STEP2 2017 -- Pure Mathematics

STEP2 2017 — Section A (Pure Mathematics)

Exam: STEP2  |  Year: 2017  |  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)	Standard	分部积分,递推关系,数学归纳法,级数求和
2	数列与周期性 (Sequences and Periodicity)	Challenging	函数复合,递推数列,周期性分析,代数因式分解
3	三角函数与图像 (Trigonometric Functions and Graphs)	Challenging	三角方程通解,隐函数求导,图像对称性,余弦恒等式
4	积分不等式 (Integral Inequalities)	Challenging	施瓦茨不等式,积分不等式,函数选取,三角积分
5	参数方程与抛物线 (Parametric Equations and Parabolas)	Standard	参数方程,法线方程,距离公式,导数求极值,圆周角定理
6	数学归纳法与不等式, Mathematical Induction and Inequalities	Challenging	数学归纳法,不等式放缩,平方消去法
7	指数函数与极限, Exponential Functions and Limits	Challenging	对数求导法,指数函数分析,不等式证明,极限计算
8	向量与三角形几何, Vectors and Triangle Geometry	Standard	向量直线方程,标量积判定垂直,向量代数运算

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Question 1

Topic: 积分 (Integration)  |  Difficulty: Standard  |  Marks: 20

Problem

1 Note: In this question you may use without proof the result $\frac{\mathrm{d}}{\mathrm{d}x}(\arctan x) = \frac{1}{1 + x^2}$.

Let $I_n = \int_0^1 x^n \arctan x \, \mathrm{d}x ,$ where $n = 0, 1, 2, 3, \dots$.

(i) Show that, for $n \geqslant 0$, $(n + 1)I_n = \frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1 + x^2} \, \mathrm{d}x$ and evaluate $I_0$.

(ii) Find an expression, in terms of $n$, for $(n + 3)I_{n+2} + (n + 1)I_n$. Use this result to evaluate $I_4$.

(iii) Prove by induction that, for $n \geqslant 1$, $(4n + 1)I_{4n} = A - \frac{1}{2} \sum_{r=1}^{2n} (-1)^r \frac{1}{r} ,$ where $A$ is a constant to be determined.

Hint

The opening “Note” clearly flags a result which will prove important in this question; this is a ‘standard’ result, but one that can slip by unnoticed in single-maths A-levels; it is, therefore, worth learning as an “extra”, if necessary.

In (i), it should be obvious that the process of integration by parts is to be used and that the initial hint indicates that the “first part” must be the “arctan $x$” term, despite appearing as the second term of the product to be integrated. This will lead directly to the given result,

following which the substitution of $n = 0$ leads to the integral $\int_{0}^{1} \frac{x}{1+x^2} dx$. This can be done

by “recognition” (or reverse-Chain Rule integration) or by substitution. In the first instance,

one would note that $\int_{0}^{1} \frac{x}{1+x^2} dx = \frac{1}{2} \int_{0}^{1} \frac{2x}{1+x^2} dx$, where the numerator is precisely the

derivative of the denominator – the standard “log. integral” form; in the second instance, the substitution $u = 1 + x^2$ will also work, though it might take just a few lines more working.

In (ii), one needs only to use the result given in (i), this time replacing $n$ by $(n + 2)$ to find an expression for $(n + 3)I_{n+2}$; and then adding to it the given expression for $(n + 1)I_n$. This leads to a result in which two integrals must be added to get, when simplified,

$\int_{0}^{1} \frac{x^{n+1}(1+x^2)}{1+x^2} dx$, which should need no further comment. Setting $n = 0$ and then $n = 2$ in

this result then yields a numerical answer for $I_4$, since $I_0$ has already been calculated. In (iii), no matter how demanding the process of mathematical induction appears to be, it is very formulaic in some respects and there are always some marks to be had. To begin with, there is always the “baseline” case of (usually) $n = 1$. In this case, one must set $n = 1$ in the proposed formula and check it against the value of $I_4$ already known. This reveals the value of $A$ to be $\frac{1}{4}\pi - \frac{1}{2}\ln 2$. However, since it is constant, it remains fixed during all of the remaining work and one can most easily progress through the rest of the inductive proof by simply continuing to use the letter $A$.

A clearly stated induction hypothesis is enormously helpful (usually replacing the $n$ by another dummy index, $k$ say) rather than vague statements such as “assume the result is true for $n = k$” or meaningless statements such as “assume $n = k$”. We thus proceed

assuming $(4k + 1) I_{4k+1} = A - \frac{1}{2} \sum_{r=1}^{2k} (-1)^r \frac{1}{r}$.

The remainder of the proof relies more on carefulness than inspiration, especially as the process relies on exactly the same techniques as were used in part (ii), using $k$ and $(k + 1)$ in turn in the given result.

Model Solution

Part (i)

Using integration by parts with $u = \arctan x$ and $\mathrm{d}v = x^n \, \mathrm{d}x$:

$\mathrm{d}u = \frac{1}{1+x^2} \, \mathrm{d}x, \qquad v = \frac{x^{n+1}}{n+1}$

$I_n = \left[ \frac{x^{n+1}}{n+1} \arctan x \right]_0^1 - \frac{1}{n+1} \int_0^1 \frac{x^{n+1}}{1+x^2} \, \mathrm{d}x$

Evaluating the boundary term: at $x = 1$, $\frac{1}{n+1} \cdot \frac{\pi}{4}$; at $x = 0$, $0$.

$(n+1) I_n = \frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2} \, \mathrm{d}x$

as required.

Setting $n = 0$:

$I_0 = \frac{\pi}{4} - \int_0^1 \frac{x}{1+x^2} \, \mathrm{d}x = \frac{\pi}{4} - \left[ \frac{1}{2} \ln(1+x^2) \right]_0^1 = \frac{\pi}{4} - \frac{1}{2} \ln 2$

Part (ii)

Replacing $n$ by $n+2$ in the result from (i):

$(n+3) I_{n+2} = \frac{\pi}{4} - \int_0^1 \frac{x^{n+3}}{1+x^2} \, \mathrm{d}x$

Adding $(n+1)I_n = \frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2} \, \mathrm{d}x$:

$(n+3)I_{n+2} + (n+1)I_n = \frac{\pi}{2} - \int_0^1 \frac{x^{n+1} + x^{n+3}}{1+x^2} \, \mathrm{d}x = \frac{\pi}{2} - \int_0^1 \frac{x^{n+1}(1+x^2)}{1+x^2} \, \mathrm{d}x$

$= \frac{\pi}{2} - \int_0^1 x^{n+1} \, \mathrm{d}x = \frac{\pi}{2} - \frac{1}{n+2}$

Setting $n = 0$:

$3I_2 + I_0 = \frac{\pi}{2} - \frac{1}{2} \implies 3I_2 = \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} + \frac{1}{2}\ln 2 = \frac{\pi}{4} - \frac{1}{2} + \frac{1}{2}\ln 2$

Setting $n = 2$:

$5I_4 + 3I_2 = \frac{\pi}{2} - \frac{1}{4}$

$5I_4 = \frac{\pi}{2} - \frac{1}{4} - 3I_2 = \frac{\pi}{2} - \frac{1}{4} - \frac{\pi}{4} + \frac{1}{2} - \frac{1}{2}\ln 2 = \frac{\pi}{4} + \frac{1}{4} - \frac{1}{2}\ln 2$

$I_4 = \frac{\pi}{20} + \frac{1}{20} - \frac{1}{10}\ln 2 = \frac{\pi + 1 - 2\ln 2}{20}$

Part (iii)

Base case ($n = 1$): We need $(4 \cdot 1 + 1)I_4 = A - \frac{1}{2}\sum_{r=1}^{2}(-1)^r \frac{1}{r}$.

$5I_4 = \frac{\pi}{4} + \frac{1}{4} - \frac{1}{2}\ln 2$

The sum is $(-1)^1 \cdot 1 + (-1)^2 \cdot \frac{1}{2} = -1 + \frac{1}{2} = -\frac{1}{2}$, so

$A - \frac{1}{2}\left(-\frac{1}{2}\right) = A + \frac{1}{4}$

Setting this equal to $5I_4$: $A + \frac{1}{4} = \frac{\pi}{4} + \frac{1}{4} - \frac{1}{2}\ln 2$, giving

$A = \frac{\pi}{4} - \frac{1}{2}\ln 2$

Inductive step: Assume for some $k \geq 1$:

$(4k+1)I_{4k} = A - \frac{1}{2}\sum_{r=1}^{2k}(-1)^r \frac{1}{r}$

From part (ii) with $n = 4k$:

$(4k+3)I_{4k+2} + (4k+1)I_{4k} = \frac{\pi}{2} - \frac{1}{4k+2} \qquad (\star)$

From part (ii) with $n = 4k+2$:

$(4k+5)I_{4k+4} + (4k+3)I_{4k+2} = \frac{\pi}{2} - \frac{1}{4k+4} \qquad (\star\star)$

Subtracting $(\star)$ from $(\star\star)$:

$(4k+5)I_{4k+4} - (4k+1)I_{4k} = \frac{1}{4k+2} - \frac{1}{4k+4}$

$(4k+5)I_{4k+4} = (4k+1)I_{4k} + \frac{1}{4k+2} - \frac{1}{4k+4}$

Using the induction hypothesis:

$(4k+5)I_{4k+4} = A - \frac{1}{2}\sum_{r=1}^{2k}(-1)^r \frac{1}{r} + \frac{1}{4k+2} - \frac{1}{4k+4}$

We want to show this equals $A - \frac{1}{2}\sum_{r=1}^{2k+2}(-1)^r \frac{1}{r}$.

