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Edexcel Core 3 - 21st June 2016 AM

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Reply 60
Original post by imran_
https://a086a5a2f39bda93734c56a63fab1d7be0a9ba38.googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/January%202014%20(IAL)%20QP%20-%20C34%20Edexcel.pdf

hi for question 4b i need help for the range. So far ive subbed 0 into it to get g(x)<= 3

but i dont understand how the other value is calculated


What's the horizontal asymptote?
Reply 61
Original post by Zacken
What's the horizontal asymptote?


it tends to 0
Reply 62
Original post by imran_
it tends to 0


Uh, no it doesn't, have a think about it.
Reply 63
Original post by Zacken
Uh, no it doesn't, have a think about it.


it tends to infinity
Reply 64
Original post by imran_
it tends to infinity


Are you even bothering or just making random guesses?
Reply 65
Original post by imran_
it tends to infinity


Substitute in a really large value of xx, what number do you get close to?
Reply 66
Original post by Zacken
Are you even bothering or just making random guesses?


from what I can see, the asymptote is going towards the x axis. Theres a reason why I asked for help on this question as i have limited understanding
Original post by imran_
from what I can see, the asymptote is going towards the x axis. Theres a reason why I asked for help on this question as i have limited understanding

Don't do it by observation. Do his method as he said above
Reply 68
Original post by imran_
from what I can see, the asymptote is going towards the x axis. Theres a reason why I asked for help on this question as i have limited understanding


Sub in a really large value of xx, what number do you get close to?
Reply 69
Original post by Zacken
Substitute in a really large value of xx, what number do you get close to?


gets closer to 0.5, is this the actual method to use? or is there an alternative approach
Reply 70
Original post by imran_
gets closer to 0.5, is this the actual method to use? or is there an alternative approach


The alternative method to use is limxx+92x+3=limx1+9x2+3x=12\displaystyle \lim_{x \to \infty} \frac{x+9}{2x+3} = \lim_{x\to \infty} \frac{1 + \frac{9}{x}}{2 + \frac{3}{x}} = \frac{1}{2}
Reply 72
Original post by boyyo
what does iygb stand for




My guess is 'if you're (a) gonad buster'? (I know it's buster gonad, but still :redface:)
(edited 7 years ago)
Reply 74
Original post by SeanFM
My guess is 'if you're (a) gonad buster'? (I know it's buster gonad, but still :redface:)


if you got...brains? :tongue:
Original post by 1 8 13 20 42
if you got...brains? :tongue:


:mmm: I fancy your suggestion better than mine :tongue:
Reply 76
Original post by SeanFM
My guess is 'if you're (a) gonad buster'? (I know it's buster gonad, but still :redface:)


Original post by 1 8 13 20 42
if you got...brains? :tongue:


LOL yh I think that makes more sense
How hard are the AEA meant to be compared to the c3/c4 exams?


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Original post by ben_urwinator
How hard are the AEA meant to be compared to the c3/c4 exams?


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Quite a bit more difficult
Reply 79
Can anyone please explain why the square root of e^-2x is e^-x Thanks :smile:

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