STEP2 2002 -- Pure Mathematics

STEP2 2002 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2002  |  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	积分 Integration	Standard	三角恒等式简化被积函数,换元积分法,积分限变换
2	多项式方程 Polynomial Equations	Standard	倒数方程性质,变量代换降次,二次方程求解
3	数论 Number Theory	Standard	数学归纳法,费马数性质,反证法,整除性论证
4	不等式与积分 Inequalities and Integration	Challenging	积分正性原理,二次函数判别式,三角代换,分部积分思想
5	递推关系与动力系统 Recurrence Relations and Dynamical Systems	Challenging	递推关系分析,二次方程求解,不动点与周期点,不等式证明
6	空间几何与三角 Spatial Geometry and Trigonometry	Challenging	三角恒等式,方向余弦,倾斜平面几何关系
7	空间向量 3D Vectors	Challenging	向量点积,方向余弦,坐标几何
8	微分方程 Differential Equations	Challenging	分离变量法,分段函数处理,连续性分析

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Question 1

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

Problem

1 Show that $\int_{\pi/6}^{\pi/4} \frac{1}{1 - \cos 2\theta} \mathrm{~d}\theta = \frac{\sqrt{3}}{2} - \frac{1}{2} .$

By using the substitution $x = \sin 2\theta$, or otherwise, show that $\int_{\sqrt{3}/2}^{1} \frac{1}{1 - \sqrt{(1 - x^2)}} \mathrm{~d}x = \sqrt{3} - 1 - \frac{\pi}{6} .$

Hence evaluate the integral $\int_{1}^{2/\sqrt{3}} \frac{1}{y(y - \sqrt{(y^2 - 1^2)})} \mathrm{~d}y .$

Model Solution

Part 1: Evaluating $\int_{\pi/6}^{\pi/4} \frac{1}{1 - \cos 2\theta} \, \mathrm{d}\theta$

Using the double-angle identity $1 - \cos 2\theta = 2\sin^2\theta$:

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

So the integral becomes:

$\int_{\pi/6}^{\pi/4} \frac{1}{1 - \cos 2\theta} \, \mathrm{d}\theta = \frac{1}{2}\int_{\pi/6}^{\pi/4} \csc^2\theta \, \mathrm{d}\theta$

Since $\int \csc^2\theta \, \mathrm{d}\theta = -\cot\theta$:

$= \frac{1}{2}\bigl[-\cot\theta\bigr]_{\pi/6}^{\pi/4} = \frac{1}{2}\bigl(-\cot(\pi/4) + \cot(\pi/6)\bigr) = \frac{1}{2}(-1 + \sqrt{3}) = \frac{\sqrt{3}}{2} - \frac{1}{2} \qquad \checkmark$

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Part 2: Evaluating $\int_{\sqrt{3}/2}^{1} \frac{1}{1 - \sqrt{1 - x^2}} \, \mathrm{d}x$

We rationalise the integrand by multiplying numerator and denominator by $1 + \sqrt{1 - x^2}$:

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

Splitting the integral:

$\int_{\sqrt{3}/2}^{1} \frac{1 + \sqrt{1-x^2}}{x^2} \, \mathrm{d}x = \int_{\sqrt{3}/2}^{1} \frac{1}{x^2} \, \mathrm{d}x + \int_{\sqrt{3}/2}^{1} \frac{\sqrt{1-x^2}}{x^2} \, \mathrm{d}x$

First integral:

$\int_{\sqrt{3}/2}^{1} x^{-2} \, \mathrm{d}x = \left[-\frac{1}{x}\right]_{\sqrt{3}/2}^{1} = -1 + \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} - 1$

Second integral: Substitute $x = \sin\phi$, so $\mathrm{d}x = \cos\phi \, \mathrm{d}\phi$ and $\sqrt{1-x^2} = \cos\phi$. When $x = \frac{\sqrt{3}}{2}$, $\phi = \frac{\pi}{3}$; when $x = 1$, $\phi = \frac{\pi}{2}$.

$\int_{\pi/3}^{\pi/2} \frac{\cos\phi}{\sin^2\phi} \cos\phi \, \mathrm{d}\phi = \int_{\pi/3}^{\pi/2} \cot^2\phi \, \mathrm{d}\phi = \int_{\pi/3}^{\pi/2} (\csc^2\phi - 1) \, \mathrm{d}\phi$

$= \bigl[-\cot\phi - \phi\bigr]_{\pi/3}^{\pi/2} = \left(-\cot\frac{\pi}{2} - \frac{\pi}{2}\right) - \left(-\cot\frac{\pi}{3} - \frac{\pi}{3}\right)$

$= \left(0 - \frac{\pi}{2}\right) - \left(-\frac{1}{\sqrt{3}} - \frac{\pi}{3}\right) = -\frac{\pi}{2} + \frac{1}{\sqrt{3}} + \frac{\pi}{3} = \frac{\sqrt{3}}{3} - \frac{\pi}{6}$

Combining:

$\int_{\sqrt{3}/2}^{1} \frac{1}{1 - \sqrt{1-x^2}} \, \mathrm{d}x = \left(\frac{2\sqrt{3}}{3} - 1\right) + \left(\frac{\sqrt{3}}{3} - \frac{\pi}{6}\right) = \sqrt{3} - 1 - \frac{\pi}{6} \qquad \checkmark$

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Part 3: Evaluating $\int_{1}^{2/\sqrt{3}} \frac{1}{y(y - \sqrt{y^2 - 1})} \, \mathrm{d}y$

We show this equals the integral from Part 2 via the substitution $y = 1/x$.

Let $y = 1/x$, so $\mathrm{d}y = -1/x^2 \, \mathrm{d}x$. When $y = 1$, $x = 1$; when $y = 2/\sqrt{3}$, $x = \sqrt{3}/2$.

Since $x \in [\sqrt{3}/2, 1]$, we have $y \geqslant 1$, so $\sqrt{y^2 - 1} = \sqrt{1/x^2 - 1} = \frac{\sqrt{1-x^2}}{x}$.

