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    Anyone up for some friendly competition to get full ums in both c3 and c4
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    (Original post by Revisionbug)
    Anyone up for some friendly competition to get full ums in both c3 and c4
    Bring it on!!


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    (Original post by Lamalam)
    Is c3 solomon paper worth doing? Coz i am afraid i dont have much time to finish them all @@




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    Do like a few of them
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    now M2 and M3 are over, solid C3 til thursday
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    Could someone help me with 7(b) on Solomon paper K.

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    What I did was this:

    (x+\frac{1}{x})^2=4sec^2\theta=4  (1+tan^2\theta) and (y+\frac{1}{y})^2=4(1+cot^2 \theta)

    tan^2\theta=\frac{1}{4}(x-\frac{1}{x})^2 and 4cot^2\theta=(y-\frac{1}{y})^2

    so \frac{4}{tan^2\theta}=16(x-\frac{1}{x})^{-2}=(y-\frac{1}{y})^{-2}

    \dfrac{4x}{x^2-1}=\dfrac{y^2-1}{y}

    The answer in the mark scheme is \dfrac{4x^2}{(x^2+1)^2}+\dfrac{4  y^2}{(y^2+1)^2}=1


    I was just wondering whether these two answers are equivalent, and if not, then where have I gone wrong. Thanks.
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    (Original post by brittanna)
    Could someone help me with 7(b) on Solomon paper K.

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    What I did was this:

    (x+\frac{1}{x})^2=4sec^2\theta=4  (1+tan^2\theta) and (y+\frac{1}{y})^2=4(1+cot^2 \theta)

    tan^2\theta=\frac{1}{4}(x-\frac{1}{x})^2 and 4cot^2\theta=(y-\frac{1}{y})^2

    so \frac{4}{tan^2\theta}=16(x-\frac{1}{x})^{-2}=(y-\frac{1}{y})^{-2}

    \dfrac{4x}{x^2-1}=\dfrac{y^2-1}{y}

    The answer in the mark scheme is \dfrac{4x^2}{(x^2+1)^2}+\dfrac{4  y^2}{(y^2+1)^2}=1


    I was just wondering whether these two answers are equivalent, and if not, then where have I gone wrong. Thanks.
    I'm pretty sure they're not equivalent.

    How did you get \tan^2 \theta=\frac{1}{4}(x-\frac{1}{x})^2 and 4\cot^2 \theta=(y-\frac{1}{y})^2 ?
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    is the c3 and c4 exam in the morning or afternoon?
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    (Original post by Anjna)
    is the c3 and c4 exam in the morning or afternoon?
    Morning I believe.


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    Hey guys wondering if anyone could help me.

    For the C3 Jan 08 paper Q)1) I used the method of equating coefficients instead of long division, and I got the first part right which is (2x^2-1).

    My questions is how do you get the second part the (dx+e) bit.

    Do I do (2x^2-1)(dx+e)=2x^4-3x^2+x+1. Then equate coefficients?

    Any guidance, tips or help is appreciated!

    Wait...I think I get it, do you do (dx+e)(1)=2x^4-3x^2+x+1.
    Thus d =1.
    But how is e=0... I though you would have done e=1?
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    (Original post by Anjna)
    is the c3 and c4 exam in the morning or afternoon?
    Both Exams are in the morning.
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    I really don't understand what you mean by equating the coefficients, but I always do long division. Once you've done the long division your remainder should be x . Because your remainder is x, you write it on top as (dx+e) So your d=1 and e=0 as there is no e. So your overall equation would be (2x^2 - 1) + x all over (x^2 - 1). Where a=2 b=0 c=-1 d=1 and e=0. Hope this helps.
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    (Original post by justinawe)
    I'm pretty sure they're not equivalent.

    How did you get \tan^2 \theta=\frac{1}{4}(x-\frac{1}{x})^2 and 4\cot^2 \theta=(y-\frac{1}{y})^2 ?
    4+4tan^2\theta=x^2+2+\frac{1}{x^  2}

    4tan^2\theta=x^2-2+\frac{1}{x^2}=(x-\frac{1}{x})^2

    \Rightarrow tan^2\theta=\frac{1}{4}(x-\frac{1}{x})^2

    I got 4cot^2\theta=(y-\frac{1}{y})^2 in the same way, but using the equation with y instead of x. It doesn't look like the two answer are equivalent, so I just want to know where I went wrong.
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    (Original post by brittanna)
    4+4tan^2\theta=x^2+2+\frac{1}{x^  2}

    4tan^2\theta=x^2-2+\frac{1}{x^2}=(x-\frac{1}{x})^2

    \Rightarrow tan^2\theta=\frac{1}{4}(x-\frac{1}{x})^2

    I got 4cot^2\theta=(y-\frac{1}{y})^2 in the same way, but using the equation with y instead of x. It doesn't look like the two answer are equivalent, so I just want to know where I went wrong.
    Brittanna thanks so much for the Tips on how to use your the Calculator to check back my answer. Surprsingly it's not even about the Calculator but how you think about Maths - verifying whether your logic is rigorous enough.

