STEP3 2004 -- Pure Mathematics

Exam: STEP3  |  Year: 2004  |  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	积分/Integration（双曲函数与代数积分）	Challenging	换元法,部分分式分解,双曲函数恒等式,反正切积分,指数换元
2	曲线分析/Curve analysis（渐近线与曲线描绘）	Standard	渐近线分析,因式分解,差平方公式,曲线描绘
3	不等式与级数/Inequalities and series（凸函数与级数估计）	Challenging	凸函数性质,积分不等式,伸缩求和法,级数估计
4	几何序列/Geometric sequences（相切圆问题）	Challenging	几何级数求和,三角恒等式,微分求极值,相切条件
5	三角方程/Trigonometric equations（辅助角公式）	Standard	辅助角公式,正切加法公式,双角公式,解的验证
6	递推序列与不变量 (Recurrence Sequences and Invariants)	Challenging	差分法, 不变量证明, 二次方程求根, 构造周期序列
7	积分序列与级数估计 (Integral Sequences and Series Estimation)	Challenging	分部积分, 裂项求和, 不等式放缩, 积分递推
8	微分方程与解曲线 (Differential Equations and Solution Curves)	Challenging	变量代换, 积分因子法, 曲线分类

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Question 1

Topic: 积分/Integration（双曲函数与代数积分）  |  Difficulty: Challenging  |  Marks: 20

Problem

1 Show that $\int_{0}^{a} \frac{\sinh x}{2 \cosh^{2} x - 1} \, dx = \frac{1}{2\sqrt{2}} \ln \left( \frac{\sqrt{2} \cosh a - 1}{\sqrt{2} \cosh a + 1} \right) + \frac{1}{2\sqrt{2}} \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)$ and find $\int_{0}^{a} \frac{\cosh x}{1 + 2 \sinh^{2} x} \, dx \, .$ Hence show that $\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2 \sinh^{2} x} \, dx = \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right) \, .$ By substituting $u = e^{x}$ in this result, or otherwise, find $\int_{1}^{\infty} \frac{1}{1 + u^{4}} \, du \, .$

Hint

The substitution $u = \cosh x$ should suggest itself (because of the factor of $\frac{du}{dx} = \sinh x$ in the numerator), and the resulting integral can be tackled by splitting the integrand into partial fractions:

$\int_{0}^{a} \frac{\sinh x}{2 \cosh^{2} x - 1} dx = \int_{1}^{\cosh a} \frac{du}{2u^{2} - 1} = \frac{1}{2} \int_{1}^{\cosh a} \frac{1}{\sqrt{2}u - 1} - \frac{1}{\sqrt{2}u + 1} du$

$= \frac{1}{2\sqrt{2}} \left[ \ln \left( \sqrt{2}u - 1 \right) - \ln \left( \sqrt{2}u + 1 \right) \right]_{1}^{\cosh a} = \frac{1}{2\sqrt{2}} \left( \ln \left( \frac{\sqrt{2} \cosh a - 1}{\sqrt{2} \cosh a + 1} \right) + \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right) \right)$

Similarly, substituting $u = \sinh x$, and then recognising an arctan integral:

$\int_{0}^{a} \frac{\cosh x}{1 + 2 \sinh^{2} x} dx = \int_{0}^{\sinh a} \frac{du}{1 + 2u^{2}} = \frac{1}{\sqrt{2}} \left[ \arctan \left( \sqrt{2}u \right) \right]_{0}^{\sinh a} = \frac{1}{\sqrt{2}} \arctan \left( \sqrt{2} \sinh a \right)$

To show that

$\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2 \sinh^{2} x} dx = \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)$

note that

a $\cosh^{2} x = \sinh^{2} x + 1$, so $2 \cosh^{2} x - 1 = 1 + 2 \sinh^{2} x$, and the integral required is the second minus the first of those calculated earlier, as $a \to \infty$.

b as $a \to \infty$, $\sinh a \to \infty$, so $\arctan \left( \sqrt{2} \sinh a \right) \to \frac{\pi}{2}$

c as $a \to \infty$, $\cosh a \to \infty$, so $\frac{\sqrt{2} \cosh a - 1}{\sqrt{2} \cosh a + 1} \to 1$ and $\ln \left( \frac{\sqrt{2} \cosh a - 1}{\sqrt{2} \cosh a + 1} \right) \to 0$.

Substituting $u = e^{x}$, so that $\cosh x = \frac{1}{2} \left( u + \frac{1}{u} \right)$ and $\sinh x = \frac{1}{2} \left( u - \frac{1}{u} \right)$:

$\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2 \sinh^{2} x} dx = \int_{1}^{\infty} \frac{\left( u + \frac{1}{u} \right) - \left( u - \frac{1}{u} \right)}{1 + \frac{1}{2} \left( u^{2} - 2 + \frac{1}{u^{2}} \right)} \frac{1}{u} du = \int_{1}^{\infty} \frac{2}{1 + u^{4}} du$

so $\int_{1}^{\infty} \frac{1}{1 + u^{4}} du = \frac{\pi}{4\sqrt{2}} - \frac{1}{4\sqrt{2}} \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)$.

Model Solution

Part 1: Show that

$\int_{0}^{a} \frac{\sinh x}{2 \cosh^{2} x - 1} \, dx = \frac{1}{2\sqrt{2}} \ln \left( \frac{\sqrt{2} \cosh a - 1}{\sqrt{2} \cosh a + 1} \right) + \frac{1}{2\sqrt{2}} \ln \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)$

Substitute $u = \cosh x$, so $du = \sinh x \, dx$. When $x = 0$, $u = 1$; when $x = a$, $u = \cosh a$.

$\int_{0}^{a} \frac{\sinh x}{2 \cosh^{2} x - 1} \, dx = \int_{1}^{\cosh a} \frac{du}{2u^{2} - 1}$

Factor the denominator: $2u^{2} - 1 = (\sqrt{2}u - 1)(\sqrt{2}u + 1)$. Using partial fractions:

$\frac{1}{2u^{2} - 1} = \frac{1}{(\sqrt{2}u - 1)(\sqrt{2}u + 1)} = \frac{A}{\sqrt{2}u - 1} + \frac{B}{\sqrt{2}u + 1}$

Multiplying through: $1 = A(\sqrt{2}u + 1) + B(\sqrt{2}u - 1)$.

Setting $u = \frac{1}{\sqrt{2}}$: $1 = 2A$, so $A = \frac{1}{2}$. Setting $u = -\frac{1}{\sqrt{2}}$: $1 = -2B$, so $B = -\frac{1}{2}$.

$\frac{1}{2u^{2} - 1} = \frac{1}{2}\left(\frac{1}{\sqrt{2}u - 1} - \frac{1}{\sqrt{2}u + 1}\right)$

Therefore:

$\int_{1}^{\cosh a} \frac{du}{2u^{2} - 1} = \frac{1}{2} \int_{1}^{\cosh a} \left(\frac{1}{\sqrt{2}u - 1} - \frac{1}{\sqrt{2}u + 1}\right) du$

$= \frac{1}{2} \left[ \frac{1}{\sqrt{2}} \ln(\sqrt{2}u - 1) - \frac{1}{\sqrt{2}} \ln(\sqrt{2}u + 1) \right]_{1}^{\cosh a}$

$= \frac{1}{2\sqrt{2}} \left[ \ln\left(\frac{\sqrt{2}u - 1}{\sqrt{2}u + 1}\right) \right]_{1}^{\cosh a}$

$= \frac{1}{2\sqrt{2}} \left( \ln\left(\frac{\sqrt{2}\cosh a - 1}{\sqrt{2}\cosh a + 1}\right) - \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right) \right)$

Since $-\ln\left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right) = \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$:

$\int_{0}^{a} \frac{\sinh x}{2 \cosh^{2} x - 1} \, dx = \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2}\cosh a - 1}{\sqrt{2}\cosh a + 1}\right) + \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \qquad \text{as required.}$

Part 2: Find $\int_{0}^{a} \frac{\cosh x}{1 + 2 \sinh^{2} x} \, dx$

Substitute $u = \sinh x$, so $du = \cosh x \, dx$. When $x = 0$, $u = 0$; when $x = a$, $u = \sinh a$.

$\int_{0}^{a} \frac{\cosh x}{1 + 2\sinh^{2} x} \, dx = \int_{0}^{\sinh a} \frac{du}{1 + 2u^{2}}$

$= \int_{0}^{\sinh a} \frac{du}{1 + (\sqrt{2}u)^{2}} = \frac{1}{\sqrt{2}} \left[ \arctan(\sqrt{2}u) \right]_{0}^{\sinh a} = \frac{1}{\sqrt{2}} \arctan(\sqrt{2}\sinh a)$

Part 3: Hence show that

$\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^{2} x} \, dx = \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$

Using the identity $\cosh^{2} x - \sinh^{2} x = 1$, we have:

$2\cosh^{2} x - 1 = 2(\sinh^{2} x + 1) - 1 = 1 + 2\sinh^{2} x$

So the two denominators are identical. We split the integral:

$\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^{2} x} \, dx = \int_{0}^{\infty} \frac{\cosh x}{1 + 2\sinh^{2} x} \, dx - \int_{0}^{\infty} \frac{\sinh x}{1 + 2\sinh^{2} x} \, dx$

Note the second integral has the same denominator as the first (since $2\cosh^{2}x - 1 = 1 + 2\sinh^{2}x$), so we can use the results from Parts 1 and 2 with $a \to \infty$.