The integrand transforms as:

$\frac{1}{y(y - \sqrt{y^2-1})} = \frac{1}{\frac{1}{x}\left(\frac{1}{x} - \frac{\sqrt{1-x^2}}{x}\right)} = \frac{1}{\frac{1}{x} \cdot \frac{1 - \sqrt{1-x^2}}{x}} = \frac{x^2}{1 - \sqrt{1-x^2}}$

Substituting into the integral (and swapping limits to absorb the minus sign from $\mathrm{d}y$):

$\int_{1}^{2/\sqrt{3}} \frac{1}{y(y-\sqrt{y^2-1})} \, \mathrm{d}y = \int_{\sqrt{3}/2}^{1} \frac{x^2}{1 - \sqrt{1-x^2}} \cdot \frac{1}{x^2} \, \mathrm{d}x = \int_{\sqrt{3}/2}^{1} \frac{1}{1 - \sqrt{1-x^2}} \, \mathrm{d}x$

This is exactly the integral from Part 2, so:

$\int_{1}^{2/\sqrt{3}} \frac{1}{y(y - \sqrt{y^2 - 1})} \, \mathrm{d}y = \sqrt{3} - 1 - \frac{\pi}{6}$

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Question 2

Topic: 多项式方程 Polynomial Equations  |  Difficulty: Standard  |  Marks: 20

Problem

2 Show that setting $z - z^{-1} = w$ in the quartic equation $z^4 + 5z^3 + 4z^2 - 5z + 1 = 0$ results in the quadratic equation $w^2 + 5w + 6 = 0$. Hence solve the above quartic equation.

Solve similarly the equation $2z^8 - 3z^7 - 12z^6 + 12z^5 + 22z^4 - 12z^3 - 12z^2 + 3z + 2 = 0 .$

Model Solution

Part 1: Reducing the quartic $z^4 + 5z^3 + 4z^2 - 5z + 1 = 0$

The quartic has palindromic coefficients (up to sign): $a_0 = a_4 = 1$, $a_1 = 5$, $a_3 = -5$. We divide through by $z^2$ (valid since $z = 0$ is not a root):

$z^2 + 5z + 4 - \frac{5}{z} + \frac{1}{z^2} = 0$

Regrouping:

$\left(z^2 + \frac{1}{z^2}\right) + 5\left(z - \frac{1}{z}\right) + 4 = 0$

Now set $w = z - z^{-1}$. Squaring: $w^2 = z^2 - 2 + z^{-2}$, so $z^2 + z^{-2} = w^2 + 2$.

Substituting:

$(w^2 + 2) + 5w + 4 = 0$

$w^2 + 5w + 6 = 0 \qquad \checkmark$

Solving for $w$:

$(w + 2)(w + 3) = 0 \qquad\Longrightarrow\qquad w = -2 \text{ or } w = -3$

Solving for $z$ from $w = -2$:

$z - \frac{1}{z} = -2 \qquad\Longrightarrow\qquad z^2 + 2z - 1 = 0$

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

Solving for $z$ from $w = -3$:

$z - \frac{1}{z} = -3 \qquad\Longrightarrow\qquad z^2 + 3z - 1 = 0$

$z = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}$

The four roots of the quartic are:

$z = -1 + \sqrt{2},\quad z = -1 - \sqrt{2},\quad z = \frac{-3 + \sqrt{13}}{2},\quad z = \frac{-3 - \sqrt{13}}{2}$

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Part 2: Solving $2z^8 - 3z^7 - 12z^6 + 12z^5 + 22z^4 - 12z^3 - 12z^2 + 3z + 2 = 0$

This is also a palindromic polynomial: the coefficients $2, -3, -12, 12, 22, 12, -12, -3, 2$ are symmetric. Since $z = 0$ is not a root, we divide by $z^4$:

$2\!\left(z^4 + \frac{1}{z^4}\right) - 3\!\left(z^3 - \frac{1}{z^3}\right) - 12\!\left(z^2 + \frac{1}{z^2}\right) + 12\!\left(z - \frac{1}{z}\right) + 22 = 0$

Expressing in terms of $w = z - z^{-1}$:

$z^2 + z^{-2} = w^2 + 2$

$z^3 - z^{-3} = (z - z^{-1})(z^2 + 1 + z^{-2}) = w(w^2 + 3) = w^3 + 3w$

$z^4 + z^{-4} = (z^2 + z^{-2})^2 - 2 = (w^2 + 2)^2 - 2 = w^4 + 4w^2 + 2$

Substituting all expressions:

$2(w^4 + 4w^2 + 2) - 3(w^3 + 3w) - 12(w^2 + 2) + 12w + 22 = 0$

Expanding:

$2w^4 + 8w^2 + 4 - 3w^3 - 9w - 12w^2 - 24 + 12w + 22 = 0$

Collecting by powers of $w$:

$2w^4 - 3w^3 + (8 - 12)w^2 + (-9 + 12)w + (4 - 24 + 22) = 0$

$2w^4 - 3w^3 - 4w^2 + 3w + 2 = 0$

Factoring the quartic in $w$:

Testing $w = 1$: $2 - 3 - 4 + 3 + 2 = 0$. So $(w - 1)$ is a factor.

Performing polynomial long division:

$2w^4 - 3w^3 - 4w^2 + 3w + 2 = (w - 1)(2w^3 - w^2 - 5w - 2)$

Testing $w = 2$ in the cubic: $16 - 4 - 10 - 2 = 0$. So $(w - 2)$ is a factor.

Dividing: $2w^3 - w^2 - 5w - 2 = (w - 2)(2w^2 + 3w + 1)$.

Factoring the quadratic: $2w^2 + 3w + 1 = (2w + 1)(w + 1)$.

Therefore:

$2w^4 - 3w^3 - 4w^2 + 3w + 2 = 2(w - 1)(w - 2)(2w + 1)(w + 1) = 0$

The four values of $w$ are:

$w = 1, \qquad w = 2, \qquad w = -\frac{1}{2}, \qquad w = -1$

Solving $z - z^{-1} = w$ for each value (each gives $z^2 - wz - 1 = 0$, so $z = \frac{w \pm \sqrt{w^2 + 4}}{2}$):

$w = 1$: $z^2 - z - 1 = 0$

$z = \frac{1 \pm \sqrt{5}}{2}$

$w = 2$: $z^2 - 2z - 1 = 0$

$z = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$

$w = -\frac{1}{2}$: $z^2 + \frac{1}{2}z - 1 = 0$, i.e. $2z^2 + z - 2 = 0$

$z = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}$

$w = -1$: $z^2 + z - 1 = 0$

$z = \frac{-1 \pm \sqrt{5}}{2}$

The eight roots of the octic are:

$z = \frac{1 + \sqrt{5}}{2},\quad \frac{1 - \sqrt{5}}{2},\quad 1 + \sqrt{2},\quad 1 - \sqrt{2},\quad \frac{-1 + \sqrt{17}}{4},\quad \frac{-1 - \sqrt{17}}{4},\quad \frac{-1 + \sqrt{5}}{2},\quad \frac{-1 - \sqrt{5}}{2}$

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Question 3

Topic: 数论 Number Theory  |  Difficulty: Standard  |  Marks: 20

Problem

3 The $n$th Fermat number, $F_n$, is defined by $F_n = 2^{2^n} + 1 , \quad n = 0, \ 1, \ 2, \ \dots \ ,$ where $2^{2^n}$ means 2 raised to the power $2^n$. Calculate $F_0$, $F_1$, $F_2$ and $F_3$. Show that, for $k = 1$, $k = 2$ and $k = 3$, $F_0 F_1 \dots F_{k-1} = F_k - 2 . \qquad \text{(*)}$

Prove, by induction, or otherwise, that $(*)$ holds for all $k \geqslant 1$. Deduce that no two Fermat numbers have a common factor greater than 1.

