STEP2 2013 -- Pure Mathematics

STEP2 2013 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2013  |  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	对数与指数函数 Logarithms and Exponentials	Challenging	隐函数求导, 对数恒等式, 函数单调性分析, 图像论证
2	积分与递推公式 Integration and Reduction Formulae	Challenging	递推关系, 代换法, Beta函数与阶乘的关系, 三角代换
3	三次方程与根的性质 Cubic Equations and Root Analysis	Challenging	韦达定理, 函数图像分析, 极值点讨论, 反例构造
4	坐标几何与轨迹 Coordinate Geometry and Loci	Challenging	联立方程, 参数消去, 轨迹方程, 圆的弦中点性质
5	函数对称性与驻点 Function Symmetry and Stationary Points	Hard	隐函数求导, 对称性分析, 曲线描绘, 代数方程求解
6	数列与极限 Sequences and Limits	Challenging	递推关系化简,数学归纳法证有界性,单调收敛定理,子序列极限论证
7	数论 Number Theory	Challenging	Pell方程递推生成解,代数变形与奇偶分析,因式分解分类讨论,丢番图方程求解
8	微积分 Calculus	Challenging	导数求极值,积分平均值函数,几何面积比较,不等式证明

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Question 1

Topic: 对数与指数函数 Logarithms and Exponentials  |  Difficulty: Challenging  |  Marks: 20

Problem

1 (i) Find the value of $m$ for which the line $y = mx$ touches the curve $y = \ln x$. If instead the line intersects the curve when $x = a$ and $x = b$, where $a < b$, show that $a^b = b^a$. Show by means of a sketch that $a < e < b$.

(ii) The line $y = mx + c$, where $c > 0$, intersects the curve $y = \ln x$ when $x = p$ and $x = q$, where $p < q$. Show by means of a sketch, or otherwise, that $p^q > q^p$.

(iii) Show by means of a sketch that the straight line through the points $(p, \ln p)$ and $(q, \ln q)$, where $e \leqslant p < q$, intersects the $y$-axis at a positive value of $y$. Which is greater, $\pi^e$ or $e^\pi$?

(iv) Show, using a sketch or otherwise, that if $0 < p < q$ and $\frac{\ln q - \ln p}{q - p} = e^{-1}$, then $q^p > p^q$.

Hint

1. The first two results, whilst not necessarily included in current A2 specifications, are standard work. Applying them, $\int_{0}^{\frac{1}{2}\pi} \frac{1}{1+a \sin x} dx = 2 \int_{0}^{1} \frac{1}{(1-a^2)+(t+a)^2} dt$, which can then be evaluated using a change of variable to give $\frac{2}{\sqrt{1-a^2}} \left( \tan^{-1} \frac{1+a}{\sqrt{1-a^2}} - \tan^{-1} \frac{a}{\sqrt{1-a^2}} \right)$. To simplify this to obtain the required result, $\tan \left( \tan^{-1} \frac{1+a}{\sqrt{1-a^2}} - \tan^{-1} \frac{a}{\sqrt{1-a^2}} \right)$ must be simplified using the relevant compound angle formula.

Model Solution

Part (i)

Finding $m$: For the line $y = mx$ to touch the curve $y = \ln x$, both the values and gradients must agree at the point of tangency:

$mx = \ln x \quad \text{and} \quad m = \frac{1}{x}$

From the gradient condition, $x = 1/m$. Substituting:

$m \cdot \frac{1}{m} = \ln \frac{1}{m} \implies 1 = -\ln m \implies m = \frac{1}{e}$

The tangent point is at $x = e$, $y = 1$.

Showing $a^b = b^a$: If $y = mx$ meets $y = \ln x$ at $x = a$ and $x = b$:

$ma = \ln a \quad \text{and} \quad mb = \ln b$

So $m = \frac{\ln a}{a} = \frac{\ln b}{b}$, giving $b \ln a = a \ln b$, hence $\ln(a^b) = \ln(b^a)$, so $a^b = b^a$. $\square$

Showing $a < e < b$: Consider $g(x) = \frac{\ln x}{x}$ for $x > 0$. Then $g'(x) = \frac{1 - \ln x}{x^2}$, so $g$ increases on $(0, e)$, decreases on $(e, \infty)$, with maximum $g(e) = 1/e$.

Since the line through the origin cuts $y = \ln x$ at two points (rather than touching), we need $m < 1/e$. Now $g(a) = g(b) = m < 1/e = g(e)$. Since $g$ is increasing on $(0, e)$ and decreasing on $(e, \infty)$, the two solutions of $g(x) = m$ must lie on either side of $e$, so $a < e < b$.

Sketch: The concave curve $y = \ln x$ with a line through the origin of gradient $m < 1/e$ intersecting it at two points straddling $x = e$.

Part (ii)

At the intersection points $x = p$ and $x = q$ ($p < q$):

$\ln p = mp + c \quad \text{and} \quad \ln q = mq + c$

Subtracting: $\ln q - \ln p = m(q - p)$, so $m = \frac{\ln q - \ln p}{q - p}$.

Solving for $c$ from the first equation:

$c = \ln p - mp = \ln p - \frac{p(\ln q - \ln p)}{q - p} = \frac{(q - p)\ln p - p(\ln q - \ln p)}{q - p} = \frac{q \ln p - p \ln q}{q - p}$

This gives $c = \frac{\ln(p^q / q^p)}{q - p}$.

Since $c > 0$ and $q - p > 0$, we need $\ln(p^q / q^p) > 0$, hence $p^q / q^p > 1$, giving $p^q > q^p$. $\square$

Part (iii)

The gradient of the chord from $(p, \ln p)$ to $(q, \ln q)$ is $m = \frac{\ln q - \ln p}{q - p}$, and the $y$-intercept is:

$c = \ln p - p \cdot \frac{\ln q - \ln p}{q - p} = \frac{q \ln p - p \ln q}{q - p} = \frac{\ln(p^q / q^p)}{q - p}$

Since $p \geqslant e$ and $q > p$, the function $g(x) = \frac{\ln x}{x}$ is strictly decreasing for $x \geqslant e$ (as $g'(x) = \frac{1 - \ln x}{x^2} < 0$ for $x > e$). Therefore $g(p) > g(q)$, i.e., $\frac{\ln p}{p} > \frac{\ln q}{q}$, which gives $q \ln p > p \ln q$, so $p^q > q^p$ and $\ln(p^q/q^p) > 0$.

Since $q - p > 0$, the $y$-intercept is positive. $\square$

Comparing $\pi^e$ and $e^\pi$: Taking $p = e$ and $q = \pi$ (since $e \geqslant e$ and $e < \pi$), the result gives $e^\pi > \pi^e$.

Therefore $e^\pi$ is greater.

Part (iv)

The condition $\frac{\ln q - \ln p}{q - p} = \frac{1}{e}$ means the chord of $y = \ln x$ from $p$ to $q$ has the same gradient as the tangent at $x = e$.

Define $h(x) = \ln x - \frac{x}{e}$. Then $h'(x) = \frac{1}{x} - \frac{1}{e}$, so $h$ has a maximum at $x = e$ with $h(e) = 1 - 1 = 0$.

From the gradient condition: $\ln q - \ln p = \frac{q - p}{e}$, rearranging to $\ln q - \frac{q}{e} = \ln p - \frac{p}{e}$, i.e., $h(p) = h(q)$.

