STEP3 2021 -- Pure Mathematics

STEP3 2021 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2021  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	纯数	Challenging	参数方程求导,三角恒等式化简,点到直线距离公式,包络线求法
2	纯数	Hard	矩阵零空间与行列式,对称多项式恒等式,不等式证明,AM-GM不等式
3	纯数	Challenging	递推关系建立,积分不等式证明,严格不等号处理
4	纯数	Challenging	向量投影公式,点积性质,角平分线条件,充要条件证明
5	纯数	Challenging	极坐标方程联立,判别式求相切条件,曲线作图技巧,对称性分析
6	纯数	Challenging	反三角函数恒等式,双曲函数求导,分段函数作图,间断点与渐近线分析
7	纯数	Hard	复数化简,纯虚数判别,垂心性质,中点分析,平移变换
8	纯数	Hard	数学归纳法,不等式放缩,发散判别,递推序列构造

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Question 1

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

1 (i) A curve has parametric equations $x = -4 \cos^3 t, \quad y = 12 \sin t - 4 \sin^3 t.$ Find the equation of the normal to this curve at the point $(-4 \cos^3 \phi, \ 12 \sin \phi - 4 \sin^3 \phi),$ where $0 < \phi < \frac{1}{2}\pi$. Verify that this normal is a tangent to the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4$ at the point $(8 \cos^3 \phi, \ 8 \sin^3 \phi)$.

(ii) A curve has parametric equations $x = \cos t + t \sin t, \quad y = \sin t - t \cos t.$ Find the equation of the normal to this curve at the point $(\cos \phi + \phi \sin \phi, \ \sin \phi - \phi \cos \phi),$ where $0 < \phi < \frac{1}{2}\pi$. Determine the perpendicular distance from the origin to this normal, and hence find the equation of a curve, independent of $\phi$, to which this normal is a tangent.

Hint

(i) $x = -4 \cos^3 t$ so $\frac{dx}{dt} = 12 \cos^2 t \sin t$ M1

$y = 12 \sin t - 4 \sin^3 t$ so $\frac{dy}{dt} = 12 \cos t - 12 \sin^2 t \cos t = 12 \cos t (1 - \sin^2 t) = 12 \cos^3 t$ M1

So $\frac{dy}{dx} = \frac{12 \cos^3 t}{12 \cos^2 t \sin t} = \cot t$ A1

Thus the equation of the normal at $(-4 \cos^3 \varphi, 12 \sin \varphi - 4 \sin^3 \varphi)$ is

$y - (12 \sin \varphi - 4 \sin^3 \varphi) = -\frac{1}{\cot \varphi} (x - -4 \cos^3 \varphi)$

M1 A1ft

This simplifies to $x \sin \varphi + y \cos \varphi = 12 \sin \varphi \cos \varphi - 4 \sin^3 \varphi \cos \varphi - 4 \sin \varphi \cos^3 \varphi$

That is $x \sin \varphi + y \cos \varphi = 8 \sin \varphi \cos \varphi$ A1 (6)

Alternative simplification $x \tan \varphi + y = 8 \sin \varphi$

For $x = 8 \cos^3 t, \frac{dx}{dt} = -24 \cos^2 t \sin t$ and for $y = 8 \sin^3 t, \frac{dy}{dt} = 24 \sin^2 t \cos t$

So $\frac{dy}{dx} = \frac{24 \sin^2 t \cos t}{-24 \cos^2 t \sin t} = -\tan t$ M1 A1ft

Thus the equation of the tangent to $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4$ at $(8 \cos^3 \varphi, 8 \sin^3 \varphi)$ is

$y - 8 \sin^3 \varphi = -\tan \varphi (x - 8 \cos^3 \varphi)$

M1

This simplifies to

$x \sin \varphi + y \cos \varphi = 8 \sin^3 \varphi \cos \varphi + 8 \sin \varphi \cos^3 \varphi = 8 \sin \varphi \cos \varphi (\sin^2 \varphi + \cos^2 \varphi)$

That is $x \sin \varphi + y \cos \varphi = 8 \sin \varphi \cos \varphi$ as required. A1 (4)

Alternative 1

the normal is a tangent to the second curve if it has the same gradient and the point $(8 \cos^3 \varphi, 8 \sin^3 \varphi)$ lies on the normal. M1

Gradient working as before M1A1ft

Substitution $x \sin \varphi + y \cos \varphi = 8 \sin \varphi \cos^3 \varphi + 8 \sin^3 \varphi \cos \varphi = 8 \sin \varphi \cos \varphi (\sin^2 \varphi + \cos^2 \varphi) = 8 \sin \varphi \cos \varphi$ as required or $x \tan \varphi + y = 8 \sin \varphi \cos \varphi (\sin^2 \varphi + \cos^2 \varphi)$ A1

Alternative 2

$\frac{2}{3} x^{-\frac{1}{3}} + \frac{2}{3} y^{-\frac{1}{3}} \frac{dy}{dx} = 0$

M1

(ii) $x = \cos t + t \sin t$ so $\frac{dx}{dt} = -\sin t + t \cos t + \sin t = t \cos t$

$y = \sin t - t \cos t$ so $\frac{dy}{dt} = \cos t - \cos t + t \sin t = t \sin t$ M1

So $\frac{dy}{dx} = \tan t$ A1

Thus the equation of the normal at $(\cos \varphi + \varphi \sin \varphi, \sin \varphi - \varphi \cos \varphi)$ is

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

M1 A1ft

This simplifies to $x \cos \varphi + y \sin \varphi = 1$ A1 (5)

Alternatives which can be followed through to perpendicular distance step, or alternative method # are

$x + y \tan \varphi = \sec \varphi$ and $x \cot \varphi + y = \csc \varphi$

The distance of $(0,0)$ from $x \cos \varphi + y \sin \varphi = 1$ is $\left| \frac{-1}{\sqrt{\cos^2 \varphi + \sin^2 \varphi}} \right| = 1$

M1 A1ft A1

Alternatively, the perpendicular to $x \cos \varphi + y \sin \varphi = 1$ through $(0,0)$ is

$y \cos \varphi - x \sin \varphi = 0$, and these two lines meet at $(\cos \varphi, \sin \varphi)$

M1 A1ft

which is a distance $\sqrt{\cos^2 \varphi + \sin^2 \varphi} = 1$ from $(0,0)$. A1

So the curve to which this normal is a tangent is a circle centre $(0,0)$, radius 1 which is thus $x^2 + y^2 = 1$ M1 A1 (5)

Model Solution

Part (i)

We differentiate the parametric equations with respect to $t$:

$\frac{dx}{dt} = -4 \cdot 3\cos^2 t \cdot (-\sin t) = 12\cos^2 t \sin t$

$\frac{dy}{dt} = 12\cos t - 4 \cdot 3\sin^2 t \cos t = 12\cos t(1 - \sin^2 t) = 12\cos^3 t$

So the gradient of the curve is

$\frac{dy}{dx} = \frac{12\cos^3 t}{12\cos^2 t \sin t} = \cot t$

At $t = \phi$, the gradient of the normal is $-\frac{1}{\cot\phi} = -\tan\phi$. The normal at $(-4\cos^3\phi,\, 12\sin\phi - 4\sin^3\phi)$ is

$y - (12\sin\phi - 4\sin^3\phi) = -\tan\phi\,(x + 4\cos^3\phi)$

Multiplying through by $\cos\phi$:

$x\sin\phi + y\cos\phi = -4\cos^3\phi\sin\phi + (12\sin\phi - 4\sin^3\phi)\cos\phi$

$= -4\cos^3\phi\sin\phi + 12\sin\phi\cos\phi - 4\sin^3\phi\cos\phi$

$= 12\sin\phi\cos\phi - 4\sin\phi\cos\phi(\cos^2\phi + \sin^2\phi)$

$= 8\sin\phi\cos\phi$

So the normal has equation

$x\sin\phi + y\cos\phi = 8\sin\phi\cos\phi \qquad \text{($\star$)}$

To verify that ($\star$) is a tangent to $x^{2/3} + y^{2/3} = 4$ at $(8\cos^3\phi,\, 8\sin^3\phi)$, we check two things.

First, the point lies on the curve:

$(8\cos^3\phi)^{2/3} + (8\sin^3\phi)^{2/3} = 4\cos^2\phi + 4\sin^2\phi = 4 \checkmark$

Second, differentiating $x^{2/3} + y^{2/3} = 4$ implicitly:

$\tfrac{2}{3}x^{-1/3} + \tfrac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}$

At $(8\cos^3\phi,\, 8\sin^3\phi)$:

$\frac{dy}{dx} = -\frac{2\sin\phi}{2\cos\phi} = -\tan\phi$

This matches the gradient of ($\star$), confirming that ($\star$) is indeed a tangent to the astroid at $(8\cos^3\phi,\, 8\sin^3\phi)$.

Part (ii)

Differentiating:

$\frac{dx}{dt} = -\sin t + \sin t + t\cos t = t\cos t$

$\frac{dy}{dt} = \cos t - \cos t + t\sin t = t\sin t$

So $\frac{dy}{dx} = \tan t$. At $t = \phi$, the normal has gradient $-\cot\phi$. Its equation at $(\cos\phi + \phi\sin\phi,\, \sin\phi - \phi\cos\phi)$ is

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

Multiply through by $\sin\phi$:

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

$= \cos^2\phi + \phi\sin\phi\cos\phi + \sin^2\phi - \phi\sin\phi\cos\phi = 1$

The normal has equation

$x\cos\phi + y\sin\phi = 1$

This is already in the form $x\cos\phi + y\sin\phi = 1$, so the perpendicular distance from the origin is

$\frac{|0 \cdot \cos\phi + 0 \cdot \sin\phi - 1|}{\sqrt{\cos^2\phi + \sin^2\phi}} = 1$

This distance is $1$ regardless of $\phi$. The normal is therefore always tangent to the circle of radius $1$ centred at the origin. The envelope curve is

$x^2 + y^2 = 1$

Examiner Notes

最受欢迎的题目（93%考生作答），平均分约15/20。考官报告指出：(i) 多数考生能正确求出法线方程，但不少人在法向梯度上忘记负号，只取倒数；隐函数求导和参数求导均可，参数求导更简洁。(ii) 约3/4考生先求法线与过原点垂线的交点再算距离，但直接用点到直线距离公式更简单；常见错误包括忽略绝对值符号、计算到曲线上点的距离而非到法线的距离；最后一问求包络线方程被不少考生忽略。

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Question 2

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

2 (i) Let $x = \frac{a}{b - c} , \quad y = \frac{b}{c - a} \text{ and } z = \frac{c}{a - b} ,$ where $a, b$ and $c$ are distinct real numbers.

Show that $\begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and use this result to deduce that $yz + zx + xy = -1$.

Hence show that $\frac{a^2}{(b - c)^2} + \frac{b^2}{(c - a)^2} + \frac{c^2}{(a - b)^2} \ge 2 .$

(ii) Let $x = \frac{2a}{b + c} , \quad y = \frac{2b}{c + a} \text{ and } z = \frac{2c}{a + b} ,$ where $a, b$ and $c$ are positive real numbers.

Using a suitable matrix, show that $xyz + yz + zx + xy = 4$.

Hence show that $(2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b + c)(c + a)(a + b) .$

Show further that $(2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b + c)(c + a)(a + b) .$

Hint

(i) $\begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} a - (b - c)x \\ b - (c - a)y \\ c - (a - b)z \end{pmatrix} = \begin{pmatrix} a - a \\ b - b \\ c - c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ as required. M1 A1*

As $a$, $b$ and $c$ are distinct, they cannot all be zero. If $M^{-1}$ exists $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = M^{-1} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ which is a contradiction.

