Author Topic: Mathematics Question Thread (Read 1626452 times) Tweet Share

RuiAce · « **Reply #2715 on:** October 06, 2017, 04:55:08 pm »

Quote from: bdobrin on October 06, 2017, 04:45:58 pm

Hi

Can someone please explain to me how the answer to this question is D.

I thought it would be 'b' because the velocity is negative meaning it moves in a negative direction and the acceleration is positive so it is picking up speed.

Thanks,
Ben

This was already addressed here

kiiaaa · « **Reply #2716 on:** October 06, 2017, 06:51:38 pm »

hi, guys, i was doing the 2012 HSC paper and was having some difficulties and was wondering if you could help me out explain and teach me how to tackle these please so I know in the future? I looked at the answers but I didn't really understand what they were doing exactly

i need help in q14a iv / q14b and q14 iii/iv

so for q14a iv - was wondering how to exactly do it as i was really lost

for q14b i dort of tried and got stuck at a dead end having no idea where I was going

and finally, for q14 iii/iv I'm really lost how to find the rate of change =/

thank you very very much and sorry for bombarding you with sooo many questions

RuiAce · « **Reply #2717 on:** October 06, 2017, 06:57:38 pm »

Quote from: kiiaaa on October 06, 2017, 06:51:38 pm

hi, guys, i was doing the 2012 HSC paper and was having some difficulties and was wondering if you could help me out explain and teach me how to tackle these please so I know in the future? I looked at the answers but I didn't really understand what they were doing exactly

i need help in q14a iv / q14b and q14 iii/iv

so for q14a iv - was wondering how to exactly do it as i was really lost

for q14b i dort of tried and got stuck at a dead end having no idea where I was going

and finally, for q14 iii/iv I'm really lost how to find the rate of change =/

thank you very very much and sorry for bombarding you with sooo many questions

$\text{b) is just}\\ \begin{align*}V&=\pi \int_0^1 \frac{9}{(x+2)^4}dx\\ &= \pi\int_0^1 9(x+2)^{-4}\,dx\\ &= \pi\left[\frac{9}{-4}(x+2)^{-3}\right]_0^1\\ &= \frac{19\pi}{72}\end{align*}$
_________________________
$\text{The rate of change is just given by the derivative }N^\prime(t)= 347 e^{0.0347t}\\ \text{The answer to part iii is just }N^\prime(12) = 347e^{0.0347\times 12}$
_________________________
$\text{For iv, note that }\textit{initially}, \, N(t)=N(0) = 1000$
$\text{We simply require when }N(t) = 100000\\ \begin{align*}100000&=1000e^{0.0347t}\\ 100 &= e^{0.0347t}\\ 0.0347t &= \ln 100\\ t&= \frac{\ln100}{0.0347}\end{align*}$

RuiAce · « **Reply #2718 on:** October 06, 2017, 07:05:30 pm »

Quote from: kiiaaa on October 06, 2017, 06:51:38 pm

hi, guys, i was doing the 2012 HSC paper and was having some difficulties and was wondering if you could help me out explain and teach me how to tackle these please so I know in the future? I looked at the answers but I didn't really understand what they were doing exactly

i need help in q14a iv / q14b and q14 iii/iv

so for q14a iv - was wondering how to exactly do it as i was really lost

for q14b i dort of tried and got stuck at a dead end having no idea where I was going

and finally, for q14 iii/iv I'm really lost how to find the rate of change =/

thank you very very much and sorry for bombarding you with sooo many questions

$\text{To understand what's going on in the first question}\\ \text{you need to understand the significance of the added }'+k'.$
$\text{In general, the graph of }y=f(x)+k\\ \text{is the graph of }y=f(x)\text{, but }\textit{translated up}\text{ by }k\text{ units.}$
e.g. y=sin(x) + 2 is just y = sin(x), but shifted up by 2 units.
$\text{What we want to do, is find the values of }k\\ \text{such that the equation }3x^4+4x^3+12x^2 + k = 0\text{ will no longer have a solution}\\ \text{This is equivalent to }y=3x^4+4x^3+12x^2+k \textbf{ not cutting the }x\textbf{-axis.}$
___________________________
$\text{From earlier calculations and your graph, you can deduce that}\\ \text{the minimum value of }f(x) = 3x^4+4x^3+12x^2\text{ is 32.}\\ \text{Now think about it intuitively.}$
$\text{If you shifted your graph up by 32 units, then the only }x\text{-intercept left}\\ \text{will be the local minimum, which is now at }(-2,0)$
$\text{The instant you shift your graph up by }\textbf{more}\text{ than 32 units}\\\text{you will no longer have an intercept left}\\ \text{and hence the equation has no solution.}$
$\therefore k > 32$

Tempestuous · « **Reply #2719 on:** October 06, 2017, 08:59:22 pm »

Quote from: RuiAce on October 06, 2017, 04:19:28 pm

I always was solving for z.

