1. Re: HSC 2017 MX1 Marathon

Originally Posted by Drsoccerball
This is even going beyond the course of MX2... It's evil to leave them wandering...

Hint: Instead of counting up count the ways you can finish.
Eh, true I guess.

You can answer after 7 days if you still want to.
Originally Posted by pikachu975
InteGrand probably will leave it alone haha

2. Re: HSC 2017 MX1 Marathon

Originally Posted by leehuan
This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions
$\noindenta_{n}=a_{n-1}+a_{n-3},\\ a_{1}=1\\a_{2}=1\\ a_{3}=2$

not 100% sure though

3. Re: HSC 2017 MX1 Marathon

Originally Posted by jathu123
$\noindenta_{n}=a_{n-1}+a_{n-3},\\ a_{1}=1\\a_{2}=1\\ a_{3}=2$

not 100% sure though
Yep. So the basic idea is which steps can I be on as I take my last step:

$a_n = a_{n-1} ( If my last step is 1 ) + _{n-3} ( If my last step is 3 )$

Extra question: Write a recurrence for the amount of ways I can climb n stairs using 1,2,3... or n step(s).

Should be easy now...

4. Re: HSC 2017 MX1 Marathon

Originally Posted by leehuan
This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions
I just constructed a diagonal line along the rows of Pascal's Triangle and proceeded to count which cells represent a set of possible paths, then sum them all together. Which is basically what si1236 did.

No complicated enumeration required.

5. Re: HSC 2017 MX1 Marathon

Originally Posted by Drsoccerball
Yep. So the basic idea is which steps can I be on as I take my last step:

$a_n = a_{n-1} ( If my last step is 1 ) + _{n-3} ( If my last step is 3 )$

Extra question: Write a recurrence for the amount of ways I can climb n stairs using 1,2,3... or n step(s).

Should be easy now...
Hence, express the general solution as the powers of the roots of the recurrence's characteristic polynomial.

6. Re: HSC 2017 MX1 Marathon

General question. Are you allowed to start on the RHS for Trig Identities? I never have done that before.

7. Re: HSC 2017 MX1 Marathon

Originally Posted by si2136
General question. Are you allowed to start on the RHS for Trig Identities? I never have done that before.
yes

8. Re: HSC 2017 MX1 Marathon

Are you allowed to use MX2 techniques in MX1?

9. Re: HSC 2017 MX1 Marathon

Originally Posted by frog1944
Are you allowed to use MX2 techniques in MX1?
No.

10. Re: HSC 2017 MX1 Marathon

Originally Posted by frog1944
Are you allowed to use MX2 techniques in MX1?
Yes

11. Re: HSC 2017 MX1 Marathon

Originally Posted by si2136
No.
Originally Posted by pikachu975
Yes
What?

12. Re: HSC 2017 MX1 Marathon

Originally Posted by eyeseeyou
What?
There's like barely any 4 unit techniques applicable to 3 unit anyway so doesn't matter

13. Re: HSC 2017 MX1 Marathon

Originally Posted by si2136
No.
Yes you can... But if you do generally the questions become harder.

14. Re: HSC 2017 MX1 Marathon

Originally Posted by Drsoccerball
Yes you can... But if you do generally the questions become harder.
I actually didn't know that o.O

15. Re: HSC 2017 MX1 Marathon

Originally Posted by pikachu975
$\\ \text{i)} \\ \text{This is legit just differentiating to find gradient then using point gradient formula...} \\ \\ \frac{dy}{dx} = \frac{x}{6} \\ \\ \text{Subbing in x = 6p} \\ \\ \frac{dy}{dx} = p \\ \\ \text{Sub into point gradient formula} \\ y - 3p^2 = p(x-6p) \\ y - 3p^2 = px - 6p^2 \\ y = px - 3p^2 \\ \\ \text{ii)} \\ \text{Find where it hits the y axis by subbing in x = 0 to the tangent equation} \\ \text{When x} = 0, y = -3p^2 \\ \text{Then use the internal division formula to find point A} \\ (\frac{1 \times 6p + 2 \times 0}{1 + 2}, \frac{1 \times 3p^2 + 2 \times -3p^2}{1+2}) \\ \\ = (2p, - p^2) \\ \text{Find point A in terms of the parameters then do the usual simultaneous equation to solve it} \\ x = 2p \\ \\ p = \frac{2}{x} \\ \\ \text{Sub into y} \\ y = -(\frac{2}{x})^2 \\ \\ y = \frac{-x^2}{4}$

This took years to type and the answer is probably wrong lol
I think you used internal division in the ratio 2:1 instead of 1:2.

16. Re: HSC 2017 MX1 Marathon

Originally Posted by si2136
I actually didn't know that o.O
Yeah the question was about inverses and areas so i thought I could just use Integration by Parts since I knew how to integrate the integral... 3 pages later gave up on the algebra and just went for the way they wanted...

