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Mathematics: Post your doubts here!

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Hello!

I am sitting the P1 paper this session and I have a doubt in this exam. It is on question 4.ii).
I could solve the previous 4.i), but this one I don't understand why that is the answer on the mark scheme.

In attachements I am sending the question paper and the mark scheme.

Thank you to anyone that can help me.
 

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  • 9709_w13_qp_11_4.ii.pdf
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  • 9709_w13_ms_11.pdf
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lol its from nov 2013 cause i just solved this question heres goes the logic:
the differential of the function in denominator is ( 12y^3) it is present in the numerator only 4 is missing so multiply/divide by four forming:
1/4 (ln(4y^3+1) )+c
how to integrate 3y^2/4y^3+1
is it by substituition ?
how do i know which method of integration i should use? really confusing at times :confused:
http://www.examsolutions.net/maths-...egration/methods/fdash(x)overf(x)/formula.gif
 
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Hello!

I am sitting the P1 paper this session and I have a doubt in this exam. It is on question 4.ii).
I could solve the previous 4.i), but this one I don't understand why that is the answer on the mark scheme.

In attachements I am sending the question paper and the mark scheme.

Thank you to anyone that can help me.
you know the x values
4 sin2x + 8 cos x − 7 = 0
x=60 and 300 (from (i)
in (ii) x=(1/2 tetha)
60=(1/2)tetha
and 300=(1/2) tetha
now we have tetha = 600,300
but the range they have given does not include 600 so tetha = 300
 
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Solve question 9 from the same paper, I was unable to do that :(
:oops: me kinda busy so i will just explain real quick
find coordinates of A and B by
7-(2x)=(x-2)^2
A(-1,9) B(3,1)
Area of shaded region = Area under the line - area under the curve
area under the line integrate (7-2x) with limits -1 and 3 you will get(7x-x^2) limits -1 and 3 answer is 20
are under the curve integrate (x-2)^2 with limits -1 and 3 you will get ((x-2)^3)/3 limits -1 and 3 answer is 28/3
Area of shaded region= 20-(28/3) =32/3=10.67
 
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