VCE Methods Question Thread!

[DSubShell]

Didnt read the question properly

My most valuable tip for methods... read the question properly! And then re-read it =P

It may seem silly though, but answering the question "you want" rather than the actual question can cause big headaches in exams. For example, find the x-value of the Intersctions when the Q asked for the full coordinates.

I always recommend a highlighter Better start getting good practices early!

[Zues]

someone show me how you get a and b and then please show me how you validate it using different numbers.

answers are a=-12 b=6 dont see this getting right results

[IndefatigableLover]

someone show me how you get a and b and then please show me how you validate it using different numbers.

answers are a=-12 b=6 dont see this getting right results

This is a really weird matrix since I can't really get a proper answer to satisfy the final matrix but I can get a=-12, b=6 (but it won't get you the final answer)... I think the question is a bit messed up :|

Basically multiply out the matrix out and you'd have gotten:

[a+9   2a+12
b-6     2b-8]

As your matrix and then you just equate them across so a+9=-3 and b-6 = 0 which you can solve for a=-12 and b=6 (though for the others it doesn't really work out which is why I think it's a bit screwed) :/

[Zues]

This is a really weird matrix since I can't really get a proper answer to satisfy the final matrix but I can get a=-12, b=6 (but it won't get you the final answer)... I think the question is a bit messed up :|

Basically multiply out the matrix out and you'd have gotten:

[a+9   2a+12
b-6     2b-8]

As your matrix and then you just equate them across so a+9=-3 and b-6 = 0 which you can solve for a=-12 and b=6 (though for the others it doesn't really work out which is why I think it's a bit screwed) :/

yeah i got same as you and also though the thing was messed up (confirmed after entering in calc)

also... the sum of 2 numbers is 10. find the sum of the squares is to be a minimum. can you show me how YOU do this

[Zues]

i know you go like

a+b=50
b=50-a

then
Sum = a^2 +(50-b)^2

but my question is, why do you just put a^2 and not 50-b in that place?

[dyii22]

Hi guys,

Heading into year 12 I was wondering which resources helped you the most? Checkpoints, all the material from the tutoring companies, Derrick Ha's notes etc?

Cheers

[theshunpo]

Hi guys,

Heading into year 12 I was wondering which resources helped you the most? Checkpoints, all the material from the tutoring companies, Derrick Ha's notes etc?

Cheers

Learn how to use your CAS inside out, I learnt a lot from:
b^3's Guide to using TI-Nspire for METHODS

[keltingmeith]

yeah i got same as you and also though the thing was messed up (confirmed after entering in calc)

also... the sum of 2 numbers is 10. find the sum of the squares is to be a minimum. can you show me how YOU do this

a+b=10 ==> a=10-b

So, let f(a,b)=a^2+b^2 - which in one dimension, is equivalently f(b)=(10-b)^2+b^2=100-20b+b^2+b^2=2(b^2-10b+50)

Now the problem is your classic min/max calc problem. However, for funsies, let's look at it from a different perspective:

We have a quadratic - and a quadratic is VERY easy to minimise. Since we have a positive quadratic, the minimum will occur at the turning point - which occurs at -b/2a=10/2=5.

So, the minimum occurs at b=5 ==> a=10-5=5, so our final answer is a=b=5. Intuitively, this also makes sense given the initial linear relationship - if we raise one, the other goes down, so we'd want to pick a point of equilibria. (if that's hard to explain, consider b=6. This implies that a must be 4 - however, if this is true, a=6 and b=4 should also satisfy, but we want one solution).

[Zues]

on the venn diagram where would

pr (not a union b) be located?

[StupidProdigy]

The moderator should change the name 'Methods [3/4] Question Thread!' to 'Zues's Question Thread!' 

[Zues]

The moderator should change the name 'Methods [3/4] Question Thread!' to 'Zues's Question Thread!' 

haha got exam coming up soon need to find all weaknesses haha

[theshunpo]

haha got exam coming up soon need to find all weaknesses haha

Use the quotient rule to work out f'(x) and then substitute in x=2. I'll do a worked solution if this isn't enough, hope it helps.

[Zues]

can you show me? i know you have to normally bring the x up and make it negative btu how so in this case

[theshunpo]

can you show me? i know you have to normally bring the x up and make it negative btu how so in this case

[IndefatigableLover]

(Image removed from quote.)

Alternatively to theshunpo, when you see fractions where the bigger power is on top of the smaller one, then you can do long division to simplify it out. Do this when you see if it applies since it would easily solve this question for you I guess.

Basically that equation would simply out to 3x-3 and if you take the derivative and sub x=2, you'd get 3 as your answer still

EDIT: Like EulerFan101's post below (except he copied the initial equation wrong  but the working out is right + right answer in the end )