ssNake's Methods 1/2 Q's.

[REBORN]

Makes sense but your use of the word 'coincidentally' makes me seem like this is one of the easiest problems on earth.

What if it wasn't such a nice problem - how would you find the range then?

[WhoTookMyUsername]

If i'm dividing by in order to find the remainder,

why am i just allowed to use

??

How is the same as

?

thanks

sorry decided to post this on the board as well, delete priveleges removeed?

[b^3]

removed post, put in other thread.

[Hutchoo]

If i'm dividing by in order to find the remainder,

why am i just allowed to use

??

Don't really know what you mean.. but using the null factor law gets the 1/2.
2x - 1 = 0
2x = 1
x = 1/2

[REBORN]

Stop hijacking.

Q: Is a dilation of factor 'a' from the x-axis same as a dilation of factor 'a' parallel to the y axis ?

[b^3]

Stop hijacking.

Q: Is a dilation of factor 'a' from the x-axis same as a dilation of factor 'a' parallel to the y axis ?

Yes it is just worded differently, try to picture it, you are pulling/streching the graph in the same direction

[REBORN]

Period and Amp of:

y=2-3Sin(pi-3x)

The Amp is 3 but period?

[b^3]

period is with n being 3 so
The negative just gives the reflection in the y-axis, so you could think of it as y=2-3sin(-(3x-pi))

[REBORN]

Ahhhhhhhhhhhhhhhhhhhhhhhhhhhh

Realisation.

Thanks.

[REBORN]

The number of solutions of the equation a = -b cos(x) where -(3pi/2)<x<2pi and a,b is an element of R+ with b>a is:

6,5,4,3 or 2? (mcq)

I have no clue on where to even start...

Apologies for the non-latex maths equation.

[b^3]

so b>a just means that cos(x)= the negative of something less than 1, so it exsits.
the negative means because it is cos there is only solutions in the 2nd and 3rd quadrants.
and since it is -3pi/2<x<2pi this means we have the one full unti circle which will have two solutions, one in the 2nd and one in the 3rd quadrant. and the other is 3/4 of a circle going anticlockwise from the starting point, to pi/2 so the 2nd, 3rd and 4th quadrants. because cos(x) = the negative there is only soultions in the 2nd and 3rd quadrants for that part, so there should be a total of 4 solutions. Is that correct?

[REBORN]

It's correct; 4 solutions.

[b^3]

ok so draw the circle out for -3pi/2 to 0 and for 0 to 2pi
see attached image, does that make any more sense?

[REBORN]

Yea.

Would this also work if I just drew out the cos graph and found where it crossed the x-axis?

[b^3]

no because -a/b is not equal to zero, you could draw it out and find the numebr of intersects of cos(x) and the line y=-a/b where -a/b is below the axis somewhere.