Since $h$ increases on $(0, e)$ and decreases on $(e, \infty)$ with maximum $h(e) = 0$, and $h(p) = h(q)$ with $p < q$, we must have $p < e < q$ (they cannot both lie on the same monotone interval).

To show $q^p > p^q$, i.e., $\frac{\ln p}{p} < \frac{\ln q}{q}$:

From $h(p) = h(q)$: $\ln q = \ln p + \frac{q - p}{e}$. Then:

$\frac{\ln p}{p} < \frac{\ln q}{q} \iff q \ln p < p \ln q = p \ln p + \frac{p(q - p)}{e} \iff (q - p)\ln p < \frac{p(q - p)}{e}$

Since $q > p$: this reduces to $\ln p < \frac{p}{e}$, i.e., $h(p) < 0$. Since $p < e$ and $h$ is increasing on $(0, e)$ with $h(e) = 0$, we have $h(p) < 0$. $\square$

Examiner Notes

This was the most popular of all of the questions. Overall part (i) of this question was well answered, although there were a number of candidates who were not able to find the tangent and intercept even in this first case. Very few attempts at part (ii) of this question involved the use of sketches. While many attempts at part (iii) recognised the link in the final part with part (ii) of the question, many of the explanations in this section were not well enough explained to gain full marks. In the final part it was pleasing to note that many candidates realised that the conditions implied that the intersection with the y-axis was at a negative value.

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Question 2

Topic: 积分与递推公式 Integration and Reduction Formulae  |  Difficulty: Challenging  |  Marks: 20

Problem

2 For $n \geqslant 0$, let $I_n = \int_0^1 x^n(1 - x)^n dx .$

(i) For $n \geqslant 1$, show by means of a substitution that $\int_0^1 x^{n-1}(1 - x)^n dx = \int_0^1 x^n(1 - x)^{n-1} dx$ and deduce that $2 \int_0^1 x^{n-1}(1 - x)^n dx = I_{n-1} .$ Show also, for $n \geqslant 1$, that $I_n = \frac{n}{n + 1} \int_0^1 x^{n-1}(1 - x)^{n+1} dx$ and hence that $I_n = \frac{n}{2(2n + 1)} I_{n-1}$.

(ii) When $n$ is a positive integer, show that $I_n = \frac{(n!)^2}{(2n + 1)!} .$

(iii) Use the substitution $x = \sin^2 \theta$ to show that $I_{\frac{1}{2}} = \frac{\pi}{8}$, and evaluate $I_{\frac{3}{2}}$.

Hint

2. This is then easy to present in the form of a proof by induction for part (ii).

The result of part (i) shows that the sequence in part (iii) is increasing and the result proved in part (ii) shows that it is bounded above. The theorem provided at the start of the question therefore shows that the sequence converges. Similarly the second sequence is bounded below and decreasing (and therefore if the terms are all multiplied by -1 a sequence will be generated which is bounded above and increasing). Therefore the second sequence also converges to a limit.

The relationship between $u_n$ and $u_{n-2}$ established in part (i) can then be used to find the value of this limit and, as it is the same for both the odd terms and the even terms, the sequence must tend to the same limit as well.

Finally, starting the sequence at 3 will still lead to the same conclusion as the next term will be between 1 and 2 and all further terms will also be within that range, so all of the arguments will still hold for this new sequence.

Model Solution

Part (i)

First result: Use the substitution $x = 1 - t$, so $dx = -dt$. When $x = 0$, $t = 1$; when $x = 1$, $t = 0$.

$\int_0^1 x^{n-1}(1 - x)^n \, dx = \int_1^0 (1 - t)^{n-1} t^n (-dt) = \int_0^1 t^n(1 - t)^{n-1} \, dt = \int_0^1 x^n(1 - x)^{n-1} \, dx \qquad \square$

Deduction: Adding the two equal integrals:

$2 \int_0^1 x^{n-1}(1 - x)^n \, dx = \int_0^1 x^{n-1}(1 - x)^n \, dx + \int_0^1 x^n(1 - x)^{n-1} \, dx$

$= \int_0^1 x^{n-1}(1 - x)^{n-1}\big[(1 - x) + x\big] \, dx = \int_0^1 x^{n-1}(1 - x)^{n-1} \, dx = I_{n-1} \qquad \square$

Second result: Integrate $I_n$ by parts with $u = x^n$ and $dv = (1 - x)^n \, dx$, so $du = nx^{n-1} dx$ and $v = -\frac{(1 - x)^{n+1}}{n + 1}$:

$I_n = \left[-\frac{x^n(1 - x)^{n+1}}{n + 1}\right]_0^1 + \frac{n}{n + 1}\int_0^1 x^{n-1}(1 - x)^{n+1} \, dx$

The boundary term vanishes (at $x = 0$, $x^n = 0$; at $x = 1$, $(1-x)^{n+1} = 0$), giving:

$I_n = \frac{n}{n + 1} \int_0^1 x^{n-1}(1 - x)^{n+1} \, dx \qquad \text{(A)}$

Deriving the reduction formula: Write $(1 - x)^{n+1} = (1 - x)^n (1 - x) = (1 - x)^n - x(1 - x)^n$:

$\int_0^1 x^{n-1}(1 - x)^{n+1} \, dx = \int_0^1 x^{n-1}(1 - x)^n \, dx - \int_0^1 x^n(1 - x)^n \, dx = \frac{I_{n-1}}{2} - I_n$

Substituting into (A):

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

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

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

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

Part (ii)

By induction on $n$.

Base case: $I_0 = \int_0^1 1 \, dx = 1 = \frac{(0!)^2}{1!}$. $\checkmark$

Inductive step: Assume $I_{n-1} = \frac{((n - 1)!)^2}{(2n - 1)!}$. Then:

$I_n = \frac{n}{2(2n + 1)} \cdot \frac{((n - 1)!)^2}{(2n - 1)!} = \frac{n \cdot ((n - 1)!)^2}{2(2n + 1)(2n - 1)!}$

Since $(n!)^2 = n^2 \cdot ((n - 1)!)^2$ and $(2n + 1)! = (2n + 1)(2n)(2n - 1)!$:

$\frac{(n!)^2}{(2n + 1)!} = \frac{n^2 \cdot ((n - 1)!)^2}{(2n + 1)(2n)(2n - 1)!} = \frac{n \cdot ((n - 1)!)^2}{2(2n + 1)(2n - 1)!}$

This matches, completing the induction. $\square$

Part (iii)

Evaluating $I_{1/2}$: With $x = \sin^2\theta$, $dx = 2\sin\theta\cos\theta \, d\theta$. When $x = 0$, $\theta = 0$; when $x = 1$, $\theta = \pi/2$.

