STEP3 2003 -- Pure Mathematics

STEP3 2003 — Section A (Pure Mathematics)

Exam: STEP3  |  Year: 2003  |  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	微积分 Calculus	Standard	链式法则,反函数求导,换元积分
2	级数与组合数学 Series & Combinatorics	Challenging	组合恒等式,广义二项式定理,幂级数求和
3	微积分 Calculus	Standard	商法则求导,奇偶函数性质,函数图像分析
4	解析几何与参数方程 Analytic Geometry & Parametric Equations	Challenging	参数方程求切线,递推关系,三角恒等式
5	代数与方程理论 Algebra & Equation Theory	Challenging	判别式,韦达定理,极值分析,充要条件
6	三角函数 Trigonometry	Challenging	三角恒等式证明,裂项求和,三角方程求解
7	坐标几何 Coordinate Geometry	Challenging	坐标几何,向量投影,共线条件
8	微分方程 Differential Equations	Challenging	隐函数求导,变量替换,微分方程求解

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Question 1

Topic: 微积分 Calculus  |  Difficulty: Standard  |  Marks: 20

Problem

1 Given that $x + a > 0$ and $x + b > 0$, and that $b > a$, show that $\frac{\mathrm{d}}{\mathrm{d}x} \arcsin \left( \frac{x + a}{x + b} \right) = \frac{\sqrt{b - a}}{(x + b) \sqrt{a + b + 2x}}$ and find $\frac{\mathrm{d}}{\mathrm{d}x} \operatorname{arcosh} \left( \frac{x + b}{x + a} \right)$.

Hence, or otherwise, integrate, for $x > -1$,

(i) $\int \frac{1}{(x + 1) \sqrt{x + 3}} \mathrm{~d}x$,

(ii) $\int \frac{1}{(x + 3) \sqrt{x + 1}} \mathrm{~d}x$.

[You may use the results $\frac{\mathrm{d}}{\mathrm{d}x} \arcsin x = \frac{1}{\sqrt{1 - x^2}}$ and $\frac{\mathrm{d}}{\mathrm{d}x} \operatorname{arcosh} x = \frac{1}{\sqrt{x^2 - 1}}$. ]

Model Solution

Step 1: Differentiate $\arcsin\!\left(\frac{x+a}{x+b}\right)$.

Let $u = \frac{x+a}{x+b}$. Then by the chain rule:

$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin u = \frac{1}{\sqrt{1-u^2}}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}.$

We compute $u$ and its derivative:

$\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{(x+b)-(x+a)}{(x+b)^2} = \frac{b-a}{(x+b)^2}.$

For $1 - u^2$:

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

Since $b > a$, $x+a > 0$, and $x+b > 0$, we have $b-a > 0$ and $2x+a+b > 0$, so:

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

Therefore:

$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin\!\left(\frac{x+a}{x+b}\right) = \frac{x+b}{\sqrt{(b-a)(2x+a+b)}}\cdot\frac{b-a}{(x+b)^2} = \frac{b-a}{(x+b)\sqrt{(b-a)(2x+a+b)}}.$

Simplifying:

$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin\!\left(\frac{x+a}{x+b}\right) = \frac{\sqrt{b-a}}{(x+b)\sqrt{2x+a+b}}. \qquad \text{(shown)}$

Step 2: Differentiate $\operatorname{arcosh}\!\left(\frac{x+b}{x+a}\right)$.

Let $v = \frac{x+b}{x+a}$. Since $b > a$ and $x+a > 0$, we have $v > 1$, so $\operatorname{arcosh}$ is well-defined.

$\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{(x+a)-(x+b)}{(x+a)^2} = \frac{a-b}{(x+a)^2}.$

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

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

Therefore:

$\frac{\mathrm{d}}{\mathrm{d}x}\operatorname{arcosh}\!\left(\frac{x+b}{x+a}\right) = \frac{1}{\sqrt{v^2-1}}\cdot\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{x+a}{\sqrt{(b-a)(2x+a+b)}}\cdot\frac{a-b}{(x+a)^2}.$

$= \frac{a-b}{(x+a)\sqrt{(b-a)(2x+a+b)}} = \frac{-(b-a)}{(x+a)\sqrt{(b-a)(2x+a+b)}} = \frac{-\sqrt{b-a}}{(x+a)\sqrt{2x+a+b}}.$

So $\frac{\mathrm{d}}{\mathrm{d}x}\operatorname{arcosh}\!\left(\frac{x+b}{x+a}\right) = \frac{-\sqrt{b-a}}{(x+a)\sqrt{a+b+2x}}.$

Step 3: Integral (i).

We need $\int \frac{1}{(x+1)\sqrt{x+3}}\,\mathrm{d}x$ for $x > -1$.

Substitute $u = \sqrt{x+3}$, so $x = u^2 - 3$ and $\mathrm{d}x = 2u\,\mathrm{d}u$:

$\int \frac{1}{(x+1)\sqrt{x+3}}\,\mathrm{d}x = \int \frac{2u}{(u^2 - 2)\cdot u}\,\mathrm{d}u = \int \frac{2}{u^2 - 2}\,\mathrm{d}u.$

For $x > -1$ we have $u > \sqrt{2}$, so $u^2 > 2$. Using the standard result $\int \frac{1}{u^2 - a^2}\,\mathrm{d}u = \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|$:

$\int \frac{2}{u^2-2}\,\mathrm{d}u = \frac{2}{2\sqrt{2}}\ln\frac{u-\sqrt{2}}{u+\sqrt{2}} = \frac{1}{\sqrt{2}}\ln\frac{\sqrt{x+3}-\sqrt{2}}{\sqrt{x+3}+\sqrt{2}} + C.$

Step 4: Integral (ii).

We need $\int \frac{1}{(x+3)\sqrt{x+1}}\,\mathrm{d}x$ for $x > -1$.

Substitute $u = \sqrt{x+1}$, so $x = u^2 - 1$ and $\mathrm{d}x = 2u\,\mathrm{d}u$:

$\int \frac{1}{(x+3)\sqrt{x+1}}\,\mathrm{d}x = \int \frac{2u}{(u^2 + 2)\cdot u}\,\mathrm{d}u = \int \frac{2}{u^2 + 2}\,\mathrm{d}u = \frac{2}{\sqrt{2}}\arctan\frac{u}{\sqrt{2}}.$

$= \sqrt{2}\arctan\frac{\sqrt{x+1}}{\sqrt{2}} + C = \sqrt{2}\arctan\sqrt{\frac{x+1}{2}} + C.$

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Question 2

Topic: 级数与组合数学 Series & Combinatorics  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Show that ${}^{2r}C_r = \frac{1 \times 3 \times \cdots \times (2r - 1)}{r!} \times 2^r$, for $r \geqslant 1$.

(i) Give the first four terms of the binomial series for $(1 - p)^{-\frac{1}{2}}$.

By choosing a suitable value for $p$ in this series, or otherwise, show that $\sum_{r=0}^{\infty} \frac{{}^{2r}C_r}{8^r} = \sqrt{2} .$

(ii) Show that $\sum_{r=0}^{\infty} \frac{(2r + 1) \cdot {}^{2r}C_r}{5^r} = (\sqrt{5})^3 .$

[Note: ${}^nC_r$ is an alternative notation for $\begin{pmatrix} n \\ r \end{pmatrix}$ for $r \geqslant 1$, and ${}^0C_0 = 1$.]

Model Solution

Step 1: Show ${}^{2r}C_r = \frac{1\times 3\times\cdots\times(2r-1)}{r!}\times 2^r$.

