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

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    (Original post by Apple24)
    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?
    hey there apple, is the maximum 5 or -5?


    thanks!
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    (Original post by sulexk)
    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)

    you are absolute amazing dude... thank you very much that has actually made me understand it :P .. thanks mate ill get bak to u soon
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    hey there, thanks man!
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    (Original post by sulexk)
    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!!
    well, the question goes like this;

    y=xe^x^2

    the book says to use the product rule and the chain rule to differentiate but its very confusing.
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    Q6 ( im working backwards :') )

    June 2007 C3.

    Part c) Solve for 0 < x < 2pi the equation

    3sinx + 2cosx = 1

    i worked out the 3sinx part to the harmonic form:
    root13Sin(x+33.7)

    i got the angles: 16.1, 163. and 376.1.

    the MS got x = 2.273 or x = 5.976
    not sure how they obtained that..
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    cool!

    imagine, u=x and v=e^(x^2)

    now the product rule tells us that:

    dy/dx=u(dv/dx)+v(du/dx)

    since u=x then du/dx=1

    but now we have something a little more complexed, with e^(x^2)

    so we have to use the chain rule.

    imagine this seperately, we have worked out the first part, now this is the second part:

    so we have e^(x^2)
    so lets say y=e^u where u=x^2

    this is seperate not related to the above

    so y=e^u and u=x^2
    now then dy/du= e^u (derivative of e^x is same as y=e^x)

    now we differentiate "u" with respect to "x":

    du/dx= 2x

    so we now multiply: dy/du x du/dx=dy/dx=2x(e^u)

    where u=x^2

    so this becomes: 2xe^(x^2)

    so now we have u=x du/dx=1
    and that v=e^(x^2) and dv/dx=2xe^(x^2)
    so then dy/dx=x(2xe^(x^2))+ e^(x^2)(1)

    and this is equal to: 2x^2(e^(x^2))+e^(x^2)

    I hope this helps!
    if you need any more with this please let me know

    thank you so much!
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    (Original post by Dyslex!c Duck)
    Q6 ( im working backwards :') )

    June 2007 C3.

    Part c) Solve for 0 < x < 2pi the equation

    3sinx + 2cosx = 1

    i worked out the 3sinx part to the harmonic form:
    root13Sin(x+33.7)

    i got the angles: 16.1, 163. and 376.1.

    the MS got x = 2.273 or x = 5.976
    not sure how they obtained that..
    they must have given their answers in radians and since 376.1 is out of interval 0<x<2pi
    It won't be included
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    (Original post by sulexk)
    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!
    Ah cool No prob. Well, you knew anyway :p:
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    (Original post by sulexk)
    they must have given their answers in radians and since 376.1 is out of interval 0<x<2pi
    It won't be included
    yeah. i converted my deg answers into radians by multiplying by pi and dividing it by 180.
    however the answers still dont match
    D:
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    (Original post by sulexk)
    they must have given their answers in radians and since 376.1 is out of interval 0<x<2pi
    It won't be included
    Actually. The angles are wrong, so even if you converted those into radians you would not get the correct answer.

    The values for the angle for x+33.7 will be : 16.1. 163.9 and 376.1

    But then you need to find the values for x-the actual angle:
    These will be -17.6, 130.2 and 342.4.

    If you look at the original range it is between 0 and 2pi. So the negative angle will not be in the range.

    When the other 2 are converted into radians, they are the correct answers as in the MS.
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    (Original post by Dyslex!c Duck)
    yeah. i converted my deg answers into radians by multiplying by pi and dividing it by 180.
    however the answers still dont match
    D:
    Look below-I have answered this
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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!!!
    Hi,

    Thank you for making this thread.

    Questions:

    1. How do I find the range of a function? Must I always sketch it or can I just subs x = 0 into the function to find the value/s of y?
    2. How do I find the domain?

    And lastly, I don't understand the direction of the inequality arrows and which way they go when listing the range/domains.

    Thanks.
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    (Original post by lethal786)
    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
    Its parallel, so they have the same grad.

    grad = 9

    therefore dy/dx = 9
    9(3x - 5)^2 = 9
    (3x - 5)^2 = 1
    x = 4/3 or x = 2 (at P)

    Wen x = 4/3, y = ((3 x 4/3) - 5)^3 = (-1)^3 = -1

    Therefore, Q(4/3, -1)
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    (Original post by sulexk)
    hey there apple, is the maximum 5 or -5?


    thanks!
    For the y co-ord it's at -5!

    And I have another Q:

    Using this as my dy/dx = -e^2 x^-2 +e^x (so in words: minus e to the power of two multiplied by x to the power of minus two plus e to the power of x)

    How do I get the gradient of 3/4e^2 like it says in the MS if I insert in x=2?

    Thank you!
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    (Original post by ViolinGirl)
    Look below-I have answered this
    i should've worked it all out in radians instead
    such a bad habit of mines.

    according to the marksheme it says 180 - their 16.1 (which got) = 163.9..
    and 360 + their 16.1 which is 376.1
    just the last mark i cant get..
    how would i go about correcting this?
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    (Original post by obviouslystudying)
    well, the question goes like this;

    y=xe^x^2

    the book says to use the product rule and the chain rule to differentiate but its very confusing.
    You do-you start off by using the product rule.

    y=uv where u= x and v = e^x^2

    when you differentiate these you should get: du/dx = 1 and dv/dx= 2xe^x^2

    Then dy/dx = vdu/dx + udv/dx

    e^x^2 + 2x^2e^x^2

    This can then be simplified.

    As far as I can see, it only really involves the product rule. I may have omitted something. What is the answer?
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    sorry sulexk- I hijacked your thread. I'll leave now...I just felt like helping. :o:
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    (Original post by Dyslex!c Duck)
    i should've worked it all out in radians instead
    such a bad habit of mines.

    according to the marksheme it says 180 - their 16.1 (which got) = 163.9..
    and 360 + their 16.1 which is 376.1
    just the last mark i cant get..
    how would i go about correcting this?
    What is the last mark? Those are the only angles you need. I was only talking about the 16.1 to explain about the range, but as the angle must be above 33.7, you don't count the 16.1 (sorry, i don't know if thats what you are asking...)
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    how do you find dy/dx of y=sin(x) ....from first principles?
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    (Original post by ViolinGirl)
    You do-you start off by using the product rule.

    y=uv where u= x and v = e^x^2

    when you differentiate these you should get: du/dx = 1 and dv/dx= 2xe^x^2

    Then dy/dx = vdu/dx + udv/dx

    e^x^2 + 2x^2e^x^2

    This can then be simplified.

    As far as I can see, it only really involves the product rule. I may have omitted something. What is the answer?
    yeah, thats how its shown, i just cant get dv/dx
    thanks alot
 
 
 
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