\sin\alpha = \cos\alpha$

$\tan\alpha = \frac{1}{k}$

Part (ii)

$A = \frac{1}{2}\int_0^\alpha [k(1 + \sin\theta)]^2 \, d\theta = \frac{k^2}{2}\int_0^\alpha (1 + \sin\theta)^2 \, d\theta$

Expanding the integrand:

$(1 + \sin\theta)^2 = 1 + 2\sin\theta + \sin^2\theta$

Using $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$:

$= \frac{3}{2} + 2\sin\theta - \frac{\cos 2\theta}{2}$

Integrating:

$\int_0^\alpha \left(\frac{3}{2} + 2\sin\theta - \frac{\cos 2\theta}{2}\right) d\theta = \left[\frac{3\theta}{2} - 2\cos\theta - \frac{\sin 2\theta}{4}\right]_0^\alpha$

$= \frac{3\alpha}{2} - 2\cos\alpha - \frac{\sin 2\alpha}{4} + 2$

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

$= \frac{3\alpha}{2} - 2\cos\alpha - \frac{\sin\alpha\cos\alpha}{2} + 2$

Therefore:

$A = \frac{k^2}{2}\left(\frac{3\alpha}{2} - 2\cos\alpha - \frac{\sin\alpha\cos\alpha}{2} + 2\right) = \frac{k^2}{4}(3\alpha - \sin\alpha\cos\alpha) + k^2(1 - \cos\alpha)$

Part (iii)

$B = \frac{1}{2}\int_\alpha^\pi (k + \cos\theta)^2 \, d\theta$

Expanding:

$(k + \cos\theta)^2 = k^2 + 2k\cos\theta + \cos^2\theta$

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:

$= k^2 + \frac{1}{2} + 2k\cos\theta + \frac{\cos 2\theta}{2}$

Integrating:

$\int_\alpha^\pi \left(k^2 + \frac{1}{2} + 2k\cos\theta + \frac{\cos 2\theta}{2}\right) d\theta = \left[k^2\theta + \frac{\theta}{2} + 2k\sin\theta + \frac{\sin 2\theta}{4}\right]_\alpha^\pi$

At $\theta = \pi$: $k^2\pi + \frac{\pi}{2} + 0 + 0$. At $\theta = \alpha$: $k^2\alpha + \frac{\alpha}{2} + 2k\sin\alpha + \frac{\sin 2\alpha}{4}$.

$B = \frac{1}{2}\left(k^2(\pi - \alpha) + \frac{\pi - \alpha}{2} - 2k\sin\alpha - \frac{\sin 2\alpha}{4}\right)$

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

$B = \frac{k^2(\pi - \alpha)}{2} + \frac{\pi - \alpha}{4} - k\sin\alpha - \frac{\sin\alpha\cos\alpha}{4}$

Part (iv)

Since $\tan\alpha = 1/k$ and $k > 1$, we have $\alpha \in (0, \pi/4)$. As $k \to \infty$, $\alpha \to 0$.

Area of $T$: Using the result from (iii) with $\alpha = 0$:

$T = \frac{1}{2}\int_0^\pi (k + \cos\theta)^2 \, d\theta = \frac{k^2\pi}{2} + \frac{\pi}{4}$

Area of $S$: Using the result from (ii) with $\alpha = \pi$:

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

Behaviour of $A$ as $k \to \infty$: Using $\sin\alpha = \frac{1}{\sqrt{k^2+1}}$ and $\cos\alpha = \frac{k}{\sqrt{k^2+1}}$:

$\frac{k^2(3\alpha - \sin\alpha\cos\alpha)}{4} = \frac{3k^2\alpha}{4} - \frac{k^3}{4(k^2+1)}$

Since $\alpha = \arctan(1/k) \sim 1/k$ as $k \to \infty$, the first term $\sim 3k/4$ and the second $\sim k/4$, so this expression $\sim k/2$.

$k^2(1 - \cos\alpha) = k^2\left(1 - \frac{k}{\sqrt{k^2+1}}\right) = \frac{k^2}{\sqrt{k^2+1}\left(\sqrt{k^2+1} + k\right)} \to \frac{1}{2}$

Therefore Area $A \sim k/2$ as $k \to \infty$.

Behaviour of $B$ as $k \to \infty$: The leading term is $\frac{k^2(\pi - \alpha)}{2} \sim \frac{k^2\pi}{2}$. The remaining terms $(\pi - \alpha)/4$, $k\sin\alpha$, and $\sin\alpha\cos\alpha/4$ are bounded. Therefore Area $B = \frac{k^2\pi}{2} + O(k)$.

The ratio $R/T$:

$R = A + B = \frac{k^2\pi}{2} + O(k)$

$\frac{R}{T} = \frac{\frac{k^2\pi}{2} + O(k)}{\frac{k^2\pi}{2} + \frac{\pi}{4}} \to 1 \text{ as } k \to \infty$

The ratio $R/S$:

$\frac{R}{S} = \frac{\frac{k^2\pi}{2} + O(k)}{k^2\left(\frac{3\pi}{4} + 2\right)} \to \frac{\pi/2}{3\pi/4 + 2} = \frac{2\pi}{3\pi + 8}$

Examiner Notes

全卷最成功的题目,平均分12+/20。第(iv)部分最难,关键观察 α→0 需用 tanα=1/k 严格论证。常见错误:未注意 α 依赖于 k,过早代入 α=0 导致错误结论如 k·sinα→0。

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Question 3

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

3 (i) Show that, if $a$ and $b$ are complex numbers, with $b \neq 0$, and $s$ is a positive real number, then the points in the Argand diagram representing the complex numbers $a + sbi$, $a - sbi$ and $a + b$ form an isosceles triangle.

Given three points which form an isosceles triangle in the Argand diagram, explain with the aid of a diagram how to determine the values of $a$, $b$ and $s$ so that the vertices of the triangle represent complex numbers $a + sbi$, $a - sbi$ and $a + b$.

(ii) Show that, if the roots of the equation $z^3 + pz + q = 0$, where $p$ and $q$ are complex numbers, are represented in the Argand diagram by the vertices of an isosceles triangle, then there is a non-zero real number $s$ such that

$\frac{p^3}{q^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}.$

(iii) Sketch the graph $y = \frac{(3x - 1)^3}{(9x + 1)^2}$, identifying any stationary points.

(iv) Show that if the roots of the equation $z^3 + pz + q = 0$ are represented in the Argand diagram by the vertices of an isosceles triangle then $\frac{p^3}{q^2}$ is a real number and $\frac{p^3}{q^2} > -\frac{27}{4}$.

Hint

(i) $sbi$ represents a vector perpendicular to the vector represented by $b$. E1 Thus, the two points represented by $a \pm sbi$ are equidistant from the point represented by $a$ E1 and they are joined to it by vectors which are perpendicular to that joining it to $C$ so they form a base of a triangle which has altitude from $a$ to $a + b$ and has two equal length sides, by Pythagoras. E1 (3)

Alternative Distance $a + b$ to $a + sbi$ is $|(a + sbi) - (a + b)| = |b||si - 1| = |b|\sqrt{s^2 + 1}$ as $s$ is real, E1 and distance $a + b$ to $a - sbi$ is $|(a - sbi) - (a + b)| = |b||-si - 1| = |b|\sqrt{s^2 + 1}$ , E1 so two equal length sides. E1

$a$ is represented by the midpoint of the base. B1 $b$ is represented by the vector joining the midpoint of the base to the other vertex. B1 $s$ is the scale factor that the magnitude of the altitude is multiplied by to obtain half the base. B1

The sketch should show the vertical asymptote at $x=-\frac{1}{9}$, the oblique asymptote $y=\frac{1}{3}x-\frac{11}{27}$, the intercepts $\left(\frac{1}{3},0\right)$ and $(0,-1)$, and the stationary maximum at $(-1,-1)$. (3)

(ii) We require complex $a$ and $b$ and real $s$ such that

$(a + sbi) + (a - sbi) + (a + b) = 0 \Rightarrow b = -3a$ M1 A1

and

$(a + sbi)(a - sbi) + (a - sbi)(a + b) + (a + b)(a + sbi) = p$

so

$a^2 + s^2b^2 + 2a(a + b) = p \Rightarrow 3a^2(3s^2 - 1) = p$

A1

and

$(a + sbi)(a - sbi)(a + b) = -q \Rightarrow -2a^3(9s^2 + 1) = -q$

Therefore

$\frac{p^3}{q^2} = \frac{[3a^2(3s^2 - 1)]^3}{[2a^3(9s^2 + 1)]^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}$

A1* (5)

(iii)

$y = \frac{(3x - 1)^3}{(9x + 1)^2}$

has $x$ intercept at $(\frac{1}{3}, 0)$, $y$ intercept at $(0, -1)$ G1 a vertical asymptote at $x = -\frac{1}{9}$ and an asymptote $y = \frac{1}{3}x - \frac{11}{27}$ as $x \to \pm\infty$. G1

$\frac{dy}{dx} = \frac{(9x + 1)^2 9(3x - 1)^2 - (3x - 1)^3 18(9x + 1)}{(9x + 1)^4}$ $= \frac{9(3x - 1)^2 (9x + 1 - 6x + 2)}{(9x + 1)^3} = \frac{27(3x - 1)^2 (x + 1)}{(9x + 1)^3}$

M1 A1

Thus, the stationary points are a maximum at $(-1, -1)$ and a point of inflection at $(\frac{1}{3}, 0)$. G1 G1

Feature	Value
$x$-intercept	$\left(\frac{1}{3}, 0\right)$
$y$-intercept	$(0, -1)$
Maximum	$(-1, -1)$
Vertical asymptote	$x = -\frac{1}{9}$
Slant asymptote	$y = \frac{1}{3}x - \frac{11}{27}$

(6)

(iv) If the roots of $z^3 + pz + q = 0$ represent the vertices of an isosceles triangle, then by (ii), $\frac{p^3}{q^2}$ must be real E1 and as $s^2 > 0$, from (iii) $\frac{p^3}{q^2} > \frac{27}{4} \times -1 = \frac{-27}{4}$ E1 as $y = \frac{(3x-1)^3}{(9x+1)^2}$ is increasing for $x > 0$. E1 (3)

Model Solution

Part (i)

Let the three points be $P = a + sbi$, $Q = a - sbi$, and $R = a + b$.

We compute the distances from $R$ to each of $P$ and $Q$:

$|R - P| = |(a + b) - (a + sbi)| = |b - sbi| = |b(1 - si)| = |b|\,|1 - si| = |b|\sqrt{1 + s^2}$

$|R - Q| = |(a + b) - (a - sbi)| = |b + sbi| = |b(1 + si)| = |b|\,|1 + si| = |b|\sqrt{1 + s^2}$

Since $|R - P| = |R - Q| = |b|\sqrt{1 + s^2}$, the triangle $PQR$ has two equal sides and is therefore isosceles. $\qquad \blacksquare$

For the explanation: given an isosceles triangle in the Argand diagram, we identify:

• $a$ is the complex number represented by the midpoint of the base (the side that is not one of the two equal sides).
• $b$ is the vector from the midpoint of the base to the apex (the vertex opposite the base).
• $s$ is the ratio: (half the base length) / (altitude from apex to base). More precisely, if the base has length $2L$ and the altitude has length $H$, then $s = L/H$.

