Author Topic: 4U Maths Question Thread (Read 665199 times) Tweet Share

amandali · « **Reply #525 on:** September 26, 2016, 08:45:05 pm »

for part d ii) do i always need to divide 3! (factorial) because there are same number of people in all the groups

how to do part c)

katherine123 · « **Reply #526 on:** September 27, 2016, 03:48:55 am »

how to do part iii)

RuiAce · « **Reply #527 on:** September 27, 2016, 08:51:46 am »

Quote from: katherine123 on September 27, 2016, 03:48:55 am

how to do part iii)

$\text{You need to do a bit of construction here.}\\ \text{The question gives you a cyclic pentagon}\\ \text{But to use part (ii) you really want a cyclic quadrilateral.}$
$\text{You can choose to either use the cyclic quadrilateral they gave you}\\\text{or make your own by treating it as a diagonal. I will make my own.}$

$\text{I have constructed two more diagonals (aka chords) in blue.}$
$\text{Now go back to part (ii). The formula they gave was}\\ BD \times AC = AD \times BC + AB \times DC\\ \text{We could label our pentagon, but I'm just going to rewrite it like this:}\\ \textit{diagonal} \times \textit{diagonal} = \textit{opposite sides multiplied} + \textit{other opposite sides multiplied}\\ \text{Do you see how?}$
$\text{Return to our pentagon. I will completely ignore that triangle on the left.}\\ \text{You'll noticed how I shaded in grey what I want you to consider.}\\ \text{That shaded area is nothing more but a cyclic quadrilateral, so we can use part (ii)}$
$\text{The formula is }\\ \textit{diagonal} \times \textit{diagonal} = \textit{opposite sides multiplied} + \textit{other opposite sides multiplied}$
$\text{Carefully subbing what we have in we get}\\ x \times x = 1 \times 1 + 1 \times x\\ \textit{Do you see why? Recall that for a regular pentagon, all sides equal}\\ \textit{and all diagonals equal.}$
$\text{Solving the quadratic gives the answer.}$

RuiAce · « **Reply #528 on:** September 27, 2016, 08:55:57 am »

Quote from: amandali on September 26, 2016, 08:45:05 pm

(Image removed from quote.)

for part d ii) do i always need to divide 3! (factorial) because there are same number of people in all the groups

(Image removed from quote.)
how to do part c)

I think the division by 3! is because of the issue with ordering here.

Suppose you had persons A, B, C, ..., L
And you had groups 1, 2, 3

Group 3 having A B C D and group A having E F G H
is the same outcome as group 3 having E F G H and group A having A B C D.

Edit: On further thought, it's probably tied down with the fact there's the same people in each group as well.
____________________________________________________
$\text{Binomial theorem expansion is just }\\ (\cos\theta+i\sin \theta)^5\\=\cos^5\theta + \binom{5}{1}i\cos^4\theta \sin\theta + \binom{5}{2}i^2\cos^3\theta \sin^2 \theta+\binom{5}{3}i^3\cos^2\theta\sin^3\theta+\binom{5}{4}i^4\cos\theta\sin^4\theta+i^5\sin^5\theta$
I can't see the rest of the question (I know where it's headed though. Just equate real or imaginary parts for part (ii), picking whichever one is appropriate.)

xXCandyDXx · « **Reply #529 on:** September 29, 2016, 01:32:15 pm »

Hi everyone !!! I know this may sound a bit basic but I am kind of confused ... With relation to resisted motion in a horizontal line and vertically in Mechanics ... When do we use -mkv and just -kv??? I can see in some working out that mass does actually make a difference even though it cancels out in some cases ... I'm just confused as to when I'm supposed to add the m in -kv or -kv^2.... Yeah that's my question thanks !!!

RuiAce · « **Reply #530 on:** September 29, 2016, 01:44:51 pm »

Quote from: xXCandyDXx on September 29, 2016, 01:32:15 pm

Hi everyone !!! I know this may sound a bit basic but I am kind of confused ... With relation to resisted motion in a horizontal line and vertically in Mechanics ... When do we use -mkv and just -kv??? I can see in some working out that mass does actually make a difference even though it cancels out in some cases ... I'm just confused as to when I'm supposed to add the m in -kv or -kv^2.... Yeah that's my question thanks !!!

They either give it to you, or they will say something along the lines of "force per unit mass".

Generally, the HSC is nice enough to just give you it and you can easily guess which to use. But this doesn't happen in textbooks.

In general, when "per unit mass" is mentioned, you are using mkv.

xXCandyDXx · « **Reply #531 on:** September 29, 2016, 02:28:58 pm »

Quote from: RuiAce on September 29, 2016, 01:44:51 pm

They either give it to you, or they will say something along the lines of "force per unit mass".

Generally, the HSC is nice enough to just give you it and you can easily guess which to use. But this doesn't happen in textbooks.

In general, when "per unit mass" is mentioned, you are using mkv.

Ohh.. Alright THANKS !!

katherine123 · « **Reply #532 on:** October 01, 2016, 08:50:41 pm »

how to do part a i)

RuiAce · « **Reply #533 on:** October 01, 2016, 10:30:49 pm »

Quote from: katherine123 on October 01, 2016, 08:50:41 pm

how to do part a i)

$\text{Note that the ordering will definitely not matter here.}\\ \text{There are }3n\text{ balls in the urn, so our total outcomes is }\binom{3n}{3}$
$\text{For our favourable outcomes, we first choose a colour out of the three.}\\ \text{Then we need to draw out 3 balls of the same colour (so order doesn't matter)}\\ \text{So we have }3\binom{n}{3}$
$\text{Thus the probability is}\\ \begin{align*}\frac{3\binom{n}{3}}{\binom{3n}{3}}&=\frac{\frac{3n!}{6(n-3)!}}{\frac{(3n)!}{6(3n-3)!}}\\ &= \frac{(n-1)(n-2)}{(3n-1)(3n-2)}\end{align*}$
$\textit{Alternatively, in particular i) can be analysed more simplistically}\\ \text{The colour of the first ball does not matter.}\\ \text{Only the second and third will matter. So we just get }\frac{n-1}{3n-1}\times\frac{n-2}{3n-2}$
Answers to the 2008 HSC exam may be found here

Neutron · « **Reply #534 on:** October 04, 2016, 05:22:49 pm »

Hey!