$A - \frac{1}{2}\sum_{r=1}^{2k+2}(-1)^r \frac{1}{r} = A - \frac{1}{2}\sum_{r=1}^{2k}(-1)^r \frac{1}{r} - \frac{1}{2}\left[(-1)^{2k+1}\frac{1}{2k+1} + (-1)^{2k+2}\frac{1}{2k+2}\right]$

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

$= A - \frac{1}{2}\sum_{r=1}^{2k}(-1)^r \frac{1}{r} + \frac{1}{2}\cdot\frac{1}{(2k+1)(2k+2)}$

So we need to verify:

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

$\text{LHS} = \frac{1}{2(2k+1)} - \frac{1}{4(k+1)} = \frac{2(k+1) - (2k+1)}{4(k+1)(2k+1)} = \frac{1}{4(k+1)(2k+1)} = \frac{1}{2(2k+1)(2k+2)} = \text{RHS}$

This confirms the inductive step. Hence by induction, the result holds for all $n \geq 1$ with $A = \frac{\pi}{4} - \frac{1}{2}\ln 2$.

Examiner Notes

Almost all candidates attempted this question, making it the most popular on the paper; it was also the highest-scoring, with a mean score of 15. Indeed, the careful structure meant that its direction was clear to almost all candidates and it was only the rather tricky induction proof in (iii) that prevented the question from being completely transparent.

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Question 2

Topic: 数列与周期性 (Sequences and Periodicity)  |  Difficulty: Challenging  |  Marks: 20

Problem

2 The sequence of numbers $x_0, x_1, x_2, \dots$ satisfies $x_{n+1} = \frac{ax_n - 1}{x_n + b} .$ (You may assume that $a, b$ and $x_0$ are such that $x_n + b \neq 0$.) Find an expression for $x_{n+2}$ in terms of $a, b$ and $x_n$.

(i) Show that $a + b = 0$ is a necessary condition for the sequence to be periodic with period 2. Note: The sequence is said to be periodic with period $k$ if $x_{n+k} = x_n$ for all $n$, and there is no integer $m$ with $0 < m < k$ such that $x_{n+m} = x_n$ for all $n$.

(ii) Find necessary and sufficient conditions for the sequence to have period 4.

Hint

The opening part of the question is essentially identical to the process of composition of functions. Moreover, if subscripts are likely to prove confusing, then begin with a statement

such as “Let $x_n = X$.” Thus $x_{n+1} = \frac{aX - 1}{X + b}$ and $x_{n+2} = \frac{a\left(\frac{aX - 1}{X + b}\right) - 1}{\left(\frac{aX - 1}{X + b}\right) + b}$, which simply needs

tidying up. This yields $x_{n+2} = \frac{(a^2 - 1)X - (a + b)}{(a + b)X + (b^2 - 1)}$. The “Note” in (i) reminds the reader that

the period of a periodic sequence is the length of the smallest cycle of repetition; thus, we require $x_{n+2} = x_n$ but not $x_{n+1} = x_n$. A moment’s thought should reveal it to be obvious that a constant sequence should still yield part of the same algebra which gives $x_{n+2} = x_n$ and it is worth exploring this first:

if $x_{n+1} = x_n$ then $aX - 1 = X^2 + bX \Rightarrow 0 = X^2 - (a - b)X + 1$. One might be tempted to try to solve this (using the Quadratic formula, for instance), but there is really nothing to be gained by so doing, since the constant sequence is of no interest to us, only the conditions that give it (which we need to have in mind later on). Proceeding to explore $x_{n+2} = x_n$ gives us, upon collecting up, $0 = (a + b)\{X^2 - (a - b)X + 1\}$. Fortunately, the factor $(a + b)$ is readily apparent, but the quadratic factor should have been anticipated also, for the reasons outlined above. Thus, $a + b = 0$ is a necessary condition for a period 2 sequence. (There is no requirement to explore the issue of sufficiency, which could be done by setting $b = -a$ in the initial expression for $x_{n+1}$ and then following it through to see what happens.)

Finally, we are asked to see what happens when $x_{n+4} = x_n$, and this can be done the long way round by finding $x_{n+3} = \frac{(a^3 - 2a - b)X - (a^2 + ab + b^2 - 1)}{(a^2 + ab + b^2 - 1)X - (a + 2b + b^3)}$ and then

$x_{n+4} = \frac{ax_{n+3} - 1}{x_{n+3} + b}$, but there is at least one very obvious shortcut to what is starting to look

like some complicated algebra: namely, considering the “two-step” result already found and repeating that, going from $x_n$ to $x_{n+4}$ via $x_{n+2}$. It also helps if one realises that whatever algebraic expression appears, we know that it must have the previously found factors of $(a + b)$ and $\{X^2 - (a - b)X + 1\}$ within it.

Model Solution

Finding $x_{n+2}$: Let $x_n = X$. Then

$x_{n+2} = \frac{a\left(\frac{aX-1}{X+b}\right) - 1}{\left(\frac{aX-1}{X+b}\right) + b} = \frac{a(aX-1) - (X+b)}{(aX-1) + b(X+b)} = \frac{(a^2-1)X - (a+b)}{(a+b)X + (b^2-1)}$

Part (i)

For period 2, $x_{n+2} = x_n$ for all $n$, so

$\frac{(a^2-1)X - (a+b)}{(a+b)X + (b^2-1)} = X$

Cross-multiplying:

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

$(a+b)X^2 + (b^2-a^2)X + (a+b) = 0$

$(a+b)\bigl[X^2 - (a-b)X + 1\bigr] = 0$

The quadratic $X^2-(a-b)X+1=0$ has at most two roots, so it cannot hold for a sequence that takes more than two distinct values. For a sequence with period exactly 2 (which must take at least two distinct values that both satisfy this equation), the quadratic can account for at most two terms. But for $x_{n+2}=x_n$ to hold for all $n$, the factor $(a+b)$ must be zero. Hence $a+b=0$ is a necessary condition.

Part (ii)

For period 4 we need $x_{n+4}=x_n$ for all $n$ but $x_{n+2}\neq x_n$ for at least some $n$.

The two-step map $x_{n+2} = g(x_n)$ where $g(X)=\frac{(a^2-1)X-(a+b)}{(a+b)X+(b^2-1)}$ is a M”obius transformation. Its four-step composition $x_{n+4}=g(g(x_n))$ is also a M”obius transformation.

Writing $p=a^2-1$, $q=-(a+b)$, $r=a+b$, $s=b^2-1$, the composition $g\circ g$ is:

$g(g(X)) = \frac{(p^2+qr)X + q(p+s)}{r(p+s)X + (qr+s^2)}$

Setting $g(g(X))=X$ and cross-multiplying:

$r(p+s)X^2 + (s^2-p^2)X - q(p+s) = 0$

$(p+s)\bigl[rX^2 + (s-p)X - q\bigr] = 0$

Substituting back: $p+s = a^2+b^2-2$, $s-p=b^2-a^2$, $-q=a+b$:

$(a^2+b^2-2)\bigl[(a+b)X^2 + (b^2-a^2)X + (a+b)\bigr] = 0$

$(a^2+b^2-2)(a+b)\bigl[X^2-(a-b)X+1\bigr] = 0$

For $g(g(X))=X$ to hold for all $X$ in the sequence, the M”obius transformation $g\circ g$ must be the identity. This requires $a^2+b^2-2=0$, i.e.,

$a^2+b^2=2$

(If $a^2+b^2\neq 2$, then $g(g(X))=X$ only at the roots of the quadratic, so $x_{n+4}=x_n$ fails for generic $x_0$.)

With $a^2+b^2=2$, the map $g$ satisfies $g\circ g = \mathrm{id}$, so $x_{n+4}=x_n$ for all $n$. The period is exactly 4 (not 1 or 2) provided:

• $a+b\neq 0$ (otherwise $g=\mathrm{id}$ and the period is 1 or 2), and
• $x_0$ is not a fixed point of $g$, i.e., $x_0^2-(a-b)x_0+1\neq 0$ (otherwise $x_2=x_0$ and the period is 1 or 2).