Since $b$ is the vector from the midpoint of the base to the apex, the apex is at $a + b$. The base is perpendicular to $b$ (rotated by $90\circ$), so the two base vertices are at $a + sbi$ and $a - sbi$, where $s$ scales the perpendicular displacement relative to $|b|$.

Part (ii)

Let the three roots of $z^3 + pz + q = 0$ be $a + sbi$, $a - sbi$, and $a + b$, forming an isosceles triangle.

By Vieta’s formulas (noting the coefficient of $z^2$ is 0):

Sum of roots:

$(a + sbi) + (a - sbi) + (a + b) = 0$

$3a + b = 0 \implies b = -3a \qquad \text{...(1)}$

Sum of products of pairs:

$(a + sbi)(a - sbi) + (a - sbi)(a + b) + (a + b)(a + sbi) = p$

The first product is $a^2 + s^2 b^2$. For the second and third products, using $b = -3a$ so $a + b = -2a$:

$(a - sbi)(-2a) + (-2a)(a + sbi) = -2a(a - sbi) - 2a(a + sbi) = -2a \cdot 2a = -4a^2$

Therefore:

$a^2 + s^2 b^2 - 4a^2 = p$

$-3a^2 + 9s^2 a^2 = p$

$p = 3a^2(3s^2 - 1) \qquad \text{...(2)}$

Product of roots:

$(a + sbi)(a - sbi)(a + b) = -q$

$(a^2 + s^2 b^2)(-2a) = -q$

$(a^2 + 9s^2 a^2)(-2a) = -q$

$q = 2a^3(1 + 9s^2) \qquad \text{...(3)}$

Now we compute $\frac{p^3}{q^2}$:

$\frac{p^3}{q^2} = \frac{[3a^2(3s^2 - 1)]^3}{[2a^3(9s^2 + 1)]^2} = \frac{27a^6(3s^2 - 1)^3}{4a^6(9s^2 + 1)^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}$

as required. $\qquad \blacksquare$

Part (iii)

Let $y = \frac{(3x - 1)^3}{(9x + 1)^2}$.

Intercepts:

• $x$-intercept: $3x - 1 = 0 \implies x = \frac{1}{3}$, so the point is $\left(\frac{1}{3},\, 0\right)$.
• $y$-intercept: $x = 0$ gives $y = \frac{(-1)^3}{1^2} = -1$, so the point is $(0,\, -1)$.

Asymptotes:

• Vertical asymptote at $9x + 1 = 0$, i.e. $x = -\frac{1}{9}$.
• As $x \to \pm\infty$: performing polynomial long division of $(3x-1)^3 = 27x^3 - 27x^2 + 9x - 1$ by $(9x+1)^2 = 81x^2 + 18x + 1$:

$\frac{27x^3 - 27x^2 + 9x - 1}{81x^2 + 18x + 1} = \frac{1}{3}x - \frac{11}{27} + \frac{\text{lower order}}{81x^2 + 18x + 1}$

So the slant asymptote is $y = \frac{1}{3}x - \frac{11}{27}$.

Stationary points:

$\frac{dy}{dx} = \frac{(9x+1)^2 \cdot 9(3x-1)^2 - (3x-1)^3 \cdot 18(9x+1)}{(9x+1)^4}$

$= \frac{9(3x-1)^2(9x+1) - 18(3x-1)^3}{(9x+1)^3}$

$= \frac{9(3x-1)^2[(9x+1) - 2(3x-1)]}{(9x+1)^3}$

$= \frac{9(3x-1)^2(9x + 1 - 6x + 2)}{(9x+1)^3} = \frac{9(3x-1)^2(3x + 3)}{(9x+1)^3} = \frac{27(3x-1)^2(x+1)}{(9x+1)^3}$

Setting $\frac{dy}{dx} = 0$: the numerator vanishes when $x = \frac{1}{3}$ (double root) or $x = -1$.

At $x = -1$: $y = \frac{(-4)^3}{(-8)^2} = \frac{-64}{64} = -1$.

Sign of $\frac{dy}{dx}$ near $x = -1$: The factor $(3x-1)^2 > 0$ and $(9x+1)^3 < 0$ near $x = -1$. For $x < -1$: $(x+1) < 0$, so $\frac{dy}{dx} = \frac{(+)(-)}{(-)} > 0$. For $-1 < x < -\frac{1}{9}$: $(x+1) > 0$, so $\frac{dy}{dx} = \frac{(+)(+)}{(-)} < 0$. So $x = -1$ is a local maximum.

At $x = \frac{1}{3}$: $y = 0$. The factor $(3x-1)^2 \geq 0$ means $\frac{dy}{dx}$ does not change sign here; this is a point of inflection.

Sketch summary:

• The curve passes through $(0, -1)$ and $\left(\frac{1}{3}, 0\right)$.
• There is a local maximum at $(-1, -1)$.
• There is a vertical asymptote at $x = -\frac{1}{9}$.
• As $x \to -\frac{1}{9}^-$, $y \to -\infty$; as $x \to -\frac{1}{9}^+$, $y \to +\infty$.
• The curve approaches the slant asymptote $y = \frac{1}{3}x - \frac{11}{27}$ for large $|x|$.

Part (iv)

From part (ii), if the roots form an isosceles triangle, then

$\frac{p^3}{q^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}$

for some non-zero real $s$. Since $s$ is real, $s^2$ is a positive real number, so the right-hand side is real. Therefore $\frac{p^3}{q^2}$ is real. $\qquad \blacksquare$

Now let $x = s^2 > 0$, so $\frac{p^3}{q^2} = \frac{27(3x - 1)^3}{4(9x + 1)^2} = \frac{27}{4} \cdot \frac{(3x - 1)^3}{(9x + 1)^2}$.

From part (iii), the stationary points of $f(x) = \frac{(3x-1)^3}{(9x+1)^2}$ occur at $x = -1$ and $x = \frac{1}{3}$.

For $x > 0$: the only stationary point is $x = \frac{1}{3}$ where $f\!\left(\frac{1}{3}\right) = 0$, and this is a point of inflection (not a maximum or minimum). Since $f(0) = -1$ and $f(x) \to +\infty$ as $x \to +\infty$, the function is increasing on $(0,\infty)$, with a horizontal tangent at $x=\frac{1}{3}$.

Therefore, for all $x > 0$:

$f(x) \geqslant f(0) = -1$

(with the infimum $-1$ approached as $x \to 0^+$ but $x > 0$). Thus $f(x)>-1$ for all $x>0$.

Hence:

$\frac{p^3}{q^2} = \frac{27}{4} f(s^2) > \frac{27}{4} \cdot (-1) = -\frac{27}{4}$

$\qquad \blacksquare$

Examiner Notes

最不受欢迎的纯数题(45%作答),平均分6/20。常见错误:将 a 和 b 误视为实数;第(iii)部分图像差异大,少有人指定无穷远处渐近行为和拐点性质;第(iv)部分未能从图中正确推导不等式。

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Question 4

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

4 Let $n$ be a positive integer. The polynomial p is defined by the identity

$p(\cos \theta) \equiv \cos ((2n + 1)\theta) + 1 .$

(i) Show that

$\cos ((2n + 1)\theta) = \sum_{r=0}^{n} \binom{2n + 1}{2r} \cos^{2n+1-2r} \theta(\cos^2 \theta - 1)^r .$

(ii) By considering the expansion of $(1 + t)^{2n+1}$ for suitable values of $t$, show that the coefficient of $x^{2n+1}$ in the polynomial $p(x)$ is $2^{2n}$.

(iii) Show that the coefficient of $x^{2n-1}$ in the polynomial $p(x)$ is $-(2n + 1)2^{2n-2}$.

(iv) It is given that there exists a polynomial q such that

$p(x) = (x + 1) [q(x)]^2$

and the coefficient of $x^n$ in $q(x)$ is greater than 0.

Write down the coefficient of $x^n$ in the polynomial $q(x)$ and, for $n \geqslant 2$, show that the coefficient of $x^{n-2}$ in the polynomial $q(x)$ is

$2^{n-2}(1 - n) .$

Hint

(i) By de Moivre,

$\cos((2n + 1)\theta) + i \sin((2n + 1)\theta) = (\cos \theta + i \sin \theta)^{2n+1}$

Expanding by the binomial theorem and equating real parts

$\cos((2n + 1)\theta) = \cos^{2n+1} \theta - \binom{2n + 1}{2} \cos^{2n-1} \theta \sin^2 \theta + \dots + (-1)^n \binom{2n + 1}{2n} \cos \theta \sin^{2n} \theta$

M1 A1

$= \cos^{2n+1} \theta + \binom{2n + 1}{2} \cos^{2n-1} \theta (\cos^2 \theta - 1) + \dots + \binom{2n + 1}{2n} \cos \theta (\cos^2 \theta - 1)^n$

M1

$= \sum_{r=0}^{n} \binom{2n + 1}{2r} \cos^{2n+1-2r} \theta (\cos^2 \theta - 1)^r$

A1* (4)

Notice, for (iv), that this expression only contains odd powers of $\cos \theta$.

(ii) The coefficient of $x^{2n+1}$ in $p(x)$ is

$\sum_{r=0}^{n} \binom{2n + 1}{2r}$

B1

$(1 + x)^{2n+1} = \sum_{r=0}^{2n+1} \binom{2n + 1}{r} x^r = \sum_{r=0}^{n} \binom{2n + 1}{2r} x^{2r} + \sum_{r=0}^{n} \binom{2n + 1}{2r + 1} x^{2r+1}$

Substituting $x = 1$,

$2^{2n+1} = \sum_{r=0}^{n} \binom{2n + 1}{2r} + \sum_{r=0}^{n} \binom{2n + 1}{2r + 1}$

and substituting $x = -1$,

M1

$0 = \sum_{r=0}^{n} \binom{2n + 1}{2r} - \sum_{r=0}^{n} \binom{2n + 1}{2r + 1}$

A1

Adding these two results,

$2^{2n+1} = 2 \sum_{r=0}^{n} \binom{2n + 1}{2r}$

and so the required coefficient is $2^{2n}$ as required. A1 (4)*

(iii) The coefficient of $x^{2n-1}$ in $p(x)$ is

$\sum_{r=0}^{n} -r \binom{2n+1}{2r}$ B1

As

$(1+x)^{2n+1} = \sum_{r=0}^{2n+1} \binom{2n+1}{r} x^r = \sum_{r=0}^{n} \binom{2n+1}{2r} x^{2r} + \sum_{r=0}^{n} \binom{2n+1}{2r+1} x^{2r+1}$

differentiating with respect to $x$

$(2n+1)(1+x)^{2n} = \sum_{r=0}^{n} 2r \binom{2n+1}{2r} x^{2r-1} + \sum_{r=0}^{n} (2r+1) \binom{2n+1}{2r+1} x^{2r}$

M1

Substituting $x=1$,

$(2n+1)2^{2n} = \sum_{r=0}^{n} 2r \binom{2n+1}{2r} + \sum_{r=0}^{n} (2r+1) \binom{2n+1}{2r+1}$ M1

and substituting $x=-1$,

$0 = -\sum_{r=0}^{n} 2r \binom{2n+1}{2r} + \sum_{r=0}^{n} (2r+1) \binom{2n+1}{2r+1}$ A1

