VCE Methods Question Thread!

[sparksfly]

When writing the final answer- do we have to make fractions a proper fraction? Or can I just leave it as an improper fraction? I don't want to waste time making my final answer "look nice"? Thoughts?

Thanks everyone for the help!

[speedy]

I did what @IndefatigableLover did, as it was only 2 marks. Aha but idk, on an exam might do what @Yacoubb did.

Also: is this wrong:


x can be negative -> since the x² makes it positive...

[BLACKCATT]

Yeah don't bother changing to a proper fraction. It doesn't matter as long as the answer is correct(unless the question says otherwise)

[IndefatigableLover]

Two conflicting statements (Edit: not really actually lol) ^^ aha. I did what @IndefatigableLover did, as it was only 2 marks. Aha but idk, on an exam might do what @Yacoubb did.

Also: is this wrong:
(Image removed from quote.)

x can be negative -> since the x² makes it positive...

Yeah I would have probably done it Yacoubb's way but so long as it's a 'Show that' question, you want to do some working to show that they're the same.. as for that picture, I don't think it's 'wrong' since in the box above the first one you outlined, it technically doesn't exist when you sub in x=-1 so I would go with the positive answer (probably want somebody else to clarify that though)..

[speedy]

Yeah I would have probably done it Yacoubb's way but so long as it's a 'Show that' question, you want to do some working to show that they're the same.. as for that picture, I don't think it's 'wrong' since in the box above the first one you outlined, it technically doesn't exist when you sub in x=-1 so I would go with the positive answer (probably want somebody else to clarify that though)..

Yeah that's true; I'm really not sure here aha

[Yacoubb]

Yeah that's true; I'm really not sure here aha

I always remember to look at the given equation before I make changes. Since substituting x= -1 would not provide a solution, x=+1

[speedy]

I always remember to look at the given equation before I make changes. Since substituting x= -1 would not provide a solution, x=+1

Mm yeah, from the given equation it doesn't work (I guess?), so go by that... But does it not work for the given equation?

yeah 2log(-1) doesn't work
but you can make it work -> log((-1)²)

Ahhh... Like it's not as if that really strays from the given equation.

[Yacoubb]

Mm yeah, from the given equation it doesn't work (I guess?), so go by that... But does it not work for the given equation?

yeah 2log(-1) doesn't work
but you can make it work -> log((-1)²)

Ahhh... Like it's not as if that really strays from the given equation.

Stick with the rule that you make substitutions into the given equation. If it doesn't work there, it cannot be a solution.

[speedy]

Stick with the rule that you make substitutions into the given equation. If it doesn't work there, it cannot be a solution.

Yeah I will go with this
Hopefully VCAA won't give something like this though...

[Reus]

How do we get 2? I understand that () = 1 thus . How does this equate to 2?
I was thinking to differentiate sec^2 but seems like I'm getting no where haha. Thanks.

[Jason12]

Could someone explain how thanks!!
Also @Jason12 what exam is this?

TSSM 2014 exam 1

[speedy]

How do we get 2? I understand that () = 1 thus . How does this equate to 2?
I was thinking to differentiate sec^2 but seems like I'm getting no where haha. Thanks.

sec(pi/4) = 1/cos(pi/4) = 1/1/root(2) = root(2)

Edit: then obviously, sec²(pi/4)/tan(pi/4) = (root(2))²/1 = 2

[Yacoubb]

(2pi +2)²=(2pi +2)(2pi +2) = 2(pi+1)*2(pi+1) = 4(pi+1)

-2/4(pi+1) = -1/2(pi+1)

[myanacondadont]

Yeah something like that needs you to employ pythagoras' theorem.

So construct a triangle, and find sinthetaband costheta. Once you've done that, refer back to 10a (acostheta and asintheta). Multiply the given value by the specified a value, and then use them to get your final answer.

Is that clear?

Ah yeah got it. So simple haha, should've gotten it. Thankyou'!

[Reus]

sec(pi/4) = 1/cos(pi/4) = 1/1/root(2) = root(2)

Ahhhh thanks! I should start to use the formula sheet more haha. However the answer is 2 not root(2)?