STEP2 1999 -- Pure Mathematics

Exam: STEP2  |  Year: 1999  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Model solutions are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	对数与大数比较 Logarithms and Large Number Comparison	Challenging	Stirling公式,对数换底,数量级分析
2	二次方程与判别式 Quadratic Equations and Discriminant	Standard	判别式,参数分类讨论,不等式推导
3	高阶导数与数学归纳法 Higher Derivatives and Mathematical Induction	Challenging	数学归纳法,乘积求导法则,多项式次数分析
4	组合恒等式 Combinatorial Identities	Challenging	二项式定理,比较系数法,组合恒等式
5	三角方程 Trigonometric Equations	Standard	辅助角公式,三角恒等式,反三角函数
6	积分与换元法 / Integration and Substitution	Challenging	线性分式换元,对数性质化简,分部积分
7	曲线分析与作图 / Curve Analysis and Sketching	Challenging	求导判驻点,参数分类讨论,渐近线分析,曲线作图
8	三角级数与求和 / Trigonometric Series and Summation	Hard	差积化简,三角求和公式,小角近似,微分求和

---

Question 1

Topic: 对数与大数比较 Logarithms and Large Number Comparison  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Let $x = 10^{100}$, $y = 10^x$, $z = 10^y$, and let $a_1 = x!, \quad a_2 = x^y, \quad a_3 = y^x, \quad a_4 = z^x, \quad a_5 = e^{xyz}, \quad a_6 = z^{1/y}, \quad a_7 = y^{z/x}.$

(i) Use Stirling’s approximation $n! \approx \sqrt{2\pi} \, n^{n+\frac{1}{2}} e^{-n}$, which is valid for large $n$, to show that $\log_{10} (\log_{10} a_1) \approx 102$.

(ii) Arrange the seven numbers $a_1, \dots, a_7$ in ascending order of magnitude, justifying your result.

Model Solution

Part (i)

Using Stirling’s approximation with $n = x = 10^{100}$:

$x! \approx \sqrt{2\pi} \, x^{x + \frac{1}{2}} e^{-x}$

Taking $\log_{10}$ of both sides:

$\log_{10}(x!) \approx \log_{10}\!\left(\sqrt{2\pi}\right) + \left(x + \tfrac{1}{2}\right)\log_{10} x - x \log_{10} e$

Since $x = 10^{100}$, we have $\log_{10} x = 100$. The first term $\log_{10}(\sqrt{2\pi}) \approx 0.40$ and the third term $x \log_{10} e \approx 0.434 \times 10^{100}$ are both negligible compared to the second term $(x + \frac{1}{2}) \cdot 100 \approx 10^{102}$. Thus:

$\log_{10}(x!) \approx 100x = 100 \times 10^{100} = 10^{102}$

Taking $\log_{10}$ again:

$\log_{10}\!\left(\log_{10}(x!)\right) \approx \log_{10}\!\left(10^{102}\right) = 102 \qquad \blacksquare$

Part (ii)

For numbers this large, we compare via $\log_{10}$, and when those are still enormous, via $\log_{10}(\log_{10})$. We use $x = 10^{100}$, $y = 10^x$, $z = 10^y$, so $\log_{10} x = 100$, $\log_{10} y = x = 10^{100}$, $\log_{10} z = y = 10^{10^{100}}$.

Step 1: compute $\log_{10} a_k$ for each $k$.

$a_6 = z^{1/y}$:

$\log_{10} a_6 = \frac{1}{y}\log_{10} z = \frac{y}{y} = 1.$

$a_1 = x!$: From part (i), $\log_{10} a_1 \approx 10^{102}$.

$a_3 = y^x$:

$\log_{10} a_3 = x \log_{10} y = x \cdot x = x^2 = 10^{200}.$

$a_5 = e^{xyz}$:

$\log_{10} a_5 = \frac{xyz}{\ln 10}.$

Here $xyz = 10^{100} \cdot 10^{10^{100}} \cdot 10^{10^{10^{100}}} = 10^{100 + 10^{100} + 10^{10^{100}}}$, so

$\log_{10} a_5 \approx 10^{10^{10^{100}}}.$

$a_2 = x^y$:

$\log_{10} a_2 = y \log_{10} x = 100y = 100 \times 10^{10^{100}} \approx 10^{10^{100} + 2}.$

$a_4 = z^x$:

$\log_{10} a_4 = x \log_{10} z = xy = 10^{100} \cdot 10^{10^{100}} = 10^{10^{100} + 100}.$

$a_7 = y^{z/x}$:

Since $z/x = 10^y / 10^{100} = 10^{y - 100}$:

$\log_{10} a_7 = \frac{z}{x}\log_{10} y = 10^{y - 100} \cdot 10^{100} = 10^{y} = 10^{10^{10^{100}}}.$

Step 2: compare the $\log_{10} a_k$ values.

$k$	$\log_{10} a_k$
6	$1$
1	$\approx 10^{102}$
3	$10^{200}$
2	$\approx 10^{10^{100} + 2}$
4	$\approx 10^{10^{100} + 100}$
5	$\approx 10^{10^{10^{100}}}$
7	$= 10^{10^{10^{100}}}$

The first four are clearly ordered: $1 \ll 10^{102} \ll 10^{200} \ll 10^{10^{100}+2}$, giving $a_6 < a_1 < a_3 < a_2$.

Next, $a_2 < a_4$ since $10^{10^{100}+2} \ll 10^{10^{100}+100}$.

For $a_4$ vs $a_5$ vs $a_7$: we have $10^{10^{100}+100} \ll 10^{10^{10^{100}}}$, so $a_4$ is smaller than both $a_5$ and $a_7$.