$I_{1/2} = \int_0^1 x^{1/2}(1 - x)^{1/2} \, dx = \int_0^{\pi/2} \sin\theta \cdot \cos\theta \cdot 2\sin\theta\cos\theta \, d\theta = 2\int_0^{\pi/2} \sin^2\theta\cos^2\theta \, d\theta$

Using $\sin^2\theta\cos^2\theta = \frac{1}{4}\sin^2 2\theta = \frac{1}{8}(1 - \cos 4\theta)$:

$I_{1/2} = \frac{1}{4}\int_0^{\pi/2}(1 - \cos 4\theta) \, d\theta = \frac{1}{4}\left[\theta - \frac{\sin 4\theta}{4}\right]_0^{\pi/2} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \qquad \square$

Evaluating $I_{3/2}$: Using the reduction formula $I_n = \frac{n}{2(2n + 1)} I_{n-1}$ with $n = 3/2$:

$I_{3/2} = \frac{3/2}{2(3 + 1)} I_{1/2} = \frac{3}{16} \cdot \frac{\pi}{8} = \frac{3\pi}{128}$

Examiner Notes

This was the second most popular question on the paper and the average score was half of the marks. Despite the instruction in the first part of the question to use a substitution a significant number of candidates chose to use integration by parts to establish the result. There were some sign errors in the integrations, but most candidates managed to reach the final result in the first part of the question. The second part of the question was found to be the hardest, with induction the most popular method, although the process was often not fully explained. The final part of the question did not appear to be too problematic for those that reached it. However, algebraic mistakes, such as factors disappearing, resulted in some marks being lost. Similarly, mistakes in the arithmetic in the final part of the question were not uncommon.

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Question 3

Topic: 三次方程与根的性质 Cubic Equations and Root Analysis  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Given that the cubic equation $x^3 + 3ax^2 + 3bx + c = 0$ has three distinct real roots and $c < 0$, show with the help of sketches that either exactly one of the roots is positive or all three of the roots are positive.

(ii) Given that the equation $x^3 + 3ax^2 + 3bx + c = 0$ has three distinct real positive roots show that $a^2 > b > 0, \quad a < 0, \quad c < 0 . \hfill (*)$

[Hint: Consider the turning points.]

(iii) Given that the equation $x^3 + 3ax^2 + 3bx + c = 0$ has three distinct real roots and that $ab < 0, \quad c > 0 ,$ determine, with the help of sketches, the signs of the roots.

(iv) Show by means of an explicit example (giving values for $a$, $b$ and $c$) that it is possible for the conditions $(*)$ to be satisfied even though the corresponding cubic equation has only one real root.

Hint

Question 3.

For it to be possible for the cubic to have three real roots it must have two stationary points. Since the coefficient of $x^3$ is positive it must have a specific shape. A sketch will show that only the two cases given will result in an intercept with the y-axis at a negative value.

In order for the cubic in part (ii) to have three positive roots, both of the turning points must be at positive values of $x$. Differentiation will allow most of the results to be established. The condition that $c < 0$ is needed to ensure that the leftmost root is also positive.

The condition $ab < 0$ implies that there must be a turning point at a positive value of $x$. The shape of the graph is as in part (i), but this time the intersection with the y-axis is at a positive value. This is sufficient to deduce the signs of the roots.

For part (iv) it is easiest to note that changing the value of $c$ does not (as long as $c$ remains negative) change whether or not the conditions of () are met. As this represents a vertical translation of the graph any example of a case satisfying () can be used to create an answer for this part by translating the graph sufficiently far downwards.

Model Solution

Part (i)

Let $f(x) = x^3 + 3ax^2 + 3bx + c$. Since the leading coefficient is positive, $f(x) \to -\infty$ as $x \to -\infty$ and $f(x) \to +\infty$ as $x \to +\infty$.

With three distinct real roots $r_1 < r_2 < r_3$, the sign pattern is:

$f(x) \begin{cases} < 0 & \text{for } x < r_1 \\ > 0 & \text{for } r_1 < x < r_2 \\ < 0 & \text{for } r_2 < x < r_3 \\ > 0 & \text{for } x > r_3 \end{cases}$

Since $f(0) = c < 0$, the value $x = 0$ must lie in a region where $f < 0$, so either $0 \in (-\infty, r_1)$ or $0 \in (r_2, r_3)$.

• If $0 \in (-\infty, r_1)$: then $0 < r_1 < r_2 < r_3$, so all three roots are positive.
• If $0 \in (r_2, r_3)$: then $r_1 < r_2 < 0 < r_3$, so exactly one root is positive.

Sketch: A cubic with positive leading coefficient has a local max then local min. With $f(0) < 0$, either the graph crosses three times for $x > 0$, or crosses twice for $x < 0$ and once for $x > 0$.

Part (ii)

Let the roots be $r_1, r_2, r_3 > 0$. By Vieta’s formulae:

$r_1 + r_2 + r_3 = -3a, \quad r_1r_2 + r_1r_3 + r_2r_3 = 3b, \quad r_1r_2r_3 = -c$

$a < 0$: $r_1 + r_2 + r_3 > 0$ gives $-3a > 0$, so $a < 0$.

$b > 0$: $r_1r_2 + r_1r_3 + r_2r_3 > 0$ gives $3b > 0$, so $b > 0$.

$c < 0$: $r_1r_2r_3 > 0$ gives $-c > 0$, so $c < 0$.

$a^2 > b$:

$a^2 > b \iff \frac{(r_1 + r_2 + r_3)^2}{9} > \frac{r_1r_2 + r_1r_3 + r_2r_3}{3}$

$\iff (r_1 + r_2 + r_3)^2 > 3(r_1r_2 + r_1r_3 + r_2r_3)$

$\iff r_1^2 + r_2^2 + r_3^2 > r_1r_2 + r_1r_3 + r_2r_3$

$\iff (r_1 - r_2)^2 + (r_1 - r_3)^2 + (r_2 - r_3)^2 > 0$

This holds since the roots are distinct. $\square$

Part (iii)

We have three distinct real roots, $c > 0$ and $ab < 0$.

Since $f(0) = c > 0$, and using the sign pattern of a cubic with positive leading coefficient, $x = 0$ must lie in a region where $f > 0$: either $0 \in (r_1, r_2)$ or $0 > r_3$.

Case 1: $a < 0, b > 0$ ($ab < 0$). Turning points of $f'(x) = 3(x^2 + 2ax + b)$ at $x = -a \pm \sqrt{a^2 - b}$. Since $a < 0$, $-a > 0$, and $\sqrt{a^2 - b} < |a| = -a$ (as $a^2 > b > 0$ for three roots), both turning points are at $x > 0$.

The function is increasing on $(-\infty, \alpha)$. Since $f(-\infty) = -\infty$ and $f(0) = c > 0$, there is one root at $x < 0$. The other two roots (where $f$ crosses the $x$-axis after the local max and before going to $+\infty$) are both positive.

Case 2: $a > 0, b < 0$ ($ab < 0$). Since $b < 0$, $a^2 - b > a^2$, so $\sqrt{a^2 - b} > a$. The turning points are at $\alpha = -a - \sqrt{a^2 - b} < 0$ and $\beta = -a + \sqrt{a^2 - b} > 0$.

With $f(0) = c > 0$, $f$ is decreasing on $(\alpha, \beta)$ with $\alpha < 0 < \beta$. Since the local min $f(\beta) < 0$ (for three real roots), there is one root in $(0, \beta)$ and one in $(\beta, +\infty)$ (both positive), plus one root in $(-\infty, \alpha)$ (negative).

In both cases: one negative root and two positive roots.

Part (iv)

Take $a = -1$, $b = \frac{1}{2}$, $c = -10$.

