HSC 2014 MX2 Marathon ADVANCED (archive) (5 Viewers)

Sy123 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
. Right idea and pretty close to being done, but we can have a pair of the k's being -1 instead of all of them having to be 1.

Ah yes my bad

$k_1, k_2, k_3 \in \{-1,1 \}$

$\\ $Case 1:$ \ \ k_1 = k_2 = k_3 = 1 \ \Rightarrow \ $(dealt with above)$ \\ \\ $Case 2:$ \ \ $2 of$ \ k_1, k_2, k_3 \ $is$ \ -1 \ $w.l.o.g$ \ k_1=k_2 = -1 \ , \ k_3 = 1 \\ \\ (2) \Rightarrow \ (c-a) = -(b-c) \Rightarrow \ a=b \ \ $but$ \ a,b \ $are distinct, a contradiction!$

$\\ \therefore \square$

Sy123 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\sum_{n=1}^{\infty} (-1)^{\frac{n}{2}(n-1)}\frac{1}{n}$

Davo_01 · Apr 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ \ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + ik} \\ \\ $Where$ \ i = \sqrt{-1}$

Nice question heres my attempt

$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n + ik}$ $= \lim_{n \to \infty} \sum_{k=1}^n \dfrac{n-ik}{n^2 + k^2}=$ $\lim_{n \to \infty} \sum_{k=1}^n \dfrac{n}{n^2 + k^2}- i\lim_{n \to \infty} \sum_{k=1}^n \dfrac{k}{n^2 + k^2}$

$=\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n}\cdot\dfrac{1}{1 + (\frac{k}{n})^2}- i\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n}\cdot \frac{\frac{k}{n}}{1 + (\frac{k}{n})^2}=$ $\int_0^1\dfrac{1}{1+x^2}\cdot dx-i\int_0^1 \dfrac{x}{1+x^2}\cdot dx$

$=\dfrac{\pi-2iln(2)}{4}$

ayecee · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

I had this idea the other day how the spokes on a wheel have no net moment on the wheel, I and wanted to prove it using maths. I liked it so much, I made a question out of it.
If you want more of a challenge, skip straight to the last part ( b ii ), otherwise, the questions should lead you through the problem.
(Let me know if I made some mistakes... I've never made a question like this before, and I'm not perfect)

Enjoy!View attachment wheel.pdf

Davo_01 · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that the average value of a function between the closed interval $[a,b]$ is given by

$\dfrac{\int_a^b f(x)\cdot dx}{b-a}$

RealiseNothing · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Prove that the average value of a function between the closed interval $[a,b]$ is given by

$\dfrac{\int_a^b f(x)\cdot dx}{b-a}$

The average value of a function is just the sum of all values of $f(x)$ in $[a,b]$ , divided by the number of values in that interval.

For simplicity's sake, I'm going to shift the interval to some new interval $[0,k]$ , this will still preserve everything.

Now we add up all value of $f(x)$ in our new interval of $[0,k]$

$\lim_{n \to \infty} \sum_{m=1}^{kn} f(\frac{m}{n})$

However, we must now divide by the amount of values within the interval, i.e.:

$\lim_{n \to \infty} \frac{1}{kn} \sum_{m=1}^{kn} f(\frac{m}{n})$

Noting the Riemann sum we now have gives:

$\frac{1}{k} \int_{0}^{k} f(x)~dx$

Shifting the interval from $[0,k]$ back to $[a,b]$ gives the result:

$\frac{1}{b-a} \int_{a}^{b} f(x)~dx$

Sy123 · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\sum_{n=1}^{\infty} (-1)^{\frac{n}{2}(n-1)}\frac{1}{n}$

Bump.

Davo_01 · Apr 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Bump.

Okay so this is what i came up with, not fully confident but here it is:

$\dfrac{n(n-1)}{2}=0+1+2+...+(n-1)$

Using the fact that an $odd+even=odd$ and $odd+odd=even$
The sum gives o, then an odd number, then another odd, then an even twice and an odd twice and so on...

