STEP2 2006 -- Pure Mathematics

STEP2 2006 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2006  |  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	递推数列与周期性 Recursive Sequences and Periodicity	Challenging	递推关系展开,多项式方程求解,因式分解,极限分析
2	指数函数与微积分 Exponentials and Calculus	Challenging	泰勒级数,阶乘与指数比较,几何级数求和,导数符号分析
3	无理数与二项式定理 Surds and Binomial Theorem	Standard	共轭根式,二项式定理,不等式放缩,数值近似
4	定积分与换元法 Definite Integrals and Substitution	Challenging	变量替换,三角恒等式,积分区间对称性,部分分式
5	取整函数与面积 Floor Functions and Area	Standard	取整函数性质,阶梯函数作图,逐段积分求面积,等差数列求和
6	向量与不等式 Vectors and Inequalities	Standard	标量积与夹角余弦,柯西-施瓦茨不等式,向量平行条件代入
7	解析几何 Analytic Geometry	Challenging	隐函数求导或参数求导,半角公式(tan(alpha/2)),椭圆切线验证
8	向量几何 Vector Geometry	Challenging	位置向量参数方程,联立求解交点,向量线性组合

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Question 1

Topic: 递推数列与周期性 Recursive Sequences and Periodicity  |  Difficulty: Challenging  |  Marks: 20

Problem

1 The sequence of real numbers $u_1, u_2, u_3, \dots$ is defined by

$u_1 = 2, \quad \text{and} \quad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n \ge 1, \qquad \text{(*)}$

where $k$ is a constant.

(i) Determine the values of $k$ for which the sequence (*) is: (a) constant; (b) periodic with period 2; (c) periodic with period 4.

(ii) In the case $k = 37$, show that $u_n \ge 2$ for all $n$. Given that in this case the sequence (*) converges to a limit $\ell$, find the value of $\ell$.

Hint

If you read through at least part (i) of the question, you will see that it is necessary to work with $u_1, u_2, u_3$ and $u_5$ (and hence, presumably - as the sequence is defined recursively - with $u_4$ also). Although it is not the only way to go about the problem, it makes sense to work each of these terms out first. Each will be an expression involving $k$ and should ideally be simplified as you go. Thus, $u_1 = 2$ gives $u_2 = k - 18$, $u_3 = k - \frac{36}{k - 18} = \frac{k^2 - 18k - 36}{k - 18}$ , etc. Then, for (a), $u_2 = 2$; for (b), $u_3 = 2$; and, for (c), $u_5 = 2$. Each result leads to a polynomial equation (of increasing orders) to be solved. Finally, you need to remember that, in the case of (c) for instance, of the four solutions given by the resulting equation, two of them must have arisen already in parts (a) and (b) - you’ll see why if you think about it for a moment. Ideally, you would see this beforehand, and then this fact will help you factorise the quartic polynomial by the factor theorem. A simple line of reasoning can be employed to establish the first result in (ii) without the need for a formal inductive proof. If $u_n \ge 2$, then $u_{n+1} = 37 - \frac{36}{u_n} \ge 37 - \frac{36}{2} = 19 > 2$. Since $u_1 = 2$, it follows that all terms of the sequence are $\ge 2$. In fact, most of them are much bigger than this. Then, for the final part of the question, the informal observation that, eventually, all terms effectively become equal is all that is required. Setting $u_{n+1} = u_n = l$ (say) leads to a quadratic, with two roots, one of which is obviously less than 2 and can therefore be rejected. Answers: (i) $k =$ (a) 20; (b) 0; (c) $\pm 6\sqrt{2}$. (ii) 36.

Model Solution

We first compute the early terms of the sequence in terms of $k$.

$u_1 = 2$

$u_2 = k - \frac{36}{u_1} = k - 18$

$u_3 = k - \frac{36}{u_2} = k - \frac{36}{k - 18} = \frac{k(k-18) - 36}{k-18} = \frac{k^2 - 18k - 36}{k-18}$

$u_4 = k - \frac{36}{u_3} = k - \frac{36(k-18)}{k^2 - 18k - 36} = \frac{k(k^2 - 18k - 36) - 36(k-18)}{k^2 - 18k - 36}$

$= \frac{k^3 - 18k^2 - 36k - 36k + 648}{k^2 - 18k - 36} = \frac{k^3 - 18k^2 - 72k + 648}{k^2 - 18k - 36}$

We will also need $u_5$ for part (c). Using $u_5 = k - 36/u_4$:

$u_5 = k - \frac{36(k^2 - 18k - 36)}{k^3 - 18k^2 - 72k + 648} = \frac{k(k^3 - 18k^2 - 72k + 648) - 36(k^2 - 18k - 36)}{k^3 - 18k^2 - 72k + 648}$

The numerator is:

$k^4 - 18k^3 - 72k^2 + 648k - 36k^2 + 648k + 1296 = k^4 - 18k^3 - 108k^2 + 1296k + 1296$

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Part (i)(a): Constant sequence

For the sequence to be constant, we need $u_2 = u_1 = 2$:

$k - 18 = 2 \implies k = 20.$

Verification: With $k = 20$, every term satisfies $u_{n+1} = 20 - 36/2 = 20 - 18 = 2$, so the sequence is indeed constant.

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Part (i)(b): Periodic with period 2

For period 2 we need $u_3 = u_1 = 2$ but $u_2 \ne 2$ (i.e. $k \ne 20$):

$\frac{k^2 - 18k - 36}{k - 18} = 2$

$k^2 - 18k - 36 = 2(k - 18) = 2k - 36$

$k^2 - 20k = 0$

$k(k - 20) = 0.$

So $k = 0$ or $k = 20$. Since $k = 20$ gives a constant sequence (period 1), the period-2 value is $k = 0$.

Verification: $u_1 = 2$, $u_2 = 0 - 18 = -18$, $u_3 = 0 - 36/(-18) = 2$, $u_4 = -18$, confirming period 2.

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Part (i)(c): Periodic with period 4

For period 4 we need $u_5 = u_1 = 2$ but $u_3 \ne 2$ and $u_2 \ne 2$ (i.e. $k \ne 0, 20$).

Setting $u_5 = 2$:

$\frac{k^4 - 18k^3 - 108k^2 + 1296k + 1296}{k^3 - 18k^2 - 72k + 648} = 2$

$k^4 - 18k^3 - 108k^2 + 1296k + 1296 = 2k^3 - 36k^2 - 144k + 1296$

$k^4 - 20k^3 - 72k^2 + 1440k = 0$

$k(k^3 - 20k^2 - 72k + 1440) = 0.$

The solution $k = 0$ gives period 2. Now factorise the cubic. Since $k = 20$ is a root of the cubic (any constant or period-2 sequence also satisfies $u_5 = u_1$):

$8000 - 8000 - 1440 + 1440 = 0. \checkmark$

Dividing the cubic by $(k - 20)$:

$k^3 - 20k^2 - 72k + 1440 = (k - 20)(k^2 - 72).$

We can verify: $(k-20)(k^2 - 72) = k^3 - 20k^2 - 72k + 1440$. $\checkmark$

So the full equation becomes:

$k(k - 20)(k^2 - 72) = 0,$

giving $k = 0$, $k = 20$, $k = \pm\sqrt{72} = \pm 6\sqrt{2}$.

Excluding $k = 0$ (period 2) and $k = 20$ (constant), the period-4 values are $k = \pm 6\sqrt{2}$.

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Part (ii): The case $k = 37$

Showing $u_n \ge 2$ for all $n$.

We prove this by induction.

Base case: $u_1 = 2 \ge 2$.