Check conditions (*):

• $a^2 = 1 > \frac{1}{2} = b > 0$ $\checkmark$
• $a = -1 < 0$ $\checkmark$
• $c = -10 < 0$ $\checkmark$

The cubic is $f(x) = x^3 - 3x^2 + \frac{3}{2}x - 10$.

$f'(x) = 3x^2 - 6x + \frac{3}{2} = 3(x^2 - 2x + \frac{1}{2})$. Discriminant: $4 - 2 = 2 > 0$, so turning points exist at $x = 1 \pm \frac{1}{\sqrt{2}}$.

The local maximum at $x = 1 - \frac{1}{\sqrt{2}} \approx 0.293$:

$f(0.293) \approx 0.025 - 0.258 + 0.440 - 10 = -9.79 < 0$

Since the local maximum is below the $x$-axis, the curve crosses the $x$-axis only once (for large positive $x$). The equation has only one real root, yet all conditions $(*)$ are satisfied. $\square$

Examiner Notes

This question was again popular and had an average score of about half of the marks. In the first part almost all candidates were able to sketch the correct shape of graph, but some did not provide suitable explanations to accompany these or included additional cases that were not asked for. A number of candidates attempting the second part of the question reached one of the results by squaring an inequality without considering the signs and many assumed that the result of part (i) implied that c must be negative. Only about half of the candidates attempted part (iii), and many of those who did did not use sketches in their solutions. Solutions to part (iv) generally involved guessing of the values of a, b and c followed by a check that the conditions were met.

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Question 4

Topic: 坐标几何与轨迹 Coordinate Geometry and Loci  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The line passing through the point $(a, 0)$ with gradient $b$ intersects the circle of unit radius centred at the origin at $P$ and $Q$, and $M$ is the midpoint of the chord $PQ$. Find the coordinates of $M$ in terms of $a$ and $b$.

(i) Suppose $b$ is fixed and positive. As $a$ varies, $M$ traces out a curve (the locus of $M$). Show that $x = -by$ on this curve. Given that $a$ varies with $-1 \leqslant a \leqslant 1$, show that the locus is a line segment of length $2b/(1 + b^2)^{\frac{1}{2}}$. Give a sketch showing the locus and the unit circle.

(ii) Find the locus of $M$ in the following cases, giving in each case its cartesian equation, describing it geometrically and sketching it in relation to the unit circle:

(a) $a$ is fixed with $0 < a < 1$, and $b$ varies with $-\infty < b < \infty$; (b) $ab = 1$, and $b$ varies with $0 < b \leqslant 1$.

Hint

Question 4.

The equations of the line and circle are easily found and so the second point of intersection (and so the coordinates of M) can be easily found. The two parts of this question then involve regarding the coordinates of M as parametric equations.

In part (i) $a$ is the parameter and is restricted so that the point that the line passes through is inside the circle. This gives a straight line between the points which result from the cases $a = -1$ and $a = 1$. The length of this line can be determined easily from the coordinates of its endpoints.

In part (ii) it is again quite easy to eliminate the parameter from the pair of equations and the shapes of the loci should be easily recognised. In part (b) however, the restriction on the values of $b$ need to be considered as the locus is not the whole shape that would be identified from the equation.

Model Solution

Finding the coordinates of $M$:

The line through $(a, 0)$ with gradient $b$: $y = b(x - a)$.

Substituting into $x^2 + y^2 = 1$:

$x^2 + b^2(x - a)^2 = 1$

$(1 + b^2)x^2 - 2ab^2 x + (a^2b^2 - 1) = 0$

The $x$-coordinate of the midpoint $M$ is the average of the roots:

$x_M = \frac{2ab^2}{2(1 + b^2)} = \frac{ab^2}{1 + b^2}$

$y_M = b(x_M - a) = b\left(\frac{ab^2}{1 + b^2} - a\right) = \frac{b(ab^2 - a - ab^2)}{1 + b^2} = \frac{-ab}{1 + b^2}$

So $M = \left(\dfrac{ab^2}{1 + b^2},\; \dfrac{-ab}{1 + b^2}\right)$.

Part (i)

With $b$ fixed and positive, dividing the coordinates:

$\frac{x_M}{y_M} = \frac{ab^2}{-ab} = -b$

Therefore $x = -by$ on the locus. $\square$

For $-1 \leqslant a \leqslant$: from $y = \frac{-ab}{1 + b^2}$ we get $a = \frac{-(1 + b^2)y}{b}$, so the constraint gives:

$-1 \leqslant \frac{-(1 + b^2)y}{b} \leqslant 1 \implies \frac{-b}{1 + b^2} \leqslant y \leqslant \frac{b}{1 + b^2}$

At $a = -1$: $y = \frac{b}{1 + b^2}$, $x = \frac{-b^2}{1 + b^2}$. At $a = 1$: $y = \frac{-b}{1 + b^2}$, $x = \frac{b^2}{1 + b^2}$.

The length of the segment:

$\sqrt{\left(\frac{2b^2}{1 + b^2}\right)^2 + \left(\frac{2b}{1 + b^2}\right)^2} = \frac{2}{1 + b^2}\sqrt{b^4 + b^2} = \frac{2b\sqrt{1 + b^2}}{1 + b^2} = \frac{2b}{\sqrt{1 + b^2}} \qquad \square$

Sketch: The locus is a line segment through the origin with gradient $-1/b$, lying inside the unit circle, connecting the two points corresponding to $a = \pm 1$.

Part (ii)(a)

$a$ fixed, $0 < a < 1$, $b \in (-\infty, \infty)$.

From $\frac{y_M}{x_M} = \frac{-1}{b}$, we get $b = \frac{-x}{y}$ (for $y \neq 0$). Substituting into $x = \frac{ab^2}{1 + b^2}$:

$x = \frac{a \cdot x^2/y^2}{1 + x^2/y^2} = \frac{ax^2}{x^2 + y^2}$

For $x \neq 0$: $x^2 + y^2 = ax$, i.e.:

$\left(x - \frac{a}{2}\right)^2 + y^2 = \left(\frac{a}{2}\right)^2$

This is a circle with centre $\left(\frac{a}{2}, 0\right)$ and radius $\frac{a}{2}$.

Since $x = \frac{ab^2}{1 + b^2}$: $x = 0$ when $b = 0$ (giving $M = (0,0)$), and $x \to a$ as $|b| \to \infty$. For $b > 0$: $y < 0$; for $b < 0$: $y > 0$. The locus is the circle $(x - a/2)^2 + y^2 = (a/2)^2$ with diameter from the origin to $(a, 0)$, excluding the point $(a, 0)$. This circle lies entirely inside the unit circle since $0 < a < 1$.

Part (ii)(b)

$ab = 1$, $0 < b \leqslant 1$. Then $a = 1/b \geqslant 1$.

$x = \frac{(1/b) \cdot b^2}{1 + b^2} = \frac{b}{1 + b^2}, \qquad y = \frac{-(1/b) \cdot b}{1 + b^2} = \frac{-1}{1 + b^2}$

From $\frac{x}{y} = -b$, so $b = -\frac{x}{y}$. Substituting into $y = \frac{-1}{1 + b^2}$:

$y = \frac{-1}{1 + x^2/y^2} = \frac{-y^2}{x^2 + y^2}$

For $y \neq 0$: $x^2 + y^2 = -y$, i.e.:

$x^2 + \left(y + \frac{1}{2}\right)^2 = \frac{1}{4}$

This is a circle with centre $(0, -1/2)$ and radius $1/2$.