So, $M^{-1}$ does not exist and thus $det \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} = 0$ , M1

i.e. $1 - xyz + xyz + yz + zx + xy = 0$ , (Sarus)

or $1(1 + yz) - -x(y - yz) + x(yz + z) = 0$ (by co-factors) M1

which simplifies to

$yz + zx + xy = -1$ A1 * (5)

$(x + y + z)^2 \geq 0$

So $x^2 + y^2 + z^2 + 2yz + 2zx + 2xy \geq 0$ M1

and so $x^2 + y^2 + z^2 \geq 2$ *A1 (2)**

(ii) $\begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 2a - (b + c)x \\ 2b - (c + a)y \\ 2c - (a + b)z \end{pmatrix} = \begin{pmatrix} 2a - 2a \\ 2b - 2b \\ 2c - 2c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

B1 M1 A1

As $a$, $b$ and $c$ are positive, they cannot all be zero. Thus as $\begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ ,

as in part (i), $det \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} = 0$ ,

i.e. $8 - xyz - xyz - 2yz - 2zx - 2xy = 0$ , that is M1 A1

$xyz + yz + zx + xy = 4$ *A1 (6)**

$(x + 1)(y + 1)(z + 1) = xyz + zx + xy + x + y + z + 1 = 4 + x + y + z + 1 > 5$

M1 A1

because as $a$, $b$, and $c$ are all positive, so are $x$, $y$ and $z$. E1

Thus $\left( \frac{2a}{b+c} + 1 \right) \left( \frac{2b}{c+a} + 1 \right) \left( \frac{2c}{a+b} + 1 \right) > 5$

Multiplying by $(b + c)(c + a)(a + b)$ , all three factors of which are positive, gives

$(2a + b + c)(a + 2b + c)(a + b + c) > 5(b + c)(c + a)(a + b)$ as required. *A1 (4)**

$x = \frac{2a}{b+c} > \frac{2a}{a+b+c}$ as a, b, and c are positive, and similarly both, $y > \frac{2b}{a+b+c}$ and $z > \frac{2c}{a+b+c}$

M1

Thus $4 + x + y + z + 1 > 4 + \frac{2a}{a+b+c} + \frac{2b}{a+b+c} + \frac{2c}{a+b+c} + 1 = 4 + \frac{2(a+b+c)}{a+b+c} + 1 = 7$

dM1

and thus following the argument used to obtain the previous result $(2a + b + c)(a + 2b + c)(a + b + c) > 7(b + c)(c + a)(a + b)$ as required.

*A1 (3)**

Model Solution

Part (i)

Step 1: Verify the matrix equation.

Compute the matrix-vector product:

$\begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} a - bx + cx \\ by + b - cy \\ -az + bz + c \end{pmatrix}$

Substituting $x = \frac{a}{b-c}$:

$a - bx + cx = a - \frac{ab}{b-c} + \frac{ac}{b-c} = a - \frac{a(b-c)}{b-c} = a - a = 0$

Substituting $y = \frac{b}{c-a}$:

$by + b - cy = \frac{b^2}{c-a} + b - \frac{bc}{c-a} = \frac{b^2 - bc}{c-a} + b = \frac{b(b-c)}{c-a} + b = \frac{b(b-c) + b(c-a)}{c-a} = \frac{b^2 - bc + bc - ab}{c-a} = \frac{b(b-a)}{c-a}$

Hmm, let me redo this more carefully.

$by + b - cy = b \cdot \frac{b}{c-a} + b - c \cdot \frac{b}{c-a} = \frac{b^2}{c-a} + b - \frac{bc}{c-a} = \frac{b^2 - bc}{c-a} + b = \frac{b(b - c)}{c-a} + b = \frac{b(b-c) + b(c-a)}{c-a} = \frac{b(b - c + c - a)}{c-a} = \frac{b(b-a)}{c-a}$

Wait, this doesn’t simplify to zero. Let me reconsider the matrix product. The second row is $(y, 1, -y)$, so:

$y \cdot a + 1 \cdot b + (-y) \cdot c = ay + b - cy = \frac{ab}{c-a} + b - \frac{bc}{c-a} = b + \frac{ab - bc}{c-a} = b + \frac{b(a-c)}{c-a} = b - b = 0$

Similarly, the third row $(-z, z, 1)$:

$-za + zb + c = z(b - a) + c = \frac{c(b-a)}{a-b} + c = -c + c = 0$

So the matrix-vector product is indeed the zero vector.

Step 2: Deduce $yz + zx + xy = -1$.

Since the matrix times the vector $(a, b, c)^T$ gives the zero vector, and $(a, b, c) \neq (0, 0, 0)$ (because $a, b, c$ are distinct), the matrix must be singular. (If the matrix were invertible, the only solution would be the zero vector, contradicting the fact that $a, b, c$ are distinct real numbers.) Therefore its determinant is zero:

$\det \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} = 0$

Expanding along the first row:

$1 \cdot \det \begin{pmatrix} 1 & -y \\ z & 1 \end{pmatrix} - (-x) \cdot \det \begin{pmatrix} y & -y \\ -z & 1 \end{pmatrix} + x \cdot \det \begin{pmatrix} y & 1 \\ -z & z \end{pmatrix}$

$= 1 \cdot (1 + yz) + x \cdot (y - yz) + x \cdot (yz + z)$

$= 1 + yz + xy - xyz + xyz + xz$

$= 1 + yz + xy + xz = 0$

Therefore $yz + zx + xy = -1$.

Step 3: Show $x^2 + y^2 + z^2 \geqslant 2$.

Since $(x + y + z)^2 \geqslant 0$:

$x^2 + y^2 + z^2 + 2(xy + yz + zx) \geqslant 0$

Substituting $xy + yz + zx = -1$:

$x^2 + y^2 + z^2 + 2(-1) \geqslant 0$

$x^2 + y^2 + z^2 \geqslant 2$

Since $x = \frac{a}{b-c}$, $y = \frac{b}{c-a}$, $z = \frac{c}{a-b}$, this is equivalent to

$\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} \geqslant 2.$

Part (ii)

Step 1: Find a suitable matrix and show $xyz + yz + zx + xy = 4$.

With $x = \frac{2a}{b+c}$, $y = \frac{2b}{c+a}$, $z = \frac{2c}{a+b}$, we have $2a = (b+c)x$, $2b = (c+a)y$, $2c = (a+b)z$. Rearranging:

$2a - bx - cx = 0, \quad 2b - ay - cy = 0, \quad 2c - az - bz = 0$

So

$\begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

Since $a, b, c > 0$, the vector $(a, b, c)^T$ is nonzero, so the matrix is singular and its determinant is zero.

Expanding the determinant along the first row:

$2 \cdot \det \begin{pmatrix} 2 & -y \\ -z & 2 \end{pmatrix} - (-x) \cdot \det \begin{pmatrix} -y & -y \\ -z & 2 \end{pmatrix} + (-x) \cdot \det \begin{pmatrix} -y & 2 \\ -z & -z \end{pmatrix}$

$= 2(4 - yz) + x(-2y - yz) - x(yz + 2z)$

$= 8 - 2yz - 2xy - xyz - xyz - 2xz$

$= 8 - 2(yz + xy + xz) - 2xyz = 0$

Dividing by 2:

$4 - (yz + xy + xz) - xyz = 0$

$xyz + yz + zx + xy = 4$

Step 2: Show $(2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b + c)(c + a)(a + b)$.

Consider $(x + 1)(y + 1)(z + 1)$:

$(x + 1)(y + 1)(z + 1) = xyz + xy + xz + yz + x + y + z + 1$

Using $xyz + yz + zx + xy = 4$:

$(x + 1)(y + 1)(z + 1) = 4 + x + y + z + 1 = 5 + x + y + z$

Since $a, b, c > 0$ and $b + c > 0$, $c + a > 0$, $a + b > 0$, we have $x, y, z > 0$. Therefore $x + y + z > 0$, giving

$(x + 1)(y + 1)(z + 1) > 5$

Substituting back:

$\left(\frac{2a}{b+c} + 1\right)\left(\frac{2b}{c+a} + 1\right)\left(\frac{2c}{a+b} + 1\right) > 5$

$\frac{2a + b + c}{b + c} \cdot \frac{2b + c + a}{c + a} \cdot \frac{2c + a + b}{a + b} > 5$

Multiplying both sides by $(b + c)(c + a)(a + b) > 0$:

$(2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b + c)(c + a)(a + b)$

Step 3: Show the stronger bound $(2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b + c)(c + a)(a + b)$.

Since $a, b, c > 0$, we have $b + c < a + b + c$, so

$x = \frac{2a}{b + c} > \frac{2a}{a + b + c}$

and similarly $y > \frac{2b}{a+b+c}$, $z > \frac{2c}{a+b+c}$. Adding:

$x + y + z > \frac{2a + 2b + 2c}{a + b + c} = 2$

Therefore

$(x + 1)(y + 1)(z + 1) = 5 + x + y + z > 5 + 2 = 7$

Following the same argument as in Step 2:

$(2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b + c)(c + a)(a + b)$

Examiner Notes

第四受欢迎（约80%作答），平均分约9/20，少有人得满分。考官报告强调：(1) 声明 det(M)=0 后必须论证为何它成立（非平凡零向量存在）；(2) 不等式涉及分式时必须注意正负性；(3) 最后一问 x+y+z>2 的证明方法多样（直接展开、AM-GM、分情况讨论均可）。忽略 hence 要求而纯代数硬算是常见扣分原因。

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Question 3

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Let $I_n = \int_0^\beta (\sec x + \tan x)^n \, dx$, where $n$ is a non-negative integer and $0 < \beta < \frac{\pi}{2}$.

For $n \geqslant 1$, show that

$\frac{1}{2}(I_{n+1} + I_{n-1}) = \frac{1}{n} \left( (\sec \beta + \tan \beta)^n - 1 \right).$

Show also that

$I_n < \frac{1}{n} \left( (\sec \beta + \tan \beta)^n - 1 \right).$

(ii) Let $J_n = \int_0^\beta (\sec x \cos \beta + \tan x)^n \, dx$, where $n$ is a non-negative integer and $0 < \beta < \frac{\pi}{2}$.

For $n \geqslant 1$, show that

$J_n < \frac{1}{n} \left( (1 + \tan \beta)^n - \cos^n \beta \right).$

Hint

(i)

$\frac{1}{2} (I_{n+1} + I_{n-1}) = \frac{1}{2} \int_{0}^{\beta} (\sec x + \tan x)^{n+1} + (\sec x + \tan x)^{n-1} dx$

$= \frac{1}{2} \int_{0}^{\beta} (\sec x + \tan x)^{n-1} ((\sec x + \tan x)^2 + 1) dx$

M1

$= \frac{1}{2} \int_{0}^{\beta} (\sec x + \tan x)^{n-1} (\sec^2 x + 2 \sec x \tan x + \tan^2 x + 1) dx$

$= \int_{0}^{\beta} (\sec x + \tan x)^{n-1} (\sec^2 x + \sec x \tan x) dx$

M1

$= \left[ \frac{1}{n} (\sec x + \tan x)^n \right]_{0}^{\beta} = \frac{1}{n} ((\sec \beta + \tan \beta)^n - 1)$

M1 A1 *A1 (5)

as required.

$\frac{1}{2} (I_{n+1} + I_{n-1}) - I_n = \frac{1}{2} (I_{n+1} - 2I_n + I_{n-1})$

$= \frac{1}{2} \int_{0}^{\beta} (\sec x + \tan x)^{n+1} - 2(\sec x + \tan x)^n + (\sec x + \tan x)^{n-1} dx$

M1

$= \frac{1}{2} \int_{0}^{\beta} (\sec x + \tan x)^{n-1} ((\sec x + \tan x) - 1)^2 dx$

M1 A1

$((\sec x + \tan x) - 1)^2 > 0$ for all $x > 0$

$\sec x \geq 1$ for $0 \leq x < \frac{\pi}{2}$ and hence for $0 \leq x < \beta$ and similarly $\tan x \geq 0$, and thus also $(\sec x + \tan x)^{n-1} > 0$. E1

Therefore, $\frac{1}{2} (I_{n+1} + I_{n-1}) - I_n > 0$, A1

and so $I_n < \frac{1}{2} (I_{n+1} + I_{n-1}) = \frac{1}{n} ((\sec \beta + \tan \beta)^n - 1)$ as required. *M1 A1 (7)

Alternative 1: it has already been shown that

$\frac{1}{2} (I_{n+1} + I_{n-1}) = \int_{0}^{\beta} (\sec x + \tan x)^{n-1} (\sec^2 x + \sec x \tan x) \, dx$

$= \int_{0}^{\beta} \sec x (\sec x + \tan x)^n \, dx$

which is greater than $I_n$ as the expression being integrated is greater than $(\sec x + \tan x)^n$ because $\sec x > 0$ over this domain.