Subtracting these two results

$(2n+1)2^{2n} = 2 \sum_{r=0}^{n} 2r \binom{2n+1}{2r} = 4 \sum_{r=0}^{n} r \binom{2n+1}{2r}$

and so the required coefficient is

$-(2n+1)2^{2n} \div 4 = -(2n+1)2^{2n-2}$ A1* (5)

Alternative

$\sum_{r=0}^{n} -r \binom{2n+1}{2r} = -\sum_{r=0}^{n} r \frac{(2n+1)!}{(2n-2r+1)! (2r)!} = -\frac{2n+1}{2} \sum_{r=1}^{n} \frac{(2n)!}{(2n-2r+1)! (2r-1)!}$

$= -\frac{2n+1}{2} \sum_{r=1}^{n} \binom{2n}{2r-1} = -\frac{2n+1}{2} \sum_{r=0}^{n-1} \binom{2n}{2r+1}$ M1

As in (ii),

$\sum_{r=0}^{n-1} \binom{2n}{2r+1} = \frac{1}{2} 2^{2n}$ A1

so

$\sum_{r=0}^{n} -r \binom{2n+1}{2r} = -\frac{2n+1}{2} \frac{1}{2} 2^{2n} = -(2n+1) 2^{2n-2}$ A1 (5)*

(iv) Suppose

$q(x) = ax^n + bx^{n-1} + cx^{n-2} + \dots$

then

$p(x) = (x+1) [ax^n + bx^{n-1} + cx^{n-2} + \dots]^2$ $= (x+1) (a^2 x^{2n} + 2ab x^{2n-1} + (b^2 + 2ac) x^{2n-2} + \dots)$ $= a^2 x^{2n+1} + (a^2 + 2ab) x^{2n} + (b^2 + 2ac + 2ab) x^{2n-1} + \dots$

M1 A1

Thus $a^2 = 2^{2n}$, $a^2 + 2ab = 0$, and $b^2 + 2ac + 2ab = -(2n+1) 2^{2n-2}$ dM1 A1

Therefore $a = 2^n$ (as $a > 0$), B1

$b = \frac{-a}{2} = -2^{n-1}$ A1

and

$2^{2n-2} + 2^{n+1}c - 2^{2n} = -(2n+1) 2^{2n-2}$

so

$2^{n-3} + c - 2^{n-1} = -(2n+1) 2^{n-3}$ $c = 2^{n-3} (4 - 1 - 2n - 1) = 2^{n-2} (1 - n)$

as required. A1(7)*

Model Solution

Part (i)

By de Moivre’s theorem:

$\cos((2n+1)\theta) + i\sin((2n+1)\theta) = (\cos\theta + i\sin\theta)^{2n+1}$

Expanding the right-hand side by the binomial theorem:

$(\cos\theta + i\sin\theta)^{2n+1} = \sum_{k=0}^{2n+1} \binom{2n+1}{k} \cos^{2n+1-k}\theta \cdot (i\sin\theta)^k$

The real part consists of terms where $k$ is even (so that $i^k$ is real). Setting $k = 2r$ for $r = 0, 1, \ldots, n$:

$\cos((2n+1)\theta) = \sum_{r=0}^{n} \binom{2n+1}{2r} \cos^{2n+1-2r}\theta \cdot (i\sin\theta)^{2r}$

Since $i^{2r} = (-1)^r$ and $\sin^{2r}\theta = (1 - \cos^2\theta)^r$, we have $(i\sin\theta)^{2r} = (-1)^r \sin^{2r}\theta = (\cos^2\theta - 1)^r$.

Therefore:

$\cos((2n+1)\theta) = \sum_{r=0}^{n} \binom{2n+1}{2r} \cos^{2n+1-2r}\theta\,(\cos^2\theta - 1)^r$

$\qquad \blacksquare$

Part (ii)

Setting $x = \cos\theta$ and $t = \cos^2\theta - 1 = x^2 - 1$, the result of part (i) gives:

$\cos((2n+1)\theta) = \sum_{r=0}^{n} \binom{2n+1}{2r} x^{2n+1-2r}\, t^r$

Note that $x^{2n+1-2r} = x \cdot x^{2(n-r)} = x \cdot (x^2)^{n-r} = x \cdot (t+1)^{n-r}$, so each term contains only odd powers of $x$. Therefore $p(x) = \cos((2n+1)\theta) + 1$ is indeed a polynomial in $x$ with the same leading terms.

The coefficient of $x^{2n+1}$ in $p(x)$ is obtained by taking the leading term from each summand, which gives:

$\text{coefficient of } x^{2n+1} = \sum_{r=0}^{n} \binom{2n+1}{2r}$

Now consider the expansion of $(1 + t)^{2n+1}$:

$(1+t)^{2n+1} = \sum_{r=0}^{2n+1} \binom{2n+1}{r} t^r$

Substituting $t = 1$:

$2^{2n+1} = \sum_{r=0}^{2n+1} \binom{2n+1}{r} = \sum_{r=0}^{n} \binom{2n+1}{2r} + \sum_{r=0}^{n} \binom{2n+1}{2r+1} \qquad \text{...(1)}$

Substituting $t = -1$:

$0 = \sum_{r=0}^{2n+1} \binom{2n+1}{r}(-1)^r = \sum_{r=0}^{n} \binom{2n+1}{2r} - \sum_{r=0}^{n} \binom{2n+1}{2r+1} \qquad \text{...(2)}$

Adding equations (1) and (2):

$2^{2n+1} = 2\sum_{r=0}^{n} \binom{2n+1}{2r}$

$\sum_{r=0}^{n} \binom{2n+1}{2r} = 2^{2n}$

Therefore the coefficient of $x^{2n+1}$ in $p(x)$ is $2^{2n}$. $\qquad \blacksquare$

Part (iii)

From part (i), each term in the sum $\sum_{r=0}^{n} \binom{2n+1}{2r} x^{2n+1-2r}(x^2 - 1)^r$ contributes to the coefficient of $x^{2n-1}$ when the expansion of $(x^2 - 1)^r$ produces a term with $x^{2(r-1)}$, i.e. from the constant in $(x^2-1)^r$ times $x^{2n+1-2r}$ we get $x^{2n+1-2r}$, and from the $x^2$ term of $(x^2-1)^r$ (which has coefficient $-r$ from choosing one factor of $x^2$ and $r-1$ factors of $-1$) times $x^{2n+1-2r}$ we get $x^{2n+3-2r}$.

More precisely, the contribution to $x^{2n-1}$ from the $r$-th term requires:

$x^{2n+1-2r} \cdot (\text{term of degree } 2r - 2 \text{ in } (x^2-1)^r)$

Consider the identity from part (i) with $t = x^2 - 1$:

$\cos((2n+1)\theta) + 1 = \sum_{r=0}^{n} \binom{2n+1}{2r} x^{2n+1-2r}(x^2 - 1)^r + 1$

The term $x^{2n+1-2r}(x^2-1)^r$ can be expanded. Its contribution to the coefficient of $x^{2n-1}$ requires $(x^2-1)^r$ to contribute $x^{2r-2}$, which means choosing exactly one factor of $(-1)$ from $r$ factors (coefficient $-r$) times $x^{2(r-1)}$ from the remaining factors. So the contribution is $\binom{2n+1}{2r} \cdot (-r)$.

Therefore the coefficient of $x^{2n-1}$ in $p(x)$ is:

$\sum_{r=0}^{n} -r\binom{2n+1}{2r}$

To evaluate this, consider the binomial expansion:

$(1+t)^{2n+1} = \sum_{k=0}^{2n+1}\binom{2n+1}{k}t^k$

Differentiating with respect to $t$:

$(2n+1)(1+t)^{2n} = \sum_{k=1}^{2n+1} k\binom{2n+1}{k}t^{k-1}$

Substituting $t = 1$:

$(2n+1)2^{2n} = \sum_{k=1}^{2n+1} k\binom{2n+1}{k} = \sum_{r=0}^{n} 2r\binom{2n+1}{2r} + \sum_{r=0}^{n}(2r+1)\binom{2n+1}{2r+1} \qquad \text{...(3)}$

Substituting $t = -1$:

$0 = \sum_{k=1}^{2n+1} k\binom{2n+1}{k}(-1)^{k-1} = \sum_{r=0}^{n} 2r\binom{2n+1}{2r}(-1)^{2r-1} + \sum_{r=0}^{n}(2r+1)\binom{2n+1}{2r+1}(-1)^{2r}$

$= -\sum_{r=0}^{n} 2r\binom{2n+1}{2r} + \sum_{r=0}^{n}(2r+1)\binom{2n+1}{2r+1} \qquad \text{...(4)}$

Subtracting (4) from (3):

$(2n+1)2^{2n} = 2\sum_{r=0}^{n} 2r\binom{2n+1}{2r} = 4\sum_{r=0}^{n} r\binom{2n+1}{2r}$

Therefore:

$\sum_{r=0}^{n} r\binom{2n+1}{2r} = \frac{(2n+1)2^{2n}}{4} = (2n+1)2^{2n-2}$

The coefficient of $x^{2n-1}$ in $p(x)$ is:

$-\sum_{r=0}^{n} r\binom{2n+1}{2r} = -(2n+1)2^{2n-2}$

$\qquad \blacksquare$

Part (iv)

From parts (ii) and (iii), $p(x) = 2^{2n}x^{2n+1} - (2n+1)2^{2n-2}x^{2n-1} + \ldots$ (only odd powers of $x$ appear, plus the constant term $+1$).

Write $q(x) = ax^n + bx^{n-1} + cx^{n-2} + \ldots$ with $a > 0$.

Then:

$[q(x)]^2 = a^2x^{2n} + 2abx^{2n-1} + (b^2 + 2ac)x^{2n-2} + \ldots$

Multiplying by $(x+1)$:

$(x+1)[q(x)]^2 = a^2x^{2n+1} + (a^2 + 2ab)x^{2n} + (b^2 + 2ac + 2ab)x^{2n-1} + \ldots$

Coefficient of $x^{2n+1}$:

$a^2 = 2^{2n} \implies a = 2^n \qquad (a > 0)$

Coefficient of $x^{2n}$:

Since $p(x)$ contains only odd powers of $x$ (from part (i), every term has odd degree in $\cos\theta$), the coefficient of $x^{2n}$ in $p(x)$ is $0$:

$a^2 + 2ab = 0 \implies 2^{2n} + 2 \cdot 2^n \cdot b = 0 \implies b = -\frac{2^{2n}}{2^{n+1}} = -2^{n-1}$

Coefficient of $x^{2n-1}$:

$b^2 + 2ac + 2ab = -(2n+1)2^{2n-2}$

Substituting $a = 2^n$ and $b = -2^{n-1}$:

$(-2^{n-1})^2 + 2 \cdot 2^n \cdot c + 2 \cdot 2^n \cdot (-2^{n-1}) = -(2n+1)2^{2n-2}$

$2^{2n-2} + 2^{n+1}c - 2^{2n} = -(2n+1)2^{2n-2}$

$2^{n+1}c = -(2n+1)2^{2n-2} - 2^{2n-2} + 2^{2n}$

$2^{n+1}c = 2^{2n-2}[-(2n+1) - 1 + 4] = 2^{2n-2}(2 - 2n)$

$c = \frac{2^{2n-2} \cdot 2(1-n)}{2^{n+1}} = 2^{2n-1-n-1}(1-n) = 2^{n-2}(1-n)$

Therefore the coefficient of $x^n$ in $q(x)$ is $\boxed{2^n}$, and the coefficient of $x^{n-2}$ in $q(x)$ is $2^{n-2}(1-n)$. $\qquad \blacksquare$

Examiner Notes

全卷最难(平均分5/20)。大多数考生仅在第(i)部分有实质性进展。第(ii)部分常见错误:写下系数后无进展或未利用代入 t=±1 的技巧。第(iv)部分几乎仅被前三部分成功的考生尝试。

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Question 5

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

5 (i) Show that if $\frac{1}{x} + \frac{2}{y} = \frac{2}{7},$ then $(2x - 7)(y - 7) = 49$.

