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Stuck on c2 question help pls URGENT EXAM TOMMOROW

http://mei.org.uk/files/papers/c2_june_2012.pdf

questions 6 and 8 i have no idea how to do, even after checking the markscheme if anyone would be kind enough to explain how to do them i would appreciate it greatly
Reply 1
exam tomorrow plz ..
Reply 2
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Original post by fsdghsdgs
http://mei.org.uk/files/papers/c2_june_2012.pdf

questions 6 and 8 i have no idea how to do, even after checking the markscheme if anyone would be kind enough to explain how to do them i would appreciate it greatly


6) Basic log graphs which should be explained in your book. You have a general equation of the line being log(y)=mlog(x)+c\log(y)=m\log(x)+c. You know two points on the graph. So you can deduce the equation in this form, yes?

Then once you have it, solving for yy is straight-forward as you simply exponentiation both sides with base 10.

8) Dunno what method you are taught to solve these types so here is one. The argument of sine 2θ2\theta while the range is only for θ\theta. We want to be for 2θ2\theta so we multiply through the range to get 0<2θ4π0<2\theta \leq 4 \pi . Then you only need to arcsin both sides to get your principal value of 2θ2\theta, call this P=arcsin(0.7)P=\arcsin(0.7) (also arcsin means sine inverse, in case you don't know). Another value is πP\pi-P. Then any further solutions would be achieved by adding on 2π2\pi onto both of these answers.
(edited 6 years ago)
Reply 4
Original post by RDKGames
6) Basic log graphs which should be explained in your book. You have a general equation of the line being log(y)=mlog(x)+c\log(y)=m\log(x)+c. You know two points on the graph. So you can deduce the equation in this form, yes?

Then once you have it, solving for yy is straight-forward as you simply exponentiation both sides with base 10.

8) Dunno what method you are taught to solve these types so here is one. The argument of sine 2θ2\theta while the range is only for θ\theta. We want to be for 2θ2\theta so we multiply through the range to get 0<2θ4π0<2\theta \leq 4 \pi . Then you only need to arcsin both sides to get your principal value of 2θ2\theta, call this P=arcsin(0.7)P=\arcsin(0.7) (also arcsin means sine inverse, in case you don't know). Another value is πP\pi-P. Then any further solutions would be achieved by adding on 2π2\pi onto both of these answers.


yeh i guess i should have got q6 just panicking and thnx for explaining q8 we learnt this stuff over a year ago and im only revisiting some of it now, thnx :smile:

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