Alternative 2:-

$I_{n+1} - I_n = \int_{0}^{\beta} (\sec x + \tan x)^n (\sec x + \tan x - 1) \, dx$

$I_n - I_{n-1} = \int_{0}^{\beta} (\sec x + \tan x)^{n-1} (\sec x + \tan x - 1) \, dx$

M1 A1 A1

For $0 < x < \beta$ , $\sec x > 1$ , $\tan x > 0$ so $\sec x + \tan x > 1$ E1 and thus $I_{n+1} - I_n > I_n - I_{n-1}$ A1

and so $I_n \le \frac{1}{2} (I_{n+1} + I_{n-1}) = \frac{1}{n} ((\sec \beta + \tan \beta)^n - 1)$ M1 *A1 (7)

(ii) $\frac{1}{2} (J_{n+1} + J_{n-1}) = \frac{1}{2} \int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n+1} + (\sec x \cos \beta + \tan x)^{n-1} dx$

$= \frac{1}{2} \int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} ((\sec x \cos \beta + \tan x)^2 + 1) \, dx$

M1

$= \frac{1}{2} \int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} (\sec^2 x \cos^2 \beta + 2 \sec x \cos \beta \tan x + \tan^2 x + 1) \, dx$

$= \frac{1}{2} \int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} (\sec^2 x (1 - \sin^2 \beta) + 2 \sec x \cos \beta \tan x + \tan^2 x + 1) \, dx$

$= \int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} ((\sec^2 x + \sec x \cos \beta \tan x) - \sec^2 x \sin^2 \beta) dx$

M1

$\int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} (\sec^2 x + \sec x \cos \beta \tan x) dx = \left[ \frac{1}{n} (\sec x \cos \beta + \tan x)^n \right]_{0}^{\beta}$

M1 $= \frac{1}{n} ((1 + \tan \beta)^n - \cos^n \beta)$ A1

$\int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} \sec^2 x \sin^2 \beta \, dx > 0$

by a similar argument to part (i), namely $\sec^2 x \sin^2 \beta > 0$ for any $x$, and $\sec x \cos \beta + \tan x > 0$ as $\sec x > 0$ and $\tan x \geq 0$ for $0 \leq x < \beta < \frac{\pi}{2}$ E1

Hence $\frac{1}{2} (J_{n+1} + J_{n-1}) < \frac{1}{n} ((1 + \tan \beta)^n - \cos^n \beta)$ A1

But $\frac{1}{2} (J_{n+1} + J_{n-1}) - J_n = = \frac{1}{2} \int_{0}^{\beta} (\sec x \cos \beta + \tan x)^{n-1} ((\sec x \cos \beta + \tan x) - 1)^2 dx > 0$

M1

as before, and thus $J_n < \frac{1}{2} (J_{n+1} + J_{n-1}) < \frac{1}{n} ((1 + \tan \beta)^n - \cos^n \beta)$ as required. *A1 (8)

Model Solution

Part (i)

We begin by computing $\frac{1}{2}(I_{n+1} + I_{n-1})$ directly:

$\frac{1}{2}(I_{n+1} + I_{n-1}) = \frac{1}{2} \int_0^\beta \left[ (\sec t + \tan t)^{n+1} + (\sec t + \tan t)^{n-1} \right] dt$

$= \frac{1}{2} \int_0^\beta (\sec t + \tan t)^{n-1} \left[ (\sec t + \tan t)^2 + 1 \right] dt$

Expanding the bracket and using $\tan^2 t + 1 = \sec^2 t$:

$(\sec t + \tan t)^2 + 1 = \sec^2 t + 2\sec t \tan t + \tan^2 t + 1 = 2\sec^2 t + 2\sec t \tan t = 2\sec t(\sec t + \tan t)$

Therefore:

$\frac{1}{2}(I_{n+1} + I_{n-1}) = \int_0^\beta \sec t \, (\sec t + \tan t)^n \, dt$

Since $\frac{d}{dt}(\sec t + \tan t) = \sec t \tan t + \sec^2 t = \sec t(\sec t + \tan t)$, the chain rule gives:

$\frac{d}{dt}\left[ \frac{1}{n}(\sec t + \tan t)^n \right] = \sec t \, (\sec t + \tan t)^n$

Hence:

$\frac{1}{2}(I_{n+1} + I_{n-1}) = \left[ \frac{1}{n}(\sec t + \tan t)^n \right]_0^\beta = \frac{1}{n}\left( (\sec\beta + \tan\beta)^n - 1 \right) \qquad \text{(*)}$

For the inequality, consider:

$\frac{1}{2}(I_{n+1} + I_{n-1}) - I_n = \frac{1}{2} \int_0^\beta \left[ (\sec t + \tan t)^{n+1} - 2(\sec t + \tan t)^n + (\sec t + \tan t)^{n-1} \right] dt$

$= \frac{1}{2} \int_0^\beta (\sec t + \tan t)^{n-1} \left[ (\sec t + \tan t) - 1 \right]^2 \, dt$

For $0 < t < \beta < \frac{\pi}{2}$, we have $\sec t > 1$ and $\tan t > 0$, so $\sec t + \tan t > 1$. Thus $(\sec t + \tan t)^{n-1} > 0$ and $[(\sec t + \tan t) - 1]^2 > 0$ on $(0, \beta)$. The integrand is strictly positive on this interval, so the integral is positive:

$\frac{1}{2}(I_{n+1} + I_{n-1}) - I_n > 0$

Combining with $(*)$:

$I_n < \frac{1}{n}\left( (\sec\beta + \tan\beta)^n - 1 \right)$

Part (ii)

Write $v(t) = \sec t \cos\beta + \tan t$. Then:

$\frac{1}{2}(J_{n+1} + J_{n-1}) = \frac{1}{2} \int_0^\beta v^{n-1}(v^2 + 1) \, dt$

Expanding $v^2 + 1$:

$v^2 + 1 = \sec^2 t \cos^2\beta + 2\sec t \cos\beta \tan t + \tan^2 t + 1$

$= \sec^2 t \cos^2\beta + 2\sec t \cos\beta \tan t + \sec^2 t$

We compute $v'(t) = \sec t \tan t \cos\beta + \sec^2 t = \sec t(\sec t + \cos\beta \tan t)$, so:

$v' = \sec^2 t + \sec t \cos\beta \tan t$

We can write:

$\frac{1}{2}(v^2 + 1) = (\sec^2 t + \sec t \cos\beta \tan t) + \frac{1}{2}\sec^2 t(\cos^2\beta - 1)$

$= v' - \frac{1}{2}\sec^2 t \sin^2\beta$

Therefore:

$\frac{1}{2}(J_{n+1} + J_{n-1}) = \int_0^\beta v^{n-1} v' \, dt - \frac{\sin^2\beta}{2} \int_0^\beta v^{n-1} \sec^2 t \, dt$

The first integral evaluates to:

$\int_0^\beta v^{n-1} v' \, dt = \left[ \frac{v^n}{n} \right]_0^\beta = \frac{1}{n}\left[ (1 + \tan\beta)^n - \cos^n\beta \right]$

For the second integral: since $0 < \beta < \frac{\pi}{2}$, we have $\sin^2\beta > 0$ and $\sec^2 t > 0$. Also $v(t) = \sec t \cos\beta + \tan t > 0$ for $0 \leqslant t \leqslant \beta$ (as $\sec t \geqslant 1$, $\cos\beta > 0$, $\tan t \geqslant 0$). So $v^{n-1} \sec^2 t > 0$ on $(0, \beta)$, making the second integral positive. Hence:

$\frac{1}{2}(J_{n+1} + J_{n-1}) < \frac{1}{n}\left[ (1 + \tan\beta)^n - \cos^n\beta \right] \qquad \text{(**)}$

For the strict inequality $J_n < \frac{1}{2}(J_{n+1} + J_{n-1})$, we proceed as in part (i):

$\frac{1}{2}(J_{n+1} + J_{n-1}) - J_n = \frac{1}{2} \int_0^\beta v^{n-1}(v - 1)^2 \, dt$

Since $v(0) = \cos\beta > 0$ and $v'(t) > 0$, the function $v$ is increasing on $[0, \beta]$ with $v(t) > 0$. Thus $v^{n-1} > 0$ on $(0, \beta)$. Moreover $v(\beta) = 1 + \tan\beta > 1$, so $(v - 1)^2 > 0$ near $t = \beta$, making the integral positive:

$\frac{1}{2}(J_{n+1} + J_{n-1}) - J_n > 0$

Combining with $(**)$:

$J_n < \frac{1}{n}\left[ (1 + \tan\beta)^n - \cos^n\beta \right]$

Examiner Notes

第二受欢迎（84%作答），平均分略低于8/20。考官报告指出：(1) 证明不等式时证大于零比直接比较更有效，即考虑 (1/2)(I_{n+1}+I_{n-1})-I_n；(2) 常见错误：积分正负性未充分论证、逻辑推导方向错误、得到弱不等式而非严格不等式、不等式与变量范围矛盾；(3) 部分考生在(ii)中忽略 sec x cos beta <= 1 的条件。

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Question 4

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

4 Let n be a vector of unit length and $\Pi$ be the plane through the origin perpendicular to n. For any vector x, the projection of x onto the plane $\Pi$ is defined to be the vector $\mathbf{x} - (\mathbf{x} \cdot \mathbf{n}) \, \mathbf{n}$.

The vectors a and b each have unit length and the angle between them is $\theta$, which satisfies $0 < \theta < \pi$. The vector m is given by $\mathbf{m} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$.

(i) Show that m bisects the angle between a and b.

(ii) The vector c also has unit length. The angle between a and c is $\alpha$, and the angle between b and c is $\beta$. Both angles are acute and non-zero.

Let $\mathbf{a}_1$ and $\mathbf{b}_1$ be the projections of a and b, respectively, onto the plane through the origin perpendicular to c. Show that $\mathbf{a}_1 \cdot \mathbf{c} = 0$ and, by considering $|\mathbf{a}_1|^2 = \mathbf{a}_1 \cdot \mathbf{a}_1$, show that $|\mathbf{a}_1| = \sin \alpha$.