Hence show that there are infinitely many prime numbers.

Model Solution

Calculating the Fermat numbers:

$F_0 = 2^{2^0} + 1 = 2^1 + 1 = 3$

$F_1 = 2^{2^1} + 1 = 2^2 + 1 = 5$

$F_2 = 2^{2^2} + 1 = 2^4 + 1 = 17$

$F_3 = 2^{2^3} + 1 = 2^8 + 1 = 257$

Verifying $F_0 F_1 \cdots F_{k-1} = F_k - 2$ for $k = 1, 2, 3$:

For $k = 1$: $F_0 = 3 = 5 - 2 = F_1 - 2$

For $k = 2$: $F_0 \cdot F_1 = 3 \times 5 = 15 = 17 - 2 = F_2 - 2$

For $k = 3$: $F_0 \cdot F_1 \cdot F_2 = 3 \times 5 \times 17 = 255 = 257 - 2 = F_3 - 2$

Proof by induction that $F_0 F_1 \cdots F_{k-1} = F_k - 2$ for all $k \geqslant 1$:

Base case: $k = 1$ gives $F_0 = F_1 - 2$, i.e. $3 = 5 - 2$. True.

Inductive step: Assume $F_0 F_1 \cdots F_{k-1} = F_k - 2$ for some $k \geqslant 1$. We need to show that $F_0 F_1 \cdots F_{k} = F_{k+1} - 2$.

Multiplying both sides of the inductive hypothesis by $F_k$:

$F_0 F_1 \cdots F_{k-1} \cdot F_k = (F_k - 2) F_k = F_k^2 - 2 F_k$

Now we compute $F_k^2 - 2F_k + 2$:

$F_k^2 - 2F_k + 2 = (2^{2^k} + 1)^2 - 2(2^{2^k} + 1) + 2$

$= 2^{2 \cdot 2^k} + 2 \cdot 2^{2^k} + 1 - 2 \cdot 2^{2^k} - 2 + 2$

$= 2^{2^{k+1}} + 1 = F_{k+1}$

Therefore $F_k^2 - 2F_k = F_{k+1} - 2$, which gives

$F_0 F_1 \cdots F_k = F_{k+1} - 2$

This completes the induction.

Deducing that no two Fermat numbers share a common factor greater than 1:

Suppose for contradiction that $d > 1$ divides both $F_m$ and $F_n$, where $m > n$. Since $d \mid F_m$ and $d \mid (F_0 F_1 \cdots F_{m-1})$ (because $F_n$ is one of the factors in the product), we have

$d \mid \bigl( F_0 F_1 \cdots F_{m-1} - (F_m - 2) \bigr)$

By the identity we just proved, $F_0 F_1 \cdots F_{m-1} = F_m - 2$, so

$d \mid (F_m - 2 - F_m + 2) = 2$

Hence $d = 2$. But $F_n = 2^{2^n} + 1$ is odd for every $n$ (since $2^{2^n}$ is even), so $d \neq 2$. Contradiction.

Infinitely many primes:

Every Fermat number $F_n$ is greater than 1, so it has at least one prime factor $p_n$. Since no two Fermat numbers share a common factor greater than 1, the primes $p_0, p_1, p_2, \ldots$ are all distinct. As there are infinitely many Fermat numbers (one for each non-negative integer $n$), there are infinitely many primes.

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Question 4

Topic: 不等式与积分 Inequalities and Integration  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Give a sketch to show that, if $f(x) > 0$ for $p < x < q$, then $\int_{p}^{q} f(x) \, dx > 0$.

(i) By considering $f(x) = ax^2 - bx + c$ show that, if $a > 0$ and $b^2 < 4ac$, then $3b < 2a + 6c$.

(ii) By considering $f(x) = a \sin^2 x - b \sin x + c$ show that, if $a > 0$ and $b^2 < 4ac$, then $4b < (a + 2c)\pi$.

(iii) Show that, if $a > 0$, $b^2 < 4ac$ and $q > p > 0$, then

$b \ln(q/p) < a \left( \frac{1}{p} - \frac{1}{q} \right) + c(q - p) \, .$

Model Solution

Sketch: If $f(x) > 0$ for $p < x < q$, then the graph of $f$ lies strictly above the $x$-axis on the open interval $(p, q)$. The definite integral $\int_p^q f(x)\,dx$ represents the area between the curve and the $x$-axis. Since the curve is strictly above the axis, this area is positive, so $\int_p^q f(x)\,dx > 0$.

(i) Let $f(x) = ax^2 - bx + c$ with $a > 0$ and $b^2 < 4ac$.

We first show that $f(x) > 0$ for all real $x$. Since $a > 0$, we complete the square:

$f(x) = a\!\left(x - \frac{b}{2a}\right)^{\!2} + c - \frac{b^2}{4a} = a\!\left(x - \frac{b}{2a}\right)^{\!2} + \frac{4ac - b^2}{4a}$

Since $a > 0$ and $4ac - b^2 > 0$ (from $b^2 < 4ac$), both terms are non-negative and cannot simultaneously be zero, so $f(x) > 0$ for all $x$.

In particular, $f(x) > 0$ for $0 < x < 1$, so by the sketch result:

$\int_0^1 f(x)\,dx > 0$

Evaluating the integral:

$\int_0^1 (ax^2 - bx + c)\,dx = \left[\frac{a}{3}x^3 - \frac{b}{2}x^2 + cx\right]_0^1 = \frac{a}{3} - \frac{b}{2} + c > 0$

Multiplying both sides by 6:

$2a - 3b + 6c > 0$

Rearranging:

$3b < 2a + 6c$

(ii) Let $f(x) = a\sin^2 x - b\sin x + c$ with $a > 0$ and $b^2 < 4ac$.

Write $f(x)$ as a quadratic in $\sin x$:

$f(x) = a\!\left(\sin x - \frac{b}{2a}\right)^{\!2} + \frac{4ac - b^2}{4a}$

Since $a > 0$ and $4ac - b^2 > 0$, both terms are non-negative for every value of $\sin x$, and they cannot simultaneously be zero. Therefore $f(x) > 0$ for all $x$.

