VCE Methods Question Thread!

[secretweapon]

Have a solution attached. If I made a mistake somewhere please let me know.
Also for the integral of 3/(5x-2), I am not sure if this is required in the methods course. I may be wrong, however.

This is a question in the cambridge senior maths methods textbook, chapter 12 tech free question 7.a., so i'm assuming it is required in the methods course?

[Lear]

This is a question in the cambridge senior maths methods textbook, chapter 12 tech free question 7.a., so i'm assuming it is required in the methods course?

Hmm I'm skeptical as Modulus has been taken out of the course. However, I have attached for you the formula you are to use if this does come up

[lzxnl]

How do you differentiate this?

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And can this be simplified any further?

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And this attachment...
I literally don't know where to start. 

Sorry, this is just getting longer and longer...

For the first one, note that all of your derivative laws only work if the base is e. Therefore, you need to write this as base e.

That's something I reckon most Methods students could do.

Hi, could you please explain with a picture if possible? Also, do you know how to find the integral of 3/(5x-2), x > 2/5
since it isn't possible to have powers of 0
Thanks

You actually can have powers of zero; x^0 = 1 for all nonzero x (if x is zero, then the expression is as well defined as 0/0; could be anything). The concern with this question is more that you cannot seem to apply

if n=-1. The simple answer is to use a log.

Hmm I'm skeptical as Modulus has been taken out of the course. However, I have attached for you the formula you are to use if this does come up

You can still write


Although I'm pretty surprised absolute values were removed from the course because quite honestly, they're arguably simpler than trigonometric and exponential functions to understand; |x-y| gives the distance between points x and y on a number line regardless of order.

[secretweapon]

y = (x)/(1-x)
find dy/dx
u = x
v = 1-x
du/dx =1
dv/dx = -1
quotient rule
((1-x)(1)-(x)(-1))(1-x)^2

= ((1-x)-(-x))/(1-x)^2

= ((1-x+x))/(1-x)^2

= 1/(1-x)^2

the answer said it was 1/(x-1)^2

Could someone please tell me where my working out is incorrect?
Thanks

[Sine]

y = (x)/(1-x)
find dy/dx
u = x
v = 1-x
du/dx =1
dv/dx = -1
quotient rule
((1-x)(1)-(x)(-1))/(1-x)^2

= ((1-x)-(-x))/(1-x)^2

= ((1-x+x))/(1-x)^2

= 1/(1-x)^2

the answer said it was 1/(x-1)^2

Could someone please tell me where my working out is incorrect?
Thanks

Those answers are equivalent

[secretweapon]

Those answers are equivalent

Thanks sine
Got another question
f(x) = 3x^2+2
if g'(x) = f'(x) and g(2) = 29,
find g(x)

g(x) = f(x) + c
g(x) = 3x^2+2+c
3(2)^2+2 + c =29
12+2+c=29
c+14=29
c=15
g(x) = 3x^2+2+15
g(x) = 3x^2+17

g'(x) = 6x
g(x) = 6x^2/2+c
g(x) = 3x^2+c
3(2)^2+c=29
c+12=29
c=17
g(x) = 3x^2+17

Which of the methods above is the better methods/would work in any situation?

[secretweapon]

Thanks sine
Got another question
f(x) = 3x^2+2
if g'(x) = f'(x) and g(2) = 29,
find g(x)

g(x) = f(x) + c
g(x) = 3x^2+2+c
3(2)^2+2 + c =29
12+2+c=29
c+14=29
c=15
g(x) = 3x^2+2+15
g(x) = 3x^2+17

g'(x) = 6x
g(x) = 6x^2/2+c
g(x) = 3x^2+c
3(2)^2+c=29
c+12=29
c=17
g(x) = 3x^2+17

Which of the methods above is the better methods/would work in any situation?

Also, which of the above methods is preferred by examiners?

[secretweapon]

Also, which of the above methods is preferred by examiners?

Bump! (24 hours later )

[RuiAce]

Bump! (24 hours later )

Can't really speak for the VCE perspective but if I were marking a paper

Would I give marks? Yes to both of them
Which would I prefer? Probably the second one because too many students get confused with the first.

[secretweapon]

Can't really speak for the VCE perspective but if I were marking a paper

Would I give marks? Yes to both of them
Which would I prefer? Probably the second one because too many students get confused with the first.

Would you give me full marks to both approaches?

[Lear]

Would you give me full marks to both approaches?

They can’t really take marks off you as long as you have a valid process, have shown proper amount of working out and have the correct answer.

[secretweapon]

They can’t really take marks off you as long as you have a valid process, have shown proper amount of working out and have the correct answer.

cheers

[RuiAce]

Would you give me full marks to both approaches?

Pretty much as above.

But for completeness, my answer is yes. Why? Because you haven't written anything mathematically incorrect.

[secretweapon]

Pretty much as above.

But for completeness, my answer is yes. Why? Because you haven't written anything mathematically incorrect.

Thanks for clarifying
Have a question, to get from velocity to acceleration, is it differentiation or anti-differentiation?

[Lear]

Thanks for clarifying
Have a question, to get from velocity to acceleration, is it differentiation or anti-differentiation?

Attached. Great to have in your bound reference