i cant read the question,attachment is not viewing full size.View attachment 41037
Could anyone help me with part (iii) of this COMPLEX no. question please.
Mark scheme
View attachment 41038
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i cant read the question,attachment is not viewing full size.View attachment 41037
Could anyone help me with part (iii) of this COMPLEX no. question please.
Mark scheme
View attachment 41038
if u cant, just post a link of this paperView attachment 41037
Could anyone help me with part (iii) of this COMPLEX no. question please.
Mark scheme
View attachment 41038
let tan*=tan inverse
Thank you. I highly appreciate your help.let tan*=tan inverse
tan*2+tan*3
=arg(1+2i) - arg(1-3i)
=arg((1+2i)/(1-3i)) =>this is because arg(a) -arg(b) = arg(a/b)
=arg(u)
=pi by 4
my pleasure to helpThank you. I highly appreciate your help.
for 1/3 tan^2 x to be convergent,http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
7)b)i)
Can anyone please explain I don't understand the method in the marking scheme and why don't we use -1<.......<1?
sweetheart its A level mathematics forumCan anyone help plz................... the ans is correct but ho to solve in steps?View attachment 41041
Can you go through that in detail please I still didn't get it?for 1/3 tan^2 x to be convergent,
it should be less then 1 n as x is greater then zero then it should be greater then zero,so this implies
0<1/3 tan^2 x<3
now make x subject
multiply through out by 3 then take square root through out n then tan inverse
this will give u the answer.
if still not clear,dont hesitate to ask
do understand the term convergent?Can you go through that in detail please I still didn't get it?
do understand the term convergent?
Yes.do understand the term convergent?
Browny kitkat <3 :P
0<1/3 tan*x>1
we want x to be subject
Multiplying by 3
0<tan*x>3
Taking under root
0< tanx>under root 3
Now tan inverse,tan inverse 0=0 and tan inverse under root 3= pi by 3
0<x>pi by 3
npThanks got it ^_^
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