In particular, $f(x) > 0$ for $0 < x < \frac{\pi}{2}$, so

$\int_0^{\pi/2} f(x)\,dx > 0$

Evaluating the integral term by term:

$\int_0^{\pi/2} a\sin^2 x\,dx = a\int_0^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = \frac{a}{2}\!\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{a}{2} \cdot \frac{\pi}{2} = \frac{a\pi}{4}$

$\int_0^{\pi/2} (-b\sin x)\,dx = \bigl[b\cos x\bigr]_0^{\pi/2} = b(0 - 1) = -b$

$\int_0^{\pi/2} c\,dx = \frac{c\pi}{2}$

Combining:

$\frac{a\pi}{4} - b + \frac{c\pi}{2} > 0$

Multiplying both sides by $\frac{4}{\pi}$:

$a - \frac{4b}{\pi} + 2c > 0$

Rearranging:

$\frac{4b}{\pi} < a + 2c$

Multiplying both sides by $\pi$:

$4b < (a + 2c)\pi$

(iii) Let $f(x) = \frac{a}{x^2} - \frac{b}{x} + c$ with $a > 0$, $b^2 < 4ac$ and $q > p > 0$.

Write $f$ as a quadratic in $\frac{1}{x}$:

$f(x) = a\!\left(\frac{1}{x} - \frac{b}{2a}\right)^{\!2} + \frac{4ac - b^2}{4a}$

Since $a > 0$ and $4ac - b^2 > 0$, both terms are non-negative for all $x \neq 0$, and they cannot simultaneously be zero. Therefore $f(x) > 0$ for all $x > 0$.

In particular, $f(x) > 0$ for $p < x < q$, so

$\int_p^q f(x)\,dx > 0$

Evaluating the integral:

$\int_p^q \left(\frac{a}{x^2} - \frac{b}{x} + c\right)dx = \left[-\frac{a}{x} - b\ln x + cx\right]_p^q$

$= \left(-\frac{a}{q} - b\ln q + cq\right) - \left(-\frac{a}{p} - b\ln p + cp\right)$

$= a\!\left(\frac{1}{p} - \frac{1}{q}\right) - b(\ln q - \ln p) + c(q - p)$

$= a\!\left(\frac{1}{p} - \frac{1}{q}\right) - b\ln\!\left(\frac{q}{p}\right) + c(q - p) > 0$

Rearranging:

$b\ln\!\left(\frac{q}{p}\right) < a\!\left(\frac{1}{p} - \frac{1}{q}\right) + c(q - p)$

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Question 5

Topic: 递推关系与动力系统 Recurrence Relations and Dynamical Systems  |  Difficulty: Challenging  |  Marks: 20

Problem

5 The numbers $x_n$, where $n = 0, 1, 2, \dots$, satisfy

$x_{n+1} = kx_n(1 - x_n) \, .$

(i) Prove that, if $0 < k < 4$ and $0 < x_0 < 1$, then $0 < x_n < 1$ for all $n$.

(ii) Given that $x_0 = x_1 = x_2 = \dots = a$, with $a \neq 0$ and $a \neq 1$, find $k$ in terms of $a$.

(iii) Given instead that $x_0 = x_2 = x_4 = \dots = a$, with $a \neq 0$ and $a \neq 1$, show that $ab^3 - b^2 + (1 - a) = 0$, where $b = k(1 - a)$. Given, in addition, that $x_1 \neq a$, find the possible values of $k$ in terms of $a$.

Model Solution

Part (i)

We prove by induction on $n$ that $0 < x_n < 1$ for all $n$.

Base case: $0 < x_0 < 1$ is given.

Inductive step: Assume $0 < x_n < 1$. Then $0 < 1 - x_n < 1$, so $x_n(1 - x_n) > 0$, giving

$x_{n+1} = kx_n(1 - x_n) > 0$

since $k > 0$. For the upper bound, we maximise $g(x) = x(1 - x)$ on $(0, 1)$. Since

$g(x) = \frac{1}{4} - \left(x - \frac{1}{2}\right)^{\!2} \leqslant \frac{1}{4} \, ,$

with equality at $x = \frac{1}{2}$, we have

$x_{n+1} = kx_n(1 - x_n) \leqslant \frac{k}{4} < \frac{4}{4} = 1$

since $k < 4$. Hence $0 < x_{n+1} < 1$, completing the induction. $\blacksquare$

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Part (ii)

If $x_0 = x_1 = x_2 = \cdots = a$, then $a$ is a fixed point of the recurrence:

$a = ka(1 - a) \, .$

Since $a \neq 0$, we divide both sides by $a$:

$1 = k(1 - a) \, .$

Since $a \neq 1$, we solve for $k$:

$\boxed{k = \frac{1}{1 - a}}$

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Part (iii)

We are given $x_0 = x_2 = x_4 = \cdots = a$ with $a \neq 0$, $a \neq 1$, and $x_1 \neq a$.

Step 1: Express $x_1$ and $x_2$ in terms of $a$.

Define $b = k(1 - a)$. Then:

$x_1 = kx_0(1 - x_0) = ka(1 - a) = ab \, .$

$x_2 = kx_1(1 - x_1) = k(ab)(1 - ab) \, .$

Step 2: Apply the condition $x_2 = a$.

Since $x_2 = a$:

$a = kab(1 - ab) \, .$

Since $a \neq 0$, divide by $a$:

$1 = kb(1 - ab) \, . \qquad \text{($\dagger$)}$

Step 3: Derive the cubic in $b$.

Since $b = k(1 - a)$ and $a \neq 1$, we have $k = \dfrac{b}{1 - a}$. Substituting into ($\dagger$):

$1 = \frac{b}{1 - a} \cdot b \cdot (1 - ab) = \frac{b^2(1 - ab)}{1 - a} \, .$

Multiply both sides by $(1 - a)$:

$1 - a = b^2 - ab^3 \, .$

Rearranging:

$\boxed{ab^3 - b^2 + (1 - a) = 0} \qquad \text{(...)}$

Step 4: Factor the cubic.

Observe that $b = 1$ is a root: $a(1) - 1 + (1 - a) = 0$. $\checkmark$

Factoring out $(b - 1)$, we perform polynomial long division of $ab^3 - b^2 + (1 - a)$ by $(b - 1)$:

$ab^3 - b^2 + (1 - a) = (b - 1)(ab^2 + (a - 1)b + (a - 1)) \, .$

Verification: Expanding the right side:

$(b - 1)(ab^2) = ab^3 - ab^2$ $(b - 1)((a - 1)b) = (a - 1)b^2 - (a - 1)b$ $(b - 1)(a - 1) = (a - 1)b - (a - 1)$

Summing: $ab^3 + (-a + a - 1)b^2 + (-(a - 1) + (a - 1))b + (-(a - 1)) = ab^3 - b^2 + (1 - a)$. $\checkmark$

Step 5: Solve for the non-trivial value of $b$.

