STEP2 2016 -- Pure Mathematics

STEP2 2016 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2016  |  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	曲线与方程 (Curves and Equations)	Standard	参数方程求导,切线方程,向量垂直条件,联立方程消参,多项式因式分解
2	代数 (Algebra)	Standard	因式定理,对称多项式分解,变量代换,方程求根
3	多项式与级数 (Polynomials and Series)	Challenging	逐项求导,多项式符号分析,反证法（无穷根矛盾）,中间值定理,无穷远处渐近行为
4	不等式与函数 (Inequalities and Functions)	Challenging	二次方程判别式,辅助角公式 R cos(theta+alpha),不等式传递性,根式化简
5	组合数学 (Combinatorics)	Challenging	负二项式展开,幂级数乘法比较系数,Vandermonde 恒等式,组合数定义扩展
6	微分方程 (Differential Equations)	Challenging	验证微分方程的解, 链式法则, 函数复合, Chebyshev多项式性质
7	积分 (Integration)	Standard	积分换元 $u=a-x$, 三角恒等式, 对数运算, 分部积分
8	级数与积分近似 (Series and Integration Approximation)	Challenging	积分近似级数, 草图论证, 误差分析, 中点近似

---

Question 1

Topic: 曲线与方程 (Curves and Equations)  |  Difficulty: Standard  |  Marks: 20

Problem

1 The curve $C_1$ has parametric equations $x = t^2$, $y = t^3$, where $-\infty < t < \infty$. Let $O$ denote the point $(0, 0)$. The points $P$ and $Q$ on $C_1$ are such that $\angle POQ$ is a right angle. Show that the tangents to $C_1$ at $P$ and $Q$ intersect on the curve $C_2$ with equation $4y^2 = 3x - 1$.

Determine whether $C_1$ and $C_2$ meet, and sketch the two curves on the same axes.

Hint

Let $P$ have parameter $t_1$ and $Q$ have parameter $t_2$ on $C_1$.

The derivative $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}$ (for $t \neq 0$).

The tangent at $P(t_1)$: $y - t_1^3 = \frac{3t_1}{2}(x - t_1^2)$, i.e. $y = \frac{3t_1}{2}x - \frac{t_1^3}{2}$.

Similarly the tangent at $Q(t_2)$: $y = \frac{3t_2}{2}x - \frac{t_2^3}{2}$.

Since $\angle POQ = 90°$, we need $\vec{OP} \cdot \vec{OQ} = 0$: $t_1^2 t_2^2 + t_1^3 t_2^3 = 0 \implies t_1^2 t_2^2(1 + t_1 t_2) = 0$

Since $P, Q \neq O$, we have $t_1 t_2 = -1$.

Setting the two tangent equations equal to find intersection: $\frac{3t_1}{2}x - \frac{t_1^3}{2} = \frac{3t_2}{2}x - \frac{t_2^3}{2}$

$\frac{3}{2}(t_1 - t_2)x = \frac{1}{2}(t_1^3 - t_2^3) = \frac{1}{2}(t_1 - t_2)(t_1^2 + t_1 t_2 + t_2^2)$

So $x = \frac{t_1^2 + t_1 t_2 + t_2^2}{3} = \frac{t_1^2 + t_2^2 - 1}{3}$.

Substituting back: $y = \frac{3t_1}{2} \cdot \frac{t_1^2 + t_2^2 - 1}{3} - \frac{t_1^3}{2} = \frac{t_1(t_2^2 - 1)}{2}$.

Using $t_2 = -1/t_1$: $x = \frac{t_1^2 + 1/t_1^2 - 1}{3}$ and $y = \frac{t_1(1/t_1^2 - 1)}{2} = \frac{1 - t_1^2}{2t_1}$.

Let $u = t_1^2$. Then $x = \frac{u + 1/u - 1}{3}$ and $y^2 = \frac{(1-u)^2}{4u} = \frac{u^2 - 2u + 1}{4u} = \frac{u + 1/u - 2}{4}$.

So $u + 1/u = 3x + 1$, giving $y^2 = \frac{3x + 1 - 2}{4} = \frac{3x - 1}{4}$, i.e. $4y^2 = 3x - 1$.

This is the curve $C_2$, as required.

Do $C_1$ and $C_2$ meet? Substituting $x = t^2, y = t^3$ into $4y^2 = 3x - 1$: $4t^6 = 3t^2 - 1 \implies 4t^6 - 3t^2 + 1 = 0$

Let $u = t^2 \geq 0$: $4u^3 - 3u + 1 = 0$. Testing $u = -1$: $4(-1) - 3(-1) + 1 = -4+3+1 = 0$, so $u+1$ is a factor.

$4u^3 - 3u + 1 = (u+1)(4u^2 - 4u + 1) = (u+1)(2u-1)^2$.

Since $u = t^2 \geq 0$, the only solution is $u = 1/2$, i.e. $t = \pm 1/\sqrt{2}$. Since $(2u-1)^2$ is a repeated root, the curves are tangent (touch but do not cross) at the points where $t = \pm 1/\sqrt{2}$.

Sketch: $C_1$ is a semicubical parabola (cuspidal curve) $y^2 = x^3$ symmetric about the $x$-axis with a cusp at the origin. $C_2$ is a rightward-opening parabola $x = \frac{4y^2+1}{3}$ with vertex at $(1/3, 0)$. They touch at two points where $x = 1/2, y = \pm 1/(2\sqrt{2})$.

Model Solution

Finding the tangent lines.

The curve $C_1$ has parametric equations $x = t^2$, $y = t^3$. We compute:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2} \quad (t \neq 0)$

The tangent to $C_1$ at a general point with parameter $t$ is:

$y - t^3 = \frac{3t}{2}(x - t^2)$

Rearranging:

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

So the tangent at $P$ (parameter $t_1$) is $y = \frac{3t_1}{2}x - \frac{t_1^3}{2}$ and the tangent at $Q$ (parameter $t_2$) is $y = \frac{3t_2}{2}x - \frac{t_2^3}{2}$.

Using the right angle condition.

Since $\angle POQ = 90^\circ$, we need $\vec{OP} \cdot \vec{OQ} = 0$:

$t_1^2 \cdot t_2^2 + t_1^3 \cdot t_2^3 = 0$

$t_1^2 t_2^2(1 + t_1 t_2) = 0$

Since $P$ and $Q$ are not the origin, $t_1 \neq 0$ and $t_2 \neq 0$, so $t_1 t_2 = -1$.

Finding the intersection point.

Setting the two tangent equations equal:

$\frac{3t_1}{2}x - \frac{t_1^3}{2} = \frac{3t_2}{2}x - \frac{t_2^3}{2}$

$\frac{3}{2}(t_1 - t_2)x = \frac{1}{2}(t_1^3 - t_2^3) = \frac{1}{2}(t_1 - t_2)(t_1^2 + t_1 t_2 + t_2^2)$

Since $t_1 \neq t_2$ (otherwise $P = Q$), we can divide by $(t_1 - t_2)$:

$x = \frac{t_1^2 + t_1 t_2 + t_2^2}{3}$

Substituting $t_1 t_2 = -1$:

$x = \frac{t_1^2 + t_2^2 - 1}{3}$

Substituting back into the tangent at $P$:

$y = \frac{3t_1}{2} \cdot \frac{t_1^2 + t_2^2 - 1}{3} - \frac{t_1^3}{2} = \frac{t_1(t_1^2 + t_2^2 - 1) - t_1^3}{2} = \frac{t_1 t_2^2 - t_1}{2} = \frac{t_1(t_2^2 - 1)}{2}$

Eliminating the parameters.

Using $t_2 = -1/t_1$:

$x = \frac{t_1^2 + \frac{1}{t_1^2} - 1}{3}, \qquad y = \frac{t_1\!\left(\frac{1}{t_1^2} - 1\right)}{2} = \frac{1 - t_1^2}{2t_1}$

Let $u = t_1^2$. Then $x = \frac{u + u^{-1} - 1}{3}$ and $y^2 = \frac{(1 - u)^2}{4u} = \frac{u^2 - 2u + 1}{4u} = \frac{u + u^{-1} - 2}{4}$.