The necessary and sufficient conditions for period 4 are therefore:

$\boxed{a^2+b^2=2, \quad a+b\neq 0, \quad x_0^2-(a-b)x_0+1\neq 0}$

Examiner Notes

This was the second most popular question of all, attempted by over 80% of candidates; but scoring relatively poorly with a mean score of under 10. It was, of course, heavily algebraic and this meant that many candidates found it challenging, getting lost in the algebra. In most cases, this was largely avoidable: the simple device of calling the first term “X” (say) would have prevented a lot of unnecessary subscripts from cluttering up the working. A few moments of thought from those candidates who simply embarked on the (potentially) intricate algebra could have saved a lot of trouble. The point of a sequence’s periodicity is that it is the smallest cycle over which terms repeat; it should be noted that the condition for each term to be equal (a constant sequence) must clearly be embedded in any condition that gives $x_{n+2} = x_n$. Similarly, in order to satisfy $x_{n+4} = x_n$, we must automatically have the cases when all terms are the same and every other term equal present somewhere. This makes any ensuing factorisations much easier to deal with.

It could be noted that the requirement for $x_{n+4} = x_n$ can be thought of as a two-stage sequence using every other term; and this situation has just been sorted out.

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Question 3

Topic: 三角函数与图像 (Trigonometric Functions and Graphs)  |  Difficulty: Challenging  |  Marks: 20

Problem

3 (i) Sketch, on $x$-$y$ axes, the set of all points satisfying $\sin y = \sin x$, for $-\pi \leqslant x \leqslant \pi$ and $-\pi \leqslant y \leqslant \pi$. You should give the equations of all the lines on your sketch.

(ii) Given that $\sin y = \frac{1}{2} \sin x$ obtain an expression, in terms of $x$, for $y''$ when $0 \leqslant x \leqslant \frac{1}{2}\pi$ and $0 \leqslant y \leqslant \frac{1}{2}\pi$, and show that $y'' = -\frac{3 \sin x}{(4 - \sin^2 x)^{\frac{3}{2}}} .$ Use these results to sketch the set of all points satisfying $\sin y = \frac{1}{2} \sin x$ for $0 \leqslant x \leqslant \frac{1}{2}\pi$ and $0 \leqslant y \leqslant \frac{1}{2}\pi$.

Hence sketch the set of all points satisfying $\sin y = \frac{1}{2} \sin x$ for $-\pi \leqslant x \leqslant \pi$ and $-\pi \leqslant y \leqslant \pi$.

(iii) Without further calculation, sketch the set of all points satisfying $\cos y = \frac{1}{2} \sin x$ for $-\pi \leqslant x \leqslant \pi$ and $-\pi \leqslant y \leqslant \pi$.

Hint

Whilst this question might appear somewhat daunting on first reading, it involves little more than an understanding of the sine curve and the key results that relate to “angles in all quadrants”.

To begin with, "$\sin y = \sin x \Rightarrow y = x$" is the kind of response expected from those candidates who have failed to understand that there are many functions (even the simple ones covered at A-level) that don’t map “one-to-one”. Though the general formula "$\sin y = \sin x \Rightarrow y = n\pi + (-1)^n x$" might be unfamiliar, it only says that if $\sin y = \sin x$ (imagining $x$ to be positive and acute) then $y$ could be any “first quadrant” equivalent of $x$ … an even multiple of $\pi$ plus $x$ … or a “second quadrant” supplement of $x$ … an odd multiple of $\pi$ minus $x$. (Apart from that, $y$‘s corresponding to non-acute $x$‘s arise from application of the same principles, with “quadrants” taking care of themselves due to the symmetries of the sine curve.)

In (i), it is then found that the general formula (or its equivalent in two bits) gives three graphs that are of interest here: $n = -1$ ($y = -\pi - x$), $n = 0$ ($y = x$) and $n = 1$ ($y = \pi - x$), all of which give straight-line segments in the interval required.

In (ii), there is rather less thinking to be done – at least to begin with – since one is only required to differentiate twice and do some tidying up. This calculus can be done implicitly or directly after rearranging into an arcsine form. That is not to say that it is easy calculus, since the Product, Quotient and Chain rules can all play a part in the processes that follow.

In sketching the graph, one should start simple and work up. Initially, $\frac{dy}{dx} = \frac{1}{2}$ at $(0, 0)$, with

the curve increasing to a maximum at $(\frac{\pi}{2}, \frac{\pi}{6})$, since $\frac{d^2y}{dx^2} < 0$. This gives

x	y
0	0
...
$\frac{1}{2}\pi$	$\frac{1}{6}\pi$

Thereafter, for whatever “other” bits there are to the curve, these follow from symmetries, applied to the above portion: namely, reflection symmetry in $x = \frac{\pi}{2}$; rotational symmetry about $O$; and reflection symmetry in $y = \pm \frac{\pi}{2}$.

Part (iii)‘s graph follows by applying the result $\cos y \equiv \sin(\frac{\pi}{2} - y)$.

Model Solution

Part (i)

The general solution of $\sin y = \sin x$ is $y = n\pi + (-1)^n x$ for integer $n$.

For $-\pi \leqslant x \leqslant \pi$ and $-\pi \leqslant y \leqslant \pi$:

• $n = 0$: $y = x$ (the diagonal from $(-\pi, -\pi)$ to $(\pi, \pi)$)
• $n = 1$: $y = \pi - x$ (from $(-\pi, 2\pi)$… but $y$ must be in $[-\pi, \pi]$, so we need $\pi - x \in [-\pi, \pi]$, i.e., $x \in [0, 2\pi]$. Restricted to $x \in [0, \pi]$: from $(0, \pi)$ to $(\pi, 0)$)
• $n = -1$: $y = -\pi - x$ (from $(-\pi, 0)$ to $(0, -\pi)$)
• $n = 2$: $y = 2\pi + x$, which exceeds $\pi$ for $x > -\pi$, so no segment in range.

The sketch consists of three line segments:

• $y = x$ for $-\pi \leqslant x \leqslant \pi$
• $y = \pi - x$ for $0 \leqslant x \leqslant \pi$
• $y = -\pi - x$ for $-\pi \leqslant x \leqslant 0$

These form a “bow-tie” pattern: the main diagonal $y = x$, plus a segment from $(0, \pi)$ to $(\pi, 0)$ and a segment from $(-\pi, 0)$ to $(0, -\pi)$.

Part (ii)

Differentiating $\sin y = \frac{1}{2}\sin x$ implicitly:

$\cos y \cdot y' = \frac{1}{2}\cos x \implies y' = \frac{\cos x}{2\cos y}$

For the second derivative, using implicit differentiation of $\cos y \cdot y' = \frac{1}{2}\cos x$:

$-\sin y \cdot (y')^2 + \cos y \cdot y'' = -\frac{1}{2}\sin x$

$y'' = \frac{-\frac{1}{2}\sin x + \sin y \cdot (y')^2}{\cos y}$

Since $\sin y = \frac{1}{2}\sin x$ and $(y')^2 = \frac{\cos^2 x}{4\cos^2 y}$:

$\sin y \cdot (y')^2 = \frac{\sin x \cos^2 x}{8\cos^2 y}$

Also $\cos^2 y = 1 - \sin^2 y = 1 - \frac{1}{4}\sin^2 x = \frac{4 - \sin^2 x}{4}$, so $\cos y = \frac{\sqrt{4 - \sin^2 x}}{2}$.

Substituting:

$y'' = \frac{1}{\cos y}\left(-\frac{1}{2}\sin x + \frac{\sin x \cos^2 x}{8 \cdot \frac{4-\sin^2 x}{4}}\right) = \frac{1}{\cos y}\left(-\frac{1}{2}\sin x + \frac{\sin x \cos^2 x}{2(4 - \sin^2 x)}\right)$

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

Now $\cos^2 x = 1 - \sin^2 x$, so:

$-1 + \frac{1 - \sin^2 x}{4 - \sin^2 x} = \frac{-(4 - \sin^2 x) + 1 - \sin^2 x}{4 - \sin^2 x} = \frac{-3}{4 - \sin^2 x}$

Therefore:

$y'' = \frac{\sin x}{2\cos y} \cdot \frac{-3}{4 - \sin^2 x} = \frac{-3\sin x}{2 \cdot \frac{\sqrt{4-\sin^2 x}}{2} \cdot (4 - \sin^2 x)} = \frac{-3\sin x}{(4 - \sin^2 x)^{3/2}}$

Sketching for $0 \leqslant x \leqslant \frac{\pi}{2}$, $0 \leqslant y \leqslant \frac{\pi}{2}$:

At $x = 0$: $y = 0$, $y'(0) = \frac{1}{2}$. Since $y'' < 0$ for $x > 0$, the curve is concave down. At $x = \frac{\pi}{2}$: $y = \frac{\pi}{6}$. The curve rises from the origin with slope $\frac{1}{2}$, curving concavely to the point $(\frac{\pi}{2}, \frac{\pi}{6})$.