Inductive step: Suppose $u_n \ge 2$. Since $u_n > 0$, we have $\frac{36}{u_n} \le \frac{36}{2} = 18$, so:

$u_{n+1} = 37 - \frac{36}{u_n} \ge 37 - 18 = 19 > 2.$

By induction, $u_n \ge 2$ for all $n$.

Finding the limit $\ell$.

Setting $u_{n+1} = u_n = \ell$ in the recurrence:

$\ell = 37 - \frac{36}{\ell}$

$\ell^2 = 37\ell - 36$

$\ell^2 - 37\ell + 36 = 0$

$(\ell - 1)(\ell - 36) = 0.$

So $\ell = 1$ or $\ell = 36$. Since $u_n \ge 2$ for all $n$, we must have $\ell \ge 2$, which rules out $\ell = 1$.

Therefore $\ell = 36$.

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Question 2

Topic: 指数函数与微积分 Exponentials and Calculus  |  Difficulty: Challenging  |  Marks: 20

Problem

2 Using the series $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots ,$

show that $e > \frac{8}{3}$.

Show that $n! > 2^n$ for $n \ge 4$ and hence show that $e < \frac{67}{24}$.

Show that the curve with equation $y = 3e^{2x} + 14 \ln(\tfrac{4}{3} - x), \quad x < \tfrac{4}{3},$

has a minimum turning point between $x = \frac{1}{2}$ and $x = 1$ and a maximum turning point between $x = 1$ and $x = \frac{5}{4}$.

Hint

The formula books give a series for $e^x$. Setting $x = 1$ then gives you $e$ as the limit of an infinite sum of positive terms, and the sum of the first four of these will then provide a lower bound to its value. In the next part, you (again) can provide a perfectly sound argument for the required result without having to resort to a formally inductive one (although one would be perfectly valid, of course). Noting firstly that $4! = 24 > 16 = 2^4$, $(n + 4)!$ consists of the product of $4!$ and $n$ positive integers, each greater than 2; while $2^{n+4}$ consists of 16 and a further $n$ factors of 2. Since each term in the first number is greater than the corresponding term in the second, the result follows. [Alternatively, $4! > 2^4$ and $n! > 2^n \Rightarrow (n + 1)! = (n + 1) \times n! > 2 \times n!$ (since $n > 4$) $> 2 \times 2^n$ (by hypothesis) $= 2^{n+1}$, and proof follows by induction.] Now, adding the terms in the expansion for $e$ beyond the cubed one, and noting that each is less than a corresponding power of $\frac{1}{2}$ using the result just established, gives $e < \frac{8}{3} +$ the sum-to-infinity of a convergent GP. There are two common methods for showing that a stationary value of a curve is a max. or a min. One involves the second derivative evaluated at the point in question. There are several drawbacks involved with this approach. One is that you have to differentiate twice (which is ok with simple functions). A second is that you need to know the exact value(s) of the variable being substituted (which isn’t the case here). Another is that the sign of $\frac{d^2y}{dx^2}$ doesn’t necessarily tell you what is happening to the curve. (Think of the graph of $y = x^4$, which has $\frac{d^2y}{dx^2} = 0$ at the origin, yet the stationary point here is a minimum!) Thus, it is the other approach that you are clearly intended to use on this occasion. This examines the sign of $\frac{dy}{dx}$ slightly to each side of the point in question. When $x = \frac{1}{2}$, using $e < \frac{67}{24}$ shows …; at $x = 1$, using $e > \frac{8}{3}$ shows …; and at $x = \frac{5}{4}$, we can use any suitable bound for $e$, such as $e < 3$ for instance, to show that … Finally, since the answers are given in the question, it is important to state carefully the reasoning that supports these answers.

Model Solution

Part 1: Showing $e > \frac{8}{3}$

Setting $x = 1$ in the Taylor series for $e^x$:

$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots$

Since every term in the series is positive, the sum of all terms is strictly greater than the sum of just the first four:

$e > 1 + 1 + \frac{1}{2} + \frac{1}{6} = \frac{6 + 6 + 3 + 1}{6} = \frac{16}{6} = \frac{8}{3}.$

Therefore $e > \frac{8}{3}$. $\checkmark$

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Part 2: Showing $n! > 2^n$ for $n \ge 4$, hence $e < \frac{67}{24}$

Proof of $n! > 2^n$ for $n \ge 4$.

Base case: $4! = 24 > 16 = 2^4$. $\checkmark$

Inductive step: Suppose $n! > 2^n$ for some $n \ge 4$. Then $n + 1 \ge 5 > 2$, so:

$(n+1)! = (n+1) \cdot n! > 2 \cdot n! > 2 \cdot 2^n = 2^{n+1}.$

By induction, $n! > 2^n$ for all $n \ge 4$. $\checkmark$

Showing $e < \frac{67}{24}$.

From Part 1, we can write:

$e = \left(1 + 1 + \frac{1}{2} + \frac{1}{6}\right) + \left(\frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \cdots\right) = \frac{8}{3} + \sum_{n=4}^{\infty} \frac{1}{n!}.$

Since $n! > 2^n$ for $n \ge 4$, we have $\frac{1}{n!} < \frac{1}{2^n}$ for each $n \ge 4$. Therefore:

$\sum_{n=4}^{\infty} \frac{1}{n!} < \sum_{n=4}^{\infty} \frac{1}{2^n} = \frac{1}{2^4} + \frac{1}{2^5} + \frac{1}{2^6} + \cdots$

This is a geometric series with first term $a = \frac{1}{16}$ and common ratio $r = \frac{1}{2}$:

$\sum_{n=4}^{\infty} \frac{1}{2^n} = \frac{1/16}{1 - 1/2} = \frac{1}{8}.$

Therefore:

$e < \frac{8}{3} + \frac{1}{8} = \frac{64 + 3}{24} = \frac{67}{24}.$

So $e < \frac{67}{24}$. $\checkmark$

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Part 3: Turning points of $y = 3e^{2x} + 14\ln\!\left(\frac{4}{3} - x\right)$

The domain is $x < \frac{4}{3}$. Differentiating:

$\frac{dy}{dx} = 6e^{2x} - \frac{14}{\frac{4}{3} - x} = 6e^{2x} - \frac{42}{4 - 3x}.$

We determine the sign of $\frac{dy}{dx}$ at $x = \frac{1}{2}$, $x = 1$, and $x = \frac{5}{4}$ using the bounds $\frac{8}{3} < e < \frac{67}{24}$.

At $x = \frac{1}{2}$: $\quad \frac{dy}{dx} = 6e - \frac{42}{5/2} = 6e - \frac{84}{5}$.

Since $e < \frac{67}{24}$: $\quad 6e < \frac{67}{4} = 16.75 < 16.8 = \frac{84}{5}$.

So $\frac{dy}{dx}\big|_{x=1/2} < 0$.

At $x = 1$: $\quad \frac{dy}{dx} = 6e^2 - 42$.

Since $e > \frac{8}{3}$: $\quad e^2 > \frac{64}{9}$, so $6e^2 > \frac{128}{3} = 42\frac{2}{3} > 42$.

So $\frac{dy}{dx}\big|_{x=1} > 0$.

At $x = \frac{5}{4}$: $\quad \frac{dy}{dx} = 6e^{5/2} - \frac{42}{1/4} = 6e^{5/2} - 168$.

Since $e < 3$: $\quad e^{5/2} < 3^{5/2} = 9\sqrt{3}$. So $6e^{5/2} < 54\sqrt{3} < 54 \times 1.733 = 93.6 < 168$.

So $\frac{dy}{dx}\big|_{x=5/4} < 0$.

Conclusion: Since $\frac{dy}{dx}$ is continuous and changes sign from negative (at $x = \frac{1}{2}$) to positive (at $x = 1$), by the intermediate value theorem there exists a turning point in $\left(\frac{1}{2}, 1\right)$ where $\frac{dy}{dx}$ changes from $-$ to $+$. This is a minimum turning point.

