STEP3 2000 -- Pure Mathematics

Exam: STEP3  |  Year: 2000  |  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	坐标几何 Coordinate Geometry	Challenging	隐函数求导,切线方程,对称性分析,几何证明
2	积分 Integration	Challenging	三角代换,定积分计算,参数化推广
3	复数 Complex Numbers	Hard	复数运算,旋转变换,等边三角形性质,共点证明
4	函数与图像 Functions and Graphs	Challenging	有理函数分析,渐近线求法,分类讨论,图像绘制
5	线性代数与向量 Linear Algebra and Vectors	Hard	行列式性质,线性变换,矩阵表示,充要条件证明
6	代数 Algebra	Challenging	待定系数法,预解三次方程,多项式因式分解
7	级数与不等式 Sequences, Series and Inequalities	Challenging	二项式定理,AM-GM不等式,无穷乘积收敛
8	递推关系与数学归纳法 Sequences and Mathematical Induction	Challenging	数学归纳法,特征方程,递推关系线性化

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Question 1

Topic: 坐标几何 Coordinate Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Sketch on the same axes the two curves $C_1$ and $C_2$, given by

$\begin{matrix} C_1 : & xy & = & 1, \\ C_2 : & x^2 - y^2 & = & 2. \end{matrix}$

The curves intersect at $P$ and $Q$. Given that the coordinates of $P$ are $(a, b)$ (which you need not evaluate), write down the coordinates of $Q$ in terms of $a$ and $b$.

The tangent to $C_1$ through $P$ meets the tangent to $C_2$ through $Q$ at the point $M$, and the tangent to $C_2$ through $P$ meets the tangent to $C_1$ through $Q$ at $N$. Show that the coordinates of $M$ are $(-b, a)$ and write down the coordinates of $N$.

Show that $PMQN$ is a square.

Model Solution

Part (i): Coordinates of $Q$

The curve $C_1 : xy = 1$ is symmetric under $(x, y) \mapsto (-x, -y)$ (a hyperbola in quadrants 1 and 3). The curve $C_2 : x^2 - y^2 = 2$ is also symmetric under $(x, y) \mapsto (-x, -y)$ (a hyperbola opening left-right). Since $P = (a, b)$ is an intersection point, $Q = (-a, -b)$ is the other.

Part (ii): Tangent equations

Tangent to $C_1$ at $P$. Differentiating $xy = 1$ implicitly: $y + xy' = 0$, so $y' = -y/x$. At $P = (a, b)$: slope $= -b/a$. Tangent:

$y - b = -\frac{b}{a}(x - a) \implies bx + ay = 2ab = 2. \qquad \text{(since } ab = 1\text{)}$

Tangent to $C_2$ at $Q$. Differentiating $x^2 - y^2 = 2$: $2x - 2yy' = 0$, so $y' = x/y$. At $Q = (-a, -b)$: slope $= (-a)/(-b) = a/b$. Tangent:

$b(y + b) = a(x + a) \implies by + b^2 = ax + a^2 \implies ax - by = b^2 - a^2.$

Since $a^2 - b^2 = 2$ (as $P$ lies on $C_2$), we get $b^2 - a^2 = -2$, so:

$ax - by = -2. \qquad \text{(tangent to } C_2 \text{ at } Q\text{)}$

Tangent to $C_2$ at $P$. At $P = (a, b)$: slope $= a/b$. Tangent: $y - b = \frac{a}{b}(x - a)$. Rearranging: $ax - by = a^2 - b^2 = 2$.

$ax - by = 2. \qquad \text{(tangent to } C_2 \text{ at } P\text{)}$

Tangent to $C_1$ at $Q$. At $Q = (-a, -b)$: slope $= -(-b)/(-a) = -b/a$. Tangent: $y + b = -\frac{b}{a}(x + a)$. Rearranging: $bx + ay = -2ab = -2$.

$bx + ay = -2. \qquad \text{(tangent to } C_1 \text{ at } Q\text{)}$

Part (iii): Show $M = (-b, a)$

$M$ is the intersection of the tangent to $C_1$ at $P$ and the tangent to $C_2$ at $Q$:

$bx + ay = 2, \qquad ax - by = -2.$

Solving: multiply the first by $b$ and the second by $a$:

Substituting $(-b,a)$ gives $b(-b)+a(a)=a^2-b^2=2$ and $a(-b)-b(a)=-2ab=-2$. Hence $M=(-b,a)$.

$\qquad \blacksquare$

Part (iv): Coordinates of $N$

$N$ is the intersection of the tangent to $C_2$ at $P$ and the tangent to $C_1$ at $Q$:

$ax - by = 2, \qquad bx + ay = -2.$

By the same verification approach (swapping the roles of the two tangent pairs), $N = (b, -a)$.

Indeed, $a(b)-b(-a)=2ab=2$ and $b(b)+a(-a)=b^2-a^2=-2$.

Therefore $N = (b, -a)$.

Part (v): Show $PMQN$ is a square

The four vertices are $P = (a, b)$, $M = (-b, a)$, $Q = (-a, -b)$, $N = (b, -a)$.

Midpoints of diagonals. Midpoint of $PQ$: $\left(\frac{a + (-a)}{2}, \frac{b + (-b)}{2}\right) = (0, 0)$. Midpoint of $MN$: $\left(\frac{-b + b}{2}, \frac{a + (-a)}{2}\right) = (0, 0)$.

The diagonals bisect each other at the origin.

Equal diagonals. $|PQ| = 2\sqrt{a^2 + b^2}$ and $|MN| = \sqrt{(-b - b)^2 + (a - (-a))^2} = \sqrt{4b^2 + 4a^2} = 2\sqrt{a^2 + b^2}$. So $|PQ| = |MN|$.

Perpendicular diagonals. Direction of $PQ$: $(-2a, -2b)$, i.e. $(a, b)$. Direction of $MN$: $(2b, -2a)$, i.e. $(b, -a)$. Dot product: $ab + b(-a) = ab - ab = 0$.

A quadrilateral whose diagonals are equal, bisect each other, and are perpendicular is a square. $\qquad \blacksquare$

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Question 2

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

Problem

2 Use the substitution $x = 2 - \cos \theta$ to evaluate the integral

$\int_{3/2}^{2} \left( \frac{x - 1}{3 - x} \right)^{\frac{1}{2}} dx.$

Show that, for $a < b$,

$\int_{p}^{q} \left( \frac{x - a}{b - x} \right)^{\frac{1}{2}} dx = \frac{(b - a)(\pi + 3\sqrt{3} - 6)}{12},$

where $p = (3a + b)/4$ and $q = (a + b)/2$.

