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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Original post by kkboyk
x2−x>x+3x \frac{x}{2-x} > \frac{x+3}{x}

I would probably do this by taking everything over to the left and looking for critical values
Hey. I was wondering if someone could clear something up for me regarding Taylor series. When using (x - a), do you sub in positive a or negative (-a) into your differentials? With the two examples in the book one does positive an the other does negative, but from most exam questions I've seen you sub in positive a. Thanks.
Reply 822
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Cpj16
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Original post by economicss
Please could anyone do me a massive favour and post a worked solution to question 16b as I always seem to just get stuck at the same stage in these types, thanks :smile:image.jpg
hi what is the answer

I tried but I don't think my answer is right. I got 8u^2 + 8v^2 =0 then gave up xD

What is the correct answer I ll have another go
Original post by Cpj16
hi what is the answer

I tried but I don't think my answer is right. I got 8u^2 + 8v^2 =0 then gave up xD

What is the correct answer I ll have another go


Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks :smile:
Reply 824
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Cpj16
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Original post by economicss
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks :smile:


I m sorry I cant get it I have tried 4 times :frown: I m so close

will try again later
Original post by economicss
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks :smile:


I'll try now.

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Original post by LikeASomebody
Hey. I was wondering if someone could clear something up for me regarding Taylor series. When using (x - a), do you sub in positive a or negative (-a) into your differentials? With the two examples in the book one does positive an the other does negative, but from most exam questions I've seen you sub in positive a. Thanks.


If its (x-a), you substitute "a", that is the positive value. If its (x+a), then it becomes "-a", since that would made it (x+a) ; this is when they ask you to expand it in ascending powers of (x-a).
Original post by techfan42
If its (x-a), you substitute "a", that is the positive value. If its (x+a), then it becomes "-a", since that would made it (x+a) ; this is when they ask you to expand it in ascending powers of (x-a).

Thank you
Reply 828
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Rkai01
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Attachment not found
image.jpg
Original post by economicss
Please could anyone do me a massive favour and post a worked solution to question 16b as I always seem to just get stuck at the same stage in these types, thanks :smile:image.jpg


Edit: can you read it alright?
(edited 7 years ago)
image.jpg529.9KB
Original post by economicss
Thank you! The answer is (u-1/2)^2 + (v-1/2)^2= 1/2 thanks :smile:


1465070002791.jpg

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Original post by Rkai01
Attachment not found
image.jpg

Edit: can you read it alright?


Thank you, that's great! How do we know that the real and imaginary parts are the same please? :smile:
Original post by Krollo


Thanks so much :smile:
Original post by economicss
Thank you, that's great! How do we know that the real and imaginary parts are the same please? :smile:


From the first part. When you remove the modulus sign and expand, you find out that x=y
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Original post by economicss
Thank you, that's great! How do we know that the real and imaginary parts are the same please? :smile:

Because locus of p is y=x and transformation maps this meaning all real parts equal imaginary parts and you use what z equals in terms of u and v then equate
Original post by Rkai01
Because locus of p is y=x and transformation maps this meaning all real parts equal imaginary parts and you use what z equals in terms of u and v then equate


I see, thanks so much! :smile:
Original post by techfan42
From the first part. When you remove the modulus sign and expand, you find out that x=y


Got it, thank you :smile:
Original post by Cpj16
I m sorry I cant get it I have tried 4 times :frown: I m so close

will try again later


Thanks so much for trying, appreciate it :smile:
Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?
Original post by fpmaniac
Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?


You could use implicit differentiation but normally I would just square root to get plus minus r and go from there like usual, hope that helps :smile:
Original post by fpmaniac
Anyone know how to solve the polar equation where its r squared and you have to find the coordinates of perpendicular or parallel line?


I think you do this

find x^2 or y^2, where x^2= r^2sin^2(theta) and y^2=r^2sin^2(theta)

then you can differentiate implicitly, so for example you'd have 2y dy/d(theta) = f'(theta) => dy/d(theta) = f'(theta)/2y, and since it equals 0 the numerator must equal zero so you just put f'(theta)=0 and forget about the 2y

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