From the expression for $x$: $u + u^{-1} = 3x + 1$, so:

$y^2 = \frac{3x + 1 - 2}{4} = \frac{3x - 1}{4}$

$4y^2 = 3x - 1$

This is the equation of $C_2$, as required.

Do $C_1$ and $C_2$ meet?

Substituting $x = t^2$, $y = t^3$ into $4y^2 = 3x - 1$:

$4t^6 = 3t^2 - 1 \implies 4t^6 - 3t^2 + 1 = 0$

Setting $u = t^2 \geq 0$: $4u^3 - 3u + 1 = 0$. Testing $u = -1$: $4(-1) - 3(-1) + 1 = -4 + 3 + 1 = 0$, so $(u + 1)$ is a factor.

$4u^3 - 3u + 1 = (u + 1)(4u^2 - 4u + 1) = (u + 1)(2u - 1)^2$

Since $u = t^2 \geq 0$, the factor $(u + 1)$ gives no solutions. The repeated factor $(2u - 1)^2 = 0$ gives $u = 1/2$, i.e. $t = \pm 1/\sqrt{2}$.

The curves meet at two points: $(1/2, \pm 1/(2\sqrt{2}))$. Since the root $u = 1/2$ is repeated, the curves are tangent at these points (they touch but do not cross).

Sketch. $C_1$: $y^2 = x^3$ is a semicubical parabola with a cusp at the origin, symmetric about the $x$-axis. $C_2$: $x = (4y^2 + 1)/3$ is a rightward-opening parabola with vertex at $(1/3, 0)$. The two curves touch at $(1/2, \pm 1/(2\sqrt{2}))$.

Examiner Notes

This was a popular question and many very good solutions were seen. The first part of the question was a relatively straightforward application of differentiation and parametric equations and was successfully completed by many of the candidates. The sketches produced were generally the correct shape for the parabola, although in some cases it was not in the correct position. The other curve caused more problems with some candidates drawing another parabola or similar shape.

---

Question 2

Topic: 代数 (Algebra)  |  Difficulty: Standard  |  Marks: 20

Problem

2 Use the factor theorem to show that $a + b - c$ is a factor of

$(a + b + c)^3 - 6(a + b + c)(a^2 + b^2 + c^2) + 8(a^3 + b^3 + c^3).$ (*)

Hence factorise (*) completely.

(i) Use the result above to solve the equation

$(x + 1)^3 - 3(x + 1)(2x^2 + 5) + 2(4x^3 + 13) = 0.$

(ii) By setting $d + e = c$, or otherwise, show that $(a + b - d - e)$ is a factor of

$(a + b + d + e)^3 - 6(a + b + d + e)(a^2 + b^2 + d^2 + e^2) + 8(a^3 + b^3 + d^3 + e^3)$

and factorise this expression completely.

Hence solve the equation

$(x + 6)^3 - 6(x + 6)(x^2 + 14) + 8(x^3 + 36) = 0.$

Hint

Showing $a+b-c$ is a factor:

Substitute $c = a+b$ into the expression: $(a+b+(a+b))^3 - 6(a+b+(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3)$

$= (2a+2b)^3 - 6(2a+2b)(a^2+b^2+a^2+2ab+b^2) + 8(a^3+b^3+a^3+3a^2b+3ab^2+b^3)$

$= 8(a+b)^3 - 12(a+b)(2a^2+2ab+2b^2) + 8(2a^3+2b^3+3a^2b+3ab^2)$

$= 8(a+b)^3 - 24(a+b)(a^2+ab+b^2) + 8(2(a+b)(a^2-ab+b^2)+3ab(a+b))$

$= 8(a+b)[(a+b)^2 - 3(a^2+ab+b^2) + 2a^2-2ab+2b^2+3ab]$

$= 8(a+b)[a^2+2ab+b^2 - 3a^2-3ab-3b^2 + 2a^2+ab+2b^2]$

$= 8(a+b) \cdot 0 = 0$

So by the factor theorem, $a+b-c$ is a factor.

Complete factorisation: By symmetry in $a, b, c$, the factors $b+c-a$ and $c+a-b$ must also be factors. Since the expression has degree 3 and each factor has degree 1, we need a constant multiplier $k$:

$(a+b+c)^3 - 6(a+b+c)(a^2+b^2+c^2) + 8(a^3+b^3+c^3) = k(a+b-c)(b+c-a)(c+a-b)$

Setting $a=1, b=0, c=0$: LHS $= 1 - 6 + 8 = 3$. RHS $= k(1)(-1)(1) = -k$. So $k = -3$.

The complete factorisation is $-3(a+b-c)(b+c-a)(c+a-b)$.

(i) $(x+1)^3 - 3(x+1)(2x^2+5) + 2(4x^3+13) = 0$.

Expanding: $(x+1)^3 = x^3+3x^2+3x+1$ $3(x+1)(2x^2+5) = 6x^3+6x^2+15x+15$ $2(4x^3+13) = 8x^3+26$

LHS $= x^3+3x^2+3x+1 - 6x^3-6x^2-15x-15 + 8x^3+26 = 3x^3-3x^2-12x+12$

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

Solutions: $x = 1, 2, -2$.

(ii) Setting $d+e = c$ in the 4-variable version. By the same factor theorem argument (substituting $e = d + a + b$ shows $a+b-d-e$ is a factor), and by symmetry, the factors $a+d-b-e$ and $a+e-b-d$ are also factors.

The complete factorisation is $-3(a+b-d-e)(a+d-b-e)(a+e-b-d)$.

For $(x+6)^3 - 6(x+6)(x^2+14) + 8(x^3+36) = 0$:

$(x+6)^3 = x^3+18x^2+108x+216$ $6(x+6)(x^2+14) = 6x^3+36x^2+84x+504$ $8(x^3+36) = 8x^3+288$

LHS $= x^3+18x^2+108x+216 - 6x^3-36x^2-84x-504 + 8x^3+288$ $= 3x^3 - 18x^2 + 24x = 3x(x^2-6x+8) = 3x(x-2)(x-4)$

Solutions: $x = 0, 2, 4$.

Model Solution

Showing $a + b - c$ is a factor.

By the factor theorem, we substitute $c = a + b$ into the expression $(a + b + c)^3 - 6(a + b + c)(a^2 + b^2 + c^2) + 8(a^3 + b^3 + c^3)$.

With $c = a + b$:

$a + b + c = 2(a + b), \qquad c^2 = (a + b)^2 = a^2 + 2ab + b^2$

$a^2 + b^2 + c^2 = 2a^2 + 2ab + 2b^2 = 2(a^2 + ab + b^2)$

$c^3 = (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

$a^3 + b^3 + c^3 = 2a^3 + 3a^2b + 3ab^2 + 2b^3 = 2(a^3 + b^3) + 3ab(a + b)$

So the expression becomes:

$[2(a+b)]^3 - 6 \cdot 2(a+b) \cdot 2(a^2+ab+b^2) + 8[2(a^3+b^3) + 3ab(a+b)]$

$= 8(a+b)^3 - 24(a+b)(a^2+ab+b^2) + 16(a^3+b^3) + 24ab(a+b)$

Factor out $(a + b)$ from the first and third terms using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$:

$= 8(a+b)^3 - 24(a+b)(a^2+ab+b^2) + 16(a+b)(a^2 - ab + b^2) + 24ab(a+b)$

$= (a+b)\left[8(a+b)^2 - 24(a^2+ab+b^2) + 16(a^2 - ab + b^2) + 24ab\right]$

Expanding the bracket:

$8(a^2 + 2ab + b^2) - 24(a^2 + ab + b^2) + 16(a^2 - ab + b^2) + 24ab$

$= (8 - 24 + 16)a^2 + (16 - 24 - 16 + 24)ab + (8 - 24 + 16)b^2$

$= 0 \cdot a^2 + 0 \cdot ab + 0 \cdot b^2 = 0$

Since substituting $c = a + b$ gives zero, $(a + b - c)$ is a factor by the factor theorem.