For $b \in (0, 1]$: $x = \frac{b}{1 + b^2} > 0$ and $y = \frac{-1}{1 + b^2} \in [-1, -1/2]$.

At $b = 1$: $(x, y) = (1/2, -1/2)$. As $b \to 0^+$: $(x, y) \to (0, -1)$.

The locus is the arc of the circle $x^2 + (y + 1/2)^2 = 1/4$ from $(0, -1)$ to $(1/2, -1/2)$, with $x > 0$. This arc lies inside the unit circle (the small circle, centred at $(0, -1/2)$ with radius $1/2$, touches the origin and $(0, -1)$, both on or inside the unit circle).

Examiner Notes

This question received a relatively small number of attempts compared to the other Pure Mathematics questions. On average candidates who attempted this question only received a quarter of the marks available. Some candidates did not manage to write down the correct equation of the line or did not appreciate that the phrase “unit radius” means that the radius is 1. Many candidates produced loci for the second part of the question without any indication of a method. In the final part of the question the significance of the restrictions on the value of b were not appreciated by many of the candidates.

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Question 5

Topic: 函数对称性与驻点 Function Symmetry and Stationary Points  |  Difficulty: Hard  |  Marks: 20

Problem

5 (i) A function $f(x)$ satisfies $f(x) = f(1 - x)$ for all $x$. Show, by differentiating with respect to $x$, that $f'(\frac{1}{2}) = 0$. If, in addition, $f(x) = f(\frac{1}{x})$ for all (non-zero) $x$, show that $f'(-1) = 0$ and that $f'(2) = 0$.

(ii) The function $f$ is defined, for $x \neq 0$ and $x \neq 1$, by

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

Show that $f(x) = f(\frac{1}{x})$ and $f(x) = f(1 - x)$.

Given that it has exactly three stationary points, sketch the curve $y = f(x)$.

(iii) Hence, or otherwise, find all the roots of the equation $f(x) = \frac{27}{4}$ and state the ranges

of values of $x$ for which $f(x) > \frac{27}{4}$.

Find also all the roots of the equation $f(x) = \frac{343}{36}$ and state the ranges of values of $x$ for which $f(x) > \frac{343}{36}$.

Hint

Question 5.

Simple applications of the chain rule lead to relationships that will allow the three cases of zero gradients to be identified in part (i).

In part (ii) the relationships follow easily from substitution and therefore the three stationary points identified in part (i) must all exist. By considering the denominator there are clearly two vertical asymptotes and the numerator is clearly always positive. Additionally, the numerator is much larger than the denominator for large values of $x$. Given this information there is only one possible shape for the graph.

In part (iii) the solutions of the first equation will already have been discovered when the coordinates of the stationary points in part (ii) were calculated. The range of values satisfying the first inequality should therefore be straightforward. One of the solutions of the second equation should be easy to spot, and consideration of the graph shows that there must be a total of six roots. Applying the two relationships about the values of $f$ will allow these other roots to be found. The solution set for the inequality then follows easily from consideration of the graph.

Model Solution

Part (i)

$f'(1/2) = 0$: Differentiating $f(x) = f(1 - x)$ with respect to $x$:

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

Setting $x = 1/2$: $f'(1/2) = -f'(1/2)$, so $2f'(1/2) = 0$, giving $f'(1/2) = 0$. $\square$

$f'(-1) = 0$: Differentiating $f(x) = f(1/x)$ with respect to $x$:

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

Setting $x = -1$: $f'(-1) = -f'(-1)$, so $2f'(-1) = 0$, giving $f'(-1) = 0$. $\square$

$f'(2) = 0$: Setting $x = 2$ in the same relation:

$f'(2) = -\frac{1}{4} f'\left(\frac{1}{2}\right) = -\frac{1}{4} \cdot 0 = 0 \qquad \square$

Part (ii)

$f(x) = f(1/x)$: Substituting $1/x$ into $f$:

$f\left(\frac{1}{x}\right) = \frac{\left(\frac{1}{x^2} - \frac{1}{x} + 1\right)^3}{\left(\frac{1}{x^2} - \frac{1}{x}\right)^2} = \frac{\left(\frac{1 - x + x^2}{x^2}\right)^3}{\left(\frac{1 - x}{x^2}\right)^2} = \frac{(x^2 - x + 1)^3 / x^6}{(1 - x)^2 / x^4} = \frac{(x^2 - x + 1)^3}{x^2(1 - x)^2}$

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

So $f(1/x) = f(x)$. $\square$

$f(x) = f(1 - x)$: Substituting $1 - x$:

$f(1 - x) = \frac{((1 - x)^2 - (1 - x) + 1)^3}{((1 - x)^2 - (1 - x))^2} = \frac{(1 - 2x + x^2 - 1 + x + 1)^3}{(1 - 2x + x^2 - 1 + x)^2} = \frac{(x^2 - x + 1)^3}{(x^2 - x)^2}$

So $f(1 - x) = f(x)$. $\square$

Sketch: From part (i), the three stationary points are $x = -1$, $x = 1/2$, and $x = 2$. Computing their values:

$f(-1) = \frac{(1 + 1 + 1)^3}{(1 + 1)^2} = \frac{27}{4}, \qquad f\left(\frac{1}{2}\right) = \frac{(3/4)^3}{(1/4)^2} = \frac{27}{4}, \qquad f(2) = \frac{(4 - 2 + 1)^3}{(4 - 2)^2} = \frac{27}{4}$

All three stationary points have $y = 27/4$.

Key features:

• Vertical asymptotes at $x = 0$ and $x = 1$.
• The numerator $(x^2 - x + 1)^3 = ((x - 1/2)^2 + 3/4)^3 > 0$ always.
• As $x \to \pm\infty$: $f(x) \sim x^2 \to +\infty$.
• Near $x = 0$ and $x = 1$: $f(x) \to +\infty$.

Each of the three intervals $(-\infty, 0)$, $(0, 1)$, $(1, +\infty)$ contains one stationary point (a minimum with value $27/4$), with the function rising to $+\infty$ at both ends.

Part (iii)

Roots of $f(x) = 27/4$:

From the analysis above, $f$ achieves the value $27/4$ exactly at its three stationary points:

$x = -1, \qquad x = \frac{1}{2}, \qquad x = 2$

Ranges where $f(x) > 27/4$:

Since $27/4$ is the minimum value of $f$ on each interval, and $f$ only equals $27/4$ at the stationary points:

$x \in (-\infty, -1) \cup (-1, 0) \cup \left(0, \frac{1}{2}\right) \cup \left(\frac{1}{2}, 1\right) \cup (1, 2) \cup (2, +\infty)$

i.e., all $x \neq 0, 1$ except $x = -1, 1/2, 2$.

