# Quod Erat Demonstrandum

## 2008/03/04

### Something to say about F.7 Pure Mathematics (II) Mock Exam (part 3)

Filed under: HKALE,Pure Mathematics — johnmayhk @ 6:54 下午

Marker’s Report on F.7 Pure Mathematics (II) Mock Examination (Short questions)

Q.1

(a) [Technique: $f < g \Rightarrow \int_a^bf \le \int_a^bg$]
For $n \in \mathbb{N}$, define $I_n = \int_0^1\frac{dx}{1 + x^n}$.
To prove {$I_n$} is an increasing sequence, some students gave $\frac{d}{dx}I_n$.

to prove a sequence {$x_n$} is increasing, we try to show $x_{n+1} \ge x_{n}$;
to prove a differentiable function $f(x)$ is increasing, we try to show $f'(x) \ge 0$ on certain interval.

A sequence may be regarded as a function (on $\mathbb{N}$) indeed.

If you do want to use differentiation, we should consider $\frac{d}{dn}I_n$ instead (i.e. differentiate with respect to $n$). But it does not work in this question.

Some students express ideas not precisely, something like

$x^n$ is decreasing."

It is better to present

“For fixed $x \in [0,1]$, {$x^n$} is decreasing as $n$ increases."

(b) [Technique: monotone sequence theorem]
To show the convergence of the sequence $\int_0^1\frac{x^ndx}{1 + x^n}$, students presented WRONG concepts like

$\because 0 \le \int_0^1\frac{x^ndx}{1 + x^n} \le 1$
$\therefore \frac{x^ndx}{1 + x^n}$ is finite and hence $\int_0^1\frac{x^ndx}{1 + x^n}$ exists.

Note $0 \le a_n \le 1$ =/=> “$\lim_{n\rightarrow \infty}{a_n}$ exists".
All we have to say is {$a_n$} is bounded, but it may not be convergent, e.g. $a_n = \sin(n)$, then $\lim_{n\rightarrow \infty}a_n$ does not exist.

Here is another reminder, the following MAY NOT hold
$lim_{n\rightarrow \infty}\int_0^1f_n(x)dx = \int_0^1\lim_{n\rightarrow \infty}f_n(x)dx$

Here are some good methods given by students.

$\spadesuit x^{n + 1} = x^nx \le x^n$ for $x \in [0 , 1] \Rightarrow x^{n + 1} \le x^n$ on [0 ,1].
$\spadesuit \int_0^1\frac{x^ndx}{1 + 1^n} \le \int_0^1\frac{x^ndx}{1 + x^n} \le \int_0^1\frac{x^ndx}{1 + 0^n} \Rightarrow \frac{1}{2(n + 1)} \le \int_0^1\frac{x^ndx}{1 + x^n} \le \frac{1}{n + 1}$ Taking limit $n \rightarrow \infty$, both upper and lower bounds are zero and by sequeezing principle, $\int_0^1\frac{x^ndx}{1 + x^n} = 0$.

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Q.2
[Technique: Showing differentiability]
To show $f(x)$ is differentiable at $x = a$, we show $\lim_{h\rightarrow 0}\frac{f(a + h) - f(a)}{h}$ exists.
To show $f(x)$ is differentiable at $x = 0$, we show $\lim_{h\rightarrow 0}\frac{f(h) - f(0)}{h}$ exists.

But some students just gave the following WRONG expressions like

$f_-(0) = \lim_{h \rightarrow 0^-}\frac{f(x + h) - f(x)}{h}$

Worse, some students thought that, to prove the differentiability, they tried to show

$\lim_{h\rightarrow 0^+}f'(x) = \lim_{h\rightarrow 0^-}f'(x)$

No! It is something about showing $f'(x)$ is continuous at $x = 0$ (if the common value of the above is $f'(0)$.

Just a reminder,

$f(x)$ is differentiable at $x = a \Rightarrow f(x)$ is continuous at $x = a$ [YES]
$f(x)$ is continuous at $x = a \Rightarrow f(x)$ is differentiable at $x = a$ [NO!]

Some used the l’ Hôpital rule wrongly by writing
$\lim_{h \rightarrow 0^+}\frac{1}{he^{1/h}} = \lim_{h \rightarrow 0^+}\frac{1}{(1/h)he^{1/h} + e^{1/h}}$ [NO!]

