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Mathematics: Post your doubts here!

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By using binomial expansion of (A + B)^n = (nC0)(A^n)(B^0) + (nC1)(A^(n-1))(B^1)+ ....... + (nCn)(A^0)(B^n) {where xCy = (x!)/(y!)(x-y)!}
Putting the corresponding values of A and B as 2 and ax respectively we get,
(2 + ax)^n = (nC0)(2^n)(ax)^0 + (nC1)(2^(n-1))(ax)^1+ (nC2)(2^(n-2))(ax)^2 +....... + (nCn)(2^0)(ax)^n
(2 + ax)^n = (2^n) + (a.x.n)(2^(n-1))+ n(n-1)(2^(n-2))(ax)^2 +.......
From the given equation we get, 2^n = 32 -> n = 5
and (a.x.n)(2^(n-1)) = -40x
-> 5.a.2^4 = -40
-> a = -0.5
and n(n-1)(2^(n-2))(ax)^2 = bx^2
-> 5.4.2^3.(0.25)x^2 = bx^2
-> b = 40
hence we get, n = 5, a = -0.5 and b = 40
 
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PLEASEEEEEEEEE! NOONE IS ANSWERING ME THIS QUESTION AND I REALLY NEED AN EXPLANATION FOR IT (HENCE THE CAPS).
Its about complex numbers, Q8(b). I get the diagram part, but dont know how to find the least value.

Here i attach the question paper and marking scheme:

Thank you in advance!
 

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f2ce2ace0b56bdbca48e9bc37dc9615e8f1e7a44b920e719f29d574d13ec3440.jpg


upload_2014-4-14_15-54-31.png
 

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looks like someone is having trouble with the same paper as me... If you know how to do Q8(b) from that paper please let me know!

This is as far as i could get, with the help of my teacher. :) its kinda blur tho, sorry.
 

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