Similarly, $\frac{dy}{dx}$ changes from positive (at $x = 1$) to negative (at $x = \frac{5}{4}$), so there exists a turning point in $\left(1, \frac{5}{4}\right)$ where $\frac{dy}{dx}$ changes from $+$ to $-$. This is a maximum turning point.

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Question 3

Topic: 无理数与二项式定理 Surds and Binomial Theorem  |  Difficulty: Standard  |  Marks: 20

Problem

3 (i) Show that $(5+\sqrt{24})^4 + \frac{1}{(5+\sqrt{24})^4}$ is an integer.

Show also that

$0.1 < \frac{1}{5+\sqrt{24}} < \frac{2}{19} < 0.11.$

Hence determine, with clear reasoning, the value of $(5+\sqrt{24})^4$ correct to four decimal places.

(ii) If $N$ is an integer greater than 1, show that $(N + \sqrt{N^2 - 1})^k$, where $k$ is a positive integer, differs from the integer nearest to it by less than $(2N - \frac{1}{2})^{-k}$.

Hint

If you fail to notice that $\frac{1}{5 + \sqrt{24}} = 5 - \sqrt{24}$, then this question is going to be a bit of a non-starter for you. The idea of conjugates, from the use of the difference of two squares, should be a familiar one. As is the binomial theorem, which you can now use to expand both $(5 + \sqrt{24})^4$ and $(5 - \sqrt{24})^4$. When you do this, you will see that all the $\sqrt{24}$ bits cancel out, to leave you with an integer. For the next part, some fairly simple inequality observations, such as $20.25 < 24 < 25 \Rightarrow 4.5 < \sqrt{24} < 5 \text{ and } 2 \times 100 = 200 < 208 = 11 \times 19 \Rightarrow \frac{2}{19} < \frac{11}{100}$ help to establish the required results. It follows that $0.1^4 < (5 - \sqrt{24})^4 < 0.11^4$ and the difference between the integer and $(5 + \sqrt{24})^4$ is this small number, which lies between … For part (ii), it is simply necessary to mimic the work of part (i) but in a general setting, again starting with the key observations that $\frac{1}{N + \sqrt{N^2 - 1}} = N - \sqrt{N^2 - 1}$ and that the binomial expansions for $(N + \sqrt{N^2 - 1})^k + (N - \sqrt{N^2 - 1})^k$ will lead to the cancelling of all surd terms, to give an integer, $M$ say. Now $(N - \sqrt{N^2 - 1})^k$ is positive, and the reciprocal of a number $> 1$, so $(N - \sqrt{N^2 - 1})^k \rightarrow 0+$ as $k \rightarrow \infty$. Also, $2N - \frac{1}{2} < N + \sqrt{N^2 - 1} < 2N \Rightarrow \frac{1}{2N - \frac{1}{2}} > N - \sqrt{N^2 - 1} > \frac{1}{2N}$ Thus $(N + \sqrt{N^2 - 1})^k = M - (N - \sqrt{N^2 - 1})^k$ differs from an integer ($M$) by less than $\left( \frac{1}{2N - \frac{1}{2}} \right)^k = (2N - \frac{1}{2})^{-k}$. Answers: (i) 9601.9999

Model Solution

Part (i)

Showing the expression is an integer.

Let $\alpha = 5 + \sqrt{24}$ and note that

$\frac{1}{5 + \sqrt{24}} = \frac{5 - \sqrt{24}}{(5 + \sqrt{24})(5 - \sqrt{24})} = \frac{5 - \sqrt{24}}{25 - 24} = 5 - \sqrt{24}.$

So we need to show that $\alpha^4 + (5 - \sqrt{24})^4$ is an integer.

We compute $\alpha^2$:

$\alpha^2 = (5 + \sqrt{24})^2 = 25 + 10\sqrt{24} + 24 = 49 + 10\sqrt{24}.$

Now we compute $\alpha^4 = (\alpha^2)^2$:

$\alpha^4 = (49 + 10\sqrt{24})^2 = 2401 + 980\sqrt{24} + 2400 = 4801 + 980\sqrt{24}.$

Similarly, $(5 - \sqrt{24})^4 = 4801 - 980\sqrt{24}$.

Therefore

$\alpha^4 + (5 - \sqrt{24})^4 = (4801 + 980\sqrt{24}) + (4801 - 980\sqrt{24}) = 9602,$

which is an integer.

Showing the inequality chain.

We have $\frac{1}{5 + \sqrt{24}} = 5 - \sqrt{24}$.

Since $24 > 20.25 = (4.5)^2$, we get $\sqrt{24} > 4.5$, so $5 - \sqrt{24} < 0.5$.

Since $24 < 25 = 5^2$, we get $\sqrt{24} < 5$, so $5 - \sqrt{24} > 0 > 0.1$.

Wait, we need $5 - \sqrt{24} > 0.1$, i.e. $\sqrt{24} < 4.9$. Since $4.9^2 = 24.01 > 24$, this gives $\sqrt{24} < 4.9$, hence $5 - \sqrt{24} > 0.1$. So $0.1 < \frac{1}{5 + \sqrt{24}}$.

Next we show $\frac{1}{5 + \sqrt{24}} < \frac{2}{19}$. Since $\frac{1}{5 + \sqrt{24}} = 5 - \sqrt{24}$, this is equivalent to $5 - \sqrt{24} < \frac{2}{19}$, i.e. $\sqrt{24} > 5 - \frac{2}{19} = \frac{93}{19}$. Squaring: $24 \stackrel{?}{>} \frac{8649}{361}$. Since $24 \times 361 = 8664 > 8649$, this holds. So $\frac{1}{5 + \sqrt{24}} < \frac{2}{19}$.

Finally, $\frac{2}{19} < 0.11 = \frac{11}{100}$ iff $200 < 209$, which is true.

Combining: $0.1 < \frac{1}{5 + \sqrt{24}} < \frac{2}{19} < 0.11$.

Determining $(5 + \sqrt{24})^4$ to four decimal places.

We have $(5 + \sqrt{24})^4 = 9602 - (5 - \sqrt{24})^4$.

From the inequality chain, $0.1 < 5 - \sqrt{24} < 0.11$, so

$0.1^4 < (5 - \sqrt{24})^4 < 0.11^4.$

Computing: $0.1^4 = 0.0001$ and $0.11^4 = 0.00014641$.

Therefore

$9602 - 0.00014641 < (5 + \sqrt{24})^4 < 9602 - 0.0001,$

$9601.99985359 < (5 + \sqrt{24})^4 < 9601.9999.$

Both bounds round to $9601.9999$ to four decimal places.

Therefore $(5 + \sqrt{24})^4 = 9601.9999$ correct to four decimal places.

Part (ii)

Let $\alpha = N + \sqrt{N^2 - 1}$ and $\beta = N - \sqrt{N^2 - 1}$.

Then $\alpha \beta = N^2 - (N^2 - 1) = 1$, so $\beta = \frac{1}{\alpha}$.

By the binomial theorem,

$\alpha^k + \beta^k = (N + \sqrt{N^2 - 1})^k + (N - \sqrt{N^2 - 1})^k = \sum_{j=0}^{k} \binom{k}{j} N^{k-j} (\sqrt{N^2-1})^j + \sum_{j=0}^{k} \binom{k}{j} N^{k-j} (-\sqrt{N^2-1})^j.$

When $j$ is odd, the terms cancel; when $j$ is even, the terms double. Every surviving term involves $(\sqrt{N^2 - 1})^{2m} = (N^2 - 1)^m$ for integer $m$, so $\alpha^k + \beta^k$ is an integer. Call it $M$.

