Difficult 2 unit questions.... (1 Viewer)

[Kutay]

Hey i was wondering if someone could help me sections of difficult 2unit questions that i am having problems with....

such as sketching y = 1 - cosx <---- does that like start at 0 and go down to -2 ??

also...
the equation x^2 - bx ax + c
----------- = ------------
p -1 p + 1

has two real roots of equal magnitude, but opposite in sign.

prove that p = a - b
--------
a +b

Thankyou in advance Kutay

 

[Dreamerish*~]

y = 1 - cosx is the same as y = -cosx + 1. So it's just the negative cosx graph, moved up one place. See attached image.

:

 

[Kutay]

Thankyou very much

 

[GaDaMIt]

might be.. and probably am wrong, but

y = 1 - cosx is the same as y = -cosx + 1. So it's just the negative cosx graph, moved up one place. See attached image.

i thought y = -cosx + 1 would change to -y = cosx -1, instead of y = cosx - 1.. Most probably wrong anyway, still in year 10 haha =\ sorry for the meaningless post if im incorrect

 

[Kutay]

could you help me solve part two of the question...?

the first question stated draw y = 3sin2x and y = 1 - cosx on same graph....

second part state determine the number of solutions to the equation y = 3sin2x +cosx = 1 in the domain of 0 to 2PI

but i dont get ti to reall = 1 on the grapt...???

 

[Dreamerish*~]

Thankyou very much

Could you rephrase the second question? We can't have multiple spaces on BoS, so I think your equations were stuffed up.

i thought y = -cosx + 1 would change to -y = cosx -1, instead of y = cosx - 1.. Most probably wrong anyway, still in year 10 haha =\ sorry for the meaningless post if im incorrect

I simply switched -cosx and 1 around.

Yes, y = -cosx + 1 does equal -y = cosx - 1, but if we want to draw a graph, -y isn't too useful.

 

[Kutay]

by the way did u have a program on the computer to draw that sketch for you if so can i get the name of it so i can download it

 

[Dreamerish*~]

could you help me solve part two of the question...?

the first question stated draw y = 3sin2x and y = 1 - cosx on same graph....

second part state determine the number of solutions to the equation y = 3sin2x +cosx = 1 in the domain of 0 to 2PI

but i dont get ti to reall = 1 on the grapt...???

See attachment for the graph, but just for clarification:

Amplitude = 3
Period = π

Now for the equation y = 3sin2x + cosx = 1:

Re-write it, and it becomes 3sin2x - (1 - cosx) = 0

Which is 3sin2x = 1 - cosx

From the graph, we can see 5 solutions (including 0 and 2π). I marked them on the attached graph.

Click here to download Winplot

:

 

[GaDaMIt]

Yes, y = -cosx + 1 does equal -y = cosx - 1, but if we want to draw a graph, -y isn't too useful.

hehe, guess ill learn next year then

 

[Kutay]

See attachment for the graph, but just for clarification:

Amplitude = 3
Period = π

Now for the equation y = 3sin2x + cosx = 1:

Re-write it, and it becomes 3sin2x - (1 - cosx) = 0

Which is 3sin2x = 1 - cosx

From the graph, we can see 5 solutions (including 0 and 2π). I marked them on the attached graph.

Click here to download Winplot

:

would it be only 4 solutions because it is between the domain of 0 to 2PI ?

 

[Dreamerish*~]

the equation x^2 - bx ax + c
----------- = ------------
p -1 p + 1

has two real roots of equal magnitude, but opposite in sign.

prove that p = a - b
--------
a +b

Thankyou in advance Kutay

I understand this question now.

(x² - bx)/(p - 1) = (ax + c)/(p + 1)

Cross multiplying:

(p + 1)(x² - bx) = (ax + c)(p - 1)

Expanding:

px² - pbx + x² - bx = pax - ax + cp - c

Collecting like-terms:

(p + 1)x² - (pb + b + pa - a)x - c(p - 1) = 0

Now we have a quadratic equation, the roots of which are α and β

α + β = (pb + b + pa - a)/(p + 1) ... 1

αβ = -(cp - c)/(p + 1) ... 2

The roots are of equal magnitude but negative signs, so α = -β.

Put that into equation 1:

β - β = (pb + b + pa - a)/(p + 1)

0 = (pb + b + pa - a)/(p + 1)

Since the denominator can't equal 0 for the equation to be valid:

pb + b + pa - a = 0

Factorising:

p(b + a) + b - a = 0

p(b + a) = a - b

p = (a - b)/(a + b)

would it be only 4 solutions because it is between the domain of 0 to 2PI ?

Well, at 0 and 2π, 3sin2x = 1 - cosx = 0.

It depends on if the sign was "equal or larger than" or "larger than".