First integral: Taking $a \to \infty$, $\sinh a \to \infty$, so $\arctan(\sqrt{2}\sinh a) \to \frac{\pi}{2}$.

$\int_{0}^{\infty} \frac{\cosh x}{1 + 2\sinh^{2} x} \, dx = \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} = \frac{\pi}{2\sqrt{2}}$

Second integral: Taking $a \to \infty$, $\cosh a \to \infty$, so:

$\frac{\sqrt{2}\cosh a - 1}{\sqrt{2}\cosh a + 1} = \frac{\sqrt{2} - \frac{1}{\cosh a}}{\sqrt{2} + \frac{1}{\cosh a}} \to 1$

and $\ln\left(\frac{\sqrt{2}\cosh a - 1}{\sqrt{2}\cosh a + 1}\right) \to \ln 1 = 0$. Thus:

$\int_{0}^{\infty} \frac{\sinh x}{2\cosh^{2} x - 1} \, dx = 0 + \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) = \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$

Therefore:

$\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^{2} x} \, dx = \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \qquad \text{as required.}$

Part 4: Find $\int_{1}^{\infty} \frac{1}{1 + u^{4}} \, du$

Substitute $u = e^{x}$, so $x = \ln u$ and $dx = \frac{du}{u}$. When $x = 0$, $u = 1$; when $x \to \infty$, $u \to \infty$.

Using the exponential definitions of hyperbolic functions:

$\cosh x = \frac{e^{x} + e^{-x}}{2} = \frac{u + u^{-1}}{2}, \qquad \sinh x = \frac{e^{x} - e^{-x}}{2} = \frac{u - u^{-1}}{2}$

The numerator:

$\cosh x - \sinh x = \frac{u + u^{-1}}{2} - \frac{u - u^{-1}}{2} = u^{-1} = \frac{1}{u}$

The denominator:

$1 + 2\sinh^{2} x = 1 + 2\left(\frac{u - u^{-1}}{2}\right)^{2} = 1 + \frac{(u - u^{-1})^{2}}{2} = 1 + \frac{u^{2} - 2 + u^{-2}}{2}$

$= \frac{2 + u^{2} - 2 + u^{-2}}{2} = \frac{u^{2} + u^{-2}}{2} = \frac{u^{4} + 1}{2u^{2}}$

So the integral becomes:

$\int_{0}^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^{2} x} \, dx = \int_{1}^{\infty} \frac{1/u}{\frac{u^{4}+1}{2u^{2}}} \cdot \frac{du}{u} = \int_{1}^{\infty} \frac{2u^{2}}{u(u^{4}+1)} \cdot \frac{1}{u} \, du = \int_{1}^{\infty} \frac{2}{u^{4}+1} \, du$

Combining with the result from Part 3:

$2 \int_{1}^{\infty} \frac{du}{1 + u^{4}} = \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$

$\int_{1}^{\infty} \frac{du}{1 + u^{4}} = \frac{\pi}{4\sqrt{2}} - \frac{1}{4\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$

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Question 2

Topic: 曲线分析/Curve analysis（渐近线与曲线描绘）  |  Difficulty: Standard  |  Marks: 20

Problem

2 The equation of a curve is $y = f(x)$ where $f(x) = x - 4 - \frac{16(2x + 1)^{2}}{x^{2}(x - 4)} \, .$ (i) Write down the equations of the vertical and oblique asymptotes to the curve and show that the oblique asymptote is a tangent to the curve.

(ii) Show that the equation $f(x) = 0$ has a double root.

(iii) Sketch the curve.

Hint

(i) Inspection of the denominator shows that the vertical asymptotes are at $x = 0$, $x = 4$, and the third term in $f(x)$ tends to zero as $|x| \rightarrow \infty$, so the oblique asymptote is just $y = x - 4$.

The oblique asymptote meets the curve when $\frac{16(2x + 1)^2}{x^2(x - 4)} = 0$ or $(2x + 1)^2 = 0$, hence there is a double root at $x = -\frac{1}{2}$ and hence the asymptote touches rather than crosses the curve at $(-\frac{1}{2}, -\frac{9}{2})$, so is a tangent there.

(ii) $f(x) = 0$ when $x^2(x - 4)^2 - 16(2x + 1)^2 = 0$.

The left hand side of this equation is a difference of two squares, so factorises to give $(x(x - 4) - 4(2x + 1))(x(x - 4) + 4(2x + 1)) = 0$; that is, $(x^2 - 12x - 4)(x + 2)^2 = 0$, which has a double root at $x = -2$.

(iii) On your sketch you should show:

• the double root at $(-2, 0)$ — the curve has a local maximum here and touches the $x$-axis;
• the remaining roots (solutions of $x^2 - 12x - 4 = 0$) at $x = 6 \pm 2\sqrt{10}$;
• the curve approaching the oblique asymptote $y = x - 4$ from below as $x \rightarrow \infty$, approaching it from above as $x \rightarrow -\infty$ and touching it at $(-\frac{1}{2}, -\frac{9}{2})$;
• $f(x) \rightarrow \infty$ as $x \rightarrow 0$ from above or below, $f(x) \rightarrow +\infty$ as $x \rightarrow 4$ from below and $f(x) \rightarrow -\infty$ as $x \rightarrow 4$ from above;
• local minima at some $x$ value with $0 < x < 4$ and with $y > 0$ and at some $x$ value with $-2 < x < -\frac{1}{2}$ and with $-\frac{9}{2} > y > x - 4$ — note that this second minimum is not at the point of tangency with the oblique asymptote.

Model Solution

Part (i): Asymptotes and tangency

$f(x) = x - 4 - \frac{16(2x+1)^{2}}{x^{2}(x-4)}$

Vertical asymptotes: The denominator $x^{2}(x-4) = 0$ when $x = 0$ or $x = 4$. Since the numerators are non-zero at these points, the vertical asymptotes are $x = 0$ and $x = 4$.

Oblique asymptote: As $|x| \to \infty$, the fraction $\frac{16(2x+1)^{2}}{x^{2}(x-4)} \to 0$ (degree 2 numerator over degree 3 denominator), so $f(x) \to x - 4$. The oblique asymptote is $y = x - 4$.

Show the oblique asymptote is a tangent: Setting $f(x) = x - 4$:

$x - 4 - \frac{16(2x+1)^{2}}{x^{2}(x-4)} = x - 4$

$\frac{16(2x+1)^{2}}{x^{2}(x-4)} = 0$

This requires $(2x+1)^{2} = 0$, i.e., $x = -\frac{1}{2}$. Since $(2x+1)^{2}$ has a double root, the curve touches the asymptote at $x = -\frac{1}{2}$ rather than crossing it.

At $x = -\frac{1}{2}$: $f(-\frac{1}{2}) = -\frac{1}{2} - 4 - 0 = -\frac{9}{2}$, and the asymptote gives $y = -\frac{1}{2} - 4 = -\frac{9}{2}$. So the point of tangency is $\left(-\frac{1}{2}, -\frac{9}{2}\right)$.

To confirm tangency, we verify the slopes match. Since $f(x) = (x-4) - g(x)$ where $g(x) = \frac{16(2x+1)^{2}}{x^{2}(x-4)}$:

$f'(x) = 1 - g'(x)$

Near $x = -\frac{1}{2}$, $g(x) = \frac{16(2x+1)^{2}}{x^{2}(x-4)}$ has a double zero in the numerator at $x = -\frac{1}{2}$ while the denominator is $\frac{1}{4} \cdot (-\frac{9}{2}) = -\frac{9}{8} \neq 0$. By the product rule, $g'(-\frac{1}{2}) = 0$ (the factor $(2x+1)^{2}$ contributes zero along with its derivative $(2x+1) \cdot 4$ evaluated at $x = -\frac{1}{2}$). Therefore $f'(-\frac{1}{2}) = 1 - 0 = 1$, which equals the slope of $y = x - 4$. The asymptote is indeed tangent at $\left(-\frac{1}{2}, -\frac{9}{2}\right)$.

Part (ii): Show $f(x) = 0$ has a double root

Setting $f(x) = 0$:

$x - 4 = \frac{16(2x+1)^{2}}{x^{2}(x-4)}$

Multiplying both sides by $x^{2}(x-4)$ (valid for $x \neq 0, 4$):

$x^{2}(x-4)^{2} = 16(2x+1)^{2}$

The left side is $[x(x-4)]^{2}$ and the right side is $[4(2x+1)]^{2}$, so this is a difference of two squares:

$[x(x-4)]^{2} - [4(2x+1)]^{2} = 0$

$[x(x-4) - 4(2x+1)][x(x-4) + 4(2x+1)] = 0$

Expanding the first factor:

$x^{2} - 4x - 8x - 4 = x^{2} - 12x - 4$

Expanding the second factor:

$x^{2} - 4x + 8x + 4 = x^{2} + 4x + 4 = (x+2)^{2}$

So the equation becomes:

$(x^{2} - 12x - 4)(x+2)^{2} = 0$

The factor $(x+2)^{2} = 0$ gives $x = -2$ as a double root. (The quadratic $x^{2} - 12x - 4 = 0$ gives two simple roots at $x = 6 \pm 2\sqrt{10}$.)

Therefore $f(x) = 0$ has a double root at $x = -2$.