Show also that the angle $\phi$ between $\mathbf{a}_1$ and $\mathbf{b}_1$ satisfies

$\cos \phi = \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta}.$

(iii) Let $\mathbf{m}_1$ be the projection of m onto the plane through the origin perpendicular to c. Show that $\mathbf{m}_1$ bisects the angle between $\mathbf{a}_1$ and $\mathbf{b}_1$ if and only if

$\alpha = \beta \quad \text{or} \quad \cos \theta = \cos(\alpha - \beta).$

Hint

(i)

$\boldsymbol{m.a} = \frac{1}{2}(\boldsymbol{a} + \boldsymbol{b}).\boldsymbol{a} = \frac{1}{2}(1 + \boldsymbol{a.b}) = m \cos \alpha$ where $\alpha$ is the non-reflex angle between a and m $\boldsymbol{m.b} = \frac{1}{2}(\boldsymbol{a} + \boldsymbol{b}).\boldsymbol{b} = \frac{1}{2}(1 + \boldsymbol{a.b}) = m \cos \beta$ where $\alpha$ is the non-reflex angle between b and m

M1 A1

Thus $\cos \alpha = \cos \beta$ and so $\alpha = \beta$ as for $0 \le \tau \le \pi$, there is only one value of $\tau$ for any given value of $\cos \tau$. E1 (3)

(ii) $\boldsymbol{a_1.c} = (\boldsymbol{a} - (\boldsymbol{a.c})\boldsymbol{c}).\boldsymbol{c} = \boldsymbol{a.c} - \boldsymbol{a.c} \ \boldsymbol{c.c} = 0$ as required. *B1 $\boldsymbol{a.c} = \cos \alpha$ , $\boldsymbol{b.c} = \cos \beta$ , $\boldsymbol{a.b} = \cos \theta$ $\boldsymbol{a_1} = \boldsymbol{a} - (\boldsymbol{a.c})\boldsymbol{c}$ and $\boldsymbol{b_1} = \boldsymbol{b} - (\boldsymbol{b.c})\boldsymbol{c}$ $|\boldsymbol{a_1}|^2 = \boldsymbol{a_1.a_1} = (\boldsymbol{a} - (\boldsymbol{a.c})\boldsymbol{c}).(\boldsymbol{a} - (\boldsymbol{a.c})\boldsymbol{c}) = \boldsymbol{a.a} - 2\boldsymbol{a.c} \ \boldsymbol{a.c} + \boldsymbol{a.c} \ \boldsymbol{a.c} \ \boldsymbol{c.c}$ $= 1 - 2 \cos^2 \alpha + \cos^2 \alpha = \sin^2 \alpha$

M1

and so. as $\alpha$ is acute, $|\boldsymbol{a_1}| = \sin \alpha$ as required. *A1

$\boldsymbol{a_1.b_1} = (\boldsymbol{a} - (\boldsymbol{a.c})\boldsymbol{c}).(\boldsymbol{b} - (\boldsymbol{b.c})\boldsymbol{c}) = \boldsymbol{a.b} - 2(\boldsymbol{a.c})(\boldsymbol{b.c}) + (\boldsymbol{a.c})(\boldsymbol{b.c})(\boldsymbol{c.c})$ $= \cos \theta - \cos \alpha \cos \beta$ M1 A1

but also, $\boldsymbol{a_1.b_1} = \sin \alpha \sin \beta \cos \phi$ B1 M1 and hence,

$\cos \phi = \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta}$

as required.

*A1 (8) (iii) $\boldsymbol{m_1} = \boldsymbol{m} - (\boldsymbol{m.c})\boldsymbol{c} = \frac{1}{2}(\boldsymbol{a} + \boldsymbol{b}) - \left(\frac{1}{2}(\boldsymbol{a} + \boldsymbol{b}).\boldsymbol{c}\right)\boldsymbol{c} = \frac{1}{2}(\boldsymbol{a_1} + \boldsymbol{b_1})$ B1 $\boldsymbol{m_1}$ bisects the angle between $\boldsymbol{a_1}$ and $\boldsymbol{b_1}$ if and only if $\frac{\boldsymbol{m_1.a_1}}{\sin \alpha} = \frac{\boldsymbol{m_1.b_1}}{\sin \beta}$ M1

Thus, multiplying through by $2 \sin \alpha \sin \beta$,

$(\boldsymbol{a_1} + \boldsymbol{b_1}).\boldsymbol{a_1} \sin \beta = (\boldsymbol{a_1} + \boldsymbol{b_1}).\boldsymbol{b_1} \sin \alpha$ A1

$(\sin^2 \alpha + \boldsymbol{a_1.b_1}) \sin \beta = (\sin^2 \beta + \boldsymbol{a_1.b_1}) \sin \alpha$

M1 A1

So $(\boldsymbol{a_1} . \boldsymbol{b_1} - \sin \alpha \sin \beta)(\sin \alpha - \sin \beta) = 0$

A1

and thus, $\sin \alpha = \sin \beta$ in which case $\alpha = \beta$ as both angles are acute, *A1

or $\cos \theta - \cos \alpha \cos \beta = \sin \alpha \sin \beta$, meaning that $\cos \theta = \cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta)$

*M1 A1 (9)

Model Solution

Part (i)

We compute the dot products of m with a and b:

$\mathbf{m} \cdot \mathbf{a} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) \cdot \mathbf{a} = \frac{1}{2}(\mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{a}) = \frac{1}{2}(1 + \cos\theta)$

$\mathbf{m} \cdot \mathbf{b} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) \cdot \mathbf{b} = \frac{1}{2}(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}) = \frac{1}{2}(\cos\theta + 1)$

So $\mathbf{m} \cdot \mathbf{a} = \mathbf{m} \cdot \mathbf{b}$. Let $\psi_a$ be the angle between m and a, and $\psi_b$ the angle between m and b. Then:

$|\mathbf{m}| \cos\psi_a = \mathbf{m} \cdot \mathbf{a} = \mathbf{m} \cdot \mathbf{b} = |\mathbf{m}| \cos\psi_b$

Since $\mathbf{m} = \frac{1}{2}(\mathbf{a}+\mathbf{b}) \neq \mathbf{0}$ (as $0 < \theta < \pi$ means $\mathbf{a} \neq -\mathbf{b}$), we have $|\mathbf{m}| \neq 0$, so $\cos\psi_a = \cos\psi_b$. Both angles lie in $[0, \pi]$, so $\psi_a = \psi_b$. Thus m bisects the angle between a and b.

Part (ii)

By definition, $\mathbf{a}_1 = \mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{c}$ and $\mathbf{b}_1 = \mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{c}$.

Showing $\mathbf{a}_1 \cdot \mathbf{c} = 0$:

$\mathbf{a}_1 \cdot \mathbf{c} = [\mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{c}] \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} - (\mathbf{a} \cdot \mathbf{c})(\mathbf{c} \cdot \mathbf{c}) = \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{c} = 0$

Showing $|\mathbf{a}_1| = \sin\alpha$:

$|\mathbf{a}_1|^2 = \mathbf{a}_1 \cdot \mathbf{a}_1 = [\mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{c}] \cdot [\mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{c}]$

$= \mathbf{a} \cdot \mathbf{a} - 2(\mathbf{a} \cdot \mathbf{c})^2 + (\mathbf{a} \cdot \mathbf{c})^2 (\mathbf{c} \cdot \mathbf{c})$

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

Since $\alpha$ is acute, $|\mathbf{a}_1| = \sin\alpha$. By identical reasoning with $\beta$, $|\mathbf{b}_1| = \sin\beta$.

Finding $\cos\phi$:

$\mathbf{a}_1 \cdot \mathbf{b}_1 = [\mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{c}] \cdot [\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{c}]$

$= \mathbf{a} \cdot \mathbf{b} - (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{c}) - (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{c}) + (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{c})(\mathbf{c} \cdot \mathbf{c})$

$= \cos\theta - 2\cos\alpha\cos\beta + \cos\alpha\cos\beta = \cos\theta - \cos\alpha\cos\beta$

Since $\mathbf{a}_1 \cdot \mathbf{b}_1 = |\mathbf{a}_1||\mathbf{b}_1|\cos\phi = \sin\alpha\sin\beta\cos\phi$:

$\cos\phi = \frac{\cos\theta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta}$

Part (iii)

By linearity of projection:

$\mathbf{m}_1 = \mathbf{m} - (\mathbf{m} \cdot \mathbf{c})\mathbf{c} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) - \frac{1}{2}[(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c}]\mathbf{c}$

$= \frac{1}{2}[\mathbf{a} - (\mathbf{a} \cdot \mathbf{c})\mathbf{c}] + \frac{1}{2}[\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{c}] = \frac{1}{2}(\mathbf{a}_1 + \mathbf{b}_1)$

m$_1$ bisects the angle between a$_1$ and b$_1$ if and only if:

$\frac{\mathbf{m}_1 \cdot \mathbf{a}_1}{|\mathbf{a}_1|} = \frac{\mathbf{m}_1 \cdot \mathbf{b}_1}{|\mathbf{b}_1|}$

We compute:

$\mathbf{m}_1 \cdot \mathbf{a}_1 = \frac{1}{2}(\mathbf{a}_1 + \mathbf{b}_1) \cdot \mathbf{a}_1 = \frac{1}{2}(|\mathbf{a}_1|^2 + \mathbf{a}_1 \cdot \mathbf{b}_1) = \frac{1}{2}(\sin^2\alpha + \cos\theta - \cos\alpha\cos\beta)$

$\mathbf{m}_1 \cdot \mathbf{b}_1 = \frac{1}{2}(\mathbf{a}_1 + \mathbf{b}_1) \cdot \mathbf{b}_1 = \frac{1}{2}(\mathbf{a}_1 \cdot \mathbf{b}_1 + |\mathbf{b}_1|^2) = \frac{1}{2}(\cos\theta - \cos\alpha\cos\beta + \sin^2\beta)$

The bisection condition becomes:

$\frac{\sin^2\alpha + \cos\theta - \cos\alpha\cos\beta}{\sin\alpha} = \frac{\cos\theta - \cos\alpha\cos\beta + \sin^2\beta}{\sin\beta}$

Cross-multiplying:

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

Writing $D = \cos\theta - \cos\alpha\cos\beta$:

$(\sin^2\alpha + D)\sin\beta = (D + \sin^2\beta)\sin\alpha$

$\sin^2\alpha\sin\beta + D\sin\beta = D\sin\alpha + \sin^2\beta\sin\alpha$

$\sin\alpha\sin\beta(\sin\alpha - \sin\beta) - D(\sin\alpha - \sin\beta) = 0$

$(\sin\alpha\sin\beta - D)(\sin\alpha - \sin\beta) = 0$

So either $\sin\alpha = \sin\beta$ (and since both angles are acute, $\alpha = \beta$), or $D = \sin\alpha\sin\beta$, i.e. $\cos\theta - \cos\alpha\cos\beta = \sin\alpha\sin\beta$, giving $\cos\theta = \cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos(\alpha - \beta)$.

Conversely: if $\alpha = \beta$, then $\mathbf{a}_1 \cdot \mathbf{b}_1 = D$ is the same in both numerators and $\sin\alpha = \sin\beta$, so both sides are equal. If $\cos\theta = \cos(\alpha-\beta)$, then $D = \sin\alpha\sin\beta$, making the left side $\frac{\sin\alpha(\sin\alpha + \sin\beta)}{\sin\alpha}$ and the right side $\frac{\sin\beta(\sin\alpha + \sin\beta)}{\sin\beta}$, both equal to $\sin\alpha + \sin\beta$. Hence the bisection condition holds if and only if

$\alpha = \beta \quad \text{or} \quad \cos\theta = \cos(\alpha - \beta).$

Examiner Notes

最不受欢迎的纯数题（仅约35%作答），平均分低于7/20，排第七。考官报告指出：(1) 题目核心是仅凭投影定义出发证明性质，不少考生直接假设投影性质（如正交性）而未加证明；(2) 几何图示中的隐含假设需要代数验证；(3) 跟随题目结构、认真做点积计算的考生通常表现较好。

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Question 5

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

5 Two curves have polar equations $r = a + 2 \cos \theta$ and $r = 2 + \cos 2\theta$, where $r \geqslant 0$ and $a$ is a constant.

(i) Show that these curves meet when

$2 \cos^2 \theta - 2 \cos \theta + 1 - a = 0 .$

Hence show that these curves touch if $a = \frac{1}{2}$ and find the other two values of $a$ for which the curves touch.

(ii) Sketch the curves $r = a + 2 \cos \theta$ and $r = 2 + \cos 2\theta$ on the same diagram in the case $a = \frac{1}{2}$. Give the values of $r$ and $\theta$ at the points at which the curves touch and justify the other features you show on your sketch.

(iii) On two further diagrams, one for each of the other two values of $a$, sketch both the curves $r = a + 2 \cos \theta$ and $r = 2 + \cos 2\theta$. Give the values of $r$ and $\theta$ at the points at which the curves touch and justify the other features you show on your sketch.