By considering the factors of 49, find all the pairs of positive integers $x$ and $y$ such that $\frac{1}{x} + \frac{2}{y} = \frac{2}{7}.$

(ii) Let $p$ and $q$ be prime numbers such that $p^2 + pq + q^2 = n^2$ where $n$ is a positive integer. Show that $(p + q + n)(p + q - n) = pq$ and hence explain why $p + q = n + 1$.

Hence find the possible values of $p$ and $q$.

(iii) Let $p$ and $q$ be positive and $p^3 + q^3 + 3pq^2 = n^3.$ Show that $p + q - n < p$ and $p + q - n < q$.

Show that there are no prime numbers $p$ and $q$ such that $p^3 + q^3 + 3pq^2$ is the cube of an integer.

Hint

(i)

$\frac{1}{x} + \frac{2}{y} = \frac{2}{7}$

$7y + 14x = 2xy$ $2xy - 7y - 14x + 49 = 49$ $(2x - 7)(y - 7) = 49$ B1*

Thus $2x - 7 = 1, y - 7 = 49$, or $2x - 7 = 7, y - 7 = 7$, or $2x - 7 = 49, y - 7 = 1$

M1

and so $(x, y) = (4, 56), (7, 14)$, or $(28, 8)$ A1 (3)

(ii)

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

B1*

$p + q + n \neq p$ and $p + q + n \neq q$ as $p, q$, and $n$ are all positive. $p + q + n > p + q - n$ so $p + q + n \neq 1$ as that would require $p + q - n = pq > 1$. M1

Thus $p + q + n = pq$ and $p + q - n = 1$ as required. A1*

Therefore $p + q + p + q - 1 = pq$ M1

$pq - 2p - 2q + 4 = 3$

$(p - 2)(q - 2) = 3$

dM1

Thus $p - 2 = 1, q - 2 = 3$, or $p - 2 = 3, q - 2 = 1$

Alternative (I)

$pq - 2p - 2q + 4 = 3$

$pq - 2p - 2q + 1 = 0$

$p(q - 2) = 2q - 1$

$p = \frac{2q - 1}{q - 2} = 2 + \frac{3}{q - 2}$

as $q \neq 2$ ($q = 2$ would yield -4+4-3=0) so $q - 2 = 1$ or $3$ E1

and so $(p, q) = (3, 5)$, or $(5, 3)$ A1 (6)

Alternative (II) $p + q + n = pq$ and $p + q - n = 1$ yield $p + q = n + 1$ and $pq = 2n + 1$

Therefore, $p$ and $q$ are solutions of $t^2 - (n + 1)t + (2n + 1) = 0$

Hence $t = \frac{(n+1) \pm \sqrt{(n+1)^2 - 4(2n+1)}}{2} = \frac{(n+1) \pm \sqrt{(n-3)^2 + 12}}{2}$

For integer $t$ we require that $(n - 3)^2 + 12$ is a perfect square (in fact an even perfect square).

Thus the difference of squares between $(n - 3)^2 + 12$ and $(n - 3)^2$ is 12. Successive squares, $z^2$ and $(z + 1)^2$ differ by $2z + 1$, which for $z \ge 6$ is $\ge 13$. Thus $(n - 3) \le 5$. Then, either by listing potential solutions exhaustively, or justifying that $(n - 3)^2 + 12$ and $(n - 3)^2$ have to be squares differing by $7 + 5$ and hence $(n - 3)^2 = 2^2$ giving $(p, q) = (3, 5)$, or $(5, 3)$. E1 A1 (6)

(iii) If $p^3 + q^3 + 3pq^2 = n^3$, and as $p, q$, and hence $n$ are all positive, then $p^3 < n^3$ and

$q^3 < n^3$ so $p < n$ and $q < n$, E1 and hence $p + q - n < p$ and $p + q - n < q$. A1*

If $p^3 + q^3 + 3pq^2 + 3p^2q = n^3 + 3p^2q$ M1 $(p + q)^3 - n^3 = 3p^2q$ dM1

$(p + q - n)((p + q)^2 + (p + q)n + n^2) = 3p^2q$ A1

As $p + q - n < p$ and $p + q - n < q$, so $p + q - n = 1$ or $3$ A1

If $p + q - n = 1$ then $(n + 1)^3 - n^3 = 3p^2q$ and hence $3n^2 + 3n + 1 = 3p^2q$ M1 which is not possible as LHS is not a multiple of 3 and RHS is. E1

If $p + q - n = 3$, then $(n + 3)^3 - n^3 = 3p^2q$ and hence $9n^2 + 27n + 27 = 3p^2q$, that is

$3(n^2 + 3n + 3) = p^2q$. M1 So $p$ or $q$ must divide 3 and hence must be 3 as $p$ and $q$ are prime. E1

If $p = 3$, then $q - n = 0$ but $q < n$ and vice versa if $q = 3$ E1* (11)

Model Solution

Part (i)

Starting from $\frac{1}{x} + \frac{2}{y} = \frac{2}{7}$, multiply both sides by $7xy$:

$7y + 14x = 2xy$

Rearranging:

$2xy - 14x - 7y = 0$

We use the factorisation technique: add 49 to both sides:

$2xy - 14x - 7y + 49 = 49$

Factor by grouping: take $2x$ from the first two terms and $-7$ from the last two:

$2x(y - 7) - 7(y - 7) = 49$

$(2x - 7)(y - 7) = 49 \qquad \blacksquare$

Now we find all positive integer solutions. Since $x$ and $y$ are positive integers, we need $2x - 7$ and $y - 7$ to be integer factors of 49. Also, since $\frac{1}{x} + \frac{2}{y} = \frac{2}{7}$ with $x, y > 0$, we need $x > \frac{7}{2}$ (i.e., $x \geq 4$) and $y > 7$ (i.e., $y \geq 8$), so $2x - 7 \geq 1$ and $y - 7 \geq 1$. Therefore both factors must be positive.

The positive factor pairs of $49 = 7^2$ are $(1, 49)$, $(7, 7)$, and $(49, 1)$.

Case 1: $2x - 7 = 1$, $y - 7 = 49$, giving $x = 4$, $y = 56$.

Case 2: $2x - 7 = 7$, $y - 7 = 7$, giving $x = 7$, $y = 14$.

Case 3: $2x - 7 = 49$, $y - 7 = 1$, giving $x = 28$, $y = 8$.

The solutions are $(x, y) = (4, 56)$, $(7, 14)$, and $(28, 8)$.

Part (ii)

Starting from $p^2 + pq + q^2 = n^2$, we add $pq$ to both sides:

$p^2 + 2pq + q^2 = n^2 + pq$

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

$(p + q + n)(p + q - n) = pq \qquad \blacksquare$

Since $p$, $q$, and $n$ are all positive, we have $p + q + n > p + q - n$ (as $n > 0$). We also note $p + q + n > p$ and $p + q + n > q$ (since $q + n > 0$ and $p + n > 0$).

Now $(p + q + n)(p + q - n) = pq$, and $pq$ is a positive integer. Since $p$ and $q$ are prime, the only positive factorisations of $pq$ are $1 \times pq$, $p \times q$, $q \times p$, and $pq \times 1$.

Since $p + q + n$ cannot equal 1 (it exceeds $p \geq 2$), and $p + q + n$ cannot equal $p$ or $q$ (it exceeds both), the only possibility is:

$p + q + n = pq \qquad \text{and} \qquad p + q - n = 1$

From the second equation: $n = p + q - 1$. Substituting into the first:

$p + q + (p + q - 1) = pq$

$2p + 2q - 1 = pq$

$pq - 2p - 2q = -1$

$(p - 2)(q - 2) = 3$

Since $p$ and $q$ are prime, $p - 2$ and $q - 2$ are integers. The factor pairs of 3 are $(1, 3)$ and $(3, 1)$ (we exclude negative pairs as they give $p \leq 0$ or $q \leq 0$).

Case 1: $p - 2 = 1$, $q - 2 = 3$, giving $p = 3$, $q = 5$.

Case 2: $p - 2 = 3$, $q - 2 = 1$, giving $p = 5$, $q = 3$.

Verification: $3^2 + 3 \cdot 5 + 5^2 = 9 + 15 + 25 = 49 = 7^2$, so $n = 7$.

The possible values are $(p, q) = (3, 5)$ or $(5, 3)$.

Part (iii)

First, we show $p + q - n < p$ and $p + q - n < q$.

Since $p$ and $q$ are positive, $p^3 > 0$ and $q^3 > 0$, so from $p^3 + q^3 + 3pq^2 = n^3$:

$n^3 = p^3 + q^3 + 3pq^2 > p^3 \implies n > p$

$n^3 = p^3 + q^3 + 3pq^2 > q^3 \implies n > q$

Therefore $p + q - n < p$ (since $q - n < 0$) and $p + q - n < q$ (since $p - n < 0$). $\qquad \blacksquare$

Now we show no prime pair works. We use the identity $(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$:

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

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

Factorising the left side:

$(p + q - n)\big[(p + q)^2 + (p + q)n + n^2\big] = 3p^2q$

Since $p + q - n$ is a positive integer dividing $3p^2q$, and $p + q - n < p$ and $p + q - n < q$, the factor $p + q - n$ cannot be divisible by $p$ or $q$. Therefore $p + q - n$ divides 3, giving $p + q - n = 1$ or $p + q - n = 3$.

Case $p + q - n = 1$: Then $n = p + q - 1$, so $(p + q)^3 - n^3 = (p+q)^3 - (p+q-1)^3$. Let $m = p + q$:

$m^3 - (m - 1)^3 = 3m^2 - 3m + 1 = 3p^2q$

The left side $3m^2 - 3m + 1 \equiv 1 \pmod{3}$, but $3p^2q \equiv 0 \pmod{3}$. Contradiction.

Case $p + q - n = 3$: Then $n = p + q - 3$, so with $m = p + q$:

$m^3 - (m-3)^3 = 9m^2 - 27m + 27 = 3(3m^2 - 9m + 9) = 3p^2q$

So $3(m^2 - 3m + 3) = p^2q$. Since 3 divides $p^2q$ and $p$, $q$ are prime, either $p = 3$ or $q = 3$.

