Author Topic: 4U Maths Question Thread (Read 665377 times) Tweet Share

massive · « **Reply #660 on:** October 20, 2016, 03:16:19 pm »

Quote from: jakesilove on October 20, 2016, 02:00:18 pm

Hey! Looks like you're mixing up your formulas a bit; I would take a look at this seriously comprehensive document HERE and come back if you have any more questions!

thanks dude

katherine123 · « **Reply #661 on:** October 20, 2016, 05:38:55 pm »

image 1: how to do part v)
image 2: how to do part ii) and iii)

Brenda0708 · « **Reply #662 on:** October 20, 2016, 05:59:39 pm »

Quote from: RuiAce on October 20, 2016, 12:20:04 pm

(1+3i) and (9+3i) are the foci.

S and S' do NOT lie on the x-axis. So the origin doesn't have to play a role here.

And of course, PS+PS' = 10 implies that the length of the major axis is 10 (hence the length of the semi-major axis is 5)

So your interpretation is correct. But your interpretation does not mean that the origin has to lie on the ellipse.
_______________________
$\text{To check: When }P\text{ is the origin }O(0,0)\\ |(0+0i)-1-3i| + |(0+0i)-9-3i| \neq 10$

Oh thank you. But then how do you sketch the locus without finding the equation? As in how do you know it lies strictly in the first quadrant? And after finding 2a, how do you find 2b without the equation?

RuiAce · « **Reply #663 on:** October 20, 2016, 06:03:24 pm »

Quote from: Brenda0708 on October 20, 2016, 05:59:39 pm

Oh thank you. But then how do you sketch the locus without finding the equation? As in how do you know it lies strictly in the first quadrant? And after finding 2a, how do you find 2b without the equation?

You don't know everything. Some things you have to infer.

2a is of course, given.

But the distance between the two foci is 2ae. Recall that SS' = 2ae
You are given the foci in the question.

Since the foci are at (1,3) and (9,3), the focal length is 4.

Hence 4 = ae
So e=4/5

Then you can use b^2=a^2(1-e^2) to find the length of the minor axis.

Note also that the ellipse's centre is also at (5,3)
Given the centre, and the length of the major and minor axes, you should be able to sketch the ellipse.

Brenda0708 · « **Reply #664 on:** October 20, 2016, 06:07:12 pm »

Quote from: RuiAce on October 20, 2016, 06:03:24 pm

You don't know everything. Some things you have to infer.

2a is of course, given.

But the distance between the two foci is 2ae. Recall that SS' = 2ae
You are given the foci in the question.

Since the foci are at (1,3) and (9,3), the focal length is 4.

Hence 4 = ae
So e=4/5

Then you can use b^2=a^2(1-e^2) to find the length of the minor axis.

Note also that the ellipse's centre is also at (5,3)
Given the centre, and the length of the major and minor axes, you should be able to sketch the ellipse.

ohhhh okay thanks

Brenda0708 · « **Reply #665 on:** October 20, 2016, 06:10:22 pm »

Haha here comes another question :'). For harder 3U, I don't quite get the logic behind proving f(x)>/= 0 for x>0. Can you please interpret this is more detail? Thanks.

RuiAce · « **Reply #666 on:** October 20, 2016, 06:11:26 pm »

Quote from: katherine123 on October 20, 2016, 05:38:55 pm

image 1: how to do part v)
image 2: how to do part ii) and iii)

$\text{For something like part }v)\text{ you need to look ahead.}\\ \text{What does }\frac{1}{n+1}\text{ remind you of?}\\ \text{The correct answer is this: }\int_0^1 {u^n}\,du$
$\text{This takes a bit of practice to spot. The idea is to think about inequalities.}\\ \text{Realise that somehow we must manipulate }\frac{u^n}{1+u^2}\\ \text{This is because the integrand of }I_n\text{ is }\frac{u^n}{1+u^2}$
$\text{It's not the most obvious thing in the world, but }1+u^2 \ge 1\text{ is all we need.}\\ \text{This allows us to deduce that }\frac{u^n}{1+u^2} \le u^n$
$\text{So we have a three-handed inequality.}\\ 0\le \frac{u^n}{1+u^2}\le u^n\\ \text{Which, if we smack the definite integral in, we get what we need.}\\ \int_0^1 0du \le \int_0^1 \frac{u^n}{1+u^2}du\le \int_0^1 u^n du$
Inequalities is the art of throwing out the garbage.

RuiAce · « **Reply #667 on:** October 20, 2016, 06:12:51 pm »

Quote from: Brenda0708 on October 20, 2016, 06:10:22 pm

Haha here comes another question :'). For harder 3U, I don't quite get the logic behind proving f(x)>/= 0 for x>0. Can you please interpret this is more detail? Thanks.

Just read those dot points again. Read them one at a time. And use a visual aid if you need it.

If it's still too hard, you might want to find a question. There are some that float around in past HSCs.

The basic idea is that if it's increasing, then it's always greater than something. So if you find the left endpoint, you'll know what it's always greater than.