17. Re: HSC 2017 MX1 Marathon

$\noindent It is given that \sin{\theta} + \cos{\theta} = r, where -\sqrt{2} \le r \le \sqrt{2} for sake of existence. \\\\ Determine the value of: \\\\ a) \sec{\theta} + \csc{\theta} \\\\ b) \tan{\theta} + \cot{\theta} \\\\c) \sqrt{\tan{\theta}} + \sqrt{\cot{\theta}} \\\\ For c), ignore the fact that the square root may not exist.$

18. Re: HSC 2017 MX1 Marathon

$\noindent It is given that \sin{\theta} + \cos{\theta} = r, where -\sqrt{2} \le r \le \sqrt{2} for sake of existence. \\\\ Determine the value of: \\\\ a) \sec{\theta} + \csc{\theta} \\\\ b) \tan{\theta} + \cot{\theta} \\\\c) \sqrt{\tan{\theta}} + \sqrt{\cot{\theta}} \\\\ For c), ignore the fact that the square root may not exist.$
$\noindent Note that (\sin \theta + \cos \theta )^2=1+2\sin \theta\cos \theta \\\Rightarrow \sin \theta \cos \theta = \frac{1}{2}(r^2-1)\\\\ a) \csc \theta + \sec \theta = \frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta }=\frac{2r}{r^2-1}\\\\b) \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}=\frac{2}{r^2-1}\\\\ c) \sqrt{\tan \theta }+\sqrt{\cot \theta}=\frac{\sin \theta+\cos \theta}{\sqrt{\sin \theta \cos \theta}}=\frac{r}{\sqrt{\frac{1}{2}(r^2-1)}} = \frac{\sqrt{2}r}{\sqrt{r^2-1}}$

not sure esp part c

19. Re: HSC 2017 MX1 Marathon

Originally Posted by Drsoccerball
Yep. So the basic idea is which steps can I be on as I take my last step:

$a_n = a_{n-1} ( If my last step is 1 ) + _{n-3} ( If my last step is 3 )$

Extra question: Write a recurrence for the amount of ways I can climb n stairs using 1,2,3... or n step(s).

Should be easy now...
$a_n = a_{n-1} + a_{n-2} + ... + a_1 + a_0$

20. Re: HSC 2017 MX1 Marathon

$\noindent Alice starts tossing a magical coin and wins if she eventually lands a heads. The coin initially starts off with some probability p (0

21. Re: HSC 2017 MX1 Marathon

Originally Posted by InteGrand
$\noindent Alice starts tossing a magical coin and wins if she eventually lands a heads. The coin initially starts off with some probability p (0
P(total) = P(H) + P(TH) + P(TTH) + P(TTTH) + ...
= p + (1-p)kp + ((1-p)^2)(k^2)p + ((1-p)^3)(k^3)p + ((1-p)^4))(k^4)p + ...
= p(1/(1-k(1-p)) (as k and (1-p) are less than one, the ratio is less than one and hence this is true)
= p/(1-k(1-p))

You can check if k=1 then the chance is 1 which makes sense, as eventually she will toss a heads no matter the probability if it is unchanging, and if k=0 then the probability is p, as she can only win if she tosses a head initially, as if she doesn't she can never win.

22. Re: HSC 2017 MX1 Marathon

Originally Posted by calamebe
P(total) = P(H) + P(TH) + P(TTH) + P(TTTH) + ...
= p + (1-p)kp + ((1-p)^2)(k^2)p + ((1-p)^3)(k^3)p + ((1-p)^4))(k^4)p + ...
= p(1/(1-k(1-p)) (as k and (1-p) are less than one, the ratio is less than one and hence this is true)
= p/(1-k(1-p))

You can check if k=1 then the chance is 1 which makes sense, as eventually she will toss a heads no matter the probability if it is unchanging, and if k=0 then the probability is p, as she can only win if she tosses a head initially, as if she doesn't she can never win.
Well done! (And good job on testing the answer, a lot of students don't seem to think to do these kinds of checks (maybe because teachers don't appear to emphasise it).)

23. Re: HSC 2017 MX1 Marathon

$\noindent It is known that if f'(x)\geq 0 and f(0)=0, then f(x)\geq 0 for x> 0.\\\\ Show that \sin x-x+\frac{x^3}{6}\geq 0 for x> 0$

24. Re: HSC 2017 MX1 Marathon

Originally Posted by jathu123
$\noindent It is known that if f'(x)\geq 0 and f(0)=0, then f(x)\geq 0 for x> 0.\\\\ Show that \sin x-x+\frac{x^3}{6}\geq 0 for x> 0$
What's f(x) ?

25. Re: HSC 2017 MX1 Marathon

Originally Posted by Drsoccerball
What's f(x) ?
That part was just a general statement.

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