HSC 2016 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Jul 17, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
If we said alternate angles as the reason for the equality, that'd mean we already know the lines are parallel, but this is what we're trying to show (we don't know it yet). We use corresponding angles in congruent triangles (proof of congruence done in first part of the question) to justify the equality, because we already know about the congruence.

So to keep it simple, if you are trying to prove parallel sides, use corresponding as the reason.

davidgoes4wce · Jul 18, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Q9 from Cambridge text and answer.

Is it ok for Q 9 to say alternate angles instead of corresponding angles as an explanation for the rhombus ?

I wanted to say something along the lines of:

$\angle ABE=\angle CDE $ (alternate angles) $$

$\angle BAE=\angle DCE $ (alternate angles) $$

$\therefore AB \parallel CD.$

$\angle ADE=\angle CBE $ (alternate angles) $$

$\angle DAE=\angle BCE $ (alternate angles) $$

$\therefore AD \parallel CB$

My only thinking of using the word (corresponding) in this example is they are 4 separate triangles within the quadrilateral. But its something im unsure of.

This is also another curious question but is it ok in this question to say

$\angle ADB=\angle CBD \? $ (instead of \angle ADE=\angle CBE) ? $$

davidgoes4wce · Jul 27, 2016

Re: HSC 2016 3U Marathon

2015 HSC Extension I paper, I had a look at this question today. (Want to put it on an open forum for discussion rather than look at the solution)

When is the particle at rest?

Am I right in saying its at rest at x=3 and x=7 cm? (when we set v=0)

Max Speed

$v=\sqrt{11} ?$

InteGrand · Jul 27, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
2015 HSC Extension I paper, I had a look at this question today. (Want to put it on an open forum for discussion rather than look at the solution)

When is the particle at rest?

Am I right in saying its at rest at x=3 and x=7 cm? (when we set v=0)

Max Speed

$v=\sqrt{11} ?$

Those are correct.

davidgoes4wce · Jul 27, 2016

Re: HSC 2016 3U Marathon

I would then say c=5 since it is shifted 5 to the right

Im also aware that this graph has nothing to do with time.

Would you solve 'a' next ? or try to solve 'n' next?

The amplitude from what I gather is a=2 and by substituting the coordinates v^2=11, a=2 and x=5 to solve for 'n', which from my calculation shows $n=\frac{\sqrt{11}}{{2}}$

InteGrand · Jul 27, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
I would then say c=5 since it is shifted 5 to the right

Im also aware that this graph has nothing to do with time.

Would you solve 'a' next ? or try to solve 'n' next?

The amplitude from what I gather is a=2 and by substituting the coordinates v^2=11, a=2 and x=5 to solve for 'n', which from my calculation shows $n=\frac{\sqrt{11}}{{2}}$

$\noindent Yes, $c$ is just $5$. It's probably easier to just get $a$ next, since it is just $2$. This is because $a$ is the amplitude, from the classic S.H.M. equation $v^2 = n^2 \left(A^2 - \left(x-x_0\right) ^2\right)$. So $a$ is just $2$. Now we just need to find $n$, which we can do by subbing in coordinates as you did.$

Paradoxica · Aug 27, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
How many ways can you place 10 different rings on 5 fingers of a hand if none of the fingers can be empty?

Stars and Bars...

14C4

davidgoes4wce · Sep 28, 2016

Re: HSC 2016 3U Marathon

This was the last question in the 2009 HSC

they put the reasoning

$OT^2=OR \times OQ $ (The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point) $$

I didn't see it first time but the external point in this case is at the foot of the wall. Could someone confirm?

InteGrand · Sep 28, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
This was the last question in the 2009 HSC

they put the reasoning

$OT^2=OR \times OQ $ (The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point) $$

I didn't see it first time but the external point in this case is at the foot of the wall. Could someone confirm?

$\noindent There's no $O$ in the diagram (and it should be $S$ rather than $R$ to use that result). So what we could say is $PT^2 = PS \times PQ$.$

$\noindent Edit: oh, yeah, if we define $O$ as the foot of the wall, then that is true.$

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Mice are placed in the centre of a maze which has five exits. Each mouse is equally likely to leave the maze through any one of the five exits. Thus, the probability of any given mouse leaving by a particular exit is 0.2.

Four mice, A,B,C, and D, are put into the maze and behave independently.

Find the probability that any three of the four mice come out the same exit, and the other comes out a different exit.

Find the probability that no more than 2 mice come out the same exit.

$1. \text{P(3 mice)} = \binom{5}{1}\cdot \binom{4}{3}\cdot (\frac{1}{5})^3\cdot \binom{4}{1}(\frac{1}{5})= \frac{16}{125}\\ \\ 2. \text{Using a complimentary relationship P(no more than 2)= 1-P(3 same or 4 same)} \\ =1- \frac{16}{125} - \left ( \binom{5}{1} \cdot \binom{4}{4}\cdot (\frac{1}{5})^4 \right ) \\ =\frac{108}{125}$

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

Might wanna use \left( \right) for brackets around the fractions with the LaTex

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

$\text{ Find}\lim_{n \to \infty}\sqrt{n^2+n}-n$

Drongoski · Sep 29, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$\text{ Find}\lim_{n \to \infty}\sqrt{n^2+n}-n$

= 1/2

But this question is outside 3U.

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drongoski said:
= 1/2

But this question is outside 3U.

It can be done using 3U methods only.

TheBoy · Sep 29, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$\text{ Find}\lim_{n \to \infty}\sqrt{n^2+n}-n$

Complete the square*√((n+0.5)^2 -0.25) - n and as n gets large, the 0.25 is negligent, so you get n+0.5 - n = 0.5. Don't care about absolute values cuz n is large in the positive direction.

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$\text{ Find}\lim_{n \to \infty}\sqrt{n^2+n}-n$

$\begin{align*}\lim_{n \to \infty} \sqrt{n^2+n}-n &= \lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}+n}\\&= \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}+1} \\&= \frac{1}{\sqrt{1+0}+1} \\&= \frac12 \end{align*}$

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

$Find the value of $ \lim_{x \to 0 } \frac{1 - \cos{x}}{x^2} $ (using 3U methods @Paradoxica)$

pikachu975 · Sep 29, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$\begin{align*}\lim_{n \to \infty} \sqrt{n^2+n}-n &= \lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}+n}\\&= \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}+1} \\&= \frac{1}{\sqrt{1+0}+1} \\&= \frac12 \end{align*}$

Just a question, this might be a 3 unit method but I haven't learnt it, but how'd you go from the original equation to the fraction with n as numerator?

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

pikachu975 said:
Just a question, this might be a 3 unit method but I haven't learnt it, but how'd you go from the original equation to the fraction with n as numerator?

It's the opposite of "rationalising" I guess - multiply the denominator and numerator by the "conjugate surd".

Drongoski · Sep 29, 2016

Re: HSC 2016 3U Marathon

pikachu975 said:
Just a question, this might be a 3 unit method but I haven't learnt it, but how'd you go from the original equation to the fraction with n as numerator?

You use:

$\sqrt {n^2 + n} - n = \frac {(\sqrt {n^2+n} - n) (\sqrt {n^2 + n} + n)}{\sqrt {n^2 + n} + n} = \frac {n}{\sqrt {n^2 + n} + n}$

HSC 2016 MX1 Marathon (archive) (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Active Member

New Member

lukewarm mess

lukewarm mess

Premium Member

lukewarm mess

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)