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# C3 Jan 09 Q8 Tweet

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1. C3 Jan 09 Q8
A question I just don't understand how to do the question, and when looking at the mark scheme, no idea why what was done was done.

Parts (c) and (d) I do not understand at all.

http://www.sci2.co.uk/maths/C3/Edexc...20Jan%20MS.pdf

Theres a link to the mark scheme, if anyone could help me, would save me having to go into college during study leave to find my maths teacher and ask them xD

Thanks!
2. Re: C3 Jan 09 Q8
You temp is given by f(t) = 10 + 5cos(x), where x = 15t - alpha

Now if you look at any cosine curve, it goes up and down, varying in value from a minimum of -1 to a maximum of +1.

So 10 + 5cos(x) will have a minimum value when cos(x) = -1 (and a maximum value when cos(x) = +1).

So f(t)_min = 10 + 5(-1) = 10 - 5 = 5
=============================

cos(x) = -1 when x = pi/2 rads or 180 degrees

i.e. x = 15t - alpha = 180
15t = 180 + 53 (alpha = 53 from part(b))
15t = 233
t = 15.5
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3. Re: C3 Jan 09 Q8
The minimum value of the cosine function is -1, hence once you've done part a, you can use your simplification of part a into part b, with theta = 15t. if the minimum value of cos (a) = -1, what can you say about the minimum of 5cos(a)?
4. Re: C3 Jan 09 Q8
Thanks a lot ppls, steve you hit the nail right on the head I get it fully now.