But z was written in terms of cos A. So I needed to get rid of it somehowIt relies on similar triangles. Quite risky to use, even though I would happily give marks for it.

Would it be safe to use in the HSC then?

RuiAce · « **Reply #2720 on:** October 06, 2017, 09:15:48 pm »

Quote from: Tempestuous on October 06, 2017, 08:59:22 pm

Would it be safe to use in the HSC then?

They’d have to be pretty evil to deduct you marks for it. But if you want to be absolutely safe, don’t

_____ · « **Reply #2721 on:** October 06, 2017, 10:08:28 pm »

Sorry to double post but I think my question might have been lost in the rush from this afternoon: https://atarnotes.com/forum/index.php?topic=164548.msg987281#msg987281

Thanks for monitoring the thread so closely Rui, it's a real help.

RuiAce · « **Reply #2722 on:** October 06, 2017, 10:18:16 pm »

Quote from: _____ on October 06, 2017, 03:52:38 pm

(Image removed from quote.)

What's the technique to solve this?

I got y' as a concave up parabola with vertex at x = (the point where the y'' graph touches the x axis), and y as a cubic with a point of inflexion at this point, which is correct. But how do I determine elevation (y value) at this point for y' and y?

---------

When I'm asked to graph say "y = SQRT(16-x^2)" do I assume this just means the positive square root and not the negative square root because it is a function? Do I only sketch the full circle if the question specifically states that it's a relation or a full circle?

Thanks

Ah yes all good, unfortunately it was lost indeed and your bump was rightly justified. I've had a busy week and my mind's been all over the place.

First bit - Seeing as though they give you no real means of handling the +C...
Whereas for y' you must draw it as you correctly did, you can draw it however high you want for y and you will still be marked correct. Your answer can only be marked on what you've been given, so your approach was fine.
__________________________
$\text{Yes, because }y=\sqrt{16-x^2}\text{ is }\textbf{always}\text{ the upper half}\\ \text{and }y=-\sqrt{16-x^2}\text{ is }\textbf{always}\text{ the lower half}$
An easy way to see this is really just a table of values. You will only get positive y-values for one of them and negative y-values for the other.
$\text{The full circle would be }x^2+y^2=r^2\\ \text{If you rearrange it, you would get }y=\pm \sqrt{16-x^2}\\ \textbf{Note that plus/minus symbol.}$

_____ · « **Reply #2723 on:** October 06, 2017, 10:50:30 pm »

Quote from: RuiAce on October 06, 2017, 10:18:16 pm

Ah yes all good, unfortunately it was lost indeed and your bump was rightly justified. I've had a busy week and my mind's been all over the place.

First bit - Seeing as though they give you no real means of handling the +C...
Whereas for y' you must draw it as you correctly did, you can draw it however high you want for y and you will still be marked correct. Your answer can only be marked on what you've been given, so your approach was fine.
__________________________
$\text{Yes, because }y=\sqrt{16-x^2}\text{ is }\textbf{always}\text{ the upper half}\\ \text{and }y=-\sqrt{16-x^2}\text{ is }\textbf{always}\text{ the lower half}$
An easy way to see this is really just a table of values. You will only get positive y-values for one of them and negative y-values for the other.
$\text{The full circle would be }x^2+y^2=r^2\\ \text{If you rearrange it, you would get }y=\pm \sqrt{16-x^2}\\ \textbf{Note that plus/minus symbol.}$

Thanks a bunch.

For the second one, I look at SQRT(16-x^2) and think that it can refer to the upper or lower half. This is probably because I've been doing questions (like rotation volume questions where you must find say x^2 = SQRT(y)) where you find the square root of something and then must test the positive and negative possibilities and choose the correct one.

Thinking about this more, the correct notation for both possibilities must always be +-SQRT(y) even though my calculator just produces SQRT(y), right? So I need to write it like this. If I (or the examiners) write SQRT(y) I'm/they're saying +SQRT(y), not +-SQRT(y). I think I understand now.

sophiegmaher · « **Reply #2724 on:** October 07, 2017, 01:47:43 pm »

How do you solve part (ii) of this question? I have confused on the simplification part of the equation you eventually get.

