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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by NotNotBatman)
    So yes?
    :yep:

    I'm laughing, but I have no idea what that means
    My exam is in a week and I should start revising is the gist of it. :lol:
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    (Original post by imran_)
    sorry I dont get it, im used to make the numerator =0 when its parallel to the x axis
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    I think you should get the gist of it all otherwise you could screw yourself over... Just two my p
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    (Original post by Ayman!)
    I would say no. Any two sets of points should work since it's pretty much proving that they're all on the same line.

    On another note, that KCN is telling me to go study chemistry. 😂
    Love the use of the smiley face to balance out the negativity from the CN-
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    (Original post by NotNotBatman)
    If I had to prove the points say,  C, M and N are collinear then could I show that \overrightarrow {CM} = k \overrightarrow {CN} and \overrightarrow {MN} = l \overrightarrow {NC} where  l , k are just constants.

    So I'm basically asking, does it matter which two set of points are used to prove that three points are collinear?
    Either one of those will suffice, not both. That is, as soon as you show CM = kCN, you're done - nothing else.
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    (Original post by NotNotBatman)
    If I had to prove the points say,  C, M and N are collinear then could I show that \overrightarrow {CM} = k \overrightarrow {CN} and \overrightarrow {MN} = l \overrightarrow {NC} where  l , k are just constants.

    So I'm basically asking, does it matter which two set of points are used to prove that three points are collinear?
    Or just prove CM and MN are parralel and since they intersect(at M) hence collinear.


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    (Original post by Ayman!)
    I would say no. Any two sets of points should work since it's pretty much proving that they're all on the same line.On another note, that KCN is telling me to go study chemistry. 😂
    (Original post by Zacken)
    Either one of those will suffice, not both. That is, as soon as you show CM = kCN, you're done - nothing else.

    (Original post by physicsmaths)
    Or just prove CM and MN are parralel and since they intersect(at M) hence collinear.


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    Thank you all.
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    (Original post by Zacken)
    That's when the gradient becomes infinite.
    Just want to ask because I'm unsure as to what that means:

    They said the tangent at Q is parallel to the y-axis, and we have just calculated dy/dx. But when the tangent is parallel to the y-axis, gradient isn't 0 (as it is when parallel to the x-axis) but infinity. And the only way for the gradient to be infinity is for the denominator of the gradient (dy/dx) to be zero.

    Does this make sense? Sorry if you've answered this before, I'm trying to get back into my maths jam after a couple weeks of science exams.
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    (Original post by Dohaeris)
    Just want to ask because I'm unsure as to what that means:

    They said the tangent at Q is parallel to the y-axis, and we have just calculated dy/dx. But when the tangent is parallel to the y-axis, gradient isn't 0 (as it is when parallel to the x-axis) but infinity. And the only way for the gradient to be infinity is for the denominator of the gradient (dy/dx) to be zero.

    Does this make sense? Sorry if you've answered this before, I'm trying to get back into my maths jam after a couple weeks of science exams.
    That's pretty much exactly right, well done.
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    (Original post by Zacken)
    That's pretty much exactly right, well done.
    Thanks.
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    does anyone know if there are grade boundaries for solomon papers???
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    (Original post by Zacken)
    That's when the gradient becomes infinite.
    When a gradient of a curve is perpendicular to an axis, and you are given an equation which is fraction, make the numerator equal to 0.

    When the gradient of a curve is parallel to an axis, and you are given an equation as a fraction, the denominator is equal to 0.
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    (Original post by Filipo)
    When a gradient of a curve is perpendicular to an axis, and you are given an equation which is fraction, make the numerator equal to 0.

    When the gradient of a curve is parallel to an axis, and you are given an equation as a fraction, the denominator is equal to 0.
    Uh, this makes 0 sense. What if the gradient of the curve is perpendicular to the x-axis, then why would I make the numerator 0? You rely too much on silly rules and too little on actual understanding.
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    I guess you could use 80% for an A and 90% for an A* like normal. It's worth just doing the questions tho coz they're good practice. Also, the edexcel bronze silver gold papers have boundaries of u want to use them.
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    (Original post by Zacken)
    Uh, this makes 0 sense. What if the gradient of the curve is perpendicular to the x-axis, then why would I make the numerator 0? You rely too much on silly rules and too little on actual understanding.
    Hi, would you advise to just think about it logically using dy/dx = change in y / change in x ?
    From that I seem to work out that if a line is parallel to the x-axis, then the numerator of the gradient (change in y) is zero and if a line is parallel to the y-axis, then the denominator of the gradient (change in x) is zero (as the gradient is infinity and 1/0=infinity). Is this correct? Fyi I'm not trying to memorise these as facts, just verifying if the way I approach it is correct.
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    Is there anything you guys can recommend doing to minimise silly errors? I get 90+ on all the papers and I've only got like 4 more for C4 and have done all of the C3 ones and a few solomons for both. For instance I've interpreted 1/root(1+x) as (1+x) to the power of 1/2 , not -1/2 in a binomial question, or I sometimes don't expand a bracket correctly. It's really infuriating for me sometimes because this can be the difference between getting an A* and an A.
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    I hope I'm not the only one that gets this.... :laugh:

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    What you think of June 2013?
    Been getting A*s then did this **** and got a C for that paper lol
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    (Original post by Pablo Picasso)
    What you think of June 2013?
    Been getting A*s then did this **** and got a C for that paper lol
    In general pretty normal. What did you find more difficult about it?
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    (Original post by 13 1 20 8 42)
    In general pretty normal. What did you find more difficult about it?
    2, 7b, 8 man
    You know those Qs were hard, why you saying 'pretty normal'
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    (Original post by yesyesyesno)
    Hi, would you advise to just think about it logically using dy/dx = change in y / change in x ?
    From that I seem to work out that if a line is parallel to the x-axis, then the numerator of the gradient (change in y) is zero and if a line is parallel to the y-axis, then the denominator of the gradient (change in x) is zero (as the gradient is infinity and 1/0=infinity). Is this correct? Fyi I'm not trying to memorise these as facts, just verifying if the way I approach it is correct.
    Yep, that's a really good way to think of it! Well done.
 
 
 
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