Author Topic: 3U Maths Question Thread (Read 1465321 times) Tweet Share

anotherworld2b · « **Reply #1410 on:** February 04, 2017, 07:57:23 pm »

Would this be correct now?
Could i also have help with q9 b please?

Quote from: RuiAce on February 03, 2017, 09:38:48 pm

a) is correct
$\text{For b), you seem to have assumed that }(1-x^2)^4=1-x^8\\ \text{This is not true. In fact}\\ (1-x^2)^4=1-4x^2+6x^4-4x^6+x^8\\ \text{The intention was for you to use the chain rule within the product rule,}$
$\frac{d}{dx}2x^3(1-x^2)^4=6x^2(1-x^2)^4+2x^3(-2x)[4(1-x^2)^3]$
c) is almost correct. Why did 2^1/2 turn into 1? 2^1/2 is approximately 1,41
d) Same problem as b). You don't expand the powers in like this.
$(a+b)^n = a^n+b^n\text{ is }\textbf{false.}\\ \text{Just like how }(x+y)^2 = x^2+y^2\text{ is false.}$
(Note: These are 2U difficulty.)

In the next set

b) is ok. Probably not the easiest approach but still correct.

Same problem with c) and d) in expanding the powers.

bsdfjnlkasn · « **Reply #1411 on:** February 04, 2017, 08:50:22 pm »

Quote from: RuiAce on February 04, 2017, 07:09:04 pm

$\sin 2u = 2\sin u \cos u\\ \text{Let }u=2x\\ \sin 4x = 2\sin 2x \cos 2x$
$\text{Double angles also work on }2x\text{'s}\\ \text{Not just }x\text{'s}$

Perfect thank you. I think there might have been a small mistake when you were evaluating as I got 1/4 which matches the answers

RuiAce · « **Reply #1412 on:** February 04, 2017, 08:53:52 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 08:50:22 pm

Perfect thank you. I think there might have been a small mistake when you were evaluating as I got 1/4 which matches the answers

Oh. My bad, sorry about that, think I misassumed cos(0) = 0

Shadowxo · « **Reply #1413 on:** February 04, 2017, 08:59:04 pm »

Quote from: anotherworld2b on February 04, 2017, 07:57:23 pm

Would this be correct now?
Could i also have help with q9 b please?

For this question, you have dy/dx already, you know dx/du = -1, and du/dt = 2
dy/dt = dy/dx * dx/dt
dx/dt = dx/du * du/dt
dy/dt = dy/dx * dx/du * du/dt
= dy/dx * -1 * 2
sub in x=0 and there's your answer

RuiAce · « **Reply #1414 on:** February 04, 2017, 09:04:06 pm »

Quote from: anotherworld2b on February 04, 2017, 07:57:23 pm

Would this be correct now?

c looks good.

d was fine until the very end. Not sure why that's 2 times 6x there and not 2 plus 6x

Chemystery · « **Reply #1415 on:** February 04, 2017, 09:43:50 pm »

Probability! Can someone please explain permutations, factorial notation and combinations to me? My teacher always rushes over concepts and doesn't explain very well so I'm on a basis where I'm close to understanding but it hasn't quite clicked!

RuiAce · « **Reply #1416 on:** February 04, 2017, 09:49:06 pm »

Quote from: Chemystery on February 04, 2017, 09:43:50 pm

Probability! Can someone please explain permutations, factorial notation and combinations to me? My teacher always rushes over concepts and doesn't explain very well so I'm on a basis where I'm close to understanding but it hasn't quite clicked!

$\text{Aside from }0!=1\\ \begin{align*}1!&=1\\2!&=2\times 1\\ 3&=3\times 2\times 1\\ 4!&=4\times3\times2\times1\\ &\vdots \\ n!&=n(n-1)(n-2)\dots 2\times 1\end{align*}\\ \text{Not sure what more there is to factorial notation.}$
Other than that, simply refer to one of Jamon's guides.

The topic is a nuisance and not easy to explain. Often (not usually but often) taught with examples. Provide some questions and we'll walk through the principles.

bsdfjnlkasn · « **Reply #1417 on:** February 04, 2017, 10:33:13 pm »

How would you differentiate y = sinx°

RuiAce · « **Reply #1418 on:** February 04, 2017, 10:34:13 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 10:33:13 pm

How would you differentiate y = sinx°

This most likely came out of Cambridge. It's just a dummy placeholder; if anything I'd ignore it and wait till radians are taught.

QC · « **Reply #1419 on:** February 04, 2017, 11:05:47 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 10:33:13 pm

How would you differentiate y = sinx°

y=sin(pix/180)
y'=pi/180cos(pix/180)

RuiAce · « **Reply #1420 on:** February 04, 2017, 11:09:11 pm »

Quote from: aroon217 on February 04, 2017, 11:05:47 pm

y=sin(pix/180)
y'=pi/180cos(pix/180)

The reason I gave my answer is because it's only a conversion. We have not addressed the greater issue of the fact that the "degree" does not have a dimension. (Edit: Deleted out reason a because that was wrong.)

In physics, we note that the units for the radian are metres per metres. This effectively cancels out to being dimensionless.
A degree doesn't follow any principles regarding dimension. So when the conversion is instead treated as a formula, what exactly are you doing?

Calculus is never done with degrees for this reason.
$\text{For calculus to make sense, }\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\text{ must make sense.}$
Questions in the HSC will never be based off this placeholder dummy.

anotherworld2b · « **Reply #1421 on:** February 04, 2017, 11:58:03 pm »

Thank you for your help Shadowxo and RuiAce

I was wondering for this question how do you obtain the same factorised answer without the assitance of a calculator?

kiwiberry · « **Reply #1422 on:** February 05, 2017, 12:04:21 am »

Quote from: anotherworld2b on February 04, 2017, 11:58:03 pm

Thank you for your help Shadowxo and RuiAce

I was wondering for this question how do you obtain the same factorised answer without the assitance of a calculator?

In the third line you accidentally copied (x-3)² as (x-2)²! From there you can factorise the 3(x-3)² out

anotherworld2b · « **Reply #1423 on:** February 05, 2017, 01:10:09 am »

I see. What can be done afterwards?
I am not sure where 12 (x+ 1) appears from

Quote from: kiwiberry on February 05, 2017, 12:04:21 am

In the third line you accidentally copied (x-3)² as (x-2)²! From there you can factorise the 3(x-3)² out

kiwiberry · « **Reply #1424 on:** February 05, 2017, 01:29:04 am »

Quote from: anotherworld2b on February 05, 2017, 01:10:09 am

I see. What can be done afterwards?
I am not sure where 12 (x+ 1) appears from

3(x-3)² is the common factor

3(x-3)²(3x+7) + 3(x-3)³
= 3(x-3)²(3x+7+x-3)
= 3(x-3)²(4x+4)
= 12(x-3)²(x+1)

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