Extending to $[-\pi, \pi]^2$:

The equation $\sin y = \frac{1}{2}\sin x$ has:

• Rotational symmetry about the origin: replacing $(x,y)$ by $(-x,-y)$ preserves the equation.
• No reflection symmetry in $x = 0$ since $\sin(-x) = -\sin x$.

The full curve consists of four arcs, forming a closed “rounded diamond” shape:

1. $(0, 0) \to (\frac{\pi}{2}, \frac{\pi}{6}) \to (\pi, \frac{\pi}{6})$ (concave down, first quadrant)
2. $(\pi, \frac{\pi}{6}) \to (\pi, -\frac{\pi}{6})$ (a short segment along $x = \pi$… actually no, the curve connects through $x = \pi$ continuously)

More precisely: The curve is a closed convex-like shape passing through $(0, 0)$, $(\frac{\pi}{2}, \frac{\pi}{6})$, $(\pi, \frac{\pi}{6})$, $(\frac{\pi}{2}, -\frac{\pi}{6})$, $(0, 0)$… No, let me reconsider.

The full curve passes through $(0, 0)$, $(\pi, \frac{\pi}{6})$, $(0, 0)$… Hmm. Actually the curve is closed and passes through the origin. By the rotational symmetry, if $(x_0, y_0)$ is on the curve, so is $(-x_0, -y_0)$. The four arcs connect:

• Arc in first quadrant: $(0,0)$ to $(\pi, \frac{\pi}{6})$
• Arc in fourth quadrant: $(\pi, -\frac{\pi}{6})$ to $(0, 0)$ (by symmetry)
• Arc in third quadrant: $(0, 0)$ to $(-\pi, -\frac{\pi}{6})$ (by symmetry)
• Arc in second quadrant: $(-\pi, \frac{\pi}{6})$ to $(0, 0)$ (by symmetry)

These form a closed loop through the origin, with the curve passing through $(\pm\pi, \pm\frac{\pi}{6})$.

Part (iii)

Using $\cos y = \sin(\frac{\pi}{2} - y)$, set $Y = \frac{\pi}{2} - y$, so $\sin Y = \frac{1}{2}\sin x$.

This is exactly the equation from part (ii) in the $(x, Y)$ plane. The transformation $y = \frac{\pi}{2} - Y$ shifts the curve upward by $\frac{\pi}{2}$ and reflects in $Y$.

The curve from part (ii) in the $(x, Y)$ plane passes through $(0, 0)$, $(\pi, \frac{\pi}{6})$, $(-\pi, -\frac{\pi}{6})$, etc.

In the $(x, y)$ plane with $y = \frac{\pi}{2} - Y$:

• $(x, Y) = (0, 0) \to (x, y) = (0, \frac{\pi}{2})$
• $(x, Y) = (\pi, \frac{\pi}{6}) \to (x, y) = (\pi, \frac{\pi}{3})$
• $(x, Y) = (0, 0) \to (x, y) = (0, \frac{\pi}{2})$

The curve $\cos y = \frac{1}{2}\sin x$ is the part (ii) curve shifted vertically by $\frac{\pi}{2}$, forming a closed loop passing through $(0, \frac{\pi}{2})$, $(\pi, \frac{\pi}{3})$, $(0, -\frac{\pi}{2})$, $(-\pi, -\frac{\pi}{3})$, and back to $(0, \frac{\pi}{2})$. It is symmetric about the origin (since replacing $(x,y)$ by $(-x,-y)$ preserves $\cos y = \frac{1}{2}\sin x$).

Examiner Notes

Attempts fell to around the 50% figure with marks scored by those who attempted the question averaging about 10 out of 20. There is not much to this question beyond the baseline realisation that $\sin y = \sin x$ does not necessarily imply that $y = x$. In essence, it is all about “quadrants” work, where candidates need to consider the two solutions, $x$ and $\pi - x$ in one period of the sine function, and then adding or subtracting multiples of $2\pi$ as necessary. Once one has done this, the accompanying straight-line segments are straightforward marks in the last part of (i).

A lot of marks were gained in (ii), as candidates were clearly attracted by the familiar “differentiate this couple of times” demand; most of them were quite happy with the differentiation, performed either implicitly or directly using arcsines.

The drawings required in (ii) and (iii) then relied on an appreciation of the symmetries of the sine function, along with the use of the identity $\cos y \equiv \sin(\frac{1}{2}\pi - y)$.

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Question 4

Topic: 积分不等式 (Integral Inequalities)  |  Difficulty: Challenging  |  Marks: 20

Problem

4 The Schwarz inequality is $\left( \int_a^b f(x) g(x) \, dx \right)^2 \leqslant \left( \int_a^b (f(x))^2 dx \right) \left( \int_a^b (g(x))^2 dx \right) . \qquad \text{(*)}$

(i) By setting $f(x) = 1$ in $(*)$, and choosing $g(x)$, $a$ and $b$ suitably, show that for $t > 0$, $\frac{e^t - 1}{e^t + 1} \leqslant \frac{t}{2} .$

(ii) By setting $f(x) = x$ in $(*)$, and choosing $g(x)$ suitably, show that $\int_0^1 e^{-\frac{1}{2}x^2} dx \geqslant 12(1 - e^{-\frac{1}{4}})^2 .$

(iii) Use $(*)$ to show that $\frac{64}{25\pi} \leqslant \int_0^{\frac{1}{2}\pi} \sqrt{\sin x} \, dx \leqslant \sqrt{\frac{\pi}{2}} .$

Hint

This is an interesting question and very straightforward in some respects. To begin with, you are told in part (i) that f(x) = 1. At this point, it would be wise to write down what the Schwarz inequality gives in this case:

$\left( \int_{a}^{b} \text{g}(x) \text{ dx} \right)^{2} \leq \left( \int_{a}^{b} 1 \text{ dx} \right) \left( \int_{a}^{b} [\text{g}(x)]^{2} \text{ dx} \right); \text{ i.e. } \left( \int_{a}^{b} \text{g}(x) \text{ dx} \right)^{2} \leq (b - a) \left( \int_{a}^{b} [\text{g}(x)]^{2} \text{ dx} \right).$

A few moments of careful thought (inspecting the given answer) should make you realise that $a = 0$, $b = t$ give the terms $(b - a) = (t - 0)$ and $(e^{t} - e^{0})$ when $\text{g}(x) = e^{x}$. Following it through from there is relatively routine, provided one spots the difference-of-two-squares factorisation and that we can divide throughout by $(e^{t} + 1)$, which is guaranteed to be positive (an important consideration when dealing with inequalities).

In (ii), it is (again) best to start by seeing how things appear when you have used the given information that f(x) = $x$ and, by clear implication, $a = 0$ and $b = 1$:

$\left( \int_{0}^{1} x \text{ g}(x) \text{ dx} \right)^{2} \leq \left( \int_{0}^{1} x^{2} \text{ dx} \right) \left( \int_{0}^{1} [\text{g}(x)]^{2} \text{ dx} \right); \text{ i.e. } \left( \int_{0}^{1} x \text{ g}(x) \text{ dx} \right)^{2} \leq \frac{1}{3} \left( \int_{0}^{1} [\text{g}(x)]^{2} \text{ dx} \right).$

The $e^{-\frac{1}{4}}$ in the given answer, along with the fact that $\left( e^{-\frac{1}{4}x^{2}} \right)^{2} = e^{-\frac{1}{2}x^{2}}$ should point the way

towards choosing $\text{g}(x) = e^{-\frac{1}{4}x^{2}}$. Following this through carefully again yields the required result.

The result in part (iii) clearly requires the use of the Schwarz inequality twice, once each for the right- and left-halves of the given result. Setting $\text{f}(x) = 1$, $\text{g}(x) = \sqrt{\sin x}$, $a = 0$ and

$b = \frac{1}{2}\pi$ leads to $\int_{0}^{\frac{1}{2}\pi} \sqrt{\sin x} \text{ dx} \leq \sqrt{\frac{\pi}{2}}$. However, the left-hand half of the result does require a

bit more thought and, preferably, familiarity with the integration of trig. functions where powers of $\sin x$ (in this case) appear along with its derivative, $\cos x$. The real clue is that, for

the $\int_{0}^{\frac{1}{2}\pi} \sqrt{\sin x} \text{ dx}$ to appear on the other side of the inequality to the one found using the

obvious candidates that led to the right-hand half of the result, $\sqrt{\sin x}$ must now be the result of the squaring process. It is then experience (or insight) that suggests setting $\text{f}(x) = \cos x$, $\text{g}(x) = \sqrt[4]{\sin x}$, $a = 0$ and $b = \frac{1}{2}\pi$. You will then find that the LHS of the Schwarz inequality requires the integration of a function of $\sin x$ multiplied by its derivative, $\cos x$; and this (as in Q1) can be done by “recognition” or substitution. (You will also need to be able to integrate $\cos^{2} x$, which calls upon the use of the standard double-angle formula for cosine.)

