Series and a trig Q from 2008 Paper (1 Viewer)

[Makro]

5.b) Consider the geometric series: 5 + 10x + 20x^2 + 40x^3 + ...
i) For what values of x does this series have a limiting sum.

I've never done a question like this and I'm a bit confused by the working out that successone gave me.

6. a) Solve 2sin²x/3 = 1 for -pi =< x =< pi.

I don't understand how dividing by 2 makes it become -pi/3 =< x =< pi/3.

Thanks for help in advance

 

[zeleboy]

I've never done a question like this and I'm a bit confused by the working out that successone gave me.


I don't understand how dividing by 2 makes it become -pi/3 =< x =< pi/3.

Thanks for help in advance

1) Limiting sum formula a/(r-1) where r is between -1 and 1
The ratio of the series is 2x, thus r = 2x
but r has to be between -1 and 1 thus -1 < 2 x < 1
-1/2 < x < 1/2.

2. I assume Solve 2sin2x/3 = 1 for -pi =< x =< pi. is meant to read

2(sin[x/3])^2 = 1
In which case the domain is divided by 3. ie -pi/3 =< x =< pi/3.
because it can only exist within those numbers

 

[boxhunter91]

|r|<1
.: |2x| <1
-1<2x<1
-1/2 < x <1/2

 

[Timothy.Siu]

a limiting sum is one that at infinite terms, it approaches that value.
this means that |r| <1 or the series would just keep getting bigger.

in this series, r=2x
for |2x| < 1

-1<2x<1
-1/2 <2x<1/2
i think that should be the answer

6. a) Solve 2sin^2x/3 = 1 for -pi =< x =< pi.

sin^2 x=3/2
sin x=+-root(3/2)
x=arcsin (+-root(3/2)) ?? (plus the other values)

seems wrong. and i dont remember doing this question like this lol.

 

[boxhunter91]

it is sin^2(x/3)=1.
I got sin x = 1/root 2 so pi/4 etc.
But that is wrong...

 

[Timothy.Siu]

oh i seeeee now.
i remember doing this last year.

yeah, u have to change the domain i guess, everything else wont fit...sorta

sin^2 (x/3)=1/2

sin (x/3)=+-(1/root2)
then i guess you could just find all the solutions, and delete those that dont fit...

 

[khorne]

sin^2 (x/3) = 1/2 -n <= x <= n

Let u = x/3, -n/3 <= u <= n/3 (adjusting limits)

sin^2 u = 1/2

sin u = +/- 1/rt(2)

Solve for u, then multiply by 3.

 

[Makro]

2. I assume Solve 2sin2x/3 = 1 for -pi =< x =< pi. is meant to read

2(sin[x/3])^2 = 1
In which case the domain is divided by 3. ie -pi/3 =< x =< pi/3.
because it can only exist within those numbers

It was written as 2sin^2(x/3) = 1, but I think we got that by the end of the discussion.

The bit I don't get is why is it divided by 3? How come it can only exist within those numbers? What makes you think "Oh, I have to divide it by 3" - this is the bit i'm not grasping.

I'm pretty sure I understand the limiting sum question, now. Thank you

 

[khorne]

Ok, this is a trig technique called substitution.

If you have a compound angle (eg, divided, added, subtracted....) then, instead of altering the equation, you substitue the angle for something, eg, u.

now, the original limit is -n <= x <= pi

If we say, let u = x/3, this means that U will be a third of x, so the limits have to be reduced, i.e, we do to the limits, what was done to x, to make sure they are the same, i.e -n/3 <= x/3 (or u) <= n/3

 

[Makro]

Thanks, guys. Finally got my head around it. Where would I find more of those trig solving questions. I don't ever remember doing an exercise on them and they just sort of seem to pop up here and there and I can never get them first go. Would my best bet be going through past papers? I have the latest SuccessOne book to look through and it looks like there's roughly one in every paper.

 

[ninetypercent]

we had heaps from jones and couchman Book 2. (i think)