how would you do that??aha(Original post by lizard54142)
You cannot define a prime number like you would an even or odd number. Make sure you know how to prove is irrational, or is irrational etc...
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OCR MEI C3 Maths June 2015 watch

Alevelstudent678
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 10062015 22:42

lizard54142
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 10062015 22:51
(Original post by Alevelstudent678)
how would you do that??aha
assume that is rational, so it can be expressed in the form where a and b share no common factors.
must be an even number.
Because a is even, we can write in the the form
must be an even number.
But this is a contradiction, because we said a and b share no common factors (and since they are both even they clearly have a factor of 2). Hence is irrational.Last edited by lizard54142; 10062015 at 22:52. 
Alevelstudent678
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 10062015 22:55
(Original post by lizard54142)
I'll do the one for you
assume that is rational, so it can be expressed in the form where a and b share no common factors.
must be an even number.
Because a is even, we can write in the the form
must be an even number.
But this is a contradiction, because we said a and b share no common factors (and since they are both even they clearly have a factor of 2). Hence is irrational. 
henrygriff28
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 10062015 22:58
(Original post by Alevelstudent678)
you know when doing proofs you can say even number is 2x, odd number is 2x+1, what others are there e.g. A prime number?
Because i really struggle on these
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lizard54142
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 10062015 23:04
(Original post by Alevelstudent678)
oh right V.clever, i'l learn those cheers mate
(Original post by henrygriff28)
Any prime number above 3 is in the form 6n+/1 but not everything in this form is a prime number, it is necessary but not sufficient; no other way to express primes
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henrygriff28
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 10062015 23:10
(Original post by lizard54142)
No problem
Is this true? Wow, I did not know this. Have you got a proof?
Only possibilities are 6n, 6n+1, 6n+2, 6n+3, 6n1, 6n2. All real integers fall into one of these categories. 6n is clearly divisible by 6, 6n+2 = 2(3n+1) so is divisible by two, 6n+3 = 3(2n+1) so is divisible by 3, 6n2 = 2(3n1) so is divisible by three so all primes must be 6+/1. Not all are though as counter examples of 35 and 25 show. QED
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lizard54142
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 10062015 23:14
(Original post by henrygriff28)
Yeah, it's by exhaustion though 😕
Only possibilities are 6n, 6n+1, 6n+2, 6n+3, 6n1, 6n2. All real integers fall into one of these categories. 6n is clearly divisible by 6, 6n+2 = 2(3n+1) so is divisible by two, 6n+3 = 3(2n+1) so is divisible by 3, 6n2 = 2(3n1) so is divisible by three so all primes must be 6+/1. Not all are though as counter examples of 35 and 25 show. QED
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Alevelstudent678
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 10062015 23:22
(Original post by henrygriff28)
Yeah, it's by exhaustion though 😕
Only possibilities are 6n, 6n+1, 6n+2, 6n+3, 6n1, 6n2. All real integers fall into one of these categories. 6n is clearly divisible by 6, 6n+2 = 2(3n+1) so is divisible by two, 6n+3 = 3(2n+1) so is divisible by 3, 6n2 = 2(3n1) so is divisible by three so all primes must be 6+/1. Not all are though as counter examples of 35 and 25 show. QED
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Supergirlxxxxxx
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 11062015 05:12
How do I do the last part to this question? I figured out its the area of the trapezium minus 25/3 but how do I do the trapezium part? ANS: 30 2/3
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Computer Geek
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 11062015 05:33
(Original post by Supergirlxxxxxx)
How do I do the last part to this question? I figured out its the area of the trapezium minus 25/3 but how do I do the trapezium part? ANS: 30 2/3
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Then subtract the integrated component and the little triangle from the big triangle.
This should give you 92/3 which is the same as 30 2/3.Last edited by Computer Geek; 11062015 at 05:44. 
Supergirlxxxxxx
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 11062015 09:15
(Original post by Computer Geek)
Don't think of it as a trapezium. Work out the area of the triangle bounded by the line y=x, the line x=11 and the x axis. Then work out the little triangle bounded by the line y=x and the line y=3 (Where point p lies).
Then subtract the integrated component and the little triangle from the big triangle.
This should give you 92/3 which is the same as 30 2/3.
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 11062015 09:44
the trig identities like sec^2x=1+tan^2x and cosec^2x=1+cot^2x, and the double angle formulae, are C4 topics right? So we shouldn't need them tomorrow??
I've been neglecting C3 revision to the point where even basic integration is a bit of a struggle right now lol, gonna have to revise like crazy 
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 11062015 10:11
(Original post by tingirl)
the trig identities like sec^2x=1+tan^2x and cosec^2x=1+cot^2x, and the double angle formulae, are C4 topics right? So we shouldn't need them tomorrow??
I've been neglecting C3 revision to the point where even basic integration is a bit of a struggle right now lol, gonna have to revise like crazy
Just remember and you can get both of them by dividing both sides by either or . 
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 11062015 10:29
(Original post by poorform)
Yeah you do need to know them for c3/c4 I think.
Just remember and you can get both of them by dividing both sides by either or . 
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 11062015 10:36
(Original post by tingirl)
Ok! Thanks, that's really helpful!! How about parametric equations  they're just a C4 topic right?
I think this is also helpful.
http://www.mathshelper.co.uk/MEI%20C...on%20Sheet.pdf
Use it to help if you are stuck on past papers. 
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 11062015 10:45
(Original post by poorform)
Yes I believe so.
I think this is also helpful.
http://www.mathshelper.co.uk/MEI%20C...on%20Sheet.pdf
Use it to help if you are stuck on past papers. 
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 11062015 12:06
Please can someone explain this to me. (See attachments)
For question three I understand that the first answer is 32x=4x
However, for the second answer I put (32x)=4x so 3+2x=4x so 3=2x so x=1.5, however according the the mark scheme I get no method marks for this even though I think it is completely correct? They put (32x)=4x, so how come you get a mark for putting a minus on the left but not the right, makes no sense??? 
lizard54142
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 11062015 12:11
(Original post by Connorbwfc)
Please can someone explain this to me. (See attachments)
For question three I understand that the first answer is 32x=4x
However, for the second answer I put (32x)=4x so 3+2x=4x so 3=2x so x=1.5, however according the the mark scheme I get no method marks for this even though I think it is completely correct? They put (32x)=4x, so how come you get a mark for putting a minus on the left but not the right, makes no sense???
is the same as 
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 11062015 12:16

ArielHaeems
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 11062015 12:16
Some notes:
Odd functions: f(x) = f(x) and these functions show rotational symmetry about the origin, order 2
Even functions: f(x) = f(x) and these functions show symmetry across the yaxis
Reflecting f(x) across y=x gives f^1(x) (the inverse function)
The domain of f(x) is the range of f^1(x), and vice versa
If the gradient of f(x) at (x,y) is a, then the gradient of f^1(x) at (y,x) is 1/a
Differentiating: sin(x) > cos(x) > sin(x) > cos(x) > sin(x)...
Differentiating sin(ax) gives acos(ax)
Integrating e^x gives e^x
Integrating f'(x)/f(x) gives ln(f(x))
integrating e^ax gives (1/a)e^ax
Integrating sin(ax) gives (1/a)cos(ax)
Integration by parts formula: uv  (Integral of)vdu
Indefinite integrations introduce +c as an unknown valueLast edited by ArielHaeems; 11062015 at 12:24.
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