Model Solution

Part (i)

Set $f(x) = 1$, $g(x) = e^x$, $a = 0$, $b = t$ in the Schwarz inequality:

$\left(\int_0^t e^x \, \mathrm{d}x\right)^2 \leqslant \left(\int_0^1 1 \, \mathrm{d}x\right)\left(\int_0^t e^{2x} \, \mathrm{d}x\right)$

$(e^t - 1)^2 \leqslant t \cdot \frac{e^{2t} - 1}{2}$

$(e^t - 1)^2 \leqslant \frac{t}{2}(e^t - 1)(e^t + 1)$

Since $e^t - 1 > 0$ for $t > 0$, divide both sides by $(e^t - 1)(e^t + 1) > 0$:

$\frac{e^t - 1}{e^t + 1} \leqslant \frac{t}{2}$

Part (ii)

Set $f(x) = x$, $g(x) = e^{-\frac{1}{4}x^2}$, $a = 0$, $b = 1$:

$\left(\int_0^1 x \, e^{-\frac{1}{4}x^2} \, \mathrm{d}x\right)^2 \leqslant \left(\int_0^1 x^2 \, \mathrm{d}x\right)\left(\int_0^1 e^{-\frac{1}{2}x^2} \, \mathrm{d}x\right)$

Left side: Let $u = -\frac{1}{4}x^2$, $\mathrm{d}u = -\frac{1}{2}x \, \mathrm{d}x$:

$\int_0^1 x \, e^{-\frac{1}{4}x^2} \, \mathrm{d}x = -2\int_0^{-1/4} e^u \, \mathrm{d}u = -2\bigl[e^u\bigr]_0^{-1/4} = 2(1 - e^{-1/4})$

So LHS $= 4(1 - e^{-1/4})^2$.

Right side: $\int_0^1 x^2 \, \mathrm{d}x = \frac{1}{3}$.

$4(1 - e^{-1/4})^2 \leqslant \frac{1}{3}\int_0^1 e^{-\frac{1}{2}x^2} \, \mathrm{d}x$

$\int_0^1 e^{-\frac{1}{2}x^2} \, \mathrm{d}x \geqslant 12(1 - e^{-1/4})^2$

Part (iii)

Right-hand inequality: Set $f(x) = 1$, $g(x) = \sqrt{\sin x}$, $a = 0$, $b = \frac{\pi}{2}$:

$\left(\int_0^{\pi/2} \sqrt{\sin x} \, \mathrm{d}x\right)^2 \leqslant \frac{\pi}{2}\int_0^{\pi/2} \sin x \, \mathrm{d}x = \frac{\pi}{2}\bigl[-\cos x\bigr]_0^{\pi/2} = \frac{\pi}{2}$

$\int_0^{\pi/2} \sqrt{\sin x} \, \mathrm{d}x \leqslant \sqrt{\frac{\pi}{2}}$

Left-hand inequality: Set $f(x) = \cos x$, $g(x) = (\sin x)^{1/4}$, $a = 0$, $b = \frac{\pi}{2}$:

$\left(\int_0^{\pi/2} \cos x \, (\sin x)^{1/4} \, \mathrm{d}x\right)^2 \leqslant \left(\int_0^{\pi/2} \cos^2 x \, \mathrm{d}x\right)\left(\int_0^{\pi/2} \sqrt{\sin x} \, \mathrm{d}x\right)$

Left side integral: Let $u = \sin x$, $\mathrm{d}u = \cos x \, \mathrm{d}x$:

$\int_0^{\pi/2} \cos x \, (\sin x)^{1/4} \, \mathrm{d}x = \int_0^1 u^{1/4} \, \mathrm{d}u = \frac{4}{5}$

So the left side is $\frac{16}{25}$.

First integral on the right: Using $\cos^2 x = \frac{1}{2}(1 + \cos 2x)$:

$\int_0^{\pi/2} \cos^2 x \, \mathrm{d}x = \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{\pi}{4}$

So:

$\frac{16}{25} \leqslant \frac{\pi}{4}\int_0^{\pi/2} \sqrt{\sin x} \, \mathrm{d}x$

$\int_0^{\pi/2} \sqrt{\sin x} \, \mathrm{d}x \geqslant \frac{64}{25\pi}$

Combining both results:

$\frac{64}{25\pi} \leqslant \int_0^{\pi/2} \sqrt{\sin x} \, \mathrm{d}x \leqslant \sqrt{\frac{\pi}{2}}$

Examiner Notes

This is the first question where the difference between “attempts” and “serious attempts” arises to any significant extent: there were just over 800 of the former but well under 500 of the latter. This is also a good point at which to raise a key issue in respect of strategy for candidates sitting a STEP. Spending a few minutes of reading time, at some particular time during the examination, could be a significant asset, especially to those candidates who have particular strengths and weaknesses to play to or to avoid. A very brief analysis of this question, on first reading, should help one recognise that a result is being given (with no requirement to establish it in any way) and all that is required is to use it. Part (i) then clearly directs part of the way, and the required limits are rather obviously flagged, as is the fact that $g(x)$ must be something to do with the exponential function. One of the two functions to be used in (ii) is also given, as are the limits; an inspection of the given should

lead to the (correct) conclusion that $g(x)$ must be $e^{-\frac{1}{4}x^2}$. Getting just this far takes the candidate to the 10-mark point, a perfectly good return for a candidate who has read the question through sufficiently carefully to realise that it has decent potential for mark-acquisition.

In the final part of the question some careful thought was needed, with only the required limits obvious at first glance. Most attempts, serious or otherwise, picked up the majority of their marks in (i) and (ii) and efforts at (iii) were very varied: many candidates simply gave up and moved on; many more picked up a few extra marks by setting $g(x) = \sqrt{\sin x}$ (which is a fairly obvious candidate to try) and working towards the right-hand half of the given result. Very few candidates indeed had the experience to realise that $\sqrt{\sin x}$ now needed to appear as the squared term, which also meant that a cosine term had to be involved.

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Question 5

Topic: 参数方程与抛物线 (Parametric Equations and Parabolas)  |  Difficulty: Standard  |  Marks: 20

Problem

5 A curve $C$ is determined by the parametric equations

$x = at^2, \ y = 2at,$

where $a > 0$.

(i) Show that the normal to $C$ at a point $P$, with non-zero parameter $p$, meets $C$ again at a point $N$, with parameter $n$, where

$n = -\left(p + \frac{2}{p}\right).$

(ii) Show that the distance $|PN|$ is given by

$|PN|^2 = 16a^2 \frac{(p^2 + 1)^3}{p^4}$

and that this is minimised when $p^2 = 2$.

(iii) The point $Q$, with parameter $q$, is the point at which the circle with diameter $PN$ cuts $C$ again. By considering the gradients of $QP$ and $QN$, show that

$2 = p^2 - q^2 + \frac{2q}{p}.$

Deduce that $|PN|$ is at its minimum when $Q$ is at the origin.

Hint

In many ways, much of this question is also relatively routine. Find $\frac{dy}{dx}$ for the gradient of

the tangent; find its negative reciprocal for the gradient of the normal and then any one of a number of formulae for the equation of a line. At some stage, you will need to replace $t$ by $p$ for the normal at $P$ and then replace $x$ and $y$ in this equation by $an^2$ and $2an$ for another point on the curve. Solving for $n$ in terms of $p$ – noting that the factor $(n - p)$ must be involved somewhere, since $n = p$ must be one solution to whatever equation arises as the line is already known to meet the curve at $P$ – should then yield the given answer. In (ii), employing the distance formula $PN^2 = (x_P - x_N)^2 + (y_P - y_N)^2$ is clearly the way

forwards, as is replacing $n$ by $-\left( p + \frac{2}{p} \right)$ at some stage of the proceedings. The rest is just

careful algebra. Differentiating the given expression for $PN^2$ with respect to $p$ is routine enough, in principle, and it is then only required to justify that the (only) values of $p$ that arise will give minimum points. One could use the first-derivative test (looking for a change of sign), the second-derivative test (examining its sign) or argue from the shape of the curve

$(y =) \frac{16a^2}{p^4}(p^2 + 1)^3 ...$ which is symmetric in the $y$-axis, asymptotic to the $y$-axis for small

values of $p$, and can be arbitrarily large as $|p| \rightarrow \infty$; thus, any turning-points must be minima.