Part (iii): Sketch of the curve

The key features are:

Asymptotes:

• Vertical: $x = 0$ and $x = 4$
• Oblique: $y = x - 4$ (tangent to the curve at $(-\frac{1}{2}, -\frac{9}{2})$)

Roots:

• Double root at $x = -2$ (curve touches $x$-axis): $f(-2) = -6 - \frac{16 \cdot 9}{4 \cdot (-6)} = -6 + 6 = 0$.
• Simple roots at $x = 6 - 2\sqrt{10} \approx -0.32$ and $x = 6 + 2\sqrt{10} \approx 12.32$.

Behaviour near asymptotes:

Near $x = 0$: $\frac{16(2x+1)^{2}}{x^{2}(x-4)} \approx \frac{16}{-4x^{2}} = \frac{-4}{x^{2}}$, so $f(x) \approx x - 4 + \frac{4}{x^{2}} \to +\infty$ from both sides.

Near $x = 4$: write $x = 4 + \epsilon$. Then $x - 4 = \epsilon$ and $\frac{16(2x+1)^{2}}{x^{2}(x-4)} \approx \frac{16 \cdot 81}{16\epsilon} = \frac{81}{\epsilon}$. So $f(x) \approx \epsilon - \frac{81}{\epsilon}$:

• As $x \to 4^{-}$ ($\epsilon \to 0^{-}$): $f(x) \approx -\frac{81}{\epsilon} \to +\infty$
• As $x \to 4^{+}$ ($\epsilon \to 0^{+}$): $f(x) \approx -\frac{81}{\epsilon} \to -\infty$

Approach to oblique asymptote:

The curve minus the asymptote is $f(x) - (x-4) = -\frac{16(2x+1)^{2}}{x^{2}(x-4)}$. The sign depends on $x^{2}(x-4)$ (since $-16(2x+1)^{2} \leqslant 0$):

• For $x > 4$: $x^{2}(x-4) > 0$, so $f(x) < x - 4$ (approaches from below)
• For $0 < x < 4$: $x^{2}(x-4) < 0$, so $f(x) > x - 4$ (approaches from above)
• For $x < 0$: $x^{2}(x-4) < 0$, so $f(x) > x - 4$ (approaches from above)

Behaviour at the double root $x = -2$:

Since $x=-2$ is a double root and a local maximum, the curve touches the $x$-axis at $(-2,0)$ from below.

Summary of the sketch:

The curve has three branches separated by the vertical asymptotes:

• Left branch ($x < 0$): Comes down from the oblique asymptote as $x \to -\infty$, crosses the $x$-axis at $x=6-2\sqrt{10}$, touches the $x$-axis from below at $(-2,0)$ (local maximum), passes through the tangent point $(-\frac{1}{2},-\frac{9}{2})$, then rises to $+\infty$ as $x \to 0^{-}$.
• Middle branch ($0 < x < 4$): Starts at $+\infty$ as $x \to 0^{+}$, stays above the $x$-axis, has a local minimum with $y>0$ somewhere in $(0,4)$, then rises to $+\infty$ as $x \to 4^{-}$.
• Right branch ($x > 4$): Comes from $-\infty$ as $x \to 4^{+}$, crosses the $x$-axis at $x = 6 + 2\sqrt{10} \approx 12.32$, then approaches $y = x - 4$ from below as $x \to +\infty$.

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Question 3

Topic: 不等式与级数/Inequalities and series（凸函数与级数估计）  |  Difficulty: Challenging  |  Marks: 20

Problem

3 Given that $f''(x) > 0$ when $a \leqslant x \leqslant b$, explain with the aid of a sketch why $(b-a) f\left(\frac{a+b}{2}\right) < \int_{a}^{b} f(x) \, dx < (b-a) \frac{f(a)+f(b)}{2} \, .$

By choosing suitable $a$, $b$ and $f(x)$, show that $\frac{4}{(2n-1)^2} < \frac{1}{n-1} - \frac{1}{n} < \frac{1}{2} \left( \frac{1}{n^2} + \frac{1}{(n-1)^2} \right) \, ,$ where $n$ is an integer greater than 1.

Deduce that $4 \left( \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \right) < 1 < \frac{1}{2} + \left( \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots \right) \, .$

Show that $\frac{1}{2} \left( \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \cdots \right) < \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots$ and hence show that $\frac{3}{2} < \sum_{n=1}^{\infty} \frac{1}{n^2} < \frac{7}{4} \, .$

Hint

The sketch should show a curve with increasing gradient: because the gradient is increasing, the curve lies below the chord joining $(a, f(a))$ and $(b, f(b))$ and above the tangent to the curve at $\left(\frac{a+b}{2}, f\left(\frac{a+b}{2}\right)\right)$. The illustration is clearer if $f(x) > 0$ for $a \leqslant x \leqslant b$: then the area of the trapezium cut off by the chord and the lines $x = a$, $x = b$ and $y = 0$, which is $(b - a) \frac{f(a) + f(b)}{2}$, is larger than the area represented by the integral and the area of the trapezium cut off by the tangent and the lines $x = a$, $x = b$ and $y = 0$, which is $(b - a) f\left(\frac{a+b}{2}\right)$, is smaller than the area represented by the integral.

Choose $f(x) = \frac{1}{x^2}$, checking that this has $f''(x) > 0$, $a = n - 1$ and $b = n$ to get the quoted result.

Take the sum from $n = 2$ to $\infty$ of each term in the inequality: the left hand sum is directly as quoted; in the middle sum, you need to notice that it telescopes, so that all the terms except the first cancel in pairs; in the right hand sum, each reciprocal square occurs twice, cancelling the factor of $\frac{1}{2}$, except the first.

For the next part, observe that $\frac{1}{(n+1)^2} < \frac{1}{n^2}$, so $\frac{1}{2} \left( \frac{1}{n^2} + \frac{1}{(n+1)^2} \right) < \frac{1}{2} \left( \frac{1}{n^2} + \frac{1}{n^2} \right) = \frac{1}{n^2}$.

Finally, combine the two previous results to get

$2 \left( \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots \right) < 1 < \frac{1}{2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots,$

so that if $S = \sum_{n=1}^{\infty} \frac{1}{n^2}$, then $2 \left( S - 1 - \frac{1}{2^2} \right) < 1 < S - \frac{1}{2}$; rearranging these inequalities gives the required bounds on $S$.

Model Solution

Part (i): The integral inequality

Since $f''(x) > 0$ on $[a, b]$, the function $f$ is strictly convex (its gradient is strictly increasing). This has two geometric consequences:

1. The curve $y = f(x)$ lies below the chord joining $(a, f(a))$ and $(b, f(b))$, so the trapezium area $(b-a)\frac{f(a)+f(b)}{2}$ exceeds the area under the curve.

2. The curve $y = f(x)$ lies above the tangent line at the midpoint $x = \frac{a+b}{2}$. The tangent at $\left(\frac{a+b}{2}, f\!\left(\frac{a+b}{2}\right)\right)$ has equation $y = f\!\left(\frac{a+b}{2}\right) + f'\!\left(\frac{a+b}{2}\right)\!\left(x - \frac{a+b}{2}\right)$. Integrating this tangent from $a$ to $b$:

$\int_{a}^{b} \left[ f\!\left(\frac{a+b}{2}\right) + f'\!\left(\frac{a+b}{2}\right)\!\left(x - \frac{a+b}{2}\right) \right] dx = (b-a)\,f\!\left(\frac{a+b}{2}\right) + f'\!\left(\frac{a+b}{2}\right) \cdot 0 = (b-a)\,f\!\left(\frac{a+b}{2}\right)$

since $\int_a^b \left(x - \frac{a+b}{2}\right) dx = 0$ by symmetry.

Therefore:

$(b-a)\,f\!\left(\frac{a+b}{2}\right) < \int_a^b f(x)\,dx < (b-a)\,\frac{f(a)+f(b)}{2} \qquad \text{(...)}$

Part (ii): Choosing $a$, $b$, $f(x)$

Take $f(x) = \frac{1}{x^2}$, $a = n-1$, $b = n$ (with $n > 1$ an integer). Then $f''(x) = \frac{6}{x^4} > 0$, so the inequality applies.

