You are Here: Home >< Maths

# FP1 Question Help

Announcements Posted on
Four hours left to win £100 of Amazon vouchers!! Don't miss out! Take our short survey to enter 24-10-2016
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
Attached Images

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

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 14, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Who is getting a uni offer this half term?

Find out which unis are hot off the mark here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams