Quick questions I had problems with.. ><

[Freitag]

I've just been going over some practice exams from 2006 and came across a problem with this question. Any help is much appreciated =/

Let u = -2 + 2i

Find a ( element of C ) such that u * a + 4i = root(2) cis (3pi / 4 )

I went through the entire question and got a = 1/2 + 0i, but when i put it back into the equation LHS didn't equal RHS. =/

EDIT: Oops, forgot to add the 4i.
Nevermind ^_^ Sorry.

[Ahmad]

How'd you solve it? I think you made an error. Try redoing it

[Freitag]

How'd you solve it? I think you made an error. Try redoing it

I think i did too. I just redid it and realised i forgot to add the 4i.

[joechan521]

I've just been going over some practice exams from 2006 and came across a problem with this question. Any help is much appreciated =/

Let u = -2 + 2i

Find a ( element of C ) such that u * a + 4i = root(2) cis (3pi / 4 )

I went through the entire question and got a = 1/2 + 0i, but when i put it back into the equation LHS didn't equal RHS. =/

i got a= 1/2  also, and it should be correct
since root(2) cis (3pi / 4 )= -1+i

[Freitag]

But the LHS has 4i added onto it too.

[BenBenMan]

I got a = -1/2 + i, it works if I sub it back into the original equation  ... I'll post my working if you want.

[Freitag]

I got a = -1/2 + i, it works if I sub it back into the original equation  ... I'll post my working if you want.

Thank you for the offer. I just came to that solution too, so it should be right.

Thanks anyways guys.

[Ahmad]

I got a = -1/2 + i, it works if I sub it back into the original equation  ... I'll post my working if you want.

This is also my answer. I just plugged the equation in my calculator -> solve. I love ti-92+

[Freitag]

I got a = -1/2 + i, it works if I sub it back into the original equation  ... I'll post my working if you want.

This is also my answer. I just plugged the equation in my calculator -> solve. I love ti-92+

The only bugger about that was that it was on a calc free paper xD

[Sheikh05]

I got a = -1/2 + i, it works if I sub it back into the original equation  ... I'll post my working if you want.

This is also my answer. I just plugged the equation in my calculator -> solve. I love ti-92+

I got the same thing. How many marks was this worth? Looks around 3 at least.

[avogarbro]

What are the acceptable methods for a proof of a rhombus?
For example the question on the 2003 exam 2.

[BenBenMan]

What are the acceptable methods for a proof of a rhombus?
For example the question on the 2003 exam 2.

A rhombus has two important properties: all the sides are equal, and the diagonals are perpendicular. You'll most likely be given one of these, and then from there you can use the dot product to show the other property.

eg. if they tell you that the diagonals are perpendicular, then if you let the top side be u and the left side be v, then the diagonals are u+v and u-v ... so considering u.v=0 (they told you that), then (u+v).(u-v)=0 ... |u|^2-|v|^2=0, so |u|=|v|. Therefore the sides are the same length, and that proves it.

[v2shan]

What are the acceptable methods for a proof of a rhombus?
For example the question on the 2003 exam 2.

I think it's show a pair of equal opposite sides and 1 pair of adjacent sides of equal magnitude

[avogarbro]

Can someone explain this MC on the 2007 MAV exam.

Question 13
The necessary and sufficient conditions for a point P to be a point of inflection are that, as the curve passes through P:

This was the soln:
 "A point of inflection occurs where the second derivative is zero and the first derivative does not change sign. So, at the point of inflection d2y/dx2 changes sign and dy/dx does not change sign."

[Daniel15]

Can someone explain this MC on the 2007 MAV exam.

Question 13
The necessary and sufficient conditions for a point P to be a point of inflection are that, as the curve passes through P:

This was the soln:
 "A point of inflection occurs where the second derivative is zero and the first derivative does not change sign. So, at the point of inflection d2y/dx2 changes sign and dy/dx does not change sign."

Basically, with a point of inflection, the double derivative at the point is equal to zero, and the gradient on either side of the point does not change sign like it does with turning points. So, the gradient either stays positive or stays negative.
I guess that's a bad explanation . Does it help you?