Author Topic: Mathematics Question Thread (Read 1626663 times) Tweet Share

RuiAce · « **Reply #2955 on:** October 21, 2017, 06:39:26 pm »

Quote from: sidzeman on October 21, 2017, 06:35:41 pm

Hey could someone help me out with this please

$\text{If }y=f[g(x)]\text{ then by the chain rule,}\\ \boxed{\frac{dy}{dx} = g^\prime(x) \times f^\prime[g(x)]}$
$\text{For }x=2\text{ we know that }g^\prime(2) = -2\\ \text{We also know that }g(2) = 1\\ \text{Hence at }x=2, \, \frac{dy}{dx} = -2 \times f^\prime(1)$
$f^\prime(1) = 3\\ \text{So }\frac{dy}{dx} = -2\times 3 = -6.$

sidzeman · « **Reply #2956 on:** October 21, 2017, 07:01:29 pm »

Quote from: RuiAce on October 21, 2017, 06:39:26 pm

$\text{If }y=f[g(x)]\text{ then by the chain rule,}\\ \boxed{\frac{dy}{dx} = g^\prime(x) \times f^\prime[g(x)]}$
$\text{For }x=2\text{ we know that }g^\prime(2) = -2\\ \text{We also know that }g(2) = 1\\ \text{Hence at }x=2, \, \frac{dy}{dx} = -2 \times f^\prime(1)$
$f^\prime(1) = 3\\ \text{So }\frac{dy}{dx} = -2\times 3 = -6.$

ahhhh I see thank you so much!

winstondarmawan · « **Reply #2957 on:** October 21, 2017, 07:51:39 pm »

Hello, just a little confused as to why Simple Harmonic Motion is in the 2009 HSC..
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22690507_1356345717824281_1978276958_o.png?oh=1e9877b120499b8eb20278f75eb94eb5&oe=59EDA08E

mxrylyn · « **Reply #2958 on:** October 21, 2017, 07:55:03 pm »

Hey.
This a some prelim revision but, How do you prove that and absolute value of a + b is greater than or equal to the absolute value of a + the absolute value of b?

RuiAce · « **Reply #2959 on:** October 21, 2017, 07:57:10 pm »

Quote from: winstondarmawan on October 21, 2017, 07:51:39 pm

Hello, just a little confused as to why Simple Harmonic Motion is in the 2009 HSC..
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22690507_1356345717824281_1978276958_o.png?oh=1e9877b120499b8eb20278f75eb94eb5&oe=59EDA08E

It's weird and it most certainly is SHM (and I said the same thing when I did 2U), but nah it's fine. It's only problematic once they start throwing stuff like $ \ddot{x} = -n^2 x $ in there, which they didn't.

So long as everything remains a function of time in the whole question (as opposed to a function of displacement), it will be fine. Period and max/min (range of trig functions) is technically still 2U material.

RuiAce · « **Reply #2960 on:** October 21, 2017, 07:59:42 pm »

Quote from: mxrylyn on October 21, 2017, 07:55:03 pm

Hey.
This a some prelim revision but, How do you prove that and absolute value of a + b is greater than or equal to the absolute value of a + the absolute value of b?

It's not. It's less than.

mxrylyn · « **Reply #2961 on:** October 21, 2017, 08:03:52 pm »

Oh yeah, sorry. I write the questions wrong.
How do you prove that it is less than?

RuiAce · « **Reply #2962 on:** October 21, 2017, 08:15:06 pm »

$\text{The only thing you have at your disposal is the definition of the absolute value}\\ |x| = \begin{cases}x&\text{if }x \ge 0\\ -x &\text{if }x < 0\end{cases}$
$\text{We note that if either }a=0\text{ or }b=0\text{ (or both)}\\ \text{Then clearly }|a+b| = |a|+|b|\text{ just because of the 0}\\ \text{so for the remainder, we assume that neither }a\text{ nor }b\text{ are 0.}$
_____________________________________