Since $N \ge 2$, we have $\alpha > 1$ and $0 < \beta < 1$, so $0 < \beta^k < 1$. Therefore

$\alpha^k = M - \beta^k,$

which means $\alpha^k$ differs from the integer $M$ by $\beta^k$ (a positive quantity less than 1), so $M$ is the nearest integer to $\alpha^k$.

We now show $\beta < \frac{1}{2N - \frac{1}{2}}$. Since $\beta = \frac{1}{\alpha}$, this is equivalent to $\alpha > 2N - \frac{1}{2}$, i.e.

$N + \sqrt{N^2 - 1} > 2N - \frac{1}{2},$

$\sqrt{N^2 - 1} > N - \frac{1}{2}.$

Squaring (both sides are positive for $N \ge 2$):

$N^2 - 1 > N^2 - N + \frac{1}{4},$

$-1 > -N + \frac{1}{4},$

$N > \frac{5}{4}.$

This holds since $N \ge 2$. Therefore $\beta < \frac{1}{2N - \frac{1}{2}}$, giving

$\beta^k < \left(\frac{1}{2N - \frac{1}{2}}\right)^k = \left(2N - \frac{1}{2}\right)^{-k}.$

Hence $(N + \sqrt{N^2 - 1})^k$ differs from the nearest integer by less than $\left(2N - \frac{1}{2}\right)^{-k}$.

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Question 4

Topic: 定积分与换元法 Definite Integrals and Substitution  |  Difficulty: Challenging  |  Marks: 20

Problem

4 By making the substitution $x = \pi - t$, show that

$\int_{0}^{\pi} x f(\sin x) dx = \frac{1}{2} \pi \int_{0}^{\pi} f(\sin x) dx,$

where $f(\sin x)$ is a given function of $\sin x$.

Evaluate the following integrals:

(i) $\int_{0}^{\pi} \frac{x \sin x}{3 + \sin^2 x} dx$;

(ii) $\int_{0}^{2\pi} \frac{x \sin x}{3 + \sin^2 x} dx$;

(iii) $\int_{0}^{\pi} \frac{x |\sin 2x|}{3 + \sin^2 x} dx$.

Hint

Using the given substitution, the initial result is established by splitting the integral into its two parts, and then making the simple observation that $\int_{0}^{\pi} x f(\sin x) dx = \int_{0}^{\pi} t f(\sin t) dt$. This result is now used directly in (i), along with a substitution (such as $c = \cos x$). The resulting integration can be avoided by referring to your formula book, or done by using partial fractions. In (ii), the integral can be split into two; one from 0 to $\pi$, the second from $\pi$ to $2\pi$. The first of these is just (i)‘s integral, and the second can be determined by using a substitution such as $y = x - \pi$ (the key here is that the limits will then match those of the initial result, which you should be looking to make use of as much as possible). In part (iii), the use of the double-angle formula for $\sin 2x$ gives an integral involving sines and cosines, but this must also count as a function of $\sin x$, since $\cos x = \sqrt{1 - \sin^2 x}$. Thus the initial result may be applied here also. Once again, the substitution $c = \cos x$ reduces the integration to a standard one. Answers: (i) $\frac{1}{4} \pi \ln 3$; (ii) $-\frac{1}{2} \pi \ln 3$; (iii) $\pi \ln \frac{4}{3}$.**

Model Solution

General identity. Let $I = \int_0^{\pi} x \, f(\sin x) \, dx$. Substituting $x = \pi - t$, so $dx = -dt$:

$I = \int_{\pi}^{0} (\pi - t) \, f(\sin(\pi - t)) \, (-dt) = \int_0^{\pi} (\pi - t) \, f(\sin t) \, dt.$

Since $\sin(\pi - t) = \sin t$, and the dummy variable can be renamed:

$I = \int_0^{\pi} (\pi - x) \, f(\sin x) \, dx = \pi \int_0^{\pi} f(\sin x) \, dx - I.$

Therefore $2I = \pi \int_0^{\pi} f(\sin x) \, dx$, giving

$I = \frac{\pi}{2} \int_0^{\pi} f(\sin x) \, dx. \qquad \text{(*)}$

Part (i)

We apply $(*)$ with $f(\sin x) = \frac{\sin x}{3 + \sin^2 x}$:

$\int_0^{\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{3 + \sin^2 x} \, dx.$

Using $\sin^2 x = 1 - \cos^2 x$, the denominator becomes $4 - \cos^2 x$. Substituting $c = \cos x$, $dc = -\sin x \, dx$:

$\int_0^{\pi} \frac{\sin x}{4 - \cos^2 x} \, dx = \int_1^{-1} \frac{-dc}{4 - c^2} = \int_{-1}^{1} \frac{dc}{4 - c^2}.$

Using partial fractions: $\frac{1}{4 - c^2} = \frac{1}{(2-c)(2+c)} = \frac{1}{4}\left(\frac{1}{2-c} + \frac{1}{2+c}\right)$.

$\int_{-1}^{1} \frac{dc}{4 - c^2} = \frac{1}{4} \left[-\ln(2-c) + \ln(2+c)\right]_{-1}^{1} = \frac{1}{4}\left[\ln\frac{2+c}{2-c}\right]_{-1}^{1}.$

At $c = 1$: $\ln\frac{3}{1} = \ln 3$. At $c = -1$: $\ln\frac{1}{3} = -\ln 3$.

So the integral equals $\frac{1}{4}(\ln 3 - (-\ln 3)) = \frac{1}{4} \cdot 2\ln 3 = \frac{\ln 3}{2}$.

Therefore

$\int_0^{\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx = \frac{\pi}{2} \cdot \frac{\ln 3}{2} = \frac{\pi \ln 3}{4}.$

Part (ii)

We split the integral:

$\int_0^{2\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx = \int_0^{\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx + \int_{\pi}^{2\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx.$

The first integral is $\frac{\pi \ln 3}{4}$ from part (i).

For the second, substitute $y = x - \pi$, so $x = y + \pi$, $dx = dy$:

$\int_{\pi}^{2\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx = \int_0^{\pi} \frac{(y + \pi) \sin(y + \pi)}{3 + \sin^2(y + \pi)} \, dy.$

Since $\sin(y + \pi) = -\sin y$ and $\sin^2(y + \pi) = \sin^2 y$:

$= \int_0^{\pi} \frac{-(y + \pi) \sin y}{3 + \sin^2 y} \, dy = -\int_0^{\pi} \frac{y \sin y}{3 + \sin^2 y} \, dy - \pi \int_0^{\pi} \frac{\sin y}{3 + \sin^2 y} \, dy.$

The first piece is $-\frac{\pi \ln 3}{4}$ from part (i). The second piece is $-\pi \cdot \frac{\ln 3}{2} = -\frac{\pi \ln 3}{2}$.

