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    (Original post by Gilo98)
    Have you got f(0) yet?
    I decided to come up with an equation of the line using c1 stuff and got y - 10 = -10/4 (x + 2) and from that got f(0)=5 bit confused now
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    (Original post by physicsmaths)
    I have noticed this year, M3 similar style to IAL, FP2 similar style to IAL not difficulty, physics today was also similar style to IAL.


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    for the jan 2015 one ?
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    (Original post by adorablegirl1202)
    I decided to come up with an equation of the line using c1 stuff and got y - 10 = -10/4 (x + 2) and from that got f(0)=5 bit confused now
    it is five. but look below at the conventional approach using similar triangles. Apply a similar method for f(5) on the RHS branch.
    Attached Images
     
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    (Original post by Gilo98)
    Can someone explain why the answer to June 2013 8b is 21/25 and not -21/25? If you see the equation as 21/25sec(x-73.74) then surely the min is negative??
    Sec = 1/cos()
    Therefore the larger cos is, the smaller sec is, so the minimum is when cos is at its maximum, which is 1



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    (Original post by Gilo98)
    Can someone explain why the answer to June 2013 8b is 21/25 and not -21/25? If you see the equation as 21/25sec(x-73.74) then surely the min is negative??
    Speed is a scalar quantity.
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    (Original post by shuff_joe)
    Sec = 1/cos()
    Therefore the larger cos is, the smaller sec is, so the minimum is when cos is at its maximum, which is 1



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    I get that, but when you look at a sketch of secx, it goes below the x axis so how can the minimum be +ve?
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    (Original post by randlemcmurphy)
    Speed is a scalar quantity.
    Thought this would be the reason why, thank you
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    Can anyone tell me if this is correct or not?
    So say the angle is 30 degrees and you found that out using arcsin.
    And if question asked to find values from -180<x<180
    (Using the CAST/Quadrant method)
    Would the solutions/values be : 30,150,-30,-150?
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    (Original post by callyx27)
    So how's everyone feeling about tomorrow?
    Quite good, although am still worried because I need to get like 90+UMS. What about you?
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    Is "proof by counter example" part of the spec? Is there a reason it crops up in the Soloman papers? It really makes me panic!


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    Can someone explain how to do b) ? I thought because x was 0 you'd use 5-2(5-2(0)) = -5 but the model answer and mark scheme both say the answer is e^2 - 4. Help ?
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    which solomon papers do you guys recommend doing?
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    Also, sorry to bug, is it 90 ums in each paper? Or 90 ums average against both papers?


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    (Original post by Nurishment)


    Can someone explain how to do b) ? I thought because x was 0 you'd use 5-2(5-2(0)) = -5 but the model answer and mark scheme both say the answer is e^2 - 4. Help ?
    which paper is this?
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    (Original post by TheAnnabelle)
    Also, sorry to bug, is it 90 ums in each paper? Or 90 ums average against both papers?


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    90 UMS average against both papers i think
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    (Original post by Gilo98)
    I get that, but when you look at a sketch of secx, it goes below the x axis so how can the minimum be +ve?
    You have to consider it as 1/cos not sec on its own, I understand your confusion but the sec graph has an indeterminable minimum y value

    So treat sec as 1/cos
    25/21×(1/cosx) (or whatever the fraction was)

    Cos needs to be it's maximum for 1/cos to be the minimum, so cos(x) = 1

    25/21 × (1/1)

    25/21

    I apologise if I got the fraction wrong, on phone and forgot what it was :P



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    (Original post by Gilo98)
    it is five. but look below at the conventional approach using similar triangles. Apply a similar method for f(5) on the RHS branch.
    Thanks, I can do this now!
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    (Original post by Nurishment)


    Can someone explain how to do b) ? I thought because x was 0 you'd use 5-2(5-2(0)) = -5 but the model answer and mark scheme both say the answer is e^2 - 4. Help ?
    You need to use the other equation because x is now bigger than for. Sub in x=5 into e^(2x-8) -4
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    (Original post by imedico10)
    which paper is this?
    June 2013 withdrawn paper
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    (Original post by TheAnnabelle)
    Also, sorry to bug, is it 90 ums in each paper? Or 90 ums average against both papers?


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    90 ums average, you need a total of 180 in c3 and c4
 
 
 
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