Model Solution

Part (i): Evaluate the integral using $x = 2 - \cos\theta$

Let $x = 2 - \cos\theta$, so $dx = \sin\theta \, d\theta$.

When $x = 3/2$: $\cos\theta = 1/2$, so $\theta = \pi/3$. When $x = 2$: $\cos\theta = 0$, so $\theta = \pi/2$.

Compute the integrand:

$\frac{x - 1}{3 - x} = \frac{(2 - \cos\theta) - 1}{3 - (2 - \cos\theta)} = \frac{1 - \cos\theta}{1 + \cos\theta}.$

Using half-angle identities: $1 - \cos\theta = 2\sin^2(\theta/2)$ and $1 + \cos\theta = 2\cos^2(\theta/2)$:

$\frac{1 - \cos\theta}{1 + \cos\theta} = \tan^2(\theta/2).$

So the integral becomes:

$\int_{\pi/3}^{\pi/2} \tan(\theta/2) \cdot \sin\theta \, d\theta.$

Since $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$:

$\int_{\pi/3}^{\pi/2} \frac{\sin(\theta/2)}{\cos(\theta/2)} \cdot 2\sin(\theta/2)\cos(\theta/2) \, d\theta = \int_{\pi/3}^{\pi/2} 2\sin^2(\theta/2) \, d\theta = \int_{\pi/3}^{\pi/2} (1 - \cos\theta) \, d\theta.$

Evaluating:

$\left[\theta - \sin\theta\right]_{\pi/3}^{\pi/2} = \left(\frac{\pi}{2} - 1\right) - \left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} - 1 + \frac{\sqrt{3}}{2} = \frac{\pi}{6} + \frac{\sqrt{3}}{2} - 1.$

Rewriting with a common denominator:

$\int_{3/2}^{2} \left(\frac{x - 1}{3 - x}\right)^{1/2} dx = \frac{\pi + 3\sqrt{3} - 6}{6}. \qquad \text{(i)}$

Part (ii): General case

For $a < b$, let $h = \frac{b - a}{2}$ (half the width of $[a, b]$). Use the substitution $x = \frac{a + b}{2} - h\cos\theta$, i.e. $x = \frac{a+b}{2} - \frac{b-a}{2}\cos\theta$.

Then $dx = h\sin\theta \, d\theta$, and:

$x - a = h(1 - \cos\theta), \qquad b - x = h(1 + \cos\theta).$

So $\frac{x - a}{b - x} = \frac{1 - \cos\theta}{1 + \cos\theta} = \tan^2(\theta/2)$, and the integrand becomes $\tan(\theta/2)$.

The limits: when $x = p = \frac{3a + b}{4} = a + \frac{h}{2}$:

$a + \frac{h}{2} = \frac{a+b}{2} - h\cos\theta \implies h\cos\theta = \frac{b-a}{2} - \frac{h}{2} = h - \frac{h}{2} = \frac{h}{2},$

so $\cos\theta = 1/2$, giving $\theta = \pi/3$.

When $x = q = \frac{a+b}{2}$:

$\frac{a+b}{2} = \frac{a+b}{2} - h\cos\theta \implies \cos\theta = 0, \quad \theta = \pi/2.$

The integral becomes:

$\int_{\pi/3}^{\pi/2} \tan(\theta/2) \cdot h\sin\theta \, d\theta = h\int_{\pi/3}^{\pi/2} (1 - \cos\theta) \, d\theta,$

using the same simplification as in Part (i). Evaluating:

$h \cdot \frac{\pi + 3\sqrt{3} - 6}{6} = \frac{b - a}{2} \cdot \frac{\pi + 3\sqrt{3} - 6}{6} = \frac{(b - a)(\pi + 3\sqrt{3} - 6)}{12}. \qquad \blacksquare$

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Question 3

Topic: 复数 Complex Numbers  |  Difficulty: Hard  |  Marks: 20

Problem

3 Given that $\alpha = e^{i\frac{\pi}{3}}$, prove that $1 + \alpha^2 = \alpha$.

A triangle in the Argand plane has vertices $A$, $B$, and $C$ represented by the complex numbers $p$, $q\alpha^2$ and $-r\alpha$ respectively, where $p$, $q$ and $r$ are positive real numbers. Sketch the triangle $ABC$.

Three equilateral triangles $ABL$, $BCM$ and $CAN$ (each lettered clockwise) are erected on sides $AB$, $BC$ and $CA$ respectively. Show that the complex number representing $N$ is $(1 - \alpha)p - \alpha^2r$ and find similar expressions for the complex numbers representing $L$ and $M$.

Show that lines $LC$, $MA$ and $NB$ all meet at the origin, and that these three line segments have the common length $p + q + r$.

Model Solution

Part (i): Prove $1 + \alpha^2 = \alpha$

Since $\alpha = e^{i\pi/3}$, we have $\alpha^2 = e^{2i\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.

Therefore $1 + \alpha^2 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i\pi/3} = \alpha$. $\qquad \blacksquare$

Note: This gives $\alpha^2 = \alpha - 1$, or equivalently $\alpha^2 - \alpha + 1 = 0$. Also $\alpha^3 = -1$ and $\alpha^6 = 1$.

Part (ii): Sketch of triangle $ABC$

$A = p$ lies on the positive real axis. $B = q\alpha^2$ lies at angle $2\pi/3$ (upper left). $C = -r\alpha = re^{i4\pi/3}$ lies at angle $4\pi/3$ (lower left). For $p = q = r$ the triangle is equilateral.

Part (iii): Find $L$, $M$, $N$

For an equilateral triangle with two given vertices $U, V$ and third vertex $W$ such that $UVW$ is labeled clockwise, we use $W = V + (U - V)\alpha$ (a $60^\circ$ counterclockwise rotation of $U$ about $V$, which produces clockwise labeling in the standard Argand diagram).

