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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Original post by economicss
Thanks so much! It says in the answers though that the answer for part c is 1, but not sure whether it's wrong or not?! :smile:


Sorry, yes. I rushed this so didn't double check! The answer is 1 because the lowest point wasn't the correct one, it was the point at which the locus is at a normal to the radii. I can draw a diagram up if you want?
Reply 981
can someone explain in words, how to solve question 8b on June 2013 (R) Paper thanks

Also is there a 2015R paper??- if there is can someone set me a link :biggrin:
link to paper June 2013 R :
https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202013%20(R)%20QP%20-%20FP2%20Edexcel.pdf

edit: I don't understand what or why we differentiate
(edited 7 years ago)
Original post by Cpj16
can someone explain in words, how to solve question 8b on June 2013 (R) Paper thanks

Also is there a 2015R paper??- if there is can someone set me a link :biggrin:
link to paper June 2013 R :
https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202013%20(R)%20QP%20-%20FP2%20Edexcel.pdf

edit: I don't understand what or why we differentiate


Find the point on the curve such that the tangent is parallel to the initial line, then calculate it's y coordinate.
Can you see what to do next?
Original post by Alby1234
Can anyone telling me what I'm doing wrong in part b? I'm separating the variables but it gives a nasty reciprocal on on of the integrals which proves difficult to solve. Am I going about it the right way?fp2 capture 5.JPG


do you have the answers?
Original post by ImJared
Sorry, yes. I rushed this so didn't double check! The answer is 1 because the lowest point wasn't the correct one, it was the point at which the locus is at a normal to the radii. I can draw a diagram up if you want?


If you could please that would be really helpful, thanks :smile:
Reply 985
Original post by EricPiphany
Find the point on the curve such that the tangent is parallel to the initial line, then calculate it's y coordinate.
Can you see what to do next?


Thanks for your help.
I don't understand what the y co-ordinate there is
because the polar co-ordinate (r, theta)
Original post by Cpj16
Also is there a 2015R paper??- if there is can someone set me a link :biggrin:


There isn't.
Original post by Cpj16
Thanks for your help.
I don't understand what the y co-ordinate there is
because the polar co-ordinate (r, theta)


We have y = r sin(θ).
(edited 7 years ago)
Original post by economicss
If you could please that would be really helpful, thanks :smile:


Just drew this up quickly - a bit scruffy but it demonstrates what I'm talking about. The lowest point is where the line from the origin to the circle is at a tangent (as the gradient no longer gets steeper). Therefore, you can use simple Pythagoras to determine the length from the origin to the point.

Reply 989
Original post by EricPiphany
We have y = r sin(θ).


Ahh thanks

Also when we have an modulus inequality question with which have the greater or equal to sign.
Then what do we do?!
Original post by Cpj16
Ahh thanks

Also when we have an modulus inequality question with which have the greater or equal to sign.
Then what do we do?!


With complex numbers or reals?
Reply 991
Original post by EricPiphany
With complex numbers or reals?


real

x+2≥3/x
(edited 7 years ago)
Reply 992
I have added an example of what I mean
Original post by Cpj16
real


Solve exactly as you do the ones with a strictly greater than sign. Then check both sides of any interval behaves as expected, and if not, remove it from the solution.
Sorry this is a really bad explanation, either someone else will help, or send an example that I can do.
Reply 994
Original post by EricPiphany
Solve exactly as you do the ones with a strictly greater than sign. Then check both sides of any interval behaves as expected, and if not, remove it from the solution.
Sorry this is a really bad explanation, either someone else will help, or send an example that I can do.


I posted one above
Original post by Cpj16
I posted one above


Capture.PNG
( I'm assuming you've been taught how to do this in the past.)T here are other ways too, you could do case work.
Reply 996
Original post by EricPiphany
Capture.PNG
( I'm assuming you've been taught how to do this in the past.)T here are other ways too, you could do case work.


thanks this was a really good explanation
(edited 7 years ago)
Reply 997
Will it always follow the same pattern
Original post by ImJared
Just drew this up quickly - a bit scruffy but it demonstrates what I'm talking about. The lowest point is where the line from the origin to the circle is at a tangent (as the gradient no longer gets steeper). Therefore, you can use simple Pythagoras to determine the length from the origin to the point.


Thanks very much! I get that the 2 is the radius but how did you find root 5 please, bit confused about how you know where to draw the lines to find the angle, thanks :smile:
Original post by economicss
Thanks very much! I get that the 2 is the radius but how did you find root 5 please, bit confused about how you know where to draw the lines to find the angle, thanks :smile:


Distance from centre to origin. Use the distance formula.

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