Complete factorisation.

The expression is symmetric in $a, b, c$. By the same argument, $(b + c - a)$ and $(c + a - b)$ are also factors. Since the expression has degree 3 and each factor has degree 1, we have:

$(a+b+c)^3 - 6(a+b+c)(a^2+b^2+c^2) + 8(a^3+b^3+c^3) = k(a+b-c)(b+c-a)(c+a-b)$

for some constant $k$. Setting $a = 1, b = 0, c = 0$:

$\text{LHS} = 1 - 6 + 8 = 3, \qquad \text{RHS} = k(1)(-1)(1) = -k$

So $k = -3$, giving the complete factorisation:

$-3(a+b-c)(b+c-a)(c+a-b)$

Part (i). We need to solve $(x+1)^3 - 3(x+1)(2x^2+5) + 2(4x^3+13) = 0$.

Expanding each term:

$(x+1)^3 = x^3 + 3x^2 + 3x + 1$

$3(x+1)(2x^2+5) = 3(2x^3 + 5x + 2x^2 + 5) = 6x^3 + 6x^2 + 15x + 15$

$2(4x^3+13) = 8x^3 + 26$

Combining:

$x^3 + 3x^2 + 3x + 1 - 6x^3 - 6x^2 - 15x - 15 + 8x^3 + 26 = 3x^3 - 3x^2 - 12x + 12$

$= 3(x^3 - x^2 - 4x + 4) = 3[x^2(x - 1) - 4(x - 1)] = 3(x - 1)(x^2 - 4) = 3(x - 1)(x - 2)(x + 2)$

Solutions: $x = 1, 2, -2$.

Part (ii). We show $(a + b - d - e)$ is a factor of the four-variable expression. Setting $e = c - d$ (i.e. $d + e = c$) reduces the four-variable expression to the three-variable expression. When $a + b = d + e = c$, we have $c = a + b$ and $e = a + b - d$, so the expression equals zero. By the factor theorem, $(a + b - d - e)$ is a factor.

By symmetry in $a, b, d, e$, the factors $(a + d - b - e)$ and $(a + e - b - d)$ are also factors. The complete factorisation is:

$-3(a + b - d - e)(a + d - b - e)(a + e - b - d)$

Solving the equation $(x+6)^3 - 6(x+6)(x^2+14) + 8(x^3+36) = 0$.

Expanding:

$(x+6)^3 = x^3 + 18x^2 + 108x + 216$

$6(x+6)(x^2+14) = 6(x^3 + 14x + 6x^2 + 84) = 6x^3 + 36x^2 + 84x + 504$

$8(x^3+36) = 8x^3 + 288$

Combining:

$x^3 + 18x^2 + 108x + 216 - 6x^3 - 36x^2 - 84x - 504 + 8x^3 + 288 = 3x^3 - 18x^2 + 24x$

$= 3x(x^2 - 6x + 8) = 3x(x - 2)(x - 4)$

Solutions: $x = 0, 2, 4$.

Examiner Notes

This question received poor responses in terms of the average mark per attempt. Application of the factor theorem to the first part often produced a neat solution. However, a number of candidates in the first part did not use the factor theorem as requested and so produced very complicated algebraic expressions that were more of a challenge to simplify. Where the result had been successfully shown and the expression factorised, many candidates were able to see the relevance to solving the equation in part (i). Those candidates who were able to follow through the method to successfully factorise the second expression were often able to complete the question.

---

Question 3

Topic: 多项式与级数 (Polynomials and Series)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 For each non-negative integer $n$, the polynomial $f_n$ is defined by

$f_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!} .$

(i) Show that $f'_n(x) = f_{n-1}(x)$ (for $n \geqslant 1$).

(ii) Show that, if $a$ is a real root of the equation

$f_n(x) = 0 , \qquad \text{(*)}$

then $a < 0$.

(iii) Let $a$ and $b$ be distinct real roots of ($*$), for $n \geqslant 2$. Show that $f'_n(a) f'_n(b) > 0$ and use a sketch to deduce that $f_n(c) = 0$ for some number $c$ between $a$ and $b$.

Deduce that ($*$) has at most one real root. How many real roots does ($*$) have if $n$ is odd? How many real roots does ($*$) have if $n$ is even?

Hint

(i) $f_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}$, so

$f_n'(x) = \sum_{k=1}^{n} \frac{kx^{k-1}}{k!} = \sum_{k=1}^{n} \frac{x^{k-1}}{(k-1)!} = \sum_{j=0}^{n-1} \frac{x^j}{j!} = f_{n-1}(x)$

(ii) If $a > 0$, then every term $\frac{a^k}{k!} > 0$ for $k \geq 0$, so $f_n(a) > 0$. If $a = 0$, then $f_n(0) = 1 \neq 0$. Therefore any real root must satisfy $a < 0$.

(iii) Suppose $a$ and $b$ are distinct real roots with $a < b$. Then $f_n(a) = f_n(b) = 0$.

From part (i), $f_n'(x) = f_{n-1}(x)$ and $f_n(x) = f_{n-1}(x) + \frac{x^n}{n!}$, so $f_n'(x) = f_n(x) - \frac{x^n}{n!}$.

At a root: $f_n'(a) = f_n(a) - \frac{a^n}{n!} = -\frac{a^n}{n!}$ and $f_n'(b) = -\frac{b^n}{n!}$.

Since $a, b < 0$ (from part (ii)), $a^n$ and $b^n$ have the same sign (both positive if $n$ is even, both negative if $n$ is odd).

So $f_n'(a) \cdot f_n'(b) = \frac{a^n b^n}{(n!)^2} = \frac{(ab)^n}{(n!)^2} > 0$ (since $ab > 0$).

This means $f_n'(a)$ and $f_n'(b)$ have the same sign, so the tangent lines at $a$ and $b$ slope in the same direction. Since $f_n$ is a polynomial (hence continuous and differentiable), and $f_n(a) = f_n(b) = 0$ with same-sign derivatives, by the graph $f_n$ must cross zero again between $a$ and $b$ — say at some $c$ with $a < c < b$.

But then $c$ is also a root, and we can repeat the argument between $a$ and $c$, and between $c$ and $b$, finding infinitely many roots. Since $f_n$ is a polynomial of degree $n$ (finite), this is a contradiction. Therefore $f_n(x) = 0$ has at most one real root.

Number of roots:

• If $n$ is odd: $f_n(x) \to +\infty$ as $x \to +\infty$ and $f_n(x) \to -\infty$ as $x \to -\infty$ (leading term $x^n/n!$ dominates). By the Intermediate Value Theorem, there is at least one real root. Combined with “at most one”, there is exactly one real root.
• If $n$ is even: $f_n(x) \to +\infty$ as $x \to \pm\infty$. Since $f_n(0) = 1 > 0$ and $f_n$ has at most one real root, and any root must be negative: if there were a root $a < 0$, then $f_n'(a) = -a^n/n! < 0$ (since $a^n > 0$ for even $n$), so $f_n$ is decreasing at $a$. But $f_n \to +\infty$ as $x \to -\infty$, so $f_n$ must increase then decrease, meaning $f_n'$ has a zero, contradicting the fact that $f_{n-1}$ (odd degree) has exactly one real root which is negative and $f_{n-1}(0) = 1 > 0$… Actually, for even $n$, $f_n$ has no real roots. For $n=2$: $f_2(x) = 1+x+x^2/2$, discriminant $= 1-2 = -1 < 0$, confirming no real roots.

Model Solution

Part (i). We have $f_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}$. Differentiating term by term:

$f_n'(x) = \sum_{k=1}^{n} \frac{kx^{k-1}}{k!} = \sum_{k=1}^{n} \frac{x^{k-1}}{(k-1)!}$

Setting $j = k - 1$:

$f_n'(x) = \sum_{j=0}^{n-1} \frac{x^j}{j!} = f_{n-1}(x)$

Part (ii). Suppose $a$ is a real root of $f_n(x) = 0$.

If $a > 0$: every term $\frac{a^k}{k!} > 0$ for $k \geq 0$, and the first term is $1 > 0$, so $f_n(a) > 0$. Contradiction.

If $a = 0$: $f_n(0) = 1 \neq 0$. Contradiction.

Therefore $a < 0$.

Part (iii). Suppose $a$ and $b$ are distinct real roots of $f_n(x) = 0$ with $n \geq 2$. From part (ii), $a < 0$ and $b < 0$.

Since $f_n(x) = f_{n-1}(x) + \frac{x^n}{n!}$, we have $f_n'(x) = f_{n-1}(x) = f_n(x) - \frac{x^n}{n!}$.

At a root, $f_n(a) = 0$, so:

$f_n'(a) = -\frac{a^n}{n!}$

Similarly $f_n'(b) = -\frac{b^n}{n!}$.

Therefore:

$f_n'(a) \cdot f_n'(b) = \frac{a^n \cdot b^n}{(n!)^2} = \frac{(ab)^n}{(n!)^2}$

Since $a < 0$ and $b < 0$, we have $ab > 0$, so $(ab)^n > 0$. Hence $f_n'(a) \cdot f_n'(b) > 0$.

Deducing at most one real root.

Since $f_n'(a)$ and $f_n'(b)$ have the same sign, the tangent lines at $a$ and $b$ slope in the same direction. Since $f_n(a) = f_n(b) = 0$ and $f_n$ is continuous:

• If $f_n'(a) > 0$ and $f_n'(b) > 0$: the function is increasing at both roots. Since $f_n(a) = 0$ and $f_n$ is increasing at $a$, $f_n$ must be negative just to the left of $a$ and positive just to the right. Since $f_n$ is increasing at $b$ too, $f_n$ must be negative just to the left of $b$ and positive just to the right. But between $a$ and $b$, $f_n$ must go from positive (right of $a$) to negative (left of $b$), so by the Intermediate Value Theorem, $f_n(c) = 0$ for some $c$ between $a$ and $b$.

• If $f_n'(a) < 0$ and $f_n'(b) < 0$: a symmetric argument (decreasing at both roots) also gives a root $c$ between $a$ and $b$.

In either case, $c$ is another distinct real root. We can now repeat the argument between $a$ and $c$, finding yet another root, and so on. This would give infinitely many roots, contradicting the fact that $f_n$ is a polynomial of degree $n$ (which has at most $n$ roots). Therefore $f_n(x) = 0$ has at most one real root.

Number of roots when $n$ is odd.

The leading term of $f_n$ is $\frac{x^n}{n!}$. When $n$ is odd:

• $f_n(x) \to +\infty$ as $x \to +\infty$
• $f_n(x) \to -\infty$ as $x \to -\infty$

Since $f_n$ is continuous, by the Intermediate Value Theorem, $f_n$ has at least one real root. Combined with “at most one”, $f_n(x) = 0$ has exactly one real root when $n$ is odd.

Number of roots when $n$ is even.

When $n$ is even, $f_n(x) \to +\infty$ as $x \to \pm \infty$. We know $f_n(0) = 1 > 0$ and any real root must be negative. Suppose $a < 0$ is a root. Then:

$f_n'(a) = -\frac{a^n}{n!}$

Since $n$ is even and $a < 0$, $a^n > 0$, so $f_n'(a) < 0$: the function is decreasing at $a$.

Since $f_n(a) = 0$, $f_n$ is positive just to the left of $a$ and negative just to the right. But $f_n(x) \to +\infty$ as $x \to -\infty$, so $f_n$ must have decreased from $+\infty$ to $0$ at $a$, meaning $f_n$ was positive for all sufficiently negative $x$. As $x$ increases past $a$, $f_n$ becomes negative. But $f_n(0) = 1 > 0$, so $f_n$ must cross zero again at some point between $a$ and $0$. This gives a second real root, contradicting “at most one”.

Therefore $f_n(x) = 0$ has no real roots when $n$ is even.

Examiner Notes

This was the most popular question on the paper, and candidates generally scored well here. The first two parts were relatively straightforward and were answered well. The final part required careful explanation from candidates and many were not able to take the initial step of rewriting the relationship between the function and its derivative in a useful way. Those who did successfully complete this part were often able to identify the number of roots in each of the two cases.

---

Question 4

Topic: 不等式与函数 (Inequalities and Functions)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Let

$y = \frac{x^2 + x \sin \theta + 1}{x^2 + x \cos \theta + 1} .$

(i) Given that $x$ is real, show that

$(y \cos \theta - \sin \theta)^2 \geqslant 4(y - 1)^2 .$

Deduce that

$y^2 + 1 \geqslant 4(y - 1)^2 ,$

and hence that

$\frac{4 - \sqrt{7}}{3} \leqslant y \leqslant \frac{4 + \sqrt{7}}{3} .$

(ii) In the case $y = \frac{4 + \sqrt{7}}{3}$, show that

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

and find the corresponding values of $x$ and $\tan \theta$.

Hint

(i) From $y = \frac{x^2 + x\sin\theta + 1}{x^2 + x\cos\theta + 1}$, rearrange:

$y(x^2 + x\cos\theta + 1) = x^2 + x\sin\theta + 1$

$(y-1)x^2 + (y\cos\theta - \sin\theta)x + (y-1) = 0$

For real $x$, the discriminant must be $\geq 0$:

$(y\cos\theta - \sin\theta)^2 - 4(y-1)^2 \geq 0$

So $(y\cos\theta - \sin\theta)^2 \geq 4(y-1)^2$.

Deduction of $y^2 + 1 \geq (y\cos\theta - \sin\theta)^2$:

Write $y\cos\theta - \sin\theta = R\cos(\theta + \alpha)$ where $R = \sqrt{y^2+1}$ and $\tan\alpha = \frac{1}{y}$. Then $(y\cos\theta - \sin\theta)^2 = (y^2+1)\cos^2(\theta+\alpha) \leq y^2+1$.

Deduction of the range: Combining: $4(y-1)^2 \leq (y\cos\theta-\sin\theta)^2 \leq y^2+1$.

So $4(y-1)^2 \leq y^2+1$, i.e. $4y^2 - 8y + 4 \leq y^2+1$, i.e. $3y^2 - 8y + 3 \leq 0$.

The roots of $3y^2 - 8y + 3 = 0$ are $y = \frac{8 \pm \sqrt{64-36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{4 \pm \sqrt{7}}{3}$.

Therefore $\frac{4-\sqrt{7}}{3} \leq y \leq \frac{4+\sqrt{7}}{3}$.

(ii) When $y = \frac{4+\sqrt{7}}{3}$:

$y - 1 = \frac{1+\sqrt{7}}{3}$, so $4(y-1)^2 = \frac{4(1+\sqrt{7})^2}{9} = \frac{4(8+2\sqrt{7})}{9} = \frac{8(4+\sqrt{7})}{9}$.

$y^2 + 1 = \frac{(4+\sqrt{7})^2}{9} + 1 = \frac{23+8\sqrt{7}+9}{9} = \frac{32+8\sqrt{7}}{9} = \frac{8(4+\sqrt{7})}{9}$.

So $y^2+1 = 4(y-1)^2$, hence $\sqrt{y^2+1} = 2(y-1)$ (positive since $y > 1$).

Finding $\tan\theta$: For the maximum to be achieved, both inequalities become equalities. From $(y^2+1)\cos^2(\theta+\alpha) = y^2+1$, we get $\cos^2(\theta+\alpha) = 1$, so $\theta + \alpha = k\pi$.

With $\tan\alpha = \frac{1}{y} = \frac{3}{4+\sqrt{7}} = \frac{3(4-\sqrt{7})}{16-7} = \frac{4-\sqrt{7}}{3}$.

Taking $\theta + \alpha = 0$: $\tan\theta = -\tan\alpha = -\frac{4-\sqrt{7}}{3} = \frac{\sqrt{7}-4}{3}$.

Finding $x$: From $(y-1)x^2 + (y\cos\theta - \sin\theta)x + (y-1) = 0$ with discriminant $= 0$: $x = -\frac{y\cos\theta - \sin\theta}{2(y-1)}$

Since $y\cos\theta - \sin\theta = \sqrt{y^2+1}\cos(\theta+\alpha) = \sqrt{y^2+1} = 2(y-1)$:

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

So $x = -1$ and $\tan\theta = \frac{\sqrt{7}-4}{3}$.

Model Solution

Part (i): Showing the discriminant inequality.

From $y = \dfrac{x^2 + x\sin\theta + 1}{x^2 + x\cos\theta + 1}$, rearrange:

$y(x^2 + x\cos\theta + 1) = x^2 + x\sin\theta + 1$

$(y - 1)x^2 + (y\cos\theta - \sin\theta)x + (y - 1) = 0 \qquad (\dagger)$

For this quadratic in $x$ to have real solutions, the discriminant must be non-negative:

$(y\cos\theta - \sin\theta)^2 - 4(y - 1)^2 \geq 0$

$(y\cos\theta - \sin\theta)^2 \geq 4(y - 1)^2 \qquad \checkmark$

Deducing $y^2 + 1 \geq 4(y-1)^2$.

Write $y\cos\theta - \sin\theta$ in the form $R\cos(\theta + \alpha)$ where $R = \sqrt{y^2 + 1}$ and $\tan\alpha = 1/y$. This is valid since:

$R\cos(\theta + \alpha) = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha)$

Setting $R\cos\alpha = y$ and $R\sin\alpha = 1$ gives $R = \sqrt{y^2 + 1}$.

Since $\cos^2(\theta + \alpha) \leq 1$:

$(y\cos\theta - \sin\theta)^2 = (y^2 + 1)\cos^2(\theta + \alpha) \leq y^2 + 1$

Combining with the discriminant inequality:

$4(y - 1)^2 \leq (y\cos\theta - \sin\theta)^2 \leq y^2 + 1$

$4(y - 1)^2 \leq y^2 + 1$

Deducing the range of $y$.

Expanding: $4y^2 - 8y + 4 \leq y^2 + 1$, so $3y^2 - 8y + 3 \leq 0$.

The roots of $3y^2 - 8y + 3 = 0$ are:

$y = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{8 \pm 2\sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3}$

Since the coefficient of $y^2$ is positive, the quadratic is non-positive between its roots:

$\frac{4 - \sqrt{7}}{3} \leq y \leq \frac{4 + \sqrt{7}}{3}$

Part (ii): The case $y = \frac{4 + \sqrt{7}}{3}$.

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

Compute $y - 1 = \frac{4 + \sqrt{7}}{3} - 1 = \frac{1 + \sqrt{7}}{3}$.

$4(y - 1)^2 = 4 \cdot \frac{(1 + \sqrt{7})^2}{9} = \frac{4(1 + 2\sqrt{7} + 7)}{9} = \frac{4(8 + 2\sqrt{7})}{9} = \frac{8(4 + \sqrt{7})}{9}$

$y^2 + 1 = \frac{(4 + \sqrt{7})^2}{9} + 1 = \frac{16 + 8\sqrt{7} + 7}{9} + 1 = \frac{23 + 8\sqrt{7} + 9}{9} = \frac{32 + 8\sqrt{7}}{9} = \frac{8(4 + \sqrt{7})}{9}$

So $y^2 + 1 = 4(y - 1)^2$. Taking square roots (both sides positive since $y > 1$):

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

Finding $\tan\theta$.

For the maximum to be achieved, both inequalities in the chain must be equalities.

From $(y^2 + 1)\cos^2(\theta + \alpha) = y^2 + 1$, we need $\cos^2(\theta + \alpha) = 1$, so $\theta + \alpha = k\pi$ for integer $k$.

With $\tan\alpha = \frac{1}{y} = \frac{3}{4 + \sqrt{7}} = \frac{3(4 - \sqrt{7})}{16 - 7} = \frac{3(4 - \sqrt{7})}{9} = \frac{4 - \sqrt{7}}{3}$.

Taking $\theta + \alpha = 0$ (so $\theta = -\alpha$):

$\tan\theta = -\tan\alpha = -\frac{4 - \sqrt{7}}{3} = \frac{\sqrt{7} - 4}{3}$

Finding $x$.

From $(\dagger)$, the discriminant is zero, so there is exactly one value of $x$:

$x = -\frac{y\cos\theta - \sin\theta}{2(y - 1)}$

Since $y\cos\theta - \sin\theta = \sqrt{y^2 + 1}\cos(\theta + \alpha) = \sqrt{y^2 + 1} \cdot (\pm 1)$. With $\theta + \alpha = 0$, $\cos(\theta + \alpha) = 1$, so:

$y\cos\theta - \sin\theta = \sqrt{y^2 + 1} = 2(y - 1)$

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

Therefore $x = -1$ and $\tan\theta = \dfrac{\sqrt{7} - 4}{3}$.

Examiner Notes

This question attracted many good responses which often successfully accomplished the first and third results of part (i) of the question, although in many cases marks were lost through incomplete explanations of some of the steps taken. Part (ii) of the question was generally answered poorly with some candidates simply stating that the square root of one expression in terms of surds was equal to the other one that they had achieved. Only a small number of candidates successfully completed the last section of this question.

---

Question 5

Topic: 组合数学 (Combinatorics)  |  Difficulty: Challenging  |  Marks: 20

Problem

5 In this question, the definition of $\begin{pmatrix} p \\ q \end{pmatrix}$ is taken to be

$\begin{pmatrix} p \\ q \end{pmatrix} = \begin{cases} \frac{p!}{q!(p - q)!} & \text{if } p \geqslant q \geqslant 0 \, , \\ 0 & \text{otherwise .} \end{cases}$

(i) Write down the coefficient of $x^n$ in the binomial expansion for $(1 - x)^{-N}$, where $N$ is a positive integer, and write down the expansion using the $\Sigma$ summation notation.

By considering $(1 - x)^{-1}(1 - x)^{-N}$, where $N$ is a positive integer, show that

$\sum_{j=0}^{n} \begin{pmatrix} N + j - 1 \\ j \end{pmatrix} = \begin{pmatrix} N + n \\ n \end{pmatrix} .$

(ii) Show that, for any positive integers $m, n$ and $r$ with $r \leqslant m + n$,

$\begin{pmatrix} m + n \\ r \end{pmatrix} = \sum_{j=0}^{r} \begin{pmatrix} m \\ j \end{pmatrix} \begin{pmatrix} n \\ r - j \end{pmatrix} .$

(iii) Show that, for any positive integers $m$ and $N$,

$\sum_{j=0}^{n} (-1)^j \begin{pmatrix} N + m \\ n - j \end{pmatrix} \begin{pmatrix} m + j - 1 \\ j \end{pmatrix} = \begin{pmatrix} N \\ n \end{pmatrix} .$

Hint

(i) The binomial expansion of $(1-x)^{-N}$ where $N$ is a positive integer:

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

The coefficient of $x^n$ is $\binom{N+n-1}{n} = \frac{(N+n-1)!}{n!(N-1)!}$.

Proving the identity: $(1-x)^{-1}(1-x)^{-N} = (1-x)^{-(N+1)}$.

$(1-x)^{-1} = \sum_{k=0}^{\infty} x^k$, so the coefficient of $x^n$ in $(1-x)^{-1}(1-x)^{-N}$ is: $\sum_{j=0}^{n} 1 \cdot \binom{N+j-1}{j} = \sum_{j=0}^{n} \binom{N+j-1}{j}$

But $(1-x)^{-(N+1)} = \sum_{n=0}^{\infty} \binom{N+n}{n} x^n$, so the coefficient of $x^n$ is $\binom{N+n}{n}$.

Equating: $\sum_{j=0}^{n} \binom{N+j-1}{j} = \binom{N+n}{n}$.

(ii) Consider $(1+x)^m(1+x)^n = (1+x)^{m+n}$.

The coefficient of $x^r$ on the left is $\sum_{j=0}^{r} \binom{m}{j}\binom{n}{r-j}$ (by the Cauchy product / Vandermonde convolution).

The coefficient of $x^r$ on the right is $\binom{m+n}{r}$.

Equating: $\binom{m+n}{r} = \sum_{j=0}^{r} \binom{m}{j}\binom{n}{r-j}$.

(iii) Consider $(1+x)^{N+m} \cdot (1+x)^{-m} = (1+x)^{N}$.

$(1+x)^{-m} = (1-(-x))^{-m} = \sum_{j=0}^{\infty} \binom{m+j-1}{j}(-x)^j = \sum_{j=0}^{\infty} (-1)^j \binom{m+j-1}{j} x^j$

$(1+x)^{N+m} = \sum_{k=0}^{N+m} \binom{N+m}{k} x^k$

The coefficient of $x^n$ in the product: $\sum_{j=0}^{n} (-1)^j \binom{m+j-1}{j} \binom{N+m}{n-j}$

And the coefficient of $x^n$ in $(1+x)^N$ is $\binom{N}{n}$.

Equating: $\sum_{j=0}^{n} (-1)^j \binom{N+m}{n-j} \binom{m+j-1}{j} = \binom{N}{n}$.

Model Solution

Part (i): Binomial expansion of $(1 - x)^{-N}$.

For a positive integer $N$, the negative binomial expansion gives:

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

The coefficient of $x^n$ is $\dbinom{N + n - 1}{n}$.

Proving the identity. Consider $(1 - x)^{-1}(1 - x)^{-N} = (1 - x)^{-(N+1)}$.

$(1 - x)^{-1} = \sum_{k=0}^{\infty} x^k$, so by the Cauchy product, the coefficient of $x^n$ in $(1 - x)^{-1}(1 - x)^{-N}$ is:

$\sum_{j=0}^{n} 1 \cdot \binom{N + j - 1}{j} = \sum_{j=0}^{n} \binom{N + j - 1}{j}$

The coefficient of $x^n$ in $(1 - x)^{-(N+1)}$ is $\dbinom{(N+1) + n - 1}{n} = \dbinom{N + n}{n}$.

Equating:

$\sum_{j=0}^{n} \binom{N + j - 1}{j} = \binom{N + n}{n}$

Part (ii): Vandermonde’s identity.

Consider $(1 + x)^m (1 + x)^n = (1 + x)^{m+n}$.

The coefficient of $x^r$ on the left-hand side, by the Cauchy product:

$\sum_{j=0}^{r} \binom{m}{j} \binom{n}{r - j}$

The coefficient of $x^r$ on the right-hand side is $\dbinom{m + n}{r}$.

Equating:

$\binom{m + n}{r} = \sum_{j=0}^{r} \binom{m}{j} \binom{n}{r - j}$

Part (iii): The alternating sum identity.

Consider $(1 + x)^{N+m} \cdot (1 + x)^{-m} = (1 + x)^N$.

The second factor can be expanded as:

$(1 + x)^{-m} = (1 - (-x))^{-m} = \sum_{j=0}^{\infty} \binom{m + j - 1}{j} (-x)^j = \sum_{j=0}^{\infty} (-1)^j \binom{m + j - 1}{j} x^j$

The first factor is:

$(1 + x)^{N+m} = \sum_{k=0}^{N+m} \binom{N + m}{k} x^k$

By the Cauchy product, the coefficient of $x^n$ in the product is:

$\sum_{j=0}^{n} (-1)^j \binom{m + j - 1}{j} \cdot \binom{N + m}{n - j}$

The coefficient of $x^n$ in $(1 + x)^N$ is $\dbinom{N}{n}$.

Equating:

$\sum_{j=0}^{n} (-1)^j \binom{N + m}{n - j} \binom{m + j - 1}{j} = \binom{N}{n}$

Examiner Notes

This was the least attempted of all of the Pure questions and one that generally produced low marks for candidates, who generally appeared to have difficulty in expressing the coefficients of the binomial expansion in the required form. In each case the desired answer was given in the question, and so successful solutions also needed to be very clear about the reasoning used to reach the answer.

---

Question 6

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

Problem

6 This question concerns solutions of the differential equation

$(1 - x^2) \left( \frac{dy}{dx} \right)^2 + k^2y^2 = k^2 \qquad \text{(*)}$

where $k$ is a positive integer.

For each value of $k$, let $y_k(x)$ be the solution of $(*)$ that satisfies $y_k(1) = 1$; you may assume that there is only one such solution for each value of $k$.

(i) Write down the differential equation satisfied by $y_1(x)$ and verify that $y_1(x) = x$.

(ii) Write down the differential equation satisfied by $y_2(x)$ and verify that $y_2(x) = 2x^2 - 1$.

(iii) Let $z(x) = 2(y_n(x))^2 - 1$. Show that

$(1 - x^2) \left( \frac{dz}{dx} \right)^2 + 4n^2z^2 = 4n^2$

and hence obtain an expression for $y_{2n}(x)$ in terms of $y_n(x)$.

(iv) Let $v(x) = y_n(y_m(x))$. Show that $v(x) = y_{mn}(x)$.

Hint

Parts (i) and (ii) only require verification in each of the cases, so simply differentiate the functions given and substitute into the differential equation to confirm that they are solutions. Remember to check as well that the boundary conditions are satisfied.

For part (iii), differentiate the given formula for $z$ and substitute into the differential equation. By observing that the new differential equation is of the same form as (*), but for $2n$ instead of $n$, the expression for $y_{2n}(x)$ can be established.

For part (iv), again differentiate the given formula, being careful about the application of the chain rule and substitute. Again, by comparing with (*) the final result should be clear.

Model Solution

Part (i): $k = 1$.

The differential equation with $k = 1$ is:

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

We verify $y_1(x) = x$: $\frac{dy_1}{dx} = 1$, and $y_1(1) = 1$. $\checkmark$

Substituting: $(1 - x^2)(1)^2 + x^2 = 1 - x^2 + x^2 = 1$. $\checkmark$

Part (ii): $k = 2$.

The differential equation with $k = 2$ is:

$(1 - x^2)\left(\frac{dy}{dx}\right)^2 + 4y^2 = 4$

We verify $y_2(x) = 2x^2 - 1$: $\frac{dy_2}{dx} = 4x$, and $y_2(1) = 2 - 1 = 1$. $\checkmark$

Substituting: $(1 - x^2)(4x)^2 + 4(2x^2 - 1)^2 = 16x^2(1 - x^2) + 4(4x^4 - 4x^2 + 1)$

$= 16x^2 - 16x^4 + 16x^4 - 16x^2 + 4 = 4$. $\checkmark$

Part (iii). Let $z(x) = 2(y_n(x))^2 - 1$.

Differentiating: $\frac{dz}{dx} = 4y_n(x) \cdot y_n'(x)$.

Squaring: $\left(\frac{dz}{dx}\right)^2 = 16y_n^2 \cdot (y_n')^2$.

From the differential equation $(*)$ for $y_n$:

$(1 - x^2)(y_n')^2 = k^2(1 - y_n^2) = n^2(1 - y_n^2)$

So:

$(1 - x^2)\left(\frac{dz}{dx}\right)^2 = 16y_n^2 \cdot (1 - x^2)(y_n')^2 = 16y_n^2 \cdot n^2(1 - y_n^2) = 16n^2 y_n^2(1 - y_n^2)$

Also: $z = 2y_n^2 - 1$, so $y_n^2 = \frac{z + 1}{2}$ and $1 - y_n^2 = 1 - \frac{z + 1}{2} = \frac{1 - z}{2}$.

