HSC 2015 MX2 Marathon (archive) (4 Viewers)

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Silly Sausage said:
University mechanics.
Given $v(t)=\frac{g}{k}(1-e^{-kt})dt$

Is this a typo? Is the dt just meant to be t? (Or should the dt just not be there?)

Silly Sausage · Oct 18, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Is this a typo? Is the dt just meant to be t? (Or should the dt just not be there?)

Yep sorry got carried away cause the full question required you to solve a differential equation beforehand

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Silly Sausage said:
Yep sorry got carried away cause the full question required you to solve a differential equation beforehand

haha thought so

kawaiipotato · Oct 18, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
That looks correct.

$Ah cool. I used some relationship that didn't even know existed. Just drew it out and inferred.$

$arg( \frac{-1}{z}) = \frac{-2 \pi}{3}$

$arg(-z) = \frac{2 \pi}{3}$

$arg(z) + \pi = arg(-z)$

$\therefore arg(z) = \frac{- \pi}{3}$

$\therefore $ z must have argument $ \frac{- \pi}{3}$

$|1 - \frac{2}{z}| = 1$

$|\frac{z-2}{z}| = 1$

$|z-2| = |z|$

$\therefore z $ lies on vertical line cutting at $ x = 1 $ with argument $ \frac{- \pi}{3}$

$z = -1 + \sqrt3 i$

omegadot · Oct 18, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
I got z = 1 - sqrt3 i?

$$That's correct. Next question. The system of equations$\\ \begin{align*}|z - 2 - 2i| &= \sqrt{23}\\|z - 8 - 5i| &= \sqrt{38},\end{align*}\\$has two solutions$\,\,z_1,z_2 \in \mathbb{C}.\,\,$Find$\,\, \displaystyle{\frac{z_1 + z_2}{2}}.$

Sy123 · Oct 19, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove for positive reals$ \ x_1,x_2, \dots , x_n \\ \\ \left(\frac{x_1 + x_2 + \dots + x_n}{n \sqrt[n]{x_1x_2\dots x_n}}\right)^n \geq \left( \frac{x_1 + x_2 + \dots + x_{n-1}}{(n-1) \sqrt[n-1]{x_1 x_2\dots x_{n-1}}} \right)^{n-1}$

For a HSC solution without invoking AM-GM:

$\\ $Let$ \ A = x_1 + x_2 + \dots + x_{n-1} \ $and$ \ B = x_1 x_2 \dots x_{n-1}$

$\\ $The inequality to be proven then becomes:$ \ \left(\frac{A + x_n}{n \sqrt[n]{Bx_n}} \right)^n \geq \left(\frac{A}{(n-1)\sqrt[n-1]{B}} \right)^{n-1}$

$\\ $Consider then$ \ f(x) = \frac{1}{n^n} (A+x)^n - \frac{1}{(n-1)^{n-1}} A^{n-1}x$

$\\ $It is sufficient then to prove that$ \ f(x) \geq 0 \ $for$ \ x > 0 \ $because that inequality transformed becomes the original inequality$$

$f'(x) = \frac{1}{n^{n-1}}(A+x)^{n-1} - \frac{1}{(n-1)^{n-1}}A^{n-1}$

$f''(x) = \frac{n-1}{n^{n-1}}(A+x)^{n-2} > 0$

$f'(x) = 0 \Rightarrow \ x = \frac{1}{n-1}A$

$\\ \therefore \left(\frac{1}{n-1}A , f\left(\frac{1}{n-1}A \right) \right) \ $is a minimum$ \\ \therefore f(x) \geq f\left(\frac{1}{n-1}A \right)$

$\\ \Rightarrow \ f(x) \geq \frac{1}{n^n} \left( A + \frac{1}{n-1}A\right)^n - \frac{1}{(n-1)^{n-1}} A^{n-1} \frac{1}{n-1}A = 0$

$\\ \Rightarrow \ f(x) \geq 0$

$\square$

Drsoccerball · Oct 19, 2015

Re: HSC 2015 4U Marathon

Silly Sausage said:
University mechanics.
Given $v(t)=\frac{g}{k}(1-e^{-kt})$
When x=0, t=0, find x(t).
Using series, prove that as k tends towards 0, $v=gt$ and $x=\frac{gt^2}{2}$ .

is v(t) not just velocity ?