Since $x_1 \neq a$, we have $b \neq 1$ (because $x_1 = ab = a$ iff $b = 1$, given $a \neq 0$). Therefore:

$ab^2 + (a - 1)b + (a - 1) = 0 \, .$

Dividing by $a$ (since $a \neq 0$):

$b^2 + \frac{a - 1}{a}b + \frac{a - 1}{a} = 0 \, .$

Applying the quadratic formula with discriminant:

$\Delta = \left(\frac{a - 1}{a}\right)^{\!2} - 4 \cdot \frac{a - 1}{a} = \frac{(a - 1)^2 - 4a(a - 1)}{a^2} = \frac{(a - 1)(a - 1 - 4a)}{a^2} = \frac{(a - 1)(-3a - 1)}{a^2}$

$= \frac{(1 - a)(3a + 1)}{a^2} = \frac{(1 - a)(1 + 3a)}{a^2} \, .$

So:

$b = \frac{-(a-1)/a \pm \sqrt{(1-a)(1+3a)}/a}{2} = \frac{(1 - a) \pm \sqrt{(1 - a)(1 + 3a)}}{2a} \, .$

Step 6: Find $k$ from $b$.

Since $k = \dfrac{b}{1 - a}$ and $1 - a \neq 0$:

$k = \frac{1}{1 - a} \cdot \frac{(1 - a) \pm \sqrt{(1 - a)(1 + 3a)}}{2a} = \frac{1 \pm \sqrt{(1 - a)(1 + 3a)}}{2a} \, .$

These are the possible values of $k$, subject to the discriminant being non-negative, i.e., $(1 - a)(1 + 3a) \geqslant 0$.

Note on the excluded case $b = 1$: Setting $b = 1$ in the quadratic gives $a + (a - 1) + (a - 1) = 3a - 2 = 0$, so $a = 2/3$. At $a = 2/3$, the quadratic has a double root at $b = 1$, meaning the 2-cycle degenerates to the fixed point. For $a \neq 2/3$, both values of $k$ yield distinct period-2 orbits (when the discriminant is non-negative).

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Question 6

Topic: 空间几何与三角 Spatial Geometry and Trigonometry  |  Difficulty: Challenging  |  Marks: 20

Problem

6 The lines $l_1, l_2$ and $l_3$ lie in an inclined plane $P$ and pass through a common point $A$. The line $l_2$ is a line of greatest slope in $P$. The line $l_1$ is perpendicular to $l_3$ and makes an acute angle $\alpha$ with $l_2$. The angles between the horizontal and $l_1, l_2$ and $l_3$ are $\pi/6, \beta$ and $\pi/4$, respectively. Show that $\cos \alpha \sin \beta = \frac{1}{2}$ and find the value of $\sin \alpha \sin \beta$. Deduce that $\beta = \pi/3$.

The lines $l_1$ and $l_3$ are rotated in $P$ about $A$ so that $l_1$ and $l_3$ remain perpendicular to each other. The new acute angle between $l_1$ and $l_2$ is $\theta$. The new angles which $l_1$ and $l_3$ make with the horizontal are $\phi$ and $2\phi$, respectively. Show that

$\tan^2 \theta = \frac{3 + \sqrt{13}}{2} \, .$

Model Solution

Key formula for a line in an inclined plane

Choose coordinates with origin at $A$: the $x$-axis runs along the line of intersection of $P$ with the horizontal, the $y$-axis is horizontal and perpendicular to this (pointing “uphill”), and the $z$-axis is vertical. The plane $P$ has equation $z = y\tan\beta$.

The line $l_2$ (greatest slope) has direction $(0, 1, \tan\beta)$, which normalises to the unit vector $(0, \cos\beta, \sin\beta)$.

A line in $P$ making angle $\theta$ with $l_2$ (measured within the plane) has unit direction:

$\mathbf{d} = \sin\theta\,\mathbf{e}_1 + \cos\theta\,\mathbf{e}_2 = (\sin\theta,\; \cos\theta\cos\beta,\; \cos\theta\sin\beta)$

where $\mathbf{e}_1 = (1, 0, 0)$ is the unit contour direction and $\mathbf{e}_2 = (0, \cos\beta, \sin\beta)$ is the unit greatest-slope direction. One can verify $|\mathbf{d}|^2 = \sin^2\theta + \cos^2\theta = 1$.

The angle $\gamma$ that $\mathbf{d}$ makes with the horizontal satisfies $\sin\gamma = |d_z|$:

$\boxed{\sin\gamma = \cos\theta\sin\beta}$

Verification: At $\theta = 0$ (greatest slope): $\sin\gamma = \sin\beta$, so $\gamma = \beta$. At $\theta = \pi/2$ (contour): $\sin\gamma = 0$, so $\gamma = 0$. Both correct. $\checkmark$

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Part (i): Show that $\cos\alpha\sin\beta = \frac{1}{2}$, find $\sin\alpha\sin\beta$, and deduce $\beta = \pi/3$.

$l_1$ makes angle $\alpha$ with $l_2$, so $\theta_1 = \alpha$ and $\gamma_1 = \pi/6$:

$\sin\frac{\pi}{6} = \cos\alpha\sin\beta \qquad \Longrightarrow \qquad \cos\alpha\sin\beta = \frac{1}{2} \qquad \text{... (A)}$

$l_3$ is perpendicular to $l_1$ in $P$. Since $l_1$ is at angle $\alpha$ from $l_2$ in $P$, the line perpendicular to $l_1$ in $P$ is at angle $\alpha + \frac{\pi}{2}$ from $l_2$. Thus $\theta_3 = \alpha + \frac{\pi}{2}$ and $\gamma_3 = \pi/4$:

$\sin\frac{\pi}{4} = \cos\!\left(\alpha + \frac{\pi}{2}\right)\sin\beta = (-\sin\alpha)\sin\beta$

Since $\gamma_3 > 0$ we need $|\sin\alpha\sin\beta|$, giving:

$\sin\alpha\sin\beta = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \qquad \text{... (B)}$

Deducing $\beta = \pi/3$:

Dividing (B) by (A):

$\tan\alpha = \frac{\sin\alpha\sin\beta}{\cos\alpha\sin\beta} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$

So $\alpha = \arctan\sqrt{2}$, giving $\sin\alpha = \dfrac{\sqrt{2}}{\sqrt{3}}$ and $\cos\alpha = \dfrac{1}{\sqrt{3}}$.