Roots of $f(x) = 343/36$:

Testing $x = 3$: $f(3) = \frac{(9 - 3 + 1)^3}{(9 - 3)^2} = \frac{7^3}{6^2} = \frac{343}{36}$. $\checkmark$

Using the symmetries from part (i) to find the remaining roots:

• $f(x) = f(1/x)$: $f(1/3) = f(3) = 343/36$
• $f(x) = f(1 - x)$: $f(-2) = f(1 - (-2)) = f(3) = 343/36$
• $f(1/x)$ on $x = -2$: $f(-1/2) = f(-2) = 343/36$
• $f(1 - x)$ on $x = 1/3$: $f(2/3) = f(1/3) = 343/36$
• $f(1 - x)$ on $x = -1/2$: $f(3/2) = f(-1/2) = 343/36$

The six roots are:

$x = -2, \quad x = -\frac{1}{2}, \quad x = \frac{1}{3}, \quad x = \frac{2}{3}, \quad x = \frac{3}{2}, \quad x = 3$

(Each of the three intervals contains two roots, consistent with $343/36 > 27/4$.)

Ranges where $f(x) > 343/36$:

Since $f$ dips below $343/36$ between each pair of roots and rises above it outside:

$x \in (-\infty, -2) \cup \left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{3}\right) \cup \left(\frac{2}{3}, 1\right) \cup \left(1, \frac{3}{2}\right) \cup (3, +\infty)$

Examiner Notes

This was one of the more successfully attempted questions on the paper and the Pure Mathematics question with the highest average mark. While some candidates struggled with the application of the chain rule throughout this question, many were able to complete the first part of the question without much difficulty. Showing that f satisfied the required conditions in part (i) was generally well done, but the sketching of the graph was found to be more difficult, with a number of

candidates not identifying the asymptotes and some thinking that part of the graph would drop below the $x$-axis. Most of the candidates who attempted part (iii) found the roots of the equation successfully, but a large number forgot to exclude the roots when solving the inequality. In the final part, many identified $x=3$ as a solution, but those who split the fraction into two equations (one for the numerator equalling 343 and one for the denominator equalling 36) did not check that the solution worked for both parts. Those who used the symmetries established in part (i) were then able to identify the other roots easily, while those who attempted algebraic solutions for the other roots were generally not successful.

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Question 6

Topic: 数列与极限 Sequences and Limits  |  Difficulty: Challenging  |  Marks: 20

Problem

6 In this question, the following theorem may be used.

Let $u_1, u_2, \dots$ be a sequence of (real) numbers. If the sequence is bounded above (that is, $u_n \leqslant b$ for all $n$, where $b$ is some fixed number) and increasing (that is, $u_n \geqslant u_{n-1}$ for all $n$), then the sequence tends to a limit (that is, converges).

The sequence $u_1, u_2, \dots$ is defined by $u_1 = 1$ and

$u_{n+1} = 1 + \frac{1}{u_n} \quad (n \geqslant 1) . \qquad \text{(*)}$

(i) Show that, for $n \geqslant 3$,

$u_{n+2} - u_n = \frac{u_n - u_{n-2}}{(1 + u_n)(1 + u_{n-2})} .$

(ii) Prove, by induction or otherwise, that $1 \leqslant u_n \leqslant 2$ for all $n$.

(iii) Show that the sequence $u_1, u_3, u_5, \dots$ tends to a limit, and that the sequence $u_2, u_4, u_6, \dots$ tends to a limit. Find these limits and deduce that the sequence $u_1, u_2, u_3, \dots$ tends to a limit.

Would this conclusion change if the sequence were defined by (*) and $u_1 = 3$?

Hint

Question 6.

The definition of the sequence can be used to find a relationship between $u_{n+2}$ and $u_n$ and therefore also a relationship between $u_n$ and $u_{n-2}$. Taking the difference of these then leads to the required result.

It is clear from the definition of the sequence that, if one term is between 1 and 2, then the next term will also be between 1 and

Model Solution

Part (i)

From the recurrence $u_{n+1} = 1 + \frac{1}{u_n}$:

$u_{n+2} = 1 + \frac{1}{u_{n+1}} = 1 + \frac{1}{1 + 1/u_n} = 1 + \frac{u_n}{u_n + 1} = \frac{2u_n + 1}{u_n + 1}$

Applying the same formula two steps back:

$u_n = \frac{2u_{n-2} + 1}{u_{n-2} + 1}$

Therefore:

$u_{n+2} - u_n = \frac{2u_n + 1}{u_n + 1} - \frac{2u_{n-2} + 1}{u_{n-2} + 1}$

$= \frac{(2u_n + 1)(u_{n-2} + 1) - (2u_{n-2} + 1)(u_n + 1)}{(u_n + 1)(u_{n-2} + 1)}$

Expanding the numerator:

$(2u_n u_{n-2} + 2u_n + u_{n-2} + 1) - (2u_n u_{n-2} + u_n + 2u_{n-2} + 1) = u_n - u_{n-2}$

So:

$u_{n+2} - u_n = \frac{u_n - u_{n-2}}{(1 + u_n)(1 + u_{n-2})} \qquad \square$

Part (ii)

By strong induction.

Base cases: $u_1 = 1 \in [1, 2]$ $\checkmark$. $u_2 = 1 + 1/1 = 2 \in [1, 2]$ $\checkmark$.

Inductive step: Assume $1 \leqslant u_k \leqslant 2$ for all $k \leqslant n$. Then:

• $u_n \geqslant 1 \implies \frac{1}{u_n} \leqslant 1 \implies u_{n+1} = 1 + \frac{1}{u_n} \leqslant 2$.
• $u_n \leqslant 2 \implies \frac{1}{u_n} \geqslant \frac{1}{2} \implies u_{n+1} = 1 + \frac{1}{u_n} \geqslant \frac{3}{2} \geqslant 1$.

So $1 \leqslant u_{n+1} \leqslant 2$. By induction, $1 \leqslant u_n \leqslant 2$ for all $n$. $\square$

Part (iii)

Convergence of the odd subsequence $u_1, u_3, u_5, \ldots$:

From part (i), $u_{n+2} - u_n$ has the same sign as $u_n - u_{n-2}$ (since the denominator $(1 + u_n)(1 + u_{n-2}) \geqslant 4 > 0$).

Computing: $u_1 = 1$ and $u_3 = 1 + 1/u_2 = 1 + 1/2 = 3/2$, so $u_3 - u_1 = 1/2 > 0$.

By the sign preservation, $u_5 - u_3 > 0$, $u_7 - u_5 > 0$, etc. The odd subsequence is increasing. By part (ii), it is bounded above by 2. By the given theorem, the odd subsequence converges to some limit $L$.

Convergence of the even subsequence $u_2, u_4, u_6, \ldots$:

$u_2 = 2$ and $u_4 = 1 + 1/u_3 = 1 + 2/3 = 5/3$, so $u_4 - u_2 = -1/3 < 0$.

By the same sign preservation, the even subsequence is decreasing. By part (ii), it is bounded below by 1. The sequence $-u_2, -u_4, -u_6, \ldots$ is increasing and bounded above, so by the given theorem, the even subsequence converges to some limit $M$.