InteGrand · Oct 19, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
is v(t) not just velocity ?

Yeah it is. It's asking to find x(t), just integrate basically. But the original Q. involved solving a differential equation to find v(t) I think. And the last part involves Maclaurin series.

kawaiipotato · Oct 19, 2015

Re: HSC 2015 4U Marathon

omegadot said:
$$That's correct. Next question. The system of equations$\\ \begin{align*}|z - 2 - 2i| &= \sqrt{23}\\|z - 8 - 5i| &= \sqrt{38},\end{align*}\\$has two solutions$\,\,z_1,z_2 \in \mathbb{C}.\,\,$Find$\,\, \displaystyle{\frac{z_1 + z_2}{2}}.$

$$ let $ z_{1} = x_{1} + iy_{1} , z_{2} = x_{2} + iy_{2}$

$The conditions given are equivalent to$

$(x-2)^2 + (y-2)^2 = 23 \longrightarrow x^2 - 4x + y^2 - 4y = 15$

$(x-8)^2 + (y-5)^2 = 38 \longrightarrow x^2 - 16x + y^2 - 10y = -51$

$Subtracting bottom equation from the top equation,$

$12x + 6y = 66$

$y = 11-2x$

$Sub. back into the top equation$

$x^2 - 4x + (11-2x)^2 - 4(11-2x) = 15$

$5x^2 - 40x + 62 = 0$

$$ has roots:$ x_{1},x_{2}$

$$ Sum of roots $ = \frac{40}{5} = 8$

$$ So $ \frac{x_{1} + x_{2}}{2} = 4$

$$ Repeat for $ y_{1}, y_{2} $ by subbing $ x = \frac{11-y}{2}$

$$ Giving $ y_{1} + y_{2} = 6$

$\therefore \frac{z_{1} + z_{2}}{2} = 4 + 3i$

glittergal96 · Oct 19, 2015

Re: HSC 2015 4U Marathon

Silly Sausage said:
University mechanics.
Given $v(t)=\frac{g}{k}(1-e^{-kt})$
When x=0, t=0, find x(t).
Using series, prove that as k tends towards 0, $v=gt$ and $x=\frac{gt^2}{2}$ .

$v=\frac{g}{k}(1-e^{-kt})=\frac{g}{k}(1-(1-kt+O(k^2t^2)))\\=\frac{g}{k}(kt+O(k^2t^2))=gt+o(1).\\ \\x=\frac{g}{k}\int_0^t 1-e^{-ks}\, ds=\frac{g}{k^2}(-1+kt+e^{-kt})\\=\frac{g}{k^2}(-1+kt+(1-kt+k^2t^2/2+O(k^3t^3)))=\frac{gt^2}{2}+o(1).$

InteGrand · Oct 19, 2015

Re: HSC 2015 4U Marathon

And of course the physical interpretation is that k represents (something proportional to) some sort of resistance force (air resistance), and as that goes to 0, the ball dropped under gravity has its motion approach that of no resistance.

Sy123 · Oct 19, 2015

Re: HSC 2015 4U Marathon

$\\ $Consider the sum$ \ P_n = \sum_{i=1}^n \frac{1}{p_i} \ $where$ \ p_i \ $is the$ \ i^{th} \ $prime number, so$ \\ p_1 = 2, \ p_2 = 3, \ p_3 = 5, \ $etc.$$

$\\ $i) Prove by mathematical induction that$ \ P_n \ $when simplified into a single fraction, is$ \ P_n = \frac{c_n}{d_n} \ $is such that$ \ c_n \ $is odd and$ \ d_n \ $is even. ($ c_n \ $and$ \ d_n \ $need not be in lowest terms)$$

$\\ $ii) Hence deduce that$ \ P_n \ $is never an integer$$

$\\ $iii) (advanced) Prove similarly or otherwise, that$ \ \sum_{i=1}^n \frac{1}{i} \ $is never an integer for$ \ n > 1$

glittergal96 · Oct 19, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
And of course the physical interpretation is that k represents (something proportional to) some sort of resistance force (air resistance), and as that goes to 0, the ball dropped under gravity has its motion approach that of no resistance.

