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Anyone need help with any *EDEXCEL* C3 and C4 past paper questions? watch

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    (Original post by sulexk)
    hey there dyslexic duck, do you have the c3 book , I could walk you through those graphs very quickly if you like?

    thanks!
    sadly no.
    im an external candidate cause i didnt manage a decent enough grade to get into uni. ):
    i needed like a minimum B but i ended up with a E..
    home problems and such.

    currently im just using my notes and whatever i can get my hands on in the internet!
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    I have a question sulexk!

    For log(base2)x, how does this become lnx/ln2 like it says on this MS I have been using?
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    (Original post by Get Cape.Wear Cape.Fly.)
    So how do I find the range of a graph? e.g. y=e^(x^2). I can sub x=0 and it works but how do you actually do it?
    The range of a function is the domain of its inverse function
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    I have learnt this rule, that a logarithm of a number can be written like this:

    logx/log(number of base)

    in this case we can write logx/log2

    you could also use ln although in logx/log2, the base is 10, in using ln, the base is "e", so since it is relative we can also use ln, since we will be using ln for bottom and top, ideally we could write it as logx/log2. But we can also write it as, lnx/ln2

    I hope that's cool!
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    (Original post by sulexk)
    Hello there,

    need help with any edexcel c3/c4 past paper questions?

    let me know

    thank you so much!!!
    A curve has the equation y = (3x − 5)3.
    (a) Find an equation for the tangent to the curve at the point P (2, 1). (4)
    The tangent to the curve at the point Q is parallel to the tangent at P.
    (b) Find the coordinates of Q. (3)

    i can do the first bit. but ive never understood if its parralell what do i do?

    thank you
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    hey there is it y=(3x-5)^3? the graph or, y=3(3x-5)?

    thanks!
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    (Original post by skiliya3)
    The range of a function is the domain of its inverse function
    lol thanks but i wasn't asking that :p:
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    Heya people (:

    im on question 7 part c. i did some working and ended up with :

    sine2theta = 2/3

    it now says to find the angles in between 0 and 360.
    how do i use my eq. to do that?

    thanks!
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    (Original post by sulexk)
    hey there is it ? the graph or, y=3(3x-5)?

    thanks!
    its to the power 3. y=(3x-5)^3
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    (Original post by Dyslex!c Duck)
    Heya people (:

    im on question 7 part c. i did some working and ended up with :

    sine2theta = 2/3

    it now says to find the angles in between 0 and 360.
    how do i use my eq. to do that?

    thanks!
    basically enter on your calculator
    Arcsin(sin^-1)2/3
    What ever answer you get, divide that by 2 to get theta
    because its sin to get the other answer, do 180- 2sintheta.
    divide that answer by two to get your second solution.

    To get your third and 4th solution I forgot the method, but a little trial and error toying around eg. 360-theta etc
    Hope I helped
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    If it's parallel, it means same gradient to the gradient you got for part a)

    and thanks sulexk!
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    also how do i approach such question
    (a) Use the identities for cos (A + B) and cos (A − B) to prove that
    2 cos A cos B ≡ cos (A + B) + cos (A − B). (2)
    (b) Hence, or otherwise, find in terms of π the solutions of the equation
    2 cos (x + π2
    ) = sec (x + π6
    ),
    for x in the interval 0 ≤ x ≤ π.


    again i can do the first bit but second bit is confusing ?
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    (Original post by sulexk)
    hey there,

    So we know the domain is (ER)- all real numbers
    Question is how do we workout the range?

    well, if we consider e^x, we know that this function, is such that it will never be less than zero, even if you input x=-10000, you will never obtain 0, so for this function(e^x), the range is simply, (e^x)>0

    but now what about for e^(x^2), well if you consider the graph, real quickly, you can figure our that it is just a parabola(U curve like x^2), so in this case the lowest value is e^0=1, so the range is

    e^(x^2)>=0
    How is it >/= to 0. Isn't it just >0?

    Erm..if you work out the inverse of e^(x^2) is that going to be sqrt(lnx)? If that is the case, then the domain of the inverse would have to be x>/= 1 But then as the domain of the inverse is the range of the original function, wouldn't the range be x>/=1.
    Sorry, I may have not thought about this properly...
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    (Original post by 9MmBulletz)
    basically enter on your calculator
    Arcsin(sin^-1)2/3
    What ever answer you get, divide that by 2 to get theta
    because its sin to get the other answer, do 180- 2sintheta.
    divide that answer by two to get your second solution.

    To get your third and 4th solution I forgot the method, but a little trial and error toying around eg. 360-theta etc
    Hope I helped
    yeah i remember noww
    for the last to you have to 360 plus and take value.
    thanks alot!
    (:
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    Sulexk,

    For a cosec graph with a minimum at (pi/2, −1), a maximum at ( 3pi/2 , −5) and an asymptote with equation x = pi.

    (a) Showing the coordinates of any stationary points, sketch the graph of y =
    Given that
    f : x → a + b cosec x, x ∈ , 0 < x < 2π, x ≠ π,
    (b) find the values of the constants a and b

    Can you explain this to me please?
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    Since they tell you that the tangent to the point Q is parallel to the tangent to the point at P, instantly you assume that both tangents have the same gradients, in this case 9(which is what I obtained from part a).

    Now, this means that we set dy/dx = 9 and since we obtained 9(3x-5)^2, for dy/dx, we set this equal to 9 since there are two points where the gradient is equal to 9, so we should get two values for x.

    now then, we have, 9(3x-5)^2=9
    ==> (3x-5)^2=1
    expand and simplify to obtain:

    9x^2-30x+25=1
    9x^2-30x+24=0

    now factor:

    9x^2-12x-18x+24=0
    3x(3x-4)-6 (3x-4)=0
    (3x-6)(3x-4)=0

    so the tangent has the same gradient when x=2 and when x=4/3

    now we were given the point P(2,1) already

    so we use x=4/3
    put this back into the original equation and we get :

    y=(3(4/3)-5)^3=(4-5)^3=(-1)^3=-1

    so Q coordinates=(4/3,-1)
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    How do you differentiate when the base has got powers thats also got powers;
    like: 7e (^x (^y)?
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    (Original post by lethal786)
    also how do i approach such question
    (a) Use the identities for cos (A + B) and cos (A − B) to prove that
    2 cos A cos B ≡ cos (A + B) + cos (A − B). (2)
    First write out cos (A+B) = cosAcosB-sinAsinB and then cos (A-B)= cosAcosB +sinAsinB.

    If we add these two equations together:

    cos(A+B)+cos(A-B) = 2cos A cos B as required
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    (Original post by ViolinGirl)
    How is it >/= to 0. Isn't it just >0?

    Erm..if you work out the inverse of e^(x^2) is that going to be sqrt(lnx)? If that is the case, then the domain of the inverse would have to be x>/= 1 But then as the domain of the inverse is the range of the original function, wouldn't the range be x>/=1.
    Sorry, I may have not thought about this properly...
    Hi there, you are absolutely correct, sorry about that it should be
    e^(x^2)>=1
    I wrote lowest value is 1 before this,

    sorry the typo!

    thank you so much!
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    (Original post by obviouslystudying)
    How do you differentiate when the base has got powers thats also got powers;
    like: 7e (^x (^y)?
    hey there is it possible if you could expand on this a little, I mean provide an example of the question you are asking?

    thank you so much!!
 
 
 
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