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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf
Question 6 please!
I stop at integrating 1/(1-y^2)
Question 6 please!
I stop at integrating 1/(1-y^2)
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Assalamu 'Alaykum May Allah Help you...ur welcome-although i was so useless
pls can any one help me in q 10 part ii of the paper...... I believe this q is only one of its kind till now..
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
Anyone?! :O
Hope you get this:
View attachment 26738
Hey, you switched your axis! You put the imaginary one in the real one!
i need helpRe: Maths help available here!!! Stuck somewhere?? Ask here!
Assalamoalaikum!!
UPDATE: Link to Sequences Help by destined007 added!
When the tangent is parallel to the y axis, the dy/dx is infinity. To express infinity= 1/0
So we equate the dy/dx=1/0
y/[x(3y^3-1)] = 1/0
cross multiplying:
x(3y^3-1)=o
3y^3-1=0
y= cube root of (1/3)
y= 0.693
Substitue this value into x(3y^3-1)=o to find the x coordinate.
ANYONE?
The issue I had with the answer to this is that instead of rounding off to 0.693 before substitution into the equation, I continue to use cuberoot of 1/3, and end up getting a different answer for the x coordinate (the difference is quite a bit actually). Confusing. :SWhen the tangent is parallel to the y axis, the dy/dx is infinity. To express infinity= 1/0
So we equate the dy/dx=1/0
y/[x(3y^3-1)] = 1/0
cross multiplying:
x(3y^3-1)=o
3y^3-1=0
y= cube root of (1/3)
y= 0.693
Substitue this value into x(3y^3-1)=o to find the x coordinate.
Dude, just multiply your two answers from part i) to get one quadratic factor and find the other one by (Ax^2 + Bx + C) inspection.ANYONE?
Never really understood what they did. I would want some help in this question as well!
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