VCE Methods Question Thread!

[soNasty]

yeah it does, thanks !

also, for this question:
 the exact solutions of the equation 2cos^2(x)+sin(x) -1=0  over -pi<=x<=pi
i rearranged it to 2(1-sin^2x)+sinx -1 =0     then i assigned u=sin(x)

so i got 2u^2-u-3
then factorised to (2u-3)(u+1)

ended up with sin(x)=3/2 and sin(x)=-1

however i cant seem to get the 3 required solutions stated as the answer

ans: -5pi/6, -pi/6, pi/2

have i mucked up somewhere?

[lzxnl]

yeah it does, thanks !

also, for this question:
 the exact solutions of the equation 2cos^2(x)+sin(x) -1=0  over -pi<=x<=pi
i rearranged it to 2(1-sin^2x)+sinx -1 =0     then i assigned u=sin(x)

so i got 2u^2-u-3
then factorised to (2u-3)(u+1)

ended up with sin(x)=3/2 and sin(x)=-1

however i cant seem to get the 3 required solutions stated as the answer

ans: -5pi/6, -pi/6, pi/2

have i mucked up somewhere?

2(1-sin^2 x) + sin x - 1=0
2 - 2u^2 + u - 1 = 0
2u^2 - u - 1 = 0
(2u+1)(u-1)
u=-1/2 or u=1

[soNasty]

oh god........... how stupid of me lmao where did i get a bloody 3 from

[idontknow2298]

oh wow, thank you so much! That really helped

[e^1]

Hi guys, how would you solve this equation

F(x)=2cos(n(x+y))
Solve for n,y using these given points
(pi/3,0) (2pi,0)



Thanks

We can use any value integers and to find the value of , which you can then use to find by substituting into [1] or [2], in one way. However, you can see you want to avoid , for we will get . Although it is technically is still a function, IT DOES not touch these two points we desire. So we can say .

Example

Let and .



This is one solution which touches the two desired points. There are many more, of course.

[Rishi97]

hi

Just wondering, how do you find the inverse of a modulus function?
Do we just treat it as a normal equation by removing the mod lines?
For eg, how do I find the equation of the following function ? I had to do it on paint cause I don't know how to write fancy math symbols in this post Pls teach me

[Yacoubb]

Find two positive numbers whose sum is 4 and such that the sum of the cube of the first and the square of the second is as small as possible.

Have no idea how to approach this ^^

Thanks

[Zealous]

Find two positive numbers whose sum is 4 and such that the sum of the cube of the first and the square of the second is as small as possible.

let x=first number
let y=second number
let f(x)=the sum of the cube of the first and the square of the second.



...substitute y into f(x).



...we now have a function for the sum of the cube of the first and square of the second. We can now use calculus to find the minimum value of this function and the corresponding x (first number) value.



...as we are only looking for two positive numbers:

[Toki]

Thanks e^1

[Frozone]

Hi guys, could you please help me solve:

Domain is between [-2pi, 2pi]



*Sorry I forgot to initially include the domain!
I really struggle with sketching trig transformations, especially when there is a translation in the x- axis. My graphs are always wrong.

[IndefatigableLover]

Hi guys, could you please help me solve:

Was there a given domain to this question or are you looking for the general solution?

@Rishi97 You could have typed up your equation from your keyboard as:

|x^2-4|-2 or as:

Code: [Select]

[tex]|x^2-4|-2[/tex]
The lines is basically when I press 'Shift + \' on your keyboard.
As for your question, you're looking for the inverses and not worried about inverse function and restricting the domain right (since inverse functions don't exist for that particular equation)..

But you could just remove the mod lines (so long as you have a plus/minus sign on the other side) and solve it like a normal inverse I guess (not really sure with this one tbh...)

What I think:



Inverse









And then when you expand it out to get all the possible inverses you'll get:






If I use my CAS and solve for the inverse I get those answers as well except it gives me domains which I'm unsure of how to go about (which is what I'm unsure about) :/

[Frozone]

Was there a given domain to this question or are you looking for the general solution?

Thanks for reminding me about the domain! 

[Yacoubb]

let x=first number
let y=second number
let f(x)=the sum of the cube of the first and the square of the second.



...substitute y into f(x).



...we now have a function for the sum of the cube of the first and square of the second. We can now use calculus to find the minimum value of this function and the corresponding x (first number) value.



...as we are only looking for two positive numbers:

Thanks so much

[IndefatigableLover]

Hi guys, could you please help me solve:

Domain is between [-2pi, 2pi]


*Sorry I forgot to initially include the domain!

Transposing:





Because it is negative, it'll appear in Quadrants 3&4

(Yes it looks a bit weird but we'll get there)





However since the last one isn't in the domain anymore you can get rid of it.

[Frozone]

(Yes it looks a bit weird but we'll get there)

I dont understand why it is and