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AS Mathematics C1 MEI 13th of May 2015

Just wondering what people think will be on this years C1 exam on wednesday! :biggrin:
Original post by alevelup
Just wondering what people think will be on this years C1 exam on wednesday! :biggrin:


Maths :wink:
Reply 2
Haha thanks for that :wink: I mean what kind of things are more likely to come up in the section B stuff, looking at other years
Original post by alevelup
Haha thanks for that :wink: I mean what kind of things are more likely to come up in the section B stuff, looking at other years


Well I sat C1 last year; you're pretty much guaranteed to get a polynomial question of sorts in section B.
Reply 4
any1 got last yrs paper?
Reply 6
All 2014 papers are now up on the AQA website. Not sure about the other exam boards. OCR seem quite slow for it..
(edited 8 years ago)
They tend to examine everything in the module so you may as well prepare for it all
Can someone help me with question 12 part (iii) ?

http://www.ocr.org.uk/Images/176221-mark-scheme-unit-4751-introduction-to-advanced-mathematics-june.pdf

How do you know what order to put the equation in? Does this affect the usage of the discriminant?

Can someone expand that equation and throw it on the other side to get the coefficient of x^2 positive and see what they get please.
Bump
Original post by Slenderman
Bump


Link the paper! You posted the mark scheme.
Original post by Slenderman
Can someone help me with question 12 part (iii) ?

http://www.ocr.org.uk/Images/176221-mark-scheme-unit-4751-introduction-to-advanced-mathematics-june.pdf

How do you know what order to put the equation in? Does this affect the usage of the discriminant?

Can someone expand that equation and throw it on the other side to get the coefficient of x^2 positive and see what they get please.


I've done it.

Make the ys equal so 1/(x-2) = -x + k

Multiply both sides by (x-2): 1 = (x-2)(-x+k)

1 = -x^2 + kx + 2x -2k

x^2 + -kx - 2x + 2k + 1 = 0

x^2 -(k+2)x + 2k + 1 = 0

Then you have to use the discriminant, so b^2 - 4ac = (k+2)^2 - (4 * 1 * (2k+1)) = 0
(you're told it's a tangent, therefore only one POI, so the discriminant = 0), where a = 1, b = k+2, c = 2k+1. That bit threw me for a bit as I didn't realise the 2k + 1 was one term.

Multiply it all out and eventually you get k^2 - 4k = 0

k(k-4) = 0

k = 0 or k = 4.
(edited 8 years ago)
Thank you, sorry for not replying earlier. Was grinding through past papers.
I can't find the exact question but if you have a circle equation and the equation of a tangent, how do you find the equation of the other parallel tangent? Obviously you have the gradient but what point would you sub in? Thanks
Original post by AlexParmenter
I can't find the exact question but if you have a circle equation and the equation of a tangent, how do you find the equation of the other parallel tangent? Obviously you have the gradient but what point would you sub in? Thanks

I'm not sure if this is allowed BUT, if the tangent meets the circle at point (x,y), the other tangent parallel to it would meet the circle at point (-x,-y) because of the symmetry of circles. Now you have a point [(-x,-y)] and the gradient [same as the other one because they are parallel], so use y-y0=m(x-x0).
Original post by gagafacea1
I'm not sure if this is allowed BUT, if the tangent meets the circle at point (x,y), the other tangent parallel to it would meet the circle at point (-x,-y) because of the symmetry of circles. Now you have a point [(-x,-y)] and the gradient [same as the other one because they are parallel], so use y-y0=m(x-x0).


Thank you! And good luck everyone !
Just did the exam. I thought that Section B was quite difficult. I have to rush it as I sent too long trying to perfect Section A. I am not great at mental artithematic so I though I'll take a long time on Section A. Even then I knew that I got one question wrong. I hope that this is the 4th time lucky but I doubt I will get anythig better than a D. :frown:

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