Hint

(i) The curves meet when $a + 2 \cos \theta = 2 + \cos 2\theta$

That is, $a + 2 \cos \theta = 2 + 2 \cos^2 \theta - 1$ or as required, B1 $2 \cos^2 \theta - 2 \cos \theta + 1 - a = 0$

The curves touch if this quadratic has coincident roots, M1 i.e. if $4 - 8(1 - a) = 0 \Rightarrow a = \frac{1}{2}$ , *A1 or if $\cos \theta = \pm 1$ , M1 in which cases $a = 1$ A1 or $a = 5$ . A1 (6)

Alternatively, for the curves to touch, they must have the same gradient, so differentiating,

$-2 \sin \theta = -2 \sin 2\theta = -4 \sin \theta \cos \theta$

M1

in which case, either $\sin \theta = 0$ giving $\cos \theta = \pm 1$ , M1 in which cases $a = 1$ A1 or $a = 5$ , A1 or $\cos \theta = \frac{1}{2}$ in which case $a = \frac{1}{2}$ . *A1 (6)

(ii) If $a = \frac{1}{2}$ then at points where they touch, $\cos \theta = \frac{1}{2}$ so $\theta = \pm \frac{\pi}{3}$ and thus $( \frac{3}{2}, \pm \frac{\pi}{3} )$ . M1A1

$r = a + 2 \cos \theta$ is symmetrical about the initial line which it intercepts at $( \frac{5}{2}, 0 )$ and has a cusp at $( 0, \pm \cos^{-1} (-\frac{1}{4}) )$ . It passes through $( \frac{1}{2}, \pm \frac{\pi}{2} )$ and only exists for $-\cos^{-1} (-\frac{1}{4}) < \theta < \cos^{-1} (-\frac{1}{4})$ .

$r = 2 + \cos 2\theta$ is symmetrical about both the initial line, and its perpendicular. It passes through

$(3,0)$ , $(3, \pi)$ , and $( 1, \pm \frac{\pi}{2} )$

Sketch G6 (8)

(iii) If $a = 1$ , then the curves meet where $2 \cos^2 \theta - 2 \cos \theta = 0$ , i.e. $\cos \theta = 1$ at $(3,0)$ where they touch, and $\cos \theta = 0$ at $( 1, \pm \frac{\pi}{2} )$

$r = a + 2 \cos \theta$ is symmetrical about the initial line which it intercepts at $( 3, 0 )$ and has a cusp at $( 0, \pm \cos^{-1} (-\frac{1}{2}) ) = ( 0, \pm \frac{2\pi}{3} )$ . It passes through $( 1, \pm \frac{\pi}{2} )$ and only exists for

$-\frac{2\pi}{3} < \theta < \frac{2\pi}{3}$ .

Sketch G3

If $a = 5$ , then the curves meet where $2 \cos^2 \theta - 2 \cos \theta - 4 = 0$ , i.e. only $\cos \theta = -1$ at $(3, \pi)$ where they touch, as $\cos \theta \neq 2$ .

$r = a + 2 \cos \theta$ is symmetrical about the initial line which it intercepts at $( 7, 0 )$ and $( 3, \pi )$ . It also passes through $( 5, \pm \frac{\pi}{2} )$ .

Sketch G3 (6)

Model Solution

Part (i)

The two curves meet when their $r$-values are equal (and non-negative):

$a + 2\cos\theta = 2 + \cos 2\theta.$

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

$a + 2\cos\theta = 2 + 2\cos^2\theta - 1 = 1 + 2\cos^2\theta.$

Rearranging:

$2\cos^2\theta - 2\cos\theta + 1 - a = 0. \qquad \text{(*)}$

This is the required condition.

Now we determine the values of $a$ for which the curves touch. Let $u = \cos\theta$, so (*) becomes

$2u^2 - 2u + (1 - a) = 0.$

The curves touch when this equation has a solution with $-1 \leqslant u \leqslant 1$ and the curves are tangent at the corresponding point.

Case 1: Double root. The quadratic in $u$ has a repeated root when the discriminant is zero:

$\Delta = 4 - 8(1 - a) = 8a - 4 = 0 \implies a = \tfrac{1}{2}.$

When $a = \frac{1}{2}$, the repeated root is $u = \frac{2}{4} = \frac{1}{2}$, i.e. $\cos\theta = \frac{1}{2}$, giving $\theta = \pm\frac{\pi}{3}$. At these points $r = \frac{1}{2} + 2 \cdot \frac{1}{2} = \frac{3}{2} > 0$, so the curves touch.

Case 2: Boundary roots $u = \pm 1$. If the quadratic has a root at $u = 1$ (i.e. $\theta = 0$):

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

At $\theta = 0$: $r = 1 + 2 = 3 > 0$ for both curves, so they meet at $(3, 0)$.

If the quadratic has a root at $u = -1$ (i.e. $\theta = \pi$):

$2(-1)^2 - 2(-1) + 1 - a = 0 \implies a = 5.$

At $\theta = \pi$: $r = 5 + 2(-1) = 3 > 0$ for the first curve and $r = 2 + \cos 2\pi = 3$ for the second, so they meet at $(3, \pi)$.

In both boundary cases, the quadratic factors as $2(u - 1)(u - \dots)$ or $2(u + 1)(u - \dots)$ with the second root outside $[-1, 1]$, so the curves meet at exactly one point and touch there.

Therefore the three values of $a$ for which the curves touch are

$a = \tfrac{1}{2}, \quad a = 1, \quad a = 5.$

Part (ii)

When $a = \frac{1}{2}$: the curves touch at $\cos\theta = \frac{1}{2}$, i.e. $\theta = \pm\frac{\pi}{3}$, with $r = \frac{3}{2}$. So the touching points are $\left(\frac{3}{2}, \pm\frac{\pi}{3}\right)$.

Curve $r = \frac{1}{2} + 2\cos\theta$: This is a limacon (inner loop type). It has $r = 0$ when $\cos\theta = -\frac{1}{4}$, i.e. at $\theta = \pm\arccos(-\frac{1}{4}) \approx \pm 104.5°$. The curve exists only for $-\arccos(-\frac{1}{4}) < \theta < \arccos(-\frac{1}{4})$, forming a single loop. Key points: intercept on the initial line at $\left(\frac{5}{2}, 0\right)$; intercept at $\left(\frac{1}{2}, \pm\frac{\pi}{2}\right)$; cusp at the origin where $\theta = \pm\arccos(-\frac{1}{4})$. The curve is symmetric about the initial line ($\theta = 0$).

Curve $r = 2 + \cos 2\theta$: This has period $\pi$ in $\theta$ and is symmetric about both the initial line and the line $\theta = \frac{\pi}{2}$. It forms a four-lobed shape. Key points: $(3, 0)$, $\left(1, \pm\frac{\pi}{2}\right)$, $(3, \pi)$, $\left(1, \pm\frac{3\pi}{2}\right)$. Since $1 \leqslant r \leqslant 3$ always, it does not pass through the origin.

The two touching points $\left(\frac{3}{2}, \pm\frac{\pi}{3}\right)$ lie on the right side of the loop of the first curve and on the “inner” portion of the four-lobed second curve.

Part (iii)

When $a = 1$: The curves meet where $2\cos^2\theta - 2\cos\theta = 0$, i.e. $\cos\theta(\cos\theta - 1) = 0$.

• $\cos\theta = 1$ ($\theta = 0$): $r = 3$. The curves touch at $(3, 0)$.
• $\cos\theta = 0$ ($\theta = \pm\frac{\pi}{2}$): $r = 1$. The curves cross at $\left(1, \pm\frac{\pi}{2}\right)$ (not touching, since the discriminant is non-zero here).

Curve $r = 1 + 2\cos\theta$: A cardioid. It has $r = 0$ when $\cos\theta = -\frac{1}{2}$, i.e. at $\theta = \pm\frac{2\pi}{3}$. It exists for $-\frac{2\pi}{3} < \theta < \frac{2\pi}{3}$. Key points: $(3, 0)$, $\left(1, \pm\frac{\pi}{2}\right)$, cusp at origin at $\theta = \pm\frac{2\pi}{3}$. Symmetric about the initial line.

Curve $r = 2 + \cos 2\theta$: As before. Both curves pass through $(3, 0)$ where they touch, and through $\left(1, \pm\frac{\pi}{2}\right)$ where they cross.

When $a = 5$: The curves meet where $2\cos^2\theta - 2\cos\theta - 4 = 0$, i.e. $(\cos\theta - 2)(\cos\theta + 1) = 0$. Since $\cos\theta \neq 2$, the only solution is $\cos\theta = -1$, i.e. $\theta = \pi$, with $r = 3$. The curves touch at $(3, \pi)$.

Curve $r = 5 + 2\cos\theta$: Since $r \geqslant 3 > 0$ for all $\theta$, this is a convex oval (no inner loop). Key points: $(7, 0)$, $(3, \pi)$, $\left(5, \pm\frac{\pi}{2}\right)$. Symmetric about the initial line.

Curve $r = 2 + \cos 2\theta$: As before. The oval lies entirely outside the four-lobed curve, and they touch only at $(3, \pi)$ on the far side.

Examiner Notes

略多于Q2的考生作答，但平均分与Q4相近。考官报告指出：(1) 几乎所有考生能得出第一个交点条件，多数能从判别式得到 a=1/2；(2) 找另外两个 a 值（1和5）时很多考生有困难，应考虑令 dr/dtheta 相等；(3) 作图常见问题：第二条曲线画成椭圆、尖点处画成角而非光滑凹陷；(4) 对称性虽被多数人感知但很少被用作论证依据。

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Question 6

Topic: 纯数  |  Difficulty: Challenging  |  Marks: 20

Problem

6 (i) For $x \neq \tan \alpha$, the function $f_\alpha$ is defined by

$f_\alpha(x) = \tan^{-1} \left( \frac{x \tan \alpha + 1}{\tan \alpha - x} \right)$

where $0 < \alpha < \frac{1}{2}\pi$.

Show that $f'_\alpha(x) = \frac{1}{1 + x^2}$.

Hence sketch $y = f_\alpha(x)$.

On a separate diagram, sketch $y = f_\alpha(x) - f_\beta(x)$ where $0 < \alpha < \beta < \frac{1}{2}\pi$.

(ii) For $0 \leqslant x \leqslant 2\pi$ and $x \neq \frac{1}{2}\pi, \frac{3}{2}\pi$, the function $g(x)$ is defined by

$g(x) = \tanh^{-1}(\sin x) - \sinh^{-1}(\tan x).$

For $\frac{1}{2}\pi < x < \frac{3}{2}\pi$, show that $g'(x) = 2 \sec x$.

Use this result to sketch $y = g(x)$ for $0 \leqslant x \leqslant 2\pi$.

Hint

(i)

$f_\alpha(x) = \tan^{-1} \left( \frac{x \tan \alpha + 1}{\tan \alpha - x} \right)$

$f'_\alpha(x) = \frac{1}{1 + \left( \frac{x \tan \alpha + 1}{\tan \alpha - x} \right)^2} \frac{(\tan \alpha - x) \tan \alpha + (x \tan \alpha + 1)}{(\tan \alpha - x)^2}$

M1 A1

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

$= \frac{\sec^2 \alpha}{\tan^2 \alpha + x^2 + x^2 \tan^2 \alpha + 1} = \frac{\sec^2 \alpha}{\sec^2 \alpha (1 + x^2)} = \frac{1}{1 + x^2}$

M1 M1 *A1 (5)

as required.

Alternative

$f_\alpha(x) = \tan^{-1} \left( \frac{x \tan \alpha + 1}{\tan \alpha - x} \right)$

$= \tan^{-1} \left( \frac{x + \cot \alpha}{1 - x \cot \alpha} \right)$

$= \tan^{-1} \left( \frac{\tan(\tan^{-1} x) + \tan \left( \frac{\pi}{2} - \alpha \right)}{1 - \tan(\tan^{-1} x) \tan \left( \frac{\pi}{2} - \alpha \right)} \right)$

M1 A1

$= \tan^{-1} \left( \tan \left( \tan^{-1} x + \frac{\pi}{2} - \alpha \right) \right)$

M1

$= \tan^{-1} x + \frac{\pi}{2} - \alpha$ if this is less than $\frac{\pi}{2}$, i.e. if $x < \tan \alpha$

or $= \tan^{-1} x - \frac{\pi}{2} - \alpha$ if $x > \tan \alpha$ M1

So $f'_\alpha(x) = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}$ *A1 (5)

Thus $f_\alpha(x) = \tan^{-1} x + c$

$f_\alpha(0) = \tan^{-1} \left( \frac{1}{\tan \alpha} \right) = \tan^{-1}(\cot \alpha) = \frac{\pi}{2} - \alpha$

$f_\alpha(x) = 0$ when $x = -\cot \alpha$

There is a discontinuity at $x = \tan \alpha$, with $f_\alpha(x)$ approaching $\frac{\pi}{2}$ from below and $-\frac{\pi}{2}$ from above.