Therefore

$\int_0^{2\pi} \frac{x \sin x}{3 + \sin^2 x} \, dx = \frac{\pi \ln 3}{4} - \frac{\pi \ln 3}{4} - \frac{\pi \ln 3}{2} = -\frac{\pi \ln 3}{2}.$

Part (iii)

We apply $(*)$ with $f(\sin x) = \frac{|\sin 2x|}{3 + \sin^2 x}$. Since $\sin 2x = 2\sin x \cos x$, the function $|\sin 2x|$ depends on $\sin x$ and $\cos x = \pm\sqrt{1 - \sin^2 x}$, so it is indeed a function of $\sin x$ (with the sign of $\cos x$ absorbed by the absolute value). Therefore

$\int_0^{\pi} \frac{x |\sin 2x|}{3 + \sin^2 x} \, dx = \frac{\pi}{2} \int_0^{\pi} \frac{|\sin 2x|}{3 + \sin^2 x} \, dx.$

Substituting $c = \cos x$, $dc = -\sin x \, dx$, and using $\sin 2x = 2\sin x \cos x$:

$\int_0^{\pi} \frac{|\sin 2x|}{3 + \sin^2 x} \, dx = \int_0^{\pi} \frac{2|\cos x| \sin x}{4 - \cos^2 x} \, dx = \int_1^{-1} \frac{-2|c|}{4 - c^2} \, dc = \int_{-1}^{1} \frac{2|c|}{4 - c^2} \, dc.$

Since the integrand is even:

$= 4\int_0^1 \frac{c}{4 - c^2} \, dc.$

Substituting $u = 4 - c^2$, $du = -2c \, dc$:

$= 4 \cdot \left[-\frac{1}{2}\ln(4 - c^2)\right]_0^1 = -2[\ln 3 - \ln 4] = 2\ln\frac{4}{3}.$

Therefore

$\int_0^{\pi} \frac{x |\sin 2x|}{3 + \sin^2 x} \, dx = \frac{\pi}{2} \cdot 2\ln\frac{4}{3} = \pi \ln\frac{4}{3}.$

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Question 5

Topic: 取整函数与面积 Floor Functions and Area  |  Difficulty: Standard  |  Marks: 20

Problem

5 The notation $[x]$ denotes the greatest integer less than or equal to the real number $x$. Thus, for example, $[\pi] = 3$, $[18] = 18$ and $[-4.2] = -5$.

(i) Two curves are given by $y = x^2 + 3x - 1$ and $y = x^2 + 3[x] - 1$. Sketch the curves for $1 \le x \le 3$, on the same axes.

Find the area between the two curves for $1 \le x \le n$, where $n$ is a positive integer.

(ii) Two curves are given by $y = x^2 + 3x - 1$ and $y = [x]^2 + 3[x] - 1$. Sketch the curves for $1 \le x \le 3$, on the same axes.

Show that the area between the two curves for $1 \le x \le n$, where $n$ is a positive integer, is

$\frac{1}{6}(n - 1)(3n + 11).$

Hint

The crucial observation here is that the integer-part (or INT or “floor”) function is a whole number. Thus, when drawing the graphs, the two curves must coincide at the left-hand (integer) endpoints of each unit range, with the second curve slowly falling behind in the first instance, and remaining at the integer level in the second. Note that the curves with the INT function-bits in them will jump at integer values, and you should not therefore join them up at the right-hand ends (to form a continuous curve). The easiest approach in (i) is not to consider $\int y_1 dx - \int y_2 dx$ (i.e. separately), but rather $\int (y_1 - y_2) dx$. This gives a multiple of $x - [x]$ to consider at each step, and this simply gives a series of “unit” right-angled triangles of area $\frac{1}{2}$ to be summed. In (ii), several possible approaches can be used, depending upon how you approached (i). If you again focus on the difference in area across a representative integer range, then you end up having to sum $k + \frac{11}{6}$ from $k = 1$ to $k = n - 1$. Otherwise, there is some integration (for the continuous curves) and some summation (for the integer-part lines) to be done, which may require the use of standard summation results for $\sum k$ and $\sum k^2$. Answers: (i) $\frac{3}{2} n(n - 1)$.**

Model Solution

Part (i)

Sketch description. The curve $y = x^2 + 3x - 1$ is a continuous parabola. The curve $y = x^2 + 3[x] - 1$ is a staircase-like curve: on each interval $[m, m+1)$ where $m$ is an integer, $[x] = m$, so the curve is $y = x^2 + 3m - 1$, which is a piece of a parabola shifted down by $3(x - [x])$ relative to the continuous one. At integer points $x = m$, both curves take the value $m^2 + 3m - 1$, so they coincide. Between integers, the second curve falls below the first.

Finding the area between the curves for $1 \le x \le n$.

The difference between the two curves is:

$f(x) - g(x) = (x^2 + 3x - 1) - (x^2 + 3[x] - 1) = 3(x - [x]) = 3\{x\},$

where $\{x\} = x - [x]$ is the fractional part of $x$.

On each interval $[m, m+1)$ for $m = 1, 2, \ldots, n-1$, we have $[x] = m$ and $\{x\} = x - m$, so:

$\int_m^{m+1} 3(x - m) \, dx = 3\left[\frac{(x-m)^2}{2}\right]_m^{m+1} = 3 \cdot \frac{1}{2} = \frac{3}{2}.$

Each unit interval contributes an area of $\frac{3}{2}$ (a right-angled triangle with base 1 and height 3).

Summing over the $n - 1$ unit intervals:

$\text{Area} = \sum_{m=1}^{n-1} \frac{3}{2} = \frac{3}{2}(n - 1).$

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Part (ii)

Sketch description. Now the second curve is $y = [x]^2 + 3[x] - 1$, which is a true staircase function: on each interval $[m, m+1)$, it takes the constant value $m^2 + 3m - 1$. The first curve $y = x^2 + 3x - 1$ is still the continuous parabola. At integer $x = m$, both curves equal $m^2 + 3m - 1$.

Showing the area is $\frac{1}{6}(n-1)(3n+11)$.

The area between the curves for $1 \le x \le n$ is:

$A = \int_1^n \left[(x^2 + 3x - 1) - ([x]^2 + 3[x] - 1)\right] dx = \int_1^n \left[x^2 + 3x - [x]^2 - 3[x]\right] dx.$

Splitting into unit intervals where $[x] = m$:

$A = \sum_{m=1}^{n-1} \int_m^{m+1} (x^2 + 3x - m^2 - 3m) \, dx.$

We compute the integral for a general interval $[m, m+1]$:

$\int_m^{m+1} (x^2 + 3x - m^2 - 3m) \, dx = \int_m^{m+1} (x^2 - m^2) \, dx + 3\int_m^{m+1} (x - m) \, dx.$

For the first part, substituting $u = x - m$:

$\int_m^{m+1} (x^2 - m^2) \, dx = \int_0^1 ((m+u)^2 - m^2) \, du = \int_0^1 (2mu + u^2) \, du = m + \frac{1}{3}.$

For the second part:

$3\int_m^{m+1} (x - m) \, dx = 3 \int_0^1 u \, du = \frac{3}{2}.$

So each interval contributes $m + \frac{1}{3} + \frac{3}{2} = m + \frac{11}{6}$.

Summing over $m = 1$ to $n - 1$:

$A = \sum_{m=1}^{n-1} \left(m + \frac{11}{6}\right) = \sum_{m=1}^{n-1} m + \frac{11}{6}(n-1).$

Using the standard formula $\sum_{m=1}^{n-1} m = \frac{(n-1)n}{2}$:

$A = \frac{(n-1)n}{2} + \frac{11(n-1)}{6} = (n-1)\left(\frac{n}{2} + \frac{11}{6}\right) = (n-1) \cdot \frac{3n + 11}{6} = \frac{(n-1)(3n+11)}{6}. \qquad \text{(as required)}$

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Question 6

Topic: 向量与不等式 Vectors and Inequalities  |  Difficulty: Standard  |  Marks: 20

Problem

6 By considering a suitable scalar product, prove that $(ax + by + cz)^2 \leqslant (a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$ for any real numbers $a, b, c, x, y$ and $z$. Deduce a necessary and sufficient condition on $a, b, c, x, y$ and $z$ for the following equation to hold: $(ax + by + cz)^2 = (a^2 + b^2 + c^2)(x^2 + y^2 + z^2).$ (i) Show that $(x + 2y + 2z)^2 \leqslant 9(x^2 + y^2 + z^2)$ for any real numbers $x, y$ and $z$, and use this result to solve the equation $(x + 56)^2 = 9(x^2 + 392)$.