If $p = 3$: then $3(m^2 - 3m + 3) = 9q$, so $q = \frac{m^2 - 3m + 3}{3}$. But also $q = m - p = m - 3$, so:

$m - 3 = \frac{m^2 - 3m + 3}{3} \implies 3m - 9 = m^2 - 3m + 3 \implies m^2 - 6m + 12 = 0$

The discriminant is $36 - 48 = -12 < 0$, so no real solution exists.

If $q = 3$: then $3(m^2 - 3m + 3) = 9p$, so $p = \frac{m^2 - 3m + 3}{3}$. But also $p = m - q = m - 3$, giving the same equation $m^2 - 6m + 12 = 0$ with no real solution.

In both sub-cases we reach a contradiction. Therefore there are no prime numbers $p$ and $q$ such that $p^3 + q^3 + 3pq^2$ is the cube of an integer. $\qquad \blacksquare$

Examiner Notes

最受欢迎的题目,平均分9/20。常见错误:处理不等式时缺乏精确性;从结果反推而未验证逻辑可逆。第(iii)部分关键:识别 p+q-n 必须为 1 或 3,但很少有考生成功排除两种情况。

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Question 6

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

6 (i) By considering the Maclaurin series for $e^x$, show that for all real $x$,

$\cosh^2 x \geqslant 1 + x^2.$

Hence show that the function f, defined for all real $x$ by $f(x) = \tan^{-1} x - \tanh x$, is an increasing function.

Sketch the graph $y = f(x)$.

(ii) Function g is defined for all real $x$ by $g(x) = \tan^{-1} x - \frac{1}{2} \pi \tanh x$.

(a) Show that g has at least two stationary points.

(b) Show, by considering its derivative, that $(1 + x^2) \sinh x - x \cosh x$ is non-negative for $x \geqslant 0$.

(c) Show that $\frac{\cosh^2 x}{1 + x^2}$ is an increasing function for $x \geqslant 0$.

(d) Hence or otherwise show that g has exactly two stationary points.

(e) Sketch the graph $y = g(x)$.

Hint

(i)

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ $\cosh x = \frac{1}{2}(e^x + e^{-x}) = \frac{1}{2}\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots\right) = 1 + \frac{x^2}{2!} + \dots$ $\cosh^2 x = \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots\right)^2 = 1 + x^2 + \frac{x^4}{3} + \dots \geq 1 + x^2$

B1*

as all terms are of even degree with positive coefficients.

Alternative

$\cosh x = \frac{1}{2}(e^x + e^{-x}) = 1 + \frac{x^2}{2!} + \dots \geq 1 + \frac{x^2}{2!}$ $\cosh^2 x \geq \left(1 + \frac{x^2}{2!}\right)^2 = 1 + x^2 + \frac{x^4}{4} \geq 1 + x^2$

B1*

$f(x) = \tan^{-1} x - \tanh x$

$f'(x) = \frac{1}{1 + x^2} - \text{sech}^2 x = \frac{\cosh^2 x - (1 + x^2)}{(1 + x^2) \cosh^2 x}$

M1

We have shown that the numerator $\cosh^2 x - (1 + x^2) \geq 0$ and the denominator is positive so $f'(x) \geq 0$ and hence the function $f$ is increasing. E1*

When $x = 0$, $f(x) = f'(x) = 0$ and for all other $x$, $f'(x) > 0$

$f(-x) = -f(x)$

G1

As $x \to \pm\infty$, $f(x) \to \pm\left(\frac{\pi}{2} - 1\right)$ respectively. G1 (5)

The sketch should show that $f$ is odd, passes through the origin with horizontal tangent, is increasing, and tends to $\pm\left(\frac{\pi}{2}-1\right)$ as $x\to\pm\infty$.

(ii) (a) $g(x) = \tan^{-1} x - \frac{1}{2} \pi \tanh x$ $g'(x) = \frac{1}{1 + x^2} - \frac{1}{2} \pi \operatorname{sech}^2 x = \frac{2 \cosh^2 x - \pi (1 + x^2)}{2 (1 + x^2) \cosh^2 x}$ M1

As in (i), the denominator is positive. When $x = 0$, the numerator $= 2 - \pi < 0$. A1

The numerator $= (2 - \pi)(1 + x^2) + 2 \left( \frac{x^4}{3} + \dots \right) \to \infty$ as $x \to \infty$. M1 Thus, there is a value of $x \neq 0$ for which $g'(x) = 0$ and as $g'(x)$ is an even function, there is also the value $-x$. E1 Hence, there are at least two stationary points for g. (4)

Alternative

$g(0) = 0$ E1 and $g(x) \to 0$ as $x \to \infty$ E1 and $g(x)$ is not identically zero E1 so there must be a stationary point for positive $x$, and similarly for negative. E1

(b) $\frac{d}{dx} [(1 + x^2) \sinh x - x \cosh x] = (1 + x^2) \cosh x + 2x \sinh x - x \sinh x - \cosh x$ M1

$= x^2 \cosh x + x \sinh x \geq 0$

for $x \geq 0$ A1

as $x^2 \geq 0$ and $\cosh x \geq 1$ for all $x$ and $\sinh x \geq 0$ for $x \geq 0$ E1

When $x = 0$, $(1 + x^2) \sinh x - x \cosh x = 0$ and we have shown $(1 + x^2) \sinh x - x \cosh x$ is increasing for $x \geq 0$, thus $(1 + x^2) \sinh x - x \cosh x$ is non-negative for $x \geq 0$. E1 (4)

(c) $\frac{d}{dx} \left[ \frac{\cosh^2 x}{1 + x^2} \right] = \frac{(1 + x^2) 2 \cosh x \sinh x - 2x \cosh^2 x}{(1 + x^2)^2} = \frac{2 \cosh x ((1 + x^2) \sinh x - x \cosh x)}{(1 + x^2)^2}$ M1 A1

$\frac{2 \cosh x}{(1 + x^2)^2} > 0$ for all $x$ and by (b) $(1 + x^2) \sinh x - x \cosh x \geq 0$ for $x \geq 0$

so $\frac{\cosh^2 x}{1 + x^2}$ is increasing for $x \geq 0$. E1 (3)

(d)

$g'(x) = \frac{1}{1 + x^2} - \frac{1}{2} \pi \operatorname{sech}^2 x = \frac{1}{\cosh^2 x} \left[ \frac{\cosh^2 x}{1 + x^2} - \frac{1}{2} \pi \right]$

By (c) , $g'$ is increasing for $x \geq 0$ , and thus there is exactly one value of $x$ for $x > 0$ that $g'(x) = 0$

Similarly, as $g'$ is an even function, there is exactly one value of $x$ for $x < 0$ that $g'(x) = 0$ Thus there are exactly two stationary points. E1 (1)

(e)

$x$	Sketch value of $y$
$-2$	$0.5$
$-1$	$1$
$0$	$0$
$1$	$-1$
$2$	$-0.5$

G3 (3)

Model Solution

Part (i)

The Maclaurin series for $e^x$ is:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

Therefore:

$\cosh x = \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots = \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}$

Squaring:

$\cosh^2 x = \left(\sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}\right)^2$

The coefficient of $x^0$ is $1 \cdot 1 = 1$. The coefficient of $x^2$ is $1 \cdot \frac{1}{2!} + \frac{1}{2!} \cdot 1 = 1$. The coefficient of $x^4$ is $1 \cdot \frac{1}{4!} + \frac{1}{2!} \cdot \frac{1}{2!} + \frac{1}{4!} \cdot 1 = \frac{1}{12} + \frac{1}{4} + \frac{1}{12} = \frac{1}{3}$.

In general, every coefficient in the expansion is a sum of products of positive terms $\frac{1}{(2k)!}$, so all coefficients are non-negative. Therefore:

$\cosh^2 x = 1 + x^2 + \frac{x^4}{3} + (\text{terms of degree } \geq 6 \text{ with non-negative coefficients})$

$\cosh^2 x \geq 1 + x^2 \qquad \text{for all real } x \qquad \blacksquare$

Now consider $f(x) = \tan^{-1} x - \tanh x$. Differentiating:

$f'(x) = \frac{1}{1 + x^2} - \operatorname{sech}^2 x = \frac{1}{1 + x^2} - \frac{1}{\cosh^2 x}$

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

The denominator $(1 + x^2)\cosh^2 x > 0$ for all $x$. By the inequality just proved, $\cosh^2 x - (1 + x^2) \geq 0$, so $f'(x) \geq 0$ for all $x$.

Moreover, $f'(x) = 0$ only when $\cosh^2 x = 1 + x^2$, which from the series expansion requires all higher-order terms to vanish, i.e., $x = 0$. So $f'(x) > 0$ for all $x \neq 0$, and $f$ is (strictly) increasing. $\qquad \blacksquare$

Sketch of $y = f(x)$:

Since $f(-x) = \tan^{-1}(-x) - \tanh(-x) = -\tan^{-1} x + \tanh x = -f(x)$, the function $f$ is odd.

At $x = 0$: $f(0) = 0$. Since $f'(0) = 0$ and $f'(x) > 0$ for $x \neq 0$, the graph passes through the origin with a horizontal tangent, then rises on both sides.

As $x \to +\infty$: $\tan^{-1} x \to \frac{\pi}{2}$ and $\tanh x \to 1$, so $f(x) \to \frac{\pi}{2} - 1 \approx 0.571$.

As $x \to -\infty$: $f(x) \to -\left(\frac{\pi}{2} - 1\right)$.

The graph is an S-shaped curve through the origin with horizontal asymptotes $y = \pm\left(\frac{\pi}{2} - 1\right)$, increasing everywhere with a flat point at the origin.

Part (ii)(a)

$g(x) = \tan^{-1} x - \frac{\pi}{2}\tanh x$

$g'(x) = \frac{1}{1 + x^2} - \frac{\pi}{2}\operatorname{sech}^2 x = \frac{2\cosh^2 x - \pi(1 + x^2)}{2(1 + x^2)\cosh^2 x}$

The denominator is positive for all $x$. Let $N(x) = 2\cosh^2 x - \pi(1 + x^2)$ be the numerator.

At $x = 0$: $N(0) = 2 - \pi < 0$.

As $x \to \infty$: using $\cosh^2 x = 1 + x^2 + \frac{x^4}{3} + \cdots$:

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

Since $(2 - \pi)(1 + x^2)$ grows like $-1.14 x^2$ while $\frac{2x^4}{3}$ grows like $0.67 x^4$, the $x^4$ term dominates for large $x$, so $N(x) \to +\infty$ as $x \to \infty$.

By the Intermediate Value Theorem, there exists $x_0 > 0$ with $N(x_0) = 0$, i.e., $g'(x_0) = 0$.

Since $g'(-x) = g'(x)$ (both $\frac{1}{1+x^2}$ and $\operatorname{sech}^2 x$ are even functions), $g'$ is even, so $g'(-x_0) = 0$ as well.

Therefore $g$ has at least two stationary points (at $x = x_0$ and $x = -x_0$). $\qquad \blacksquare$

Part (ii)(b)

Let $h(x) = (1 + x^2)\sinh x - x\cosh x$.