Comparing $a_5$ and $a_7$: $\log_{10} a_5 = \frac{xyz}{\ln 10}$ and $\log_{10} a_7 = 10^y$. Now

$\frac{xyz}{\ln 10} = \frac{10^{100} \cdot 10^{10^{100}} \cdot 10^{10^{10^{100}}}}{\ln 10} = \frac{10^{10^{10^{100}}} \cdot 10^{10^{100} + 100}}{\ln 10}.$

Since $\frac{10^{10^{100}+100}}{\ln 10} \gg 1$, we get $\log_{10} a_5 \gg \log_{10} a_7 = 10^{10^{10^{100}}}$, so $a_5 > a_7$.

Ascending order:

$\boxed{a_6 \;<\; a_1 \;<\; a_3 \;<\; a_2 \;<\; a_4 \;<\; a_7 \;<\; a_5}$

---

Question 2

Topic: 二次方程与判别式 Quadratic Equations and Discriminant  |  Difficulty: Standard  |  Marks: 20

Problem

2 Consider the quadratic equation $nx^2 + 2x\sqrt{(pn^2 + q)} + rn + s = 0, \qquad \text{(*)}$ where $p > 0$, $p \neq r$ and $n = 1, 2, 3, \dots$.

(i) For the case where $p = 3$, $q = 50$, $r = 2$, $s = 15$, find the set of values of $n$ for which equation ($*$) has no real roots.

(ii) Prove that if $p < r$ and $4q(p - r) > s^2$, then ($*$) has no real roots for any value of $n$.

(iii) If $n = 1$, $p - r = 1$ and $q = s^2/8$, show that ($*$) has real roots if, and only if, $s \leqslant 4 - 2\sqrt{2}$ or $s \geqslant 4 + 2\sqrt{2}$.

Model Solution

Part (i)

With $p = 3$, $q = 50$, $r = 2$, $s = 15$, the equation becomes

$nx^2 + 2x\sqrt{3n^2 + 50} + 2n + 15 = 0.$

For no real roots, the discriminant must be negative:

$D = 4(3n^2 + 50) - 4n(2n + 15) < 0$

$12n^2 + 200 - 8n^2 - 60n < 0$

$4n^2 - 60n + 200 < 0$

$n^2 - 15n + 50 < 0$

$(n - 5)(n - 10) < 0$

This holds when $5 < n < 10$. Since $n$ is a positive integer, the set of values is $n \in \{6, 7, 8, 9\}$.

Part (ii)

The discriminant of $nx^2 + 2x\sqrt{pn^2 + q} + rn + s = 0$ is

$D = 4(pn^2 + q) - 4n(rn + s) = 4\left[(p - r)n^2 - sn + q\right].$

We need to show $D < 0$ for all positive integers $n$ when $p < r$ and $4q(p - r) > s^2$.

Since $p < r$, we have $p - r < 0$. Write $p - r = -(r - p)$ where $r - p > 0$. Then:

$D = 4\left[-(r - p)n^2 - sn + q\right]$

We want to show $(r - p)n^2 + sn - q > 0$ for all positive integers $n$. Consider the quadratic

$g(n) = (r - p)n^2 + sn - q$

as a function of real $n$. Its discriminant is

$\Delta_g = s^2 + 4q(r - p).$

The condition $4q(p - r) > s^2$ rearranges to $s^2 < 4q(p - r) = -4q(r - p)$, so

$s^2 + 4q(r - p) < 0,$

meaning $\Delta_g < 0$. Since the leading coefficient $r - p > 0$ and the discriminant is negative, $g(n) > 0$ for all real $n$. In particular, $g(n) > 0$ for all positive integers $n$, so $D < 0$ and the equation has no real roots for any value of $n$. $\qquad \blacksquare$

Part (iii)

With $n = 1$ and $p - r = 1$ (so $p = r + 1$), the equation becomes

$x^2 + 2x\sqrt{p + q} + r + s = 0.$

For real roots, the discriminant must satisfy:

$D = 4(p + q) - 4(r + s) \geqslant 0$

$p + q - r - s \geqslant 0$

Since $p - r = 1$:

$1 + q - s \geqslant 0$

Substituting $q = s^2/8$:

$1 + \frac{s^2}{8} - s \geqslant 0$

Multiplying through by 8:

$s^2 - 8s + 8 \geqslant 0$

Using the quadratic formula to find the roots of $s^2 - 8s + 8 = 0$:

$s = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = \frac{8 \pm 4\sqrt{2}}{2} = 4 \pm 2\sqrt{2}$

Since the quadratic $s^2 - 8s + 8$ opens upward, it is non-negative outside the roots:

$s^2 - 8s + 8 \geqslant 0 \iff s \leqslant 4 - 2\sqrt{2} \quad \text{or} \quad s \geqslant 4 + 2\sqrt{2}$

Therefore the equation has real roots if and only if $s \leqslant 4 - 2\sqrt{2}$ or $s \geqslant 4 + 2\sqrt{2}$. $\qquad \blacksquare$

---

Question 3

Topic: 高阶导数与数学归纳法 Higher Derivatives and Mathematical Induction  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Let $S_n(x) = e^{x^3} \frac{d^n}{dx^n} \left( e^{-x^3} \right).$ Show that $S_2(x) = 9x^4 - 6x$ and find $S_3(x)$.

Prove by induction on $n$ that $S_n(x)$ is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient of the highest power of $x$.

Show also that if $\frac{dS_n}{dx} = 0$ for some value $a$ of $x$, then $S_n(a)S_{n+1}(a) \leqslant 0$.

Model Solution

We write $f(x) = e^{-x^3}$, so $S_n(x) = e^{x^3} f^{(n)}(x)$.

