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q.4(ii) .. initial value is not given so choose eg, pi/4 and plug it in the equation given .
then you 1.32 ( radians measure ) , plug this in equation and you get value , and plug in eqn again till you get a constant value of 1.33.
q.5(ii) we got dy/dx = 2sin^2 t *cos^2 t .. so to find gradient of tangent we need to substitute t value in dy/dx ..
we are given x - value ( 0 ) .. so find t value by substituting x=o in x=ln(tant) ,, we get t =1/4 pi .
. substitute in dy/dx to get gradient of tanget .. so get gradient = 0.5 .. use y=mx +c to find C .. hence you get eqn .
q.7(b) for |z| < 2 draw circle with centre at origin and radius =2
for |z − 2 − 2i|. arrange to make z - ( ... ) so becomes z - ( 2 +2i ) .. draw at 2+2i ..real 2 and imaginery 2 ..and join to origin ...
thats where i can help
ooh so we can use any value of x in Q4??
and umm i didn't quite get the 7th question.... i got the circle part ... but wen u rearranged the second equation where is the |z| in :
|z|<|z-2-2i|
?? u just aaranged the r.h.s .... is this a perpendicular bisector or a half line?? I'm sorry but i'm really out of practice with the p3
Thanks a lot for the help!!