VCE Methods Question Thread!

[Stevensmay]

Change of base rule can apply to everything, with respect to the normal domains for logarithms.

[Alwin]

From a few posts/pages/aeons back (sorry ahat!)

The graph of is transformed to , where k is a positive real number.

i) Find the value of k such that the equation    has only one solution in the domain and that the solution is at .

ii) Find the values of k such that the equation   has only one solution in the domain .

iii) Find the values of k such that the equation   has exactly three solutions in the domain .

Help on this question would be appreciated. My teacher used trial/error, but surely there's a better method?

Thanks

Solutions

Spoiler

Using a purely graphical approach since I didn't do such a good job algebraically:

Look at this demonstration and pla
y around with the value of k
Note that the domain is the area between the orange and purple lines




a. For one solution at 6pi

So clearly we is the solution at 6pi



b. For on solution between -pi and 6pi
From part a, we can see that if then
This is because you as 'squishing' the graph since your dilating by a greater factor from the x axis

But, if we have a problem when k is too big, as there are now two solutions in the domain.

So, we have to make this second solution outside of the domain. In other words,





c. For three solutions between -pi to 3 pi

Clearly, we want:




Hopefully it makes more sense this time!

[shadows]

If a question says for example

find f'(x) in terms of a... Only pro-numerals of a are allowed?

Heffernan 2013 Exam 1 question...  there's a question like that, but the answer is in terms of a and x....

Someone clarify?

When a question states Use Calculus in Exam 2 and its worth 3 or 4 marks, my teacher says you need to show working out by hand, (proof of subbing in for integration), or would putting it in calculator and getting numerical answers straight for calculator sufficient

How about use Algebra, everything needs to be worked out by hand? I am kind of confused about these questions. Sometimes if the numbers start getting weird, I get confused what steps I need to show and what I can just put in my calculator

How strict are the examiners with all this stuff? Does it depend on the examiner you get? My teacher is super strict with marking etc, but I don't know if VCAA will be this strict?

[Stevensmay]

When a question states Use Calculus in Exam 2 and its worth 3 or 4 marks, my teacher says you need to show working out by hand, (proof of subbing in for integration), or would putting it in calculator and getting numerical answers straight for calculator sufficient

If you write just a numerical answer you would most likely get 1/4 marks. You need to have one answer and three pieces of working out. For most of these questions an integral, the result of that integral with terminals and then evaluating with the terminals is sufficient


VCAA is very strict with regards to showing enough working out, and you will lose marks if you do not have enough answer material written down.

[shadows]

Sorry I don't think I explained this clearly...

I've give an example

Say, I had to find area of 2x from 4 to 1 using calculus


(2x)dx (with terminals etc.)

=[x^2] (I show the antideriviative, but I just use cas to get it)


=[16] - [1] (I just evaluated these numbers by cas)  (My teacher says you need to show a line of subbing and evaluating etc. You can't just skip to this)

=15

This is what I meant. (Obviously it is more applicable when the question is more complicated

[ahat]

Hopefully it makes more sense this time!

Alwin, I can't even express my gratitude, thanks so much, you explained it really well. I can die happy now

As fate would have it though, I am back yet again with another question

X ~ Bi(n,p)
it can be shown that
For a certain binomial distribution with 8 independent trials, Pr(X = 5) is the mode. Find the values of and where
Sort of unsure where to start.
P.S. is there a quick code for text wrap (for writing maths) rather than having to press the pi button?

@Shadows, I really don't see why you'd need any more than that

[abcdqd]

This is what I meant. (Obviously it is more applicable when the question is more complicated

yep need to show subbing in!

[Stevensmay]

This question would be worth around three marks, maybe two.

Find the area bound by the x and y axes and the curve




s.q units.

Critical things to watch out for here are whether we need to give units, and also making sure we have dx where appropriate. A mark has been taken off in the past if dx's are omitted.

[Thu Thu Train]

This question would be worth around three marks, maybe two.

Find the area bound by the x and y axes and the curve




s.q units.

Critical things to watch out for here are whether we need to give units, and also making sure we have dx where appropriate. A mark has been taken off in the past if dx's are omitted.

>Talks about not forgetting dx
>Forgets dx
>Not sure if intentional.

[Stevensmay]

>Talks about not forgetting dx
>Forgets dx
>Not sure if intentional.

Don't do maths and watch TV kids!

[~T]

Just a wording query here. Insight 2012 exam 2, multiple choice question 18. Basically a normal distribution question, asks for "the minimum accepted height, correct to the nearest centimetre" for Australia's Next Top Model entry.

Answer with calc comes out as 176.46cm. 176cm and 177cm are both alternatives. The answers say 177cm because 176cm wouldn't be accepted. I don't think that's what is implied in the question though?? It's asking primarily for "the minimum accepted height" which you evaluate "correct to the nearest centimetre", not asking for the minimum whole centimetre height which would be accepted.

Anyway, agree? Disagree?
__________________________________

Another question: I've completed the VCAA past exams for both methods and spec, and have just been doing various other company exams since. Even though I went quite well on them all, would it be worth doing the VCAA ones again closer to the exam?