Computing $S_2(x)$. Differentiate $f(x) = e^{-x^3}$:

$f'(x) = -3x^2 \, e^{-x^3}.$

Differentiate again using the product rule:

$f''(x) = \bigl(-6x + 9x^4\bigr) e^{-x^3}.$

Therefore

$S_2(x) = e^{x^3} f''(x) = 9x^4 - 6x. \qquad \checkmark$

Computing $S_3(x)$. Differentiate $f''(x) = (-6x + 9x^4)e^{-x^3}$ using the product rule:

$f'''(x) = \bigl(-6 + 36x^3\bigr) e^{-x^3} + \bigl(-6x + 9x^4\bigr)\bigl(-3x^2\bigr) e^{-x^3}.$

Expanding the second product:

$(-6x + 9x^4)(-3x^2) = 18x^3 - 27x^6.$

Combining:

$f'''(x) = \bigl(-6 + 36x^3 + 18x^3 - 27x^6\bigr) e^{-x^3} = \bigl(-27x^6 + 54x^3 - 6\bigr) e^{-x^3}.$

Therefore

$S_3(x) = e^{x^3} f'''(x) = -27x^6 + 54x^3 - 6.$

Recurrence relation. From $S_n(x) = e^{x^3} f^{(n)}(x)$, differentiating with the product rule:

$S_n'(x) = 3x^2 \, e^{x^3} f^{(n)}(x) + e^{x^3} f^{(n+1)}(x) = 3x^2 S_n(x) + S_{n+1}(x).$

Rearranging:

$S_{n+1}(x) = S_n'(x) - 3x^2 S_n(x). \qquad (\dagger)$

Induction proof. We prove by induction that $S_n(x)$ is a polynomial of degree $2n$ with leading coefficient $(-3)^n$.

Base case. $S_0(x) = e^{x^3} \cdot e^{-x^3} = 1$, a polynomial of degree $0 = 2 \times 0$ with leading coefficient $1 = (-3)^0$.

Inductive step. Suppose $S_n(x)$ is a polynomial of degree $2n$ with leading coefficient $(-3)^n$, so

$S_n(x) = (-3)^n x^{2n} + \text{lower-order terms}.$

Then $S_n'(x) = 2n(-3)^n x^{2n-1} + \cdots$ is a polynomial of degree $2n - 1$.

Also $3x^2 S_n(x) = 3(-3)^n x^{2n+2} + \cdots$, a polynomial of degree $2n + 2$.

Since $3(-3)^n = 3 \cdot (-1)^n \cdot 3^n = (-1)^n \cdot 3^{n+1}$ and $(-3)^{n+1} = (-1)^{n+1} \cdot 3^{n+1}$, we have

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

So $3x^2 S_n(x)$ has leading term $-(-3)^{n+1} x^{2n+2}$.

From $(\dagger)$: the term $S_n'(x)$ has degree $2n - 1$, which is less than $2n + 2$, so the leading term of $S_{n+1}(x)$ comes entirely from $-3x^2 S_n(x)$:

$S_{n+1}(x) = \underbrace{S_n'(x)}_{\text{degree } 2n-1} - \underbrace{3x^2 S_n(x)}_{\text{degree } 2n+2} = (-3)^{n+1} x^{2n+2} + \cdots$

Hence $S_{n+1}(x)$ is a polynomial of degree $2(n+1)$ with leading coefficient $(-3)^{n+1}$. This completes the induction.

Conclusion. $S_n(x)$ is a polynomial of degree $2n$ with leading coefficient $(-3)^n$.

Sign property. Suppose $S_n'(a) = 0$ for some value $a$. Substituting into $(\dagger)$:

$S_{n+1}(a) = S_n'(a) - 3a^2 S_n(a) = 0 - 3a^2 S_n(a) = -3a^2 S_n(a).$

Therefore

$S_n(a) \, S_{n+1}(a) = S_n(a) \cdot \bigl(-3a^2 S_n(a)\bigr) = -3a^2 \bigl[S_n(a)\bigr]^2.$

Since $a^2 \geqslant 0$ and $[S_n(a)]^2 \geqslant 0$, we conclude

$S_n(a) \, S_{n+1}(a) = -3a^2 [S_n(a)]^2 \leqslant 0. \qquad \blacksquare$

---

Question 4

Topic: 组合恒等式 Combinatorial Identities  |  Difficulty: Challenging  |  Marks: 20

Problem

4 By considering the expansions in powers of $x$ of both sides of the identity $(1 + x)^n(1 + x)^n \equiv (1 + x)^{2n},$ show that $\sum_{s=0}^{n} \binom{n}{s}^2 = \binom{2n}{n},$ where $\binom{n}{s} = \frac{n!}{s!(n-s)!}$.

By considering similar identities, or otherwise, show also that:

(i) if $n$ is an even integer, then $\sum_{s=0}^{n} (-1)^s \binom{n}{s}^2 = (-1)^{n/2} \binom{n}{n/2};$

(ii) $\sum_{t=1}^{n} 2t \binom{n}{t}^2 = n \binom{2n}{n}$.

Model Solution

Main identity. Expand both sides of $(1+x)^n(1+x)^n = (1+x)^{2n}$.

The left side is

$(1+x)^n \cdot (1+x)^n = \left(\sum_{s=0}^{n} \binom{n}{s} x^s\right)\left(\sum_{t=0}^{n} \binom{n}{t} x^t\right).$

The coefficient of $x^n$ in this product is obtained by pairing terms where $s + t = n$, i.e. $t = n - s$:

$[x^n]: \sum_{s=0}^{n} \binom{n}{s} \binom{n}{n-s} = \sum_{s=0}^{n} \binom{n}{s}^2,$

using $\binom{n}{n-s} = \binom{n}{s}$.

The right side gives $(1+x)^{2n}$, whose coefficient of $x^n$ is $\binom{2n}{n}$.

Equating coefficients of $x^n$:

$\sum_{s=0}^{n} \binom{n}{s}^2 = \binom{2n}{n}. \qquad \blacksquare$

---

Part (i). Consider the identity

$(1+x)^n(1-x)^n = (1-x^2)^n.$

The left side expands as

$\left(\sum_{s=0}^{n} \binom{n}{s} x^s\right)\left(\sum_{t=0}^{n} \binom{n}{t} (-x)^t\right) = \left(\sum_{s=0}^{n} \binom{n}{s} x^s\right)\left(\sum_{t=0}^{n} (-1)^t \binom{n}{t} x^t\right).$

The coefficient of $x^n$ on the left side is

$\sum_{s=0}^{n} \binom{n}{s} \cdot (-1)^{n-s} \binom{n}{n-s} = (-1)^n \sum_{s=0}^{n} (-1)^{-s} \binom{n}{s}^2 = (-1)^n \sum_{s=0}^{n} (-1)^{-s} \binom{n}{s}^2.$

Since $(-1)^{-s} = (-1)^s$ (as $(-1)^{-1} = -1$), this becomes

$(-1)^n \sum_{s=0}^{n} (-1)^s \binom{n}{s}^2.$

The right side is $(1-x^2)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{2k}$.