Therefore:

$(1 - x^2)\left(\frac{dz}{dx}\right)^2 = 16n^2 \cdot \frac{z + 1}{2} \cdot \frac{1 - z}{2} = 4n^2(1 - z^2)$

$(1 - x^2)\left(\frac{dz}{dx}\right)^2 + 4n^2 z^2 = 4n^2$

This is the differential equation $(*)$ with $k = 2n$. Also $z(1) = 2(y_n(1))^2 - 1 = 2(1)^2 - 1 = 1$.

Since the solution satisfying $y(1) = 1$ is unique, $z(x) = y_{2n}(x)$.

Therefore:

$y_{2n}(x) = 2(y_n(x))^2 - 1$

Part (iv). Let $v(x) = y_n(y_m(x))$. We show $v = y_{mn}$.

By the chain rule:

$\frac{dv}{dx} = y_n'(y_m(x)) \cdot y_m'(x)$

Squaring:

$\left(\frac{dv}{dx}\right)^2 = (y_n')^2 \big|_{y_m(x)} \cdot (y_m')^2$

From the differential equation for $y_n$ (with $k = n$, evaluated at $y_m(x)$, using $y_m(x) \in [-1, 1]$):

$(1 - y_m^2)(y_n')^2\big|_{y_m} = n^2(1 - y_n^2(y_m)) = n^2(1 - v^2)$

From the differential equation for $y_m$ (with $k = m$):

$(1 - x^2)(y_m')^2 = m^2(1 - y_m^2)$

Therefore:

$(1 - x^2)\left(\frac{dv}{dx}\right)^2 = (1 - x^2)(y_m')^2 \cdot (y_n')^2\big|_{y_m}$

$= m^2(1 - y_m^2) \cdot (y_n')^2\big|_{y_m}$

$= m^2 \cdot \frac{n^2(1 - v^2)}{1 - y_m^2} \cdot (1 - y_m^2) = m^2 n^2(1 - v^2)$

So:

$(1 - x^2)\left(\frac{dv}{dx}\right)^2 + m^2 n^2 v^2 = m^2 n^2$

This is the differential equation $(*)$ with $k = mn$. Also $v(1) = y_n(y_m(1)) = y_n(1) = 1$.

By uniqueness of the solution with $y(1) = 1$:

$y_n(y_m(x)) = y_{mn}(x)$

Examiner Notes

Many candidates did not appear to read the question carefully enough and so attempted to solve the differential equations in parts (i) and (ii) rather than simply verifying the results given. Where candidates moved on to parts (iii) and (iv) they were generally successful if they were able to complete the differentiation of the new function.

---

Question 7

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

Problem

7 Show that

$\int_0^a f(x) dx = \int_0^a f(a - x) dx, \qquad \text{(*)}$

where $f$ is any function for which the integrals exist.

(i) Use $(*)$ to evaluate $\int_0^{\frac{1}{2}\pi} \frac{\sin x}{\cos x + \sin x} dx .$

(ii) Evaluate $\int_0^{\frac{1}{4}\pi} \frac{\sin x}{\cos x + \sin x} dx .$

(iii) Evaluate $\int_0^{\frac{1}{4}\pi} \ln(1 + \tan x) dx .$

(iv) Evaluate $\int_0^{\frac{1}{4}\pi} \frac{x}{\cos x (\cos x + \sin x)} dx .$

Hint

The first result can be shown by using a substitution into the integral, being careful to explain the change of sign when the limits of the integral are switched.

Simple application of knowledge of trigonometric graphs once the substitution has been made can be used to show that twice the integral is equivalent to integrating the function 1 over the interval.

Similarly, the remaining integrals can all be rearranged using standard trigonometric identities and knowledge of logarithms into forms that can be integrated from standard results once the substitution from (*) has been made.

Model Solution

Proving $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.

Substitute $u = a - x$, so $du = -dx$. When $x = 0$, $u = a$; when $x = a$, $u = 0$.

$\int_0^a f(a - x)\,dx = \int_a^0 f(u)(-du) = \int_0^a f(u)\,du = \int_0^a f(x)\,dx$

Part (i). Let $I = \int_0^{\pi/2} \frac{\sin x}{\cos x + \sin x}\,dx$.

By the result $(*)$ with $a = \pi/2$:

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

Adding:

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

$I = \frac{\pi}{4}$

Part (ii). Let $J = \int_0^{\pi/4} \frac{\sin x}{\cos x + \sin x}\,dx$.

By $(*)$ with $a = \pi/4$:

$J = \int_0^{\pi/4} \frac{\sin(\pi/4 - x)}{\cos(\pi/4 - x) + \sin(\pi/4 - x)}\,dx$

Using $\sin(\pi/4 - x) = \frac{1}{\sqrt{2}}(\cos x - \sin x)$ and $\cos(\pi/4 - x) = \frac{1}{\sqrt{2}}(\cos x + \sin x)$:

$J = \int_0^{\pi/4} \frac{\frac{1}{\sqrt{2}}(\cos x - \sin x)}{\frac{1}{\sqrt{2}}(\cos x + \sin x) + \frac{1}{\sqrt{2}}(\cos x - \sin x)}\,dx = \int_0^{\pi/4} \frac{\cos x - \sin x}{2\cos x}\,dx$

$= \frac{1}{2}\int_0^{\pi/4} \left(1 - \frac{\sin x}{\cos x}\right)dx = \frac{1}{2}\int_0^{\pi/4} (1 - \tan x)\,dx$

$= \frac{1}{2}\left[x + \ln|\cos x|\right]_0^{\pi/4} = \frac{1}{2}\left[\frac{\pi}{4} + \ln\frac{1}{\sqrt{2}} - 0 - \ln 1\right]$

$= \frac{1}{2}\left[\frac{\pi}{4} - \frac{1}{2}\ln 2\right] = \frac{\pi}{8} - \frac{\ln 2}{4}$

Part (iii). Let $K = \int_0^{\pi/4} \ln(1 + \tan x)\,dx$.

By $(*)$ with $a = \pi/4$:

$K = \int_0^{\pi/4} \ln\!\left(1 + \tan\!\left(\frac{\pi}{4} - x\right)\right)dx$

Using $\tan(\pi/4 - x) = \frac{1 - \tan x}{1 + \tan x}$:

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

So $K = \int_0^{\pi/4} \ln\!\left(\frac{2}{1 + \tan x}\right)dx = \int_0^{\pi/4} [\ln 2 - \ln(1 + \tan x)]\,dx$.

$K = \frac{\pi}{4}\ln 2 - K$

$2K = \frac{\pi}{4}\ln 2$

$K = \frac{\pi}{8}\ln 2$

Part (iv). Let $L = \int_0^{\pi/4} \frac{x}{\cos x(\cos x + \sin x)}\,dx$.

Using $\frac{1}{\cos x(\cos x + \sin x)} = \frac{1}{\cos^2 x(1 + \tan x)} = \frac{\sec^2 x}{1 + \tan x}$.

Note that $\frac{d}{dx}\ln(1 + \tan x) = \frac{\sec^2 x}{1 + \tan x}$. So we integrate by parts with $u = x$, $dv = \frac{\sec^2 x}{1 + \tan x}\,dx$:

$L = \left[x\ln(1 + \tan x)\right]_0^{\pi/4} - \int_0^{\pi/4} \ln(1 + \tan x)\,dx$

$= \frac{\pi}{4}\ln 2 - 0 - \frac{\pi}{8}\ln 2 = \frac{\pi}{8}\ln 2$

Examiner Notes

This was the second most popular and one of the best-answered questions on the paper, with many candidates scoring very high marks. In many cases the initial result was explained clearly and then applied successfully to the first example. Candidates were generally able to follow through the calculations where they were able to see the way in which the result could be achieved, and so most of the attempts that followed a correct method only lost marks through occasional errors in calculation.

