VCE Methods Question Thread!

[Insa]

Thanks man! May I ask why you need to divide [1] by [2]?

[tony3272]

It's to cancel out the A's so you only have one variable to find

[Insa]

Oh thank you

[ICECOLD]

Hi.

Find lim x approach 0 for (1-root(1+3x)) / x

I thought undef as divide by 0 but CAS gives an answer :-S

[pi]

Hi.

Find lim x approach 0 for (1-root(1+3x)) / x

I thought undef as divide by 0 but CAS gives an answer :-S

I'm pretty sure this is way beyond the methods course. For a step-by-step solution, type this into wolfram alpha: lim(x->0) ((1-sqrt(1+3x)) / x)

[paulsterio]

you can just graph the graph of y = 1-sqrt(1+3x)/x, that's as far as i can think of

but in other terms, i guess you could apply some first year calculus and use L'Hopital's Rule

basically this states that if you have a limit which evaluates to 0/0 or infinity/infinity at x = a, then lim(x->a) of f(x) = lim(x->a) of f'(x)
i can't prove it up the top of my head, so just trust that it works

you'll have to find the derivative of ((1-sqrt(1+3x)) / x) and then find lim (x->0) of that derivative, shouldn't be too hard

[#1procrastinator]

What's the difference between something like [-pi, pi] and [0, 2pi] when you're working with circular functions? Isn't it just the same interval?

I got confused with this when working with complex numbers in specialist but I think it might be more of a trig thing

If you have then some of the angles within the restriction are negative, such as
whereas for these angles would be given as respectively.
To alternate between the two you just add or subtract

Could you please give a couple of different examples? With the unit circle or something...it's just not clicking me with yet :p
Seems a little random at the moment lol

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For what values do the equations  (m-1)x + 5y = 7 and 3x + (m-3)y = 0.7m have infinitely many solutions?

I rearranged it and solved for m (setting the gradients equal) and got m = 6 and m = -2. Why is 6 an answer but -2 isn't (I plotted it and it gave me parallel lines, but why?)
 

Ok, I solved for the constant and the gradient separately. For the constant I got m = 6, and for the gradient, I ended up with the same quadratic which gave me m = -2 and m = 6.  So is this interpreted as when m = -2, the gradient is the same but the constant is different (any number other than 6?), therefore there are no solutions? And for one unique solution would be any values of m other than those two?

Is it ok to solve it by finding the gradient and constant separately? It's a lot quicker than setting the two equations equal to each other as I did initially...and I ended up with the same answer anyway.

Also, do you prefer to solve these type of problems algebraically or using matrices?

[Phy124]

This is how I used to do it, many people use different methods, some of which are easier, but this may still help.

























These two lines have the same gradient but different "c" values, which means they are parallel (no solutions) and thus this is not the answer we are looking for.











These two equations have the same gradient and "c" value (infinite solutions) therefore this is the answer we are looking for.

[oliverk94]

Hey guy's having trouble understanding this question

f(x)= 3x-1/x+1, convert into f(x)= a/x+1 + b, find the values of a and b.

I don't seem to understand how you change that into the other form. Can someone please tell me how to solve these types of equations. I think I lost 1-3 marks on the SAC because of this.

[pi]

It just wants you to simplify it (long division is one such method, but there are much easier ways too), my way:

f(x)= (3x-1)/(x+1) = (3(x+1)-4)/(x+1) = 3(x+1)/(x+1) - 4/(x+1) = 3 - 4/(x+1)

Therefore a=-4, b=3

[oliverk94]

Thanks a lot. Is there a lot of these types of questions in the exam?

[pi]

Thanks a lot. Is there a lot of these types of questions in the exam?

Usually not, but it might turn up in a small 1 mark part to a larger question

Come exam time, you'll find these questions to be a breeze Just practice!

[ICECOLD]

Hi.

How do we solve for x in:

(4x^2 -1)^6 > (4x^2 -1)^5

Thankyou

[xZero]

(4x^2-1)^6 - (4x^2-1)^5 > 0

(4x^2-1)^5 * ( (4x^2-1) -1) > 0

Either (4x^2-1)^5 > 0 and ( (4x^2-1) -1) > 0

or (4x^2-1)^5 < 0 and ( (4x^2-1) -1) < 0

Solve for these to scenario and you'll have your answer

[ICECOLD]

But y/where did you know that you had > and <

Couldn't I just divide both sides by (4x^2-1)^5 to get

(4x^2-1) > 1