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C3 Help

Having trouble with 2. b) in getting the last mark.

http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC3-W-QP-JUN11.PDF

http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC3-W-MS-JUN11.PDF

Cheers.
Reply 1
Avatar for raheem94
raheem94
13
Original post by mcfc2009


6cos3x+6(1+cos3x)2=6(cos3x+1)(1+cos3x)2=61+cos3x \displaystyle \frac{6cos3x + 6}{(1+cos3x)^2} = \frac{6(cos3x + 1)}{(1+cos3x)^2} = \frac6{1+cos3x}

Was this the confusion?
Reply 2
Avatar for mcfc2009
mcfc2009
OP
Original post by raheem94
6cos3x+6(1+cos3x)2=6(cos3x+1)(1+cos3x)2=61+cos3x \displaystyle \frac{6cos3x + 6}{(1+cos3x)^2} = \frac{6(cos3x + 1)}{(1+cos3x)^2} = \frac6{1+cos3x}

Was this the confusion?


Thanks, obvious!
Reply 3
Avatar for opiumJ
opiumJ
Do you mean from:

6cos3x+6/(1+cos3x)^2

to

6/1+cos3x



Because if that's the last mark then 6cos3x+6 can be re-written as 6+6k (let's say cos3x is k).

so 6+6k/(1+k)^2 which is equal to:

6(1+k)/(1+k)(1+k)

take away a (1+k) from top and bottom and you are left with 6/1+k which is:

6/1+cos3x




Somehow, if you made it that far, I doubt that is the step you were stuck at so apologies if I just explained something basic.
Reply 4
Avatar for mcfc2009
mcfc2009
OP
Can anyone help me with 7b
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC3-W-MS-JAN10.PDF
http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC3-W-QP-JAN10.PDF

cheers
Reply 5
Avatar for GOT111
GOT111
4
Original post by mcfc2009


Try to rearrange the dy/dx into p(1+(tan4x)(tan4x))
This can then be used in the product rule (or chain, i can't remember which is which.)

Once you find that rearrange and remember y=tan4x

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