HSC 2015 MX2 Marathon (archive) (1 Viewer)

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

relativity1 said:
Please dont tell me this is a typical 4u question...

Don't worry, it's not.

Glyde · Oct 17, 2015

Re: HSC 2015 4U Marathon

relativity1 said:
Please dont tell me this is a typical 4u question...

Lmao I was thinking the same thing

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

Here is a my HSC style solution:

$\\ $Watch the clock from time$ \ k : 00 \ $to$ \ k:np-1 \ $with$ \ k = 0, 1, \dots, p - 1$

$\\ $At the start of the watch, the hour-hand points to the$ \ nk^{th} \ $mark on the clock, and the minute-hand points to the$ \ 0^{th} \ $mark on the clock.$$

$\\ $Let the amount of time in hours that have passed since the beginning of our clock watching be$ \ t \ $so$ \ 0 \leq t < 1$

$\\ $So then at time$ \ t \ $the hour-hand has progressed$ \ nt \ $marks, and the minute hand has progressed$ \ npt \ $marks clockwise.$$

$\\ $The span between the hour-hand and minute-hand is a half-clock, so$ \ |(nk + nt) - npt| = \frac{np}{2}$

$\\ $Thus,$ \ t = \frac{p+2k}{2(p-1)} \ $for$ \ 0 \leq k < \frac{p}{2} \\ \\ t = \frac{2k-p}{2(p-1)} \ $for$ \ \frac{p}{2} \leq k < p \\ \\ $Note, since$ \ 0 \leq t < 1 \ $then$ \ k = \frac{p}{2} - 1 \ $if$ \ p \ $is even does not yield a valid solution$$

$\\ $At this time, then$ \ nk + nt \ $(the mark the hour-hand is pointing at) and$ \ npt \ $are both integers, implies that the time is straight$$

$\\ $Given conditions in$ \ \alpha \ $then$ \ nk + nt \ $and$ \ npt \ $are all integers for each integer$ \ k$

$\\ $Since there are then$ \ p-1 \ $values of$ \ k \ $(as$ \ \frac{p}{2} - 1 \ $is not valid) then there must be$ \ p - 1 \ $solutions$$

$\\ $Therefore there are$ \ p - 1 \ $times that the time is straight$$

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

relativity1 said:
Please dont tell me this is a typical 4u question...

I think at least part (i) could be, and maybe part (ii) if split into different parts

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

An easier question:

$\\ $The polynomial$ \ z^3 - Az^2 + Az - 1 = 0 \ $has three roots of equal modulus, find the range of real values$ \ A \ $can take.$$

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

$\\ $i) Prove that$ \ e^x \geq x + 1 \ $for$ \ x \geq 0$

$\\ $ii) Use the above inequality to prove that$ \ \sum_{k=1}^n \frac{1}{k} > \ln (1 + n)$

$\\ $iii) Hence prove that (where$ \ a \neq 0) \ \sum_{n=1}^{N} \frac{1}{an + b} \to \infty \ $as$ \ N \ $increases without bound$$

Drsoccerball · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
An easier question:

$\\ $The polynomial$ \ z^3 - Az^2 + Az - 1 = 0 \ $has three roots of equal modulus, find the range of real values$ \ A \ $can take.$$

$$ Since the equation has equal modulus and throught the product of roots it is safe to assume that $ 1, \alpha \frac{1}{\alpha} $ are roots where $ \alpha = cis\theta$

$\sum{\alpha} = A = 1+2cos\theta$

$-1 \leq A \leq 3$

Drsoccerball · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Prove that$ \ e^x \geq x + 1 \ $for$ \ x \geq 0$

$\\ $ii) Use the above inequality to prove that$ \ \sum_{k=1}^n \frac{1}{k} > \ln (1 + n)$

$\\ $iii) Hence prove that (where$ \ a \neq 0) \ \sum_{n=1}^{N} \frac{1}{an + b} \to \infty \ $as$ \ N \ $increases without bound$$

I can't do ii

i and iii straightfoward what do you do for ii ?

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I can't do ii i and iii straightfoward what do you do for ii ?

Transform the inequality in part (i), as it is in the form now, it is useless for part (ii)

Drsoccerball · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Transform the inequality in part (i), as it is in the form now, it is useless for part (ii)

I tried doing that but i get :

$\sum_{k=1}^n{\frac{1}{k}} \geq \sum_{k=1}^n{\frac{1}{e^k-1}}$ Am i doing something wrong?

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ a,b \ $be positive real numbers so that$ \ a + b = 1 \\ \\ $Prove that$ \ \frac{a}{1+a} + \frac{b}{1+b} \leq \frac{2}{3}$

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I tried doing that but i get :

$\sum_{k=1}^n{\frac{1}{k}} \geq \sum_{k=1}^n{\frac{1}{e^k-1}}$ Am i doing something wrong?