• Midpoint value: $f\!\left(\frac{a+b}{2}\right) = f\!\left(n - \frac{1}{2}\right) = \frac{1}{\left(n - \frac{1}{2}\right)^2} = \frac{4}{(2n-1)^2}$

• Integral: $\int_{n-1}^{n} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{n-1}^{n} = -\frac{1}{n} + \frac{1}{n-1} = \frac{1}{n-1} - \frac{1}{n}$

• Trapezium estimate: $(b-a)\frac{f(a)+f(b)}{2} = 1 \cdot \frac{1}{2}\!\left(\frac{1}{(n-1)^2} + \frac{1}{n^2}\right)$

Substituting into (…):

$\frac{4}{(2n-1)^2} < \frac{1}{n-1} - \frac{1}{n} < \frac{1}{2}\!\left(\frac{1}{(n-1)^2} + \frac{1}{n^2}\right) \qquad \text{(...)}$

Part (iii): Deduce the series inequality

Sum the left inequality of (**) from $n = 2$ to $N$:

$\sum_{n=2}^{N} \frac{4}{(2n-1)^2} < \sum_{n=2}^{N} \left(\frac{1}{n-1} - \frac{1}{n}\right)$

The right side telescopes:

$\sum_{n=2}^{N} \left(\frac{1}{n-1} - \frac{1}{n}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N-1} - \frac{1}{N}\right) = 1 - \frac{1}{N}$

As $N \to \infty$, the right side tends to $1$, so:

$4\!\left(\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots\right) \leqslant 1$

In fact the inequality is strict (each term of the left inequality of (**) is strict), giving:

$4\!\left(\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots\right) < 1 \qquad \text{(...)}$

Now sum the right inequality of (**) from $n = 2$ to $N$:

$\sum_{n=2}^{N} \left(\frac{1}{n-1} - \frac{1}{n}\right) < \sum_{n=2}^{N} \frac{1}{2}\!\left(\frac{1}{(n-1)^2} + \frac{1}{n^2}\right)$

The right side: each reciprocal square $\frac{1}{k^2}$ for $2 \leqslant k \leqslant N-1$ appears twice (once from $n = k$ giving $\frac{1}{(n-1)^2}$, once from $n = k+1$ giving $\frac{1}{n^2}$), while $\frac{1}{1^2}$ appears once (from $n=2$) and $\frac{1}{N^2}$ appears once (from $n=N$). So:

$\frac{1}{2}\!\left(1 + \frac{1}{N^2} + 2\sum_{k=2}^{N-1}\frac{1}{k^2}\right) = \frac{1}{2} + \frac{1}{2N^2} + \sum_{k=2}^{N-1}\frac{1}{k^2}$

Taking $N \to \infty$ and using the left side $\to 1$:

$1 \leqslant \frac{1}{2} + \sum_{k=2}^{\infty}\frac{1}{k^2} = \frac{1}{2} + \left(\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots\right)$

Again the inequality is strict:

$1 < \frac{1}{2} + \left(\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots\right) \qquad \text{(...)}$

Part (iv): Prove the final bounds on $\sum \frac{1}{n^2}$

We show $\frac{1}{2}\!\left(\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots\right) < \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots$

It suffices to show $\frac{1}{2}\!\left(\frac{1}{k^2} + \frac{1}{(k+1)^2}\right) < \frac{1}{k^2}$ for each odd $k \geqslant 3$, since the left side pairs consecutive terms while the right picks out the odd ones.

This holds because $\frac{1}{(k+1)^2} < \frac{1}{k^2}$, so $\frac{1}{2}\!\left(\frac{1}{k^2} + \frac{1}{(k+1)^2}\right) < \frac{1}{2}\!\left(\frac{1}{k^2} + \frac{1}{k^2}\right) = \frac{1}{k^2}$.

Writing this more explicitly: group the left side as

$\frac{1}{2}\!\left(\frac{1}{3^2} + \frac{1}{4^2}\right) + \frac{1}{2}\!\left(\frac{1}{5^2} + \frac{1}{6^2}\right) + \frac{1}{2}\!\left(\frac{1}{7^2} + \frac{1}{8^2}\right) + \cdots$

Each pair satisfies $\frac{1}{2}\!\left(\frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2}\right) < \frac{1}{(2m+1)^2}$, so the sum of the pairs is less than $\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots$, as required.

Now let $S = \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$

From (***):

$4\!\left(\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots\right) < 1$

Using the inequality just proved, $4 \cdot \frac{1}{2}\!\left(\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots\right) < 4\!\left(\frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots\right) < 1$, so:

$2\!\left(\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots\right) < 1$

Since $\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots = S - 1 - \frac{1}{4}$, we get $2\!\left(S - \frac{5}{4}\right) < 1$, giving $S < \frac{5}{4} + \frac{1}{2} = \frac{7}{4}$.

From (****):

$1 < \frac{1}{2} + \left(\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots\right) = \frac{1}{2} + S - 1$

so $S > 1 + \frac{1}{2} = \frac{3}{2}$.

Therefore:

$\frac{3}{2} < \sum_{n=1}^{\infty} \frac{1}{n^2} < \frac{7}{4}$

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Question 4

Topic: 几何序列/Geometric sequences（相切圆问题）  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The triangle $OAB$ is isosceles, with $OA = OB$ and angle $AOB = 2\alpha$ where $0 < \alpha < \frac{\pi}{2}$. The semi-circle $C_0$ has its centre at the midpoint of the base $AB$ of the triangle, and the sides $OA$ and $OB$ of the triangle are both tangent to the semi-circle. $C_1, C_2, C_3, \dots$ are circles such that $C_n$ is tangent to $C_{n-1}$ and to sides $OA$ and $OB$ of the triangle.

Let $r_n$ be the radius of $C_n$. Show that $\frac{r_{n+1}}{r_n} = \frac{1 - \sin \alpha}{1 + \sin \alpha} \, .$

Let $S$ be the total area of the semi-circle $C_0$ and the circles $C_1, C_2, C_3, \dots$. Show that $S = \frac{1 + \sin^2 \alpha}{4 \sin \alpha} \pi r_0^2 \, .$

Show that there are values of $\alpha$ for which $S$ is more than four fifths of the area of triangle $OAB$.

Hint

If circle $n$ has centre $O_n$ then $OO_n = \frac{r_n}{\sin \alpha}$, $OO_{n+1} = \frac{r_{n+1}}{\sin \alpha}$ and $OO_n - OO_{n+1} = r_n + r_{n+1}$.

Substituting and multiplying by $\sin \alpha$ gives $r_n - r_{n+1} = \sin \alpha(r_n + r_{n+1})$ which simplifies to $\frac{r_{n+1}}{r_n} = \frac{1 - \sin \alpha}{1 + \sin \alpha}$.

This result then implies that $r_n = \left( \frac{1 - \sin \alpha}{1 + \sin \alpha} \right)^n r_0$, so the total area is

$S = \frac{1}{2} \pi r_0^2 + \pi \left( \left( \frac{1 - \sin \alpha}{1 + \sin \alpha} \right) r_0 \right)^2 + \pi \left( \left( \frac{1 - \sin \alpha}{1 + \sin \alpha} \right)^2 r_0 \right)^2 + \pi \left( \left( \frac{1 - \sin \alpha}{1 + \sin \alpha} \right)^3 r_0 \right)^2 + \dots$

which is almost a geometric series with common ratio $\left( \frac{1 - \sin \alpha}{1 + \sin \alpha} \right)^2$, so

$S = \frac{\pi r_0^2}{1 - \left( \frac{1 - \sin \alpha}{1 + \sin \alpha} \right)^2} - \frac{1}{2} \pi r_0^2 = \left( \frac{(1 + \sin \alpha)^2}{4 \sin \alpha} - \frac{1}{2} \right) \pi r_0^2 = \frac{1 + \sin^2 \alpha}{4 \sin \alpha} \pi r_0^2.$

Area $T$ of triangle $OAB = \frac{1}{2} AB \times OO_0 = \frac{r_0}{\cos \alpha} \frac{r_0}{\sin \alpha}$,

so $\frac{S}{T} = \frac{\pi}{4} \cos \alpha (1 + \sin^2 \alpha) = \frac{\pi}{4} \cos \alpha (2 - \cos^2 \alpha)$.

By differentiation, the maximum $\frac{S}{T}$ occurs where $2 - 3 \cos^2 \alpha = 0$ (not $\sin \alpha = 0$) and equals

$\frac{\pi}{4} \sqrt{\frac{2}{3}} \left( 2 - \frac{2}{3} \right) = \frac{\pi}{3} \sqrt{\frac{2}{3}} > \sqrt{\frac{2}{3}} = \sqrt{\frac{16}{24}} > \sqrt{\frac{16}{25}} = \frac{4}{5}.$

Model Solution

Part (i): Show that $\frac{r_{n+1}}{r_n} = \frac{1 - \sin \alpha}{1 + \sin \alpha}$

Let $O$ be the apex of the isosceles triangle, and let $O_n$ be the centre of circle $C_n$. Since $C_n$ is tangent to both sides $OA$ and $OB$, and the angle between these sides is $2\alpha$, the centre $O_n$ lies on the angle bisector from $O$.

The distance from $O$ to $O_n$ can be found by considering the right triangle formed by $O$, the point of tangency on $OA$, and $O_n$. The angle at $O$ is $\alpha$ and the opposite side is $r_n$, so:

$OO_n = \frac{r_n}{\sin \alpha}$

Similarly, $OO_{n+1} = \frac{r_{n+1}}{\sin \alpha}$.

Since $C_n$ and $C_{n+1}$ are externally tangent to each other, the distance between their centres equals the sum of their radii. Also, $O_{n+1}$ lies between $O$ and $O_n$ (since $C_{n+1}$ is the smaller circle closer to $O$), so:

$OO_n - OO_{n+1} = r_n + r_{n+1}$

Substituting:

$\frac{r_n}{\sin \alpha} - \frac{r_{n+1}}{\sin \alpha} = r_n + r_{n+1}$

Multiplying both sides by $\sin \alpha$:

$r_n - r_{n+1} = \sin \alpha \, (r_n + r_{n+1})$

$r_n - r_{n+1} = r_n \sin \alpha + r_{n+1} \sin \alpha$

$r_n(1 - \sin \alpha) = r_{n+1}(1 + \sin \alpha)$

$\frac{r_{n+1}}{r_n} = \frac{1 - \sin \alpha}{1 + \sin \alpha} \qquad \text{as required.}$

Part (ii): Show that $S = \frac{1 + \sin^2 \alpha}{4 \sin \alpha} \pi r_0^2$

From Part (i), the ratio $k = \frac{1 - \sin \alpha}{1 + \sin \alpha}$ is constant, so $r_n = k^n r_0$ for $n \geqslant 1$.

