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Edexcel Core 3 - 21st June 2016 AM Watch

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    (Original post by SeanFM)
    Domain - consider the function and look for x values that will give you 'undefined' values. You can often compare it to normal functions and work from there.
    What I mean by that is, for example,

    If you had ln(2x+5) then you know that for something simple like lnx, x has to be >0 so you know that in ln(2x+5), 2x+5 > 0 and so x> -5/2.

    Also for something like tanf(x) where f(x) is any function of x, f(x) must not be equal to any multiple of pi/2 as it is undefined (easy way to remember - tanx = sinx/cosx, and cosx = 0 when x = pi/2, hence tanx is .../0 = undefined.

    And lastly for fractions, you can't divide by 0 - so if you had \frac{x}{x-3} for example, then x can take any value but 3

    For the range, it just takes practice. Once you know the domain, it may help to put in numbers close to the domain to see if you can spot the range. If, for example,
    \frac{x}{x-3} had the domain  4 \leq x \leq 7 (for whatever reason, rather than just x = anything but 3) then you could find f(4) = 4 and f(7) = (7/4) and deduce that it is a decreasing function. You could also write the function as \frac{x-3+3}{x-3}  = 1 - \frac{1}{x-3} to see it more easily.

    Also, test out large values of x. If you had \frac{3x + 1}{x-2} then as x gets really large (eg 10000, but ideally infinity) y becomes \frac{3\times 10000 - 1}{10000 - 2} which is pretty much equal to 3*10000/10000 = 3, and you can tell that it's always going to be greater than 3 as x gets larger as the numerator is slightly greater than 3x, and the numerator is slightly less than 1x so it will always be 3.0000'..12321566 or whatever and you can deduce that y>3.
    Ahh ... right thanks a lot
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    (Original post by SeanFM)
    Domain - consider the function and look for x values that will give you 'undefined' values. You can often compare it to normal functions and work from there.
    What I mean by that is, for example,

    If you had ln(2x+5) then you know that for something simple like lnx, x has to be >0 so you know that in ln(2x+5), 2x+5 > 0 and so x> -5/2.

    Also for something like tanf(x) where f(x) is any function of x, f(x) must not be equal to any multiple of pi/2 as it is undefined (easy way to remember - tanx = sinx/cosx, and cosx = 0 when x = pi/2, hence tanx is .../0 = undefined.

    And lastly for fractions, you can't divide by 0 - so if you had \frac{x}{x-3} for example, then x can take any value but 3.

    For the range, it just takes practice. Once you know the domain, it may help to put in numbers close to the domain to see if you can spot the range. If, for example,
    \frac{x}{x-3} had the domain  4 \leq x \leq 7 (for whatever reason, rather than just x = anything but 3) then you could find f(4) = 4 and f(7) = (7/4) and deduce that it is a decreasing function. You could also write the function as \frac{x-3+3}{x-3}  = 1 + \frac{3}{x-3} to see it more easily.

    Also, test out large values of x. If you had \frac{3x + 1}{x-2} then as x gets really large (eg 10000, but ideally infinity) y becomes \frac{3\times 10000 - 1}{10000 - 2} which is pretty much equal to 3*10000/10000 = 3, and you can tell that it's always going to be greater than 3 as x gets larger as the numerator is slightly greater than 3x, and the numerator is slightly less than 1x so it will always be 3.0000'..12321566 or whatever and you can deduce that y>3.
    Nice explanation. Thanks!
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    Are there any proofs we need to remember for C3? Tyyy in advance
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    (Original post by physicalgraffiti)
    Are there any proofs we need to remember for C3? Tyyy in advance
    As far as I know, you have to remember the identities from chap 6 and from chap 7,
    the Sin(A plus/minus B), cos(A plus/minus B) and tan(A plus/minus B).
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    If cot2x=0 then why are the x values 45 and 135


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    (Original post by Supermanxxxxxx)
    If cot2x=0 then why are the x values 45 and 135


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    I'm assuming the range is 0 < x < 360.

    cot2x = 1/tan2x

    so 1/tan2x = 0

    When you flip it over to get tan2x on top, you are doing 0/0 ---> 0/0 = undefined. So you're basically looking at wherever there are asymptotes on the y = tanx graph. This is where x = 90 and 270 for tanx. So, for tan2x, it's 45 and 135 degrees because you divide 90 and 270 by two
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    (Original post by Don Pedro K.)
    I'm assuming the range is 0 < x < 360.

    cot2x = 1/tan2x

    so 1/tan2x = 0

    When you flip it over to get tan2x on top, you are doing 0/0 ---> 0/0 = undefined. So you're basically looking at wherever there are asymptotes on the y = tanx graph. This is where x = 90 and 270 for tanx. So, for tan2x, it's 45 and 135 degrees because you divide 90 and 270 by two
    Wouod Tanx=0/0 not just be 0,180 and 360 though


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    (Original post by Supermanxxxxxx)
    Wouod Tanx=0/0 not just be 0,180 and 360 though


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    No because 0/0 does not equal 0 Try it on your calculator; you get "MATH ERROR".
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    (Original post by Don Pedro K.)
    No because 0/0 does not equal 0 Try it on your calculator; you get "MATH ERROR".
    Thanks that's just me being retarded


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    (Original post by Supermanxxxxxx)
    Thanks that's just me being retarded


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    Hey don't worry about it we all have our moments xD Like me in M1... l0l
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    If u get below an E (40ums) does that mean u get automatically 0 UMS (U) and it counts as if u've never done the exam?
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    (Original post by Sophie Wilczek)
    If u get below an E (40ums) does that mean u get automatically 0 UMS (U) and it counts as if u've never done the exam?
    No, you still get UMS, provided it is at least 1 UMS.
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    (Original post by NotNotBatman)
    No, you still get UMS, provided it is at least 1 UMS.
    Thank you!
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    Hi! Would someone help out in 9(b) and (c)?
    http://www.madasmaths.com/archive/iy...apers/c3_d.pdf
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    (Original post by sabahshahed294)
    Hi! Would someone help out in 9(b) and (c)?
    http://www.madasmaths.com/archive/iy...apers/c3_d.pdf
    b this is when denominator = 0
    c just use the Rcos form you derived
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    (Original post by 13 1 20 8 42)
    b this is when denominator = 0
    c just use the Rcos form you derived
    Oh. Okay, Thank you!
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    (Original post by 13 1 20 8 42)
    b this is when denominator = 0
    c just use the Rcos form you derived
    Could you explain b a bit further? Is it when the denominator = 0 because anything / 0 = undefined (i.e. asymptote?)
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    (Original post by Don Pedro K.)
    Could you explain b a bit further? Is it when the denominator = 0 because anything / 0 = undefined (i.e. asymptote?)
    Pretty much, yeah. The magnitude of the function is shooting off to infinity, as it gets arbitrarily large the closer the denominator gets to 0, and so the line of the function will "tend to" the line x = beta, i.e. it will become almost vertical. The key thing about an asymptote is the graph starting to behave more and more like it the closer and closer it gets but never touching it, but in practice when actually finding them you can just check where the function isn't defined.
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    So for the equation y = (3x + 5)/(6x + 2) you would find the asymptote by setting 6x + 2 = 0 and solve for x?
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    (Original post by Don Pedro K.)
    So for the equation y = (3x + 5)/(6x + 2) you would find the asymptote by setting 6x + 2 = 0 and solve for x?
    Indeed

    (vertical asymptotes are found this way anyhow. In general horizontal are more complicated)
 
 
 
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