$h'(x) = \frac{d}{dx}\big[(1+x^2)\sinh x\big] - \frac{d}{dx}\big[x\cosh x\big]$

$= 2x\sinh x + (1+x^2)\cosh x - (\cosh x + x\sinh x)$

$= 2x\sinh x + (1+x^2)\cosh x - \cosh x - x\sinh x$

$= x\sinh x + x^2\cosh x$

For $x \geq 0$: $x^2 \geq 0$ and $\cosh x \geq 1 > 0$, so $x^2\cosh x \geq 0$. Also $x \geq 0$ and $\sinh x \geq 0$ for $x \geq 0$, so $x\sinh x \geq 0$.

Therefore $h'(x) \geq 0$ for $x \geq 0$.

At $x = 0$: $h(0) = (1)(0) - 0 = 0$.

Since $h(0) = 0$ and $h$ is increasing for $x \geq 0$, we have $h(x) \geq 0$ for $x \geq 0$.

Therefore $(1 + x^2)\sinh x - x\cosh x \geq 0$ for $x \geq 0$. $\qquad \blacksquare$

Part (ii)(c)

$\frac{d}{dx}\left[\frac{\cosh^2 x}{1 + x^2}\right] = \frac{2\cosh x \sinh x \cdot (1 + x^2) - \cosh^2 x \cdot 2x}{(1 + x^2)^2}$

$= \frac{2\cosh x\big[(1 + x^2)\sinh x - x\cosh x\big]}{(1 + x^2)^2}$

For $x \geq 0$: $\cosh x > 0$ and $(1 + x^2)^2 > 0$. By part (b), $(1 + x^2)\sinh x - x\cosh x \geq 0$.

Therefore $\frac{d}{dx}\left[\frac{\cosh^2 x}{1 + x^2}\right] \geq 0$ for $x \geq 0$, so $\frac{\cosh^2 x}{1 + x^2}$ is an increasing function for $x \geq 0$. $\qquad \blacksquare$

Part (ii)(d)

We can write:

$g'(x) = \frac{1}{1 + x^2} - \frac{\pi}{2}\cdot\frac{1}{\cosh^2 x} = \frac{1}{\cosh^2 x}\left[\frac{\cosh^2 x}{1 + x^2} - \frac{\pi}{2}\right]$

Since $\frac{1}{\cosh^2 x} > 0$ for all $x$, the sign of $g'(x)$ is determined by $\frac{\cosh^2 x}{1 + x^2} - \frac{\pi}{2}$.

By part (c), $\frac{\cosh^2 x}{1 + x^2}$ is increasing for $x \geq 0$. At $x = 0$: $\frac{\cosh^2 0}{1 + 0} = 1 < \frac{\pi}{2}$, so $g'(0) < 0$.

As $x \to \infty$: $\frac{\cosh^2 x}{1 + x^2} \to \infty$ (exponential growth dominates polynomial growth), so eventually $\frac{\cosh^2 x}{1 + x^2} > \frac{\pi}{2}$ and $g'(x) > 0$.

Since $\frac{\cosh^2 x}{1 + x^2}$ is continuous and strictly increasing for $x \geq 0$, there is exactly one $x_0 > 0$ where $\frac{\cosh^2 x_0}{1 + x_0^2} = \frac{\pi}{2}$, giving exactly one positive stationary point.

Since $g'$ is an even function, there is exactly one negative stationary point at $x = -x_0$.

Therefore $g$ has exactly two stationary points. $\qquad \blacksquare$

Part (ii)(e)

$g$ is an odd function: $g(-x) = -g(x)$.

At $x = 0$: $g(0) = 0$ and $g'(0) = 1 - \frac{\pi}{2} < 0$.

As $x \to +\infty$: $\tan^{-1} x \to \frac{\pi}{2}$ and $\tanh x \to 1$, so $g(x) \to \frac{\pi}{2} - \frac{\pi}{2} = 0$. The approach is from below since $g'(0) < 0$ and $g$ must dip below zero.

As $x \to -\infty$: $g(x) \to 0$ from above (by odd symmetry).

Key features:

• $g$ passes through the origin with negative slope.
• There is a local minimum at $x = x_0 > 0$ with $g(x_0) < 0$.
• By odd symmetry, there is a local maximum at $x = -x_0 < 0$ with $g(-x_0) = -g(x_0) > 0$.
• For $x > 0$: $g$ decreases from 0, reaches a minimum at $x_0$, then increases back toward 0.
• For $x < 0$: $g$ increases from 0, reaches a maximum at $-x_0$, then decreases back toward 0.

The graph is an odd, wave-like curve that starts from $0^-$ at $-\infty$, rises to a positive peak, passes through the origin with negative slope, falls to a negative trough, then returns to $0^+$ at $+\infty$.

Examiner Notes

第四受欢迎(75%作答),第二成功(平均分10+/20)。第(ii)(a)最难,常见错误:未论证 g(x) 或 g’(x) 在无穷远处的行为。第(d)部分常见错误:未考虑 cosh²x/(1+x²) 的对称性。第(e)部分常见错误:图像反射,x>0 时 g(x) 应为负值。

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Question 7

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

7 (i) Let $f$ be a continuous function defined for $0 \leqslant x \leqslant 1$. Show that

$\int_{0}^{1} f(\sqrt{x}) \, dx = 2 \int_{0}^{1} xf(x) \, dx \, .$

(ii) Let $g$ be a continuous function defined for $0 \leqslant x \leqslant 1$ such that

$\int_{0}^{1} (g(x))^2 \, dx = \int_{0}^{1} g(\sqrt{x}) \, dx - \frac{1}{3} \, .$

Show that $\int_{0}^{1} (g(x) - x)^2 \, dx = 0$ and explain why $g(x) = x$ for $0 \leqslant x \leqslant 1$.

(iii) Let $h$ be a continuous function defined for $0 \leqslant x \leqslant 1$ with derivative $h'$ such that

$\int_{0}^{1} (h'(x))^2 \, dx = 2h(1) - 2 \int_{0}^{1} h(x) \, dx - \frac{1}{3} \, .$

Given that $h(0) = 0$, find $h$.

(iv) Let $k$ be a continuous function defined for $0 \leqslant x \leqslant 1$ and $a$ be a real number, such that

$\int_{0}^{1} e^{ax} (k(x))^2 \, dx = 2 \int_{0}^{1} k(x) \, dx + \frac{e^{-a}}{a} - \frac{1}{a^2} - \frac{1}{4} \, .$

Show that $a$ must be equal to 2 and find $k$.

Hint

(i)

Let $x = u^2$, $\frac{dx}{du} = 2u$, $\sqrt{x} = u$ M1 $\int_{0}^{1} f(\sqrt{x}) \ dx = \int_{0}^{1} f(u) \ 2u \ du = 2 \int_{0}^{1} x f(x) \ dx$

as required. A1* (2)

(ii) $\int_{0}^{1} (g(x) - x)^2 \ dx = \int_{0}^{1} (g(x))^2 \ dx - 2 \int_{0}^{1} x \ g(x) \ dx + \int_{0}^{1} x^2 \ dx$

$= \int_{0}^{1} g(\sqrt{x}) \ dx - \frac{1}{3} - 2 \int_{0}^{1} x \ g(x) \ dx + \int_{0}^{1} x^2 \ dx$ M1 $= 2 \int_{0}^{1} x \ g(x) \ dx - \frac{1}{3} - 2 \int_{0}^{1} x \ g(x) \ dx + \left[ \frac{x^3}{3} \right]_{0}^{1}$

M1 $= 0 - \frac{1}{3} + \frac{1}{3} = 0$ A1*

$(g(x) - x)^2 \geq 0$

So, the area under the graph of $y = (g(x) - x)^2 \geq 0$, and the area can only equal zero if $(g(x) - x)^2 = 0$ for $0 \leq x \leq 1$, that is $g(x) = x$. E1 (4)

(iii) $\int_{0}^{1} (h'(x) - x)^2 \ dx = \int_{0}^{1} (h'(x))^2 - 2xh'(x) + x^2 \ dx$

M1

We are given that

$\int_{0}^{1} (h'(x))^2 = 2h(1) - 2 \int_{0}^{1} h(x) \ dx - \frac{1}{3}$

Integrating by parts

$\int_{0}^{1} 2xh'(x) \ dx = [2xh(x)]_{0}^{1} - 2 \int_{0}^{1} h(x) \ dx = 2h(1) - 2 \int_{0}^{1} h(x) \ dx$ M1 A1

and

$\int_{0}^{1} x^{2} dx = \left[ \frac{x^{3}}{3} \right]_{0}^{1} = \frac{1}{3}$

So,

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

A1

As in (ii) with g, $h'(x) = x$. Thus $h(x) = \frac{1}{2}x^{2} + c$ but $h(0) = 0$ so $c = 0$ and thus $h(x) = \frac{1}{2}x^{2}$

E1 M1 A1 A1 (8)

(iv)

$\int_{0}^{1} \left( e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} \right)^{2} dx = \int_{0}^{1} e^{ax} (k(x))^{2} - 2k(x) + e^{-ax} dx$

M1 dM1

$= 2 \int_{0}^{1} k(x) dx + \frac{e^{-a}}{a} - \frac{1}{a^{2}} - \frac{1}{4} - 2 \int_{0}^{1} k(x) dx - \left[ \frac{e^{-a}}{a} \right]_{0}^{1}$

$= \frac{e^{-a}}{a} - \frac{1}{a^{2}} - \frac{1}{4} - \frac{e^{-a}}{a} + \frac{1}{a} = -\frac{1}{a^{2}} + \frac{1}{a} - \frac{1}{4} = -\frac{4 - 4a + a^{2}}{4a^{2}} = -\frac{(2 - a)^{2}}{4a^{2}}$

A1 A1

As before, $\int_{0}^{1} \left( e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} \right)^{2} dx \geq 0$ but $-\frac{(2 - a)^{2}}{4a^{2}} \leq 0$ A1

Therefore, $\int_{0}^{1} \left( e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} \right)^{2} dx = 0$ and $\frac{(2 - a)^{2}}{4a^{2}} = 0$ E1

Thus $e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} = 0$ and $2 - a = 0$

So $a = 2$ and $k(x) = e^{-ax} = e^{-2x}$ A1 (6)

Model Solution

Part (i)

Substitute $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$. When $x = 0$, $u = 0$; when $x = 1$, $u = 1$.

$\int_0^1 f(\sqrt{x}) \, dx = \int_0^1 f(u) \cdot 2u \, du = 2 \int_0^1 u f(u) \, du = 2 \int_0^1 x f(x) \, dx$

where the last step relabels the dummy variable $u$ as $x$. $\qquad \blacksquare$

Part (ii)

Expand the integrand:

$\int_0^1 (g(x) - x)^2 \, dx = \int_0^1 (g(x))^2 \, dx - 2 \int_0^1 x \, g(x) \, dx + \int_0^1 x^2 \, dx \qquad \text{...(1)}$

From the given equation:

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

By part (i), $\int_0^1 g(\sqrt{x}) \, dx = 2 \int_0^1 x \, g(x) \, dx$, so:

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

Substituting into (1):

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

$= -\frac{1}{3} + \left[ \frac{x^3}{3} \right]_0^1 = -\frac{1}{3} + \frac{1}{3} = 0 \qquad \blacksquare$

Since $g$ is continuous, $(g(x) - x)^2$ is a continuous non-negative function on $[0, 1]$. A non-negative continuous function whose integral is zero must be identically zero: if $(g(x_0) - x_0)^2 > 0$ at some point $x_0$, then by continuity $(g(x) - x)^2 > 0$ in a neighbourhood of $x_0$, making the integral strictly positive, a contradiction.