The coefficient of $x^n$ on the right side is $0$ when $n$ is odd (since $2k$ is always even). When $n$ is even, the term $x^n$ arises from $2k = n$, i.e. $k = n/2$:

$[x^n]_{\text{right}} = \begin{cases} 0 & \text{if } n \text{ is odd}, \\ (-1)^{n/2} \binom{n}{n/2} & \text{if } n \text{ is even}. \end{cases}$

For even $n$, equating coefficients of $x^n$:

$(-1)^n \sum_{s=0}^{n} (-1)^s \binom{n}{s}^2 = (-1)^{n/2} \binom{n}{n/2}.$

Since $n$ is even, $(-1)^n = 1$, so

$\sum_{s=0}^{n} (-1)^s \binom{n}{s}^2 = (-1)^{n/2} \binom{n}{n/2}. \qquad \blacksquare$

---

Part (ii). We use the product $(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$ and differentiate both sides with respect to $x$.

Differentiating: $2n(1+x)^n(1+x)^{n-1} = 2n(1+x)^{2n-1}$.

Equivalently, writing $f(x) = (1+x)^n$:

$f'(x) \cdot f(x) + f(x) \cdot f'(x) = 2n(1+x)^{2n-1},$

which simplifies to $2 \cdot n(1+x)^{n-1}(1+x)^n = 2n(1+x)^{2n-1}$.

Now extract the coefficient of $x^{n-1}$ from each side.

Left side. We have

$2 \left(\sum_{s=0}^{n} s \binom{n}{s} x^{s-1}\right)\left(\sum_{t=0}^{n} \binom{n}{t} x^t\right).$

The coefficient of $x^{n-1}$ is obtained when $(s-1) + t = n-1$, i.e. $t = n - s$:

$2 \sum_{s=0}^{n} s \binom{n}{s} \binom{n}{n-s} = 2 \sum_{s=0}^{n} s \binom{n}{s}^2.$

Right side. The coefficient of $x^{n-1}$ in $2n(1+x)^{2n-1}$ is

$2n \binom{2n-1}{n-1}.$

Equating:

$2 \sum_{s=0}^{n} s \binom{n}{s}^2 = 2n \binom{2n-1}{n-1}.$

Using the identity $\binom{2n-1}{n-1} = \frac{n}{2n} \binom{2n}{n} = \frac{1}{2}\binom{2n}{n}$:

$2 \sum_{s=0}^{n} s \binom{n}{s}^2 = 2n \cdot \frac{1}{2} \binom{2n}{n} = n\binom{2n}{n}.$

Relabelling $s$ as $t$:

$\sum_{t=1}^{n} 2t \binom{n}{t}^2 = n \binom{2n}{n}. \qquad \blacksquare$

---

Question 5

Topic: 三角方程 Trigonometric Equations  |  Difficulty: Standard  |  Marks: 20

Problem

5 Show that if $\alpha$ is a solution of the equation $5\cos x + 12\sin x = 7,$ then either $\cos \alpha = \frac{35 - 12\sqrt{120}}{169}$ or $\cos \alpha$ has one other value which you should find.

Prove carefully that if $\pi/2 < \alpha < \pi$, then $\alpha < 3\pi/4$.

Model Solution

Finding the possible values of $\cos\alpha$. Since $5\cos\alpha + 12\sin\alpha = 7$, we have

$\sin\alpha = \frac{7 - 5\cos\alpha}{12}. \qquad (1)$

Substituting into $\sin^2\alpha + \cos^2\alpha = 1$:

$\left(\frac{7 - 5\cos\alpha}{12}\right)^2 + \cos^2\alpha = 1.$

Expanding:

$\frac{49 - 70\cos\alpha + 25\cos^2\alpha}{144} + \cos^2\alpha = 1.$

Multiply through by $144$:

$49 - 70\cos\alpha + 25\cos^2\alpha + 144\cos^2\alpha = 144.$

$169\cos^2\alpha - 70\cos\alpha - 95 = 0. \qquad (2)$

Applying the quadratic formula with $a = 169$, $b = -70$, $c = -95$:

$\cos\alpha = \frac{70 \pm \sqrt{4900 + 4 \times 169 \times 95}}{338} = \frac{70 \pm \sqrt{4900 + 64220}}{338} = \frac{70 \pm \sqrt{69120}}{338}.$

Simplifying $\sqrt{69120}$: since $69120 = 576 \times 120$, we get $\sqrt{69120} = 24\sqrt{120}$. Therefore

$\cos\alpha = \frac{70 \pm 24\sqrt{120}}{338} = \frac{35 \pm 12\sqrt{120}}{169}.$

So the two possible values are

$\cos\alpha = \frac{35 - 12\sqrt{120}}{169} \qquad \text{or} \qquad \cos\alpha = \frac{35 + 12\sqrt{120}}{169}. \qquad \checkmark$

Checking both values give valid solutions. Note $\sqrt{120} \approx 10.954$, so:

• $\cos\alpha = \frac{35 - 12\sqrt{120}}{169} \approx \frac{35 - 131.45}{169} \approx -0.571$: this gives $\sin\alpha = \frac{7 - 5(-0.571)}{12} = \frac{9.855}{12} > 0$, consistent with $\alpha$ in the second quadrant.

• $\cos\alpha = \frac{35 + 12\sqrt{120}}{169} \approx \frac{166.45}{169} \approx 0.985$: this gives $\sin\alpha = \frac{7 - 5(0.985)}{12} = \frac{2.075}{12} > 0$, consistent with $\alpha$ in the first quadrant.