Finding $N$ (equilateral triangle $CAN$, clockwise):

$N = A + (C - A)\alpha = p + (-r\alpha - p)\alpha = p - r\alpha^2 - p\alpha = (1 - \alpha)p - r\alpha^2.$

Using $\alpha^2 = \alpha - 1$: coefficient of $r$ is $-(\alpha - 1) = 1 - \alpha$. So:

$N = (1 - \alpha)p - \alpha^2 r. \qquad \blacksquare$

Finding $L$ (equilateral triangle $ABL$, clockwise):

$L = B + (A - B)\alpha = q\alpha^2 + (p - q\alpha^2)\alpha = q\alpha^2 + p\alpha - q\alpha^3.$

Since $\alpha^3 = -1$:

$L = p\alpha + q\alpha^2 + q = p\alpha + q(\alpha^2 + 1) = p\alpha + q\alpha = (p + q)\alpha.$

Finding $M$ (equilateral triangle $BCM$, clockwise):

$M = C + (B - C)\alpha = -r\alpha + (q\alpha^2 + r\alpha)\alpha = -r\alpha + q\alpha^3 + r\alpha^2.$

Since $\alpha^3 = -1$ and $\alpha^2 = \alpha - 1$:

$M = -r\alpha - q + r(\alpha - 1) = -r\alpha - q + r\alpha - r = -q - r.$

So $M = -(q + r)$, lying on the negative real axis.

$L = (p+q)\alpha$, $M = -(q+r)$, $N = (1-\alpha)p - \alpha^2 r$.

Part (iv): Show $LC$, $MA$, $NB$ meet at the origin with common length $p + q + r$

Collinearity with the origin. A point $Z$ lies on the line through the origin if and only if $\arg(Z)$ is constant, equivalently $Z/\bar{Z}$ is real (or $Z = 0$). We show each pair of points is collinear with $O$.

$L$ and $C$ are collinear with $O$: $L = (p+q)\alpha$ and $C = -r\alpha$. Both are real multiples of $\alpha$, so they lie on the ray at angle $\pi/3$ and its extension. Hence $O$, $L$, $C$ are collinear.

$M$ and $A$ are collinear with $O$: $M = -(q+r)$ and $A = p$. Both are real, so they lie on the real axis. Hence $O$, $M$, $A$ are collinear.

$N$ and $B$ are collinear with $O$: We need $N/B$ to be real.

$\frac{N}{B} = \frac{(1-\alpha)p - \alpha^2 r}{q\alpha^2}.$

Since $\alpha^2 - \alpha + 1 = 0$, we have $1-\alpha=-\alpha^2$. So

$N = -\alpha^2 p - \alpha^2 r = -\alpha^2(p + r).$

Therefore $\frac{N}{B} = \frac{-\alpha^2(p+r)}{q\alpha^2} = -\frac{p+r}{q}$, which is real and negative. So $N$ and $B$ lie on the same line through the origin.

Common length. We compute $|LC| = |L - C|$:

$L - C = (p+q)\alpha - (-r\alpha) = (p + q + r)\alpha.$

So $|LC| = (p + q + r)|\alpha| = p + q + r$.

$|MA| = |M - A| = |-(q+r) - p| = p + q + r$.

$|NB| = |N - B| = |-\alpha^2(p+r) - q\alpha^2| = |\alpha^2| \cdot |p + q + r| = p + q + r$.

Therefore all three line segments have the common length $p + q + r$. $\qquad \blacksquare$

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Question 4

Topic: 函数与图像 Functions and Graphs  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The function $f(x)$ is defined by

$f(x) = \frac{x(x - 2)(x - a)}{x^2 - 1} .$

Prove algebraically that the line $y = x + c$ intersects the curve $y = f(x)$ if $|a| \geqslant 1$, but there are values of $c$ for which there are no points of intersection if $|a| < 1$.

Find the equation of the oblique asymptote of the curve $y = f(x)$. Sketch the graph in the two cases (i) $a < -1$; and (ii) $-1 < a < -\frac{1}{2}$. (You need not calculate the turning points.)

Model Solution

Part (i): Show $y = x + c$ intersects $y = f(x)$ if $|a| \geq 1$

Setting $f(x) = x + c$:

$\frac{x(x-2)(x-a)}{x^2 - 1} = x + c.$

For $x \neq \pm 1$, cross-multiplying:

$x(x-2)(x-a) = (x + c)(x^2 - 1).$

Expand the left side: $x(x^2 - (a+2)x + 2a) = x^3 - (a+2)x^2 + 2ax$.

Expand the right side: $x^3 - x + cx^2 - c = x^3 + cx^2 - x - c$.

Setting them equal and cancelling $x^3$ gives

$-(a+2)x^2 + 2ax = cx^2 - x - c.$

Rearranging, the possible intersection abscissae are the roots of the quadratic

$(a+2+c)x^2 - (2a+1)x - c = 0. \qquad (*)$

A quadratic $Ax^2 + Bx + C = 0$ has a real root that is not $\pm 1$ if its discriminant is non-negative and at least one root is not excluded by the vertical asymptotes.

The discriminant is:

$\Delta = (2a+1)^2 + 4c(a+2+c) = (2a+1)^2 + 4c(a+2) + 4c^2.$

$= 4a^2 + 4a + 1 + 4ac + 8c + 4c^2$

$\Delta = 4c^2 + 4(a+2)c + (2a+1)^2$

Viewing as quadratic in $c$: $\Delta = 4c^2 + 4(a+2)c + (2a+1)^2$.

Discriminant of this in $c$: $16(a+2)^2 - 16(2a+1)^2 = 16[(a+2)^2 - (2a+1)^2] = 16[(a+2-2a-1)(a+2+2a+1)] = 16[(1-a)(3a+3)] = 48(1-a)(a+1)$.

For $\Delta \geq 0$ for all $c$, we need the quadratic in $c$ to be non-negative for all $c$. Since the leading coefficient is $4 > 0$, this requires the discriminant (in $c$) $\leq 0$:

$48(1-a)(a+1) \leq 0 \iff (1-a)(a+1) \leq 0 \iff a \leq -1 \text{ or } a \geq 1 \iff |a| \geq 1.$

So when $|a| \geq 1$, $\Delta \geq 0$ for all $c$, meaning ($*$) always has real roots.

We must also check that the roots are not at $x = \pm 1$ (where $f$ is undefined). If $x = 1$ is a root of ($*$):

$(a+2+c) - (2a+1) - c = 0 \implies a+2+c-2a-1-c = 0 \implies 1 - a = 0 \implies a = 1$.