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Q.3
(a) [Technique: conormal to parabola]
In general, students can derive equation of normal at $(at_i^2,2at_i)$ to the parabola $y^2 = 4ax$.However, when it comes to show $t_1, t_2, t_3$ are roots of $at^3 + (2a - h)t - k = 0$, students gave

$\because at_i^3 + 2at_i - t_ix - y = 0$ (for $i = 1,2,3$) – – – (*)
$\therefore t_1, t_2, t_3$ are roots of $at^3 + (2a - h)t - k = 0$

NO! (*) is NOT a cubic equation in $t$.

(b) [Relation among roots and trigonometry]
Not many students could observe the following
If $tan(270^o) = \frac{A}{B}$ where $A \ne 0$, then $B = 0$

We come up with the conclusion

$t_1 + t_2 + t_3 = 0$

too often for conormals, but in this question, we need

$t_1t_2 + t_2t_3 + t_3t_1$ and $t_1t_2t_3$ as well, also, we need the following relation

$\tan(x + y + z) = \frac{\tan(x) + \tan(y) + \tan(z) - \tan(x)\tan(y)\tan(z)}{1 - \tan(x)\tan(x) - \tan(y)\tan(z) - \tan(z)\tan(x)}$

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Q.4
(a) [Fundamental theorem of integral calculus]
Many students know that

$\frac{d}{dx}\int_0^xf(t)dt = f(x)$ – – – (*)

However, when the integrand involves something $xt$ or $\frac{x}{t}$, it is better to let $u = xt$ or $u = \frac{x}{t}$ to convert the stuff into something involves $u$ only.

In (*), the $x$ in $\int_0^xf(t)dt$ should be regarded as a constant (with respect to $t$), hence $u = xt \Rightarrow du = xdt$ (not $xdt + tdx$)

Serious problems found, like

$\lim_{x\rightarrow 0}\frac{1}{x^2}\int_0^{2x}\frac{sin(xt)}{t}dt = \lim_{x\rightarrow 0}\frac{1}{x^2}\int_0^{2x}1dt$ [NO!]

At least, $\lim_{x\rightarrow 0}\int f$ may not be equal to $\int\lim_{x\rightarrow 0}f$

(b) [Rationalization]
It is WRONG to write
$\lim_{n \rightarrow \infty}[\sqrt{n + 2} - \sqrt{n + 1}] = \lim_{n \rightarrow \infty}\sqrt{n + 2} - \lim_{n \rightarrow \infty}\sqrt{n + 1}$ [NO!],

Instead, we should use “Rationalization", namely
$\lim_{n \rightarrow \infty}[\sqrt{n + 2} - \sqrt{n + 1}] = \lim_{n \rightarrow \infty}[\sqrt{n + 2} - \sqrt{n + 1}][\frac{\sqrt{n + 2} + \sqrt{n + 1}}{\sqrt{n + 2} + \sqrt{n + 1}}] = \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}$

In using the hint, some wrote

$\lim_{n\rightarrow \infty}\sin^2(\pi\sqrt{n^2 + n + 1})$
$= \lim_{n\rightarrow \infty}\sin^2(\pi\sqrt{n^2 + n + 1} - m)$ for $m \in \mathbb{Z}$ (instead of $\lim_{n\rightarrow \infty}\sin^2(\pi\sqrt{n^2 + n + 1} - n)$) and thus they failed to obtain correct answer.

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Q.5
(a) Few students forget to “+C" for the indefinite integration.
(b) [Technique: Taking logarithm, Riemann sum]
Let $y = (\frac{_{3n}C_{n}}{_{2n}C_{n}})^{\frac{1}{n}}$, students are required to evaluate $\lim_{n\rightarrow \infty}y$.

Many students tried to take logarithm and problems occured, including

$\ln(y) = \frac{1}{n}(\frac{_{3n}C_{n}}{_{2n}C_{n}})$ [NO!]
$\ln(y) = \frac{1}{n} \frac{\ln(3) + \ln(3 - \frac{1}{n}) + \dots + \ln(2 + \frac{1}{n})}{\ln(2) + \ln(2 - \frac{1}{n}) + \dots + \ln(1 + \frac{1}{n})}$ [NO!] [Note: $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$, NOT $\frac{\ln(a)}{\ln(b)}$]

Many students tried to use l’ Hôpital rule at the beginning and obtained nothing correct.

