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    http://www.examsolutions.net/maths-r...utorial-1a.php

    He uses the squaring method here
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    (Original post by PhyM23)
    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.
    So if we combine the condition that X>0 they gave we see that X can be > 0 but less then 1 and be valid
    So we include and the answer is 0<X<root2
    But when we say X>0 we are including 1 right and that can't be right because 1 is not a valid solution in our calculations? So our answer could be 0<x<root2 where x is not equal to 1?
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    (Original post by NatoHeadshot)
    So if we combine the condition that X>0 they gave we see that X can be > 0 but less then 1 and be valid
    So we include and the answer is 0<X<root2
    But when we say X>0 we are including 1 right and that can't be right because 1 is not a valid solution in our calculations? So our answer could be 0<x<root2 where x is not equal to 1?
    1 is a valid solution, series is although not diverging at 1 is still equal to infinity (1+1+1+1+1+1... >3)
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    I am having trouble (I always have troubles ) with following problem. The question is interesting. You might want to attempt it before looking at the answer if you haven't already.The problem:Name:  question.png
Views: 120
Size:  30.1 KBAttachment 562508The solution and the trouble: and solution : http://www.physicsandmathstutor.com/pat/solutions-2010/ question 11.
    Spoiler:
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    Name:  solution.png
Views: 127
Size:  39.9 KB
    How did he know that A=1/6 and B=6/1 but not 2/5?
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    (Original post by tangotangopapa2)
    I am having trouble (I always have troubles ) with following problem. The question is interesting. You might want to attempt it before looking at the answer if you haven't already.The problem:Attachment 562508The solution and the trouble:
    Spoiler:
    Show
    Name:  solution.png
Views: 127
Size:  39.9 KB
    How did he know that A=1/6 and B=6/1 but not 2/5?
    The attachment is invalid it says. Wont let me open it
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    (Original post by hellomynameisr)
    The attachment is invalid it says. Wont let me open it
    I am sorry. I was having trouble with attachments. Edited the post.
    Solution : http://www.physicsandmathstutor.com/pat/solutions-2010/ question 11

    How did he know that A and B were 6/1 and 1/6 and not 2/5 and 5/2?
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    (Original post by tangotangopapa2)
    I am sorry. I was having trouble with attachments. Edited the post.
    Solution : http://www.physicsandmathstutor.com/pat/solutions-2010/ question 11

    How did he know that A and B were 6/1 and 1/6 and not 2/5 and 5/2?
    I think since it says that there is an equal possibility to get all the scores, it wouldnt matter if it was 1/6 or 2/5 as you get the same answer with either.

    Perhaps it was just the person that done the answers quickest and easiest way to approach it.
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    (Original post by hellomynameisr)
    I think since it says that there is an equal possibility to get all the scores, it wouldnt matter if it was 1/6 or 2/5 as you get the same answer with either.

    Perhaps it was just the person that done the answers quickest and easiest way to approach it.
    If ye take 2/5 then.
    2 d1 + 5d2 - 7 = S
    does not have equal probability for all the scores.

    For instant d1 = 1 and d2 = 3 gives S=10 and
    d1 = 6 and d2 = 1 gives S =10.

    There is no combination that gives S=34.
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    (Original post by tangotangopapa2)
    I am having trouble (I always have troubles ) with following problem. The question is interesting. You might want to attempt it before looking at the answer if you haven't already.The problem:Name:  question.png
Views: 120
Size:  30.1 KBAttachment 562508The solution and the trouble: and solution : http://www.physicsandmathstutor.com/pat/solutions-2010/ question 11.
    Spoiler:
    Show
    Name:  solution.png
Views: 127
Size:  39.9 KB
    How did he know that A=1/6 and B=6/1 but not 2/5?
    The important part is the equal probability part. If we look at the list of numbers for 1 and 6 we have 1-6 and 6,12...36. As you can see there is only one way to get any particular outcome as the only overlap is the 6,6. If we looked at 2 and 5 as a solution then we get some issues with overlap e.g 27 i could have 25 and 2 or 15 and 12 so this condition isnt met
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    (Original post by samb1234)
    The important part is the equal probability part. If we look at the list of numbers for 1 and 6 we have 1-6 and 6,12...36. As you can see there is only one way to get any particular outcome as the only overlap is the 6,6. If we looked at 2 and 5 as a solution then we get some issues with overlap e.g 27 i could have 25 and 2 or 15 and 12 so this condition isnt met
    Thanks for the reply.
    I see that 1/6 is the only way to satisfy all the conditions but how would you figure out that A and B are 1/6 and 6/1 in the first place? They could be any real numbers.