When $a = 1$ and $x = 1$ is a root, the quadratic becomes $(3+c)x^2 - 3x - c = 0$ with $x = 1$ a root: $3 + c - 3 - c = 0$, confirmed. The other root is $x = -c/(3+c)$ by Vieta’s (product of roots $= -c/(3+c)$, one root is 1, so other is $-c/(3+c)$). This root equals $-1$ when $-c = -(3+c)$, i.e. $3 = 0$, impossible. So the other root is valid.

If $x = -1$ is a root of ($*$):

$(a+2+c) + (2a+1) - c = 0 \implies 3a + 3 = 0 \implies a = -1$.

When $a = -1$ and $x = -1$ is a root, the quadratic becomes $(1+c)x^2 + x - c = 0$ with $x = -1$: $1+c-1-c = 0$, confirmed. The other root is $x = c/(1+c)$ by Vieta’s. This equals $1$ when $c = 1+c$, impossible. So the other root is valid.

In both edge cases, at least one valid intersection exists. Therefore for $|a| \geq 1$ and any $c$, the line $y = x + c$ intersects $y = f(x)$. $\qquad \blacksquare$

Part (ii): No intersection for some $c$ when $|a| < 1$

When $|a| < 1$, the discriminant $\Delta$ (as a quadratic in $c$) can be negative for some $c$. Specifically, $\Delta < 0$ when $c$ lies between the two roots of $\Delta = 0$ (viewed as quadratic in $c$). For such $c$, the quadratic ($*$) has no real roots, so there is no intersection.

For example, when $a = 0$: $\Delta = 4c^2 + 8c + 1$. Setting $\Delta = 0$: $c = \frac{-8 \pm \sqrt{64 - 16}}{8} = \frac{-8 \pm 4\sqrt{3}}{8} = \frac{-2 \pm \sqrt{3}}{2}$. For $c$ in the interval $\left(\frac{-2 - \sqrt{3}}{2}, \frac{-2 + \sqrt{3}}{2}\right)$, $\Delta < 0$ and there are no points of intersection. $\qquad \blacksquare$

Part (iii): Oblique asymptote

Perform polynomial long division of $x(x-2)(x-a) = x^3 - (a+2)x^2 + 2ax$ by $x^2 - 1$:

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

Check: $x(x^2 - 1) = x^3 - x$. Subtract: $-(a+2)x^2 + (2a+1)x$. Then $-(a+2)(x^2 - 1) = -(a+2)x^2 + (a+2)$. Subtract: $(2a+1)x - (a+2) = (2a+3)x - (a+2)$.

As $x \to \pm\infty$, the remainder tends to $0$, so the oblique asymptote is:

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

Part (iv): Sketches

Both cases share: vertical asymptotes at $x = \pm 1$; oblique asymptote $y = x - (a+2)$; roots of numerator at $x = 0$, $x = 2$, $x = a$.

(i) $a < -1$:

Roots in order: $a < -1 < 0 < 1 < 2$. The numerator $N(x)=x(x-2)(x-a)$ has sign pattern:

• $x < a$: $-$
• $a < x < 0$: $+$
• $0 < x < 2$: $-$
• $x > 2$: $+$

The sign of $D(x) = x^2 - 1$:

• $|x| > 1$: $+$
• $|x| < 1$: $-$

Sign of $f = N/D$:

• $x < a$: $-$.
• $a < x < -1$: $+$. $f \to +\infty$ as $x \to -1^-$.
• $-1 < x < 0$: $-$. $f \to -\infty$ as $x \to -1^+$.
• $0 < x < 1$: $+$. $f \to +\infty$ as $x \to 1^-$.
• $1 < x < 2$: $-$. $f \to -\infty$ as $x \to 1^+$.
• $x > 2$: $+$.

The curve has three branches between the asymptotes: $(a, -1)$, $(-1, 1)$, and $(1, \infty)$. The branch in $(-1, 1)$ passes through the origin and goes from $-\infty$ to $+\infty$. The branch in $(1, 2)$ goes from $-\infty$, crosses zero at $x = 2$, and approaches the asymptote $y = x - (a+2)$ from below for large $x$.

(ii) $-1 < a < -1/2$:

Roots in order: $-1 < a < 0 < 1 < 2$. (Since $a < 0$.)

Sign of $N(x) = x(x-2)(x-a)$:

• $x < a$: $-$ (three negative factors: $x < 0$, $x - 2 < 0$, $x - a < 0$; product $= -\cdot -\cdot - = -$).
• $a < x < 0$: $+$ ($x < 0$, $x - 2 < 0$, $x - a > 0$; product $= +$).
• $0 < x < 2$: $-$ ($x > 0$, $x - 2 < 0$, $x - a > 0$; product $= -$).
• $x > 2$: $+$.

Sign of $D(x)$: same as before.

Sign of $f$:

• $x < -1$: $-/+\; = -$.
• $-1 < x < a$: $-/ -\; = +$. $f \to +\infty$ as $x \to -1^+$.
• $a < x < 0$: $+/ -\; = -$. $f$ crosses zero at $x = a$ (going from $+$ to $-$).
• $0 < x < 1$: $-/ -\; = +$. $f$ crosses zero at $x = 0$.
• $1 < x < 2$: $-/+\; = -$. $f \to -\infty$ as $x \to 1^+$.
• $x > 2$: $+/+\; = +$. $f$ crosses zero at $x = 2$.

The branch in $(-1, 1)$ now has two sign changes: it comes from $+\infty$ at $x = -1^+$, crosses zero at $x = a$ (going negative), crosses zero again at $x = 0$ (going positive), and heads to $+\infty$ as $x \to 1^-$. This creates a “dip” below the axis between $x = a$ and $x = 0$.

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Question 5

Topic: 线性代数与向量 Linear Algebra and Vectors  |  Difficulty: Hard  |  Marks: 20

Problem

5 Given two non-zero vectors $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$ we define $\Delta(\mathbf{a}, \mathbf{b})$ by $\Delta(\mathbf{a}, \mathbf{b}) = a_1b_2 - a_2b_1$.

Let $A$, $B$ and $C$ be points with position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, respectively, no two of which are parallel. Let $P$, $Q$ and $R$ be points with position vectors $\mathbf{p}$, $\mathbf{q}$ and $\mathbf{r}$, respectively, none of which are parallel.

(i) Show that there exists a $2 \times 2$ matrix $\mathbf{M}$ such that $P$ and $Q$ are the images of $A$ and $B$ under the transformation represented by $\mathbf{M}$.

