Edexcel FP3 June 2015 - Official Thread

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Reply 440

chemlover12

Original post by nayilgervinho

The mark scheme said something different, could you do it?

the second part of my working out isnt uploading.

Reply 441

mdemda

Once more, if it is possible, can someone explain, why, if we have transformation and plane in vector notation, and we need to find normal to transformed plane, we first cannot find normal to plane and then transform it?

I got ninjasinjps explanation, but I always thought that angle between lines is conserved in linear transformation.

Reply 442

Pri2506

Original post by ninjasinpjs

Hey bud, that was a very challenging question because of the trigonometry stuff, however the concepts are true if you keep to them you can do the question.

1) Find the gradient between the two points
2) Use the equation of the line
3) Use algebra manipulation by the factor formula in the c3 section of the book [ this is the tricky part]
4) Sub the factor formula into the question of your line, and then at the very last part apply double angle formula HINT [ Cos ( a - b) formula is used]

Hope this helps

Thank you! :smile:

Reply 443

chemlover12

Original post by nayilgervinho

The mark scheme said something different, could you do it?

Finally. 2nd part

20150619_200534.jpg482.8KB

Reply 444

ninjasinpjs

Original post by 1 8 13 20 42

As the denominator gets smaller x gets larger; thus the largest denominator (in terms of magnitude) gives the smallest magnitude x values, hence |x| >/= a rather than the other way round

I do not understand, why lxl >/= a, i think i do not understand because of the modulus notation, ill assume that is means x >/= a and x</= -a.

The smallest value of the denominator is +/- 1, oh i SEEEEE, so if you got to 0.5, then x is larger than a. Can't believe i missed this point, thanks bud

Reply 445

bwr19

Sorry to keep asking questions guys, but what on earth has happened in these two stages (highlighted) for this reduction formula? I can see where the nI_n-1 comes from, but how've they managed to get out 2nI_n from all that?

Thanks!

Reply 446

ninjasinpjs

Original post by bwr19

By parts and some algebra, did you want help with the algebra?

Posted from TSR Mobile

(edited 8 years ago)

Reply 447

bwr19

Original post by ninjasinpjs

By parts and some algebra, did you want help with the algebra?

Posted from TSR Mobile

Well basically I've got this:

Unparseable latex formula:

\displaymode 5^n \times 3 - 1 - \int^5_1 nx^{n-1} (2x-1)(2x-1)^{-\frac{1}{2}} dx

But where do I go from there? Any help is much appreciated!

I can see how they got the nI_n-1 out of there, but how did they work with n(2x-1) to make it 2nI_n?

(edited 8 years ago)

Reply 448

luckystar1900

Can anyone help me out on part b of Q57 on review excersise 1 in the edexcel book?? Like seriously how is anyone meant to think of that I am very confused :frown:

Reply 449

Boop.

Hey has anyone done June 2013 Question 2a, why do they multiply the answer by 0.5 after they have used the formula?

FP3 - JUN13 - QP.pdf125.1KB

FP3 - JUN13 - MS.pdf849.4KB

Reply 450

math42

Original post by Boop.

Hey has anyone done June 2013 Question 2a, why do they multiply the answer by 0.5 after they have used the formula?

The formula is for x^2, in this case you have (2x)^2, you need to multiply by a half as when you differentiate you get an extra 2 from the chain rule if that makes sense..try differentiating to convince yourself of this
This is true in general; if it was like 1/root(16x^2 +9) your integral would be (1/4)arsinh(4x/3) for instance

Reply 451

chemlover12

Original post by bwr19

Well basically I've got this:

Unparseable latex formula:

\displaymode 5^n \times 3 - 1 - \int^5_1 nx^{n-1} (2x-1)(2x-1)^{-\frac{1}{2}} dx

But where do I go from there? Any help is much appreciated!

I can see how they got the nI_n-1 out of there, but how did they work with n(2x-1) to make it 2nI_n?

Reply 452

Boop.

Original post by 1 8 13 20 42

Thanks mate

Reply 453

Hazllye

Hi can anyone help with question 5a from the June 2013 paper please?
I'm stuck with how to use the j+k and i-k in the eigen vectors and where they've got the (0,1,1) and (1,0,-1) column vectors from on the mark scheme

mark scheme https://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20MS%20-%20FP3%20Edexcel.pdf
paper
https://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20QP%20-%20FP3%20Edexcel.pdf

Reply 454

math42

Original post by Hazllye

Those vectors are just j + k and i - k in column vector form
You want to use the relationship Ax = lambda(x) with A the matrix, x the eigenvector and lambda the eigenvalue