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there was a link on xtreme papers for papers of oct nov 2011where did you get the mark scheme from, It's not on xtremepapers, neither is the paper for oct/nov 11. any other source?
Dude, there will be 2 corresponding values of x for y=1, the one you have found out is for point "A" as he has pointed out in the Pic.how did u get 5pie/12??
i mean it is simple .. lemme solve here
y=p+3tan3x
u got p= -2
so=> y=-2+3tan3x
put point (q,1)
1=-2+3tan3x
1+2=3tan3x
3/3=tan3x
3x=tan(inverse) (1)
3x=pie/4
x=pie/12
ok itx nt the required value, bt y ADD pi/3??Dude, there will be 2 corresponding values of x for y=1, the one you have found out is for point "A" as he has pointed out in the Pic.
For q, you have to add (1/3) pi. This will get you the x-coordinate for point "B".
Look, The period is 120 degrees, which is 2/3 pi, and for point B, the tan curve hasn't completed 120 degrees, its at 60 degrees.ok itx nt the required value, bt y ADD pi/3??
You do know how I got the period, right?ok itx nt the required value, bt y ADD pi/3??
oh sorry!! didn't notice the period...Look, The period is 120 degrees, which is 2/3 pi, and for point B, the tan curve hasn't completed 120 degrees, its at 60 degrees.
so divide 2/3 pi by 2 and you will have 1/3 pi. To get the x-value for point B add this to 1/12 pi (which corresponds to point A).
Hope you got it nowoh sorry!! didn't notice the period...
found it myself.Did you find this error ureself? or you got it frm somebody else...
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