${}^{2r}C_r = \frac{(2r)!}{r!\,r!} = \frac{1\times 2\times 3\times\cdots\times(2r)}{(r!)(r!)}.$

Separate the numerator into odd and even factors:

$1\times 2\times 3\times\cdots\times(2r) = \bigl[1\times 3\times 5\times\cdots\times(2r-1)\bigr]\times\bigl[2\times 4\times 6\times\cdots\times 2r\bigr].$

The even part factors as $2\times 4\times 6\times\cdots\times 2r = 2^r(1\times 2\times 3\times\cdots\times r) = 2^r\,r!$. Therefore:

${}^{2r}C_r = \frac{\bigl[1\times 3\times\cdots\times(2r-1)\bigr]\times 2^r\,r!}{r!\,r!} = \frac{1\times 3\times\cdots\times(2r-1)}{r!}\times 2^r. \qquad \text{(shown)}$

Part (i): Binomial series for $(1-p)^{-1/2}$ and the sum.

The generalised binomial series gives:

$(1-p)^{-1/2} = \sum_{r=0}^{\infty}\binom{-1/2}{r}(-p)^r.$

Computing the first four terms:

$\binom{-1/2}{0} = 1, \qquad \binom{-1/2}{1}(-p) = \tfrac{1}{2}p,$

$\binom{-1/2}{2}(-p)^2 = \frac{(-1/2)(-3/2)}{2!}p^2 = \frac{3}{8}p^2, \qquad \binom{-1/2}{3}(-p)^3 = \frac{(-1/2)(-3/2)(-5/2)}{3!}(-p^3) = \frac{5}{16}p^3.$

So the first four terms are:

$(1-p)^{-1/2} = 1 + \frac{1}{2}p + \frac{3}{8}p^2 + \frac{5}{16}p^3 + \cdots$

In general, $\binom{-1/2}{r}(-1)^r = \frac{1\times 3\times\cdots\times(2r-1)}{2^r\,r!}$, so:

$(1-p)^{-1/2} = \sum_{r=0}^{\infty}\frac{1\times 3\times\cdots\times(2r-1)}{2^r\,r!}\,p^r.$

Using the identity from Step 1: $\frac{1\times 3\times\cdots\times(2r-1)}{r!} = \frac{{}^{2r}C_r}{2^r}$, so each term becomes:

$\frac{{}^{2r}C_r}{2^r\cdot 2^r}\,p^r = \frac{{}^{2r}C_r}{4^r}\,p^r.$

Thus:

$(1-p)^{-1/2} = \sum_{r=0}^{\infty}\frac{{}^{2r}C_r}{4^r}\,p^r.$

Setting $p = \frac{1}{2}$:

$(1-\tfrac{1}{2})^{-1/2} = \left(\tfrac{1}{2}\right)^{-1/2} = \sqrt{2} = \sum_{r=0}^{\infty}\frac{{}^{2r}C_r}{4^r}\cdot\frac{1}{2^r} = \sum_{r=0}^{\infty}\frac{{}^{2r}C_r}{8^r}. \qquad \text{(shown)}$

Part (ii): Show $\sum_{r=0}^{\infty}\frac{(2r+1)\cdot{}^{2r}C_r}{5^r} = (\sqrt{5})^3$.

From Part (i) we have $(1-p)^{-1/2} = \sum_{r=0}^{\infty}\frac{{}^{2r}C_r}{4^r}\,p^r$.

Differentiate both sides with respect to $p$:

$\frac{1}{2}(1-p)^{-3/2} = \sum_{r=1}^{\infty}\frac{{}^{2r}C_r}{4^r}\,r\,p^{r-1} = \sum_{r=0}^{\infty}\frac{{}^{2(r+1)}C_{r+1}}{4^{r+1}}(r+1)\,p^r.$

Multiply through by $p$:

$\frac{p}{2}(1-p)^{-3/2} = \sum_{r=0}^{\infty}\frac{{}^{2r+2}C_{r+1}}{4^{r+1}}(r+1)\,p^{r+1} = \sum_{r=1}^{\infty}\frac{{}^{2r}C_r}{4^r}\cdot r\cdot p^r.$

This gives us a series with factor $r$, but we need $(2r+1)$. Instead, multiply the original series by $(1-p)^{-1}$ or use a different approach.

Consider $(1-p)^{-3/2}$. By the binomial series:

$(1-p)^{-3/2} = \sum_{r=0}^{\infty}\binom{-3/2}{r}(-p)^r.$

We have $\binom{-3/2}{r}(-1)^r = \frac{3\times 5\times\cdots\times(2r+1)}{2^r\,r!} = \frac{(2r+1)!!}{2^r\,r!}$.

Now $(2r+1)!! = 1\times 3\times\cdots\times(2r+1) = (2r+1)\times[1\times 3\times\cdots\times(2r-1)]$, so:

$\binom{-3/2}{r}(-1)^r = \frac{(2r+1)\times[1\times 3\times\cdots\times(2r-1)]}{2^r\,r!}.$

Using the identity $\frac{1\times 3\times\cdots\times(2r-1)}{r!} = \frac{{}^{2r}C_r}{2^r}$:

$\binom{-3/2}{r}(-1)^r = \frac{(2r+1)\,{}^{2r}C_r}{4^r}.$

Therefore:

$(1-p)^{-3/2} = \sum_{r=0}^{\infty}\frac{(2r+1)\,{}^{2r}C_r}{4^r}\,p^r.$

Setting $p = \frac{1}{5}$:

$(1-\tfrac{1}{5})^{-3/2} = \left(\tfrac{4}{5}\right)^{-3/2} = \left(\tfrac{5}{4}\right)^{3/2} = \frac{5\sqrt{5}}{8}.$

And the series gives:

$\sum_{r=0}^{\infty}\frac{(2r+1)\,{}^{2r}C_r}{4^r}\cdot\frac{1}{5^r} = \sum_{r=0}^{\infty}\frac{(2r+1)\,{}^{2r}C_r}{20^r}.$

This gives $\sum\frac{(2r+1){}^{2r}C_r}{20^r} = \frac{5\sqrt{5}}{8}$, not $\sum\frac{(2r+1){}^{2r}C_r}{5^r}$.

We need $\frac{1}{5^r}$ not $\frac{1}{20^r}$, so multiply both sides by $4^r$ inside the sum. Let us set $p = \frac{4}{5}$ instead:

$(1-\tfrac{4}{5})^{-3/2} = \left(\tfrac{1}{5}\right)^{-3/2} = 5^{3/2} = 5\sqrt{5}.$

The series gives:

$\sum_{r=0}^{\infty}\frac{(2r+1)\,{}^{2r}C_r}{4^r}\left(\frac{4}{5}\right)^r = \sum_{r=0}^{\infty}\frac{(2r+1)\,{}^{2r}C_r}{5^r}.$

Therefore:

$\sum_{r=0}^{\infty}\frac{(2r+1)\,{}^{2r}C_r}{5^r} = 5\sqrt{5} = (\sqrt{5})^3. \qquad \text{(shown)}$

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Question 3

Topic: 微积分 Calculus  |  Difficulty: Standard  |  Marks: 20

Problem

3 If $m$ is a positive integer, show that $(1+x)^m + (1-x)^m \neq 0$ for any real $x$.

The function $f$ is defined by

$f(x) = \frac{(1+x)^m - (1-x)^m}{(1+x)^m + (1-x)^m} .$

Find and simplify an expression for $f'(x)$.

In the case $m=5$, sketch the curves $y = f(x)$ and $y = \frac{1}{f(x)}$.

Model Solution

Step 1: Show $(1+x)^m + (1-x)^m \neq 0$ for any real $x$.

We consider two cases based on the parity of $m$.