(ii) Find real numbers $p, q$ and $r$ that satisfy both $p^2 + 4q^2 + 9r^2 = 729 \text{ and } 8p + 8q + 3r = 243.$

Hint

Q6 The two vectors to be used are clearly $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$. The inequality arises when you

note that $\cos^2 \theta \le 1$. The statement is an equality (equation) when $\cos \theta = \pm 1$, in which case the two vectors must be parallel, so that one is a (non-zero) multiple of the other. [The question cites an example of a result widely known as the Cauchy-Schwarz Inequality.] The equality case of the inequality is then used in the two following parts; simply in (i) – since we must have $y = z = \dots$, from which it follows that $x = \frac{1}{2}$ this. In (ii), you should check that this is indeed an equality case of the inequality when the two vectors are $\dots$ and $\dots$. The parallel condition (one being a multiple of the other) now gives $p$, $q$ and $r$ in terms of some parameter (say $\lambda$), and you can substitute them into the linear equation (of the two given this is clearly the more straightforward one to use), find $\lambda$, and then deduce $p$, $q$ and $r$; these values actually being unique.

Answers: $x = \lambda a$, $y = \lambda b$ and $z = \lambda c$; (i) $x = 7$; (ii) $p = 24$, $q = 6$, $r = 1$.

Model Solution

General result: Cauchy-Schwarz inequality.

Let $\mathbf{u} = (a, b, c)$ and $\mathbf{v} = (x, y, z)$ be vectors in $\mathbb{R}^3$. Their scalar (dot) product is:

$\mathbf{u} \cdot \mathbf{v} = ax + by + cz.$

By definition, $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$, where $\theta$ is the angle between the vectors. Therefore:

$(ax + by + cz)^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 \cos^2\theta.$

Since $\cos^2\theta \le 1$ for all $\theta$, we have:

$(ax + by + cz)^2 \le |\mathbf{u}|^2 |\mathbf{v}|^2 = (a^2 + b^2 + c^2)(x^2 + y^2 + z^2). \qquad \text{(as required)}$

Equality condition. Equality holds if and only if $\cos^2\theta = 1$, i.e. $\theta = 0$ or $\theta = \pi$. This means $\mathbf{u}$ and $\mathbf{v}$ are parallel, so one is a scalar multiple of the other (assuming neither is the zero vector):

$(x, y, z) = \lambda(a, b, c) \quad \text{for some scalar } \lambda \ne 0,$

i.e. $x = \lambda a$, $y = \lambda b$, $z = \lambda c$.

---

Part (i)

We apply the Cauchy-Schwarz inequality with $(a, b, c) = (1, 2, 2)$ and the general vector $(x, y, z)$:

$(x + 2y + 2z)^2 \le (1^2 + 2^2 + 2^2)(x^2 + y^2 + z^2) = 9(x^2 + y^2 + z^2). \qquad \text{(as required)}$

Solving $(x + 56)^2 = 9(x^2 + 392)$.

We rewrite the equation to identify it as an equality case of the above inequality. Note that $56 = 2 \cdot 14 + 2 \cdot 14$ and $392 = 14^2 + 14^2 = 196 + 196$. So:

$(x + 56)^2 = (x + 2 \cdot 14 + 2 \cdot 14)^2 = (1 \cdot x + 2 \cdot 14 + 2 \cdot 14)^2,$

$9(x^2 + 392) = 9(x^2 + 14^2 + 14^2).$

The equation becomes $(1 \cdot x + 2 \cdot 14 + 2 \cdot 14)^2 = 9(x^2 + 14^2 + 14^2)$, which is exactly the equality case of the Cauchy-Schwarz inequality with $\mathbf{u} = (1, 2, 2)$ and $\mathbf{v} = (x, 14, 14)$.

By the equality condition, $\mathbf{v} = \lambda \mathbf{u}$ for some scalar $\lambda$, so:

$x = \lambda, \quad 14 = 2\lambda, \quad 14 = 2\lambda.$

From the second equation, $\lambda = 7$, giving $x = 7$.

Verification: $(7 + 56)^2 = 63^2 = 3969$ and $9(49 + 392) = 9 \times 441 = 3969$.

Therefore $x = 7$.

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Part (ii)

We have the system:

$p^2 + 4q^2 + 9r^2 = 729 \qquad \text{and} \qquad 8p + 8q + 3r = 243.$

We rewrite the linear equation to set up Cauchy-Schwarz. Note that $8p + 8q + 3r = 8 \cdot p + 4 \cdot (2q) + 1 \cdot (3r)$. Applying Cauchy-Schwarz with $\mathbf{u} = (p, 2q, 3r)$ and $\mathbf{v} = (8, 4, 1)$:

$(8p + 4(2q) + 1(3r))^2 \le (p^2 + (2q)^2 + (3r)^2)(8^2 + 4^2 + 1^2),$

$(8p + 8q + 3r)^2 \le (p^2 + 4q^2 + 9r^2)(64 + 16 + 1),$

$243^2 \le 729 \times 81.$

Computing both sides: $243^2 = 59049$ and $729 \times 81 = 59049$. So equality holds!

By the equality condition, $(p, 2q, 3r) = \lambda(8, 4, 1)$ for some scalar $\lambda$:

$p = 8\lambda, \quad 2q = 4\lambda \implies q = 2\lambda, \quad 3r = \lambda \implies r = \frac{\lambda}{3}.$

Substituting into the linear equation:

$8(8\lambda) + 8(2\lambda) + 3 \cdot \frac{\lambda}{3} = 243,$

$64\lambda + 16\lambda + \lambda = 243,$

$81\lambda = 243 \implies \lambda = 3.$

Therefore:

$p = 24, \quad q = 6, \quad r = 1.$

Verification: $p^2 + 4q^2 + 9r^2 = 576 + 144 + 9 = 729$ and $8p + 8q + 3r = 192 + 48 + 3 = 243$.

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Question 7

Topic: 解析几何 Analytic Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

7 An ellipse has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are positive. Show that the equation of the tangent at the point $(a \cos \alpha, b \sin \alpha)$ is $y = -\frac{b \cot \alpha}{a} x + b \csc \alpha.$

The point $A$ has coordinates $(-a, -b)$. The point $E$ has coordinates $(-a, 0)$ and the point $P$ has coordinates $(a, kb)$, where $0 < k < 1$. The line through $E$ parallel to $AP$ meets the line $y = b$ at the point $Q$. Show that the line $PQ$ is tangent to the above ellipse at the point given by $\tan(\alpha/2) = k$.

Determine by means of sketches, or otherwise, whether this result holds also for $k = 0$ and $k = 1$.

Hint

Q7 This is a reasonably routine question to begin with. The general gradient to the curve can be found by differentiating either implicitly or parametrically. Finding the gradient and equation of line $AP$ is also standard enough; as is setting $y = b$ in order

to find the coordinates of $Q$: $\left( \frac{(1 - k)a}{(1 + k)}, b \right)$. The equation of line $PQ$ follows a similar

line of working, to get $y = \left( \frac{-(1 - k^2)b}{2ka} \right)x + \frac{b(1 + k^2)}{2k}$. If you are not familiar with the

$t = \tan \frac{1}{2}$-angle identities, the next part should still not prove too taxing, as you should be able to quote, or derive (from the formula for $\tan(A + B)$ in the formula books), the formula for $\tan 2A$ soon enough; and the widely known, “Pythagorean”, identity $\csc^2 A = 1 + \cot^2 A$ will help you sort out the gradient and intercept of $PQ$ to show that the two forms of this line are indeed the same when $k = \tan(\frac{1}{2} \alpha)$.

A sketch of the ellipse, though not explicitly asked-for, should be made (at least once) so that you can draw on the lines $PQ$ in the cases $k = 0$ and $k = 1$.