(ii) Show that $\Delta(\mathbf{a}, \mathbf{b}) \mathbf{c} + \Delta(\mathbf{c}, \mathbf{a}) \mathbf{b} + \Delta(\mathbf{b}, \mathbf{c}) \mathbf{a} = \mathbf{0}$. Hence, or otherwise, prove that a necessary and sufficient condition for the points $P$, $Q$, and $R$ to be the images of points $A$, $B$ and $C$ under the transformation represented by some $2 \times 2$ matrix $\mathbf{M}$ is that

$\Delta(\mathbf{a}, \mathbf{b}) : \Delta(\mathbf{b}, \mathbf{c}) : \Delta(\mathbf{c}, \mathbf{a}) = \Delta(\mathbf{p}, \mathbf{q}) : \Delta(\mathbf{q}, \mathbf{r}) : \Delta(\mathbf{r}, \mathbf{p}) .$

Model Solution

Part (i)

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, they are linearly independent and form a basis for $\mathbb{R}^2$. The matrix $[\mathbf{a} \; \mathbf{b}]$ (with $\mathbf{a}$ and $\mathbf{b}$ as columns) is therefore invertible, with

$\det[\mathbf{a} \; \mathbf{b}] = a_1 b_2 - a_2 b_1 = \Delta(\mathbf{a}, \mathbf{b}) \neq 0.$

We seek $\mathbf{M}$ with $\mathbf{M}\mathbf{a} = \mathbf{p}$ and $\mathbf{M}\mathbf{b} = \mathbf{q}$, i.e.

$\mathbf{M}[\mathbf{a} \; \mathbf{b}] = [\mathbf{p} \; \mathbf{q}].$

Since $[\mathbf{a} \; \mathbf{b}]$ is invertible, the unique solution is

$\mathbf{M} = [\mathbf{p} \; \mathbf{q}][\mathbf{a} \; \mathbf{b}]^{-1} = \frac{1}{\Delta(\mathbf{a}, \mathbf{b})} \begin{pmatrix} p_1 & q_1 \\ p_2 & q_2 \end{pmatrix} \begin{pmatrix} b_2 & -b_1 \\ -a_2 & a_1 \end{pmatrix}.$

Hence such a matrix $\mathbf{M}$ exists. $\qquad \blacksquare$

Part (ii)

Step 1: Verify the identity.

We check the first component of $\Delta(\mathbf{a}, \mathbf{b})\mathbf{c} + \Delta(\mathbf{c}, \mathbf{a})\mathbf{b} + \Delta(\mathbf{b}, \mathbf{c})\mathbf{a}$:

$(a_1 b_2 - a_2 b_1)c_1 + (c_1 a_2 - c_2 a_1)b_1 + (b_1 c_2 - b_2 c_1)a_1$

$= a_1 b_2 c_1 - a_2 b_1 c_1 + c_1 a_2 b_1 - c_2 a_1 b_1 + b_1 c_2 a_1 - b_2 c_1 a_1 = 0.$

The second component follows identically (by the symmetry of the cyclic structure). Hence

$\Delta(\mathbf{a}, \mathbf{b})\mathbf{c} + \Delta(\mathbf{c}, \mathbf{a})\mathbf{b} + \Delta(\mathbf{b}, \mathbf{c})\mathbf{a} = \mathbf{0}. \qquad \blacksquare$

Step 2: Express $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Rearranging the identity:

$\Delta(\mathbf{a}, \mathbf{b})\mathbf{c} = -\Delta(\mathbf{c}, \mathbf{a})\mathbf{b} - \Delta(\mathbf{b}, \mathbf{c})\mathbf{a} = \Delta(\mathbf{a}, \mathbf{c})\mathbf{b} + \Delta(\mathbf{c}, \mathbf{b})\mathbf{a}.$

Since $\Delta(\mathbf{a}, \mathbf{b}) \neq 0$:

$\mathbf{c} = \frac{\Delta(\mathbf{c}, \mathbf{b})}{\Delta(\mathbf{a}, \mathbf{b})}\mathbf{a} + \frac{\Delta(\mathbf{a}, \mathbf{c})}{\Delta(\mathbf{a}, \mathbf{b})}\mathbf{b}. \qquad (\star)$

Step 3: Necessity ($\Longrightarrow$).

Suppose $\mathbf{M}$ exists with $\mathbf{M}\mathbf{a} = \mathbf{p}$, $\mathbf{M}\mathbf{b} = \mathbf{q}$, $\mathbf{M}\mathbf{c} = \mathbf{r}$. For any vectors $\mathbf{u}, \mathbf{v}$:

$\Delta(\mathbf{M}\mathbf{u}, \mathbf{M}\mathbf{v}) = (\mathbf{M}\mathbf{u})_1(\mathbf{M}\mathbf{v})_2 - (\mathbf{M}\mathbf{u})_2(\mathbf{M}\mathbf{v})_1 = \det(\mathbf{M})(u_1 v_2 - u_2 v_1) = \det(\mathbf{M})\Delta(\mathbf{u}, \mathbf{v}).$

Applying this:

$\Delta(\mathbf{p}, \mathbf{q}) = \det(\mathbf{M})\Delta(\mathbf{a}, \mathbf{b}), \quad \Delta(\mathbf{q}, \mathbf{r}) = \det(\mathbf{M})\Delta(\mathbf{b}, \mathbf{c}), \quad \Delta(\mathbf{r}, \mathbf{p}) = \det(\mathbf{M})\Delta(\mathbf{c}, \mathbf{a}).$

Dividing through by $\det(\mathbf{M})$ (which is nonzero since $\mathbf{a}, \mathbf{b}$ are independent and $\mathbf{p}, \mathbf{q}$ are independent):

$\Delta(\mathbf{a}, \mathbf{b}) : \Delta(\mathbf{b}, \mathbf{c}) : \Delta(\mathbf{c}, \mathbf{a}) = \Delta(\mathbf{p}, \mathbf{q}) : \Delta(\mathbf{q}, \mathbf{r}) : \Delta(\mathbf{r}, \mathbf{p}).$

Step 4: Sufficiency ($\Longleftarrow$).

Suppose the ratio condition holds. By Part (i), there exists $\mathbf{M}$ with $\mathbf{M}\mathbf{a} = \mathbf{p}$ and $\mathbf{M}\mathbf{b} = \mathbf{q}$. We must show $\mathbf{M}\mathbf{c} = \mathbf{r}$.