Yep

.

omegadot · Oct 19, 2015

Re: HSC 2015 4U Marathon

$$Let$\,\,u,v \in \mathbb{C}\,\,$be the roots of the equation$\\z + \frac{1}{z} = 2 (\cos \alpha + i \sin \alpha), \quad 0 < \alpha < \pi.\\$(a) Show that$\,\,u + i\,\,$and$\,\,v + i\,\,$have the same argument and that$\,\,u - i\,\,$and$\,\,v - i\,\,$have the same modulus.\\(b) Find the locus of the roots$\,\,u,v\,\,$when$\,\,\alpha\,\,$varies from 0 to $\,\,\pi.$

godofindolence · Oct 19, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
$Ah cool. I used some relationship that didn't even know existed. Just drew it out and inferred.$

$arg( \frac{-1}{z}) = \frac{-2 \pi}{3}$

$arg(-z) = \frac{2 \pi}{3}$

$arg(z) + \pi = arg(-z)$

$\therefore arg(z) = \frac{- \pi}{3}$

$\therefore $ z must have argument $ \frac{- \pi}{3}$

$|1 - \frac{2}{z}| = 1$

$|\frac{z-2}{z}| = 1$

$|z-2| = |z|$

$\therefore z $ lies on vertical line cutting at $ x = 1 $ with argument $ \frac{- \pi}{3}$

$z = -1 + \sqrt3 i$

Just wondering if this is remotely correct:

$arg( \frac{-1}{z}) = \frac{-2 \pi}{3}$

from this -1/z = -r (1+ isqrt 3) [where r is modulus of -1/z]
rz = 1/ (1+ isqrt3)
r(x+ iy) = (1- isqrt3)/4
but since we know that x =1 can we assume that r = 1/4 and hence y = sqrt 3 ?
thus z = 1- sqrt 3

InteGrand · Oct 19, 2015

Re: HSC 2015 4U Marathon

godofindolence said:
Just wondering if this is remotely correct:

$arg( \frac{-1}{z}) = \frac{-2 \pi}{3}$

from this -1/z = -r (1+ isqrt 3) [where r is modulus of -1/z]
rz = 1/ (1+ isqrt3)
r(x+ iy) = (1- isqrt3)/4
but since we know that x =1 can we assume that r = 1/4 and hence y = sqrt 3 ?
thus z = 1- sqrt 3

This isn't correct. The second line of the working is an incorrect conclusion from the first line; something we could conclude from the first line is that -1/z = r.cis(-2pi/3), where r is the modulus of -1/z (which is the reciprocal of the modulus of z).

godofindolence · Oct 19, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
This isn't correct. The second line of the working is an incorrect conclusion from the first line; something we could conclude from the first line is that -1/z = r.cis(-2pi/3), where r is the modulus of -1/z (which is the reciprocal of the modulus of z).

Ahh cool, so from -1/z = r.cis(-2pi/3), you could get
-1/z = r.[cos(-2pi/3) + isin(-2pi/3)]
which would net you -1/z = -[r(1+ sqrt3)]/2
and re-arrange to get rz =2/(1+ sqrt 3)

Is that and using that fact that we know x= 1 to find r and y correct?
Thank you so much btw!!

