OCR MEI FP2 Thread - AM 27th June 2016

OCR MEI FP2 Thread - AM 27th June 2016

I didn't see any other 2016 threads for this module, please let me know if one is already established!



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Also, I've written an Enhanced Examination Paper for this module, specifically including MEI questions to help you all

It is meant to be a significantly more challenging examination of general concepts specified by MEI. It is fully doable for students with knowledge of standard A-level mathematics and further mathematics.

Being able to make a good start on all of the questions is a good indicator that you are on track to a strong A* in this module.

https://www.furthermathstutor.co.uk/fp2_mei.pdf

Let me know if you have any questions on the paper.

Does anybody have the 2015 FP2 paper?

Could anyone help with part ii) of this question please? From June 2010.

Need help on june 09 2ii, ive worked out the eigenvector but I dunno how to solve the equation? Could anyone help? have no idea

Posted from TSR Mobile

This is the method I usually use when asked to solve equations using eigenvectors:

The eigenvector you worked out is (-1,6,1) (or a multiple of it). The matrix multiplied by the eigenvector equals the eigenvector multiplied by its eigenvalue. If v is the eigenvector and x is the eigenvalue:

Mv = xv

In our equation, x is -1, so Mv = -v.

The right hand side of the above equation, -v, is negative the eigenvector, so (1,-6,-1). The right hand side of the equation we have to solve is (-0.1,0.6,0.1), which is -v * (-1/10). So multiple by the above equation by (-1/10):

Mv*(-1/10) = (-0.1,0.6,0.1)

So we can see from the left hand side that the solution is v*(-1/10), which is (0.1, -0.6, 0.1). x = 0.1, y = -0.6, z = -0.1.

still confused I dont see how xyz equals that. I know Ms=Ys but dont understand what happebs next

hmmm actually I think I doo, not entirely sure though. Is there another example I can do?

Posted from TSR Mobile

Think back to solving simultaneous equations with matrices in FP1 - for example, if you have a matrix M with elements:

1, 2, 3
4, 5, 6
7, 8, 9

And you multiply it by a 1x3 matrix (x,y,z) and set it equal to a 1x3 vector with values (a,b,c), then it is equivalent to writing:

1x + 2y + 3z = a
4x + 5y + 6z = b
7x + 8y + 9z = c

So finding the 1x3 matrix (x,y,z) means you have found the values for x, y and z in the equations. Let's call the 1x3 matrix (x,y,z) "T", and the solution matrix (a,b,c) "R". We have MT = R. We know the numerical values of R, so we need to find T.

In your question, let's compare the final stage of working to the above:

Mv*(-1/10) = (-0.1,0.6,0.1)
MT = R

So M is M, R is (-0.1,0.6,0.1), and T is v*(-1/10). So T is -1/10 times the eigenvector, which is -1/10 * (-1,6,1), which is (0.1,-0.6,-0.1), so these are the values for x, y and z.

Just wanna check this properly; is an eigenvector with all the elements' signs swapped compared to what's on the mark scheme still a valid answer to give? Is it in any way incorrect?

June 10, 3a!!)

EvaluateM^2(1,-1,1/3)
wouldnt it be 4/3 times the eigenvector??? the mark scheme only multiplies it by 4

Posted from TSR Mobile

The eigenvalue is 2, so it is 2^2 * v which is 4 * v.