The ratio condition means there exists $k \neq 0$ such that

$\Delta(\mathbf{p}, \mathbf{q}) = k\Delta(\mathbf{a}, \mathbf{b}), \quad \Delta(\mathbf{q}, \mathbf{r}) = k\Delta(\mathbf{b}, \mathbf{c}), \quad \Delta(\mathbf{r}, \mathbf{p}) = k\Delta(\mathbf{c}, \mathbf{a}).$

From $(\star)$ and linearity of $\mathbf{M}$:

$\Delta(\mathbf{a}, \mathbf{b}) \cdot \mathbf{M}\mathbf{c} = \Delta(\mathbf{c}, \mathbf{b})\mathbf{p} + \Delta(\mathbf{a}, \mathbf{c})\mathbf{q}. \qquad (\star\star)$

Now apply the cyclic identity to $\mathbf{p}, \mathbf{q}, \mathbf{r}$:

$\Delta(\mathbf{p}, \mathbf{q})\mathbf{r} + \Delta(\mathbf{r}, \mathbf{p})\mathbf{q} + \Delta(\mathbf{q}, \mathbf{r})\mathbf{p} = \mathbf{0}.$

Substituting the ratio conditions:

$k\Delta(\mathbf{a}, \mathbf{b})\mathbf{r} + k\Delta(\mathbf{c}, \mathbf{a})\mathbf{q} + k\Delta(\mathbf{b}, \mathbf{c})\mathbf{p} = \mathbf{0}.$

Dividing by $k$:

$\Delta(\mathbf{a}, \mathbf{b})\mathbf{r} = -\Delta(\mathbf{b}, \mathbf{c})\mathbf{p} - \Delta(\mathbf{c}, \mathbf{a})\mathbf{q} = \Delta(\mathbf{c}, \mathbf{b})\mathbf{p} + \Delta(\mathbf{a}, \mathbf{c})\mathbf{q}. \qquad (\star\star\star)$

Comparing $(\star\star)$ and $(\star\star\star)$, we see $\Delta(\mathbf{a}, \mathbf{b}) \cdot \mathbf{M}\mathbf{c} = \Delta(\mathbf{a}, \mathbf{b})\mathbf{r}$, and since $\Delta(\mathbf{a}, \mathbf{b}) \neq 0$:

$\mathbf{M}\mathbf{c} = \mathbf{r}.$

Therefore $\mathbf{M}$ maps $A \mapsto P$, $B \mapsto Q$, $C \mapsto R$, and the ratio condition is both necessary and sufficient. $\qquad \blacksquare$

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Question 6

Topic: 代数 Algebra  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Given that

$x^4 + px^2 + qx + r = (x^2 - ax + b)(x^2 + ax + c),$

express $p$, $q$ and $r$ in terms of $a$, $b$ and $c$.

Show also that $a^2$ is a root of the cubic equation

$u^3 + 2pu^2 + (p^2 - 4r)u - q^2 = 0.$

Explain why this equation always has a non-negative root, and verify that $u = 9$ is a root in the case $p = -1, q = -6, r = 15$.

Hence, or otherwise, express

$y^4 - 8y^3 + 23y^2 - 34y + 39$

as a product of two quadratic factors.

Model Solution

Part (i): Express $p$, $q$, $r$ in terms of $a$, $b$, $c$

Expand $(x^2 - ax + b)(x^2 + ax + c)$:

$= x^4 + ax^3 + cx^2 - ax^3 - a^2x^2 - acx + bx^2 + abx + bc$

$= x^4 + (b + c - a^2)x^2 + a(b - c)x + bc.$

Matching coefficients with $x^4 + px^2 + qx + r$:

$p = b + c - a^2, \qquad q = a(b - c), \qquad r = bc. \qquad \text{(i)}$

Part (ii): Show $a^2$ is a root of $u^3 + 2pu^2 + (p^2 - 4r)u - q^2 = 0$

From (i): $b + c = p + a^2$ and $b - c = q/a$ (for $a \neq 0$).

Therefore $b = \frac{p + a^2 + q/a}{2}$ and $c = \frac{p + a^2 - q/a}{2}$.

From $r = bc$:

$r = \frac{(p + a^2)^2 - (q/a)^2}{4} = \frac{(p + a^2)^2 - q^2/a^2}{4}.$

So $4r = (p + a^2)^2 - q^2/a^2$, giving:

$(p + a^2)^2 - 4r = \frac{q^2}{a^2}.$

Multiply both sides by $a^2$:

$a^2(p + a^2)^2 - 4ra^2 = q^2.$

Expand: $a^2(p^2 + 2pa^2 + a^4) - 4ra^2 = q^2$, so:

$a^6 + 2pa^4 + p^2 a^2 - 4ra^2 - q^2 = 0.$

Setting $u = a^2$:

$u^3 + 2pu^2 + (p^2 - 4r)u - q^2 = 0. \qquad \blacksquare$

Part (iii): Non-negative root

The product of the roots of $u^3 + 2pu^2 + (p^2 - 4r)u - q^2 = 0$ equals $q^2$ (by Vieta’s, the product is $-(-q^2)/1 = q^2 \geq 0$).

If all three roots were negative, their product would be negative, contradicting $q^2 \geq 0$. Therefore at least one root is non-negative. (When $q = 0$, $u = 0$ is a root.) $\qquad \blacksquare$

Part (iv): Verify $u = 9$ for $p = -1$, $q = -6$, $r = 15$

$9^3 + 2(-1)(81) + (1 - 60)(9) - 36 = 729 - 162 - 531 - 36 = 0.$

Part (v): Factor $y^4 - 8y^3 + 23y^2 - 34y + 39$

First, eliminate the cubic term with $y = x + 2$ (since $-8/4 = -2$):

$(x+2)^4 - 8(x+2)^3 + 23(x+2)^2 - 34(x+2) + 39.$

Expanding: $(x+2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16$; $-8(x+2)^3 = -8x^3 - 48x^2 - 96x - 64$; $23(x+2)^2 = 23x^2 + 92x + 92$; $-34(x+2) = -34x - 68$; $+39$.

Collecting:

• $x^4$: coefficient $1$.
• $x^3$: $8 - 8 = 0$.
• $x^2$: $24 - 48 + 23 = -1$.
• $x$: $32 - 96 + 92 - 34 = -6$.
• constant: $16 - 64 + 92 - 68 + 39 = 15$.