Ekman · Oct 20, 2015

Re: HSC 2015 4U Marathon

Next Question (relatively easy one):

$$ For $ x > 0, $ prove: $ x-\frac{x^3}{3} < \tan^{-1}{x} < x-\frac{x^3}{3} + \frac{x^5}{3}$

Sy123 · Oct 20, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Consider the sum$ \ P_n = \sum_{i=1}^n \frac{1}{p_i} \ $where$ \ p_i \ $is the$ \ i^{th} \ $prime number, so$ \\ p_1 = 2, \ p_2 = 3, \ p_3 = 5, \ $etc.$$

$\\ $i) Prove by mathematical induction that$ \ P_n \ $when simplified into a single fraction, is$ \ P_n = \frac{c_n}{d_n} \ $is such that$ \ c_n \ $is odd and$ \ d_n \ $is even. ($ c_n \ $and$ \ d_n \ $need not be in lowest terms)$$

$\\ $ii) Hence deduce that$ \ P_n \ $is never an integer$$

$\\ $iii) (advanced) Prove similarly or otherwise, that$ \ \sum_{i=1}^n \frac{1}{i} \ $is never an integer for$ \ n > 1$

$\\ $i)$ \ P_1 = \frac{1}{2} = \frac{c_1}{d_1} = \frac{\text{odd}}{\text{even}} \ \Rightarrow \ $true for$ \ n = 1$

$\\ $Assume true of$ \ P_k = \frac{c_k}{d_k}$

$\\ P_{k+1} = P_k + \frac{1}{p_{k+1}} = \frac{c_k p_{k+1} + d_k}{d_k p_{k+1}}$

$\\ $Since$ \ p_{k+1} \ $is odd for$ \ k \geq 1 \ $then$ \ p_{k+1} c_k \ $is odd (as$ \ c_k \ $is odd by assumption), and$ \ d_k \ $is even (by assumption), so$ \ d_k + c_k p_{k+1} = c_{k+1} \ $is odd$$

$\\ d_k p_{k+1} = d_{k+1} \ $is even as$ \ d_k \ $is even (by assumption)$$

$\\ \therefore P_{k+1} = \frac{c_{k+1}}{d_{k+1}}$

--------------

$\\ $ii) Since$ \ \frac{c_{k}}{d_k} \ $is of the form odd over, even, it can never simplify to$ \ \frac{N}{1} \ $as$ \ c_k \ $is not divisible by$ \ 2 \ $whereas$ \ d_k \ $is, so$ \ c_k \ $cannot be divisible by$ \ d_k$

-------------

$\\ $iii) In similar fashion to part (i) we are going to infer that$ \ H_n = \sum_{i=1}^n \frac{1}{i} \ $when simplified, is of the form$ \ \frac{\text{odd}}{\text{even}} \ $and use the same inference in part (ii) to conclude that$ \ H_n \ $is never an integer for$ \ n > 1$

$\\ $Moving by induction,$ \ H_2 = \frac{1}{2} \ $which is of the desired form$$

$\\ $Assume true for$ \ H_k = \frac{c_k}{d_k}$

$\\ H_{k+1} = \frac{c_k}{d_k} + \frac{1}{k+1} = \frac{(k+1)c_k + d_k}{(k+1)d_k}$

$\\ $If$ \ (k+1) \ $is odd, then$ \ H_{k+1} \ $is of the desired form (same inference to part (i))$$

$\\ $if$ \ (k+1) \ $is even, then there is some positive integer$ \ p \ $such that$ \ \frac{k+1}{2^p} \ $is an integer, whereas$ \ \frac{k+1}{2^{p+1}} \ $is not (i.e.$ \ p \ $is the amount of times 2 can go into$ \ k+1 $).$$

$\\ $Since$ \ d_k \ $is even, similarly there is a positive integer$ \ q \ $which is the number of times$ \ 2 \ $can go into$ \ d_k$

$\\ $Suppose that$ \ m = \text{min} (p,q)$

$\\ $Then, simplify$ \ H_{k+1} = \frac{(k+1)c_k + d_k}{(k+1)d_k} \Rightarrow H_{k+1} = \frac{\frac{k+1}{2^m} c_k + \frac{d_k}{2^m}}{\frac{(k+1)d_k}{2^m}}$