The total area is:

$S = \frac{1}{2}\pi r_0^2 + \sum_{n=1}^{\infty} \pi r_n^2 = \frac{1}{2}\pi r_0^2 + \pi r_0^2 \sum_{n=1}^{\infty} k^{2n}$

The geometric series sums to $\sum_{n=1}^{\infty} k^{2n} = \frac{k^2}{1 - k^2}$, so:

$S = \pi r_0^2 \left( \frac{1}{2} + \frac{k^2}{1 - k^2} \right) = \pi r_0^2 \left( \frac{1 - k^2 + 2k^2}{2(1 - k^2)} \right) = \pi r_0^2 \cdot \frac{1 + k^2}{2(1 - k^2)}$

Now compute $k^2$ and $1 - k^2$:

$k^2 = \frac{(1 - \sin \alpha)^2}{(1 + \sin \alpha)^2}$

$1 - k^2 = \frac{(1 + \sin \alpha)^2 - (1 - \sin \alpha)^2}{(1 + \sin \alpha)^2} = \frac{4 \sin \alpha}{(1 + \sin \alpha)^2}$

$1 + k^2 = \frac{(1 + \sin \alpha)^2 + (1 - \sin \alpha)^2}{(1 + \sin \alpha)^2} = \frac{2(1 + \sin^2 \alpha)}{(1 + \sin \alpha)^2}$

Therefore:

$\frac{1 + k^2}{2(1 - k^2)} = \frac{2(1 + \sin^2 \alpha)}{(1 + \sin \alpha)^2} \cdot \frac{(1 + \sin \alpha)^2}{2 \cdot 4 \sin \alpha} = \frac{1 + \sin^2 \alpha}{4 \sin \alpha}$

$S = \frac{1 + \sin^2 \alpha}{4 \sin \alpha} \, \pi r_0^2 \qquad \text{as required.}$

Part (iii): Show that $S > \frac{4}{5} T$ for some values of $\alpha$

The area of triangle $OAB$ is $T = \frac{1}{2} \cdot AB \cdot h$ where $h$ is the height from $O$ to $AB$.

The base $AB = 2 \cdot \frac{r_0}{\cos \alpha}$ (from the right triangle at the base, where the tangent from $A$ to the semicircle gives half-base $= \frac{r_0}{\cos \alpha}$).

The height from $O$ to the base $AB$ is $\frac{r_0}{\sin \alpha}$ (the distance from $O$ to the centre of $C_0$ along the angle bisector, which equals $\frac{r_0}{\sin \alpha}$).

$T = \frac{1}{2} \cdot \frac{2r_0}{\cos \alpha} \cdot \frac{r_0}{\sin \alpha} = \frac{r_0^2}{\sin \alpha \cos \alpha}$

The ratio:

$\frac{S}{T} = \frac{1 + \sin^2 \alpha}{4 \sin \alpha} \cdot \pi r_0^2 \cdot \frac{\sin \alpha \cos \alpha}{r_0^2} = \frac{\pi \cos \alpha (1 + \sin^2 \alpha)}{4}$

Since $\sin^2 \alpha = 1 - \cos^2 \alpha$:

$\frac{S}{T} = \frac{\pi \cos \alpha (2 - \cos^2 \alpha)}{4}$

Let $c = \cos \alpha$ and define $g(c) = c(2 - c^2) = 2c - c^3$ for $0 < c < 1$. Then:

$g'(c) = 2 - 3c^2 = 0 \implies c = \sqrt{\frac{2}{3}}$

At this critical point:

$g\!\left(\sqrt{\frac{2}{3}}\right) = \sqrt{\frac{2}{3}}\left(2 - \frac{2}{3}\right) = \sqrt{\frac{2}{3}} \cdot \frac{4}{3} = \frac{4}{3}\sqrt{\frac{2}{3}} = \frac{4\sqrt{6}}{9}$

So the maximum ratio is:

$\frac{S}{T} = \frac{\pi}{4} \cdot \frac{4\sqrt{6}}{9} = \frac{\pi\sqrt{6}}{9}$

Now $\frac{\pi\sqrt{6}}{9} > \frac{4}{5}$ is equivalent to $5\pi\sqrt{6} > 36$, i.e., $150\pi^2 > 1296$, i.e., $\pi^2 > 8.64$. Since $\pi^2 \approx 9.87 > 8.64$, this holds.

Therefore there exist values of $\alpha$ (specifically, $\alpha = \arccos\sqrt{\frac{2}{3}}$) for which $S$ exceeds $\frac{4}{5}$ of the area of triangle $OAB$.

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Question 5

Topic: 三角方程/Trigonometric equations（辅助角公式）  |  Difficulty: Standard  |  Marks: 20

Problem

5 Show that if $\cos(x - \alpha) = \cos \beta$ then either $\tan x = \tan(\alpha + \beta)$ or $\tan x = \tan(\alpha - \beta)$. By choosing suitable values of $x$, $\alpha$ and $\beta$, give an example to show that if $\tan x = \tan(\alpha + \beta)$, then $\cos(x - \alpha)$ need not equal $\cos \beta$.

Let $\omega$ be the acute angle such that $\tan \omega = \frac{4}{3}$.

(ii) For $0 \leqslant x \leqslant 2\pi$, solve the equation

$\cos x - 7 \sin x = 5$

giving both solutions in terms of $\omega$.

(i) For $0 \leqslant x \leqslant 2\pi$, solve the equation

$2 \cos x + 11 \sin x = 10$

showing that one solution is twice the other and giving both in terms of $\omega$.

Hint

If $\cos(x - \alpha) = \cos \beta$ then $x - \alpha = 2n\pi \pm \beta$ so $x = \alpha \pm \beta + 2n\pi$ so $\tan x = \tan(\alpha \pm \beta)$ however, for example, $x = \pi$, $\alpha = \beta = 0$ has $\tan x = \tan \pi = \tan 0 = \tan(\alpha + \beta)$ but $\cos(x - \alpha) = \cos \pi = -1 \neq 1 = \cos \beta$.

a Writing $\cos x - 7 \sin x = R \cos(x - \alpha)$ requires $R = \sqrt{50} = 5\sqrt{2}$ and $\tan \alpha = -7$, so $\cos(x - \alpha) = \cos \beta$, where $\cos \beta = \frac{1}{\sqrt{2}}$, so we can take $\tan \beta = 1$.

$\text{Hence } \tan x = \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{-7 \pm 1}{1 \pm 7} = -\frac{3}{4} \text{ or } \frac{4}{3}.$

The first of these gives $x = \frac{1}{2}\pi + \omega$ or $x = \frac{3}{2}\pi + \omega$ (since $\arctan \frac{3}{4} = \frac{\pi}{2} - \arctan \frac{4}{3}$) and the second $x = \omega$ or $x = \pi + \omega$. However, the first solution in each case does not satisfy the original equation (both have $\sin x > 0$, so $\cos x - 7 \sin x < 1$), so $x = \frac{3}{2}\pi + \omega$ or $\pi + \omega$.

b proceeding as in (i), $\cos(x - \alpha) = \cos \beta$, where $\tan \alpha = \frac{11}{2}$ and $R = 5\sqrt{5}$, so $\cos \beta = \frac{2}{\sqrt{5}}$ and so $\tan \beta = \frac{1}{2}$. Hence $\tan x = \frac{22 \pm 2}{4 \mp 11} = -\frac{24}{7}$ or $\frac{4}{3}$.

Notice that $\tan 2\omega = \frac{2 \tan \omega}{1 - \tan^2 \omega} = -\frac{24}{7}$ so the solutions are $x = \omega$ and $x = 2\omega$, again eliminating the other two possibilities, $\omega + \pi$ and $2\omega + \pi$, by checking in the original equation.

Model Solution

Part (i): Show the trigonometric implication and provide a counterexample

If $\cos(x - \alpha) = \cos \beta$, then $x - \alpha = 2n\pi \pm \beta$ for some integer $n$.

Case 1: $x - \alpha = 2n\pi + \beta$, so $x = \alpha + \beta + 2n\pi$, giving $\tan x = \tan(\alpha + \beta)$.

Case 2: $x - \alpha = 2n\pi - \beta$, so $x = \alpha - \beta + 2n\pi$, giving $\tan x = \tan(\alpha - \beta)$.

Therefore either $\tan x = \tan(\alpha + \beta)$ or $\tan x = \tan(\alpha - \beta)$.

Counterexample: Take $x = \pi$, $\alpha = 0$, $\beta = 0$. Then $\tan x = \tan \pi = 0 = \tan(0 + 0) = \tan(\alpha + \beta)$, but $\cos(x - \alpha) = \cos \pi = -1 \neq 1 = \cos \beta$. So $\tan x = \tan(\alpha + \beta)$ does not imply $\cos(x - \alpha) = \cos \beta$.

Part (ii): Solve $\cos x - 7 \sin x = 5$ for $0 \leqslant x \leqslant 2\pi$

Write $\cos x - 7 \sin x = R \cos(x - \alpha)$ where $R=\sqrt{50}=5\sqrt2$. Comparing coefficients in $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$, we need $R\cos\alpha=1$ and $R\sin\alpha=-7$, so $\tan\alpha=-7$.

The equation becomes $5\sqrt{2}\cos(x - \alpha) = 5$, so $\cos(x - \alpha) = \frac{1}{\sqrt{2}}$.