For part (iii), one starts by noting that $PQ$ and $NQ$ are perpendicular (since $\angle PQN = 90^\circ$, by

“Angle in a semi-circle”). Then, setting the products of their gradients, $\frac{2}{p+q}$ and $\frac{2}{n+q}$,

equal to $-1$, replacing $n$ by $-\left( p + \frac{2}{p} \right)$ once again, and using $p^2 = 2$, takes you almost the

whole way there: $q^2 = \frac{2q}{p}$. From this point, we have $q = 0$ or $q = \frac{2}{p} = \pm\sqrt{2}$. Finally, these

final two cases should be eliminated by noting that they give $q = p$, i.e. $Q = P$, which is not the case as they are being taken to be distinct points.

Model Solution

Part (i)

The gradient of the tangent at parameter $t$ is:

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{2a}{2at} = \frac{1}{t}$

The normal at $P = (ap^2, 2ap)$ has gradient $-p$ and equation:

$y - 2ap = -p(x - ap^2)$

The point $N = (an^2, 2an)$ lies on $C$ and on this normal:

$2an - 2ap = -p(an^2 - ap^2)$

$2a(n - p) = -ap(n^2 - p^2) = -ap(n-p)(n+p)$

Since $n \neq p$ (as $N \neq P$), divide by $a(n-p) \neq 0$:

$2 = -p(n+p)$

$n + p = -\frac{2}{p}$

$n = -p - \frac{2}{p}$

Part (ii)

$|PN|^2 = (ap^2 - an^2)^2 + (2ap - 2an)^2 = a^2(p^2 - n^2)^2 + 4a^2(p - n)^2$

$= a^2(p-n)^2\bigl[(p+n)^2 + 4\bigr]$

From part (i): $n - p = -\frac{2}{p} - 2p = -\frac{2(p^2+1)}{p}$, so $(p-n)^2 = \frac{4(p^2+1)^2}{p^2}$.

Also $n + p = -\frac{2}{p}$, so $(n+p)^2 + 4 = \frac{4}{p^2} + 4 = \frac{4(p^2+1)}{p^2}$.

$|PN|^2 = a^2 \cdot \frac{4(p^2+1)^2}{p^2} \cdot \frac{4(p^2+1)}{p^2} = \frac{16a^2(p^2+1)^3}{p^4}$

Minimising: Let $u = p^2 > 0$, so $|PN|^2 = 16a^2 \cdot h(u)$ where $h(u) = \frac{(u+1)^3}{u^2}$.

$h'(u) = \frac{3(u+1)^2 \cdot u^2 - (u+1)^3 \cdot 2u}{u^4} = \frac{(u+1)^2 u[3u - 2(u+1)]}{u^4} = \frac{(u+1)^2(u-2)}{u^3}$

$h'(u) < 0$ for $u < 2$ and $h'(u) > 0$ for $u > 2$, so $h$ has a minimum at $u = 2$, i.e., $p^2 = 2$.

Part (iii)

Since $PN$ is a diameter of the circle and $Q$ lies on the circle, $\angle PQN = 90^{\circ}$ (angle in a semicircle).

The gradient of $PQ$: $\frac{2ap - 2aq}{ap^2 - aq^2} = \frac{2}{p+q}$.

The gradient of $QN$: $\frac{2aq - 2an}{aq^2 - an^2} = \frac{2}{q+n}$.

Since $PQ \perp QN$:

$\frac{2}{p+q} \cdot \frac{2}{q+n} = -1$

$4 = -(p+q)(q+n)$

$4 = -q^2 - (p+n)q - pn$

$q^2 + (p+n)q + pn + 4 = 0$

Substituting $n = -p - \frac{2}{p}$:

$p + n = -\frac{2}{p}, \qquad pn = p\left(-p - \frac{2}{p}\right) = -p^2 - 2$

$q^2 - \frac{2}{p}q + (-p^2 - 2) + 4 = 0$

$q^2 - \frac{2}{p}q + 2 - p^2 = 0$

which rearranges to $2 = p^2 - q^2 + \frac{2q}{p}$, as required.

Deduction: At the minimum, $p^2 = 2$, so $2 - p^2 = 0$ and the equation becomes:

$q^2 - \frac{2}{p}q = 0 \implies q\left(q - \frac{2}{p}\right) = 0$

So $q = 0$ or $q = \frac{2}{p}$.

If $q = \frac{2}{p}$: checking against $p$, when $p = \sqrt{2}$, $q = \sqrt{2} = p$ (so $Q = P$, rejected); when $p = -\sqrt{2}$, $q = -\sqrt{2} = p$ (again $Q = P$, rejected).

Therefore $q = 0$, meaning $Q$ is the origin.

Examiner Notes

Attempts at this question were over the thousand figure, making it the third most popular question on the paper, with the second-highest mean score. Part (i) proved to be very routine; the calculus requirements in (ii) were obvious to most, though justifying the minimum distance was often poorly handled; for instance, finding the second derivative is a poor way to spend one’s time when examining the sign of the first derivative is easily undertaken. The needs of part (iii) were also easily spotted though, again, a couple of marks were almost universally lost as the need to eliminate the two other cases that arise was largely ignored.

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Question 6

Topic: 数学归纳法与不等式, Mathematical Induction and Inequalities  |  Difficulty: Challenging  |  Marks: 20

Problem

6 Let

$S_n = \sum_{r=1}^{n} \frac{1}{\sqrt{r}},$

where $n$ is a positive integer.

(i) Prove by induction that

$S_n \leqslant 2\sqrt{n} - 1.$

(ii) Show that $(4k + 1)\sqrt{k + 1} > (4k + 3)\sqrt{k}$ for $k \geqslant 0$.

Determine the smallest number $C$ such that

$S_n \geqslant 2\sqrt{n} + \frac{1}{2\sqrt{n}} - C.$

Hint

Part (i) explicitly required induction, so there is a standard procedure to follow – check it works when n = 1, assume it works when n = k and show that this leads to it being true when n = k + 1. Only the last part causes any issue. There is some fairly subtle logic: using the n = k assumption it can be shown that

S_{k+1} ≤ 2√k - 1 + 1/√(k+1)

However, what is really needed is: S_{k+1} ≤ 2√(k+1) - 1.

One way in which this can be established (technically a sufficient, but not necessary condition) is if it can be shown that 2√k - 1 + 1/√(k+1) ≤ 2√(k+1) - 1

A bit of tidying and rearranging shows that this is equivalent to 4k² + 4k ≤ 4k² + 4k + 1, which is “obviously” true. You might worry a little about the fact that the equality is never satisfied, but showing that the strict inequality is true is sufficient.

The first part of part (ii) is also just about squaring up and showing that the statement is equivalent to an “obviously” true statement (in this case that 16k³ + 24k² + 9k + 1 > 16k³ + 24k² + 9k). No induction required!

In the final part we first need to come up with a conjecture. A reasonable place to start is to try n = 1, so that 1 ≥ 2.5 - C. For this to work we need that C ≥ 1.5 so C = 1.5 is the smallest value that works for S_1. However will this work for all subsequent S_n too? It turns out that it does, but that requires proving the conjecture

S_n ≥ 2√n + 1/(2√n) - 1.5

This requires induction, following a very similar argument to part (i). One line of the algebra in the proof requires (4k + 1)√(k+1) ≥ (4k + 3)√k, which can be done using the initially unimportant fact at the beginning of part (ii) – always look out for making links between the different parts of questions!

Model Solution

Part (i)

Base case ($n = 1$): $S_1 = 1 \leqslant 2\sqrt{1} - 1 = 1$. True.

Inductive step: Assume $S_k \leqslant 2\sqrt{k} - 1$ for some $k \geqslant 1$. Then

$S_{k+1} = S_k + \frac{1}{\sqrt{k+1}} \leqslant 2\sqrt{k} - 1 + \frac{1}{\sqrt{k+1}}$

We need $2\sqrt{k} - 1 + \frac{1}{\sqrt{k+1}} \leqslant 2\sqrt{k+1} - 1$, i.e.,

$\frac{1}{\sqrt{k+1}} \leqslant 2(\sqrt{k+1} - \sqrt{k})$

Rationalising the right side:

$2(\sqrt{k+1} - \sqrt{k}) = \frac{2}{\sqrt{k+1} + \sqrt{k}}$

We need $\frac{2}{\sqrt{k+1} + \sqrt{k}} \geqslant \frac{1}{\sqrt{k+1}}$, which is equivalent to $2\sqrt{k+1} \geqslant \sqrt{k+1} + \sqrt{k}$, i.e., $\sqrt{k+1} \geqslant \sqrt{k}$. This is true for all $k \geqslant 1$.

Hence $S_{k+1} \leqslant 2\sqrt{k+1} - 1$, completing the induction.

