Edexcel A2 C3 Mathematics 12th June 2015 Watch

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toddmcnugget
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#1961
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#1961
(Original post by tinoah)
Any particularly tough papers anyone can recommend? (I've done all the Gold Edexcel papers, Solomon papers, Zig Zag papers, and S&T and U-V IYGB papers on madasmaths)
Enjoy (I guess)

Paper:
https://07a69ccf283966549a9350d1a669...% 20Edexcel.pdf

Mark scheme:
https://07a69ccf283966549a9350d1a669...% 20Edexcel.pdf

If you've done all those I don't think you need any more work hahah
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frozo123
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#1962
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#1962
I would rep you but ran out, thanks
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Kevin De Bruyne
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#1963
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#1963
(Original post by frozo123)
https://07a69ccf283966549a9350d1a669...%20Edexcel.pdf

sean 5b?

I got 1/ ( 6x^2(x^2-1)^1/2 )
how do I go from there?
Doesn't look like you can go anywhere from there - what did you get for 5a?
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TheWrongTrick
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#1964
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#1964
Is there a quicker foolproof way of working all the solutions to a trigonometric equation?
eg, for tan, you just keep adding 180 degrees
I know for sin, you can take the value away from 180 and get another solution
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Maham88
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#1965
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#1965
(Original post by frozo123)
I would rep you but ran out, thanks
haha it's ok
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Oxyfrost
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#1966
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#1966
Hello, can anyone help me? I'm stuck on C3 Solomon paper C Question four. Attached is the question and mark scheme. It's a trig proof question. What I can't figure out, is for example, the numerator is sin2x + sin2y. The first step of the mark scheme sets the numerator as
2sin(x + y)cos(x-y)
How did they get to this? It's driving me insane! Really thrown my confidence, day before exam :/
Attached files
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frozo123
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#1967
(Original post by SeanFM)
Doesn't look like you can go anywhere from there - what did you get for 5a?
Nevermind I misread thanks
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Hudl
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#1968
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#1968
(Original post by Tiwa)
Could someone help me with question 4b? Never seen this question come up before.Attachment 425929
Proof by contradiction is a method of proof whereby you assume the conclusion is false, and then show this assumption leads to something which can't be true (e.g. 1=0 or "2 is odd").A number is rational if it is in the form \dfrac{p}{q}, where p,q are integers (q \ne 0).Piecing this together, we want to show that \log_2 3 is irrational; i.e. that it can't be written in the form \dfrac{p}{q} for any integers p,q. So, we start our proof by assuming that there exist integers p,q (q nonzero) such that \log_2 3 = \dfrac{p}{q}.By the definition of logarithms, this gives 3 = 2^{p/q}, and raising both sides to the power of q gives 3^q=2^p. This can only happen if p=q=0, but we can't have q=0 so this can't be true, so the assumption can't be true, so it must be false; hence the proposition is true.
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frankthebunny
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#1969
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#1969
Guys what are the tips tricks and **** like that for tomorrow im ****tng myself
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frozo123
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#1970
(Original post by Maham88)
haha it's ok
how did you find the chem yesterday?
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tinoah
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#1971
(Original post by toddmcnugget)
Enjoy (I guess)

Paper:
https://07a69ccf283966549a9350d1a669...% 20Edexcel.pdf

Mark scheme:
https://07a69ccf283966549a9350d1a669...% 20Edexcel.pdf

If you've done all those I don't think you need any more work hahah
All the exams I've had thus far haven't gone as well as I'd have hoped even with all the work so I'm kind of in panic mode but thanks for the paper!
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Bustamove
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#1972
(Original post by jf1994)
Do we need to know cos(-x) = cos(x) and sin(-x) = -sin(x)?
wait whaaa?
what formula is that? When is it used?
thanks
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Andy98
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#1973
(Original post by Bustamove)
wait whaaa?
what formula is that? When is it used?
thanks
I figure it's transformation stuff

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radhikagulati
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#1974
Is the range of f-1(x) the same as the domain of f(x) and the domain of the inverse the same as the range of the normal function?
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JizzaStanger
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#1975
(Original post by lightningdoritos)
Just had another go at it, not sure what you would do after putting the discriminant to <0 because surely you would have to square root a negative value?
My apologies, I was trying to tell you how to do it without having done it myself! If you do it as I suggest you will indeed have to take the square root of a negative number!

First do a sketch to determine roughly where the graphs touch. Because the one of the functions is |f(x)|, you use the sketch to determine whether you're in the section where |f(x)| = -f(x) or the section where |f(x)| = f(x).

Now you've determined that with a sketch (from the fact that you would otherwise have to take the square root of a negative number, it's in the region where |f(x)| = -f(x) ). You should now set -f(x)=g(x), and follow the method again. It will get you k = +/- 2a.

You can determine whether k = 2a or -2a by using your sketch again. The question tells you that g(x) cuts |f(x)| at another point, Q. You'll find that one value of k makes it impossible for the curves to cross at Q and that one makes it possible. The one the makes it possible is, of course, 2a.

I hope this clears it up!
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anndz3007
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#1976
(Original post by Oxyfrost)
Hello, can anyone help me? I'm stuck on C3 Solomon paper C Question four. Attached is the question and mark scheme. It's a trig proof question. What I can't figure out, is for example, the numerator is sin2x + sin2y. The first step of the mark scheme sets the numerator as
2sin(x + y)cos(x-y)
How did they get to this? It's driving me insane! Really thrown my confidence, day before exam :/
it's in the formula booklet
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Kevin De Bruyne
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#1977
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#1977
(Original post by Oxyfrost)
Hello, can anyone help me? I'm stuck on C3 Solomon paper C Question four. Attached is the question and mark scheme. It's a trig proof question. What I can't figure out, is for example, the numerator is sin2x + sin2y. The first step of the mark scheme sets the numerator as
2sin(x + y)cos(x-y)
How did they get to this? It's driving me insane! Really thrown my confidence, day before exam :/
Addition formulae, P124 of the textbook.

(Original post by radhikagulati)
Is the range of f-1(x) the same as the domain of f(x) and the domain of the inverse the same as the range of the normal function?
Correct. Imagine a factory, and you have a production line controlled by a function. You have the domain on one side, and the range on the other. When you put an object in you get a car out. Now imagine the function works backwards. You put a car in and get an object out. You can't put an object in the reverse direction, it must be car.
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Raj_B961
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#1978
quick question, how do we find the maximum or minimum value in trig questions??????????????????????? ???????????????????????????????? ???????????????????????????????? ??????????????????
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frozo123
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#1979
(Original post by SeanFM)
Addition formulae, P124 of the textbook.
It's really nice of you spending your evenings during exam week to help people out
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radhikagulati
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#1980
Is the range/domain of an inverse function the same as the domain/range respectively of the normal function?
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