Answers: Yes; $PQ$ is the vertical tangent to the ellipse. Yes; $PQ$ is the horizontal tangent to the ellipse.

Model Solution

Part 1: Deriving the tangent equation

The ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Differentiating implicitly with respect to $x$:

$\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}.$

At the point $(a\cos\alpha,\, b\sin\alpha)$:

$\frac{dy}{dx} = -\frac{b^2 \cdot a\cos\alpha}{a^2 \cdot b\sin\alpha} = -\frac{b\cos\alpha}{a\sin\alpha} = -\frac{b\cot\alpha}{a}.$

Using point-slope form $y - y_0 = m(x - x_0)$:

$y - b\sin\alpha = -\frac{b\cot\alpha}{a}(x - a\cos\alpha).$

Expanding the right side:

$y = -\frac{b\cot\alpha}{a}\,x + b\cot\alpha\cos\alpha + b\sin\alpha.$

Simplifying the constant term:

$b\cot\alpha\cos\alpha + b\sin\alpha = b\!\left(\frac{\cos^2\alpha}{\sin\alpha} + \sin\alpha\right) = b\!\left(\frac{\cos^2\alpha + \sin^2\alpha}{\sin\alpha}\right) = \frac{b}{\sin\alpha} = b\csc\alpha.$

Therefore the tangent is:

$y = -\frac{b\cot\alpha}{a}\,x + b\csc\alpha. \qquad \text{(as required)}$

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Part 2: Showing $PQ$ is tangent at the point where $\tan(\alpha/2) = k$

We have $A = (-a, -b)$, $E = (-a, 0)$, $P = (a, kb)$ with $0 < k < 1$.

Gradient of $AP$:

$m_{AP} = \frac{kb - (-b)}{a - (-a)} = \frac{(1+k)b}{2a}.$

Finding $Q$: The line through $E$ parallel to $AP$ has the same gradient. Its equation is:

$y - 0 = \frac{(1+k)b}{2a}(x + a).$

Setting $y = b$:

$b = \frac{(1+k)b}{2a}(x_Q + a) \implies 1 = \frac{(1+k)}{2a}(x_Q + a) \implies x_Q + a = \frac{2a}{1+k}.$

$x_Q = \frac{2a}{1+k} - a = \frac{2a - a(1+k)}{1+k} = \frac{a(1-k)}{1+k}.$

So $Q = \left(\dfrac{(1-k)a}{1+k},\, b\right)$.

Gradient of $PQ$:

$\Delta x = a - \frac{(1-k)a}{1+k} = \frac{a(1+k) - a(1-k)}{1+k} = \frac{2ak}{1+k}.$

$\Delta y = kb - b = (k-1)b.$

$m_{PQ} = \frac{(k-1)b}{\dfrac{2ak}{1+k}} = \frac{(k-1)(1+k)b}{2ak} = \frac{(k^2-1)b}{2ak}.$

Equation of $PQ$: Using point $P(a, kb)$:

$y - kb = \frac{(k^2-1)b}{2ak}(x - a).$

$y = \frac{(k^2-1)b}{2ak}\,x - \frac{(k^2-1)b}{2k} + kb = \frac{(k^2-1)b}{2ak}\,x + \frac{b(k^2+1)}{2k}. \qquad (\star)$

Verifying with $\tan(\alpha/2) = k$: We need to show $(\star)$ matches $y = -\dfrac{b\cot\alpha}{a}\,x + b\csc\alpha$.

Using the half-angle identities with $t = \tan(\alpha/2) = k$:

$\sin\alpha = \frac{2t}{1+t^2} = \frac{2k}{1+k^2}, \qquad \cos\alpha = \frac{1-t^2}{1+t^2} = \frac{1-k^2}{1+k^2}.$

Slope check:

$\cot\alpha = \frac{\cos\alpha}{\sin\alpha} = \frac{1-k^2}{2k}, \qquad -\frac{b\cot\alpha}{a} = -\frac{b(1-k^2)}{2ak} = \frac{(k^2-1)b}{2ak}. \checkmark$

Intercept check:

$\csc\alpha = \frac{1}{\sin\alpha} = \frac{1+k^2}{2k}, \qquad b\csc\alpha = \frac{b(1+k^2)}{2k}. \checkmark$

Both the slope and intercept match, so the line $PQ$ is indeed tangent to the ellipse at the point where $\tan(\alpha/2) = k$.

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Part 3: Cases $k = 0$ and $k = 1$

Case $k = 0$: Then $P = (a, 0)$ (the right vertex of the ellipse). The gradient of $AP$ is $\dfrac{b}{2a}$, and $Q = (a, b)$. The line $PQ$ passes through $(a, 0)$ and $(a, b)$, so it is the vertical line $x = a$, which is the tangent at the right vertex $(a, 0)$. This is consistent with $\tan(\alpha/2) = 0$, giving $\alpha = 0$ and the point $(a\cos 0, b\sin 0) = (a, 0)$. So the result holds.

Case $k = 1$: Then $P = (a, b)$. The gradient of $AP$ is $\dfrac{b}{a}$, and as $k \to 1$, $x_Q = \dfrac{(1-k)a}{1+k} \to 0$, so $Q \to (0, b)$. The line $PQ$ passes through $(a, b)$ and $(0, b)$, which is the horizontal line $y = b$, the tangent at the top vertex $(0, b)$. This is consistent with $\tan(\alpha/2) = 1$, giving $\alpha = \pi/2$ and the point $(a\cos(\pi/2), b\sin(\pi/2)) = (0, b)$. So the result also holds.

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Question 8

Topic: 向量几何 Vector Geometry  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Show that the line through the points with position vectors $\mathbf{x}$ and $\mathbf{y}$ has equation $\mathbf{r} = (1 - \alpha)\mathbf{x} + \alpha\mathbf{y},$ where $\alpha$ is a scalar parameter.

The sides $OA$ and $CB$ of a trapezium $OABC$ are parallel, and $OA > CB$. The point $E$ on $OA$ is such that $OE : EA = 1 : 2$, and $F$ is the midpoint of $CB$. The point $D$ is the intersection of $OC$ produced and $AB$ produced; the point $G$ is the intersection of $OB$ and $EF$; and the point $H$ is the intersection of $DG$ produced and $OA$. Let $\mathbf{a}$ and $\mathbf{c}$ be the position vectors of the points $A$ and $C$, respectively, with respect to the origin $O$.

(i) Show that $B$ has position vector $\lambda\mathbf{a} + \mathbf{c}$ for some scalar parameter $\lambda$.

(ii) Find, in terms of $\mathbf{a}, \mathbf{c}$ and $\lambda$ only, the position vectors of $D, E, F, G$ and $H$. Determine the ratio $OH : HA$.

Hint

Q8 I’m afraid that this question involves but a single idea: namely, that of intersecting lines. The first two parts are simple “bookwork” tasks, requiring nothing more than an explanation of the vector form of a line equation as $\mathbf{r} = \text{p.v. of any point on the line} + \text{some scalar multiple of any vector (such as } \mathbf{y} - \mathbf{x}, \text{ in this case) parallel to the line}$; then the basic observation that $CB \parallel OA \implies \vec{CB} = \lambda \mathbf{a}$ to justify the second result.

Thereafter, it is simply a case, with (admittedly) increasingly complicated looking position vectors coming into play, of equating a’s and c’s in pairs of lines to find out the position vector of the point where they intersect. If the final part is to be answered numerically, then the parameter $\lambda$ must cancel somewhere before the end.

Answers: (ii) $\mathbf{d} = \left( \frac{1}{1 - \lambda} \right) \mathbf{c}$ ; $\mathbf{e} = \frac{1}{3} \mathbf{a}$ ; $\mathbf{f} = \mathbf{c} + \frac{1}{2} \lambda \mathbf{a}$ ; $\mathbf{g} = \left( \frac{2\lambda}{2 + 3\lambda} \right) \mathbf{a} + \left( \frac{2}{2 + 3\lambda} \right) \mathbf{c}$ ; $\mathbf{h} = \frac{2}{5} \mathbf{a}$.