By the result from Part (i) with $\beta$ such that $\cos\beta = \frac{1}{\sqrt{2}}$ (so $\tan\beta = 1$):

$\tan x = \tan(\alpha + \beta) \quad \text{or} \quad \tan x = \tan(\alpha - \beta)$

Using the tangent addition formula with $\tan\alpha = -7$ and $\tan\beta = 1$:

$\tan(\alpha + \beta) = \frac{-7 + 1}{1 + 7} = \frac{-6}{8} = -\frac{3}{4}$

$\tan(\alpha - \beta) = \frac{-7 - 1}{1 - 7} = \frac{-8}{-6} = \frac{4}{3}$

Since $\tan\omega = \frac{4}{3}$ with $\omega$ acute, and $\tan(\pi + \omega) = \tan\omega = \frac{4}{3}$.

For $\tan x = \frac{4}{3}$: $x = \omega$ or $x = \pi + \omega$.

For $\tan x = -\frac{3}{4}$: since $\arctan\frac{3}{4} = \frac{\pi}{2} - \omega$ (complementary angles), $\tan\left(\frac{\pi}{2} - \omega\right) = \frac{3}{4}$, so $\tan x = -\frac{3}{4}$ gives $x = \pi - \left(\frac{\pi}{2} - \omega\right) = \frac{\pi}{2} + \omega$ or $x = 2\pi - \left(\frac{\pi}{2} - \omega\right) = \frac{3\pi}{2} + \omega$.

We have four candidates: $\omega$, $\pi + \omega$, $\frac{\pi}{2} + \omega$, $\frac{3\pi}{2} + \omega$. We must check which satisfy the original equation (the tangent result only gives necessary conditions).

Note $\cos\alpha = \frac{1}{5\sqrt{2}}$ and $\sin\alpha = \frac{-7}{5\sqrt{2}}$, so $\alpha$ is in the fourth quadrant. For $\cos(x - \alpha) = \frac{1}{\sqrt{2}}$, we need $x - \alpha = \pm\frac{\pi}{4} + 2n\pi$.

Checking the four candidates in the original equation:

• $x = \omega$: $\cos\omega = \frac{3}{5}$, $\sin\omega = \frac{4}{5}$, so $\frac{3}{5} - \frac{28}{5} = -5 \neq 5$. Not a solution.
• $x = \pi + \omega$: $\cos(\pi + \omega) = -\frac{3}{5}$, $\sin(\pi + \omega) = -\frac{4}{5}$, so $-\frac{3}{5} + \frac{28}{5} = 5$. Solution.
• $x = \frac{\pi}{2} + \omega$: $\cos(\frac{\pi}{2} + \omega) = -\sin\omega = -\frac{4}{5}$, $\sin(\frac{\pi}{2} + \omega) = \cos\omega = \frac{3}{5}$, so $-\frac{4}{5} - \frac{21}{5} = -5 \neq 5$. Not a solution.
• $x = \frac{3\pi}{2} + \omega$: $\cos(\frac{3\pi}{2} + \omega) = \sin\omega = \frac{4}{5}$, $\sin(\frac{3\pi}{2} + \omega) = -\cos\omega = -\frac{3}{5}$, so $\frac{4}{5} + \frac{21}{5} = 5$. Solution.

The solutions are $x = \pi + \omega$ and $x = \frac{3\pi}{2} + \omega$.

Part (iii): Solve $2 \cos x + 11 \sin x = 10$ for $0 \leqslant x \leqslant 2\pi$

Write $2\cos x + 11\sin x = R\cos(x - \alpha)$ where $R = \sqrt{4 + 121} = \sqrt{125} = 5\sqrt{5}$ and $R\cos\alpha = 2$, $R\sin\alpha = 11$, so $\tan\alpha = \frac{11}{2}$.

The equation becomes $5\sqrt{5}\cos(x - \alpha) = 10$, so $\cos(x - \alpha) = \frac{2}{\sqrt{5}}$.

By Part (i), with $\cos\beta = \frac{2}{\sqrt{5}}$ (so $\tan\beta = \frac{1}{2}$):

$\tan x = \tan(\alpha + \beta) \quad \text{or} \quad \tan x = \tan(\alpha - \beta)$

$\tan(\alpha + \beta) = \frac{\frac{11}{2} + \frac{1}{2}}{1 - \frac{11}{2} \cdot \frac{1}{2}} = \frac{6}{1 - \frac{11}{4}} = \frac{6}{-\frac{7}{4}} = -\frac{24}{7}$

$\tan(\alpha - \beta) = \frac{\frac{11}{2} - \frac{1}{2}}{1 + \frac{11}{2} \cdot \frac{1}{2}} = \frac{5}{1 + \frac{11}{4}} = \frac{5}{\frac{15}{4}} = \frac{4}{3}$

Now $\tan\omega = \frac{4}{3}$, so $\tan(\alpha - \beta) = \tan\omega$, giving $x = \omega$ or $x = \pi + \omega$.

For $\tan(\alpha + \beta) = -\frac{24}{7}$: note that $\tan 2\omega = \frac{2\tan\omega}{1 - \tan^2\omega} = \frac{8/3}{1 - 16/9} = \frac{8/3}{-7/9} = -\frac{24}{7}$.

So $\tan x = \tan 2\omega$, giving $x = 2\omega$ or $x = \pi + 2\omega$.

We have four candidates: $\omega$, $\pi + \omega$, $2\omega$, $\pi + 2\omega$. Checking each:

• $x = \omega$: $2\cos\omega + 11\sin\omega = 2 \cdot \frac{3}{5} + 11 \cdot \frac{4}{5} = \frac{6 + 44}{5} = 10$. Solution.
• $x = 2\omega$: $\cos 2\omega = \frac{1 - \tan^2\omega}{1 + \tan^2\omega} = \frac{1 - 16/9}{1 + 16/9} = \frac{-7/9}{25/9} = -\frac{7}{25}$, $\sin 2\omega = \frac{2\tan\omega}{1 + \tan^2\omega} = \frac{8/3}{25/9} = \frac{24}{25}$. Then $2(-\frac{7}{25}) + 11 \cdot \frac{24}{25} = \frac{-14 + 264}{25} = \frac{250}{25} = 10$. Solution.
• $x = \pi + \omega$: $2\cos(\pi + \omega) + 11\sin(\pi + \omega) = -2\cos\omega - 11\sin\omega = -10 \neq 10$. Not a solution.
• $x = \pi + 2\omega$: similarly gives $-10 \neq 10$. Not a solution.

The solutions are $x = \omega$ and $x = 2\omega$, where one solution is twice the other.

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Question 6

Topic: 递推序列与不变量 (Recurrence Sequences and Invariants)  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Given a sequence $w_0, w_1, w_2, \dots$, the sequence $F_1, F_2, \dots$ is defined by

$F_n = w_n^2 + w_{n-1}^2 - 4w_n w_{n-1} .$

Show that $F_n - F_{n-1} = (w_n - w_{n-2})(w_n + w_{n-2} - 4w_{n-1})$ for $n \geqslant 2$.

(i) The sequence $u_0, u_1, u_2, \dots$ has $u_0 = 1$, and $u_1 = 2$ and satisfies

$u_n = 4u_{n-1} - u_{n-2} \quad (n \geqslant 2) .$

Prove that $u_n^2 + u_{n-1}^2 = 4u_n u_{n-1} - 3$ for $n \geqslant 1$.

(ii) A sequence $v_0, v_1, v_2, \dots$ has $v_0 = 1$ and satisfies

$v_n^2 + v_{n-1}^2 = 4v_n v_{n-1} - 3 \quad (n \geqslant 1) . \qquad \text{(*)}$

(a) Find $v_1$ and prove that, for each $n \geqslant 2$, either $v_n = 4v_{n-1} - v_{n-2}$ or $v_n = v_{n-2}$.

(b) Show that the sequence, with period 2, defined by

$v_n = \begin{cases} 1 & \text{for } n \text{ even} \\ 2 & \text{for } n \text{ odd} \end{cases}$

satisfies (*).

(c) Find a sequence $v_n$ with period 4 which has $v_0 = 1$, and satisfies (*).

Hint

6 $F_n - F_{n-1} = w_n^2 + w_{n-1}^2 - 4w_n w_{n-1} - w_{n-1}^2 - w_{n-2}^2 + 4w_{n-1} w_{n-2} = w_n^2 - w_{n-2}^2 - 4w_{n-1}(w_n - w_{n-2})$ $= (w_n - w_{n-2})(w_n + w_{n-2} - 4w_{n-1}.)$ (+)

(i) Let $w_n$ be $u_n$; then $u_n + u_{n-2} - 4u_{n-1} = 0$, so $F_n - F_{n-1} = 0$ for $n \geq 2$, by (+), but $F_1 = u_1^2 + u_0^2 - 4u_1 u_0 = -3$ so $F_n = -3$ for $n \geq 1$

(ii) In this part, let $w_n$ be $v_n$.

(a) $v_1^2 + 1 = 4v_1 - 3 \Rightarrow (v_1 - 2)^2 = 0 \Rightarrow v_1 = 2$ $F_n = v_n^2 + v_{n-1}^2 - 4v_n v_{n-1} = -3$ for $n \geq 1$

$\Rightarrow v_n - v_{n-2} = 0$ or $v_n + v_{n-2} - 4v_{n-1} = 0$, for $n \geq 2$, by (+).