Few students could find the value of $\lim_{n\rightarrow \infty}\ln(y)$ correctly, but they forgot to evaluate $\lim_{n\rightarrow \infty}y$.

Not many students could write down the corresponding definite integral for evaluating the limit in question correctly.

Some wrote

$\int_0^3\ln(x)dx = [x\ln(x)]_0^3 - 3 = 3\ln(3) - 3$

Here, we should pay attention to the fact that $x\ln(x)$ is undefined at $x = 0$.
To obtain $\int_0^3\ln(x)dx$, we need to write
$\int_0^3\ln(x)dx = \int_{0^+}^3\ln(x)dx = \lim_{h\rightarrow 0^+}[x\ln(x)]_h^3 - 3$
and $lim_{h\rightarrow 0^+}x\ln(x) = 0$ (easily verified by l’ Hôpital rule)
[Note: the above is a kind of improper integral which is out-of-syllabus already; but, in the suggested solution, it is not necessary to come up with that improper integral.]

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Q.6
(a) [Technique: For differentiable $g(x)$, $g(x)$ is increasing iff $g'(x) \ge 0$]

The major problem was

letting $g(t) = pf(t) + (1 - p)f(y) - f(pt + (1 - p)y)$, then many students wrote
$g'(t) = pf'(t) + (1 - p)f'(y) - f'(pt + (1 - p)y)$ [NO!]

(1) “$g(t) = pf(t) + (1 - p)f(y) - f(pt + (1 - p)y)$" means “$t$ is the ONLY variable", that is, $y$ is a constant with respect to $t$. Thus, $\frac{d}{dt}f(y) = 0$
(2) $g'(t)$ means differenting $g(t)$ with respect to $t$, hence, the third term on the R.H.S. of the equation should be $\frac{d}{dt}f(pt + (1 - p)y)$ and we need Chain Rule to proceed the calculation, namely

$\frac{d}{dt}f(pt + (1 - p)y) = f'(pt + (1 - p)y)\frac{d(pt + (1 - p)y)}{dt} = pf'(pt + (1 - p)y)$

More explicitly,

$f'(pt + (1 - p)y) = \frac{d}{d(pt + (1 - p)y)}f(pt + (1 - p)y)$

Thus,

$g'(t) = pf'(t) - pf'(pt + (1-p)y)$

Some careless and minor mistakes included

$g''(x) \ge 0 \Rightarrow g(x)$ is increasing. [NO!]
$\frac{d}{dt}f(pt + (1 - p)y) = f'((pt + (1 - p)y)p) = f'(p^2t + p(1 - p)y)$ [NO!]

(b) Many students could give correct injective function, however, when using $x^2$ (say) as an example, they should pay attention to the domain. If the domain is $(-1,1)$ (say), the example is not correct. Hence it may be better to give something like $x^{2n - 1}$ ($n \in \mathbb{N}$) to fit the general domain $(a , b)$.

## 5 則迴響 »

1. I read some super-easy questions here, alrite…
by the way you must have heard about the ‘stuff’ on 13th of March at lunchtime, you think you’re joining it?
Thanks for your fifty ^^ Dunno if you wanna come?

迴響 由 Ed — 2008/03/04 @ 7:23 下午 | 回應

2. by the way are you coming to see our class typing competition tomorrow, there would be many of our classes… see if you have time

迴響 由 Ed — 2008/03/04 @ 7:31 下午 | 回應

3. sorry, it should be the day after tmr

迴響 由 Ed — 2008/03/04 @ 7:59 下午 | 回應

4. OIC, the fifty, yes, I just wanna give chance, but not everyone can receive that fifty…

The ‘saying goodbye’ gathering on 13/3? Is it convenient for me to stay? F.2V may have lot of ‘sharing’ with the substitution-Chan, right?

May be I have no time to join the typing competition, just try the best!

迴響 由 johnmayhk — 2008/03/04 @ 9:56 下午 | 回應

5. Could you upload your mock paper which is being discussed here???

迴響 由 Jonny — 2011/03/21 @ 12:00 上午 | 回應