$\therefore S=1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}+...$

$= [x-\dfrac{x^2}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{4}+\dfrac{x^5}{5}-\dfrac{x^6}{6}-\dfrac{x^7}{7}+... ]_0^1$

$=1-\int_0^1 (x+x^2-x^3-x^4+x^5+x^6-x^7-x^8+...)\cdot dx$

$=1-\int_0^1 \dfrac{x+x^2}{1+x^2}\cdot dx$

$=\dfrac{\pi-2ln(2)}{4}$

Sy123 · Apr 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Okay so this is what i came up with, not fully confident but here it is:

$\dfrac{n(n-1)}{2}=0+1+2+...+(n-1)$

Using the fact that an $odd+even=odd$ and $odd+odd=even$
The sum gives o, then an odd number, then another odd, then an even twice and an odd twice and so on...

$\therefore S=1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}+...$

$= [x-\dfrac{x^2}{2}-\dfrac{x^3}{3}+\dfrac{x^4}{4}+\dfrac{x^5}{5}-\dfrac{x^6}{6}-\dfrac{x^7}{7}+... ]_0^1$

$=1-\int_0^1 (x+x^2-x^3-x^4+x^5+x^6-x^7-x^8+...)\cdot dx$

$=1-\int_0^1(x+x^2+x^5+x^6+x^9+x^{10}+...)-(x^3+x^4+x^7+x^8+x^{11}+x^{12}+...)\cdot dx$

$=1-\int_0^1 \dfrac{x+x^2}{1-x^4}-\dfrac{x^3+x^4}{1-x^4}\cdot dx$

$=1-\int_0^1 \dfrac{x^2+x}{1+x^2}\cdot dx$

$=\dfrac{\pi-2ln(2)}{4}$

Yes well done

seanieg89 · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

1. Do there exist functions f and g defined on the real numbers such that:

$f(g(x))=x^2\textrm{ and } g(f(x))=x^3\textrm{ for all }x.$

2. Do there exist functions f and g defined on the real numbers such that:

$f(g(x))=x^2\textrm{ and } g(f(x))=x^4\textrm{ for all }x.$

Justify your answers.

RealiseNothing · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
1. Do there exist functions f and g defined on the real numbers such that:

$f(g(x))=x^2\textrm{ and } g(f(x))=x^3\textrm{ for all }x.$

Here's my attempt.

Assume that there does exists real functions defined as above. Then:

$f(g(f(x))) = [f(x)]^2$

Since:

$g(f(x)) = x^3$

We can deduce:

$f(x^3) = [f(x)]^2$

Let $x=-1,~0,~1$

$f(-1) = [f(-1)]^2$

$f(0) = [f(0)]^2$

$f(1) = [f(1)]^2$

These are true IFF:

$f(-1) = 0~or~1$

$f(0) = 0~or~1$

$f(1) = 0~or~1$

Now if we consider our original definition of our function:

$g(f(x)) = x^3$

Then we arrive at:

$g(f(-1)) = -1$

So either:

$g(0) = -1 ~or~ g(1) = -1$

Similarly:

$g(f(0)) = 0$

So either:

$g(0) = 0 ~or~ g(1) = 0$

Also:

$g(f(1)) = 1$

So either:

$g(0) = 1 ~or~ g(1) = 1$

Arbitrarily set $g(0) = -1$ and $g(1) = 0$ .

Now from the above possibilities, either $g(0)=1 ~or~ g(1)=1$ .

In these cases, we will reach either:

$g(0) = -1 ~and~ g(0) = 1$ or $g(1) = 0 ~and~ g(1) = 1$

In both cases there is a contradiction as this says that $g(x)$ is NOT a function as for some x-value there exists two y-values.

Thus there does not exist real valued functions such that the given definitions hold.

seanieg89 · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Correct.

You should have a crack at the second

.

RealiseNothing · Apr 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Correct.

You should have a crack at the second .

Soon (if not later tonight then tomorrow).

Enough procrastination for now.