(b) Since $1, 2, 1, 2, \dots$ satisfies $v_n - v_{n-2} = 0$ for $n \geq 2$, $F_n$ is constant, by (+) and since $v_0 = 1, v_1 = 2$ that constant is $-3$, so the sequence satisfies (*).

(c) The sequence $1, 2, 7, 2, \dots$, with period 4, satisfies $v_n - v_{n-2} = 0$ for odd $n \geq 2$ and $v_n + v_{n-2} - 4v_{n-1} = 0$ for even $n \geq 2$, so $F_n$ is constant, by (+), and since $v_0 = 1, v_1 = 2$ that constant is $-3$, so the sequence satisfies (*).

Model Solution

Preliminary: Show that $F_n - F_{n-1} = (w_n - w_{n-2})(w_n + w_{n-2} - 4w_{n-1})$

$F_n - F_{n-1} = (w_n^2 + w_{n-1}^2 - 4w_n w_{n-1}) - (w_{n-1}^2 + w_{n-2}^2 - 4w_{n-1} w_{n-2})$

$= w_n^2 - w_{n-2}^2 - 4w_{n-1}(w_n - w_{n-2})$

$= (w_n - w_{n-2})(w_n + w_{n-2}) - 4w_{n-1}(w_n - w_{n-2})$

$= (w_n - w_{n-2})(w_n + w_{n-2} - 4w_{n-1}) \qquad \text{(...)}$

Part (i): Prove that $u_n^2 + u_{n-1}^2 = 4u_n u_{n-1} - 3$ for $n \geqslant 1$

Define $F_n = u_n^2 + u_{n-1}^2 - 4u_n u_{n-1}$. Since $u_n = 4u_{n-1} - u_{n-2}$, we have $u_n + u_{n-2} - 4u_{n-1} = 0$.

By the result (…), $F_n - F_{n-1} = (u_n - u_{n-2})(u_n + u_{n-2} - 4u_{n-1}) = 0$ for all $n \geqslant 2$.

So $F_n$ is constant for $n \geqslant 1$. Computing $F_1$:

$F_1 = u_1^2 + u_0^2 - 4u_1 u_0 = 4 + 1 - 8 = -3$

Therefore $F_n = -3$ for all $n \geqslant 1$, i.e.:

$u_n^2 + u_{n-1}^2 - 4u_n u_{n-1} = -3$

$u_n^2 + u_{n-1}^2 = 4u_n u_{n-1} - 3 \qquad \text{for } n \geqslant 1. \qquad \text{(...)}$

Part (ii)(a): Find $v_1$ and prove the disjunction

Setting $n = 1$ in (*): $v_1^2 + v_0^2 = 4v_1 v_0 - 3$.

$v_1^2 + 1 = 4v_1 - 3$

$v_1^2 - 4v_1 + 4 = 0$

$(v_1 - 2)^2 = 0$

$v_1 = 2$

Now let $F_n = v_n^2 + v_{n-1}^2 - 4v_n v_{n-1}$. From (*), $F_n = -3$ for all $n \geqslant 1$.

For $n \geqslant 2$, using (…):

$F_n - F_{n-1} = (v_n - v_{n-2})(v_n + v_{n-2} - 4v_{n-1}) = 0$

since both $F_n = -3$ and $F_{n-1} = -3$. Therefore:

$(v_n - v_{n-2})(v_n + v_{n-2} - 4v_{n-1}) = 0$

So either $v_n = v_{n-2}$ or $v_n = 4v_{n-1} - v_{n-2}$ for each $n \geqslant 2$.

Part (ii)(b): Show the period-2 sequence satisfies (*)

The sequence is $1, 2, 1, 2, 1, 2, \ldots$ with $v_0 = 1$, $v_1 = 2$.

For all $n \geqslant 2$, $v_n = v_{n-2}$ (this is the definition of period 2). By the result (…), $F_n - F_{n-1} = 0$, so $F_n$ is constant. Since:

$F_1 = v_1^2 + v_0^2 - 4v_1 v_0 = 4 + 1 - 8 = -3$

we have $F_n = -3$ for all $n \geqslant 1$, which is exactly (*).

Part (ii)(c): Find a period-4 sequence with $v_0 = 1$ satisfying (*)

We need a sequence where the pattern of choices (either $v_n = v_{n-2}$ or $v_n = 4v_{n-1} - v_{n-2}$) repeats with period 4. We already know $v_0 = 1$ and $v_1 = 2$.

To get period 4, we need $v_2 \neq v_0$ (otherwise it would be period 2). So at $n = 2$, choose $v_2 = 4v_1 - v_0 = 8 - 1 = 7$.

For period 4, we need $v_3 = v_1 = 2$ (using $v_n = v_{n-2}$ at $n = 3$). Then $v_4 = v_0 = 1$ and $v_5 = v_1 = 2$, confirming period 4.

The sequence is $1, 2, 7, 2, 1, 2, 7, 2, \ldots$.

Check (*): $v_n^2 + v_{n-1}^2 = 4v_n v_{n-1} - 3$ for each $n \geqslant 1$.

• $n = 1$: $4 + 1 = 8 - 3 = 5$.
• $n = 2$: $49 + 4 = 56 - 3 = 53$.
• $n = 3$: $4 + 49 = 56 - 3 = 53$.
• $n = 4$: $1 + 4 = 8 - 3 = 5$.

For the recurrence choices: at $n=2$, $v_2=4v_1-v_0=7$; at $n=3$, $v_3=v_1=2$; at $n=4$, $v_4=4v_3-v_2=1$; and at $n=5$, $v_5=v_3=2$. The same four choices repeat, so the sequence has period 4 and satisfies (*) throughout.

Thus $v_n=1,2,7,2,1,2,7,2,\ldots$ is a period-4 sequence with $v_0=1$ satisfying (*).

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Question 7

Topic: 积分序列与级数估计 (Integral Sequences and Series Estimation)  |  Difficulty: Challenging  |  Marks: 20

Problem

7 For $n = 1, 2, 3, \dots$, let

$I_n = \int_0^1 \frac{t^{n-1}}{(t+1)^n} \, dt \, .$

By considering the greatest value taken by $\frac{t}{t+1}$ for $0 \leqslant t \leqslant 1$ show that $I_{n+1} < \frac{1}{2} I_n$.

Show also that $I_{n+1} = -\frac{1}{n \, 2^n} + I_n$.

Deduce that $I_n < \frac{1}{n \, 2^{n-1}}$.

Prove that

$\ln 2 = \sum_{r=1}^n \frac{1}{r \, 2^r} + I_{n+1}$

and hence show that $\frac{2}{3} < \ln 2 < \frac{17}{24}$.

Hint

7 On $0 \leqslant t \leqslant 1$, the integrand is non-negative and $0 \leqslant \frac{t}{t+1} = 1 - \frac{1}{t+1} \leqslant \frac{1}{2}$, so

$I_{n+1} = \int_{0}^{1} \frac{t^n}{(t+1)^{n+1}} dt = \int_{0}^{1} \frac{t}{t+1} \frac{t^{n-1}}{(t+1)^n} dt < \frac{1}{2} \int_{0}^{1} \frac{t^{n-1}}{(t+1)^n} dt = \frac{1}{2} I_n.$

Integration by parts gives

$I_{n+1} = \left[ -\frac{t^n}{n(t+1)^n} \right]_{0}^{1} + \int_{0}^{1} \frac{nt^{n-1}}{n(t+1)^n} dt = -\frac{1}{n2^n} + I_n,$

so $I_n > 2I_{n+1} = -\frac{1}{n2^{n-1}} + 2I_n \implies I_n < \frac{1}{n2^{n-1}} \quad (*).$

Since $\frac{1}{2^r} = I_r - I_{r+1}$, $\sum_{r=1}^{n} \frac{1}{r2^r} = (I_1 - I_2) + (I_2 - I_3) + \dots + (I_n - I_{n+1}) = I_1 - I_{n+1}$, and

$I_1 = \int_{0}^{1} \frac{1}{t+1} dt = \ln 2, \text{ so } \ln 2 = \sum_{r=1}^{n} \frac{1}{r2^r} + I_{n+1}.$

Hence $\ln 2 > \sum_{r=1}^{3} \frac{1}{r2^r} = \frac{2}{3}$ and, by inequality $(*)$, $\ln 2 = \sum_{r=1}^{2} \frac{1}{r2^r} + I_3 < \sum_{r=1}^{2} \frac{1}{r2^r} + \frac{1}{3 \cdot 2^2} = \frac{17}{24}.$

Model Solution

Part (i): Show that $I_{n+1} < \frac{1}{2} I_n$

On the interval $0 \leqslant t \leqslant 1$, the function $\frac{t}{t+1} = 1 - \frac{1}{t+1}$ is increasing, so its greatest value is attained at $t = 1$:

$\frac{t}{t+1} \leqslant \frac{1}{2}$

Since the integrand of $I_{n+1}$ is non-negative for $t \in [0,1]$, we write:

$I_{n+1} = \int_0^1 \frac{t^n}{(t+1)^{n+1}} \, dt = \int_0^1 \frac{t}{t+1} \cdot \frac{t^{n-1}}{(t+1)^n} \, dt$

Because $\frac{t}{t+1} \leqslant \frac{1}{2}$ on $[0,1]$ with strict inequality on $[0,1)$:

$I_{n+1} < \frac{1}{2} \int_0^1 \frac{t^{n-1}}{(t+1)^n} \, dt = \frac{1}{2} I_n$

Part (ii): Show that $I_{n+1} = -\frac{1}{n \, 2^n} + I_n$

We apply integration by parts to $I_{n+1} = \int_0^1 \frac{t^n}{(t+1)^{n+1}} \, dt$.