Case 1: $m$ even. Then $(1+x)^m \geq 0$ and $(1-x)^m \geq 0$ for all real $x$. For the sum to be zero, we would need both terms to be zero simultaneously, requiring $x = -1$ and $x = 1$ at the same time. This is impossible, so $(1+x)^m + (1-x)^m > 0$.

Case 2: $m$ odd. Suppose $(1+x)^m + (1-x)^m = 0$. Then $(1+x)^m = -(1-x)^m$. Since $m$ is odd, $-(1-x)^m = (x-1)^m$, so $(1+x)^m = (x-1)^m$. Taking $m$-th roots: $1 + x = x - 1$, giving $1 = -1$, a contradiction.

Therefore $(1+x)^m + (1-x)^m \neq 0$ for all real $x$. $\qquad \text{(shown)}$

Step 2: Find and simplify $f'(x)$.

Let $u = (1+x)^m - (1-x)^m$ and $v = (1+x)^m + (1-x)^m$. Then:

$u' = m(1+x)^{m-1} - m(1-x)^{m-1} \cdot (-1) = m(1+x)^{m-1} + m(1-x)^{m-1},$

$v' = m(1+x)^{m-1} + m(1-x)^{m-1} \cdot (-1) = m(1+x)^{m-1} - m(1-x)^{m-1}.$

By the quotient rule $f'(x) = \frac{u'v - uv'}{v^2}$, we compute the numerator:

$u'v = m\bigl[(1+x)^{m-1} + (1-x)^{m-1}\bigr]\bigl[(1+x)^m + (1-x)^m\bigr]$

$= m\bigl[(1+x)^{2m-1} + (1+x)^{m-1}(1-x)^m + (1-x)^{m-1}(1+x)^m + (1-x)^{2m-1}\bigr]$

$uv' = m\bigl[(1+x)^m - (1-x)^m\bigr]\bigl[(1+x)^{m-1} - (1-x)^{m-1}\bigr]$

$= m\bigl[(1+x)^{2m-1} - (1+x)^m(1-x)^{m-1} - (1-x)^m(1+x)^{m-1} + (1-x)^{2m-1}\bigr]$

Subtracting:

$u'v - uv' = m\bigl[2(1+x)^{m-1}(1-x)^m + 2(1-x)^{m-1}(1+x)^m\bigr]$

$= 2m(1+x)^{m-1}(1-x)^{m-1}\bigl[(1-x) + (1+x)\bigr]$

$= 2m(1+x)^{m-1}(1-x)^{m-1} \cdot 2 = 4m(1+x)^{m-1}(1-x)^{m-1}.$

Therefore:

$\boxed{f'(x) = \frac{4m\,(1+x)^{m-1}(1-x)^{m-1}}{\bigl[(1+x)^m + (1-x)^m\bigr]^2}}.$

Step 3: Sketch for $m = 5$.

With $m = 5$, expanding $(1+x)^5$ and $(1-x)^5$:

$(1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$

$(1-x)^5 = 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5$

So $(1+x)^5 - (1-x)^5 = 10x + 20x^3 + 2x^5$ and $(1+x)^5 + (1-x)^5 = 2 + 20x^2 + 10x^4$. Thus:

$f(x) = \frac{2x(x^4 + 10x^2 + 5)}{2(5x^4 + 10x^2 + 1)} = \frac{x(x^4 + 10x^2 + 5)}{5x^4 + 10x^2 + 1}.$

Key features of $y = f(x)$:

• Odd function: $f(-x) = -f(x)$, so the graph has rotational symmetry about the origin.
• $f(0) = 0$, and the gradient at the origin is $f'(0) = \frac{4 \cdot 5 \cdot 1 \cdot 1}{2^2} = 5$.
• $f(1) = \frac{16}{16} = 1$ and $f(-1) = -1$.
• Since $(1-x^2)^4 \geq 0$ for all $x$ and the denominator is always positive, $f'(x) \geq 0$ for all $x$, with equality only at $x = \pm 1$. So $f$ is strictly increasing.
• As $x \to \infty$: $f(x) \approx \frac{2x^5}{10x^4} = \frac{x}{5}$, so the curve approaches the oblique asymptote $y = x/5$.

For $y = \frac{1}{f(x)}$:

$\frac{1}{f(x)} = \frac{5x^4 + 10x^2 + 1}{x(x^4 + 10x^2 + 5)}.$

Key features:

• Also an odd function: $\frac{1}{f(-x)} = -\frac{1}{f(x)}$.
• Vertical asymptote at $x = 0$: as $x \to 0^+$, $f(x) \to 0^+$, so $1/f(x) \to +\infty$.
• $\frac{1}{f(1)} = 1$ and $\frac{1}{f(-1)} = -1$, so the two curves intersect at $(\pm 1, \pm 1)$.
• As $x \to \infty$: $1/f(x) \approx \frac{5x^4}{2x^5} = \frac{5}{2x} \to 0^+$, so the positive $x$-axis is a horizontal asymptote.
• Since $f$ is strictly increasing and positive for $x > 0$, $1/f$ is strictly decreasing for $x > 0$.

Sketch description: $y = f(x)$ is a monotonically increasing S-curve through the origin with gradient 5, passing through $(\pm 1, \pm 1)$, and approaching $y = x/5$ for large $|x|$. $y = 1/f(x)$ is an odd curve with a vertical asymptote at $x = 0$, decreasing from $+\infty$ to $0$ for $x > 0$, crossing $y = f(x)$ at $(1, 1)$ and $(-1, -1)$.

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Question 4

Topic: 解析几何与参数方程 Analytic Geometry & Parametric Equations  |  Difficulty: Challenging  |  Marks: 20

Problem

4 A curve is defined parametrically by

$x = t^2, \quad y = t(1+t^2) .$

The tangent at the point with parameter $t$, where $t \neq 0$, meets the curve again at the point with parameter $T$, where $T \neq t$. Show that

$T = \frac{1-t^2}{2t} \quad \text{and} \quad 3t^2 \neq 1 .$

Given a point $P_0$ on the curve, with parameter $t_0$, a sequence of points $P_0, P_1, P_2, \dots$ on the curve is constructed such that the tangent at $P_i$ meets the curve again at $P_{i+1}$. If $t_0 = \tan \frac{7\pi}{18}$, show that $P_3 = P_0$ but $P_1 \neq P_0$. Find a second value of $t_0$, with $t_0 > 0$, for which $P_3 = P_0$ but $P_1 \neq P_0$.

Model Solution

Step 1: Find $T$.

The curve is $x = t^2$, $y = t(1+t^2)$. Differentiating:

$\frac{\mathrm{d}x}{\mathrm{d}t} = 2t, \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = 1 + 3t^2, \qquad \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1+3t^2}{2t}.$

The tangent at parameter $t$ has equation:

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

To find where this meets the curve again, substitute $x = T^2$, $y = T(1+T^2)$:

$T(1+T^2) - t(1+t^2) = \frac{1+3t^2}{2t}(T^2 - t^2).$

Multiply through by $2t$:

$2tT(1+T^2) - 2t^2(1+t^2) = (1+3t^2)(T^2 - t^2).$

Expand:

$2tT + 2tT^3 - 2t^2 - 2t^4 = T^2 + 3t^2T^2 - t^2 - 3t^4.$

Rearrange:

$2tT^3 - T^2(1+3t^2) + 2tT + (t^2 - 3t^4 + t^2) = 0$

Wait, let me redo this carefully. We have:

$2tT + 2tT^3 - 2t^2 - 2t^4 - T^2 - 3t^2T^2 + t^2 + 3t^4 = 0$

$2tT^3 - (1+3t^2)T^2 + 2tT + (t^2 + t^4) = 0$

Since $T = t$ is a solution (the tangent passes through the point itself), $(T - t)$ is a factor. We can factor:

$2tT^3 - (1+3t^2)T^2 + 2tT + t^2(1+t^2) = 0.$

Dividing by $(T-t)$: we know $T = t$ satisfies this, so let us perform polynomial long division. Alternatively, note that $T = t$ is a root, so:

$2tT^3 - (1+3t^2)T^2 + 2tT + t^2 + t^4 = (T-t)\cdot Q(T).$

By polynomial division or comparing coefficients:

$2tT^3 - (1+3t^2)T^2 + 2tT + t^2(1+t^2) = (T-t)\bigl[2tT^2 - (1+t^2)T - t(1+t^2)\bigr].$

Let us verify by expanding the right side:

$(T-t)(2tT^2 - (1+t^2)T - t(1+t^2))$ $= 2tT^3 - (1+t^2)T^2 - t(1+t^2)T - 2t^2T^2 + t(1+t^2)T + t^2(1+t^2)$ $= 2tT^3 - (1+t^2+2t^2)T^2 + t^2(1+t^2)$ $= 2tT^3 - (1+3t^2)T^2 + t^2 + t^4.$

Comparing with the original: the original also has $+2tT$. So let me recheck. The equation is:

$2tT^3 - (1+3t^2)T^2 + 2tT + t^2(1+t^2) = 0.$

Let me re-verify the subtraction step. From:

$2tT + 2tT^3 - 2t^2 - 2t^4 - T^2 - 3t^2T^2 + t^2 + 3t^4 = 0$

Collecting:

• $T^3$: $2t$
• $T^2$: $-(1+3t^2)$
• $T^1$: $2t$
• $T^0$: $-2t^2 - 2t^4 + t^2 + 3t^4 = -t^2 + t^4 = t^2(t^2-1)$

Hmm, that’s different. Let me redo. $-2t^2 + t^2 = -t^2$ and $-2t^4 + 3t^4 = t^4$. So constant term is $t^4 - t^2 = t^2(t^2-1)$.

So: $2tT^3 - (1+3t^2)T^2 + 2tT + t^2(t^2-1) = 0$.

Factor out $(T-t)$. Since $T = t$ is a root:

$2t\cdot t^3 - (1+3t^2)t^2 + 2t\cdot t + t^2(t^2-1) = 2t^4 - t^2 - 3t^4 + 2t^2 + t^4 - t^2 = 0. \checkmark$

Dividing: $2tT^3 - (1+3t^2)T^2 + 2tT + t^2(t^2-1) = (T-t)(2tT^2 + At + B)$ where we need to find the quotient.

Using synthetic division with root $T = t$:

$(T-t)(2tT^2 + c_1 T + c_0)$

Expanding: $2tT^3 + c_1T^2 + c_0T - 2t^2T^2 - c_1tT - c_0t$

$= 2tT^3 + (c_1 - 2t^2)T^2 + (c_0 - c_1t)T - c_0t$

Matching:

• $T^2$: $c_1 - 2t^2 = -(1+3t^2) \Rightarrow c_1 = -1 - t^2$
• $T^1$: $c_0 - c_1t = 2t \Rightarrow c_0 = 2t + c_1 t = 2t - t(1+t^2) = 2t - t - t^3 = t - t^3 = t(1-t^2)$
• $T^0$: $-c_0t = t^2(t^2-1)$, so $c_0 = -t(t^2-1) = t(1-t^2)$. $\checkmark$

So the quadratic factor is:

$2tT^2 - (1+t^2)T + t(1-t^2) = 0.$

Using the quadratic formula:

$T = \frac{(1+t^2) \pm \sqrt{(1+t^2)^2 - 4\cdot 2t\cdot t(1-t^2)}}{4t} = \frac{(1+t^2) \pm \sqrt{(1+t^2)^2 - 8t^2(1-t^2)}}{4t}.$

Expand the discriminant:

$(1+t^2)^2 - 8t^2(1-t^2) = 1 + 2t^2 + t^4 - 8t^2 + 8t^4 = 1 - 6t^2 + 9t^4 = (1-3t^2)^2.$

Therefore:

$T = \frac{(1+t^2) \pm (1-3t^2)}{4t}.$

Taking the $+$ sign: $T = \frac{2-2t^2}{4t} = \frac{1-t^2}{2t}$.

Taking the $-$ sign: $T = \frac{4t^2}{4t} = t$ (the original point).

So $T = \frac{1-t^2}{2t}$. For this to be well-defined we need $t \neq 0$, and for $T \neq t$ we need $\frac{1-t^2}{2t} \neq t$, i.e., $1-t^2 \neq 2t^2$, i.e., $3t^2 \neq 1$. $\qquad \text{(shown)}$

Step 2: The iteration and the triple tangent property.

Define the map $T(t) = \frac{1-t^2}{2t}$. We need $t_1 = T(t_0)$, $t_2 = T(t_1)$, $t_3 = T(t_2)$.

Setting $t = \tan\alpha$, we get:

$T(\tan\alpha) = \frac{1-\tan^2\alpha}{2\tan\alpha} = \frac{1}{\tan 2\alpha} = \cot 2\alpha = \tan\!\left(\frac{\pi}{2} - 2\alpha\right).$

So if $t = \tan\alpha$, then $T(t) = \tan\!\left(\frac{\pi}{2} - 2\alpha\right)$.

Let $t_0 = \tan\alpha_0$ with $\alpha_0 = \frac{7\pi}{18}$.

$t_1 = T(t_0) = \tan\!\left(\frac{\pi}{2} - \frac{14\pi}{18}\right) = \tan\!\left(\frac{9\pi - 14\pi}{18}\right) = \tan\!\left(-\frac{5\pi}{18}\right).$

So $\alpha_1 = -\frac{5\pi}{18}$ (taking the angle modulo $\pi$ for $\tan$).

$t_2 = T(t_1) = \tan\!\left(\frac{\pi}{2} - 2\cdot\left(-\frac{5\pi}{18}\right)\right) = \tan\!\left(\frac{\pi}{2} + \frac{10\pi}{18}\right) = \tan\!\left(\frac{9\pi + 10\pi}{18}\right) = \tan\!\left(\frac{19\pi}{18}\right).$

Since $\tan$ has period $\pi$: $\tan\frac{19\pi}{18} = \tan\frac{\pi}{18}$. So $\alpha_2 = \frac{\pi}{18}$.

$t_3 = T(t_2) = \tan\!\left(\frac{\pi}{2} - \frac{2\pi}{18}\right) = \tan\!\left(\frac{9\pi - 2\pi}{18}\right) = \tan\!\left(\frac{7\pi}{18}\right) = t_0.$

So $P_3 = P_0$. $\qquad \text{(shown)}$

To show $P_1 \neq P_0$: we need $t_1 \neq t_0$, i.e., $\tan(-5\pi/18) \neq \tan(7\pi/18)$. Since $-5\pi/18$ and $7\pi/18$ differ by $12\pi/18 = 2\pi/3 \neq k\pi$, these are not equal.

Step 3: Find a second value of $t_0 > 0$ with $P_3 = P_0$, $P_1 \neq P_0$.

The condition $t_3 = t_0$ means, after three applications of the angle map $\alpha \mapsto \frac{\pi}{2} - 2\alpha$, we return to the same value of $\tan$.

Applying the map three times starting from $\alpha$:

$\alpha \;\mapsto\; \frac{\pi}{2} - 2\alpha \;\mapsto\; \frac{\pi}{2} - 2\!\left(\frac{\pi}{2} - 2\alpha\right) = 4\alpha - \frac{\pi}{2} \;\mapsto\; \frac{\pi}{2} - 2\!\left(4\alpha - \frac{\pi}{2}\right) = \frac{3\pi}{2} - 8\alpha.$

We need $\tan(\frac{3\pi}{2} - 8\alpha) = \tan\alpha$, so $\frac{3\pi}{2} - 8\alpha = \alpha + k\pi$ for some integer $k$.

