VCE Methods Question Thread!

[Dayman]

Well since you are just looking for the turning points the modulus signs are redundant.

let g(x)=sin²(x)-cos(x), therefore f(x) and g(x) have their turning points at the same x-values
-->g'(x)=2*cos(x)*sin(x)+sin(x)
g'(x)=0
2*cos(x)*sin(x)+sin(x)=0
sin(x)(2*cos(x)+1)=0

sin(x)=0  v  cos(x)=-1/2

-->x=n*pi , n e Z  v  x=2pi/3 + 2pi *n , n e Z   v  x= 4pi/3 + 2pi*n , n e Z

I hope this is kinda correct lol :p

Cheers florian your answer was "kinda correct" but you misinterpreted my question maybe due to my ugly lay out.

But you should've wrote g(x)=sin^2(x)-cos(x)sin(x)

Then g'(x)= (after a bit of simplification with trig identities) sin(2x)-cos(2x)

Then I either used your method or banged  out g'(x)=0 on calc and I got an answer of

X=((4n-3)/8))pi n e z.

Thanks for your answer it gave me an epiphany.

[rhinwarr]

f(x) = x^3 – 6x^2 + 9x – 4
Find the real values of h for which only one of the solutions of the equation f(x + h) = 0 is positive.

Could someone explain this to me please? The answer is 1<h<4 which are values equal to the values of x between the x-intercepts.

[abcdqd]

f(x) = x^3 – 6x^2 + 9x – 4
Find the real values of h for which only one of the solutions of the equation f(x + h) = 0 is positive.

Could someone explain this to me please? The answer is 1<h<4 which are values equal to the values of x between the x-intercepts.

f(x + h) is translation to the left by h units if h>0 and translation to the right if h<0. if only one solution to f(x+h)=0 is positive, there is only one positive x-intercept. since the two x-intercepts of f(x) are 1 and 4, h cannot be negative: if it is moved to the right there will always be 2 x-ints. if 0<h<1 (moving left by a value smaller than one), graph still has 2 positive xintercepts; and if h>4(moving left by more than 4 units), graph has 2 negative intercepts. so if 1<h<4 there will be one x-intercept.
but i think the answer should be because when h=1 the xints are at 0 and 3...which means theres only one positive.

[shadows]

f(x) = x^3 – 6x^2 + 9x – 4
Find the real values of h for which only one of the solutions of the equation f(x + h) = 0 is positive.

Could someone explain this to me please? The answer is 1<h<4 which are values equal to the values of x between the x-intercepts.

Try graph f(x) with a cas, it might clear things up.

You essentially need to find the translation of how many units (indicated by value of h), to ensure that for x is bigger than 0, (becarwful, this doesn't include zero), f(x) only cuts or touches it once.

Edit. Lol beaten. Wtf how come abcdqd, you posted 5 mins ago? I swear it only took me 1 minute to type up this post...

[rhinwarr]

Oh I forgot that f(x+h) was translation. Thats easy then. Thanks.

[zvezda]

For graphing the modulus of a truncus with range (-infinity,3) for example; when graphing the modulus of the function, is it required to draw the whole asymptote or just partially where relevant? Because technically you cant have a graph passing through an asymptote?
Thanks in advance

[abeybaby]

For graphing the modulus of a truncus with range (-infinity,3) for example; when graphing the modulus of the function, is it required to draw the whole asymptote or just partially where relevant? Because technically you cant have a graph passing through an asymptote?
Thanks in advance

draw entire asymptotes, because the graph only tends towards that value as x (or y) tends to +infinity or -infinity, but not for values in between. 

Think of it like this: whats the definition of an asymptote? It's a value to which x tends as y becomes very large (or very small), or a value to which y tends as x becomes very large (or very small). So it doesnt matter what x and y do when theyre NOT very small, it only matters what happens as they do become very large or small.

Hopefully that makes sense

[DetteAmelie]

Um, does anyone remember how to perform long division of polynomials on the calculator?

[rhinwarr]

I don't think the calculator has anything that displays the actual long division process but you can simply type it into the calculator. Theres also some alternatives like factorise, find the roots and a whole bunch of other things. If you have ti-nspire it will be under 'Polynomial tools' (3,8).

[abeybaby]

its the command 'expand'

[DetteAmelie]

its the command 'expand'

Woo! Thanks so much Abey

[abeybaby]

Woo! Thanks so much Abey

That will be $3.50, or your first born child. I accept cash or card.

Hahahaha sorry, couldnt resist. no problem

[rhinwarr]

Does anyone know if it is necessary to show all the working when doing chain rule questions in the short answer section? From what I've seen, they're usually only 1-2 marks. I usually skip out the steps like 'Let u=?, Let v=?' etc so my working doesn't have any u's and v's. Is that alright cos I'm out of practise with writing out the whole thing.

[abeybaby]

Does anyone know if it is necessary to show all the working when doing chain rule questions in the short answer section? From what I've seen, they're usually only 1-2 marks. I usually skip out the steps like 'Let u=?, Let v=?' etc so my working doesn't have any u's and v's. Is that alright cos I'm out of practise with writing out the whole thing.

You don't need to show an entire substitution - to be safe, I wrote 'by the chain rule', and then just did it the quick way (without actually showing a substitution)

[abcdqdxD]

how much time should I be spending on the MC section of the exam?