Sy123 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $i) Show that for a convex function$ \ f \ $then$ \\ \\ w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2) \\ \\ $Where$ \ \ w_1 + w_2 = 1 \ \ $for non-negative$ \ w_1, w_2$

$\\ $ii) Hence prove that for some positive integer$ \ n \\ \\ \sum_{j=1}^n w_j f(x_j) \geq f \left(\sum_{j=1}^n w_j x_j \right) \\ \\ $For$ \ \ \sum_{j=1}^n w_j = 1 \ \ \ $for non-negative$ \ \ w_j$

$\\ $iii) Hence prove for positive$ \ \ a_j \\ \\ \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \dots \cdot a_{n-1} \cdot a_n}$

seanieg89 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Assuming a,b,c are positive, then b^2(a^2+c^2), just by expanding and comparing terms.

Sy123 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Assuming a,b,c are positive, then b^2(a^2+c^2), just by expanding and comparing terms.

Yea it was supposed to go in the normal level haha

Davo_01 · May 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $i) Show that for a convex function$ \ f \ $then$ \\ \\ w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2) \\ \\ $Where$ \ \ w_1 + w_2 = 1 \ \ $for non-negative$ \ w_1, w_2$

$\\ $ii) Hence prove that for some positive integer$ \ n \\ \\ \sum_{j=1}^n w_j f(x_j) \geq f \left(\sum_{j=1}^n w_j x_j \right) \\ \\ $For$ \ \ \sum_{j=1}^n w_j = 1 \ \ \ $for non-negative$ \ \ w_j$

$\\ $iii) Hence prove for positive$ \ \ a_j \\ \\ \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \dots \cdot a_{n-1} \cdot a_n}$

A property of convex functions is that the line passing through the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is greater than or equal to the function at every point within the closed interval $[x_1,x_2]$ .

$\therefore y=f(x_1)+\dfrac{f(x_2)-f(x_1)}{x_2-x_1} (x-x_1) \geq f(x)$ in the closed interval $[x_1,x_2]$ .

Substitute $x=\omega_1 x_1+\omega_2 x_2$ where $\omega_1+\omega_2=1$

Simplification gives:

$w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2)$

ii) Using the inequality for convex functions

$f(a)+f(b) \geq 2f(\frac{a+b}{2})$ and showing that:

$w_1 f(x_1) + w_2 f(x_2)+f(x_3) + 2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq f(w_1 x_1+w_2 x_2)+ f(x_3) +2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 2f(\frac{w_1 x_1+w_2 x_2+x_3}{2})+2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 4f(\frac{w_1 x_1+w_2 x_2+x_3+2x_4}{4})+4f(x_5)+...+2^{n-3}f(x_n)$

$...\geq 2^{n-2}f( \frac{w_1 x_1+w_2 x_2+x_3+2x_4+...+2^{n-3}x_n}{2^{n-2}})$

Replacing coefficients of $x_j$ with $\omega_j$

$\therefore \sum_{j=1}^n \omega_j =\dfrac{w_1+w_2+1+2+4+...+2^{n-3}}{2^{n-2}}=1$

$\therefore \sum_{j=1}^n \omega_j f(x_j) \geq f \left(\sum_{j=1}^n \omega_j x_j \right)$

iii) let $f(x)=e^x$ (A convex function)
Define $a_j=e^{x_j}$
Also let $w_j=\dfrac{1}{n}$

Applying part ii

$\therefore \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n}$ $\geq e^{\frac{x_1 + x_2 + \dots + x_{n-1} + x_n}{n}}$ $=\sqrt[n]{e^{x_1}e^{x_2}e^{x_3}...e^{x_n}} = \sqrt[n]{a_1 a_2 a_3 ... a_n}$

edit: Ok i noticed a problem with part ii, what i did only verifies the inequality but does not prove it. Another problem is I let w=1/n in part 3 which isnt consistent with part ii. any suggestions?

Sy123 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
A property of convex functions is that the line passing through the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is greater than or equal to the function at every point within the closed interval $[x_1,x_2]$ .

$\therefore y=f(x_1)+\dfrac{f(x_2)-f(x_1)}{x_2-x_1} (x-x_1) \geq f(x)$ in the closed interval $[x_1,x_2]$ .