Let $u = t^n$ and $dv = (t+1)^{-(n+1)} \, dt$. Then $du = n t^{n-1} \, dt$ and $v = -\frac{1}{n(t+1)^n}$.

$I_{n+1} = \left[ -\frac{t^n}{n(t+1)^n} \right]_0^1 + \int_0^1 \frac{n t^{n-1}}{n(t+1)^n} \, dt$

Evaluating the boundary term:

$\left[ -\frac{t^n}{n(t+1)^n} \right]_0^1 = -\frac{1}{n \cdot 2^n} - 0 = -\frac{1}{n \, 2^n}$

The remaining integral is:

$\int_0^1 \frac{t^{n-1}}{(t+1)^n} \, dt = I_n$

Therefore:

$I_{n+1} = -\frac{1}{n \, 2^n} + I_n \qquad \text{(...)}$

Part (iii): Deduce that $I_n < \frac{1}{n \, 2^{n-1}}$

From the recurrence (…), $I_n = I_{n+1} + \frac{1}{n \, 2^n}$. From Part (i), $I_{n+1} < \frac{1}{2} I_n$, so:

$I_n = I_{n+1} + \frac{1}{n \, 2^n} < \frac{1}{2} I_n + \frac{1}{n \, 2^n}$

Subtracting $\frac{1}{2} I_n$ from both sides:

$\frac{1}{2} I_n < \frac{1}{n \, 2^n}$

$I_n < \frac{1}{n \, 2^{n-1}} \qquad \text{(**)}$

Part (iv): Prove that $\ln 2 = \sum_{r=1}^n \frac{1}{r \, 2^r} + I_{n+1}$

From the recurrence (…):

$I_r - I_{r+1} = \frac{1}{r \, 2^r}$

Summing from $r = 1$ to $n$:

$\sum_{r=1}^n \frac{1}{r \, 2^r} = \sum_{r=1}^n (I_r - I_{r+1}) = (I_1 - I_2) + (I_2 - I_3) + \cdots + (I_n - I_{n+1}) = I_1 - I_{n+1}$

This telescopes, leaving only $I_1 - I_{n+1}$. Now:

$I_1 = \int_0^1 \frac{1}{t+1} \, dt = \left[ \ln(t+1) \right]_0^1 = \ln 2$

Therefore:

$\ln 2 = \sum_{r=1}^n \frac{1}{r \, 2^r} + I_{n+1}$

Part (v): Show that $\frac{2}{3} < \ln 2 < \frac{17}{24}$

Since $I_{n+1} > 0$ (the integrand is positive on $[0,1]$), we have:

$\ln 2 = \sum_{r=1}^n \frac{1}{r \, 2^r} + I_{n+1} > \sum_{r=1}^n \frac{1}{r \, 2^r}$

Taking $n = 3$:

$\ln 2 > \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 4} + \frac{1}{3 \cdot 8} = \frac{1}{2} + \frac{1}{8} + \frac{1}{24} = \frac{12 + 3 + 1}{24} = \frac{16}{24} = \frac{2}{3}$

For the upper bound, take $n = 2$:

$\ln 2 = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 4} + I_3 = \frac{1}{2} + \frac{1}{8} + I_3 = \frac{5}{8} + I_3$

By inequality (**), $I_3 < \frac{1}{3 \cdot 2^2} = \frac{1}{12}$, so:

$\ln 2 < \frac{5}{8} + \frac{1}{12} = \frac{15}{24} + \frac{2}{24} = \frac{17}{24}$

Therefore $\frac{2}{3} < \ln 2 < \frac{17}{24}$.

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Question 8

Topic: 微分方程与解曲线 (Differential Equations and Solution Curves)  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Show that if

$\frac{dy}{dx} = f(x)y + \frac{g(x)}{y}$

then the substitution $u = y^2$ gives a linear differential equation for $u(x)$.

Hence or otherwise solve the differential equation

$\frac{dy}{dx} = \frac{y}{x} - \frac{1}{y} \, .$

Determine the solution curves of this equation which pass through $(1, 1)$, $(2, 2)$ and $(4, 4)$ and sketch graphs of all three curves on the same axes.

Hint

8 If $u = y^2$ then $\frac{du}{dx} = 2y \frac{dy}{dx} = 2f(x)y^2 + 2g(x) = 2f(x)u + 2g(x)$,

which is a linear differential equation for $u(x)$.

In this case, $f(x) = \frac{1}{x}$, $g(x) = -1$ so the differential equation is $\frac{du}{dx} = \frac{2u}{x} - 2$.

The integrating factor is $e^{\int \frac{2}{x} dx} = e^{2 \ln x} = \frac{1}{x^2}$, giving $\frac{1}{x^2} \frac{du}{dx} - \frac{2u}{x^3} = \frac{d}{dx} \left( \frac{u}{x^2} \right) = \frac{-2}{x^2}$

so that $\frac{u}{x^2} = \int \frac{-2}{x^2} dx = \frac{2}{x} + c$ or $u = y^2 = cx^2 + 2x$.

The solution curves which pass through $(1, 1)$, $(2, 2)$ and $(4, 4)$ are $y^2 + (x - 1)^2 = 1$, $y^2 = 2x$ and $(x + 2)^2 - 2y^2 = 4$ respectively. In drawing these curves it should be made clear that all of them pass through the origin, and that this is their only point of intersection; that the first is a circle with centre $(1, 0)$, the second a parabola and the third an hyperbola with centre $(-2, 0)$ and asymptotes $y = \pm \frac{x+2}{\sqrt{2}}$.

Model Solution

Part (i): Show that $u = y^2$ gives a linear differential equation

Given:

$\frac{dy}{dx} = f(x)y + \frac{g(x)}{y}$

Set $u = y^2$. Then $\frac{du}{dx} = 2y \frac{dy}{dx}$, so:

$\frac{du}{dx} = 2y \left( f(x)y + \frac{g(x)}{y} \right) = 2f(x)y^2 + 2g(x) = 2f(x)u + 2g(x)$

This is a linear first-order ODE for $u(x)$:

$\frac{du}{dx} - 2f(x)u = 2g(x)$

Part (ii): Solve $\frac{dy}{dx} = \frac{y}{x} - \frac{1}{y}$

Here $f(x) = \frac{1}{x}$ and $g(x) = -1$. With $u = y^2$:

$\frac{du}{dx} = \frac{2u}{x} - 2$

Rearranging:

$\frac{du}{dx} - \frac{2u}{x} = -2$

The integrating factor is:

$e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln x} = \frac{1}{x^2}$

Multiplying both sides by $\frac{1}{x^2}$:

$\frac{1}{x^2} \frac{du}{dx} - \frac{2u}{x^3} = -\frac{2}{x^2}$

The left side is $\frac{d}{dx}\!\left(\frac{u}{x^2}\right)$, so:

$\frac{d}{dx}\!\left(\frac{u}{x^2}\right) = -\frac{2}{x^2}$

Integrating:

$\frac{u}{x^2} = \int -\frac{2}{x^2} \, dx = \frac{2}{x} + c$

Therefore:

$u = cx^2 + 2x$

Substituting back $u = y^2$:

$y^2 = cx^2 + 2x$

Part (iii): Solution curves through $(1, 1)$, $(2, 2)$ and $(4, 4)$

Curve through $(1, 1)$: Substituting $x = 1$, $y = 1$:

$1 = c + 2 \implies c = -1$

So $y^2 = -x^2 + 2x$, i.e., $x^2 + y^2 - 2x = 0$. Completing the square:

$(x - 1)^2 + y^2 = 1$

This is a circle centred at $(1, 0)$ with radius $1$. Since $y^2 = 2x - x^2 = x(2-x) \geqslant 0$, we need $0 \leqslant x \leqslant 2$.

Curve through $(2, 2)$: Substituting $x = 2$, $y = 2$:

$4 = 4c + 4 \implies c = 0$

So $y^2 = 2x$. This is a parabola opening to the right, valid for $x \geqslant 0$.

Curve through $(4, 4)$: Substituting $x = 4$, $y = 4$:

$16 = 16c + 8 \implies c = \frac{1}{2}$

So $y^2 = \frac{1}{2}x^2 + 2x$, i.e., $2y^2 = x^2 + 4x$. Completing the square:

$2y^2 = (x + 2)^2 - 4 \implies \frac{(x+2)^2}{4} - \frac{y^2}{2} = 1$

This is a hyperbola centred at $(-2, 0)$ with asymptotes $y = \pm \frac{x+2}{\sqrt{2}}$. The right branch passes through the origin and $(4, 4)$.

Sketch description: All three curves pass through the origin (setting $y = 0$ in $y^2 = cx^2 + 2x$ gives $x = 0$ for any $c$). The origin is their only common point. The sketch should show:

• The circle $(x-1)^2 + y^2 = 1$ (a complete circle, symmetric about the $x$-axis, passing through $(0,0)$ and $(2,0)$).
• The parabola $y^2 = 2x$ opening rightward, passing through the origin and $(2,2)$.
• The right branch of the hyperbola $\frac{(x+2)^2}{4} - \frac{y^2}{2} = 1$, passing through the origin and $(4,4)$, with asymptotes $y = \pm \frac{x+2}{\sqrt{2}}$.