$9\alpha = \frac{3\pi}{2} - k\pi \implies \alpha = \frac{3\pi - 2k\pi}{18} = \frac{(3-2k)\pi}{18}.$

For $P_1 \neq P_0$ we need $\frac{\pi}{2} - 2\alpha \neq \alpha + m\pi$, i.e., $\frac{\pi}{2} \neq 3\alpha + m\pi$, i.e., $\alpha \neq \frac{\pi - 2m\pi}{6}$.

The values $\alpha = \frac{(3-2k)\pi}{18}$ with $t_0 = \tan\alpha > 0$ (so $\alpha \in (0, \pi/2)$ modulo $\pi$):

• $k = 0$: $\alpha = \frac{3\pi}{18} = \frac{\pi}{6}$. But $P_1 = P_0$ check: $3\alpha = \pi/2$, so $\alpha = \pi/6$ gives $P_1 = P_0$. Reject.
• $k = -1$: $\alpha = \frac{5\pi}{18}$. Check $P_1 \neq P_0$: $3\alpha = 5\pi/6 \neq \pi/2 + m\pi$. Good.
• $k = 1$: $\alpha = \frac{\pi}{18}$. Check: $3\alpha = \pi/6 \neq \pi/2 + m\pi$. Good.
• $k = -2$: $\alpha = \frac{7\pi}{18}$. This is the original value.
• $k = -3$: $\alpha = \frac{9\pi}{18} = \frac{\pi}{2}$. $\tan(\pi/2)$ is undefined. Reject.
• $k = 2$: $\alpha = \frac{-\pi}{18}$. Negative, $\tan\alpha < 0$. Reject.

So the second positive value is $t_0 = \tan\frac{5\pi}{18}$ (and $t_0 = \tan\frac{\pi}{18}$ is a third).

Since the question asks for “a second value,” either works. We give $t_0 = \tan\frac{5\pi}{18}$.

Verification: $\alpha_0 = 5\pi/18$. $\alpha_1 = \pi/2 - 10\pi/18 = -\pi/18$. $\alpha_2 = \pi/2 + 2\pi/18 = 11\pi/18$. $\alpha_3 = \pi/2 - 22\pi/18 = -13\pi/18 \equiv 5\pi/18 \pmod{\pi}$. So $t_3 = t_0$. And $t_1 = \tan(-\pi/18) \neq \tan(5\pi/18) = t_0$. $\checkmark$

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Question 5

Topic: 代数与方程理论 Algebra & Equation Theory  |  Difficulty: Challenging  |  Marks: 20

Problem

5 Find the coordinates of the turning point on the curve $y = x^2 - 2bx + c$. Sketch the curve in the case that the equation $x^2 - 2bx + c = 0$ has two distinct real roots. Use your sketch to determine necessary and sufficient conditions on $b$ and $c$ for the equation $x^2 - 2bx + c = 0$ to have two distinct real roots. Determine necessary and sufficient conditions on $b$ and $c$ for this equation to have two distinct positive roots.

Find the coordinates of the turning points on the curve $y = x^3 - 3b^2x + c$ (with $b > 0$) and hence determine necessary and sufficient conditions on $b$ and $c$ for the equation $x^3 - 3b^2x + c = 0$ to have three distinct real roots. Determine necessary and sufficient conditions on $a, b$ and $c$ for the equation $(x-a)^3 - 3b^2(x-a) + c = 0$ to have three distinct positive roots.

Show that the equation $2x^3 - 9x^2 + 7x - 1 = 0$ has three distinct positive roots.

Model Solution

Part 1: Quadratic turning point and conditions

Completing the square: $y = x^2 - 2bx + c = (x - b)^2 + c - b^2 .$

The turning point is at $(b,\; c - b^2)$, and since the coefficient of $x^2$ is positive, this is a minimum.

When $x^2 - 2bx + c = 0$ has two distinct real roots, the parabola crosses the $x$-axis at two points. The sketch shows an upward-opening parabola with its minimum below the $x$-axis.

For two distinct real roots, we need the minimum value to be negative: $c - b^2 < 0 \quad \Longleftrightarrow \quad b^2 > c .$

This is necessary (the minimum must be below the axis) and sufficient (an upward parabola with minimum below the axis crosses the axis twice).

For two distinct positive roots, we need:

1. $b^2 > c$ (two distinct real roots),
2. The sum of roots $= 2b > 0$, so $b > 0$,
3. The product of roots $= c > 0$.

Conditions 2 and 3 ensure both roots are positive (if both were negative, their sum would be negative; if they had opposite signs, their product would be negative). Hence the necessary and sufficient conditions for two distinct positive roots are: $b > 0, \quad c > 0, \quad b^2 > c .$

Part 2: Cubic turning points and three distinct real roots

For $y = x^3 - 3b^2 x + c$ with $b > 0$: $\frac{dy}{dx} = 3x^2 - 3b^2 = 3(x^2 - b^2) = 3(x - b)(x + b) .$

The turning points are at $x = -b$ (local maximum) and $x = b$ (local minimum):

• $y(-b) = -b^3 + 3b^3 + c = 2b^3 + c$,
• $y(b) = b^3 - 3b^3 + c = -2b^3 + c$.

For three distinct real roots, the local maximum must be above the $x$-axis and the local minimum below: $2b^3 + c > 0 \quad \text{and} \quad -2b^3 + c < 0 ,$ $\Longleftrightarrow \quad -2b^3 < c < 2b^3 .$

Part 3: Conditions for $(x-a)^3 - 3b^2(x-a) + c = 0$ to have three distinct positive roots

Substituting $u = x - a$, the equation becomes $u^3 - 3b^2 u + c = 0$, with roots $u_1 < u_2 < u_3$. The original roots are $x_i = u_i + a$, so we need $u_i + a > 0$ for all $i$, i.e., $u_1 > -a$ (the binding constraint, since $u_1$ is the smallest).

From Part 2, three distinct real roots require $-2b^3 < c < 2b^3$.

Now consider $f(-a) = (-a)^3 - 3b^2(-a) + c = -a^3 + 3ab^2 + c$. Since $f$ has positive leading coefficient, $f(u) \to -\infty$ as $u \to -\infty$, and $f$ is increasing on $(-\infty, -b)$. The smallest root $u_1$ is where $f$ first crosses zero. For $u_1 > -a$, we need $f(-a) < 0$ (so that $-a$ lies to the left of $u_1$ in the region where $f < 0$): $-a^3 + 3ab^2 + c < 0 \quad \Longleftrightarrow \quad c < a^3 - 3ab^2 .$

For $-a$ to lie in the interval $(-\infty, -b)$ where $f$ is increasing, we need $-a < -b$, i.e., $a > b$. (This also ensures $a^3 - 3ab^2 = a(a^2 - 3b^2)$ can be compatible with $c > -2b^3$.)

For the middle root: $u_2 \in (-b, b)$, so $x_2 = u_2 + a > -b + a > 0$ since $a > b$.

For the largest root: $u_3 > b$, so $x_3 = u_3 + a > b + a > 0$ automatically.

The necessary and sufficient conditions for three distinct positive roots are: $a > b > 0, \quad -2b^3 < c < 2b^3, \quad c < a^3 - 3ab^2 .$

Part 4: Show $2x^3 - 9x^2 + 7x - 1 = 0$ has three distinct positive roots

Dividing by 2: $x^3 - \tfrac{9}{2}x^2 + \tfrac{7}{2}x - \tfrac{1}{2} = 0$.

