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September 19, 2020, 07:11:46 pm

Author Topic: QCE Maths Methods Questions Thread  (Read 3279 times)

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A.Rose

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Re: QCE Maths Methods Questions Thread
« Reply #15 on: April 27, 2020, 05:56:57 pm »
+1
Hi!
I'm stuck on a question about Bernoulli sequences and the binomial distribution from Unit 4 Topic 3 Methods.
If it says 'at least' does this mean P(X>9)? And could you calculate that by doing P(X>9)=1-P(0≤X≤9)? I tried that and I didn't achieve the answer in the textbook.
Thank you!

Bri MT

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Re: QCE Maths Methods Questions Thread
« Reply #16 on: April 27, 2020, 08:35:51 pm »
+2
Hi!
I'm stuck on a question about Bernoulli sequences and the binomial distribution from Unit 4 Topic 3 Methods.
If it says 'at least' does this mean P(X>9)? And could you calculate that by doing P(X>9)=1-P(0≤X≤9)? I tried that and I didn't achieve the answer in the textbook.
Thank you!

Hi!

Your thinking is good here except that "at least" means greater than or equal to - not just greater than :)

So in your subtraction you shouldn't include the case where x=9

RuiAce

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Re: QCE Maths Methods Questions Thread
« Reply #17 on: May 27, 2020, 06:20:44 pm »
+3
Hi!
I have a question I'm stuck on from Unit 4 Topic 2 Methods. It's either really simple and I'm missing something or its generally a bit tricky. First time I read the question I thought of a right triangle but you can't assume that can you? Plus this Unit 4 methods, not Unit 1.  ;D
So if you can't assume it is a right triangle then how do you find the height? It reminded me of the ambiguous case but I'm not sure how/if to use that property to answer the question. I would greatly appreciate the help as I am doing a revision of Unit 4 Topic 1&2 in class this week.
Thank you!

At first glance, I actually think that it is really simple. Technically speaking, "the distance from the foot of the cliff to a boat", and "the height of the cliff above sea level", are exactly the ingredients of a right-angled triangle formulation!

I am surprised that this is considered a Unit 4 question instead of Unit 1. For all I know I could be missing something too, but my logic came from the wording of the question entirely.

Perhaps there's follow-up questions to this problem that are more Unit 4 based? Or perhaps there's an answer to aim for?

orla007

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Re: QCE Maths Methods Questions Thread
« Reply #18 on: July 09, 2020, 05:31:19 pm »
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Hi!

I am a little confused as to what I need to do to complete this question- as it is a proof the textbook doesn't have a solution.

Thanks in advance :)

Bri MT

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Re: QCE Maths Methods Questions Thread
« Reply #19 on: July 09, 2020, 05:49:59 pm »
+1
Hi!

I am a little confused as to what I need to do to complete this question- as it is a proof the textbook doesn't have a solution.

Thanks in advance :)

Hi!

Welcome to the forums!

It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.

Once you have part a part b should be pretty straightforward.


Hope this helps and please let me know if this clarifies things enough :)

orla007

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Re: QCE Maths Methods Questions Thread
« Reply #20 on: July 09, 2020, 05:59:17 pm »
+1
Hi!

Welcome to the forums!

It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.

Once you have part a part b should be pretty straightforward.


Hope this helps and please let me know if this clarifies things enough :)


Ahhhh thank you!!!

Bri MT

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Re: QCE Maths Methods Questions Thread
« Reply #21 on: July 09, 2020, 06:04:41 pm »
0

Ahhhh thank you!!!

No worries! I was actually teaching the sine rule on Monday so it was very fresh in my memory :)

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Re: QCE Maths Methods Questions Thread
« Reply #22 on: July 09, 2020, 06:15:24 pm »
+1
Hi!

Welcome to the forums!

It's hard to describe much of how you approach this without giving it away (which I will happily do if you're still confused but I think it's more rewarding and you learn more from help than answers). That being said:
- try starting by writing out the sine rule
- then look at c/sinC and either b/sinB or a/sinA (i.e. work with only 2 fractions at a time)
- Rearrange it so that you get [something]/c = [something]/sinC
- Do again but for whichever of b/sinB and a/sinA you didn't use
- Then combine the two equations you get together for the result you need.

Once you have part a part b should be pretty straightforward.


Hope this helps and please let me know if this clarifies things enough :)
a+b/c can be separated into a/c +b/c
sinA/a=sinC/c
So sinA/sinC= a/c
sinB/b=sinC/c
So sinB/sinC=b/c
So sinB/sinC+ sinA/sinC=( sinA + sinB)/sinC

From then you should be able to get the second one.

matthew hay

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Re: QCE Maths Methods Questions Thread
« Reply #23 on: August 26, 2020, 06:42:21 pm »
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Hi there,

Do you or anyone know if there is an annotated syllabus? or even annotated Units 4 syllabus?? Please reply no if not (it saves me the extravagant search!

Thanks and kind regards, Matt

Bri MT

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Re: QCE Maths Methods Questions Thread
« Reply #24 on: August 26, 2020, 07:48:03 pm »
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Hey,

I'm pretty sure there aren't any annotated QCE methods syllabuses. There are other methods resources listed here but an annotated syllabus isn't one of them.

snr.mmorris4.19

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Re: QCE Maths Methods Questions Thread
« Reply #25 on: September 15, 2020, 06:07:31 pm »
0
whats the process behind solving these? Never seen them before i'm not sure where to start

P is a movable point on the line y=7-x
Find the coordinates of P when it is closest to the origin (0,0).

keltingmeith

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Re: QCE Maths Methods Questions Thread
« Reply #26 on: September 15, 2020, 06:12:36 pm »
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whats the process behind solving these? Never seen them before i'm not sure where to start

P is a movable point on the line y=7-x
Find the coordinates of P when it is closest to the origin (0,0).

This is a really interesting question! What I'm going to do is give you some starting points, then why don't you let us know how far you get?

Firstly, you may remember that the straight-line distance between any two points is given by the equation:



which is secretly just Pythagoras' theorem in disguise. Because you don't know the exact points of P, you could actually pretend that it lies on the point x_2=x and y_2=y.

Another thing to think about is that closest and furthest have some synonyms that would normally raise alarm bells to any student studying differentiation - maybe you can use differentiation in some way, here?


EDIT: See below for some pointers on using a non-calculus approach - I tried to generalise for other curves, not just for straight lines, but their approach is WAY simpler and still worth knowing. Particularly if you're not up to differentiation, yet
« Last Edit: September 15, 2020, 06:25:18 pm by keltingmeith »
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Re: QCE Maths Methods Questions Thread
« Reply #27 on: September 15, 2020, 06:17:54 pm »
+3
In general, the shortest distance from a point to a line will be mapped by a line through the point and the line such that the two lines are perpendicular. For curves, we use the same idea but choose a line through the point and the curve such that the line is perpendicular to the tangent at the point of intersection (for further application that I'm certain will come up at some point down the line). Other problems similar to this are called locus problems - you might want to have a further look at these (there are all sorts, but they can all be more or less divvied up reasonably distinctly).

There are multiple ways of solving this - the two I'd recommend you have a look at are a) using the perpendicular distance formula or b) finding a perpendicular line passing through the given point, then finding the point of intersection to determine the coordinates of P. Expressing the distance as a polynomial function to exploit it as a max/min problem seems excessive for a question that involves a point and a line, but is worth exploring as suggested by keltingmeith as it will provide practice for tougher locus problems :)


EDIT: beaten by keltingmeith
« Last Edit: September 15, 2020, 06:19:57 pm by fun_jirachi »
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