Therefore $(g(x) - x)^2 = 0$ for all $0 \leqslant x \leqslant 1$, which gives $g(x) = x$. $\qquad \blacksquare$

Part (iii)

Expand the integrand:

$\int_0^1 (h'(x) - x)^2 \, dx = \int_0^1 (h'(x))^2 \, dx - 2 \int_0^1 x \, h'(x) \, dx + \int_0^1 x^2 \, dx \qquad \text{...(2)}$

Middle term (integration by parts, using $h(0) = 0$):

$\int_0^1 2x \, h'(x) \, dx = \left[ 2x \, h(x) \right]_0^1 - \int_0^1 2h(x) \, dx = 2h(1) - 2 \int_0^1 h(x) \, dx$

Last term:

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

Substituting into (2) along with the given equation $\int_0^1 (h'(x))^2 \, dx = 2h(1) - 2\int_0^1 h(x)\,dx - \frac{1}{3}$:

$\int_0^1 (h'(x) - x)^2 \, dx = \left( 2h(1) - 2 \int_0^1 h(x)\,dx - \frac{1}{3} \right) - \left( 2h(1) - 2 \int_0^1 h(x)\,dx \right) + \frac{1}{3} = 0$

Since $(h'(x) - x)^2$ is continuous and non-negative with zero integral, the same argument as in (ii) gives $h'(x) = x$ for all $0 \leqslant x \leqslant 1$.

Integrating: $h(x) = \frac{x^2}{2} + C$. Using $h(0) = 0$ gives $C = 0$, so:

$h(x) = \frac{x^2}{2} \qquad \blacksquare$

Part (iv)

Consider the integral of a perfect square:

$\int_0^1 \left( e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} \right)^2 dx$

Expanding the square:

$= \int_0^1 \left[ e^{ax}(k(x))^2 - 2 e^{\frac{1}{2}ax} k(x) \cdot e^{-\frac{1}{2}ax} + e^{-ax} \right] dx$

$= \int_0^1 e^{ax}(k(x))^2 \, dx - 2 \int_0^1 k(x) \, dx + \int_0^1 e^{-ax} \, dx$

Substituting the given equation for $\int_0^1 e^{ax}(k(x))^2 \, dx$:

$= \left( 2 \int_0^1 k(x) \, dx + \frac{e^{-a}}{a} - \frac{1}{a^2} - \frac{1}{4} \right) - 2 \int_0^1 k(x) \, dx + \int_0^1 e^{-ax} \, dx$

$= \frac{e^{-a}}{a} - \frac{1}{a^2} - \frac{1}{4} + \int_0^1 e^{-ax} \, dx$

Evaluating the remaining integral ($a \neq 0$):

$\int_0^1 e^{-ax} \, dx = \left[ -\frac{1}{a} e^{-ax} \right]_0^1 = -\frac{e^{-a}}{a} + \frac{1}{a} = \frac{1 - e^{-a}}{a}$

Therefore:

$\int_0^1 \left( e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} \right)^2 dx = \frac{e^{-a}}{a} - \frac{1}{a^2} - \frac{1}{4} + \frac{1 - e^{-a}}{a}$

$= \frac{1}{a} - \frac{1}{a^2} - \frac{1}{4} = \frac{4a - 4 - a^2}{4a^2} = -\frac{(a - 2)^2}{4a^2}$

The left-hand side is the integral of a non-negative continuous function, so it is $\geqslant 0$. The right-hand side is $-\frac{(a-2)^2}{4a^2} \leqslant 0$ for all $a \neq 0$.

Therefore both sides equal zero. From the right-hand side: $(a - 2)^2 = 0$, so $a = 2$. $\qquad \blacksquare$

From the left-hand side: $e^{\frac{1}{2}ax} k(x) - e^{-\frac{1}{2}ax} = 0$ for all $x \in [0, 1]$, giving $k(x) = e^{-ax} = e^{-2x}$.

$\boxed{a = 2, \quad k(x) = e^{-2x}}$

Examiner Notes

第三受欢迎,平均分9.4/20。第(ii)部分常见错误:未使用 (g(x)−x)²≥0 或错误声称严格大于零。第(iii)部分:对 xh’(x) 分部积分的考生普遍成功。第(iv)部分常见错误:未能分解关于 1/a 的二次式,或无理由地令二次式为零。

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Question 8

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

8 If $y = \begin{cases} k_1(x) & x \leqslant b \\ k_2(x) & x \geqslant b \end{cases}$ with $k_1(b) = k_2(b)$, then $y$ is said to be continuously differentiable at $x = b$ if $k'_1(b) = k'_2(b)$.

(i) Let $f(x) = xe^{-x}$. Verify that, for all real $x$, $y = f(x)$ is a solution to the differential equation $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0$ and that $y = 0$ and $\frac{dy}{dx} = 1$ when $x = 0$. Show that $f'(x) \geqslant 0$ for $x \leqslant 1$.

(ii) You are given the differential equation $\frac{d^2y}{dx^2} + 2\left| \frac{dy}{dx} \right| + y = 0$ where $y = 0$ and $\frac{dy}{dx} = 1$ when $x = 0$. Let $y = \begin{cases} g_1(x) & x \leqslant 1 \\ g_2(x) & x \geqslant 1 \end{cases}$ be a solution of the differential equation which is continuously differentiable at $x = 1$. Write down an expression for $g_1(x)$ and find an expression for $g_2(x)$.

(iii) State the geometrical relationship between the curves $y = g_1(x)$ and $y = g_2(x)$.

(iv) Prove that if $y = k(x)$ is a solution of the differential equation $\frac{d^2y}{dx^2} + p\frac{dy}{dx} + qy = 0$ in the interval $r \leqslant x \leqslant s$, where $p$ and $q$ are constants, then, in a suitable interval which you should state, $y = k(c - x)$ satisfies the differential equation $\frac{d^2y}{dx^2} - p\frac{dy}{dx} + qy = 0 \, .$

(v) You are given the differential equation

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2 \left| \frac{\mathrm{d}y}{\mathrm{d}x} \right| + 2y = 0$

where $y = 0$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ when $x = 0$.

Let $h(x) = e^{-x} \sin x$. Show that $h'(\frac{1}{4}\pi) = 0$.

It is given that $y = h(x)$ satisfies the differential equation in the interval $-\frac{3}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi$ and that $h'(x) \geqslant 0$ in this interval.

In a solution to the differential equation which is continuously differentiable at $(n + \frac{1}{4})\pi$ for all $n \in \mathbb{Z}$, find $y$ in terms of $x$ in the intervals

(a) $\frac{1}{4}\pi \leqslant x \leqslant \frac{5}{4}\pi$,

(b) $\frac{5}{4}\pi \leqslant x \leqslant \frac{9}{4}\pi$.

Hint

(i) $y = xe^{-x}$ $\frac{dy}{dx} = e^{-x} - xe^{-x} = (1 - x)e^{-x}$ $\frac{d^2y}{dx^2} = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$ $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = (x - 2)e^{-x} + 2(1 - x)e^{-x} + xe^{-x} = 0$

M1 A1

$x = 0, \quad y = xe^{-x} = 0, \quad \frac{dy}{dx} = (1 - x)e^{-x} = 1$

B1*

For $x \le 1$, $(1 - x) \ge 0$, $e^{-x} > 0$, so $\frac{dy}{dx} = (1 - x)e^{-x} \ge 0$ E1 (4)

(ii) From (i), $g_1(x) = xe^{-x}$

B1

Consider $y = g_2(x) = (a + bx)e^x$ for $x \ge 1$ B1

Then $g_2$ must be a solution of $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$, $g_1(1) = g_2(1)$, and $g'_1(1) = g'_2(1)$ $\frac{dy}{dx} = be^x + (a + bx)e^x = ((a + b) + bx)e^x$ $\frac{d^2y}{dx^2} = be^x + ((a + b) + bx)e^x = ((a + 2b) + bx)e^x$ $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = ((a + 2b) + bx)e^x - 2((a + b) + bx)e^x + (a + bx)e^x = 0$

as required.

$g_1(1) = g_2(1) \Rightarrow e^{-1} = (a + b)e$ $g'_1(1) = g'_2(1) \Rightarrow 0 = (a + 2b)e$

M1 A1

So $a = -2b$ and thus $b = -e^{-2}$

Hence, $g_2(x) = (2e^{-2} - e^{-2}x)e^x = (2 - x)e^{x-2}$ A1ft (5)

(iii) $y = g_2(x)$ is a reflection of $y = g_1(x)$ in $x = 1$, B1 which can be justified by substituting for $x$ using $x' = 2 - x$ in $y = g_1(x)$, $xe^{-x} = (2 - x')e^{x'-2}$ as expected. E1 (2)

(iv) If $y = k(c - x)$, then $\frac{dy}{dx} = -k'(c - x)$, and $\frac{d^2y}{dx^2} = k''(c - x)$ M1

So $\frac{d^2y}{dx^2} - p\frac{dy}{dx} + qy = k''(c - x) + pk'(c - x) + qk(c - x) = 0$ A1 provided that $r \leq c - x \leq s$

i.e. $c - s \leq x \leq c - r$ B1 (3)

(v) If $h(x) = e^{-x} \sin x$, then $h'(x) = -e^{-x} \sin x + e^{-x} \cos x$,

so $h'\left(\frac{\pi}{4}\right) = -e^{-\frac{\pi}{4}} \sin \frac{\pi}{4} + e^{-\frac{\pi}{4}} \cos \frac{\pi}{4} = -\frac{1}{\sqrt{2}}e^{-\frac{\pi}{4}} + \frac{1}{\sqrt{2}}e^{-\frac{\pi}{4}} = 0$ as required. B1*

(a) Using (iv), the solution for $\frac{1}{4}\pi \leq x \leq \frac{5}{4}\pi$ must be $y = e^{-(c-x)} \sin(c - x)$ M1 where

$-\frac{3}{4}\pi \leq (c - x) \leq \frac{1}{4}\pi$. M1 That is $c = \frac{1}{2}\pi$. So $y = e^{x-\frac{1}{2}\pi} \cos x$ A1

(b) Similarly, the solution for $\frac{5}{4}\pi \leq x \leq \frac{9}{4}\pi$ must be $y = e^{c-x-\frac{1}{2}\pi} \cos(c - x)$ where

$\frac{1}{4}\pi \leq (c - x) \leq \frac{5}{4}\pi$. That is $c = \frac{5}{2}\pi$. B1 So $y = e^{2\pi-x} \sin x$ B1 (6)

Model Solution

Part (i)

Let $f(x) = xe^{-x}$. We compute the derivatives:

$f'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}$

$f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$

Substituting into the left-hand side of the differential equation:

$f''(x) + 2f'(x) + f(x) = (x - 2)e^{-x} + 2(1 - x)e^{-x} + xe^{-x}$

$= e^{-x}\big[(x - 2) + 2(1 - x) + x\big] = e^{-x}[x - 2 + 2 - 2x + x] = 0$

So $y = xe^{-x}$ satisfies the differential equation for all real $x$. $\qquad \blacksquare$

At $x = 0$:

$y = 0 \cdot e^{0} = 0, \qquad \frac{dy}{dx} = (1 - 0)e^{0} = 1$

as required. $\qquad \blacksquare$

For $x \leqslant 1$: we have $(1 - x) \geqslant 0$ and $e^{-x} > 0$ for all real $x$. Therefore:

$f'(x) = (1 - x)e^{-x} \geqslant 0 \qquad \text{for } x \leqslant 1 \qquad \blacksquare$

Part (ii)

Since $y'(0) = 1 > 0$, the solution initially has $y' > 0$, so $|y'| = y'$ and the differential equation becomes:

$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0$

This is the same equation as in part (i), with the same initial conditions $y(0) = 0$ and $y'(0) = 1$. By part (i), $g_1(x) = xe^{-x}$ is the solution while $y' \geqslant 0$.