As $x \to \pm\infty$, $f_\alpha(x) \to \tan^{-1}(-\tan \alpha) = -\alpha$

So $f_{\alpha}(x) = \tan^{-1} x + \frac{\pi}{2} - \alpha$ for $x < \tan \alpha$ and $f_{\alpha}(x) = \tan^{-1} x - \frac{\pi}{2} - \alpha$ for $x > \tan \alpha$

Sketch G1 G1 G1 (3)

$y = f_{\alpha}(x) - f_{\beta}(x) =$

$(\frac{\pi}{2} - \alpha) - (\frac{\pi}{2} - \beta) = \beta - \alpha$ for $x < \tan \alpha$

$(-\frac{\pi}{2} - \alpha) - (\frac{\pi}{2} - \beta) = \beta - \alpha - \pi$ for $\tan \alpha < x < \tan \beta$

and $(-\frac{\pi}{2} - \alpha) - (-\frac{\pi}{2} - \beta) = \beta - \alpha$ for $x > \tan \beta$

Sketch G1 G1 G1 (3)

(ii) $g(x) = \tanh^{-1}(\sin x) - \sinh^{-1}(\tan x)$

$g'(x) = \frac{1}{1 - \sin^2 x} \cos x - \frac{1}{\sqrt{1 + \tan^2 x}} \sec^2 x$

M1 A1 A1

$= \frac{\cos x}{\cos^2 x} - \frac{\sec^2 x}{|\sec x|} = \sec x - \frac{\sec^2 x}{-\sec x} = 2 \sec x$

M1 *A1 (5)

as required, for $\sec x < 0$, i.e. for $\frac{\pi}{2} < x < \frac{3\pi}{2}$.

(For $\sec x > 0$, $g'(x) = 0$)

Sketch G1 G1 G1 G1 (4)

Model Solution

Part (i)

We differentiate $f_\alpha(x) = \tan^{-1}\!\left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right)$ using the chain rule. Let $u = \frac{x\tan\alpha + 1}{\tan\alpha - x}$, so

$f_\alpha'(x) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}.$

First, compute $\frac{du}{dx}$ by the quotient rule:

$\frac{du}{dx} = \frac{\tan\alpha \cdot (\tan\alpha - x) - (x\tan\alpha + 1) \cdot (-1)}{(\tan\alpha - x)^2} = \frac{\tan^2\alpha - x\tan\alpha + x\tan\alpha + 1}{(\tan\alpha - x)^2} = \frac{\tan^2\alpha + 1}{(\tan\alpha - x)^2} = \frac{\sec^2\alpha}{(\tan\alpha - x)^2}.$

Next, compute $1 + u^2$:

$1 + u^2 = 1 + \frac{(x\tan\alpha + 1)^2}{(\tan\alpha - x)^2} = \frac{(\tan\alpha - x)^2 + (x\tan\alpha + 1)^2}{(\tan\alpha - x)^2}.$

Expand the numerator:

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

$= \tan^2\alpha + 1 + x^2(1 + \tan^2\alpha) = \sec^2\alpha + x^2\sec^2\alpha = \sec^2\alpha(1 + x^2).$

Therefore:

$f_\alpha'(x) = \frac{1}{\frac{\sec^2\alpha(1+x^2)}{(\tan\alpha - x)^2}} \cdot \frac{\sec^2\alpha}{(\tan\alpha - x)^2} = \frac{(\tan\alpha - x)^2}{\sec^2\alpha(1+x^2)} \cdot \frac{\sec^2\alpha}{(\tan\alpha - x)^2} = \frac{1}{1+x^2}.$

Since $f_\alpha'(x) = \frac{1}{1+x^2} = \frac{d}{dx}\tan^{-1}x$, we have $f_\alpha(x) = \tan^{-1}x + c$ for some constant $c$ (on each continuous branch).

Sketch of $y = f_\alpha(x)$:

To find $c$, evaluate at $x = 0$:

$f_\alpha(0) = \tan^{-1}\!\left(\frac{1}{\tan\alpha}\right) = \tan^{-1}(\cot\alpha) = \tan^{-1}\!\left(\tan\!\left(\frac{\pi}{2} - \alpha\right)\right) = \frac{\pi}{2} - \alpha,$

since $0 < \frac{\pi}{2} - \alpha < \frac{\pi}{2}$. So $c = \frac{\pi}{2} - \alpha$ and $f_\alpha(x) = \tan^{-1}x + \frac{\pi}{2} - \alpha$ on the branch containing $x = 0$, i.e. for $x < \tan\alpha$.

There is a discontinuity at $x = \tan\alpha$. As $x \to (\tan\alpha)^-$, the argument of $\tan^{-1}$ goes to $+\infty$, so $f_\alpha(x) \to \frac{\pi}{2}$. As $x \to (\tan\alpha)^+$, the argument goes to $-\infty$, so $f_\alpha(x) \to -\frac{\pi}{2}$.

For $x > \tan\alpha$, $f_\alpha(x) = \tan^{-1}x - \frac{\pi}{2} - \alpha$ (matching the $-\frac{\pi}{2}$ jump at the discontinuity).

Additional features:

• $x$-intercept: $f_\alpha(x) = 0 \implies \tan^{-1}x = \alpha - \frac{\pi}{2} \implies x = -\cot\alpha$ (on the left branch).
• As $x \to \pm\infty$: $f_\alpha(x) \to -\alpha$.
• Value at $x = 0$: $y = \frac{\pi}{2} - \alpha$.

The graph consists of two branches of the arctangent curve (shifted vertically by $\frac{\pi}{2} - \alpha$ on the left and $-\frac{\pi}{2} - \alpha$ on the right), joined by a jump discontinuity at $x = \tan\alpha$ from $\frac{\pi}{2}$ to $-\frac{\pi}{2}$.

Sketch of $y = f_\alpha(x) - f_\beta(x)$ where $0 < \alpha < \beta < \frac{\pi}{2}$:

Since $f_\alpha'(x) = f_\beta'(x) = \frac{1}{1+x^2}$, the difference is piecewise constant:

• For $x < \tan\alpha$: $\left(\frac{\pi}{2} - \alpha\right) - \left(\frac{\pi}{2} - \beta\right) = \beta - \alpha$.
• For $\tan\alpha < x < \tan\beta$: $\left(-\frac{\pi}{2} - \alpha\right) - \left(\frac{\pi}{2} - \beta\right) = \beta - \alpha - \pi$.
• For $x > \tan\beta$: $\left(-\frac{\pi}{2} - \alpha\right) - \left(-\frac{\pi}{2} - \beta\right) = \beta - \alpha$.

The graph is a step function: constant at $\beta - \alpha$ on the left and right, dropping by $\pi$ to $\beta - \alpha - \pi$ in the middle interval, with jump discontinuities at $x = \tan\alpha$ and $x = \tan\beta$.

Part (ii)

We compute $g'(x) = \frac{d}{dx}\!\left[\tanh^{-1}(\sin x) - \sinh^{-1}(\tan x)\right]$ for $\frac{\pi}{2} < x < \frac{3\pi}{2}$.

First term:

$\frac{d}{dx}\tanh^{-1}(\sin x) = \frac{1}{1 - \sin^2 x} \cdot \cos x = \frac{\cos x}{\cos^2 x} = \sec x.$

Second term:

$\frac{d}{dx}\sinh^{-1}(\tan x) = \frac{1}{\sqrt{1 + \tan^2 x}} \cdot \sec^2 x = \frac{\sec^2 x}{\sqrt{\sec^2 x}} = \frac{\sec^2 x}{|\sec x|}.$

For $\frac{\pi}{2} < x < \frac{3\pi}{2}$, we have $\cos x < 0$, so $\sec x < 0$ and $|\sec x| = -\sec x$. Therefore:

$\frac{\sec^2 x}{|\sec x|} = \frac{\sec^2 x}{-\sec x} = -\sec x.$

Combining:

$g'(x) = \sec x - (-\sec x) = 2\sec x.$

Sketch of $y = g(x)$ for $0 \leqslant x \leqslant 2\pi$:

For $0 < x < \frac{\pi}{2}$ and $\frac{3\pi}{2} < x < 2\pi$, we have $\sec x > 0$, so $|\sec x| = \sec x$ and

$g'(x) = \sec x - \frac{\sec^2 x}{\sec x} = \sec x - \sec x = 0.$

Therefore $g(x)$ is constant on each of these intervals. Since $g(0) = \tanh^{-1}(0) - \sinh^{-1}(0) = 0$, we have $g(x) = 0$ for $0 \leqslant x < \frac{\pi}{2}$ and $\frac{3\pi}{2} < x \leqslant 2\pi$.

For $\frac{\pi}{2} < x < \frac{3\pi}{2}$, $g'(x) = 2\sec x$, so

$g(x) = 2\ln|\sec x + \tan x| + C.$

At $x = \pi$: $g(\pi) = \tanh^{-1}(0) - \sinh^{-1}(0) = 0$, and $2\ln|\sec\pi + \tan\pi| = 2\ln|{-1} + 0| = 0$, so $C = 0$.

The curve $g(x) = 2\ln|\sec x + \tan x|$ on $(\frac{\pi}{2}, \frac{3\pi}{2})$. Using the identity $\sec x + \tan x = \frac{1 + \sin x}{\cos x}$:

At $x = \pi$: $g = 2\ln|{-1} + 0| = 0$ (the midpoint).

As $x \to (\frac{\pi}{2})^+$: $1 + \sin x \to 2$ and $\cos x \to 0^-$, so $\sec x + \tan x \to -\infty$. Thus $|\sec x + \tan x| \to +\infty$ and $g(x) \to +\infty$.

As $x \to (\frac{3\pi}{2})^-$: $\sin x \to -1^+$ so $1 + \sin x \to 0^+$. $\cos x \to 0^-$ (since $\cos(\frac{3\pi}{2} - \epsilon) = -\sin\epsilon < 0$). So $\frac{1 + \sin x}{\cos x} \to \frac{0^+}{0^-} = 0^-$, meaning $|\sec x + \tan x| \to 0^+$ and $g(x) \to -\infty$.

So the curve goes from $+\infty$ at $x = (\frac{\pi}{2})^+$, decreases, passes through $0$ at $x = \pi$, then continues to $-\infty$ as $x \to (\frac{3\pi}{2})^-$.

Shape: $g'(x) = 2\sec x < 0$ on $(\frac{\pi}{2}, \frac{3\pi}{2})$, so $g$ is strictly decreasing throughout. $g''(x) = 2\sec x \tan x$: for $\frac{\pi}{2} < x < \pi$, $\sec x < 0$ and $\tan x < 0$, so $g'' > 0$ (concave up); for $\pi < x < \frac{3\pi}{2}$, $\sec x < 0$ and $\tan x > 0$, so $g'' < 0$ (concave down).

Summary of the sketch:

• On $[0, \frac{\pi}{2})$ and $(\frac{3\pi}{2}, 2\pi]$: $g(x) = 0$ (flat, horizontal line on the $x$-axis).
• On $(\frac{\pi}{2}, \frac{3\pi}{2})$: $g$ is continuous and strictly decreasing from $+\infty$ to $-\infty$, passing through $0$ at $x = \pi$. It is concave up on $(\frac{\pi}{2}, \pi)$ and concave down on $(\pi, \frac{3\pi}{2})$, forming an S-shaped curve with an inflection point at $x = \pi$. The graph is antisymmetric about the point $(\pi, 0)$, since $g(2\pi - x) = -g(x)$.