Because this is not a 4U course, the only approach we can take is dealing with cases
$\text{Case 1: }\boxed{a > 0\text{ and }b > 0}\\ \text{Note that the sum of two positive numbers is positive, so }a+b > 0\\ \text{Then, }|a+b| = a+b\\ \text{and }|a|+|b| = a+b\\ \text{so for this case, }\boxed{|a+b| = |a|+|b|}$
$\text{Case 2: }\boxed{a < 0\text{ and }b < 0}\\ \text{Similar to the above case, }a-b < 0\\ \text{so }|a+b| = -(a+b) = -a-b\\ \text{and }|a|+|b| = (-a) + (-b) = -a-b\\ \text{so for this case, }\boxed{|a+b| = |a|+|b|}\text{ as well}$
_____________________________________
$\text{Case 3: }\boxed{a > 0\text{ and }b < 0}\\ \text{This will require subcases}$
$\text{First, suppose that }\boxed{|a| > |b|}\\ \text{If }|a| > |b|\textbf{ and }b < 0 < a\\ \text{we can deduce that }a + b = a - (-b) > 0$
$\text{So for this subcase, we have}\\ |a+b| = a+b\\ \text{but }|a|+|b| = a + (-b) = a-b$
$\text{This is the tricky part:}\\ \textbf{Because }b\textbf{ is negative}\\ \text{we can deduce that }\boxed{a + b < a - b}$
$\text{Hence, for this subcase, }|a+b| < |a| + |b|$
____
$\text{The other subcase is when }|b| > |a|\\ \text{Under these conditions, we see that } a + b < 0.$
$\text{So we will now have}\\ |a + b| = -(a+b) = -a-b\\ \text{and }|a| + |b| = a-b$
$\text{Since }a\text{ is positive, it is clear that }\boxed{-a-b < a-b}\\ \text{so for this case, }|a+b| < |a|+|b|\text{ as well}$
$\text{Regrouping all subcases, it appears as though for case 3}\\ \text{indeed we have }|a+b| < |a| + |b|$
_____________________________________
$\text{Case 4: }\boxed{a < 0\text{ and }b > 0}\\ \text{This is just case 3 but with the roles of }a\text{ and }b\text{ swapped}\\ \text{so similarly to case 3, we can guarantee that }|a+b| < |a|+|b|.$
_____________________________________
$\textbf{Recollecting every case,}\text{ it follows that}\\ |a+b| \le |a|+|b|$

mxrylyn · « **Reply #2963 on:** October 21, 2017, 08:53:47 pm »

THANK YOU!
If this as a 4Unit question, would I have to answer it differently?

RuiAce · « **Reply #2964 on:** October 21, 2017, 09:05:00 pm »

Quote from: mxrylyn on October 21, 2017, 08:53:47 pm

THANK YOU!
If this as a 4Unit question, would I have to answer it differently?

The 4U method is actually in the 4U thread right now.

It relies on something you can't really do in 2U, because the number line is what's called "degenerate".

sidzeman · « **Reply #2965 on:** October 21, 2017, 09:37:58 pm »

Can someone please help with part ii and iii

hobocop · « **Reply #2966 on:** October 21, 2017, 09:53:23 pm »

Hi, could I get an explanation on why you need to integrate 't' for part ii of this question?
This is Q9 b of 2011 HSC paper.

Thanks

RuiAce · « **Reply #2967 on:** October 21, 2017, 09:56:31 pm »

Quote from: sidzeman on October 21, 2017, 09:37:58 pm

Can someone please help with part ii and iii

$\text{This is a growing annuity.}\\ \text{Each successive payment will be }\textbf{1.6 times greater}\\ \text{than the previous one.}$
Note that 6% is also the interest rate. So some really nice things happen in the computations and we dodge the need for a geometric series.
$\text{Let }A_n\text{ be the amount after }n\text{ years}\\ \text{Note that the payment itself is at the }\textbf{start}\text{ of the year.}$
The payment at the start of year 1 is 1000
The payment at the start of year 2 is 1000(1.06)
The payment at the start of year 3 is 1000(1.06)^2
...so...
\begin{align*}A_1 &= 1000(1.06)\\ A_2& = [A_1 + 1000(1.06)](1.06)\\ &=1000(1.06)^2 + 1000(1.06)^2 \\ A_3 &=[A_2 + 1000(1.06)^2](1.06)\\ &= 1000(1.06)^3+1000(1.06)^3+1000(1.06)^3 \end{align*}
$\text{Notice the pattern?}\\ A_n = \underbrace{1000(1.06)^n + \dots + 1000(1.06)^n}_{n\text{ terms}} = \boxed{1000n(1.06)^n}$

RuiAce · « **Reply #2968 on:** October 21, 2017, 10:01:50 pm »

Quote from: hobocop on October 21, 2017, 09:53:23 pm

Hi, could I get an explanation on why you need to integrate 't' for part ii of this question?
This is Q9 b of 2011 HSC paper.

Thanks

$t\text{ was the answer to part i).}$
$t\text{ measured the }\textbf{difference}\text{ between the volumes' rates}\\ \text{i.e. it measured the }\textbf{ rate of change of the difference in the volumes}\\\text{ for the tanks}$
$\text{When you integrate the }\textbf{rate of change of the difference}\\ \text{you integrate back to }\textbf{the difference itself.}$
Of course, you could've just integrated the rates independently. But good luck trying to integrate $2 +\frac{t^2}{t+1} $

blasonduo · « **Reply #2969 on:** October 21, 2017, 10:53:46 pm »

Hey! Just a quick question

For questions such as these, I always normally answer the question with decimal places, however, the answers normally give it in exact form.

My question is if they don't specify the exact value, and I put an approximate answer (to 3 d.p for example) will I lose a mark?

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