Both values are genuine solutions of the original equation.

---

Proving $\alpha < 3\pi/4$ when $\pi/2 < \alpha < \pi$. If $\pi/2 < \alpha < \pi$, then $\sin\alpha > 0$ and $\cos\alpha < 0$, so the relevant root is

$\cos\alpha = \frac{35 - 12\sqrt{120}}{169}.$

Since $\cos$ is strictly decreasing on $(\pi/2, \pi)$, the condition $\alpha < 3\pi/4$ is equivalent to $\cos\alpha > \cos(3\pi/4) = -\frac{\sqrt{2}}{2}$.

We need to show:

$\frac{35 - 12\sqrt{120}}{169} > -\frac{\sqrt{2}}{2}.$

Multiply both sides by $338 = 2 \times 169 > 0$:

$70 - 24\sqrt{120} > -169\sqrt{2}.$

Rearranging:

$70 + 169\sqrt{2} > 24\sqrt{120}.$

Since both sides are positive, square both sides:

$\text{LHS}^2 = 4900 + 2 \times 70 \times 169\sqrt{2} + 28561 \times 2 = 4900 + 23660\sqrt{2} + 57122 = 62022 + 23660\sqrt{2}.$

$\text{RHS}^2 = 576 \times 120 = 69120.$

So we need $62022 + 23660\sqrt{2} > 69120$, i.e. $23660\sqrt{2} > 7098$.

Since $\sqrt{2} > 1.4$, we have $23660\sqrt{2} > 23660 \times 1.4 = 33124 > 7098$. This confirms the inequality.

Therefore $\cos\alpha > -\frac{\sqrt{2}}{2}$, which gives $\alpha < \frac{3\pi}{4}$. $\qquad \blacksquare$

---

Question 6

Topic: 积分与换元法 / Integration and Substitution  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Find $\frac{dy}{dx}$ if $y = \frac{ax + b}{cx + d}. \qquad \text{(*)}$

By using changes of variable of the form $(*)$, or otherwise, show that $\int_{0}^{1} \frac{1}{(x + 3)^2} \ln \left( \frac{x + 1}{x + 3} \right) dx = \frac{1}{6} \ln 3 - \frac{1}{4} \ln 2 - \frac{1}{12},$ and evaluate the integrals $\int_{0}^{1} \frac{1}{(x + 3)^2} \ln \left( \frac{x^2 + 3x + 2}{(x + 3)^2} \right) dx \quad \text{and} \quad \int_{0}^{1} \frac{1}{(x + 3)^2} \ln \left( \frac{x + 1}{x + 2} \right) dx.$

Model Solution

Part 1: Finding $\frac{dy}{dx}$

For $y = \dfrac{ax + b}{cx + d}$, apply the quotient rule:

$\frac{dy}{dx} = \frac{a(cx + d) - c(ax + b)}{(cx + d)^2} = \frac{acx + ad - acx - bc}{(cx + d)^2} = \frac{ad - bc}{(cx + d)^2}. \qquad (\dagger)$

This shows that the derivative of a linear fractional function takes the form $\dfrac{k}{(cx+d)^2}$ where $k = ad - bc$ is a constant — exactly the factor appearing in each integral below.

---

Part 2: Evaluating $I_1 = \displaystyle\int_{0}^{1} \frac{1}{(x+3)^2} \ln\!\left(\frac{x+1}{x+3}\right) dx$

The form of $(\dagger)$ suggests the substitution $u = \dfrac{x+1}{x+3}$ (a linear fractional function with $a = 1, b = 1, c = 1, d = 3$).

By $(\dagger)$:

$\frac{du}{dx} = \frac{1 \cdot 3 - 1 \cdot 1}{(x+3)^2} = \frac{2}{(x+3)^2},$

so $\dfrac{1}{(x+3)^2}\, dx = \dfrac{1}{2}\, du$.

Changing limits: when $x = 0$, $u = \frac{1}{3}$; when $x = 1$, $u = \frac{1}{2}$.

$I_1 = \frac{1}{2}\int_{1/3}^{1/2} \ln u\, du.$

Integrating by parts ($\int \ln u\, du = u\ln u - u$):

$I_1 = \frac{1}{2}\Big[u\ln u - u\Big]_{1/3}^{1/2}.$

At $u = \frac{1}{2}$:

$\frac{1}{2}\ln\frac{1}{2} - \frac{1}{2} = -\frac{1}{2}\ln 2 - \frac{1}{2}.$

At $u = \frac{1}{3}$:

$\frac{1}{3}\ln\frac{1}{3} - \frac{1}{3} = -\frac{1}{3}\ln 3 - \frac{1}{3}.$

Difference:

$\left(-\frac{1}{2}\ln 2 - \frac{1}{2}\right) - \left(-\frac{1}{3}\ln 3 - \frac{1}{3}\right) = \frac{1}{3}\ln 3 - \frac{1}{2}\ln 2 - \frac{1}{6}.$

Therefore:

$I_1 = \frac{1}{2}\left(\frac{1}{3}\ln 3 - \frac{1}{2}\ln 2 - \frac{1}{6}\right) = \frac{1}{6}\ln 3 - \frac{1}{4}\ln 2 - \frac{1}{12}. \qquad \text{(shown)}$

---

Part 3: Evaluating $I_2 = \displaystyle\int_{0}^{1} \frac{1}{(x+3)^2} \ln\!\left(\frac{x^2+3x+2}{(x+3)^2}\right) dx$

Factor $x^2 + 3x + 2 = (x+1)(x+2)$ and use $\ln\dfrac{AB}{C^2} = \ln\dfrac{A}{C} + \ln\dfrac{B}{C}$:

$\ln\frac{(x+1)(x+2)}{(x+3)^2} = \ln\frac{x+1}{x+3} + \ln\frac{x+2}{x+3}.$

So $I_2 = I_1 + J$ where

$J = \int_{0}^{1} \frac{1}{(x+3)^2}\ln\frac{x+2}{x+3}\, dx.$

For $J$, substitute $v = \dfrac{x+2}{x+3}$ (with $a=1, b=2, c=1, d=3$). By $(\dagger)$:

$\frac{dv}{dx} = \frac{3 - 2}{(x+3)^2} = \frac{1}{(x+3)^2},$

so $\dfrac{1}{(x+3)^2}\, dx = dv$.