From (A):

$\sin\beta = \frac{1}{2\cos\alpha} = \frac{1}{2/\sqrt{3}} = \frac{\sqrt{3}}{2}$

Hence $\beta = \dfrac{\pi}{3}$. $\blacksquare$

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Part (ii): Show that $\tan^2\theta = \dfrac{3 + \sqrt{13}}{2}$.

After rotation, $l_1$ makes angle $\theta$ with $l_2$, and $l_3$ (perpendicular to $l_1$ in $P$) is at angle $\theta + \frac{\pi}{2}$ from $l_2$. Their inclinations to the horizontal are $\phi$ and $2\phi$.

Using the key formula with $\beta = \pi/3$ (so $\sin\beta = \frac{\sqrt{3}}{2}$):

$\sin\phi = \cos\theta \cdot \frac{\sqrt{3}}{2} \qquad \text{... (C)}$

$\sin 2\phi = \sin\theta \cdot \frac{\sqrt{3}}{2} \qquad \text{... (D)}$

From (C): $\cos\theta = \dfrac{2\sin\phi}{\sqrt{3}}$.

From (D) and the double angle formula $\sin 2\phi = 2\sin\phi\cos\phi$:

$2\sin\phi\cos\phi = \frac{\sqrt{3}}{2}\sin\theta \qquad \Longrightarrow \qquad \sin\theta = \frac{4\sin\phi\cos\phi}{\sqrt{3}}$

Applying $\cos^2\theta + \sin^2\theta = 1$:

$\frac{4\sin^2\phi}{3} + \frac{16\sin^2\phi\cos^2\phi}{3} = 1$

Multiply through by 3:

$4\sin^2\phi + 16\sin^2\phi\cos^2\phi = 3$

$4\sin^2\phi\bigl(1 + 4\cos^2\phi\bigr) = 3$

Substitute $\cos^2\phi = 1 - \sin^2\phi$. Let $s = \sin^2\phi$:

$4s(1 + 4(1 - s)) = 3$

$4s(5 - 4s) = 3$

$20s - 16s^2 = 3$

$16s^2 - 20s + 3 = 0$

By the quadratic formula:

$s = \frac{20 \pm \sqrt{400 - 192}}{32} = \frac{20 \pm \sqrt{208}}{32} = \frac{20 \pm 4\sqrt{13}}{32} = \frac{5 \pm \sqrt{13}}{8}$

Since $s = \sin^2\phi \leqslant 1$ and $\dfrac{5 + \sqrt{13}}{8} \approx 1.076 > 1$, we reject the positive root:

$\sin^2\phi = \frac{5 - \sqrt{13}}{8}$

Now compute $\tan^2\theta$:

$\cos^2\theta = \frac{4\sin^2\phi}{3} = \frac{4}{3} \cdot \frac{5 - \sqrt{13}}{8} = \frac{5 - \sqrt{13}}{6}$

$\sin^2\theta = 1 - \frac{5 - \sqrt{13}}{6} = \frac{6 - 5 + \sqrt{13}}{6} = \frac{1 + \sqrt{13}}{6}$

$\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta} = \frac{1 + \sqrt{13}}{5 - \sqrt{13}}$

Rationalise by multiplying numerator and denominator by $5 + \sqrt{13}$:

$\tan^2\theta = \frac{(1 + \sqrt{13})(5 + \sqrt{13})}{(5 - \sqrt{13})(5 + \sqrt{13})} = \frac{5 + \sqrt{13} + 5\sqrt{13} + 13}{25 - 13} = \frac{18 + 6\sqrt{13}}{12} = \frac{3 + \sqrt{13}}{2}$

$\boxed{\tan^2\theta = \frac{3 + \sqrt{13}}{2}}$

$\blacksquare$

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Question 7

Topic: 空间向量 3D Vectors  |  Difficulty: Challenging  |  Marks: 20

Problem

7 In 3-dimensional space, the lines $m_1$ and $m_2$ pass through the origin and have directions $\mathbf{i} + \mathbf{j}$ and $\mathbf{i} + \mathbf{k}$, respectively. Find the directions of the two lines $m_3$ and $m_4$ that pass through the origin and make angles of $\pi/4$ with both $m_1$ and $m_2$. Find also the cosine of the acute angle between $m_3$ and $m_4$.

The points $A$ and $B$ lie on $m_1$ and $m_2$ respectively, and are each at distance $\lambda\sqrt{2}$ units from $O$. The points $P$ and $Q$ lie on $m_3$ and $m_4$ respectively, and are each at distance 1 unit from $O$. If all the coordinates (with respect to axes $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$) of $A, B, P$ and $Q$ are non-negative, prove that:

(i) there are only two values of $\lambda$ for which $AQ$ is perpendicular to $BP$;

(ii) there are no non-zero values of $\lambda$ for which $AQ$ and $BP$ intersect.

Model Solution

Finding directions of $m_3$ and $m_4$

The lines $m_1$ and $m_2$ have direction vectors $\mathbf{d}_1 = (1, 1, 0)$ and $\mathbf{d}_2 = (1, 0, 1)$.

Let the direction of a line through the origin making angle $\pi/4$ with both $m_1$ and $m_2$ be $\mathbf{d} = (a, b, c)$.

For the angle with $m_1$:

$\cos\frac{\pi}{4} = \frac{|\mathbf{d} \cdot \mathbf{d}_1|}{|\mathbf{d}|\,|\mathbf{d}_1|} = \frac{|a + b|}{\sqrt{a^2 + b^2 + c^2}\cdot\sqrt{2}} = \frac{1}{\sqrt{2}}$

$\Longrightarrow\quad |a + b| = \sqrt{a^2 + b^2 + c^2}$

Squaring both sides:

$(a + b)^2 = a^2 + b^2 + c^2 \quad\Longrightarrow\quad a^2 + 2ab + b^2 = a^2 + b^2 + c^2 \quad\Longrightarrow\quad 2ab = c^2 \qquad \text{(*)}$

For the angle with $m_2$:

$\cos\frac{\pi}{4} = \frac{|\mathbf{d} \cdot \mathbf{d}_2|}{|\mathbf{d}|\,|\mathbf{d}_2|} = \frac{|a + c|}{\sqrt{a^2 + b^2 + c^2}\cdot\sqrt{2}} = \frac{1}{\sqrt{2}}$

Squaring:

$(a + c)^2 = a^2 + b^2 + c^2 \quad\Longrightarrow\quad 2ac = b^2 \qquad \text{(**)}$

From $(\ast\ast)$: $b^2 = 2ac$. From $(\ast)$: $c^2 = 2ab$. Dividing these (assuming $b, c \neq 0$):

$\frac{b^2}{c^2} = \frac{2ac}{2ab} = \frac{c}{b} \quad\Longrightarrow\quad b^3 = c^3 \quad\Longrightarrow\quad b = c$

Substituting $b = c$ into $(\ast)$: $2ab = b^2$, so either $b = 0$ or $b = 2a$.