Finding the limits: From $u_{n+1} = 1 + 1/u_n$, taking limits along the odd and even subsequences:

$L = 1 + \frac{1}{M} \qquad \text{and} \qquad M = 1 + \frac{1}{L}$

Substituting the second into the first:

$L = 1 + \frac{1}{1 + 1/L} = 1 + \frac{L}{L + 1} = \frac{2L + 1}{L + 1}$

$L(L + 1) = 2L + 1 \implies L^2 - L - 1 = 0 \implies L = \frac{1 + \sqrt{5}}{2}$

(taking the positive root since $L \geqslant 1$). Then:

$M = 1 + \frac{1}{L} = 1 + \frac{2}{1 + \sqrt{5}} = 1 + \frac{\sqrt{5} - 1}{2} = \frac{1 + \sqrt{5}}{2} = L$

Both subsequences converge to the same limit $\phi = \frac{1 + \sqrt{5}}{2}$ (the golden ratio). Since the odd and even subsequences both converge to $\phi$, the full sequence $u_1, u_2, u_3, \ldots$ converges to $\phi$. $\square$

If $u_1 = 3$: Then $u_2 = 1 + 1/3 = 4/3$, which lies in $[1, 2]$. From $u_2$ onwards, the same induction argument shows all terms lie in $[1, 2]$. The analysis above applies to the sequence starting at $u_2$, so both subsequences converge to $\phi$.

Therefore the conclusion does not change: the sequence still converges to $\phi = \frac{1 + \sqrt{5}}{2}$.

Examiner Notes

The algebra required for the first part of the question proved to be quite challenging for a number of candidates, but most were able to reach the required answer. The proof by induction in the second part of the question was generally well done, although a number of candidates did not write up the process clearly. In the final part of the question it was clear that many candidates had identified the relationship between the sequences and Fibonacci numbers and some candidates therefore stated that the limit would be the golden ratio, but without any supporting calculations. In the final part there were few responses which clearly explained that the new sequence would still satisfy the conditions required if it were started at a later term.

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Question 7

Topic: 数论 Number Theory  |  Difficulty: Challenging  |  Marks: 20

Problem

7 (i) Write down a solution of the equation $x^2 - 2y^2 = 1 , \qquad \text{(*)}$ for which $x$ and $y$ are non-negative integers. Show that, if $x = p, y = q$ is a solution of $(*)$, then so also is $x = 3p + 4q, y = 2p + 3q$. Hence find two solutions of $(*)$ for which $x$ is a positive odd integer and $y$ is a positive even integer.

(ii) Show that, if $x$ is an odd integer and $y$ is an even integer, $(*)$ can be written in the form $n^2 = \frac{1}{2}m(m + 1) ,$ where $m$ and $n$ are integers.

(iii) The positive integers $a, b$ and $c$ satisfy $b^3 = c^4 - a^2 ,$ where $b$ is a prime number. Express $a$ and $c^2$ in terms of $b$ in the two cases that arise. Find a solution of $a^2 + b^3 = c^4$, where $a, b$ and $c$ are positive integers but $b$ is not prime.

Hint

Question 7.

A solution of the equation should be easy to spot and a simple substitution will establish the new solution that can be generated from an existing one. This therefore allows two further solutions to be found easily by repeated application of this result.

In part (ii) write $x = 2m + 1$ and $y = 2n$ and then substitute into (*). With some simplification the required relationship will be established.

Since $b$ is a prime number there is only two ways in which it can be split into a product of two numbers ($1 \times b^3$ and $b \times b^2$). The right hand side of the equation is clearly a difference of two squares and therefore a pair of simultaneous equations can be solved to give expressions for $a$ and $c^2$. Finally, the expression for $c^2$ is similar to the relationship established in part (ii), so solutions to the original equation can be used to generate values of $a$, $b$ and $c$ which satisfy this equation.

Model Solution

Part (i)

A solution with $x, y$ non-negative integers is $x = 1, y = 0$, since $1^2 - 2(0)^2 = 1$.

Now suppose $x = p, y = q$ satisfies $(*)$, so $p^2 - 2q^2 = 1$. We check that $x = 3p + 4q, y = 2p + 3q$ is also a solution:

$(3p + 4q)^2 - 2(2p + 3q)^2 = 9p^2 + 24pq + 16q^2 - 2(4p^2 + 12pq + 9q^2)$ $= 9p^2 + 24pq + 16q^2 - 8p^2 - 24pq - 18q^2 = p^2 - 2q^2 = 1 . \checkmark$

Starting from the trivial solution $(p, q) = (1, 0)$:

• First application: $x = 3(1) + 4(0) = 3$, $y = 2(1) + 3(0) = 2$. Check: $9 - 8 = 1$. $\checkmark$
• Second application: $x = 3(3) + 4(2) = 17$, $y = 2(3) + 3(2) = 12$. Check: $289 - 288 = 1$. $\checkmark$

The two solutions with $x$ a positive odd integer and $y$ a positive even integer are $(x, y) = (3, 2)$ and $(x, y) = (17, 12)$.

Part (ii)

Let $x = 2m + 1$ (odd) and $y = 2n$ (even), where $m, n$ are integers. Substituting into $(*)$:

$(2m+1)^2 - 2(2n)^2 = 1$ $4m^2 + 4m + 1 - 8n^2 = 1$ $4m^2 + 4m = 8n^2$ $4m(m + 1) = 8n^2$ $m(m + 1) = 2n^2$

Therefore $n^2 = \dfrac{1}{2}m(m+1)$, where $m$ and $n$ are integers. $\square$

Part (iii)

We have $b^3 = c^4 - a^2 = (c^2 - a)(c^2 + a)$, where $b$ is prime and $a, c$ are positive integers.

Since $b$ is prime, $b^3$ has exactly four positive divisors: $1, b, b^2, b^3$. Also $c^2 + a > c^2 - a > 0$ (since $a > 0$). So there are exactly two ways to factorise $b^3$ into two positive factors with the larger one second:

Case 1: $c^2 - a = 1$ and $c^2 + a = b^3$.

Adding: $2c^2 = b^3 + 1$, so $c^2 = \dfrac{b^3 + 1}{2}$. Subtracting: $2a = b^3 - 1$, so $a = \dfrac{b^3 - 1}{2}$.

Case 2: $c^2 - a = b$ and $c^2 + a = b^2$.

Adding: $2c^2 = b^2 + b = b(b + 1)$, so $c^2 = \dfrac{b(b + 1)}{2}$. Subtracting: $2a = b^2 - b = b(b - 1)$, so $a = \dfrac{b(b - 1)}{2}$.

Finding a solution where $b$ is not prime:

We need positive integers $a, b, c$ with $b$ not prime satisfying $a^2 + b^3 = c^4$.

From Case 2, $c^2 = \frac{b(b+1)}{2}$. If $b$ is a perfect square, say $b = k^2$, then $c^2 = \frac{k^2(k^2+1)}{2}$, which requires $\frac{k^2+1}{2}$ to be a perfect square.

Taking $k = 7$: $b = 49$, $\frac{49+1}{2} = 25 = 5^2$, so $c^2 = 49 \times 25 = 1225$ and $c = 35$.

Then $a = \frac{49 \times 48}{2} = 1176$.

Verification: $1176^2 + 49^3 = 1382976 + 117649 = 1500625 = 35^4$. $\checkmark$

Since $b = 49 = 7^2$ is not prime, the solution is $a = 1176, \; b = 49, \; c = 35$.