Limits: $x = 0 \Rightarrow v = \frac{2}{3}$; $x = 1 \Rightarrow v = \frac{3}{4}$.

$J = \int_{2/3}^{3/4} \ln v\, dv = \Big[v\ln v - v\Big]_{2/3}^{3/4}.$

At $v = \frac{3}{4}$:

$\frac{3}{4}\ln\frac{3}{4} - \frac{3}{4} = \frac{3}{4}\ln 3 - \frac{3}{4}\ln 2 - \frac{3}{4}.$

At $v = \frac{2}{3}$:

$\frac{2}{3}\ln\frac{2}{3} - \frac{2}{3} = \frac{2}{3}\ln 2 - \frac{2}{3}\ln 3 - \frac{2}{3}.$

Difference:

$\left(\frac{3}{4}\ln 3 - \frac{3}{4}\ln 2 - \frac{3}{4}\right) - \left(\frac{2}{3}\ln 2 - \frac{2}{3}\ln 3 - \frac{2}{3}\right)$

$= \left(\frac{3}{4} + \frac{2}{3}\right)\ln 3 - \left(\frac{3}{4} + \frac{2}{3}\right)\ln 2 - \left(\frac{3}{4} - \frac{2}{3}\right)$

$= \frac{17}{12}\ln 3 - \frac{17}{12}\ln 2 - \frac{1}{12}.$

Therefore:

$I_2 = I_1 + J = \left(\frac{1}{6}\ln 3 - \frac{1}{4}\ln 2 - \frac{1}{12}\right) + \left(\frac{17}{12}\ln 3 - \frac{17}{12}\ln 2 - \frac{1}{12}\right)$

$= \left(\frac{2}{12} + \frac{17}{12}\right)\ln 3 - \left(\frac{3}{12} + \frac{17}{12}\right)\ln 2 - \frac{2}{12} = \frac{19}{12}\ln 3 - \frac{5}{3}\ln 2 - \frac{1}{6}.$

---

Part 4: Evaluating $I_3 = \displaystyle\int_{0}^{1} \frac{1}{(x+3)^2} \ln\!\left(\frac{x+1}{x+2}\right) dx$

Use $\ln\dfrac{x+1}{x+2} = \ln\dfrac{x+1}{x+3} - \ln\dfrac{x+2}{x+3}$, so $I_3 = I_1 - J$:

$I_3 = \left(\frac{1}{6}\ln 3 - \frac{1}{4}\ln 2 - \frac{1}{12}\right) - \left(\frac{17}{12}\ln 3 - \frac{17}{12}\ln 2 - \frac{1}{12}\right)$

$= \left(\frac{2}{12} - \frac{17}{12}\right)\ln 3 + \left(-\frac{3}{12} + \frac{17}{12}\right)\ln 2 + \left(-\frac{1}{12} + \frac{1}{12}\right)$

$= -\frac{5}{4}\ln 3 + \frac{7}{6}\ln 2.$

---

Summary of results:

$\int_{0}^{1} \frac{1}{(x+3)^2} \ln\!\left(\frac{x+1}{x+3}\right) dx = \frac{1}{6}\ln 3 - \frac{1}{4}\ln 2 - \frac{1}{12}$

$\int_{0}^{1} \frac{1}{(x+3)^2} \ln\!\left(\frac{x^2+3x+2}{(x+3)^2}\right) dx = \frac{19}{12}\ln 3 - \frac{5}{3}\ln 2 - \frac{1}{6}$

$\int_{0}^{1} \frac{1}{(x+3)^2} \ln\!\left(\frac{x+1}{x+2}\right) dx = -\frac{5}{4}\ln 3 + \frac{7}{6}\ln 2$

---

Question 7

Topic: 曲线分析与作图 / Curve Analysis and Sketching  |  Difficulty: Challenging  |  Marks: 20

Problem

7 The curve $C$ has equation $y = \frac{x}{\sqrt{(x^2 - 2x + a)}} ,$ where the square root is positive. Show that, if $a > 1$, then $C$ has exactly one stationary point.

Sketch $C$ when (i) $a = 2$ and (ii) $a = 1$.

Model Solution

We have $y = \dfrac{x}{\sqrt{x^2 - 2x + a}}$, where the square root is positive.

Finding stationary points. Let $u = x^2 - 2x + a$, so $y = x \, u^{-1/2}$. By the product rule and chain rule:

$\frac{dy}{dx} = u^{-1/2} + x \cdot \left(-\frac{1}{2}\right) u^{-3/2} \cdot (2x - 2) = \frac{1}{\sqrt{u}} - \frac{x(x-1)}{u^{3/2}}$

Combining over the common denominator $u^{3/2}$:

$\frac{dy}{dx} = \frac{u - x(x-1)}{u^{3/2}} = \frac{(x^2 - 2x + a) - (x^2 - x)}{u^{3/2}} = \frac{a - x}{(x^2 - 2x + a)^{3/2}}$

Stationary points require $dy/dx = 0$, so $a - x = 0$, giving $x = a$. We must also check that this value lies in the domain, i.e., $u = x^2 - 2x + a > 0$.

When $a > 1$, the discriminant of $x^2 - 2x + a$ is $4 - 4a < 0$, so $u > 0$ for all $x$. Thus $x = a$ is in the domain. Since $dy/dx = 0$ only when $x = a$ (the denominator is always positive), $C$ has exactly one stationary point.