Case 1: $b = c = 0$. Then $\mathbf{d} = (a, 0, 0)$, giving direction $(1, 0, 0)$.

Case 2: $b = c = 2a$. Then $\mathbf{d} = (a, 2a, 2a) = a(1, 2, 2)$, giving direction $(1, 2, 2)$.

Verification. For $(1, 0, 0)$: angle with $m_1$ gives $\frac{|1|}{1\cdot\sqrt{2}} = \frac{1}{\sqrt{2}}$, and angle with $m_2$ gives $\frac{|1|}{1\cdot\sqrt{2}} = \frac{1}{\sqrt{2}}$. Both are $\pi/4$.

For $(1, 2, 2)$: angle with $m_1$ gives $\frac{|3|}{3\cdot\sqrt{2}} = \frac{1}{\sqrt{2}}$, and angle with $m_2$ gives $\frac{|3|}{3\cdot\sqrt{2}} = \frac{1}{\sqrt{2}}$. Both are $\pi/4$.

So $m_3$ has direction $\mathbf{i}$ (i.e. $(1, 0, 0)$) and $m_4$ has direction $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ (i.e. $(1, 2, 2)$), or vice versa.

Cosine of the acute angle between $m_3$ and $m_4$:

$\cos\theta = \frac{|(1,0,0)\cdot(1,2,2)|}{1\cdot 3} = \frac{1}{3}$

Setting up coordinates

Since all coordinates are non-negative, we take the positive directions:

• $A$ lies on $m_1$ at distance $\lambda\sqrt{2}$: direction $(1,1,0)$ has magnitude $\sqrt{2}$, so $A = \lambda\sqrt{2}\cdot\frac{(1,1,0)}{\sqrt{2}} = (\lambda, \lambda, 0)$.
• $B$ lies on $m_2$ at distance $\lambda\sqrt{2}$: similarly $B = (\lambda, 0, \lambda)$.
• $P$ lies on $m_3 = (1,0,0)$ at distance 1: $P = (1, 0, 0)$.
• $Q$ lies on $m_4 = (1,2,2)$ at distance 1: $|{(1,2,2)}| = 3$, so $Q = \frac{1}{3}(1, 2, 2) = \left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$.

(i) Two values of $\lambda$ for which $AQ \perp BP$

$\overrightarrow{AQ} = Q - A = \left(\tfrac{1}{3} - \lambda,\; \tfrac{2}{3} - \lambda,\; \tfrac{2}{3}\right)$

$\overrightarrow{BP} = P - B = (1 - \lambda,\; 0,\; -\lambda)$

Setting $\overrightarrow{AQ} \cdot \overrightarrow{BP} = 0$:

$\left(\tfrac{1}{3} - \lambda\right)(1 - \lambda) + \left(\tfrac{2}{3} - \lambda\right)(0) + \tfrac{2}{3}(-\lambda) = 0$

Expanding the first term:

$\tfrac{1}{3} - \tfrac{\lambda}{3} - \lambda + \lambda^2 - \tfrac{2\lambda}{3} = 0$

$\lambda^2 - \lambda\!\left(\tfrac{1}{3} + 1 + \tfrac{2}{3}\right) + \tfrac{1}{3} = 0$

$\lambda^2 - 2\lambda + \tfrac{1}{3} = 0$

Multiplying through by 3:

$3\lambda^2 - 6\lambda + 1 = 0$

By the quadratic formula:

$\lambda = \frac{6 \pm \sqrt{36 - 12}}{6} = \frac{6 \pm \sqrt{24}}{6} = \frac{6 \pm 2\sqrt{6}}{6} = \frac{3 \pm \sqrt{6}}{3}$

Since $\sqrt{6} < 3$, both roots are positive (hence valid). There are exactly two values of $\lambda$ for which $AQ \perp BP$.

(ii) No non-zero values of $\lambda$ for which $AQ$ and $BP$ intersect

The line through $A$ and $Q$ has parametric form:

$\mathbf{r} = A + s(Q - A) = \left(\lambda + s\!\left(\tfrac{1}{3} - \lambda\right),\;\; \lambda + s\!\left(\tfrac{2}{3} - \lambda\right),\;\; \tfrac{2s}{3}\right)$

The line through $B$ and $P$ has parametric form:

$\mathbf{r} = B + t(P - B) = \left(\lambda + t(1 - \lambda),\;\; 0,\;\; \lambda(1 - t)\right)$

For an intersection, we need all three components equal:

$\lambda + s\!\left(\tfrac{1}{3} - \lambda\right) = \lambda + t(1 - \lambda) \qquad \text{(I)}$

$\lambda + s\!\left(\tfrac{2}{3} - \lambda\right) = 0 \qquad \text{(II)}$

$\tfrac{2s}{3} = \lambda(1 - t) \qquad \text{(III)}$

From (II): $s = \dfrac{-\lambda}{\frac{2}{3} - \lambda} = \dfrac{-3\lambda}{2 - 3\lambda}$ (valid when $\lambda \neq 2/3$).

From (III): $t = 1 - \dfrac{2s}{3\lambda}$ (valid when $\lambda \neq 0$).

Substituting $t$ into (I):

$s\!\left(\tfrac{1}{3} - \lambda\right) = \left(1 - \frac{2s}{3\lambda}\right)(1 - \lambda)$

$s\!\left(\tfrac{1}{3} - \lambda\right) = (1 - \lambda) - \frac{2s(1 - \lambda)}{3\lambda}$

$s\!\left(\tfrac{1}{3} - \lambda\right) + \frac{2s(1 - \lambda)}{3\lambda} = 1 - \lambda$

Combining the left-hand side over a common denominator $3\lambda$:

$s \cdot \frac{\lambda(1 - 3\lambda) + 2(1 - \lambda)}{3\lambda} = 1 - \lambda$

$s \cdot \frac{\lambda - 3\lambda^2 + 2 - 2\lambda}{3\lambda} = 1 - \lambda$

$s \cdot \frac{-3\lambda^2 - \lambda + 2}{3\lambda} = 1 - \lambda$

Factor the numerator: $-3\lambda^2 - \lambda + 2 = -(3\lambda - 2)(\lambda + 1)$.

