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1. (Original post by tangotangopapa2)

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Hey I saw you posted some snail question last night it looked interesting could you post it here or the link to the book?
Hey I saw you posted some snail question last night it looked interesting could you post it here or the link to the book?
Sure.
Link to the post : http://www.thestudentroom.co.uk/show....php?t=4198061

Link to the book : I am not sure if I am allowed post the link to the book this way. I found it in the google search. Anyway : https://docs.google.com/file/d/0B4II...t?pref=2&pli=1

And the problem :
3. Is the following answer presented by physicsandmathstutor not fully correct or is there a flaw in my reasoning?
4. (Original post by tangotangopapa2)
Is the following answer presented by physicsandmathstutor not fully correct or is there a flaw in my reasoning?
I see what you mean but i don't understand why he multiplied the LHS of the inequality by (x-1)^2 instead of just (x-1) ?
5. We don't know weather or not is (x-1) positive. If x-1 is positive then we don't flip inequality sign if it is negative then we flip. But if we multiply both sides by (x-1)^2 then we are multiplying by positive number so we don't have to flip the inequality sign.

(Original post by hellomynameisr)
I see what you mean but i don't understand why he multiplied the LHS of the inequality by (x-1)^2 instead of just (x-1) ?
6. (Original post by hellomynameisr)
I see what you mean but i don't understand why he multiplied the LHS of the inequality by (x-1)^2 instead of just (x-1) ?
common technique with inequalities, need to square denominators before multiplying to ensure that its positive in this case as otherwise multiplying through flips the inequality (in this case, if the denominator is always positive then thats fine)
7. (Original post by tangotangopapa2)
We don't know weather or not is (x-1) positive. If x-1 is positive then we don't flip inequality sign if it is negative then we flip. But if we multiply both sides by (x-1)^2 then we are multiplying by positive number so we don't have to flip the inequality sign.
When you test between 0 and 1 in the factorised cubic you get a positive answer which doesn't agree with the inequality.
8. (Original post by PhyM23)
When you test between 0 and 1 in the factorised cubic you get a positive answer which doesn't agree with the inequality.
The formula a/(1-r) is used in the solution which is true only if -1 < r < 1 i.e if x > 1. So, we can not conclude weather or not the given inequality is satisfied when x < 1 based on calculation using this formula.

If you take x = 0.5 (to the inequality of the question), L.H.S x + 2 is 2.5 but RHS is infinitely large. Clearly x + 2 is less than R.H.S.
9. (Original post by tangotangopapa2)
The formula a/(1-r) is used in the solution which is true only if -1 < r < 1 i.e if x > 1. So, we can not conclude weather or not the given inequality is satisfied when x < 1 based on calculation using this formula.

If you take x = 0.5 (to the inequality of the question), L.H.S x + 2 is 2.5 but RHS is infinitely large. Clearly x + 2 is less than R.H.S.
correct, for x<1 sequence diverges so sum is infinitely large
10. (Original post by samb1234)
correct, for x<1 sequence diverges so sum is infinitely large
Thanks
So, the ans is 0 < x < root 2, isn't it?
11. (Original post by tangotangopapa2)
Thanks
So, the ans is 0 < x < root 2, isn't it?
should be
12. (Original post by samb1234)
common technique with inequalities, need to square denominators before multiplying to ensure that its positive in this case as otherwise multiplying through flips the inequality (in this case, if the denominator is always positive then thats fine)
(Original post by tangotangopapa2)
We don't know weather or not is (x-1) positive. If x-1 is positive then we don't flip inequality sign if it is negative then we flip. But if we multiply both sides by (x-1)^2 then we are multiplying by positive number so we don't have to flip the inequality sign.
Ah okay I see. Thanks guys
13. You're very welcome.
(Original post by hellomynameisr)
Ah okay I see. Thanks guys
14. (Original post by tangotangopapa2)
Thanks
So, the ans is 0 < x < root 2, isn't it?
I thought the sign was the other way round for some reason so I edited my reply. I feel there is a crucial property of infinite series we're failing to see. I will quote someone who knows more about this than me.

Zacken

Please could you take a look at this. Have we been taught an over-simplified version of this topic in school which doesn't explain series like this? It is post 163. Thanks
15. (Original post by hellomynameisr)
Ah okay I see. Thanks guys
It's not strictly necessary for something like this, by using the fact that r<1 i can easily conclude that x>1 so i can safely not square and restrict the domain to x>1 as that is always positive
16. I was actually about to post up thto same questions a couple of days ago couldn't find the year so what year is it?? I feel like there's a vital this in the series we are not seeing as well,hope someone with a better understanding replies on this

17. Guys this is another guy who writes solutions here was his comment correcting and saying the answe was 0 to root2
18. Ah yes I get what tango was pointing out now (sorry I missed this). I was researching earlier and there are different types of summations. I think I was trying to overcomplicate it though.
19. http://www.examsolutions.net/maths-r...utorial-1b.php

Taking a look at this, then the original answer of 1<x<root2 is correct isn't it?
http://www.examsolutions.net/maths-r...utorial-1b.php

Taking a look at this, then the original answer of 1<x<root2 is correct isn't it?
To use S = a/(1-r) , -1<r<1 , which doesn't hold if 0<x<1 as the series diverges, so the formula can't apply to this set of values.

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