[Alwin]

Just a wording query here. Insight 2012 exam 2, multiple choice question 18. Basically a normal distribution question, asks for "the minimum accepted height, correct to the nearest centimetre" for Australia's Next Top Model entry.

Answer with calc comes out as 176.46cm. 176cm and 177cm are both alternatives. The answers say 177cm because 176cm wouldn't be accepted. I don't think that's what is implied in the question though?? It's asking primarily for "the minimum accepted height" which you evaluate "correct to the nearest centimetre", not asking for the minimum whole centimetre height which would be accepted.

Anyway, agree? Disagree?
__________________________________

Another question: I've completed the VCAA past exams for both methods and spec, and have just been doing various other company exams since. Even though I went quite well on them all, would it be worth doing the VCAA ones again closer to the exam?

For the rounding issue, I would say round up because it is the minimum height.

You found that they had to be 176.46cm tall at least, so someone 176.45cm (after carefully patting down her hair and measuring very precisely) would not be accepted. So, following this logic someone 176cm would not be accepted either, so out of the option 177 should be the right answer.

I think VCAA has done quite a few questions like this before, but usually with people: for example you were required to realise that you had to round 15.1 to 16 people for whatever reason. But, there was one "tricky" question with Tasmania Jones asking for the maximum time he can take to get to the antidote (or whatever). You had to round down this time because you couldn't round up to the nearest whole number because by then Tasmania Jones would be dead even tho the answer was ___.52 or something.

Hope it kind of makes sense

[LOLs99]

Hey, can someone please help me on Q4 of vcaa 2012 exam1- http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012mmcas1-w.pdf.  pr(total 4calls | calls on both Monday,Tuesday)= pr(total 4 calls [intersect] calls on both mon,tues)/ pr(calls on both mon,tues) so how would i write the numerator next?
and how do i know when to use conditional probability?

[Alwin]

Hey, can someone please help me on Q4 of vcaa 2012 exam1- http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012mmcas1-w.pdf.  pr(total 4calls | calls on both Monday,Tuesday)= pr(total 4 calls [intersect] calls on both mon,tues)/ pr(calls on both mon,tues) so how would i write the numerator next?
and how do i know when to use conditional probability?

I hate this question so much now. Only mark I lost on Exam 1 coz of the wording AND when I was LaTeXing my solution up freaking internet died and lost everything I was typing -.-
So sorry, no pretty looking LaTeX:

> Pr(total 4 calls | calls on Mon and Tue) = Pr(total 4 calls n calls on Mon and Tue) / Pr(calls on Mon and Tue)
   ^ By applying the conditional probability formula

> Pr(calls on Mon and Tue) = Pr(X > 0 on Mon and Tue) = Pr(X>0) x Pr(X>0) = 0.8 x 0.8 = 0.64
   ^ Finding the denominator

> Pr(total 4 calls n calls on Mon and Tue) == probability of ways for calls to be on both days that sum to 4
>> Pr(1 on Mon and 3 on Tue) x Pr(2 on Mon and 2 on Tue) x Pr(3 on Mon and 1 on Tue) = 0.1x0.2 + 0.5x0.5 + 0.1x0.2 = 0.29
      ^Finding the numerator

> Pr(total 4 calls | calls on Mon and Tue) = 0.29/0.64 = 29/64
   ^Subbing the numerator and denominator into the formula

Hope it makes sense!

[LOLs99]

I hate this question so much now. Only mark I lost on Exam 1 coz of the wording AND when I was LaTeXing my solution up freaking internet died and lost everything I was typing -.-
So sorry, no pretty looking LaTeX:

> Pr(total 4 calls | calls on Mon and Tue) = Pr(total 4 calls n calls on Mon and Tue) / Pr(calls on Mon and Tue)
   ^ By applying the conditional probability formula

> Pr(calls on Mon and Tue) = Pr(X > 0 on Mon and Tue) = Pr(X>0) x Pr(X>0) = 0.8 x 0.8 = 0.64
   ^ Finding the denominator

> Pr(total 4 calls n calls on Mon and Tue) == probability of ways for calls to be on both days that sum to 4
>> Pr(1 on Mon and 3 on Tue) x Pr(2 on Mon and 2 on Tue) x Pr(3 on Mon and 1 on Tue) = 0.1x0.2 + 0.5x0.5 + 0.1x0.2 = 0.29
      ^Finding the numerator

> Pr(total 4 calls | calls on Mon and Tue) = 0.29/0.64 = 29/64
   ^Subbing the numerator and denominator into the formula

Hope it makes sense!

Same here.I wasn't aware of the conditional probability when I was doing it so bascially I did the numerator part only.
Your solution is just too perfect !! I totally get it now. Thanks but is there a technique to look out whether it is a conditional probability or not?

Another question from vcaa 2012 exam 2 Q2e. http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012mmcas2-w.pdf
If they want u to go back to original, dont u use T,R,D instead of dilation-->reflection-->translation because a lecturer told us this. and I don't get how their second solution from the start.