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.

Please be reminded that,

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.

===================================================================

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!]

===================================================================

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)}

===================================================================

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)

Also read
https://johnmayhk.wordpress.com/2007/10/15/alpm-be-careful-with-the-fundamental-theorem-of-integral-calculus/

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.

===================================================================

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.]

===================================================================

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!]

Please be reminded that

(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 | 回覆

  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 | 回覆

  3. sorry, it should be the day after tmr

    迴響 由 Ed — 2008/03/04 @ 7:59 下午 | 回覆

  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 | 回覆

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

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


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