Examiner Notes

This question was attempted by a large number of candidates, only slightly fewer than question 2, and was one of the more successful ones with an average score above half of the marks. While some candidates proved the converse of the required result, part (i) of the question was generally done well, although a surprising number of candidates did not write down the numerical solutions when asked. Those students who realised the way to write $x$ and $y$ in terms of $m$ and $n$ reached the result of part (ii) easily, while others sometimes spent a lot of effort on this making little or no progress. In part (iii) many candidates spotted the difference of two squares, but some did not realise that there would be two ways to factorise $b^3$. Only very few students were able to solve the final part of the question.

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Question 8

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

Problem

8 The function $f$ satisfies $f(x) > 0$ for $x \geqslant 0$ and is strictly decreasing (which means that $f(b) < f(a)$ for $b > a$).

(i) For $t \geqslant 0$, let $A_0(t)$ be the area of the largest rectangle with sides parallel to the coordinate axes that can fit in the region bounded by the curve $y = f(x)$, the $y$-axis and the line $y = f(t)$. Show that $A_0(t)$ can be written in the form $A_0(t) = x_0 (f(x_0) - f(t)) ,$ where $x_0$ satisfies $x_0f'(x_0) + f(x_0) = f(t)$.

(ii) The function $g$ is defined, for $t > 0$, by $g(t) = \frac{1}{t} \int_0^t f(x)dx .$ Show that $tg'(t) = f(t) - g(t)$. Making use of a sketch show that, for $t > 0$, $\int_0^t (f(x) - f(t)) dx > A_0(t)$ and deduce that $-t^2g'(t) > A_0(t)$.

(iii) In the case $f(x) = \frac{1}{1 + x}$, use the above to establish the inequality $\ln \sqrt{1 + t} > 1 - \frac{1}{\sqrt{1 + t}} ,$ for $t > 0$.

Hint

Question 8.

Begin by calculating the largest area of a rectangle with a given width and then maximize this function as the width of the rectangle is varied. The definition of $x_0$ can be reached by setting the derivative of the area function to

Model Solution

Part (i)

The region is bounded by $y = f(x)$ (above), the $y$-axis (left), and the horizontal line $y = f(t)$ (below). Since $f$ is strictly decreasing with $f(x) > 0$, the curve lies above the line $y = f(t)$ for $0 \leqslant x < t$.

A rectangle inscribed in this region with its base on $y = f(t)$ and its top edge touching the curve at $x = x_0$ (where $0 < x_0 < t$) has:

• width $= x_0$
• height $= f(x_0) - f(t)$

So its area is $A(x_0) = x_0\bigl(f(x_0) - f(t)\bigr)$.

To maximise, differentiate with respect to $x_0$:

$\frac{dA}{dx_0} = \bigl(f(x_0) - f(t)\bigr) + x_0 f'(x_0) = x_0 f'(x_0) + f(x_0) - f(t) .$

Setting $\frac{dA}{dx_0} = 0$ at $x_0$ gives $x_0 f'(x_0) + f(x_0) = f(t)$, and the maximum area is

$A_0(t) = x_0\bigl(f(x_0) - f(t)\bigr) . \qquad \square$

Part (ii)

We have $g(t) = \frac{1}{t}\int_0^t f(x)\,dx$. Differentiating by the quotient rule:

$g'(t) = \frac{t \cdot f(t) - \int_0^t f(x)\,dx}{t^2} = \frac{f(t)}{t} - \frac{1}{t^2}\int_0^t f(x)\,dx .$

Multiplying by $t$:

$tg'(t) = f(t) - \frac{1}{t}\int_0^t f(x)\,dx = f(t) - g(t) . \qquad \square$

Geometric inequality. Consider the region $R$ between the curve $y = f(x)$ and the line $y = f(t)$, for $0 \leqslant x \leqslant t$. The inscribed rectangle with maximum area $A_0(t)$ lies entirely within $R$, and $R$ also contains regions above the rectangle (between the curve and the top edge of the rectangle). Since these additional regions have positive area:

$\int_0^t \bigl(f(x) - f(t)\bigr)\,dx > A_0(t) . \qquad \square$

Deducing $-t^2 g'(t) > A_0(t)$. From $tg'(t) = f(t) - g(t)$:

$-tg'(t) = g(t) - f(t) = \frac{1}{t}\int_0^t f(x)\,dx - f(t) = \frac{1}{t}\int_0^t \bigl(f(x) - f(t)\bigr)\,dx .$

Multiplying both sides by $t > 0$:

$-t^2 g'(t) = \int_0^t \bigl(f(x) - f(t)\bigr)\,dx > A_0(t) . \qquad \square$

Part (iii)

Let $f(x) = \frac{1}{1+x}$. This is positive and strictly decreasing for $x \geqslant 0$, so the results above apply.

Computing $-t^2 g'(t)$:

$g(t) = \frac{1}{t}\int_0^t \frac{1}{1+x}\,dx = \frac{\ln(1+t)}{t} .$

From part (ii), $-t^2 g'(t) = t\bigl(g(t) - f(t)\bigr)$:

$-t^2 g'(t) = t\left(\frac{\ln(1+t)}{t} - \frac{1}{1+t}\right) = \ln(1+t) - \frac{t}{1+t} .$

Since $\frac{t}{1+t} = 1 - \frac{1}{1+t}$:

$-t^2 g'(t) = \ln(1+t) - 1 + \frac{1}{1+t} .$

Computing $A_0(t)$:

The area function is $A(x) = x\left(\frac{1}{1+x} - \frac{1}{1+t}\right)$. Setting $A'(x) = 0$:

$\frac{1}{1+x} - \frac{1}{1+t} + x \cdot \frac{-1}{(1+x)^2} = 0$

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

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

So $(1+x)^2 = 1+t$, giving $x_0 = \sqrt{1+t} - 1$.

At $x_0$: $1 + x_0 = \sqrt{1+t}$, so $f(x_0) = \frac{1}{\sqrt{1+t}}$. Then:

$A_0(t) = (\sqrt{1+t}-1)\left(\frac{1}{\sqrt{1+t}} - \frac{1}{1+t}\right) = (\sqrt{1+t}-1) \cdot \frac{\sqrt{1+t}-1}{1+t} = \frac{(\sqrt{1+t}-1)^2}{1+t} .$

Expanding: $A_0(t) = \frac{1+t - 2\sqrt{1+t} + 1}{1+t} = 1 - \frac{2}{\sqrt{1+t}} + \frac{1}{1+t}$.

Applying the inequality $-t^2 g'(t) > A_0(t)$:

$\ln(1+t) - 1 + \frac{1}{1+t} > 1 - \frac{2}{\sqrt{1+t}} + \frac{1}{1+t}$

The $\frac{1}{1+t}$ terms cancel:

$\ln(1+t) - 1 > 1 - \frac{2}{\sqrt{1+t}}$

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

Dividing by 2:

$\frac{1}{2}\ln(1+t) > 1 - \frac{1}{\sqrt{1+t}}$

$\ln\sqrt{1+t} > 1 - \frac{1}{\sqrt{1+t}} . \qquad \square$

Examiner Notes

Candidates attempting question 8 generally received either a very low or a very high score. Many attempts did not progress further than an attempt to sketch the graph and identify the rectangle to be used. There were also some attempts that confused the line $y = f(t)$ with a transformation of the curve $y = f(x)$. In the second part of the question there were some difficulties with the differentiation of $g(t)$, but those candidates who successfully completed this section did not in general have any difficulties with the remainder of the question.