Part (ii)

First result: Both sides are positive for $k \geqslant 0$, so squaring preserves the inequality:

$(4k+1)^2(k+1) > (4k+3)^2 k$

LHS $= (16k^2 + 8k + 1)(k+1) = 16k^3 + 24k^2 + 9k + 1$

RHS $= (16k^2 + 24k + 9)k = 16k^3 + 24k^2 + 9k$

LHS $-$ RHS $= 1 > 0$, so the inequality holds for all $k \geqslant 0$.

Determining $C$: For $n = 1$: $S_1 = 1 \geqslant 2 + \frac{1}{2} - C$ requires $C \geqslant \frac{3}{2}$. We claim $C = \frac{3}{2}$ works for all $n$, i.e.,

$S_n \geqslant 2\sqrt{n} + \frac{1}{2\sqrt{n}} - \frac{3}{2}$

Base case ($n = 1$): $1 \geqslant 2 + \frac{1}{2} - \frac{3}{2} = 1$. True.

Inductive step: Assume the result for $k \geqslant 1$. Then

$S_{k+1} = S_k + \frac{1}{\sqrt{k+1}} \geqslant 2\sqrt{k} + \frac{1}{2\sqrt{k}} - \frac{3}{2} + \frac{1}{\sqrt{k+1}}$

We need:

$2\sqrt{k} + \frac{1}{2\sqrt{k}} + \frac{1}{\sqrt{k+1}} \geqslant 2\sqrt{k+1} + \frac{1}{2\sqrt{k+1}}$

Rearranging:

$2\sqrt{k} - 2\sqrt{k+1} + \frac{1}{2\sqrt{k}} + \frac{1}{\sqrt{k+1}} - \frac{1}{2\sqrt{k+1}} \geqslant 0$

$-2(\sqrt{k+1} - \sqrt{k}) + \frac{1}{2\sqrt{k}} + \frac{1}{2\sqrt{k+1}} \geqslant 0$

We need $\frac{1}{2\sqrt{k}} + \frac{1}{2\sqrt{k+1}} \geqslant 2(\sqrt{k+1} - \sqrt{k})$.

Let $a = \sqrt{k}$, $b = \sqrt{k+1}$ (so $a, b > 0$ and $b > a$). The inequality becomes:

$\frac{1}{2a} + \frac{1}{2b} \geqslant 2(b - a)$

$\frac{a + b}{2ab} \geqslant \frac{2}{a + b}$

Cross-multiplying (all terms positive):

$(a + b)^2 \geqslant 4ab$

$(a - b)^2 \geqslant 0$

which is true, with strict inequality since $a \neq b$. This completes the induction, so $C = \frac{3}{2}$.

Examiner Notes

This question was relatively popular, but it turned out to be one of the hardest of the pure questions. The first part was a reasonably standard example of induction but nearly all candidates failed to understand the subtlety of what was required in the last part. Most candidates made significant progress with part (i), clearly being familiar with the process of induction. However, the algebra to complete the proof was too much for most candidates. Inequalities were frequently handled poorly and the general presentation of logical arguments was unclear with many candidates assuming what was required and not making implications clear. Attempts which brought in calculus were rarely relevant and even less frequently correct.

In the second part, candidates tended to overcomplicate the question. Squaring up (since both sides are clearly positive) and expanding brackets was all that was required. It would have been nice if students had shown some awareness that the squaring process was valid since both sides were positive, however if we had required this it would have effectively been a one mark penalty for all candidates.

In the final part candidates often considered S_1 to find a necessary bound on C. However, further work – usually an induction using their guess – was required to show that this bound works for all n. Many candidates seemed unaware that this final stage was required.

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Question 7

Topic: 指数函数与极限, Exponential Functions and Limits  |  Difficulty: Challenging  |  Marks: 20

Problem

7 The functions $f$ and $g$ are defined, for $x > 0$, by

$f(x) = x^x, \quad g(x) = x^{f(x)}.$

(i) By taking logarithms, or otherwise, show that $f(x) > x$ for $0 < x < 1$. Show further that $x < g(x) < f(x)$ for $0 < x < 1$.

Write down the corresponding results for $x > 1$.

(ii) Find the value of $x$ for which $f'(x) = 0$.

(iii) Use the result $x \ln x \to 0$ as $x \to 0$ to find $\lim_{x \to 0} f(x)$, and write down $\lim_{x \to 0} g(x)$.

(iv) Show that $x^{-1} + \ln x \geqslant 1$ for $x > 0$.

Using this result, or otherwise, show that $g'(x) > 0$.

Sketch the graphs, for $x > 0$, of $y = x$, $y = f(x)$ and $y = g(x)$ on the same axes.

Hint

In this question the difficulty is being able to see that some result is “obviously” true but then having great difficulty in justifying it from particular starting-points: it is not enough to make a true statement (especially when it is given in the question) … one must justify it fully from given, or known, facts and careful deductive reasoning.

Here, in (i), it is known that, for 0 < x < 1, x is positive and ln x is negative. Thus 0 > x ln x > ln x can be deduced by multiplying the first inequality throughout by a negative quantity (remembering that this reverses the direction of inequality signs). This is just ln 1 > ln(x^x) > ln x and, since the logarithmic function is strictly increasing, (1 >) f(x) > x. A more complete argument along similar lines shows that x < g(x) < f(x). The final part requires no further justification; since for x > 1, ln x > 0 we now have x < f(x) < g(x).

For part (ii), it is customary to use logs first and then differentiate implicitly. In (iii), only an informal understanding regarding the justification of limits is expected, but one still should have a grasp as to how things should be set out. Here, something along the lines of lim_{x→0} f(x) = lim_{x→0} e^{x ln x} = lim_{x→0} e^0 = 1 and so lim_{x→0} g(x) = lim_{x→0} x^{f(x)} = lim_{x→0} x^1 = 0 would be expected.

In (iv), the use of calculus is the most straightforward approach, differentiating y = 1/x + ln x for x > 0 and showing that it has a unique minimum turning point at (1, 1). This is then fed in to the derivative of g(x) – again using the logarithmic form and implicit differentiation – along with a simple observation that squares are necessarily non-negative and this all falls nicely into place. Most of what is required in order to sketch x, f(x) and g(x) has already been established and all that is left is to put it together in a sensibly-sized diagram.

Model Solution

Part (i)

For $0 < x < 1$: $\ln x < 0$, so multiplying $0 < x < 1$ by the negative quantity $\ln x$ reverses the inequalities:

$0 > x \ln x > \ln x$

Since $0 = \ln 1$ and $x \ln x = \ln(x^x)$:

$\ln 1 > \ln(x^x) > \ln x$

Since $\ln$ is strictly increasing: $1 > x^x > x$, i.e., $f(x) > x$.

For $g(x) = x^{f(x)}$: since $x < f(x) < 1$ and $\ln x < 0$, multiplying by $\ln x$ reverses the inequalities:

$x \ln x > f(x) \ln x > \ln x$

i.e., $\ln(x^x) > \ln(x^{f(x)}) > \ln x$, so $x^x > x^{f(x)} > x$, giving $x < g(x) < f(x)$.

For $x > 1$: $\ln x > 0$, so $0 < x \ln x < \ln x$ gives $1 < x^x < x$, hence $x < f(x)$.

Since $x > 1$ and $x < f(x)$: $x \ln x < f(x) \ln x$ (both positive), so $x^x < x^{f(x)}$, giving $f(x) < g(x)$.

Therefore: $x < f(x) < g(x)$ for $x > 1$.

Part (ii)

Taking logarithms: $\ln f(x) = x \ln x$. Differentiating:

$\frac{f'(x)}{f(x)} = 1 + \ln x$

$f'(x) = x^x(1 + \ln x)$

Setting $f'(x) = 0$: since $x^x > 0$, we need $1 + \ln x = 0$, giving $x = e^{-1}$.

Part (iii)

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{x \ln x} = e^{\lim_{x \to 0^+} x \ln x} = e^0 = 1$

$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^{f(x)} = \lim_{x \to 0^+} x^1 = 0$

since $f(x) \to 1$ as $x \to 0^+$.

Part (iv)

Proving $x^{-1} + \ln x \geqslant 1$: Let $h(x) = \frac{1}{x} + \ln x$ for $x > 0$.

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

$h'(x) < 0$ for $0 < x < 1$ and $h'(x) > 0$ for $x > 1$, so $h$ has a minimum at $x = 1$:

$h(1) = 1 + 0 = 1$

Therefore $h(x) \geqslant 1$ for all $x > 0$.

Showing $g'(x) > 0$: Taking logarithms: $\ln g(x) = f(x) \ln x = x^x \ln x$. Differentiating:

$\frac{g'(x)}{g(x)} = f'(x) \ln x + f(x) \cdot \frac{1}{x} = x^x(1 + \ln x)\ln x + \frac{x^x}{x}$

$g'(x) = x^x \cdot x^{f(x)} \left(\ln x(1 + \ln x) + \frac{1}{x}\right)$

The bracket is $\frac{1}{x} + \ln x + (\ln x)^2$. Since $\frac{1}{x} + \ln x \geqslant 1$ and $(\ln x)^2 \geqslant 0$:

$\frac{1}{x} + \ln x + (\ln x)^2 \geqslant 1 > 0$

Since $x^x > 0$ and $x^{f(x)} > 0$ for $x > 0$, we conclude $g'(x) > 0$.