So the depressed quartic is $x^4 - x^2 - 6x + 15$, with $p = -1$, $q = -6$, $r = 15$.

The resolvent cubic is $u^3 - 2u^2 - 59u - 36 = 0$. From Part (iv), $u = 9$ is a root.

So $a^2 = 9$, giving $a = 3$. Then:

$b + c = p + a^2 = -1 + 9 = 8, \qquad c - b = q/a = -6/3 = -2.$

Together with $bc=15$, these equations give $b=3$ and $c=5$.

So the factorization is:

$(x^2 - 3x + 3)(x^2 + 3x + 5).$

Indeed, $(x^2 - 3x + 3)(x^2 + 3x + 5) = x^4 - x^2 - 6x + 15$.

Substituting back $x = y - 2$:

$(y - 2)^2 - 3(y - 2) + 3 = y^2 - 4y + 4 - 3y + 6 + 3 = y^2 - 7y + 13.$

$(y - 2)^2 + 3(y - 2) + 5 = y^2 - 4y + 4 + 3y - 6 + 5 = y^2 - y + 3.$

Therefore:

$y^4 - 8y^3 + 23y^2 - 34y + 39 = (y^2 - 7y + 13)(y^2 - y + 3). \qquad \blacksquare$

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Question 7

Topic: 级数与不等式 Sequences, Series and Inequalities  |  Difficulty: Challenging  |  Marks: 20

Problem

7 Given that $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{r!} + \cdots ,$ use the binomial theorem to show that $\left( 1 + \frac{1}{n} \right)^n < e$ for any positive integer $n$.

The product $P(n)$ is defined, for any positive integer $n$, by $P(n) = \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{9}{8} \cdot \ldots \cdot \frac{2^n + 1}{2^n} .$ Use the arithmetic-geometric mean inequality, $\frac{a_1 + a_2 + \cdots + a_n}{n} \geqslant (a_1 \cdot a_2 \cdot \ldots \cdot a_n)^{\frac{1}{n}} ,$ to show that $P(n) < e$ for all $n$.

Explain briefly why $P(n)$ tends to a limit as $n \to \infty$. Show that this limit, $L$, satisfies $2 < L \leqslant e$.

Model Solution

Part 1: Show that $\left(1 + \frac{1}{n}\right)^n < e$

By the binomial theorem,

$\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k}.$

For $k \geqslant 1$, the $k$-th term is

$\binom{n}{k} \frac{1}{n^k} = \frac{n!}{k!(n-k)!} \cdot \frac{1}{n^k} = \frac{1}{k!} \cdot \frac{n(n-1)(n-2) \cdots (n-k+1)}{n^k}.$

The fraction on the right is a product of $k$ factors:

$\frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{n-k+1}{n} = 1 \cdot \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k-1}{n}\right).$

For $k \geqslant 2$, at least one factor satisfies $1 - \frac{j}{n} < 1$ (with $j \geqslant 1$), and for $k = 1$ the product is just $1$. Thus for each $k = 1, 2, \ldots, n$:

$\binom{n}{k} \frac{1}{n^k} \leqslant \frac{1}{k!},$

with strict inequality for $k \geqslant 2$. Therefore

$\left(1 + \frac{1}{n}\right)^n = 1 + \sum_{k=1}^{n} \binom{n}{k} \frac{1}{n^k} < 1 + \sum_{k=1}^{n} \frac{1}{k!} < 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots = e.$

Part 2: Show that $P(n) < e$ using AM-GM

Write $P(n) = \prod_{k=1}^{n} a_k$ where $a_k = \frac{2^k + 1}{2^k} = 1 + \frac{1}{2^k}$.

Since not all $a_k$ are equal (for instance $a_1 = \frac{3}{2} \neq \frac{5}{4} = a_2$), the AM-GM inequality is strict:

$P(n)^{1/n} = \left(\prod_{k=1}^{n} a_k\right)^{1/n} < \frac{1}{n} \sum_{k=1}^{n} a_k = \frac{1}{n} \sum_{k=1}^{n} \left(1 + \frac{1}{2^k}\right) = 1 + \frac{1}{n} \sum_{k=1}^{n} \frac{1}{2^k}.$

The geometric series gives

$\sum_{k=1}^{n} \frac{1}{2^k} = \frac{1}{2} \cdot \frac{1 - (1/2)^n}{1 - 1/2} = 1 - \frac{1}{2^n}.$

Hence

$P(n)^{1/n} < 1 + \frac{1 - 2^{-n}}{n}.$

Raising both sides to the power $n$ (valid since both sides are positive):

$P(n) < \left(1 + \frac{1 - 2^{-n}}{n}\right)^n.$

Since $1 - 2^{-n} < 1$, we have $\frac{1 - 2^{-n}}{n} < \frac{1}{n}$, so

$\left(1 + \frac{1 - 2^{-n}}{n}\right)^n < \left(1 + \frac{1}{n}\right)^n < e,$

where the last step uses Part 1. Therefore $P(n) < e$ for all $n$.

Part 3: Convergence and bounds for $L$

$P(n)$ is strictly increasing because each factor $a_k = 1 + \frac{1}{2^k} > 1$, so $P(n+1) = P(n) \cdot a_{n+1} > P(n)$.

$P(n)$ is bounded above by $e$ (Part 2). A bounded monotone increasing sequence converges, so $P(n) \to L$ for some limit $L$.

Since $P(n)$ is increasing and $P(2) = \frac{3}{2} \cdot \frac{5}{4} = \frac{15}{8} > 2$, we have $P(n) > 2$ for all $n \geqslant 2$, giving $L \geqslant 2$. In fact $L > 2$: if $L = 2$ then $P(n) \leqslant 2$ for all $n$, but $P(2) = \frac{15}{8} > 2$, a contradiction.

Since $P(n) < e$ for all $n$, we have $L \leqslant e$.

Therefore $2 < L \leqslant e$.

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Question 8

Topic: 递推关系与数学归纳法 Sequences and Mathematical Induction  |  Difficulty: Challenging  |  Marks: 20

Problem

8 The sequence $a_n$ is defined by $a_0 = 1$, $a_1 = 1$, and $a_n = \frac{1 + a_{n-1}^2}{a_{n-2}} \quad (n \geqslant 2).$ Prove by induction that $a_n = 3a_{n-1} - a_{n-2} \quad (n \geqslant 2).$ Hence show that $a_n = \frac{\alpha^{2n-1} + \alpha^{-(2n-1)}}{\sqrt{5}} \quad (n \geqslant 1),$ where $\alpha = \frac{1 + \sqrt{5}}{2}$.