Examiner Notes

第七受欢迎（约70%作答），平均分接近8/20，排第四。考官报告指出：(1) 求导大多正确；(2) 作图常见失误：忽略 x=tan alpha 处的间断点导致曲线有两支、忘记值域为 (-pi/2, pi/2)、未标注截距/间断点/渐近线/值域；(3) (ii)中求导时负号处理错误是主要问题，导致只画出中间段而遗漏两侧常数段。

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Question 7

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

7 (i) Let $z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}} ,$ where $\theta$ and $\phi$ are real, and $\theta - \phi \neq 2n\pi$ for any integer $n$. Show that $z = i \cot \left( \frac{1}{2}(\phi - \theta) \right)$ and give expressions for the modulus and argument of $z$.

(ii) The distinct points $A$ and $B$ lie on a circle with radius 1 and centre $O$. In the complex plane, $A$ and $B$ are represented by the complex numbers $a$ and $b$, and $O$ is at the origin. The point $X$ is represented by the complex number $x$, where $x = a + b$ and $a + b \neq 0$. Show that $OX$ is perpendicular to $AB$.

If the distinct points $A$, $B$ and $C$ in the complex plane, which are represented by the complex numbers $a$, $b$ and $c$, lie on a circle with radius 1 and centre $O$, and $h = a + b + c$ represents the point $H$, then $H$ is said to be the orthocentre of the triangle $ABC$.

(iii) The distinct points $A$, $B$ and $C$ lie on a circle with radius 1 and centre $O$. In the complex plane, $A$, $B$ and $C$ are represented by the complex numbers $a$, $b$ and $c$, and $O$ is at the origin.

Show that, if the point $H$, represented by the complex number $h$, is the orthocentre of the triangle $ABC$, then either $h = a$ or $AH$ is perpendicular to $BC$.

(iv) The distinct points $A$, $B$, $C$ and $D$ (in that order, anticlockwise) all lie on a circle with radius 1 and centre $O$. The points $P$, $Q$, $R$ and $S$ are the orthocentres of the triangles $ABC$, $BCD$, $CDA$ and $DAB$, respectively. By considering the midpoint of $AQ$, show that there is a single transformation which maps the quadrilateral $ABCD$ on to the quadrilateral $QRSP$ and describe this transformation fully.

Hint

$z = \frac{e^{i\theta} + e^{i\varphi}}{e^{i\theta} - e^{i\varphi}}$ $= \frac{\cos \theta + i \sin \theta + \cos \varphi + i \sin \varphi}{\cos \theta + i \sin \theta - \cos \varphi - i \sin \varphi}$

M1

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

M1 A1 A1

$= \frac{2 \cos \frac{\theta - \varphi}{2} \left( \cos \frac{\theta + \varphi}{2} + i \sin \frac{\theta + \varphi}{2} \right)}{2 \sin \frac{\theta - \varphi}{2} \left( i \cos \frac{\theta + \varphi}{2} - \sin \frac{\theta + \varphi}{2} \right)}$ $= -i \cot \frac{\theta - \varphi}{2}$ $= i \cot \frac{\varphi - \theta}{2}$ *A1 (5)

as required.

Alternatively,

$z = \frac{e^{i\theta} + e^{i\varphi}}{e^{i\theta} - e^{i\varphi}} = \frac{e^{i\left(\frac{\theta - \varphi}{2}\right)} + e^{-i\left(\frac{\theta - \varphi}{2}\right)}}{e^{i\left(\frac{\theta - \varphi}{2}\right)} - e^{-i\left(\frac{\theta - \varphi}{2}\right)}} = \frac{2 \cos \frac{\theta - \varphi}{2}}{2i \sin \frac{\theta - \varphi}{2}} = -i \cot \frac{\theta - \varphi}{2} = i \cot \frac{\varphi - \theta}{2}$

M1

M1 A1 A1

*A1 (5)

$|z| = \left| \cot \frac{\theta - \varphi}{2} \right|$

M1 A1

$|\arg z| = \frac{\pi}{2}$

[or $\arg z = \frac{\pi}{2}$ or $\frac{3\pi}{2}$]

M1 A1 (4)

(ii) Let $a = e^{i\alpha}$ and $b = e^{i\beta}$ M1 then $x = a + b = e^{i\alpha} + e^{i\beta}$ and $AB = b - a = e^{i\beta} - e^{i\alpha}$

$\arg x - \arg AB = \arg \frac{x}{AB} = \arg \frac{e^{i\alpha} + e^{i\beta}}{e^{i\beta} - e^{i\alpha}}$

so using (i), $|\arg x - \arg AB| = \frac{\pi}{2}$ A1 and thus OX and AB are perpendicular, since $x = a + b \neq 0$ and $a \neq b$ as A and B are distinct. E1 (3)

Alternative:- $0, a, a + b, b$ define a rhombus OAXB as $|a| = |b| = 1$. Diagonals of a rhombus are perpendicular (and bisect one another).

(iii) $h = a + b + c$ so $AH = a + b + c - a = b + c$ and $BC = c - b$ and thus

$\frac{AH}{BC} = \frac{b + c}{c - b}$ B1

as $c - b \neq 0$

From (ii),

$\left| \arg \frac{AH}{BC} \right| = \frac{\pi}{2}$

so BC is perpendicular to AH E1

unless $b + c = 0$ E1 in which case $h = a$ E1 (4)

(iv) $p = a + b + c \quad q = b + c + d \quad r = c + d + a \quad s = d + a + b$

The midpoint of AQ is $\frac{a+q}{2} = \frac{a+b+c+d}{2}$ and so by its symmetry it is also the midpoint of BR, CS, and DP, B1 E1

and thus ABCD is transformed to PQRS by a rotation of $\pi$ radians about midpoint of AQ. E1 B1 (4)

Alternatively, ABCD is transformed to PQRS by an enlargement scale factor -1, centre of enlargement midpoint of AQ.

Model Solution

Part (i)

We have $z = \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} - e^{i\phi}} .$

Factor out $e^{i(\theta+\phi)/2}$ from numerator and denominator: $z = \frac{e^{i(\theta+\phi)/2}\left(e^{i(\theta-\phi)/2} + e^{-i(\theta-\phi)/2}\right)}{e^{i(\theta+\phi)/2}\left(e^{i(\theta-\phi)/2} - e^{-i(\theta-\phi)/2}\right)} .$

Using $e^{i\alpha} + e^{-i\alpha} = 2\cos\alpha$ and $e^{i\alpha} - e^{-i\alpha} = 2i\sin\alpha$: $z = \frac{2\cos\frac{\theta-\phi}{2}}{2i\sin\frac{\theta-\phi}{2}} = \frac{1}{i}\cot\frac{\theta-\phi}{2} = -i\cot\frac{\theta-\phi}{2} = i\cot\frac{\phi-\theta}{2} .$

This is the required result.

Since $z = i\cot\frac{\phi-\theta}{2}$ is purely imaginary: $|z| = \left|\cot\frac{\phi-\theta}{2}\right| .$

For the argument: when $\cot\frac{\phi-\theta}{2} > 0$, we have $z$ on the positive imaginary axis, so $\arg z = \frac{\pi}{2}$; when $\cot\frac{\phi-\theta}{2} < 0$, $z$ is on the negative imaginary axis, so $\arg z = \frac{3\pi}{2}$. In summary: $\arg z = \frac{\pi}{2} \quad \text{or} \quad \frac{3\pi}{2} ,$ equivalently $|\arg z| = \frac{\pi}{2}$ (mod $\pi$).

Part (ii)

Since $A$ and $B$ lie on the unit circle centred at the origin, we can write $a = e^{i\alpha}$ and $b = e^{i\beta}$ for some real $\alpha, \beta$ with $\alpha \neq \beta$ (since $A$ and $B$ are distinct).

The direction of $OX$ is represented by $x = a + b = e^{i\alpha} + e^{i\beta}$, and the direction of $AB$ is represented by $b - a = e^{i\beta} - e^{i\alpha}$.

Consider the ratio: $\frac{x}{b - a} = \frac{e^{i\alpha} + e^{i\beta}}{e^{i\beta} - e^{i\alpha}} .$

Setting $\theta = \alpha$ and $\phi = \beta$ in the result of part (i): $\frac{x}{b - a} = i\cot\frac{\beta - \alpha}{2} .$

Since $\alpha \neq \beta$ (as $A$ and $B$ are distinct), $\cot\frac{\beta-\alpha}{2}$ is a nonzero real number, so the ratio is purely imaginary. Therefore: $\arg x - \arg(b - a) = \pm\frac{\pi}{2} ,$

which means $OX$ is perpendicular to $AB$.

(We can divide by $b - a$ since $a \neq b$; and $x = a + b \neq 0$ by hypothesis.)

Part (iii)

The orthocentre is $H$ with $h = a + b + c$.

We need to show either $h = a$ or $AH$ is perpendicular to $BC$.

The vector from $A$ to $H$ is: $h - a = (a + b + c) - a = b + c .$

The vector from $B$ to $C$ is: $c - b .$

If $b + c = 0$, then $h - a = 0$, so $h = a$.

If $b + c \neq 0$, consider: $\frac{h - a}{c - b} = \frac{b + c}{c - b} .$

Since $b$ and $c$ lie on the unit circle, we can set $b = e^{i\beta}$, $c = e^{i\gamma}$ and apply part (ii) with $\alpha = \beta$, $\phi = \gamma$: $\frac{b + c}{c - b} = \frac{e^{i\beta} + e^{i\gamma}}{e^{i\gamma} - e^{i\beta}} = i\cot\frac{\gamma - \beta}{2} ,$

which is purely imaginary (and nonzero since $b \neq c$). Therefore: $\arg(h - a) - \arg(c - b) = \pm\frac{\pi}{2} ,$

so $AH$ is perpendicular to $BC$.

Part (iv)

The orthocentres are: $p = a + b + c, \quad q = b + c + d, \quad r = c + d + a, \quad s = d + a + b .$

The midpoint of $AQ$ is: $\frac{a + q}{2} = \frac{a + b + c + d}{2} .$

Similarly: $\frac{b + r}{2} = \frac{b + c + d + a}{2} = \frac{a + b + c + d}{2} ,$ $\frac{c + s}{2} = \frac{a + b + c + d}{2} ,$ $\frac{d + p}{2} = \frac{a + b + c + d}{2} .$

All four midpoints coincide at the single point $M$ represented by $m = \frac{a+b+c+d}{2}$.

This means $M$ is the midpoint of each of the segments $AQ$, $BR$, $CS$, and $DP$. Therefore the mapping $A \mapsto Q$, $B \mapsto R$, $C \mapsto S$, $D \mapsto P$ is a half-turn (rotation by $\pi$ radians) about the point $M$, i.e., the point with complex coordinate $\frac{a+b+c+d}{2}$.

This is a single isometry: it maps $ABCD$ to $QRSP$.

Examiner Notes

纯数题中最不成功的（平均分低于6/20）。考官报告指出：(1) (i)一般表现良好，但模的表达式漏绝对值、辐角漏第二可能值是常见扣分；(2) (ii)用(i)的结果时除以非零量未加说明、向量法和复数法混淆不清、斜率法未处理水平/垂直特殊情况；(3) (iii)多数人能正确应用(ii)，但遗漏 b+c=0 时 h=a 的情况；(4) (iv)不少考生未作答，常见错误是将变换描述为关于点的反射（这不是单一变换），中点计算也常出错。

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Question 8

Topic: 纯数  |  Difficulty: Hard  |  Marks: 20

Problem

8 A sequence $x_1, x_2, \dots$ of real numbers is defined by $x_{n+1} = x_n^2 - 2$ for $n \geqslant 1$ and $x_1 = a$.

(i) Show that if $a > 2$ then $x_n \geqslant 2 + 4^{n-1}(a - 2)$.

(ii) Show also that $x_n \to \infty$ as $n \to \infty$ if and only if $|a| > 2$.

