Mathematics: Post your doubts here!

LMGD33

LMGD33

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For who ever is having trouble viewing my pictures, please PM me you email address and I'll send you the images, other than that GOOD LUCK FOR ALL OF US TOMORROW!!
 
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LindaKim123

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Ashique said:
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.
Click to expand...

how did u do i) part? I did but I didn't get the answer
 
Jaf

Jaf

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Ashique said:
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.
Click to expand...
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. :S
 
Jaf

Jaf

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WayneRooney10 said:
ANYONE?
Click to expand...
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.
That is, the roots you get in i) are 1 - √2 i and 1 + √2 i. So the factors become (x-1+√2 i)(x-1-√2 i). You multiply the brackets and you'll see everything in terms of 'i' gets cancelled to give you a quadratic factor of x^2 - 2x +3.

If you still don't get it, let me know I'll post a screenshot.
 
minie23

minie23

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Ashique said:
Never really understood what they did. I would want some help in this question as well!
Click to expand...

November 2012 P3, was the hardest paper which CIE released since 2002 ! :eek:
Lets hope we get one which will be much easier tomorrow so that we all can ace Insha Allah :)
 
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