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# Step I 1987 Q1 watch

1. It's been a while since I have done some maths, so have lost my confidence a bit. Just thought I'd ask TSR to check if this is correct

Question

Differentiated to get

Stationary points at

Then work out f(x) at stationary points, eventually got

Then I looked at the next stationary point ie with k+1 instead of k. This gave

Therefore the next stationary point gives a value for f that is the previous one multiplied by which implies a GP with a common ratio .
2. Looks all good to me... differentiate using product rule, factorise and set = 0 (presuming the solution is not to be considered as a is not specified as positive or negative) then sub result back in and use cos(A+B) formula with the sin part going to 0. Reasoning in final part seems sound also - wouldn't expect anything less with someone of your grades though!
3. (Original post by miml)
It's been a while since I have done some maths, so have lost my confidence a bit. Just thought I'd ask TSR to check if this is correct

Question

Differentiated to get

Stationary points at

Then work out f(x) at stationary points, eventually got

Then I looked at the next stationary point ie with k+1 instead of k. This gave

Therefore the next stationary point gives a value for f that is the previous one multiplied by which implies a GP with a common ratio .
That's completely sound [as you've shown that, for the sequence , is a constant].
4. Yes, Putting f(x-(2pi/b)) as f(x) shows that the original function returns , but multiplied by a factor independent of x.

So each cycle of the graph is identical to the previous one, but multiplied by a scale factor.
5. Oh, and don't have things like cos (arctan a/b).

If tan z = a/b , then what is cos (arctan a/b)?

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