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    (Original post by tangotangopapa2)
    Thanks for the reply.
    I see that 1/6 is the only way to satisfy all the conditions but how would you figure out that A and B are 1/6 and 6/1 in the first place? They could be any real numbers.

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    technically there is a second solution. Since the sign of the constants isn't disclosed in the question, either a and b are both positive or both negative (we can't have one of each as otherwise our simultaneous eq isnt solvable for c). taking the negative form gives you a similar solution, with a+b=-7 and both a and b being negative. The 1 and 6 solution (or -1, -6 ) is easily spotted by looking at the early numbers -it works because using the numbers on the one dice added to the 6 dice i can make every single number in between them, so 7,8,9,10,11,12 and so on meaning i can create every single number between 7 and 42. In comparison with say 5 and 2 or 4 and 3 i am taking jumps, so taking the start i can make 7,9,11,12,13,14 but there isn't anyway for me to make 8 or 10 (as in this case to make numbers in between numbers on the big dice i sometimes have to add to a preceding number, e.g. to make 11 which is between 10 and 15 i have to go back to 5 and add 6 which i obviously cant do at the very start or at the end of the series
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    Question 18 here:
    https://www2.physics.ox.ac.uk/sites/..._pdf_10581.pdf
    Solution:
    https://oxfordpat.wordpress.com/oxfo...1/#comment-349

    When the switch is connected to e, does the capcitor not get charged so should the graph not look like this instead of what was drawn on the first graph
    (attached)
    or basically should the graphs not be switched around? Because when teh switch is connected to the other end it gets no voltage so it discharges right?
    Also, any ideas how he estimated "signifcant time"
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    (Original post by NatoHeadshot)
    Question 18 here:
    https://www2.physics.ox.ac.uk/sites/..._pdf_10581.pdf
    Solution:
    https://oxfordpat.wordpress.com/oxfo...1/#comment-349

    When the switch is connected to e, does the capcitor not get charged so should the graph not look like this instead of what was drawn on the first graph
    (attached)
    or basically should the graphs not be switched around? Because when teh switch is connected to the other end it gets no voltage so it discharges right?
    Also, any ideas how he estimated "signifcant time"
    Significant time is 5 times RC.

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    (Original post by tangotangopapa2)
    Significant time is 5 times RC.

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    is that a standard thing?
    Also would you mind clearing my graph question?
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    (Original post by NatoHeadshot)
    is that a standard thing?
    Also would you mind clearing my graph question?
    5 RC or 5 times time constant is usually considered to be time to fully charge a capacitor. The first graph of the solution is current-time graph. Initially, there is no charge so there is high rate of increase of charge I.e. high current. The rate of increase of charge decreases exponentially.

    Yours is voltage-time graph and is correct.
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    wait wait, Ive done C1, C2, M1, M2, FP1 and S1.So do I have to self teach C3 and C4 this summer omds fml
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    (Original post by CaiusMartius)
    wait wait, Ive done C1, C2, M1, M2, FP1 and S1.So do I have to self teach C3 and C4 this summer omds fml
    Precisely.
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    (Original post by tangotangopapa2)
    5 RC or 5 times time constant is usually considered to be time to fully charge a capacitor. The first graph of the solution is current-time graph. Initially, there is no charge so there is high rate of increase of charge I.e. high current. The rate of increase of charge decreases exponentially.

    Yours is voltage-time graph and is correct.
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    So the Order of the graphs is right in the solution? When the capacitor is connected the battery circuit the current decreases exponentially and when it's one without the battery it increases???
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    (Original post by NatoHeadshot)
    So the Order of the graphs is right in the solution? When the capacitor is connected the battery circuit the current decreases exponentially and when it's one without the battery it increases???
    You are right about the first case.
    In the second case i.e. while discharging, the current decreases exponentially too. Since, the direction of flow of current is opposite to that when charging, this current has been shown negative.
    It is still decreasing nonetheless.

    Yes, the order of graph is right.


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    Hi, does anyone know what I should learn in prep for the exam? So far I've only learnt AS Physics and Maths, some core 3 and some A2 physics. I'm already planning on learning all of A2 physics and maths this summer. Which F.Maths modules should I learn (I haven't done any of it so far.) And also, which BPhO papers should I do?
 
 
 
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