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Mathematics: Post your doubts here!

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Lostsoul

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I have just solved 2008 papers onwards. Sone of which I get stuck, I havr piled them and gonna et them answered tomorrow.
I hope its enough to score me an A
Is it?
 
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unique111

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salvatore said:
Thank you for your reply.
The marking scheme got the value of C as 3 (I got 6). They also got the value of K as 0.4 (I got 0.4). You must've noticed that I got half their values.. so I've messed up in one tiny bit of the question. I'll appreciate if you could take a closer look at my solution
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I continued your work, and got the answer, that's why. Idk about the value of c though. I tried doing the integrating part directly without bringing about the constant 'c' and still got the value of k=0.8 and the final answer still turns out to be 63.
 
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unique111

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airyhat said:
N10/33/Q10(ii)(b) plz solve!!!!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_33.pdf
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P(z^2)=0 that means z^2= -2, and the z^2=two other complex roots from 10a.
Solving for z, you get z= -2^.5 which will give you plus, minus i2^.5 for the first one.
For the other complex roots, assume that their root is a+bi where a and b are real numbers. Equate them simultaneously and you get a^2-b^2= -2 and ab=-3^.5
From here find the values of a and b and replace them in a+bi and you get two more roots+ two more.
 
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airyhat

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unique111 said:
P(z^2)=0 that means z^2= -2, and the z^2=two other complex roots from 10a.
Solving for z, you get z= -2^.5 which will give you plus, minus i2^.5 for the first one.
For the other complex roots, assume that their root is a+bi where a and b are real numbers. Equate them simultaneously and you get a^2-b^2= -2 and ab=-3^.5
From here find the values of a and b and replace them in a+bi and you get two more roots+ two more.
Click to expand...
I didnt understand
 
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