Substitute $x=\omega_1 x_1+\omega_2 x_2$ where $\omega_1+\omega_2=1$

Simplification gives:

$w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2)$

ii) Using the inequality for convex functions

$f(a)+f(b) \geq 2f(\frac{a+b}{2})$ and showing that:

$w_1 f(x_1) + w_2 f(x_2)+f(x_3) + 2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq f(w_1 x_1+w_2 x_2)+ f(x_3) +2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 2f(\frac{w_1 x_1+w_2 x_2+x_3}{2})+2f(x_4)+4f(x_5)+...+2^{n-3}f(x_n)$

$\geq 4f(\frac{w_1 x_1+w_2 x_2+x_3+2x_4}{4})+4f(x_5)+...+2^{n-3}f(x_n)$

$...\geq 2^{n-2}f( \frac{w_1 x_1+w_2 x_2+x_3+2x_4+...+2^{n-3}x_n}{2^{n-2}})$

Replacing coefficients of $x_j$ with $\omega_j$

$\therefore \sum_{j=1}^n \omega_j =\dfrac{w_1+w_2+1+2+4+...+2^{n-3}}{2^{n-2}}=1$

$\therefore \sum_{j=1}^n \omega_j f(x_j) \geq f \left(\sum_{j=1}^n \omega_j x_j \right)$

iii) let $f(x)=e^x$ (A convex function)
Define $a_j=e^{x_j}$
Also let $w_j=\dfrac{1}{n}$

Applying part ii

$\therefore \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n}$ $\geq e^{\frac{x_1 + x_2 + \dots + x_{n-1} + x_n}{n}}$ $=\sqrt[n]{e^{x_1}e^{x_2}e^{x_3}...e^{x_n}} = \sqrt[n]{a_1 a_2 a_3 ... a_n}$

edit: Ok i noticed a problem with part ii, what i did only verifies the inequality but does not prove it. Another problem is I let w=1/n in part 3 which isnt consistent with part ii. any suggestions?

Yea I quickly wrote the question up, I thought with part(ii) it would be straightforward induction, but I don't think it is.

Here is one (I did this one)

$\\ $Let$ \ a,b,c \ $be non-negative reals such that$ \ a^2+b^2+c^2 = 8 \\ \\ $Prove that$ \ (4-a)(4-b)(4-c) + abc \geq 16$

RealiseNothing · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Here is one (I did this one)

$\\ $Let$ \ a,b,c \ $be non-negative reals such that$ \ a^2+b^2+c^2 = 8 \\ \\ $Prove that$ \ (4-a)(4-b)(4-c) + abc \geq 16$

Taking the 16 to the LHS, then expanding and simplifying gives:

$(ab+bc+ca)-4(a+b+c) + 12$

However:

$a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) = 8$

$(ab+bc+ca) = \frac{(a+b+c)^2-8}{2}$

Substituting this in and letting $a+b+c=x$ gives:

$\frac{x^2-8}{2} - 4x + 12$

$x^2-8x+16$

$(x-4)^2$

Which is a square number and is thus always positive, hence the required result.

An observation would be to see equality occurs when:

$x=a+b+c=4$

Trebla · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $i) Show that for a convex function$ \ f \ $then$ \\ \\ w_1 f(x_1) + w_2f(x_2) \geq f(w_1x_1 + w_2x_2) \\ \\ $Where$ \ \ w_1 + w_2 = 1 \ \ $for non-negative$ \ w_1, w_2$

$\\ $ii) Hence prove that for some positive integer$ \ n \\ \\ \sum_{j=1}^n w_j f(x_j) \geq f \left(\sum_{j=1}^n w_j x_j \right) \\ \\ $For$ \ \ \sum_{j=1}^n w_j = 1 \ \ \ $for non-negative$ \ \ w_j$

$\\ $iii) Hence prove for positive$ \ \ a_j \\ \\ \frac{a_1 + a_2 + \dots + a_{n-1} + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \dots \cdot a_{n-1} \cdot a_n}$

Sy123, I just wanna say SCREW YOU because this question (worded slightly differently) was a potential consideration for this year's BoS trials

HSC 2014 MX2 Marathon ADVANCED (archive) (5 Viewers)

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