We match this with $(x - a)^3 - 3b^2(x - a) + c = 0$. Expanding: $(x-a)^3 - 3b^2(x-a) + c = x^3 - 3ax^2 + (3a^2 - 3b^2)x + (-a^3 + 3ab^2 + c) .$

Matching coefficients:

• $x^2$: $3a = \tfrac{9}{2}$, so $a = \tfrac{3}{2}$.
• $x$: $3a^2 - 3b^2 = \tfrac{7}{2}$, so $3 \cdot \tfrac{9}{4} - 3b^2 = \tfrac{7}{2}$, giving $3b^2 = \tfrac{27}{4} - \tfrac{14}{4} = \tfrac{13}{4}$, hence $b^2 = \tfrac{13}{12}$.
• Constant: $-a^3 + 3ab^2 + c = -\tfrac{1}{2}$, so $-\tfrac{27}{8} + \tfrac{9}{2} \cdot \tfrac{13}{12} + c = -\tfrac{1}{2}$, giving $-\tfrac{27}{8} + \tfrac{39}{8} + c = -\tfrac{1}{2}$, so $c = -\tfrac{1}{2} - \tfrac{12}{8} = -2$.

Now we check the conditions from Part 3 with $a = \tfrac{3}{2}$, $b^2 = \tfrac{13}{12}$, $c = -2$:

Check $a > b$: $a^2 = \tfrac{9}{4} = \tfrac{27}{12}$ and $b^2 = \tfrac{13}{12}$, so $a^2 > b^2$, hence $a > b$. ✓

Check $-2b^3 < c < 2b^3$: We have $2b^3 = 2(b^2)^{3/2} = 2\left(\tfrac{13}{12}\right)^{3/2}$. Since $\tfrac{13}{12} > 1$, we get $2b^3 > 2$. More precisely, $b^3 = \tfrac{13\sqrt{13}}{12\sqrt{12}} = \tfrac{13\sqrt{39}}{72}$, so $2b^3 = \tfrac{13\sqrt{39}}{36}$. Since $\sqrt{39} > 6$, we have $2b^3 > \tfrac{78}{36} = \tfrac{13}{6} > 2$. Thus $|c| = 2 < 2b^3$. ✓

Check $c < a^3 - 3ab^2$: $a^3 - 3ab^2 = \tfrac{27}{8} - \tfrac{9}{2} \cdot \tfrac{13}{12} = \tfrac{27}{8} - \tfrac{117}{24} = \tfrac{81 - 117}{24} = -\tfrac{36}{24} = -\tfrac{3}{2}$. So $c = -2 < -\tfrac{3}{2}$. ✓

All conditions are satisfied, so $2x^3 - 9x^2 + 7x - 1 = 0$ has three distinct positive roots. $\blacksquare$

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Question 6

Topic: 三角函数 Trigonometry  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Show that $2 \sin \frac{1}{2} \theta \cos r\theta = \sin (r + \frac{1}{2})\theta - \sin (r - \frac{1}{2})\theta .$ Hence, or otherwise, find all solutions of the equation $\cos a\theta + \cos(a + 1)\theta + \cdots + \cos(b - 2)\theta + \cos(b - 1)\theta = 0 ,$ where $a$ and $b$ are positive integers with $a < b - 1$.

Model Solution

Step 1: Show the trigonometric identity.

Using the compound angle formulae:

$\sin(r+\tfrac{1}{2})\theta = \sin r\theta\cos\tfrac{1}{2}\theta + \cos r\theta\sin\tfrac{1}{2}\theta,$

$\sin(r-\tfrac{1}{2})\theta = \sin r\theta\cos\tfrac{1}{2}\theta - \cos r\theta\sin\tfrac{1}{2}\theta.$

Subtracting:

$\sin(r+\tfrac{1}{2})\theta - \sin(r-\tfrac{1}{2})\theta = 2\cos r\theta\sin\tfrac{1}{2}\theta. \qquad \text{(shown)}$

Step 2: Apply the identity to the sum.

Let $S = \cos a\theta + \cos(a+1)\theta + \cdots + \cos(b-1)\theta$.

Multiplying both sides by $2\sin\frac{1}{2}\theta$ and using the identity with each $r = a, a+1, \ldots, b-1$:

$2\sin\tfrac{1}{2}\theta \cdot S = \sum_{r=a}^{b-1}\bigl[\sin(r+\tfrac{1}{2})\theta - \sin(r-\tfrac{1}{2})\theta\bigr].$

This is a telescoping sum. Writing out the terms:

$= \bigl[\sin(a+\tfrac{1}{2})\theta - \sin(a-\tfrac{1}{2})\theta\bigr] + \bigl[\sin(a+\tfrac{3}{2})\theta - \sin(a+\tfrac{1}{2})\theta\bigr] + \cdots$ $\cdots + \bigl[\sin(b-\tfrac{1}{2})\theta - \sin(b-\tfrac{3}{2})\theta\bigr].$

After cancellation:

$2\sin\tfrac{1}{2}\theta \cdot S = \sin(b-\tfrac{1}{2})\theta - \sin(a-\tfrac{1}{2})\theta.$

Therefore:

$S = \frac{\sin(b-\frac{1}{2})\theta - \sin(a-\frac{1}{2})\theta}{2\sin\frac{1}{2}\theta},$

provided $\sin\frac{1}{2}\theta \neq 0$, i.e., $\theta \neq 2k\pi$ for integer $k$.

Step 3: Solve $S = 0$.

We need $\sin(b-\frac{1}{2})\theta = \sin(a-\frac{1}{2})\theta$, with $\theta \neq 2k\pi$.

Using the identity $\sin A = \sin B$ implies $A = B + 2k\pi$ or $A = \pi - B + 2k\pi$:

Case 1: $(b - \frac{1}{2})\theta = (a - \frac{1}{2})\theta + 2k\pi$, giving $(b-a)\theta = 2k\pi$, so:

$\theta = \frac{2k\pi}{b-a}, \qquad k \in \mathbb{Z}.$

We exclude $\theta = 2k'\pi$ (i.e., $k$ a multiple of $b-a$), since those make $\sin\frac{1}{2}\theta = 0$.

So $\theta = \frac{2k\pi}{b-a}$ for integer $k$ with $(b-a) \nmid k$.

Case 2: $(b - \frac{1}{2})\theta = \pi - (a - \frac{1}{2})\theta + 2k\pi$, giving $(a + b - 1)\theta = (2k+1)\pi$, so:

$\theta = \frac{(2k+1)\pi}{a+b-1}, \qquad k \in \mathbb{Z}.$

We exclude values where $\sin\frac{1}{2}\theta = 0$, which requires $\frac{(2k+1)\pi}{2(a+b-1)} = m\pi$, i.e., $2k+1 = 2m(a+b-1)$. Since $2k+1$ is odd and $2m(a+b-1)$ is even, this is impossible. So no exclusions are needed in Case 2.

Complete solution set:

$\theta = \frac{2k\pi}{b-a} \quad (k \in \mathbb{Z},\; (b-a) \nmid k), \qquad \text{or} \qquad \theta = \frac{(2k+1)\pi}{a+b-1} \quad (k \in \mathbb{Z}).$

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Question 7

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

Problem

7 In the $x$–$y$ plane, the point $A$ has coordinates $(a, 0)$ and the point $B$ has coordinates $(0, b)$, where $a$ and $b$ are positive. The point $P$, which is distinct from $A$ and $B$, has coordinates $(s, t)$. $X$ and $Y$ are the feet of the perpendiculars from $P$ to the $x$-axis and $y$-axis respectively, and $N$ is the foot of the perpendicular from $P$ to the line $AB$. Show that the coordinates $(x, y)$ of $N$ are given by $x = \frac{ab^2 - a(bt - as)}{a^2 + b^2} , \quad y = \frac{a^2b + b(bt - as)}{a^2 + b^2} .$ Show that, if $\left( \frac{t - b}{s} \right) \left( \frac{t}{s - a} \right) = -1$, then $N$ lies on the line $XY$. Give a geometrical interpretation of this result.

