# C4 solomon paper question. Slightly confused....

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#1
So basically im not sure how to attempt question 4 B).
Do i go about making t---->infinity or something.
Cant really get my head around it. Only part of the paper i got wrong.
Any explanations would be great for that and similar questions thanks!

https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf
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5 years ago
#2
(Original post by ThatGuyRik)
So basically im not sure how to attempt question 4 B).
Do i go about making t---->infinity or something.
Cant really get my head around it. Only part of the paper i got wrong.
Any explanations would be great for that and similar questions thanks!

https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf
Essentially you have to show that as t gets very large, it tends to a value lower than 3.
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#3
(Original post by Jai Sandhu)
Essentially you have to show that as t gets very large, it tends to a value lower than 3.
is that it? so what your saying is i just put in any large number to show T still stays below 3?
0
5 years ago
#4
(Original post by ThatGuyRik)
So basically im not sure how to attempt question 4 B).
Do i go about making t---->infinity or something.
Cant really get my head around it. Only part of the paper i got wrong.
Any explanations would be great for that and similar questions thanks!

https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf
Yeah, that'll do it. You should find that the limit as t->inf is 3, and because the function is strictly increasing, that means you can never hit 3 in finite time.
0
5 years ago
#5
(Original post by ThatGuyRik)
is that it? so what your saying is i just put in any large number to show T still stays below 3?
No, you need to show that *for all* large numbers, T<3.
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#6
(Original post by Smaug123)
Yeah, that'll do it. You should find that the limit as t->inf is 3, and because the function is strictly increasing, that means you can never hit 3 in finite time.
okay so i did make t=infinite(which is 1 mark) but how would i go about showing that T still stays below 3(second mark)?
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5 years ago
#7
In part (a), you should get something like ln((x-6)/(x-3))=6t + ln2.
So if you sub in x=3, you get ln (-3/0) = 6t + ln2. It is impossible to do ln(-3/0), so x cannot be 3
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#8
(Original post by sunsri101)
In part (a), you should get something like ln((x-6)/(x-3))=6t + ln2.
So if you sub in x=3, you get ln (-3/0) = 6t + ln2. It is impossible to do ln(-3/0), so x cannot be 3
yeah i figured it out. I was being stupid lol. Thanks anyway!
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5 years ago
#9
(Original post by ThatGuyRik)
okay so i did make t=infinite(which is 1 mark) but how would i go about showing that T still stays below 3(second mark)?
I solved the DE to get .

. Does that help?
1
#10
(Original post by Smaug123)
I solved the DE to get .

. Does that help?
yeah i got it from what you said earlier, thanks!. My differential equation isn't like that though as i kept it as ln. But yeah its still correct just checked the MS.

Thanks again!
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5 years ago
#11
The mark scheme substitutes in x=3 and shows that the resulting expression is

and states...

as x → 3, t → ∞ ∴ cannot make 3 g

Some important things to note are that t=0 at x=0 and t tends to infinity as x approaches 3, since at this point the expression inside (ln) is switching from positive to negative (and therefore does not exist in the real domain).

(Here's a link to the mark scheme: https://goo.gl/th4Vft)

And here is a link to the graph of the final equation with t on the y axis and x on the x axis. You can see that a value for t does not exist for x>=3, since there is a vertical asymptote at x=3. Thought this might be good to help you visualise what's going on:

https://goo.gl/x2d58f
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