$s \cdot \frac{-(3\lambda - 2)(\lambda + 1)}{3\lambda} = 1 - \lambda$

$s = \frac{3\lambda(\lambda - 1)}{(3\lambda - 2)(\lambda + 1)} \qquad \text{from (I)}$

From (II): $s = \dfrac{3\lambda}{3\lambda - 2}$.

Setting these equal (for $\lambda \neq 0$ and $\lambda \neq 2/3$):

$\frac{3\lambda}{3\lambda - 2} = \frac{3\lambda(\lambda - 1)}{(3\lambda - 2)(\lambda + 1)}$

Dividing both sides by $3\lambda/(3\lambda - 2)$ (which is non-zero):

$1 = \frac{\lambda - 1}{\lambda + 1}$

$\lambda + 1 = \lambda - 1 \quad\Longrightarrow\quad 1 = -1$

This is a contradiction, so the system has no solution for any $\lambda \neq 0$.

Checking $\lambda = 2/3$: Equation (II) becomes $\frac{2}{3} + s \cdot 0 = 0$, i.e. $\frac{2}{3} = 0$, which is impossible.

Checking $\lambda = 0$: $A = B = O$, so $AQ$ is the line $m_4$ and $BP$ is the line $m_3$, which intersect at $O$.

Therefore, for all non-zero values of $\lambda$, the lines $AQ$ and $BP$ do not intersect.

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Question 8

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

Problem

8 Find $y$ in terms of $x$, given that:

$$\begin{aligned} \text{for } x < 0, \quad \frac{\mathrm{d}y}{\mathrm{d}x} &= -y \quad \text{and} \quad y = a \text{ when } x = -1; \\ \text{for } x > 0, \quad \frac{\mathrm{d}y}{\mathrm{d}x} &= y \quad \text{and} \quad y = b \text{ when } x = 1. \end{aligned}$$

Sketch a solution curve. Determine the condition on $a$ and $b$ for the solution curve to be continuous (that is, for there to be no ‘jump’ in the value of $y$) at $x = 0$.

Solve the differential equation

$\frac{\mathrm{d}y}{\mathrm{d}x} = |e^x - 1|y$

given that $y = e^e$ when $x = 1$ and that $y$ is continuous at $x = 0$. Write down the following limits:

(i) $\lim_{x \to +\infty} y \exp(-e^x)$; (ii) $\lim_{x \to -\infty} y e^{-x}$.

Model Solution

Part 1: Solving the piecewise ODE

For $x < 0$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -y$.

Separating variables: $\dfrac{\mathrm{d}y}{y} = -\mathrm{d}x$, so $\ln|y| = -x + C$, giving $y = Ae^{-x}$.

Applying the condition $y = a$ at $x = -1$: $a = Ae^{1}$, so $A = ae^{-1}$.

$y = ae^{-(x+1)} \qquad (x < 0)$

For $x > 0$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = y$.

Separating variables: $\dfrac{\mathrm{d}y}{y} = \mathrm{d}x$, so $\ln|y| = x + C$, giving $y = Be^{x}$.

Applying the condition $y = b$ at $x = 1$: $b = Be^{1}$, so $B = be^{-1}$.

$y = be^{x-1} \qquad (x > 0)$

Sketch. The curve $y = ae^{-(x+1)}$ for $x < 0$ is a decaying exponential (if $a > 0$) passing through $(-1, a)$ and tending to $+\infty$ as $x \to -\infty$. At $x = 0^-$, $y = a/e$. The curve $y = be^{x-1}$ for $x > 0$ is a growing exponential passing through $(1, b)$ with $y = b/e$ at $x = 0^+$. If $a \neq b$, there is a jump discontinuity at $x = 0$.

Continuity at $x = 0$:

$\lim_{x \to 0^-} y = ae^{-1} = \frac{a}{e}, \qquad \lim_{x \to 0^+} y = be^{-1} = \frac{b}{e}$

For continuity at $x = 0$ we need $\dfrac{a}{e} = \dfrac{b}{e}$, i.e.

$a = b$

Part 2: Solving $\dfrac{\mathrm{d}y}{\mathrm{d}x} = |e^x - 1|\,y$

We split into two regions based on the sign of $e^x - 1$.

For $x < 0$: $e^x < 1$, so $|e^x - 1| = 1 - e^x$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = (1 - e^x)\,y$

Separating variables: $\dfrac{\mathrm{d}y}{y} = (1 - e^x)\,\mathrm{d}x$.

$\ln|y| = x - e^x + C_1 \qquad\Longrightarrow\qquad y = A_1 e^{x - e^x}$

For $x > 0$: $e^x > 1$, so $|e^x - 1| = e^x - 1$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = (e^x - 1)\,y$

Separating variables: $\dfrac{\mathrm{d}y}{y} = (e^x - 1)\,\mathrm{d}x$.

$\ln|y| = e^x - x + C_2 \qquad\Longrightarrow\qquad y = A_2 e^{e^x - x}$

Using $y = e^e$ at $x = 1$:

$A_2\,e^{e^1 - 1} = e^e \qquad\Longrightarrow\qquad A_2\,e^{e-1} = e^e \qquad\Longrightarrow\qquad A_2 = \frac{e^e}{e^{e-1}} = e$

So for $x > 0$: $y = e \cdot e^{e^x - x} = e^{1 + e^x - x}$.

Continuity at $x = 0$:

$\lim_{x \to 0^+} y = e^{1 + e^0 - 0} = e^{1 + 1} = e^2$

$\lim_{x \to 0^-} y = A_1\,e^{0 - e^0} = A_1\,e^{-1}$

Setting $A_1 e^{-1} = e^2$ gives $A_1 = e^3$.

So for $x < 0$: $y = e^3 \cdot e^{x - e^x} = e^{3 + x - e^x}$.

Summary:

$y = \begin{cases} e^{3 + x - e^x} & x < 0 \\[4pt] e^{1 + e^x - x} & x > 0 \end{cases}$

(Note that both expressions give $y = e^2$ at $x = 0$, confirming continuity.)

(i) $\displaystyle\lim_{x \to +\infty} y\exp(-e^x)$:

For $x > 0$:

$y\,e^{-e^x} = e^{1 + e^x - x}\cdot e^{-e^x} = e^{1 - x}$

$\lim_{x \to +\infty} e^{1-x} = 0$

(ii) $\displaystyle\lim_{x \to -\infty} y\,e^{-x}$:

For $x < 0$:

$y\,e^{-x} = e^{3 + x - e^x}\cdot e^{-x} = e^{3 - e^x}$

As $x \to -\infty$, $e^x \to 0$, so:

$\lim_{x \to -\infty} e^{3 - e^x} = e^{3 - 0} = e^3$