$\\ $Either$ \ p \neq q \ $or$ \ p = q \ $taking the first case, then:$$

$\\ $In,$ \ H_{k+1} = \frac{\frac{k+1}{2^m} c_k + \frac{d_k}{2^m}}{\frac{(k+1)d_k}{2^m}} \ $the numerator is an integer, that is odd, since either$ \frac{k+1}{2^m} \ $is even AND$ \ \frac{d_k}{2^m} \ $is odd, in which case the numerator must be odd (and the denominator stays even), OR$ \ \frac{k+1}{2^m} \ $is odd AND$ \ \frac{d_k}{2^m} \ $is even, in which case, the numerator is still odd, and the denominator stays even$$

$\\ $Therefore, in the case$ \ p \neq q \ H_{k+1} = \frac{c_{k+1}}{d_{k+1}}$

$\\ $In the case$ \ p = q \ $then$ \ \frac{k+1}{2^m} c_k + \frac{d_k}{2^m} \ $is even, let the amount of times 2 can go into it be$ \ r$

$\\ $Thus, we can simplify$ \ H_{k+1} \ $into$ \ \frac{\frac{1}{2^r} \left( \frac{k+1}{2^m} c_k + \frac{d_k}{2^m} \right) }{\frac{(k+1) d_k}{2^{r+m}}}$

$\\ $Now, it must be the case that the numerator is odd, therefore the only question remaining is if the denominator remains even. The denominator, originally being$ \ d_k (k+1) \ $then 2 can go into it exactly$ \ p + q \ $times, so, if$ \ r < p,q \ $then$ \ \frac{d_k (k+1)}{2^{m+r}} \ $remains even$$

$\\ $Suppose to the contrary, that$ \ r \geq p$

$\\ $Then,$ \ \frac{1}{2^r} \left( \frac{k+1}{2^m} c_k + \frac{d_k}{2^m} \right) = \frac{1}{2^{r+m}} ((k+1)c_k + d_k) \ $is an integer$$

$\\ $The only way this is possible, is if at least one of$ \ (k+1)c_k \ $or$ \ d_k \ $prime factorisation must have the term$ \ 2^{r+m} \ $in it, it is impossible for$ \ d_k \ $to be like this, as it is originally supposed that 2 goes into$ \ d_k \ $exactly$\ q = p = m \ $times, and$ \ r + m > m$

$\\ $It also cannot be$ \ (k+1)c_k \ $as$ \ c_k \ $is odd, and 2 goes into$ \ k+1 \ $exactly$ \ p=q=m \ $times, thus 2 goes into$ \ (k+1)c_k \ $exactly$ \ m \ $times, and$ \ r +m > m$

$\\ $Therefore, it could not be the case that$ \ r \geq p \ $therefore,$ \ r < p \ $therefore (as shown above) the denominator remains even, even when the numerator is odd, therefore for the case$ \ p =q \ $then$ \ H_{k+1} = \frac{c_{k+1}}{d_{k+1}} \ $and since this has been proven for$ \ p \neq q \ $then it is proven for all cases, and the inductive step is complete$$

$\square$

Paradoxica · Oct 20, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Consider the sum$ \ P_n = \sum_{i=1}^n \frac{1}{p_i} \ $where$ \ p_i \ $is the$ \ i^{th} \ $prime number, so$ \\ p_1 = 2, \ p_2 = 3, \ p_3 = 5, \ $etc.$$

$\\ $i) Prove by mathematical induction that$ \ P_n \ $when simplified into a single fraction, is$ \ P_n = \frac{c_n}{d_n} \ $is such that$ \ c_n \ $is odd and$ \ d_n \ $is even. ($ c_n \ $and$ \ d_n \ $need not be in lowest terms)$$

$\\ $ii) Hence deduce that$ \ P_n \ $is never an integer$$

$\\ $iii) (advanced) Prove similarly or otherwise, that$ \ \sum_{i=1}^n \frac{1}{i} \ $is never an integer for$ \ n > 1$

$Hence, prove $H_n$ diverges.$

HSC 2015 MX2 Marathon (archive) (4 Viewers)

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