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# FP1 Question Help Watch

1. Very easy question however I don't get the right answer. I've asked my friends who also do FP1 and they get the same as me.

Question 4: http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

I got

x= π/8 + (πn)/2
x= (11π)/24 + (πn)/2

However the mark scheme suggests:

x= π/8 + (πn)/2
x= (-π)/24 + (πn)/2
2. QUOTE=2014_GCSE;65789279]Very easy question however I don't get the right answer. I've asked my friends who also do FP1 and they get the same as me.

Question 4: http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

I got

x= π/8 + (πn)/2
x= (11π)/24 + (πn)/2

However the mark scheme suggests:

x= π/8 + (πn)/2
x= (-π)/24 + (πn)/2[ QUOTE]

Just tried it and got the solution from the markscheme - want me to send over a pic? :P
3. (Original post by Chickenslayer69)
QUOTE=2014_GCSE;65789279]Very easy question however I don't get the right answer. I've asked my friends who also do FP1 and they get the same as me.

Question 4: http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

I got

x= π/8 + (πn)/2
x= (11π)/24 + (πn)/2

However the mark scheme suggests:

x= π/8 + (πn)/2
x= (-π)/24 + (πn)/2[ QUOTE]

Just tried it and got the solution from the markscheme - want me to send over a pic? :P
YES PLEASE! Basically, I got P.V as -1/6pi and so you do 2pi(n) - 1/6pi and then 2pi(n) + pi - -1/6pi which gives you 2pi(n) + 7/6pi.

4. if you dont understand something ive done just ask sry for handwriting lool
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5. (Original post by Chickenslayer69)

if you dont understand something ive done just ask sry for handwriting lool
We always learned it, that you get the two values by the the initial value (in this case -1/6pi) and then 2pi - the initial value (which in this case would be 7/6pi). So why doesn't this work? Is it because you have to do it within a certain domain?
6. QUOTE=2014_GCSE;65790239]We always learned it, that you get the two values by the the initial value (in this case -1/6pi) and then 2pi - the initial value (which in this case would be 7/6pi). So why doesn't this work? Is it because you have to do it within a certain domain?[ QUOTE]

Drawing the sin graph in this case works - you can see from the graph I drew that the initial value (-1/2) occurs at -pi + 1/6pi = -5/6pi, and the other value is 1/6pi as you have already worked out. 7/6pi would be impossible since it is out of the pi range either way (values for general solution can only be from -pi and pi unless you are given an interval). So you always look for the values the initial value (in this case -1/2) gives in the range from -pi to pi.

Hope this helps

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Updated: June 15, 2016
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