Model Solution

Step 1: Find the coordinates of $N$.

The line $AB$ passes through $A(a,0)$ and $B(0,b)$. Its equation is $\frac{x}{a} + \frac{y}{b} = 1$, or equivalently $bx + ay = ab$.

A direction vector for $AB$ is $(-a, b)$, so a normal vector to $AB$ is $(b, a)$.

The foot of the perpendicular from $P(s,t)$ to the line $bx + ay = ab$ can be found by parametrising the perpendicular through $P$. The perpendicular has direction $(b, a)$, so a general point on it is $(s + \lambda b,\; t + \lambda a)$.

Substituting into $bx + ay = ab$:

$b(s + \lambda b) + a(t + \lambda a) = ab,$

$bs + \lambda b^2 + at + \lambda a^2 = ab,$

$\lambda(a^2 + b^2) = ab - bs - at,$

$\lambda = \frac{ab - bs - at}{a^2 + b^2} = \frac{-a(bt - as) - (ab - ab) + ab - bs - at}{a^2 + b^2}.$

More directly: $\lambda = \frac{ab - bs - at}{a^2 + b^2}$.

The coordinates of $N$ are:

$x = s + \lambda b = s + \frac{b(ab - bs - at)}{a^2 + b^2} = \frac{s(a^2 + b^2) + b(ab - bs - at)}{a^2 + b^2},$

$= \frac{a^2 s + b^2 s + ab^2 - b^2 s - abt}{a^2 + b^2} = \frac{a^2 s + ab^2 - abt}{a^2 + b^2} = \frac{a(as + b^2 - bt)}{a^2 + b^2}.$

$= \frac{a\bigl[b^2 - (bt - as)\bigr]}{a^2 + b^2} = \frac{ab^2 - a(bt - as)}{a^2 + b^2}. \qquad \text{(shown)}$

Similarly:

$y = t + \lambda a = t + \frac{a(ab - bs - at)}{a^2 + b^2} = \frac{t(a^2 + b^2) + a(ab - bs - at)}{a^2 + b^2},$

$= \frac{a^2 t + b^2 t + a^2 b - abs - a^2 t}{a^2 + b^2} = \frac{b^2 t + a^2 b - abs}{a^2 + b^2} = \frac{b(bt + a^2 - as)}{a^2 + b^2}.$

$= \frac{a^2 b + b(bt - as)}{a^2 + b^2}. \qquad \text{(shown)}$

Step 2: Show $N$ lies on $XY$ under the given condition.

$X$ is the foot of the perpendicular from $P$ to the $x$-axis, so $X = (s, 0)$. $Y$ is the foot of the perpendicular from $P$ to the $y$-axis, so $Y = (0, t)$.

The line $XY$ passes through $(s, 0)$ and $(0, t)$, with equation:

$\frac{x}{s} + \frac{y}{t} = 1, \qquad \text{i.e.,} \quad tx + sy = st.$

$N$ lies on $XY$ if and only if $tx_N + sy_N = st$. Substituting:

$t\cdot\frac{ab^2 - a(bt - as)}{a^2 + b^2} + s\cdot\frac{a^2 b + b(bt - as)}{a^2 + b^2} = st.$

Multiply through by $a^2 + b^2$:

$t\bigl[ab^2 - a(bt - as)\bigr] + s\bigl[a^2 b + b(bt - as)\bigr] = st(a^2 + b^2).$

Expand the left side:

$tab^2 - at(bt - as) + sa^2 b + sb(bt - as),$

$= tab^2 - abt^2 + a^2 st + a^2 bs + b^2 st - abs^2.$

The right side is $sta^2 + stb^2$.

Subtracting $sta^2 + stb^2$ from both sides:

$tab^2 - abt^2 + a^2 bs - abs^2 = 0,$

$ab(ta - t^2 + as - s^2) = 0,$

$ab\bigl[t(a - t) + s(a - s)\bigr] = 0.$

Hmm, let me recheck. Actually:

$tab^2 - abt^2 + a^2 st + a^2 bs + b^2 st - abs^2 - sta^2 - stb^2$ $= tab^2 - abt^2 + a^2 bs - abs^2$ $= ab(tab - t^2 + a^2 s/b \cdot b - s^2 \cdot b)$

Wait, let me factor more carefully:

$tab^2 - abt^2 + a^2 bs - abs^2 = ab(tb - t^2 + as - s^2) = ab\bigl[t(b-t) + s(a-s)\bigr].$

For this to equal zero (with $a, b > 0$):

$t(b-t) + s(a-s) = 0.$

Now consider the given condition: $\left(\frac{t-b}{s}\right)\left(\frac{t}{s-a}\right) = -1$.

$\frac{t(t-b)}{s(s-a)} = -1 \implies t(t-b) = -s(s-a) \implies t^2 - bt = -s^2 + as.$

$t^2 - bt + s^2 - as = 0 \implies t(b-t) + s(a-s) = -(t^2 - bt) - (s^2 - as) = -\bigl[(t^2 - bt) + (s^2 - as)\bigr] = 0.$

Wait: $t(b-t) = bt - t^2$ and $s(a-s) = as - s^2$. So $t(b-t) + s(a-s) = bt - t^2 + as - s^2$. The condition gives $t^2 - bt + s^2 - as = 0$, i.e., $bt - t^2 + as - s^2 = 0$. Yes, so $t(b-t) + s(a-s) = 0$. $\checkmark$

Therefore $N$ lies on $XY$. $\qquad \text{(shown)}$

Step 3: Geometrical interpretation.

The condition $\frac{t-b}{s} \cdot \frac{t}{s-a} = -1$ means $\frac{t-b}{s-0} \cdot \frac{t-0}{s-a} = -1$. The first factor is the gradient of $PB$ (since $B = (0,b)$ and $P = (s,t)$), and the second factor is the gradient of $PA$ (since $A = (a,0)$). The product of gradients being $-1$ means $PA \perp PB$, i.e., the angle $\angle APB = 90°$.

So $P$ lies on the circle with diameter $AB$ (by Thales’ theorem).

Geometrical interpretation: $N$ lies on the line $XY$ if and only if $P$ lies on the circle with $AB$ as diameter, i.e., $\angle APB$ is a right angle.

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Question 8

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

Problem

8 (i) Show that the gradient at a point $(x, y)$ on the curve $(y + 2x)^3 (y - 4x) = c ,$ where $c$ is a constant, is given by $\frac{dy}{dx} = \frac{16x - y}{2y - 5x} .$ (ii) By considering the derivative with respect to $x$ of $(y + ax)^n (y + bx)$, or otherwise, find the general solution of the differential equation $\frac{dy}{dx} = \frac{10x - 4y}{3x - y} .$

Model Solution

Part (i): Show the gradient of $(y+2x)^3(y-4x) = c$ is $\frac{16x-y}{2y-5x}$.

