VCE Methods Question Thread!

[psyxwar]

Be careful of using that logic. The function f(x) = e^(-1/(x^2)) when x isn't 0 and f(x) = 0 when x = 0 is continuous everywhere on the real numbers. Furthermore, it is infinitely differentiable, despite there being a zero in the denominator of the argument of the exponential.

Similarly, y = sin x/x when x isn't 0 and 1 when x = 0 is also infinitely differentiable across the real numbers, even though if you sub in x = 0, you get 0/0. Your reasoning may work for VCE, but not in higher level maths.

A more appropriate way to say why it doesn't exist is because there is no unique tangent at x = 0. It's pointy there so you can't define a tangent.

wait what, isn't a function necessarily continuous at a point if it is differentiable at that point? So even though d/dx(sin x/x) exists for x=0, it is still not differentiable at this point?

correct me if I'm wrong but I thought the criteria for differentiability at point a were:

lim (x-> a+) f'(x) = lim (x-> a-) f'(x) and the continuity of the function at a

O__O

[keltingmeith]

wait what, isn't a function necessarily continuous at a point if it is differentiable at that point? So even though d/dx(sin x/x) exists for x=0, it is still not differentiable at this point?

correct me if I'm wrong but I thought the criteria for differentiability at point a were:

lim (x-> a+) f'(x) = lim (x-> a-) f'(x) and the continuity of the function at a

O__O

No, you're right - a function is only differentiable if the gradient is the same on either side and it is continuous at that point. |zxn| described functions that do fit that criteria, but aren't necesarily continuous themselves. For example, consider the hybrid function:



In this function, itself is not defined for x=0, but if you put it in this hybrid function, and all of a sudden it is defined for x=0 - and, better yet, is now continuous for all real numbers. So, now our function is differentiable at every point.

[lzxnl]

wait what, isn't a function necessarily continuous at a point if it is differentiable at that point? So even though d/dx(sin x/x) exists for x=0, it is still not differentiable at this point?

correct me if I'm wrong but I thought the criteria for differentiability at point a were:

lim (x-> a+) f'(x) = lim (x-> a-) f'(x) and the continuity of the function at a

O__O

There are several definitions of differentiability. One is that the limit (f(x+h) - f(x)/h) exists as h approaches zero. You'll find that this limit will exist for my function when x = 0. Let's try it.

The derivative at x = 0 is going to be the limit of (f(h) - f(0))/h as h => 0 (sorry I dunno how to use limit latex)
f(h) = sin h/h, f(0) = 1 by definition
So our limit is (sin h/h - 1)/h = (sin h - h)/h^2 as h approaches zero
It is then possible to prove that this limit exists and is equal to 0

No, you're right - a function is only differentiable if the gradient is the same on either side and it is continuous at that point. |zxn| described functions that do fit that criteria, but aren't necesarily continuous themselves. For example, consider the hybrid function:



In this function, itself is not defined for x=0, but if you put it in this hybrid function, and all of a sudden it is defined for x=0 - and, better yet, is now continuous for all real numbers. So, now our function is differentiable at every point.

The functions I suggested ARE continuous because of the extra hybrid definitions. f(x) = sin x / x, without having a definition at x = 0, couldn't strictly be held to be differentiable there.
Maybe it wasn't clear that I was defining them as hybrid functions.

[knightrider]

How would you solve the following equation for x?

[ikiwi]

How would you solve the following equation for x?

Split it into two parts:
       when x>=-2
      when x<-2

Then solve each equation separately.

[knightrider]

If the price of a product is 50 cents and it says it is 76% off what was the original price

[Vividdreamer]

If the price of a product is 50 cents and it says it is 76% off what was the original price

Original Price = New Price * (100/100 - rate))
Therefore, original price = (0.50 * (100/(100-76)) = $2.08

[knightrider]

Original Price = New Price * (100/100 - rate))
Therefore, original price = (0.50 * (100/(100-76)) = $2.08

Thanks Vividdreamer but why did you use 100/100 - rate how do you get this (logic behind it)

[Vividdreamer]

Thanks Vividdreamer but why did you use 100/100 - rate how do you get this (logic behind it)

I use this formula sheet, the left hand side is for stuff that is being decreased while the right hand side is for stuff being increased.
https://www.dropbox.com/s/wbetw6jbunbz2um/2014-10-27%2021.09.21%20HDR.jpg?dl=0
Hope that helps. You just have to select the right formula and plug in the right values.

[psyxwar]

If the price of a product is 50 cents and it says it is 76% off what was the original price

Rather than using formula, think about it logically. If a product is 76 percent off it means it's (100-76)=24% of its original price.

To find 1% of its initial price, we divide it by 24. (so $0.50/24)
To find 100% of its initial price we then multiply it by 100 (so 100($0.5)/24 = $50/24 = $25/12)

[Phy124]

O small question.

The answer to a question that I did says d element of (-5, 27)

I wrote:

d <27 union d > -5

Are those the same thing? (pretty sure it's not, but not enitrely sure)

If not, how many marks do I stand to lose if it's a 2 marker

Notation aside, you would need to write the intersection rather than the union for it to be correct.

You're saying d < 27 or d > -5, which is the same as d , because you only need to satisfy one of the criteria e.g. d = 100 is greater than -5 therefore satisfies the criteria or d = -100 is less than 27 so satisfies the criteria.

Alternatively the intersection states that d < 27 and d > -5, otherwise written -5 < d < 27 or d , any of these is the correct response.

If in an alternative question the solution was d then we could write the union of d < -5 and d > 27.

Assuming all working up to this point was correct and your only mistake was writing union rather than intersection (or "or" rather than "and" etc.) then my assumption would be you would lose only the mark for having a correct final answer.

For further reading this might help. (explanation of union and intersection and correct notation when using them)

[Professor_Oak]

Do we need to define the probability distribution again for every subsequent question even if its the same as a previous question? Came across this in VCAA 2009 (attached to post).

[LiquidPaperz]

just a quick question.

when trying to solve for CQ, why cannot we go CQ/65 = 120/130  which gives us CQ of 60mm..instead, why have solutions gone drawing a right triangle of OYQ  and solved OQ to be 25 and QC being 40mm (which i get) but why is the first approach wrong?

[bobisnotmyname]

for binomial questions, for the 2 mark questions do we need the (8Cr6)*(0.^6*(0.2)^2 part of
X ~ Bi(8, 0.
Pr(X = 6) =(8Cr6)*(0.^6*(0.2)^2= 0.2936
to get the full marks for working?
thanks in advance

[Robert123]

for binomial questions, for the 2 mark questions do we need the (8Cr6)*(0.^6*(0.2)^2 part of
X ~ Bi(8, 0.
Pr(X = 6) =(8Cr6)*(0.^6*(0.2)^2= 0.2936
to get the full marks for working?
thanks in advance

I highly doubt that you would need the full thing as many schools don't say to use that way in your working out. If it was a requirement, you would also expect it to be mention in the exam reports since only a minority would use that method. Anyway, the exam is there to test your mathematical abilities, not just subbing numbers into a formula.
If you want to play it safe, then use the full thing but it shouldn't be ba requirement.