Value at the stationary point. At $x = a$:

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

Asymptotic behaviour. For large $|x|$:

$y = \frac{x}{\sqrt{x^2 - 2x + a}} = \frac{x}{|x|\sqrt{1 - 2/x + a/x^2}} \to \frac{x}{|x|}$

So $y \to 1$ as $x \to +\infty$ and $y \to -1$ as $x \to -\infty$.

Part (i): $a = 2$. Then $y = \dfrac{x}{\sqrt{x^2 - 2x + 2}}$, defined for all $x$ since $x^2 - 2x + 2 = (x-1)^2 + 1 > 0$.

The stationary point is at $x = 2$, $y = \sqrt{2} \approx 1.41$. Checking the sign of $dy/dx = (2-x)/u^{3/2}$: positive for $x < 2$ (increasing) and negative for $x > 2$ (decreasing), so this is a maximum.

For $x < 0$: $x^2 - 2x + 2 > x^2$ (since $-2x > 0$), so $|y| < 1$. The curve approaches $y = -1$ from above as $x \to -\infty$ and rises through the origin.

For $x > 0$ and large: $y = \frac{x}{\sqrt{x^2-2x+2}} > 1$ (since $x^2 > x^2 - 2x + 2$ when $x > 1$), and $y \to 1$ from above as $x \to +\infty$. The curve rises from the origin, reaches a maximum of $\sqrt{2}$ at $x = 2$, then decreases toward $y = 1$ from above.

Sketch for $a = 2$: the curve passes through the origin, rises to a maximum of $\sqrt{2}$ at $x = 2$, then decreases toward the horizontal asymptote $y = 1$ (approached from above) as $x \to \infty$. For negative $x$, the curve approaches the asymptote $y = -1$ from above as $x \to -\infty$ and then rises through the origin.

Part (ii): $a = 1$. Then $y = \dfrac{x}{\sqrt{x^2 - 2x + 1}} = \dfrac{x}{\sqrt{(x-1)^2}} = \dfrac{x}{|x - 1|}$, defined for $x \neq 1$.

For $x < 1$: $y = \dfrac{x}{1 - x} = -1 + \dfrac{1}{1 - x}$, which is increasing (from $y \to -1$ as $x \to -\infty$ to $y \to -\infty$ as $x \to 1^-$).

For $x > 1$: $y = \dfrac{x}{x - 1} = 1 + \dfrac{1}{x - 1}$, which is decreasing (from $y \to +\infty$ as $x \to 1^+$ to $y \to 1$ as $x \to +\infty$).

There is a vertical asymptote at $x = 1$, with $y \to -\infty$ from the left and $y \to +\infty$ from the right. The horizontal asymptotes are $y = -1$ (as $x \to -\infty$) and $y = 1$ (as $x \to +\infty$). There are no stationary points (the derivative $\frac{1}{(x-1)^2} > 0$ on each piece). The curve passes through the origin.

---

Question 8

Topic: 三角级数与求和 / Trigonometric Series and Summation  |  Difficulty: Hard  |  Marks: 20

Problem

8 Prove that $\sum_{k=0}^{n} \sin k\theta = \frac{\cos \frac{1}{2}\theta - \cos\bigl((n + \frac{1}{2})\theta\bigr)}{2 \sin \frac{1}{2}\theta} . \qquad \text{(*)}$

(i) Deduce that, when $n$ is large, $\sum_{k=0}^{n} \sin \left( \frac{k\pi}{n} \right) \approx \frac{2n}{\pi} .$

(ii) By differentiating $(*)$ with respect to $\theta$, or otherwise, show that, when $n$ is large, $\sum_{k=0}^{n} k \sin^2 \left( \frac{k\pi}{2n} \right) \approx \left( \frac{1}{4} + \frac{1}{\pi^2} \right) n^2 .$

[The approximations, valid for small $\theta$, $\sin \theta \approx \theta$ and $\cos \theta \approx 1 - \frac{1}{2}\theta^2$ may be assumed.]

Model Solution

Proof of the summation formula. Multiply both sides of the identity by $2\sin\frac{\theta}{2}$ (assuming $\sin\frac{\theta}{2} \neq 0$). The left side becomes

$2\sin\tfrac{\theta}{2} \sum_{k=0}^{n} \sin k\theta = \sum_{k=0}^{n} 2\sin k\theta \sin\tfrac{\theta}{2}.$

Using the product-to-sum identity $2\sin A\sin B = \cos(A-B) - \cos(A+B)$:

$2\sin k\theta \sin\tfrac{\theta}{2} = \cos\!\bigl((k - \tfrac{1}{2})\theta\bigr) - \cos\!\bigl((k + \tfrac{1}{2})\theta\bigr).$

The sum telescopes:

$k = 0: \quad \cos\tfrac{\theta}{2} - \cos\tfrac{\theta}{2} = 0 \quad (\text{since } \cos(-\tfrac{\theta}{2}) = \cos\tfrac{\theta}{2})$ $k = 1: \quad \cos\tfrac{\theta}{2} - \cos\tfrac{3\theta}{2}$ $k = 2: \quad \cos\tfrac{3\theta}{2} - \cos\tfrac{5\theta}{2}$ $\vdots$ $k = n: \quad \cos\!\bigl((n - \tfrac{1}{2})\theta\bigr) - \cos\!\bigl((n + \tfrac{1}{2})\theta\bigr)$

All intermediate terms cancel, leaving

$2\sin\tfrac{\theta}{2} \sum_{k=0}^{n} \sin k\theta = \cos\tfrac{\theta}{2} - \cos\!\bigl((n + \tfrac{1}{2})\theta\bigr).$

Dividing both sides by $2\sin\frac{\theta}{2}$:

$\sum_{k=0}^{n} \sin k\theta = \frac{\cos\frac{\theta}{2} - \cos\!\bigl((n + \frac{1}{2})\theta\bigr)}{2\sin\frac{\theta}{2}}. \qquad \blacksquare$

---

Part (i)

Set $\theta = \dfrac{\pi}{n}$ in the formula:

$\sum_{k=0}^{n} \sin\frac{k\pi}{n} = \frac{\cos\frac{\pi}{2n} - \cos\!\bigl((n + \frac{1}{2})\frac{\pi}{n}\bigr)}{2\sin\frac{\pi}{2n}}.$

Since $\bigl(n + \frac{1}{2}\bigr)\frac{\pi}{n} = \pi + \frac{\pi}{2n}$, we have

$\cos\!\bigl(\pi + \tfrac{\pi}{2n}\bigr) = -\cos\tfrac{\pi}{2n},$

so the numerator becomes $\cos\frac{\pi}{2n} - \bigl(-\cos\frac{\pi}{2n}\bigr) = 2\cos\frac{\pi}{2n}$, giving

$\sum_{k=0}^{n} \sin\frac{k\pi}{n} = \frac{2\cos\frac{\pi}{2n}}{2\sin\frac{\pi}{2n}} = \cot\frac{\pi}{2n}. \qquad (\dagger)$

For large $n$, $\frac{\pi}{2n}$ is small. Using $\sin\phi \approx \phi$ and $\cos\phi \approx 1 - \frac{\phi^2}{2}$ with $\phi = \frac{\pi}{2n}$:

$\cot\frac{\pi}{2n} = \frac{\cos\frac{\pi}{2n}}{\sin\frac{\pi}{2n}} \approx \frac{1 - \frac{\pi^2}{8n^2}}{\frac{\pi}{2n}} = \frac{2n}{\pi}\!\left(1 - \frac{\pi^2}{8n^2}\right) \approx \frac{2n}{\pi}. \qquad \blacksquare$

---

Part (ii)

We need $\displaystyle\sum_{k=0}^{n} k\sin^2\frac{k\pi}{2n}$. Using $\sin^2\phi = \frac{1 - \cos 2\phi}{2}$ with $\phi = \frac{k\pi}{2n}$:

$\sum_{k=0}^{n} k\sin^2\frac{k\pi}{2n} = \frac{1}{2}\sum_{k=0}^{n} k - \frac{1}{2}\sum_{k=0}^{n} k\cos\frac{k\pi}{n}. \qquad (\ddagger)$

First sum.

$\sum_{k=0}^{n} k = \frac{n(n+1)}{2} \approx \frac{n^2}{2}.$

Second sum. We compute $\sum_{k=0}^{n} k\cos k\theta$ by differentiating $(*)$ with respect to $\theta$. Write

$S(\theta) = \sum_{k=0}^{n} \sin k\theta = \frac{\cos\frac{\theta}{2} - \cos\bigl((n+\frac{1}{2})\theta\bigr)}{2\sin\frac{\theta}{2}}.$

Then $S'(\theta) = \sum_{k=0}^{n} k\cos k\theta$. For the right side, let $N = \cos\frac{\theta}{2} - \cos\bigl((n+\frac{1}{2})\theta\bigr)$ and $D = 2\sin\frac{\theta}{2}$:

$N' = -\tfrac{1}{2}\sin\tfrac{\theta}{2} + (n+\tfrac{1}{2})\sin\bigl((n+\tfrac{1}{2})\theta\bigr), \qquad D' = \cos\tfrac{\theta}{2},$

$S'(\theta) = \frac{N'D - ND'}{D^2}.$

Now evaluate at $\theta = \frac{\pi}{n}$, where $(n+\frac{1}{2})\theta = \pi + \frac{\pi}{2n}$, giving $\sin\bigl((n+\frac{1}{2})\theta\bigr) = -\sin\frac{\pi}{2n}$ and $\cos\bigl((n+\frac{1}{2})\theta\bigr) = -\cos\frac{\pi}{2n}$.

At $\theta = \frac{\pi}{n}$:

$N = \cos\tfrac{\pi}{2n} - (-\cos\tfrac{\pi}{2n}) = 2\cos\tfrac{\pi}{2n},$

$N' = -\tfrac{1}{2}\sin\tfrac{\pi}{2n} + (n+\tfrac{1}{2})(-\sin\tfrac{\pi}{2n}) = -(n+1)\sin\tfrac{\pi}{2n},$

$D = 2\sin\tfrac{\pi}{2n}, \qquad D' = \cos\tfrac{\pi}{2n}.$

Computing the pieces:

$N'D = -2(n+1)\sin^2\tfrac{\pi}{2n}, \qquad ND' = 2\cos^2\tfrac{\pi}{2n}, \qquad D^2 = 4\sin^2\tfrac{\pi}{2n}.$

Therefore

$\sum_{k=0}^n k\cos\frac{k\pi}{n} = \frac{-2(n+1)\sin^2\frac{\pi}{2n} - 2\cos^2\frac{\pi}{2n}}{4\sin^2\frac{\pi}{2n}} = -\frac{n+1}{2} - \frac{1}{2}\cot^2\frac{\pi}{2n}. \qquad (\S')$

Approximating for large $n$. From part (i), $\cot\frac{\pi}{2n} \approx \frac{2n}{\pi}$, so $\cot^2\frac{\pi}{2n} \approx \frac{4n^2}{\pi^2}$. Thus

$\sum_{k=0}^n k\cos\frac{k\pi}{n} \approx -\frac{n}{2} - \frac{2n^2}{\pi^2}.$

Final assembly. Substituting into $(\ddagger)$:

$\sum_{k=0}^n k\sin^2\frac{k\pi}{2n} = \frac{1}{2}\sum_{k=0}^n k - \frac{1}{2}\sum_{k=0}^n k\cos\frac{k\pi}{n}$

$\approx \frac{n^2}{4} - \frac{1}{2}\!\left(-\frac{n}{2} - \frac{2n^2}{\pi^2}\right) = \frac{n^2}{4} + \frac{n}{4} + \frac{n^2}{\pi^2} \approx \frac{n^2}{4} + \frac{n^2}{\pi^2} = \left(\frac{1}{4} + \frac{1}{\pi^2}\right)n^2. \qquad \blacksquare$