Model Solution

Part 1: Prove by induction that $a_n = 3a_{n-1} - a_{n-2}$ for $n \geqslant 2$

We first establish a helper identity.

Lemma. $a_n a_{n-2} - a_{n-1}^2 = 1$ for all $n \geqslant 2$.

Proof. From the definition, $a_n a_{n-2} = 1 + a_{n-1}^2$, so $a_n a_{n-2} - a_{n-1}^2 = 1$. $\square$

We also record the key algebraic identity we need:

$a_n^2 - 3a_n a_{n-1} + a_{n-1}^2 + 1 = 0 \quad \text{for all } n \geqslant 1. \qquad (\dagger)$

We prove $(\dagger)$ by induction.

Base case ($n = 1$). $a_1^2 - 3a_1 a_0 + a_0^2 + 1 = 1 - 3 + 1 + 1 = 0$.

Inductive step. Suppose $(\dagger)$ holds for some $n \geqslant 1$. From the lemma, $a_{n+1} = \frac{1 + a_n^2}{a_{n-1}}$, so

$a_{n+1}^2 - 3a_{n+1}a_n + a_n^2 + 1 = \frac{(1+a_n^2)^2}{a_{n-1}^2} - \frac{3a_n(1+a_n^2)}{a_{n-1}} + a_n^2 + 1.$

Put everything over $a_{n-1}^2$:

$= \frac{(1+a_n^2)^2 - 3a_n a_{n-1}(1+a_n^2) + (a_n^2+1)a_{n-1}^2}{a_{n-1}^2}.$

Factor out $(1 + a_n^2)$ from the numerator:

$= \frac{(1+a_n^2)\bigl[(1+a_n^2) - 3a_n a_{n-1} + a_{n-1}^2\bigr]}{a_{n-1}^2}.$

By the inductive hypothesis, $(1+a_n^2) - 3a_n a_{n-1} + a_{n-1}^2 = a_n^2 - 3a_n a_{n-1} + a_{n-1}^2 + 1 = 0$. Hence $(\dagger)$ holds for $n+1$.

By induction, $(\dagger)$ holds for all $n \geqslant 1$. $\square$

Now we prove the main claim $a_n = 3a_{n-1} - a_{n-2}$ by induction.

Base cases. Direct computation from the definition $a_n = \frac{1 + a_{n-1}^2}{a_{n-2}}$:

$a_2 = \frac{1 + 1}{1} = 2, \quad a_3 = \frac{1 + 4}{1} = 5, \quad a_4 = \frac{1 + 25}{2} = 13.$

These also satisfy $a_2 = 3(1) - 1 = 2$ and $a_3 = 3(2) - 1 = 5$.

Inductive step. Suppose $a_k = 3a_{k-1} - a_{k-2}$ for all $2 \leqslant k \leqslant n$. We show $a_{n+1} = 3a_n - a_{n-1}$.

From the definition, $a_{n+1} = \frac{1 + a_n^2}{a_{n-1}}$. By $(\dagger)$:

$1 + a_n^2 = 3a_n a_{n-1} - a_{n-1}^2 = a_{n-1}(3a_n - a_{n-1}).$

Therefore

$a_{n+1} = \frac{a_{n-1}(3a_n - a_{n-1})}{a_{n-1}} = 3a_n - a_{n-1}.$

By induction, $a_n = 3a_{n-1} - a_{n-2}$ for all $n \geqslant 2$. $\qquad \blacksquare$

Part 2: Show that $a_n = \frac{\alpha^{2n-1} + \alpha^{-(2n-1)}}{\sqrt{5}}$ for $n \geqslant 1$

The recurrence $a_n = 3a_{n-1} - a_{n-2}$ has characteristic equation $t^2 - 3t + 1 = 0$, with roots

$t = \frac{3 \pm \sqrt{5}}{2}.$

Since $\alpha = \frac{1+\sqrt{5}}{2}$ satisfies $\alpha^2 = \alpha + 1 = \frac{3+\sqrt{5}}{2}$ and $\alpha^{-2} = 2 - \alpha = \frac{3-\sqrt{5}}{2}$, the two roots are $\alpha^2$ and $\alpha^{-2}$.

The general solution is

$a_n = A\alpha^{2n} + B\alpha^{-2n}.$

Using the initial conditions $a_0 = 1$ and $a_1 = 1$:

$A + B = 1, \qquad A\alpha^2 + B\alpha^{-2} = 1.$

From the first equation, $B = 1 - A$. Substituting into the second:

$A\alpha^2 + (1-A)\alpha^{-2} = 1$ $A(\alpha^2 - \alpha^{-2}) = 1 - \alpha^{-2}.$

Now $\alpha^2 - \alpha^{-2} = \frac{3+\sqrt{5}}{2} - \frac{3-\sqrt{5}}{2} = \sqrt{5}$ and $1 - \alpha^{-2} = 1 - \frac{3-\sqrt{5}}{2} = \frac{\sqrt{5}-1}{2}$, so

$A = \frac{\sqrt{5}-1}{2\sqrt{5}}.$

Since $\alpha = \frac{1+\sqrt{5}}{2}$, we have $\alpha - 1 = \frac{\sqrt{5}-1}{2} = \alpha^{-1}$, giving $A = \frac{\alpha^{-1}}{\sqrt{5}}$.

Then $B = 1 - \frac{\alpha^{-1}}{\sqrt{5}} = \frac{\sqrt{5} - \alpha^{-1}}{\sqrt{5}}$. Since $\sqrt{5} = \alpha + \alpha^{-1}$ (as $\alpha + \alpha^{-1} = \frac{1+\sqrt{5}}{2} + \frac{\sqrt{5}-1}{2} = \sqrt{5}$), we get $B = \frac{\alpha}{\sqrt{5}}$.

Therefore

$a_n = \frac{\alpha^{-1} \cdot \alpha^{2n} + \alpha \cdot \alpha^{-2n}}{\sqrt{5}} = \frac{\alpha^{2n-1} + \alpha^{-(2n-1)}}{\sqrt{5}} \quad (n \geqslant 1). \qquad \blacksquare$