    Just did C3 Jan 2013, surprisingly the Calculator stopped me from making horrendously horrible errors. I think I was a bit had a bit of Anxiety whilst I was doing it; but definitely great I did it today: Felt like an exam.

    If anyone sees a question they don't get 100% remember to just Keep on Going.

    It sounds silly but what I do for Chemistry and Physics, is when I'm stuck I flip to an Easy Question which is worth like 10 Marks and do that in 3-5 minutes and then go back again. It's what I just did on C3, and it definitely worked.

    From doing this it is very easy to maintain momentum.

    How to Solve it: A New Aspect of Mathematical Method - George Polya
    http://www.math.utah.edu/~pa/math/polya.html
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    (Original post by Better)
    Brittanna thanks so much for the Tips on how to use your the Calculator to check back my answer. Surprsingly it's not even about the Calculator but how you think about Maths - verifying whether your logic is rigorous enough.

    Just did C3 Jan 2013, surprisingly the Calculator stopped me from making horrendously horrible errors. I think I was a bit had a bit of Anxiety whilst I was doing it; but definitely great I did it today: Felt like an exam.

    If anyone sees a question they don't get 100% remember to just Keep on Going.

    It sounds silly but what I do for Chemistry and Physics, is when I'm stuck I flip to an Easy Question which is worth like 10 Marks and do that in 3-5 minutes and then go back again. It's what I just did on C3, and it definitely worked.

    From doing this it is very easy to maintain momentum.
    That's kind of what I do in physics. Although for me it's more when my hand starts aching, i'll go and do some of the calculations (which are also usually easier) .

    But yeah, although it isn't good to become too dependant on calculators, they can be a really big help as well.
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    (Original post by brittanna)
    That's kind of what I do in physics. Although for me it's more when my hand starts aching, i'll go and do some of the calculations (which are also usually easier) .

    But yeah, although it isn't good to become too dependant on calculators, they can be a really big help as well.
    I'm glad I'm not the only one then.

    I may just start off with the Essay Style Questions on Wednesday, before I do anything else!

    But back to work, thanks again - time for me to do Chem 5 Jan 13, then Phys 5 Jan 13.....then argh chem 4 work after that al night...
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    Has anyone got that black thick A2 core for edexcel textbook, if so question 5a on the sample exam paper I'm stuck (C3)
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    (Original post by brittanna)
    4+4tan^2\theta=x^2+2+\frac{1}{x^  2}

    4tan^2\theta=x^2-2+\frac{1}{x^2}=(x-\frac{1}{x})^2

    \Rightarrow tan^2\theta=\frac{1}{4}(x-\frac{1}{x})^2

    I got 4cot^2\theta=(y-\frac{1}{y})^2 in the same way, but using the equation with y instead of x. It doesn't look like the two answer are equivalent, so I just want to know where I went wrong.
    Actually, after subbing in values, I think they may actually be equivalent. But I'm not sure how to justify it.

    I'm not sure if you can just square root both sides like you did though, you need to account for negative roots as well.
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    hey guys, I am quite confused about the edexcel c3 june 2006 q5a, Name:  c3.gif
Views: 165
Size:  11.2 KB and here is how I worl the questionName:  image.jpg
Views: 182
Size:  501.6 KB, am I correct?

    i am confused about the cancelling of cos^2 (2K), why can it be eliminated?
    my teacher told me that
    if x^2 = 2x , I cannot cross out anything but putting the 2x on the same side as x^2 and carry on working.
    why in the q5, the cos^2 (2k) can be eliminated? will this cause a loss of answer?
    in what situation can a quantity be eliminated?

    thank you soooo much!! please help!
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    Stuck on this, any help is appreciated! Stuck on the whole question!
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    (Original post by Delusional09)
    Stuck on this, any help is appreciated! Stuck on the whole question!
    For part (a), use the identity in the question and the fact that cos3x=cos(2x+x).

    For part (b), try expressing the two fractions as a single fraction.
 
 
 
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