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    (Original post by papayarama)
    Is there a quick way or rule to know when to use integration by substitution or by parts? Thanks
    Integration by substitution is usually done to fractions and it is also obvious if the question says something like 'using the substitution u=x+1 or otherwise.....' . Also, when it is a 'function of a function'. Integration by parts is used to integrate products.
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    Boom! after hours and hours spending till late at night trying to do well in C3 so i can accomplish a B (shh dont judge me :P ) i finally have mastered Section A to a grade A level... Just a C in section B and i should get that B but omg i do not like section B. Is it weird that i like C4 section A? sometimes....
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    Thanks! Also, what is the period of a function / how do you find it?
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    (Original post by papayarama)


    Thanks! Also, what is the period of a function / how do you find it?
    The period is the amount of x before the function repeats. Eg, for f(x) = sinx the period is 360 degrees (or 2 pi radians). The same for cosx, while f(x) = tanx has a period of 180 degrees.

    If the question asks for the period of another function they'll probably have a sketch, and then just take the difference between the start of a 'cycle' and end (like wavelength in Physics if that helps)

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    Does anyone have a link for last years paper for more revision? Ant find it anywhere.
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    These type of questions get me every time - can someone explain the answer this to me?
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    (Original post by Hody421)
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    These type of questions get me every time - can someone explain the answer this to me?
    I'm assuming you understand the concept of the range? If not, it's the range (literally) of values that f(x) can take (always for a one - one function).

    For this question, you want the range of f(x) = 0.5 atanx in terms of radians.

    If g(x) = atanx then you can imagine it's simply the tanx graph on it's side. The domain of the one - one tan function is -pi to +pi and so you can just transform this for f(x) - the minimum and maximum are going to be 1/2 of g(x), so the range is -pi/2 <= f(x) <= pi/2

    If you have any more questions just ask!

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    (Original post by emarshy97)
    Does anyone have a link for last years paper for more revision? Ant find it anywhere.
    Your teacher should. I don't think the 2014 papers have been released publicly yet, although they probably will be so soon.

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    (Original post by papayarama)
    Is there a quick way or rule to know when to use integration by substitution or by parts? Thanks
    Integration by substitution will always work when the integral looks like a derivative (think the chain rule - if something does look like this, it's also easy to do in your head).

    Integration by parts is for when you've got a product and it doesn't look like a derivative/you can't see the substation easily.

    Generally the question will tell you if you have to use substation - if it doesn't tell you then I guess think about the above.

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    (Original post by Duskstar)
    I'm assuming you understand the concept of the range? If not, it's the range (literally) of values that f(x) can take (always for a one - one function).

    For this question, you want the range of f(x) = 0.5 atanx in terms of radians.

    If g(x) = atanx then you can imagine it's simply the tanx graph on it's side. The domain of the one - one tan function is -pi to +pi and so you can just transform this for f(x) - the minimum and maximum are going to be 1/2 of g(x), so the range is -pi/2 <= f(x) <= pi/2

    If you have any more questions just ask!

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    But from the mark scheme:
    −π/2 < arctan x < π/2⇒ −π/4 < f(x) < π/4⇒ range is −π/4 to π/4
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    Really looking forward to getting this out of the way! My class finished the work for it about 3/4 months ago, so for 3/4 months all we've been doing is papers. Likewise for C4. I just want these to end... they're boring me now :/
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    (Original post by Hody421)
    But from the mark scheme:
    −π/2 < arctan x < π/2⇒ −π/4 < f(x) < π/4⇒ range is −π/4 to π/4
    Oh yeah that's my fault sorry - that's correct; I haven't done maths in a while lol.

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    Oh boy, with this M1 that idk if I'm looking at an A or B... I need to 90%+ this exam.
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    (Original post by lizard54142)
    x
    (Original post by Leechayy)
    x
    http://www.mei.org.uk/files/papers/c307ja_02c76.pdf

    Q1(ii) how does that work? x
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    (Original post by QueenAryela)
    http://www.mei.org.uk/files/papers/c307ja_02c76.pdf

    Q1(ii) how does that work? x
    Q is the point where |x| = |x-2| + 1. It says verify so you can just use y=1.5
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    (Original post by QueenAryela)
    http://www.mei.org.uk/files/papers/c307ja_02c76.pdf

    Q1(ii) how does that work? x
    Notice that the right hand branch of y=|x| is simply the line y=x.

    Notice that the left hand branch of y=|x-2|+1 is the line y=2-x+1. (since x<2 in this region)

    So you can solve easily for x and y to verify.
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    (Original post by lizard54142)
    Q is the point where |x| = |x-2| + 1. It says verify so you can just use y=1.5
    did that, didnt know how to open the modulus


    (Original post by poorform)
    Notice that the right hand branch of y=|x| is simply the line y=x.

    Notice that the left hand branch of y=|x-2|+1 is the line y=2-x+1. (since x<2 in this region)

    So you can solve easily for x and y to verify.
    That makes sense. I was wondering why it became essentially x = -(x-2)+1

    So it's not possible to work out without the graph?
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    (Original post by QueenAryela)
    did that, didnt know how to open the modulus




    That makes sense. I was wondering why it became essentially x = -(x-2)+1

    So it's not possible to work out without the graph?
    Notice at Q x > 0, so you can just use y=x to find x. You have to then verify this with the other equation
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    (Original post by QueenAryela)
    did that, didnt know how to open the modulus




    That makes sense. I was wondering why it became essentially x = -(x-2)+1

    So it's not possible to work out without the graph?
    Lizard's solution is quicker. You don't need to anything really. You know that the point of intersection must lie on the line y=x so consequently y=x=1.5 so just sub x=1.5 into both sides, show that the equation holds and you are done.

    I suppose you could work it out without the graph if you wanted it's given but doesn't really help that much
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    If anybody wants the 2014 C3 past paper, I will attach both the question paper and the mark scheme/ examiners report to this post.
    Attached Images
  1. File Type: pdf C3-2014-June-Papers.1-4.pdf (192.1 KB, 124 views)
  2. File Type: pdf C3-2014-June-Papers.21-35.pdf (331.2 KB, 232 views)
 
 
 
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