Thus $OH : HA = 2 : 3$ (as $H$ lies two-fifths of the way along the line $OA$).

Model Solution

Part 1: Deriving the line equation

A point on the line through $\mathbf{x}$ and $\mathbf{y}$ can be written as a starting point plus a scalar multiple of the direction vector:

$\mathbf{r} = \mathbf{x} + t(\mathbf{y} - \mathbf{x}),$

where $t$ is a scalar parameter. Setting $\alpha = t$ and rearranging:

$\mathbf{r} = (1 - \alpha)\mathbf{x} + \alpha\mathbf{y}. \qquad \text{(as required)}$

Note: when $\alpha = 0$, $\mathbf{r} = \mathbf{x}$; when $\alpha = 1$, $\mathbf{r} = \mathbf{y}$, confirming that both points lie on the line.

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Part (i): Position vector of $B$

In trapezium $OABC$, $OA \parallel CB$ and $OA > CB$. Since $O$ is the origin, the position vector of $A$ is $\mathbf{a}$, so $\vec{OA} = \mathbf{a}$. The position vector of $C$ is $\mathbf{c}$.

Since $CB \parallel OA$, we have $\vec{CB} = \lambda \mathbf{a}$ for some scalar $\lambda$. Since $OA > CB$, we have $0 < \lambda < 1$.

Therefore the position vector of $B$ is:

$\mathbf{b} = \vec{OB} = \vec{OC} + \vec{CB} = \mathbf{c} + \lambda\mathbf{a} = \lambda\mathbf{a} + \mathbf{c}. \qquad \text{(as required)}$

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Part (ii): Finding position vectors of $D, E, F, G, H$

Point $E$: On $OA$ with $OE : EA = 1 : 2$, so $E$ is one-third of the way from $O$ to $A$:

$\mathbf{e} = \frac{1}{3}\mathbf{a}.$

Point $F$: Midpoint of $CB$:

$\mathbf{f} = \frac{\mathbf{c} + \mathbf{b}}{2} = \frac{\mathbf{c} + \lambda\mathbf{a} + \mathbf{c}}{2} = \frac{\lambda\mathbf{a} + 2\mathbf{c}}{2} = \frac{\lambda}{2}\mathbf{a} + \mathbf{c}.$

Point $D$: Intersection of $OC$ produced and $AB$ produced.

Line $OC$: $\mathbf{r} = t\mathbf{c}$ for scalar $t$.

Line $AB$: $\mathbf{r} = \mathbf{a} + s(\mathbf{b} - \mathbf{a}) = \mathbf{a} + s(\lambda\mathbf{a} + \mathbf{c} - \mathbf{a}) = (1 - s + s\lambda)\mathbf{a} + s\mathbf{c}$.

At the intersection, equating coefficients of the independent vectors $\mathbf{a}$ and $\mathbf{c}$:

From $\mathbf{a}$: $\quad 0 = 1 - s + s\lambda = 1 - s(1 - \lambda) \implies s = \dfrac{1}{1 - \lambda}.$

From $\mathbf{c}$: $\quad t = s = \dfrac{1}{1 - \lambda}.$

Therefore:

$\mathbf{d} = \frac{1}{1 - \lambda}\,\mathbf{c}.$

Point $G$: Intersection of $OB$ and $EF$.

Line $OB$: $\mathbf{r} = u(\lambda\mathbf{a} + \mathbf{c}) = u\lambda\mathbf{a} + u\mathbf{c}$.

Line $EF$: $\mathbf{r} = \mathbf{e} + v(\mathbf{f} - \mathbf{e})$.

Computing $\mathbf{f} - \mathbf{e}$:

$\mathbf{f} - \mathbf{e} = \left(\frac{\lambda}{2}\mathbf{a} + \mathbf{c}\right) - \frac{1}{3}\mathbf{a} = \left(\frac{\lambda}{2} - \frac{1}{3}\right)\mathbf{a} + \mathbf{c} = \frac{3\lambda - 2}{6}\,\mathbf{a} + \mathbf{c}.$

So line $EF$: $\mathbf{r} = \frac{1}{3}\mathbf{a} + v\!\left(\frac{3\lambda - 2}{6}\,\mathbf{a} + \mathbf{c}\right) = \left(\frac{1}{3} + \frac{v(3\lambda - 2)}{6}\right)\mathbf{a} + v\mathbf{c}.$

Equating coefficients at the intersection:

From $\mathbf{c}$: $\quad u = v.$

From $\mathbf{a}$: $\quad u\lambda = \frac{1}{3} + \frac{u(3\lambda - 2)}{6}.$

Multiplying through by 6:

$6u\lambda = 2 + u(3\lambda - 2) = 2 + 3u\lambda - 2u.$

$6u\lambda - 3u\lambda + 2u = 2 \implies u(3\lambda + 2) = 2 \implies u = \frac{2}{2 + 3\lambda}.$

Therefore:

$\mathbf{g} = u\lambda\mathbf{a} + u\mathbf{c} = \frac{2\lambda}{2 + 3\lambda}\,\mathbf{a} + \frac{2}{2 + 3\lambda}\,\mathbf{c}.$

Point $H$: Intersection of $DG$ produced and $OA$.

Line $OA$: $\mathbf{r} = w\mathbf{a}$ (points on $OA$ have zero $\mathbf{c}$-component).

Line $DG$: $\mathbf{r} = \mathbf{d} + t(\mathbf{g} - \mathbf{d})$.

$\mathbf{g} - \mathbf{d} = \frac{2\lambda}{2 + 3\lambda}\,\mathbf{a} + \frac{2}{2 + 3\lambda}\,\mathbf{c} - \frac{1}{1 - \lambda}\,\mathbf{c} = \frac{2\lambda}{2 + 3\lambda}\,\mathbf{a} + \left(\frac{2}{2 + 3\lambda} - \frac{1}{1 - \lambda}\right)\mathbf{c}.$

Computing the $\mathbf{c}$-coefficient:

$\frac{2}{2 + 3\lambda} - \frac{1}{1 - \lambda} = \frac{2(1 - \lambda) - (2 + 3\lambda)}{(2 + 3\lambda)(1 - \lambda)} = \frac{2 - 2\lambda - 2 - 3\lambda}{(2 + 3\lambda)(1 - \lambda)} = \frac{-5\lambda}{(2 + 3\lambda)(1 - \lambda)}.$

So line $DG$ has $\mathbf{c}$-component:

$\frac{1}{1 - \lambda} + t \cdot \frac{-5\lambda}{(2 + 3\lambda)(1 - \lambda)}.$

Setting this to zero (since $H$ lies on $OA$):

$\frac{1}{1 - \lambda} = \frac{5t\lambda}{(2 + 3\lambda)(1 - \lambda)} \implies 1 = \frac{5t\lambda}{2 + 3\lambda} \implies t = \frac{2 + 3\lambda}{5\lambda}.$

The $\mathbf{a}$-component of $H$ is:

$t \cdot \frac{2\lambda}{2 + 3\lambda} = \frac{2 + 3\lambda}{5\lambda} \cdot \frac{2\lambda}{2 + 3\lambda} = \frac{2}{5}.$

Therefore:

$\mathbf{h} = \frac{2}{5}\,\mathbf{a}.$

Ratio $OH : HA$: Since $H$ lies on $OA$ with position vector $\frac{2}{5}\mathbf{a}$, the point $H$ is two-fifths of the way from $O$ to $A$. Thus:

$OH : HA = 2 : 3.$