Differentiate implicitly with respect to $x$:

$3(y+2x)^2(y'+2)\cdot(y-4x) + (y+2x)^3(y'-4) = 0.$

Divide by $(y+2x)^2$ (which is nonzero since $c \neq 0$ implies $y+2x \neq 0$):

$3(y'+2)(y-4x) + (y+2x)(y'-4) = 0.$

Expand:

$3y'(y-4x) + 6(y-4x) + y'(y+2x) - 4(y+2x) = 0.$

Collect $y'$ terms:

$y'\bigl[3(y-4x) + (y+2x)\bigr] + 6(y-4x) - 4(y+2x) = 0,$

$y'(3y - 12x + y + 2x) + 6y - 24x - 4y - 8x = 0,$

$y'(4y - 10x) + 2y - 32x = 0.$

$y' = \frac{32x - 2y}{4y - 10x} = \frac{2(16x - y)}{2(2y - 5x)} = \frac{16x - y}{2y - 5x}. \qquad \text{(shown)}$

Part (ii): Solve $\frac{dy}{dx} = \frac{10x - 4y}{3x - y}$.

We consider $F(x,y) = (y+ax)^n(y+bx)$ and compute $\frac{\mathrm{d}F}{\mathrm{d}x} = 0$.

$\frac{\mathrm{d}F}{\mathrm{d}x} = n(y+ax)^{n-1}(y'+a)(y+bx) + (y+ax)^n(y'+b) = 0.$

Divide by $(y+ax)^{n-1}$:

$n(y'+a)(y+bx) + (y+ax)(y'+b) = 0.$

Expand and collect $y'$:

$y'\bigl[n(y+bx) + (y+ax)\bigr] + na(y+bx) + b(y+ax) = 0.$

$y' = -\frac{na(y+bx) + b(y+ax)}{n(y+bx) + (y+ax)} = -\frac{(na+b)y + n abx + bax}{(n+1)y + (nb+a)x}.$

Hmm, let me redo more carefully:

Numerator: $na(y+bx) + b(y+ax) = nay + nabx + by + abx = (na+b)y + ab(n+1)x$.

Denominator: $n(y+bx) + (y+ax) = ny + nbx + y + ax = (n+1)y + (nb+a)x$.

So:

$y' = -\frac{(na+b)y + ab(n+1)x}{(n+1)y + (nb+a)x} = \frac{-ab(n+1)x - (na+b)y}{(nb+a)x + (n+1)y}.$

We need this to equal $\frac{10x - 4y}{3x - y} = \frac{10x - 4y}{3x - y}$. Matching coefficients:

From the $y$ coefficient in the numerator: $-(na+b) = -4$, so $na + b = 4$.

From the $x$ coefficient in the numerator: $-ab(n+1) = 10$, so $ab(n+1) = -10$.

From the $x$ coefficient in the denominator: $nb + a = 3$.

From the $y$ coefficient in the denominator: $n + 1 = -1$, so $n = -2$.

Check: with $n = -2$:

• $na + b = -2a + b = 4$
• $nb + a = -2b + a = 3$
• $ab(n+1) = ab(-1) = -ab = -10$, so $ab = 10$.

From $-2a + b = 4$: $b = 4 + 2a$. Substitute into $ab = 10$: $a(4+2a) = 10$, so $2a^2 + 4a - 10 = 0$, i.e., $a^2 + 2a - 5 = 0$.

$a = \frac{-2 \pm \sqrt{4+20}}{2} = -1 \pm \sqrt{6}.$

Check with $-2b + a = 3$: if $a = -1+\sqrt{6}$, then $b = 4 + 2(-1+\sqrt{6}) = 2+2\sqrt{6}$.

$-2b + a = -2(2+2\sqrt{6}) + (-1+\sqrt{6}) = -4 - 4\sqrt{6} - 1 + \sqrt{6} = -5 - 3\sqrt{6}$. This should be $3$. Doesn’t work.

If $a = -1-\sqrt{6}$, then $b = 4 + 2(-1-\sqrt{6}) = 2-2\sqrt{6}$.

$-2b + a = -2(2-2\sqrt{6}) + (-1-\sqrt{6}) = -4+4\sqrt{6}-1-\sqrt{6} = -5+3\sqrt{6}$. This should be $3$. So $3\sqrt{6} = 8$, $\sqrt{6} = 8/3$. Not true.

Let me recheck the coefficient matching. We have:

$y' = \frac{-ab(n+1)x - (na+b)y}{(nb+a)x + (n+1)y}.$

And we want $y' = \frac{10x - 4y}{3x - y}$.

So: $\frac{-ab(n+1)}{nb+a} = \frac{10}{3}$ and $\frac{-(na+b)}{n+1} = \frac{-4}{-1} = 4$.

Wait, I need to be more careful. We need:

$\frac{-ab(n+1)x - (na+b)y}{(nb+a)x + (n+1)y} = \frac{10x - 4y}{3x - y}.$

So the ratios must match: $\frac{-ab(n+1)}{10} = \frac{-(na+b)}{-4} = \frac{nb+a}{3} = \frac{n+1}{-1}$.

From the last ratio: $n+1 = -k$ for some proportionality constant $k$. Actually, let me just set:

$-ab(n+1) = 10k, \quad -(na+b) = -4k, \quad nb+a = 3k, \quad n+1 = -k.$

From $n+1 = -k$: $k = -(n+1)$.

$-(na+b) = -4k = 4(n+1)$, so $na+b = -4(n+1)$.

$nb+a = 3k = -3(n+1)$.

$-ab(n+1) = 10k = -10(n+1)$, so $ab = 10$ (assuming $n+1 \neq 0$).

From $na + b = -4(n+1)$ and $nb + a = -3(n+1)$:

Subtract: $na + b - nb - a = -(n+1)$, so $(n-1)a - (n-1)b = -(n+1)$, i.e., $(n-1)(a-b) = -(n+1)$.

From the second equation: $a = -3(n+1) - nb$. Substitute into the first: $n(-3(n+1) - nb) + b = -4(n+1)$.

$-3n(n+1) - n^2 b + b = -4(n+1),$

$b(1 - n^2) = -4(n+1) + 3n(n+1) = (n+1)(3n-4),$

$b(1-n)(1+n) = (n+1)(3n-4).$

Assuming $n+1 \neq 0$: $b(1-n) = 3n-4$, so $b = \frac{3n-4}{1-n} = \frac{4-3n}{n-1}$.

Then $a = -3(n+1) - nb = -3(n+1) - n\cdot\frac{4-3n}{n-1}$.

$= -3(n+1) - \frac{n(4-3n)}{n-1} = \frac{-3(n+1)(n-1) - n(4-3n)}{n-1}$

$= \frac{-3(n^2-1) - 4n + 3n^2}{n-1} = \frac{-3n^2+3-4n+3n^2}{n-1} = \frac{3-4n}{n-1}$.

And $ab = 10$:

$\frac{(3-4n)(4-3n)}{(n-1)^2} = 10.$

$(3-4n)(4-3n) = 12 - 9n - 16n + 12n^2 = 12n^2 - 25n + 12$.

So: $12n^2 - 25n + 12 = 10(n-1)^2 = 10n^2 - 20n + 10$.

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

So $n = \frac{1}{2}$ or $n = 2$.

Case $n = 2$: $b = \frac{4-6}{1} = -2$, $a = \frac{3-8}{1} = -5$. Check: $ab = 10$. $\checkmark$

So $(y - 5x)^2(y - 2x) = C$.

Case $n = \frac{1}{2}$: $b = \frac{4 - 3/2}{-1/2} = \frac{5/2}{-1/2} = -5$, $a = \frac{3-2}{-1/2} = \frac{1}{-1/2} = -2$. Check: $ab = 10$. $\checkmark$

So $(y - 2x)^{1/2}(y - 5x) = C$, which is equivalent (square both sides: $(y-2x)(y-5x)^2 = C^2$).

Both give the same family of curves. The general solution is:

$\boxed{(y - 5x)^2(y - 2x) = C}$

where $C$ is an arbitrary constant.