---

Question 8

Topic: 级数与积分近似 (Series and Integration Approximation)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Evaluate the integral

$\int_{m-\frac{1}{2}}^{\infty} \frac{1}{x^2} \, dx \quad (m > \frac{1}{2}) .$

Show by means of a sketch that

$\sum_{r=m}^{n} \frac{1}{r^2} \approx \int_{m-\frac{1}{2}}^{n+\frac{1}{2}} \frac{1}{x^2} \, dx , \qquad \text{(*)}$

where $m$ and $n$ are positive integers with $m < n$.

(i) You are given that the infinite series $\sum_{r=1}^{\infty} \frac{1}{r^2}$ converges to a value denoted by $E$. Use $(*)$ to obtain the following approximations for $E$:

$E \approx 2; \quad E \approx \frac{5}{3}; \quad E \approx \frac{33}{20} .$

(ii) Show that, when $r$ is large, the error in approximating $\frac{1}{r^2}$ by $\int_{r-\frac{1}{2}}^{r+\frac{1}{2}} \frac{1}{x^2} \, dx$ is approximately $\frac{1}{4r^4}$.

Given that $E \approx 1.645$, show that $\sum_{r=1}^{\infty} \frac{1}{r^4} \approx 1.08$.

Hint

The integral required at the start of the question should be a straightforward one to evaluate. When making a sketch to illustrate the result in the second part, ensure that the sum is indicated by a series of rectangles, with the graph of the curve passing through the midpoints of the tops.

In part (i), the integral that would match the sum given results in an answer of 2, so this is the first of the estimates. The remaining estimates arise from using the integral to estimate most of the sum, but taking the first few terms as the exact values (so in each case the integration is taken from a different lower limit).

For part (ii), evaluate the integral for one particular term of the sum and note that it is approximately rac{1}{4r^4}. Finally, using the most accurate estimate for E \left( rac{33}{20} ight) the sum from $r = 3$ onwards can be calculated and then the first two values of rac{1}{r^4} can be added to achieve the desired result.

Model Solution

Evaluating the integral.

$\int_{m - 1/2}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{m - 1/2}^{\infty} = 0 - \left(-\frac{1}{m - 1/2}\right) = \frac{1}{m - 1/2} = \frac{2}{2m - 1}$

Sketch argument for $(*)$.

Draw the curve $y = 1/x^2$ for $x > 0$. For each integer $r$ from $m$ to $n$, draw a rectangle of width 1 centred at $x = r$, i.e. from $x = r - 1/2$ to $x = r + 1/2$, with height $1/r^2$. The curve $y = 1/x^2$ passes through the midpoint of the top of each rectangle.

Since $1/x^2$ is a smooth, convex, decreasing function, the area of each rectangle (equal to $1/r^2$) is approximately equal to the area under the curve over $[r - 1/2, r + 1/2]$. The errors at adjacent intervals partially cancel (the curve is above the rectangle on one side and below on the other), so:

$\sum_{r=m}^{n} \frac{1}{r^2} \approx \int_{m - 1/2}^{n + 1/2} \frac{1}{x^2}\,dx$

Part (i): Obtaining the three approximations.

The integral gives:

$\int_{m - 1/2}^{n + 1/2} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{m - 1/2}^{n + 1/2} = \frac{1}{m - 1/2} - \frac{1}{n + 1/2} = \frac{2}{2m - 1} - \frac{2}{2n + 1}$

As $n \to \infty$, the sum $\sum_{r=m}^{n} 1/r^2$ tends to $E - \sum_{r=1}^{m-1} 1/r^2$, and the integral tends to $\frac{2}{2m - 1}$.

First approximation ($m = 1$): Taking $m = 1$ and $n \to \infty$:

$E \approx \frac{2}{2(1) - 1} = 2$

Second approximation ($m = 2$): Taking $m = 2$ and $n \to \infty$. The sum from $r = 2$ to $\infty$ is $E - 1$:

$E - 1 \approx \frac{2}{2(2) - 1} = \frac{2}{3}$

$E \approx 1 + \frac{2}{3} = \frac{5}{3}$

Third approximation ($m = 3$): Taking $m = 3$ and $n \to \infty$. The sum from $r = 3$ to $\infty$ is $E - 1 - 1/4 = E - 5/4$:

$E - \frac{5}{4} \approx \frac{2}{2(3) - 1} = \frac{2}{5}$

$E \approx \frac{5}{4} + \frac{2}{5} = \frac{25 + 8}{20} = \frac{33}{20}$

Part (ii): Error approximation.

For large $r$, the error in approximating $\frac{1}{r^2}$ by $\int_{r - 1/2}^{r + 1/2} \frac{1}{x^2}\,dx$ is:

$\frac{1}{r^2} - \int_{r - 1/2}^{r + 1/2} \frac{1}{x^2}\,dx = \frac{1}{r^2} - \left[\frac{1}{r - 1/2} - \frac{1}{r + 1/2}\right]$

$= \frac{1}{r^2} - \frac{(r + 1/2) - (r - 1/2)}{(r - 1/2)(r + 1/2)} = \frac{1}{r^2} - \frac{1}{r^2 - 1/4}$

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

For large $r$, $r^2 - 1/4 \approx r^2$, so the error is approximately:

$\frac{-1/4}{r^4} = -\frac{1}{4r^4}$

The magnitude of the error is approximately $\frac{1}{4r^4}$.

Finding $\sum_{r=1}^{\infty} 1/r^4$.

From the approximation $(*)$ with $m = 1$ and $n \to \infty$, the total error is:

$E - \int_{1/2}^{\infty} \frac{1}{x^2}\,dx = E - 2$

This total error is the sum of the individual errors $\sum_{r=1}^{\infty} \left(\frac{1}{r^2} - \int_{r - 1/2}^{r + 1/2} \frac{1}{x^2}\,dx\right)$.

For large $r$, each error is approximately $-1/(4r^4)$. The errors for small $r$ also contribute, but using the approximation for all $r$:

$E - 2 \approx -\frac{1}{4}\sum_{r=1}^{\infty} \frac{1}{r^4}$

With $E \approx 1.645$:

$1.645 - 2 = -0.355 \approx -\frac{1}{4}\sum_{r=1}^{\infty} \frac{1}{r^4}$

$\sum_{r=1}^{\infty} \frac{1}{r^4} \approx 4 \times 0.355 = 1.42$

However, the approximation $\frac{1}{4r^4}$ is not very accurate for small $r$. A more careful approach: use the most accurate estimate $E \approx 33/20$ to find the sum from $r = 3$ onwards, then add the first two terms exactly.

From $E - 5/4 \approx 2/5$, the error in this approximation is $33/20 - 5/4 - 2/5 = 33/20 - 25/20 - 8/20 = 0$. So the approximation $E \approx 33/20$ absorbs the error from $r = 1$ and $r = 2$ exactly.

Using the error formula: the total error from $r = 3$ onwards is approximately $-\frac{1}{4}\sum_{r=3}^{\infty} \frac{1}{r^4}$.

The actual value of $\sum_{r=3}^{\infty} 1/r^2 = E - 1 - 1/4 = E - 5/4$ and the integral estimate is $2/5$. The error is:

$\left(E - \frac{5}{4}\right) - \frac{2}{5} \approx -\frac{1}{4}\sum_{r=3}^{\infty} \frac{1}{r^4}$

With $E \approx 1.645$: $E - 5/4 = 0.395$, and $0.395 - 0.4 = -0.005$.

$-0.005 \approx -\frac{1}{4}\sum_{r=3}^{\infty} \frac{1}{r^4}$

$\sum_{r=3}^{\infty} \frac{1}{r^4} \approx 0.02$

Adding the first two terms: $1 + 1/16 = 1.0625$.

$\sum_{r=1}^{\infty} \frac{1}{r^4} \approx 1.0625 + 0.02 = 1.08$

Examiner Notes

The first task in this question was generally well answered, although sketches were often unclear or difficult to interpret. In part (i) many candidates were able to obtain the first approximation, but then could not see how to achieve the other two, often offering sums of a non-integer number of terms which gives the correct approximation when substituted into the formula. Those who successfully completed part (i) were often able to approximate the error in part (ii) and see how it applies to the final sum.