(i) In the following series, for what values of x will a sum to infinity exist?
(1-x)^1/2 + (1-x) + (1-x)^3/2+...

(ii) Find the value of x, if the series in part (i) has a limiting sum of 2root3 + 3

RuiAce · « **Reply #2725 on:** October 07, 2017, 02:24:31 pm »

Quote from: sophiegmaher on October 07, 2017, 01:47:43 pm

How do you solve part (ii) of this question? I have confused on the simplification part of the equation you eventually get.

(i) In the following series, for what values of x will a sum to infinity exist?
(1-x)^1/2 + (1-x) + (1-x)^3/2+...

(ii) Find the value of x, if the series in part (i) has a limiting sum of 2root3 + 3

$\text{Since the series has first turn }(1-x)^{\frac12}\text{ and ratio }(1-x)^{\frac12}\\ \text{its limiting sum is given by }S_{\infty} = \frac{(1-x)^{\frac12}}{1-(1-x)^{\frac12}}$
$\text{If this is equal to }2\sqrt{3}+3\text{, then}\\ \begin{align*}\frac{\sqrt{1-x}}{1-\sqrt{1-x}}&= 2\sqrt{3}+3\\ \sqrt{1-x}&= (2\sqrt3+3)(1-\sqrt{1-x})\end{align*}$
$\text{There are at least two possible approaches at this point}\\ \text{One possible approach is to semi-expand the RHS and move anything}\\ \text{involving }x\text{ to the LHS, for cleanliness}$
$\begin{align*}\sqrt{1-x}&=(2\sqrt3+3) - (2\sqrt3+3)\sqrt{1-x}\\ \sqrt{1-x} + (2\sqrt3+3)\sqrt{1-x} &= 2\sqrt3+3\\ \big(1+(2\sqrt3+3)\big)&=2\sqrt3+3\tag{upon factorising}\\ \sqrt{1-x}&=\frac{2\sqrt3+3}{2\sqrt3+4}\end{align*}$
$\text{Hence }1-x = \left(\frac{2\sqrt{3}+3}{2\sqrt{3}+4}\right)^2 \implies \boxed{x =1- \left(\frac{2\sqrt{3}+3}{2\sqrt{3}+4}\right)^2}\\ \text{You can rearrange this however you want, including rationalise the denominator}$

Edit: Mistake in sign

sophiegmaher · « **Reply #2726 on:** October 07, 2017, 02:53:17 pm »

I worked it out by expanding the brackets, but I'm not sure how to simplify it like the answers I have do, like how they go from line 3 of the answers to line 4 of the answers:

Shadowxo · « **Reply #2727 on:** October 07, 2017, 03:10:49 pm »

Quote from: sophiegmaher on October 07, 2017, 02:53:17 pm

I worked it out by expanding the brackets, but I'm not sure how to simplify it like the answers I have do, like how they go from line 3 of the answers to line 4 of the answers:

They just skipped a few steps. All they did was move all the sqrt(1-x) terms to one side
$\begin{align*}\sqrt{1-x}=&2\sqrt{3}-2\sqrt{3}\sqrt{1-x}-3\sqrt{1-x}+3\\ \sqrt{1-x}=&2\sqrt{3}-(2\sqrt{3}+3)\sqrt{1-x}+3\\ \sqrt{1-x}+(2\sqrt{3}+3)\sqrt{1-x}=&2\sqrt{3}+3\\ (2\sqrt{3}+4)\sqrt{1-x}=&2\sqrt{3}+3 \end{align*}$

sophiegmaher · « **Reply #2728 on:** October 07, 2017, 04:24:52 pm »

Quote from: Shadowxo on October 07, 2017, 03:10:49 pm

They just skipped a few steps. All they did was move all the sqrt(1-x) terms to one side
$\begin{align*}\sqrt{1-x}=&2\sqrt{3}-2\sqrt{3}\sqrt{1-x}-3\sqrt{1-x}+3\\ \sqrt{1-x}=&2\sqrt{3}-(2\sqrt{3}+3)\sqrt{1-x}+3\\ \sqrt{1-x}+(2\sqrt{3}+3)\sqrt{1-x}=&2\sqrt{3}+3\\ (2\sqrt{3}+4)\sqrt{1-x}=&2\sqrt{3}+3 \end{align*}$

Thank you!! That makes so much more sense, I was just getting confused with the factorising

hobocop · « **Reply #2729 on:** October 08, 2017, 05:42:20 pm »

Hi! Could I get some help with part iii) please?

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