Rather use $x \geq \ln (1+x)$
Intuition behind this transformation is that we want an inequality with an ln() function in it because the RHS has a ln() function

Kaido · Oct 18, 2015

Re: HSC 2015 4U Marathon

Show that it's greater than the sigma notation taylor series for e^x?

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I tried doing that but i get :

$\sum_{k=1}^n{\frac{1}{k}} \geq \sum_{k=1}^n{\frac{1}{e^k-1}}$ Am i doing something wrong?

$Note that$

$\ln (n+1) = \ln\left(\frac{2}{1}\times\frac{3}{2}\times\frac{4}{3}\times \cdots \times \frac{n}{n-1}\times \frac{n+1}{n}\right)$

$=\ln \left( \prod _{k=1}^n \frac{k+1}{k}\right)$

$= \ln \left(\prod _{k=1}^n \left(\frac{1}{k}+1\right)\right)$

$$<\ln\left( \prod _{k=1}^n \exp \left(\frac{1}{k} \right)\right)$ (by (i))$

$$= \ln \left(\exp \left(\sum_{k=1} ^n\frac{1}{k} \right)\right)$ (index laws)$

$$=\sum_{k=1} ^n\frac{1}{k}$, as required.$

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Note that$

$\ln (n+1) = \ln\left(\frac{2}{1}\times\frac{3}{2}\times\frac{4}{3}\times \cdots \times \frac{n}{n-1}\times \frac{n+1}{n}\right)$

$=\ln \left( \prod _{k=1}^n \frac{k+1}{k}\right)$

$= \ln \left(\prod _{k=1}^n \left(\frac{1}{k}+1\right)\right)$

$$<\ln\left( \prod _{k=1}^n \exp \left(\frac{1}{k} \right)\right)$ (by (i))$

$$= \ln \left(\exp \left(\sum_{k=1} ^n\frac{1}{k} \right)\right)$ (index laws)$

$$=\sum_{k=1} ^n\frac{1}{k}$, as required.$

Yes, or a more intuitive way of looking at it:

$x \geq \ln (1+x)$

$\frac{1}{k} \geq \ln \left( 1 + \frac{1}{k} \right)$

$\frac{1}{k} \geq \ln \left(\frac{1+k}{k} \right) = \ln ( k+1) - \ln k$

$\sum_{k=1}^{n} \frac{1}{k} > \ln 2 - \ln 1 + \ln 3 - \ln 2 + \dots + \ln (n+1) - \ln n = \ln (1 + n)$

It should be a HSC standard question

Ekman · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Rather use $x \geq \ln (1+x)$
Intuition behind this transformation is that we want an inequality with an ln() function in it because the RHS has a ln() function

After that all I did was:

$$ Let $ x=\frac{1}{k} \therefore \frac{1}{k} \geq \ln{\left(1+\frac{1}{k}\right)} \Rightarrow \frac{1}{k} \geq \ln{(1+k)} - \ln{(k)}$

And you sum both sides to get:

$\sum_{k=1}^{n} \frac{1}{k} \geq \ln{(n+1)} - \ln{(1)}$

Which gets you to the final inequality.

Edit: Nvm Sy did it above

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

With these inequalities, if you can't see how to do it, it may help to start from the statement you want to prove and use biconditional steps (basically if and only if steps) to simplify it:

$\sum_{k=1}^n \frac{1}{k} > \ln (1+n) \Longleftrightarrow \exp \left( \sum_{k=1}^n \frac{1}{k} \right)> 1+n \Longleftrightarrow \prod _{k=1}^n \exp \left( \frac{1}{k} \right)> 1+n$

and this latest equivalent inequality may be easier for you to prove using part (i).

Ekman · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Yes, or a more intuitive way of looking at it:

$x \geq \ln (1+x)$

$\frac{1}{k} \geq \ln \left( 1 + \frac{1}{k} \right)$

$\frac{1}{k} \geq \ln \left(\frac{1+k}{k} \right) = \ln ( k+1) - \ln k$

$\sum_{k=1}^{n} \frac{1}{k} > \ln 2 - \ln 1 + \ln 3 - \ln 2 + \dots + \ln (n+1) - \ln n = \ln (1 + n)$

It should be a HSC standard question

Just out of curiosity, when making the substitution x=1/k, would we need to state conditions for k in order for the inequality to exist? For example, in this question, would we need to state $k \geq 1$

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Just out of curiosity, when making the substitution x=1/k, would we need to state conditions for k in order for the inequality to exist? For example, in this question, would we need to state $k \geq 1$

Better safe than sorry

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Just out of curiosity, when making the substitution x=1/k, would we need to state conditions for k in order for the inequality to exist? For example, in this question, would we need to state $k \geq 1$

$It is assumed from the summation index that $k\geq 1$, so you won't lose marks if you don't state it.$

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