Could you guys please help me with the last question of the 2011 paper? Thanks, I don't even understand the question haha

Neutron

RuiAce · « **Reply #535 on:** October 04, 2016, 07:34:03 pm »

Quote from: Neutron on October 04, 2016, 05:22:49 pm

Hey!

Could you guys please help me with the last question of the 2011 paper? Thanks, I don't even understand the question haha

Neutron

$\text{This was quite a challenging question that they asked in 2011.}\\ \text{A significant minority got out c) (i) so they may have been too scared to continue.}$
$\text{Recall the triangle inequality }|a_1+a_2|\le |a_1|+|a_2|\\ \text{As a matter of fact, this inequality can be }\textit{generalised to more variables}\\ |a_1+a_2+a_3+\dots+a_n| \le |a_1|+|a_2|+|a_3|+\dots+|a_n|$
$\text{Since }\beta\text{ is a zero of }P(z)\text{, we can sub in and equate with 0.}\\ \begin{align*}\beta^n+a_{n-1}\beta^{n-1}+\dots+a_1\beta+a_0&=0\\ \beta^n&=-(a_{n-1}\beta^{n-1}+a_{n-2}\beta^{n-2}+\dots+a_1\beta+a_0) \end{align*}$
$\text{Taking absolute values and applying the triangle inequality we have:}\\ \begin{align*}|\beta|^n&= |a_{n-1}\beta^{n-1}+a_{n-2}\beta^{n-2}+\dots+a_1\beta+a_0|\\ & \le |a_{n-1}|\beta|^{n-1}+|a_{n-2}||\beta|^{n-2}+\dots+|a_1||\beta|+|a_0|\end{align*}\\ \text{Notes: }|x|^n=|x^n|, \, |ab|=|a||b|$
$M\text{ being the maximum value of }|a_{n-1}|, \dots, |a_0|\text{ means that}\\ \text{for ALL values of }k=0,1,\dots,n-1\\ M \ge |a_k|$
$\text{So if we substitute this in, we have}\\ \begin{align*}|\beta|^n&\le M|\beta|^{n-1}+M|\beta|^{n-2}+\dots+M|\beta|+M\\ &=M(|\beta|^{n-1}+|\beta|^{n-2}+\dots+|\beta|+1)\end{align*}$
_____________________________________________________
$\text{Take a long hard look at the expression in (i).}\\ \text{What do we do when we see a }\textbf{geometric progression}?\\ \text{We're tempted to sum it.}$
$\begin{align*}|\beta|^n& \le M\left(\frac{|\beta|^n-1}{|\beta|-1}\right)\\ &< \frac{M|\beta|^n}{|\beta|-1}\end{align*}\\ \text{Here I removed the }-1\text{ before cross multiplying and dividing. You may do it in a different}\\ \text{order if it makes more sense to you that way.}$
$\text{Obviously }\beta \neq 0\text{ so }1 < \frac{M}{|\beta| -1}\\ \implies |\beta| - 1 < M$
$\textit{Remark: We multiplied }|\beta|-1\textit{ assuming that }|\beta|-1 > 0\\ \textit{However, if }|\beta| -1 < 0\textit{, then }|\beta| - 1 < M\\ \textit{ is obvious as }M> 0$

RuiAce · « **Reply #536 on:** October 04, 2016, 07:46:50 pm »

aoifera · « **Reply #537 on:** October 04, 2016, 08:57:15 pm »

Would I be able to get some help with this question:
"A stone of mass 6kg is tied at one end of a 3 metre long strong. The other end is fixed to a point O. Find the tension in the string when the mass is rotating at 40 revolutions per minute"
Thank you

jakesilove · « **Reply #538 on:** October 04, 2016, 10:07:00 pm »

Quote from: aoifera on October 04, 2016, 08:57:15 pm

Would I be able to get some help with this question:
"A stone of mass 6kg is tied at one end of a 3 metre long strong. The other end is fixed to a point O. Find the tension in the string when the mass is rotating at 40 revolutions per minute"
Thank you

Hey! You just need to find the force exerted by the stone. To do this, we calculate centripetal force, which has formula mv^2/r. We have all values except velocity, but this is something you can easily figure out by knowing the radius of the circle, and how many revolutions occur each minute! If you need any further help, post your working out and I'll see what I can do

Jake

nay103 · « **Reply #539 on:** October 05, 2016, 11:33:26 pm »

Hi! I'm a bit confused about this 2011 HSC question:
8(b) A bag contains seven balls numbered from 1 to 7. A ball is chosen at random
and its number is noted. The ball is then returned to the bag. This is done a total
of seven times.
(i) What is the probability that each ball is selected exactly once?
(ii) What is the probability that at least one ball is not selected?
(iii) What is the probability that exactly one of the balls is not selected?

I am fine with parts i and ii, but iii had me a bit confused, and I'm not really sure I quite understand the given answer. My answer was
7/7x6/7x5/7x4/7x3/7x2/7x6/7 -> i.e. probability of six balls being selected exactly once, then one of those six balls being selected again. I don't see what is incorrect with this, if you could explain that would be great!

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amandali

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