(iii) When $a > 2$, a second sequence $y_1, y_2, \dots$ is defined by

$y_n = \frac{Ax_1x_2 \cdots x_n}{x_{n+1}},$

where $A$ is a positive constant and $n \geqslant 1$.

Prove that, for a certain value of $a$, with $a > 2$, which you should find in terms of $A$,

$y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}}$

for all $n \geqslant 1$.

Determine whether, for this value of $a$, the second sequence converges.

Hint

(i) Suppose $x_k \geq 2 + 4^{k-1}(a - 2)$ for some particular integer k (and this is positive as $a > 2$)

E1

Then $x_{k+1} = {x_k}^2 - 2 \geq [2 + 4^{k-1}(a - 2)]^2 - 2 = 4 + 4^k(a - 2) + 4^{2k-2}(a - 2)^2 - 2$ $= 2 + 4^k(a - 2) + 4^{2k-2}(a - 2)^2$ $> 2 + 4^k(a - 2)$

M1 A1

which is the required result for $k + 1$.

For $n = 1$, $2 + 4^{n-1}(a - 2) = 2 + a - 2 = a$ so in this case, $x_n = 2 + 4^{n-1}(a - 2)$ B1 and thus by induction $x_n \geq 2 + 4^{n-1}(a - 2)$ for positive integer n. E1 (5)

(ii) If $|x_k| \leq 2$, then $0 \leq |x_k|^2 \leq 4$, so $-2 \leq |x_k|^2 - 2 \leq 2$, that is $-2 \leq x_{k+1} \leq 2$. M1A1

If $|a| \leq 2$, $|x_1| \leq 2$ and thus by induction $-2 \leq x_n \leq 2$, that is $x_n \not\to \infty$ E1

Whether $a = \pm\alpha$, $x_2$ would equal the same value, namely $\alpha^2 - 2$. E1

So to consider $|a| \geq 2$, we only need consider $a > 2$ to discuss the behaviour of all terms after the first. Therefore, from part (i), we know $x_n \geq 2 + 4^{n-1}(|a| - 2)$ for $n \geq 2$, and thus $x_n \to \infty$ as $n \to \infty$; B1 hence we have shown $x_n \to \infty$ as $n \to \infty$ if and only if $|a| \geq 2$. (5)

(iii)

$y_k = \frac{Ax_1x_2 \cdots x_k}{x_{k+1}}$

$y_{k+1} = \frac{Ax_1x_2 \cdots x_{k+1}}{x_{k+2}} = \frac{{x_{k+1}}^2}{x_{k+2}}y_k$

M1

Suppose that

$y_k = \frac{\sqrt{{x_{k+1}}^2 - 4}}{x_{k+1}}$

for some positive integer k, E1 then

$y_{k+1} = \frac{{x_{k+1}}^2}{x_{k+2}} \frac{\sqrt{{x_{k+1}}^2 - 4}}{x_{k+1}} = \frac{x_{k+1}\sqrt{{x_{k+1}}^2 - 4}}{x_{k+2}}$

As $x_{k+2} = {x_{k+1}}^2 - 2$, $x_{k+1} = \sqrt{x_{k+2} + 2}$, and $\sqrt{{x_{k+1}}^2 - 4} = \sqrt{x_{k+2} - 2}$,

and thus,

$y_{k+1} = \frac{\sqrt{x_{k+2} + 2}\sqrt{x_{k+2} - 2}}{x_{k+2}} = \frac{\sqrt{{x_{k+2}}^2 - 4}}{x_{k+2}}$

M1 A1

which is the required result for $k + 1$.

$y_1 = \frac{Ax_1}{x_2}$

and also we wish to have

$y_1 = \frac{\sqrt{{x_2}^2 - 4}}{x_2}$

M1

then $Ax_1 = \sqrt{{x_2}^2 - 4}$ , that is $A^2{x_1}^2 = {x_2}^2 - 4$ , and as $x_1 = a$ , $x_2 = {x_1}^2 - 2 = a^2 - 2$

so $A^2a^2 = (a^2 - 2)^2 - 4 = a^4 - 4a^2$ , $A^2 = a^2 - 4$ , and thus $a = \sqrt{A^2 + 4}$ , as $a \neq 0$ nor $-\sqrt{A^2 + 4}$ because $a > 2$. A1 E1

So as the result is true for $y_1$ , and we have shown it to be true for $y_{k+1}$ if it is true for $y_k$ , it is true by induction for all positive integer $n$ that

$y_n = \frac{\sqrt{{x_{n+1}}^2 - 4}}{x_{n+1}}$

E1 (8)

As $a > 2$ from (ii) $x_n \rightarrow \infty$ as $n \rightarrow \infty$ M1 and thus using result just proved, $y_n \rightarrow 1$ as $n \rightarrow \infty$ , i.e. the sequence converges. *A1 (2)

Model Solution

Part (i)

We prove by induction on $n$ that $x_n \geqslant 2 + 4^{n-1}(a - 2)$ for all $n \geqslant 1$, given $a > 2$.

Base case ($n = 1$): $x_1 = a$ and $2 + 4^{0}(a - 2) = 2 + a - 2 = a$, so $x_1 = a = 2 + 4^{0}(a - 2)$. The result holds with equality.

Inductive step: Suppose $x_k \geqslant 2 + 4^{k-1}(a - 2)$ for some $k \geqslant 1$. Since $a > 2$, the right-hand side satisfies $2 + 4^{k-1}(a-2) > 2 > 0$, so $x_k > 0$ and in particular $x_k \geqslant 2$.

Now: $x_{k+1} = x_k^2 - 2 \geqslant \left(2 + 4^{k-1}(a - 2)\right)^2 - 2 .$

Expanding the square: $\left(2 + 4^{k-1}(a - 2)\right)^2 = 4 + 2 \cdot 2 \cdot 4^{k-1}(a - 2) + 4^{2(k-1)}(a - 2)^2 = 4 + 4^{k}(a - 2) + 4^{2k - 2}(a - 2)^2 .$

Therefore: $x_{k+1} \geqslant 4 + 4^{k}(a - 2) + 4^{2k - 2}(a - 2)^2 - 2 = 2 + 4^{k}(a - 2) + 4^{2k - 2}(a - 2)^2 .$

Since $a > 2$, the term $4^{2k-2}(a-2)^2 > 0$, so: $x_{k+1} > 2 + 4^{k}(a - 2) .$

This is the required result for $n = k + 1$. By induction, $x_n \geqslant 2 + 4^{n-1}(a - 2)$ for all $n \geqslant 1$.

Part (ii)

Forward direction: We show that if $|a| \leqslant 2$ then $x_n$ does not tend to infinity.

If $|a| \leqslant 2$, then $x_1 = a$ satisfies $|x_1| \leqslant 2$.

Suppose $|x_k| \leqslant 2$ for some $k \geqslant 1$. Then $0 \leqslant x_k^2 \leqslant 4$, so: $-2 \leqslant x_k^2 - 2 \leqslant 2 ,$ that is, $-2 \leqslant x_{k+1} \leqslant 2$.

By induction, $-2 \leqslant x_n \leqslant 2$ for all $n \geqslant 1$, so the sequence is bounded and $x_n \not\to \infty$.

Converse: We show that if $|a| > 2$ then $x_n \to \infty$.

First observe that $x_2 = a^2 - 2$. Since $|a| > 2$, we have $a^2 > 4$, so $x_2 = a^2 - 2 > 2$.

Also, $x_3 = x_2^2 - 2$. Since $x_2 > 2 > 0$, the subsequent terms $x_3, x_4, \ldots$ depend only on $x_2$. By part (i) applied with $a$ replaced by $x_2$ (which is $> 2$): $x_n \geqslant 2 + 4^{n-2}(x_2 - 2) \quad \text{for } n \geqslant 2 .$

Since $x_2 - 2 > 0$, the right-hand side tends to infinity as $n \to \infty$, so $x_n \to \infty$.

Combining both directions: $x_n \to \infty$ as $n \to \infty$ if and only if $|a| > 2$.

Part (iii)

We seek $a > 2$ such that $y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}}$ for all $n \geqslant 1$, where $y_n = \frac{Ax_1 x_2 \cdots x_n}{x_{n+1}}$.

Base case ($n = 1$): We need: $\frac{Ax_1}{x_2} = \frac{\sqrt{x_2^2 - 4}}{x_2} .$

Since $a > 2$ implies $x_2 = a^2 - 2 > 2 > 0$, both sides have the same (positive) denominator, so this simplifies to: $Ax_1 = \sqrt{x_2^2 - 4} .$

Squaring both sides (both sides are positive): $A^2 x_1^2 = x_2^2 - 4 .$

With $x_1 = a$ and $x_2 = a^2 - 2$: $A^2 a^2 = (a^2 - 2)^2 - 4 = a^4 - 4a^2 + 4 - 4 = a^4 - 4a^2 = a^2(a^2 - 4) .$

Since $a > 2$ implies $a \neq 0$, we can divide by $a^2$: $A^2 = a^2 - 4 \implies a^2 = A^2 + 4 \implies a = \sqrt{A^2 + 4} ,$

taking the positive root since $a > 0$.

Note $a = \sqrt{A^2 + 4} > \sqrt{4} = 2$, so the condition $a > 2$ is satisfied.

Inductive step: Suppose $y_k = \frac{\sqrt{x_{k+1}^2 - 4}}{x_{k+1}}$ for some $k \geqslant 1$.

From the definitions: $y_{k+1} = \frac{Ax_1 x_2 \cdots x_{k+1}}{x_{k+2}} = \frac{x_{k+1}^2}{x_{k+2}} \cdot \frac{Ax_1 x_2 \cdots x_k}{x_{k+1}} = \frac{x_{k+1}^2}{x_{k+2}} \cdot y_k .$

Using the inductive hypothesis: $y_{k+1} = \frac{x_{k+1}^2}{x_{k+2}} \cdot \frac{\sqrt{x_{k+1}^2 - 4}}{x_{k+1}} = \frac{x_{k+1}\sqrt{x_{k+1}^2 - 4}}{x_{k+2}} .$

Now, from the recurrence $x_{k+2} = x_{k+1}^2 - 2$: $x_{k+1}^2 = x_{k+2} + 2 \implies x_{k+1} = \sqrt{x_{k+2} + 2} \quad (\text{since } x_{k+1} > 0) ,$ $x_{k+1}^2 - 4 = x_{k+2} - 2 \implies \sqrt{x_{k+1}^2 - 4} = \sqrt{x_{k+2} - 2} \quad (\text{since } x_{k+2} > 2) .$

Substituting: $y_{k+1} = \frac{\sqrt{x_{k+2} + 2} \cdot \sqrt{x_{k+2} - 2}}{x_{k+2}} = \frac{\sqrt{(x_{k+2} + 2)(x_{k+2} - 2)}}{x_{k+2}} = \frac{\sqrt{x_{k+2}^2 - 4}}{x_{k+2}} .$

This is the required result for $k + 1$. By induction, $y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}}$ for all $n \geqslant 1$, with $a = \sqrt{A^2 + 4}$.

Convergence: Since $a = \sqrt{A^2 + 4} > 2$, by part (ii) we have $x_n \to \infty$ as $n \to \infty$. Therefore $x_{n+1} \to \infty$, and: $y_n = \frac{\sqrt{x_{n+1}^2 - 4}}{x_{n+1}} = \sqrt{1 - \frac{4}{x_{n+1}^2}} \to \sqrt{1 - 0} = 1 \quad \text{as } n \to \infty .$

The sequence $(y_n)$ converges, and its limit is $1$.

Examiner Notes

第五受欢迎（77%作答），平均分约6.5/20。考官报告指出：(1) 归纳法执行一般较好，但逻辑细节常丢失分；(2) (i)中平方不等式需注意下界非负性，几乎无人注意到；基始步骤也有困难（有人误认为 4^0=0）；(3) (ii)中证明 |a|<2 时序列有界，常见错误是未排除 x_2<-2 导致发散的情况；(4) (iii)多数人从目标反推 a 值，但未验证 a>2 且步骤可逆（因为可能除以零）；收敛性判断大多非正式但正确。