Since $g_1'(x) = (1 - x)e^{-x}$, we have $g_1'(x) \geqslant 0$ for $x \leqslant 1$ and $g_1'(1) = 0$. For $x > 1$, $g_1'(x) < 0$, so the sign of $y'$ changes at $x = 1$.

Therefore $g_1(x) = xe^{-x}$ for $x \leqslant 1$.

For $x \geqslant 1$, we have $y' \leqslant 0$ (by the continuously differentiable condition, $y'(1) = 0$ and $y'$ becomes negative), so $|y'| = -y'$ and the differential equation becomes:

$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$

The auxiliary equation is $r^2 - 2r + 1 = 0$, giving $(r - 1)^2 = 0$, so $r = 1$ is a repeated root. The general solution is:

$g_2(x) = (A + Bx)e^x$

We determine $A$ and $B$ from the continuous differentiability conditions at $x = 1$.

Matching function values $g_1(1) = g_2(1)$:

$1 \cdot e^{-1} = (A + B)e \implies A + B = e^{-2} \qquad \text{...(1)}$

Matching derivatives $g_1'(1) = g_2'(1)$. First:

$g_2'(x) = Be^x + (A + Bx)e^x = (A + B + Bx)e^x$

$g_2'(1) = (A + 2B)e$

Since $g_1'(1) = 0$:

$(A + 2B)e = 0 \implies A + 2B = 0 \implies A = -2B \qquad \text{...(2)}$

Substituting (2) into (1):

$-2B + B = e^{-2} \implies -B = e^{-2} \implies B = -e^{-2}$

$A = -2(-e^{-2}) = 2e^{-2}$

Therefore:

$g_2(x) = (2e^{-2} - e^{-2}x)e^x = (2 - x)e^{x - 2}$

We verify that $g_2'(x) < 0$ for $x > 1$ to confirm the equation remains $y'' - 2y' + y = 0$:

$g_2'(x) = (1 - x)e^{x - 2}$

For $x > 1$: $(1 - x) < 0$ and $e^{x-2} > 0$, so $g_2'(x) < 0$. This confirms $|y'| = -y'$ for $x > 1$.

$\boxed{g_1(x) = xe^{-x}, \qquad g_2(x) = (2 - x)e^{x - 2}}$

Part (iii)

The curve $y = g_2(x)$ is the reflection of $y = g_1(x)$ in the vertical line $x = 1$.

To see this, substitute $x' = 2 - x$ (which reflects in $x = 1$) into $g_1$:

$g_1(2 - x) = (2 - x)e^{-(2 - x)} = (2 - x)e^{x - 2} = g_2(x)$

So replacing $x$ by $2 - x$ in $g_1$ gives $g_2$, confirming the reflection. $\qquad \blacksquare$

Part (iv)

Let $y = k(x)$ be a solution of $y'' + py' + qy = 0$ on $[r, s]$, and let $m(x) = k(c - x)$.

Differentiating using the chain rule:

$m'(x) = -k'(c - x)$

$m''(x) = k''(c - x)$

Substituting into $y'' - py' + qy$:

$m''(x) - pm'(x) + qm(x) = k''(c - x) + pk'(c - x) + qk(c - x)$

Since $k$ satisfies $k'' + pk' + qk = 0$ for all values of its argument in $[r, s]$, and $c - x$ is the argument, we need $c - x \in [r, s]$.

This requires $r \leqslant c - x \leqslant s$, i.e., $c - s \leqslant x \leqslant c - r$.

Therefore $y = k(c - x)$ satisfies $y'' - py' + qy = 0$ in the interval $c - s \leqslant x \leqslant c - r$. $\qquad \blacksquare$

Part (v)

Showing $h'(\frac{1}{4}\pi) = 0$:

$h(x) = e^{-x}\sin x$

$h'(x) = -e^{-x}\sin x + e^{-x}\cos x = e^{-x}(\cos x - \sin x)$

$h'\left(\frac{\pi}{4}\right) = e^{-\pi/4}\left(\cos\frac{\pi}{4} - \sin\frac{\pi}{4}\right) = e^{-\pi/4}\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = 0 \qquad \blacksquare$

(a) Finding $y$ for $\frac{1}{4}\pi \leqslant x \leqslant \frac{5}{4}\pi$:

On the interval $[-\frac{3}{4}\pi, \frac{1}{4}\pi]$, $y = h(x) = e^{-x}\sin x$ satisfies $y'' + 2y' + 2y = 0$ (since $h'(x) \geqslant 0$, so $|y'| = y'$).

At $x = \frac{1}{4}\pi$, $h'(\frac{1}{4}\pi) = 0$. For $x$ just beyond $\frac{1}{4}\pi$, $\cos x - \sin x < 0$, so $y'$ becomes negative. Therefore on $[\frac{1}{4}\pi, \frac{5}{4}\pi]$, $|y'| = -y'$ and the equation becomes:

$y'' - 2y' + 2y = 0$

By part (iv) with $p = 2$: if $y = k(x)$ solves $y'' + 2y' + 2y = 0$, then $y = k(c - x)$ solves $y'' - 2y' + 2y = 0$ on the interval $c - s \leqslant x \leqslant c - r$.

We need $k(c - x)$ to use the known solution $h$ on $[-\frac{3}{4}\pi, \frac{1}{4}\pi]$, so we require $c - x \in [-\frac{3}{4}\pi, \frac{1}{4}\pi]$ for $x \in [\frac{1}{4}\pi, \frac{5}{4}\pi]$.

At $x = \frac{1}{4}\pi$: $c - \frac{1}{4}\pi \leqslant \frac{1}{4}\pi$, so $c \leqslant \frac{1}{2}\pi$.

At $x = \frac{5}{4}\pi$: $c - \frac{5}{4}\pi \geqslant -\frac{3}{4}\pi$, so $c \geqslant \frac{1}{2}\pi$.

Therefore $c = \frac{1}{2}\pi$ and the solution is:

$y = h\left(\frac{\pi}{2} - x\right) = e^{-(\pi/2 - x)}\sin\left(\frac{\pi}{2} - x\right) = e^{x - \pi/2}\cos x$

Verification of continuous differentiability at $x = \frac{1}{4}\pi$:

From the left: $h(\frac{\pi}{4}) = e^{-\pi/4}\sin\frac{\pi}{4} = \frac{e^{-\pi/4}}{\sqrt{2}}$.

From the right: $e^{\pi/4 - \pi/2}\cos\frac{\pi}{4} = e^{-\pi/4} \cdot \frac{1}{\sqrt{2}} = \frac{e^{-\pi/4}}{\sqrt{2}}$. Values match. $\checkmark$

From the left: $h'(\frac{\pi}{4}) = 0$.

From the right: $\frac{d}{dx}[e^{x - \pi/2}\cos x] = e^{x - \pi/2}(\cos x - \sin x)$, at $x = \frac{\pi}{4}$: $e^{-\pi/4} \cdot 0 = 0$. Derivatives match. $\checkmark$

(b) Finding $y$ for $\frac{5}{4}\pi \leqslant x \leqslant \frac{9}{4}\pi$:

On $[\frac{1}{4}\pi, \frac{5}{4}\pi]$, the solution is $m(x) = e^{x - \pi/2}\cos x$, satisfying $y'' - 2y' + 2y = 0$ (since $m'(x) = e^{x-\pi/2}(\cos x - \sin x) \leqslant 0$ there).

At $x = \frac{5}{4}\pi$: $m'(\frac{5}{4}\pi) = e^{3\pi/4}(\cos\frac{5\pi}{4} - \sin\frac{5}{4}\pi) = e^{3\pi/4}(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 0$.

For $x$ just beyond $\frac{5}{4}\pi$: $\cos x - \sin x > 0$ (e.g., at $x = \frac{3}{2}\pi$: $\cos\frac{3\pi}{2} - \sin\frac{3\pi}{2} = 0 + 1 = 1$), so $y'$ becomes positive. Therefore $|y'| = y'$ and the equation returns to $y'' + 2y' + 2y = 0$.

By part (iv) with $p = -2$: if $y = m(x)$ solves $y'' - 2y' + 2y = 0$, then $y = m(c - x)$ solves $y'' + 2y' + 2y = 0$ on $[c - \frac{5}{4}\pi, c - \frac{1}{4}\pi]$.

We need $[c - \frac{5}{4}\pi, c - \frac{1}{4}\pi] = [\frac{5}{4}\pi, \frac{9}{4}\pi]$:

$c - \frac{5}{4}\pi = \frac{5}{4}\pi \implies c = \frac{5}{2}\pi$

Check: $c - \frac{1}{4}\pi = \frac{5}{2}\pi - \frac{1}{4}\pi = \frac{9}{4}\pi$. $\checkmark$

The solution is:

$y = m\left(\frac{5\pi}{2} - x\right) = e^{(5\pi/2 - x) - \pi/2}\cos\left(\frac{5\pi}{2} - x\right) = e^{2\pi - x}\cos\left(\frac{5\pi}{2} - x\right)$

Since $\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$:

$\cos\left(\frac{5\pi}{2} - x\right) = \cos\left(2\pi + \frac{\pi}{2} - x\right) = \cos\left(\frac{\pi}{2} - x\right) = \sin x$

Therefore:

$\boxed{y = e^{2\pi - x}\sin x}$

Verification of continuous differentiability at $x = \frac{5}{4}\pi$:

From the left: $e^{5\pi/4 - \pi/2}\cos\frac{5\pi}{4} = e^{3\pi/4} \cdot \left(-\frac{1}{\sqrt{2}}\right)$.

From the right: $e^{2\pi - 5\pi/4}\sin\frac{5\pi}{4} = e^{3\pi/4} \cdot \left(-\frac{1}{\sqrt{2}}\right)$. Values match. $\checkmark$

From the left: $m'(\frac{5\pi}{4}) = 0$ (computed above).

From the right: $\frac{d}{dx}[e^{2\pi - x}\sin x] = e^{2\pi - x}(\cos x - \sin x)$, at $x = \frac{5\pi}{4}$: $e^{3\pi/4}(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 0$. Derivatives match. $\checkmark$

Examiner Notes

中等受欢迎(略低于Q2),平均分7.3/20。第(ii)部分常见错误:直接翻转 g₁ 的符号而非重新求解。第(iii)部分极差。第(v)(b)部分常见错误:在 −π/4 而非 5π/4 处匹配函数值。