RuiAce · « **Reply #668 on:** October 20, 2016, 06:24:31 pm »

Quote from: katherine123 on October 20, 2016, 05:38:55 pm

image 1: how to do part v)
image 2: how to do part ii) and iii)

$\text{Part (iii) is a nifty trick. It's just the fact that }-1 \le \cos \theta \le 1\\ \text{Inspiration to use this: }\theta\text{ vanishes, but leaves behind an inequality, not an equation.}$
$\text{Since }0 \le \theta < \frac{\pi}{2}\\ \text{If }\theta \neq 0,\text{ then firstly }\cos \theta \neq 1\\ \text{Also, if }\theta \neq 0,\text{ then from the question }\cos \theta = \frac{g}{R\omega^2}$
$\text{So we have}\\ \begin{align*}1& > \frac{g}{R\omega^2}\\ \omega^2 &> \frac{g}{R}\\ \omega &> \sqrt{\frac{g}{R}}\end{align*}\\ \text{noting }\omega > 0$

Brenda0708 · « **Reply #669 on:** October 20, 2016, 06:24:42 pm »

Can you please walk me through this question

RuiAce · « **Reply #670 on:** October 20, 2016, 06:30:07 pm »

Quote from: Brenda0708 on October 20, 2016, 06:24:42 pm

(Image removed from quote.)
Can you please walk me through this question

$\text{I'll just use part (i) as a demo. Try doing etc. by yourself and post up whatever you have.}$
$\text{Note that we are basically proving }x-\sin x > 0\\ \text{We move everything to one side to make it easier on ourselves.}\\ \text{That way, we only have to compute 1 derivative, not 2.}$
$\text{Consider }g(x) = x-\sin x\text{ defined for all }x \ge 0\\ \text{Then }g^\prime(x) = 1-\cos x$
$\text{Now, }1-\cos x \ge 0\text{ always.}\\ \text{This is simply due to the fact that }\cos x \le 1$
$\text{So }g^\prime(x) \ge 0\text{ for all }x\\ \text{This means, by definition, }g(x)\text{ is monotonic increasing.}$
$\text{If it's monotonic increasing, then }g(x)\text{ never goes down. It either sits or always goes up.}\\ \text{So where does it start?}\\ \text{Find }g(0) = 0 - \sin 0 = 0$
$\text{So }g(x)\text{ starts increasing, from }x=0, y=0\\ \text{Hence }g(x) \ge 0 \text{ for all }x \ge 0\\ \text{i.e. }x - \sin x \ge 0 \text{ for all }x\ge 0$
$\text{Then the }\ge \text{ becomes }> \text{ cause we really just care about }x> 0$

jakesilove · « **Reply #671 on:** October 20, 2016, 06:31:56 pm »

Quote from: RuiAce on October 20, 2016, 06:30:07 pm

$\text{I'll just use part (i) as a demo. Try doing etc. by yourself and post up whatever you have.}$
$\text{Note that we are basically proving }x-\sin x > 0\\ \text{We move everything to one side to make it easier on ourselves.}\\ \text{That way, we only have to compute 1 derivative, not 2.}$
$\text{Consider }g(x) = x-\sin x\text{ defined for all }x \ge 0\\ \text{Then }g^\prime(x) = 1-\cos x$
$\text{Now, }1-\cos x \ge 0\text{ always.}\\ \text{This is simply due to the fact that }\cos x \le 1$
$\text{So }g^\prime(x) \ge 0\text{ for all }x\\ \text{This means, by definition, }g(x)\text{ is monotonic increasing.}$
$\text{If it's monotonic increasing, then }g(x)\text{ never goes down. It either sits or always goes up.}\\ \text{So where does it start?}\\ \text{Find }g(0) = 0 - \sin 0 = 0$
$\text{So }g(x)\text{ starts increasing, from }x=0, y=0\\ \text{Hence }g(x) \ge 0 \text{ for all }x \ge 0\\ \text{i.e. }x - \sin x \ge 0 \text{ for all }x\ge 0$
$\text{Then the }\ge \text{ becomes }> \text{ cause we really just care about }x> 0$

I think it'd be easy to do this part graphically as well

RuiAce · « **Reply #672 on:** October 20, 2016, 06:36:17 pm »

Quote from: jakesilove on October 20, 2016, 06:31:56 pm

I think it'd be easy to do this part graphically as well

Would they let that go in 4U? Cause people mess up drawing to scale graphs.

Also it's not a 1 marker

jakesilove · « **Reply #673 on:** October 20, 2016, 06:38:52 pm »

Quote from: RuiAce on October 20, 2016, 06:36:17 pm

Would they let that go in 4U? Cause people mess up drawing to scale graphs.

Also it's not a 1 marker

I'd definitely back a graphical solution to an inequality as full marks, if it's draw to scale. Smart, easy solution; the sort of this 4U markers love. Your solution is kind of a formal 'proof' of the graphical solution; definitely better maths, but like bruh. Cmon. Easy answers = good answers.

RuiAce · « **Reply #674 on:** October 20, 2016, 06:40:30 pm »

Quote from: jakesilove on October 20, 2016, 06:38:52 pm

I'd definitely back a graphical solution to an inequality as full marks, if it's draw to scale. Smart, easy solution; the sort of this 4U markers love. Your solution is kind of a formal 'proof' of the graphical solution; definitely better maths, but like bruh. Cmon. Easy answers = good answers.

If you really wanted to be easy then "by inspection" is the best answer please.

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