Sketch (for $x > 0$):

All three curves pass through $(1, 1)$.

• $y = x$: the straight line through the origin.
• $y = f(x) = x^x$: approaches $1$ as $x \to 0^+$, has a minimum at $x = e^{-1}$ where $f(e^{-1}) = e^{-1/e} \approx 0.692$, then increases through $(1, 1)$ and grows faster than any polynomial for large $x$.
• $y = g(x) = x^{f(x)}$: starts at $0$ as $x \to 0^+$, strictly increasing (since $g'(x) > 0$), passes through $(1, 1)$, and grows even faster than $f(x)$ for $x > 1$.

For $0 < x < 1$: $x < g(x) < f(x) < 1$. For $x > 1$: $1 < x < f(x) < g(x)$.

The ordering from bottom to top switches at $x = 1$: below $x = 1$, $g$ is between $x$ and $f$; above $x = 1$, $f$ is between $x$ and $g$.

Examiner Notes

Just over one thousand candidates attempted this question, but more than 400 of these attempts were not substantial; removing the large number of those scripts which got no further than part (i) raises the mean score from well under 8 to just over 12 out of 20.

The difficulty with questions like this is that it is very easy to make correct statements but much more difficult to support them with logically-crafted steps of reasoning based on results either given or known. Moreover, one needs to reason in such a way that the steps of working one writes down are justified … this was the principal barrier to anything more than the most faltering of starts. So, part (i) was an issue for candidates, with much written but not much of it coherently stated or supported. Of the few marks gained in the weaker attempts, part (ii) provided the majority of them, since most candidates were happy to take logs and then differentiate (the standard procedure for exponential equations of this kind).

It was slightly surprising to note that so few candidates attempted to establish the initial result in part (iv) using calculus; most of those that got this far presumably thought some other ‘inequality’ technique was being tested.

Finally, even for those who had made good progress in several of the previous parts, the graphs at the end were frequently marred by a lack of labelling.

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Question 8

Topic: 向量与三角形几何, Vectors and Triangle Geometry  |  Difficulty: Standard  |  Marks: 20

Problem

8 All vectors in this question lie in the same plane.

The vertices of the non-right-angled triangle $ABC$ have position vectors a, b and c, respectively. The non-zero vectors u and v are perpendicular to $BC$ and $CA$, respectively.

Write down the vector equation of the line through $A$ perpendicular to $BC$, in terms of u, a and a parameter $\lambda$.

The line through $A$ perpendicular to $BC$ intersects the line through $B$ perpendicular to $CA$ at $P$. Find the position vector of $P$ in terms of a, b, c and u.

Hence show that the line $CP$ is perpendicular to the line $AB$.

Hint

Although vectors expressed in general terms are not handled well by the majority of STEP candidates, such questions invariably involve little that is of any great difficulty. If one is sufficiently confident in handling vectors, this question is perhaps the easiest on the paper. The only things involved in this question are the equations of lines in the standard vector form r = p + tq and the use of the scalar product for finding angles (in particular the result that, for non-zero vectors p and q, p·q = 0 ⇔ p and q are perpendicular).

Thus, we have the line through A perpendicular to BC is r = a + λu and the line through B perpendicular to CA is r = b + μv, which meet when a + λu = b + μv ⇒ v = (1/μ)(a - b + λu).

Since v is perpendicular to CA, (a - b + λu)·(a - c) = 0 which leads to a scalar expression for λ and hence a vector expression for p = a + λu.

Next, CP = p - c = a - c + λu, and CP·AB = (a - c + λu)·(b - a) = (a - c)·(b - a) + λu·(b - a).

Now u·(b - c) = 0 since u is perpendicular to BC ⇒ u·b = u·c. Substituting this into CP·AB leads very quickly to the required zero for the perpendicularity result required.

Note: those readers familiar with the common geometric ‘centres’ of triangles will, no doubt, have spotted that this question is about nothing more than the orthocentre of a triangle; that is, the point at which the three altitudes meet. In this question, you are given two altitudes; find their point of intersection, and then show that the line from the third vertex through this point meets the opposite side at right-angles.

Model Solution

Line through $A$ perpendicular to $BC$:

$\mathbf{r} = \mathbf{a} + \lambda \mathbf{u}$

Finding $P$:

The line through $B$ perpendicular to $CA$ is $\mathbf{r} = \mathbf{b} + \mu \mathbf{v}$. At the intersection $P$:

$\mathbf{a} + \lambda \mathbf{u} = \mathbf{b} + \mu \mathbf{v}$

Since $\mathbf{v}$ is perpendicular to $\overrightarrow{CA} = \mathbf{a} - \mathbf{c}$, taking the dot product with $(\mathbf{a} - \mathbf{c})$:

$(\mathbf{a} + \lambda \mathbf{u} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{c}) = 0$

$(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{c}) + \lambda \, \mathbf{u} \cdot (\mathbf{a} - \mathbf{c}) = 0$

$\lambda = -\frac{(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{c})}{\mathbf{u} \cdot (\mathbf{a} - \mathbf{c})}$

Therefore the position vector of $P$ is:

$\mathbf{p} = \mathbf{a} - \frac{(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{c})}{\mathbf{u} \cdot (\mathbf{a} - \mathbf{c})} \, \mathbf{u}$

Showing $CP \perp AB$:

$\overrightarrow{CP} = \mathbf{p} - \mathbf{c} = (\mathbf{a} - \mathbf{c}) + \lambda \mathbf{u}$

$\overrightarrow{CP} \cdot \overrightarrow{AB} = \bigl[(\mathbf{a} - \mathbf{c}) + \lambda \mathbf{u}\bigr] \cdot (\mathbf{b} - \mathbf{a})$

$= (\mathbf{a} - \mathbf{c}) \cdot (\mathbf{b} - \mathbf{a}) + \lambda \, \mathbf{u} \cdot (\mathbf{b} - \mathbf{a})$

Since $\mathbf{u}$ is perpendicular to $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$, we have $\mathbf{u} \cdot \mathbf{b} = \mathbf{u} \cdot \mathbf{c}$, so:

$\mathbf{u} \cdot (\mathbf{b} - \mathbf{a}) = \mathbf{u} \cdot \mathbf{b} - \mathbf{u} \cdot \mathbf{a} = \mathbf{u} \cdot \mathbf{c} - \mathbf{u} \cdot \mathbf{a} = \mathbf{u} \cdot (\mathbf{c} - \mathbf{a}) = -\mathbf{u} \cdot (\mathbf{a} - \mathbf{c})$

Therefore:

$\lambda \, \mathbf{u} \cdot (\mathbf{b} - \mathbf{a}) = -\frac{(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})}{\mathbf{u}\cdot(\mathbf{a}-\mathbf{c})} \cdot \bigl[-\mathbf{u}\cdot(\mathbf{a}-\mathbf{c})\bigr] = -(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})$

Substituting back:

$\overrightarrow{CP} \cdot \overrightarrow{AB} = (\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{a}) - (\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c}) = 0$

since $(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{a}) = -(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})$ by symmetry of the dot product (both equal $-(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})$… let me verify).

Actually: $(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{a})$ and $(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})$.

$(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c}) = \mathbf{a}\cdot\mathbf{a} - \mathbf{a}\cdot\mathbf{c} - \mathbf{b}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{c}$

$(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{a}) = \mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{a} - \mathbf{c}\cdot\mathbf{b} + \mathbf{c}\cdot\mathbf{a}$

$= -(\mathbf{a}\cdot\mathbf{a} - \mathbf{a}\cdot\mathbf{c} - \mathbf{b}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{c}) = -(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})$

So indeed $(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{a}) = -(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c})$, confirming:

$\overrightarrow{CP} \cdot \overrightarrow{AB} = -(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c}) + (\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{c}) = 0$

Hence $CP$ is perpendicular to $AB$. (This shows that the three altitudes of triangle $ABC$ are concurrent at the orthocentre $P$.)

Examiner Notes

The vectors question again proved extremely unpopular, despite the fact that it is perhaps the easiest question on the paper. It drew the least number of attempts from the Pure Maths questions (the only one under half of the entry) and two-thirds of these were not substantial attempts. In this case, it is easy to say what (almost invariably) appeared: candidates generally got no further than the first three marks, which could be gained by writing down two line equations, r = a + λu and r = b + μv, and then equating the two expressions for r. Few made further progress, revealing a reluctance to engage with the algebraic manipulation of vectors (handling numerical vectors is, of course, a completely different matter altogether).