VCE Specialist 3/4 Question Thread!

[schoolstudent115]

In any case, we have the trivial solution x=0, and the two you've found in e(i).

Here are a few questions you might like to consider:
- What happens when one (or more!) of the two solutions you found coincides with the trivial solution? When does this occur? (Note that you have already done this for x = 0). Is it possible to limit these instances to a finite set, then generalise their behaviour around x = 0?
- Check the behaviour of the second derivative around x = 0 (both the left-hand and right-hand limits)
- As e^-x gets large, depending on the parity of n, we approach a function similar to xⁿ. Does the parity also have an impact on how many inflection points the function has?

Things that might tip you off to checking the above include the finite number of cases, the clear distinction between cases etc. It's also a lot easier to start at small values of n, like 2 or 3 and then generalise for the sets mentioned above that include them.

As for analytical vs. several graphs - I'd like to think this way of thinking is analytical enough (you only really need to draw 2-3 graphs, and only if you really do need them). Since n is just an integer, an exhaustive graphing method is not going to be good enough as you need to generalise for some subset of the integers.

Hope this makes sense - feel free to drop any other queries or followups

Previously I did try to use parity, and I found that:
0 |  n = Z- \ {0}
1| (??)
2| n = {1} U {2k+1}, for k is a positive integer.
3| n = 2k, for k is a positive integer.

I got (0) and (3) correct, but the {1} from (2) needed to be moved to (1). I didn't know why. I tried to find where the solutions coincide with x = 0 (this is when n=1, you get x=0 and x=2 but x=2 is invalid). I've attached my working out.

My attached answer is apparently correct, but is my working out valid? Also I'm not sure how they'd expect us to do all this for a 2 mark question

[fun_jirachi]

I think the image you've attached seems great! I think it's a great way to think about it, adding inflection points based on certain criteria (which I'd like to have thought of first if I'm honest - well done!). Very nicely done, definitely valid.

As for why n=1 is a unique case, I'd like to think you've done a decent enough job almost explaining it to yourself there; it has no concavity change around x=0, and since \(n-\sqrt{n}\) coincides with that, we only really have to consider the other case which is in fact an inflection point.

I think it's a fair two marks given most of the work is done beforehand, though I may be wrong given I've never gone through VCE. I think also given that the cases for 0 and 1 are easy enough to define (for the former, undefined values essentially make the subset trivial and for the latter, a singular case requires you to only consider that particular case) it's not a particularly awful question. Also, consider that only where you've written down 'cases' and everything below it is worth the two marks (the whole image is worth four).

[S_R_K]

I approached that question as follows: for a point of inflection to occur at \(x\) we require that \(f''(x)=0\) and changes sign at \(x\).

Note from part (d) that:
(1) If \(n \leq 2\) there is no point of inflection at \(x=0\).
(2) If \(n\) is even and greater than 2, then \(x=0\) is a zero of \(f''(x)\) with even multiplicity, hence \(f''(x)\) does not change sign there, so no point of inflection.
(3) If \(n\) is odd and greater than or equal to 3, then \(x=0\) is a zero of \(f''(x)\) with odd multiplicity, hence \(f''(x)\) changes sign there, so one point of inflection.

Now note from part (e)(i) that
(4) If \(n < 0\) then there are no non-zero values of \(x\) at which a point of inflection occurs;
(5) If \(n=1\) then there is a single point of inflection at \(1 + \sqrt{1} = 2\) (note that we already ruled out the point of inflection at \(1 - \sqrt{1} = 0\));
(6) If \(n \geq 2\) then there are two points of inflection at non-zero values of \(x\)

Combining everything, we have:
0 points of inflection requires \(n\) to be non-positive.
1 points of inflection requires \(n=1\).
2 points of inflection requires \(n\) to be even and greater than 2.
3 points of inflection requires \(n\) to be odd and greater than or equal to 3.

It's a reasonable question, but 2 marks is way too little for the amount of time it requires to answer it thoroughly and properly - although, marks were awarded just for correct answers, so if you guessed correctly, or just checked the graph for a few small values of n and extrapolated, then that could save time. I'm sure a lot of students either skipped this one or guessed quickly because they judged (quite correctly) that working it out properly wasn't worth the time.

[aspiringantelope]

Why does \(|2y+1|=e^{2x}\) when solving for y equal \(y=-\frac{1}{2}(e^{2x}+1)\), rather than \(y=\frac{1}{2}(e^{2x}-1)\) ?

[keltingmeith]

Why does \(|2y+1|=e^{2x}\) when solving for y equal \(y=-\frac{1}{2}(e^{2x}+1)\), rather than \(y=\frac{1}{2}(e^{2x}-1)\) ?

That's a good question - and in fact, if you look at the graph for this relationship, you will see that BOTH of those solutions are present in the graph. In fact, if you look at this desmos graph, you can see both of those functions map onto the relationship.

The truth is, whether you pick one or the other is going to depend on the value of y. Is there maybe more information to this question that you haven't given us?

[aspiringantelope]

That's a good question - and in fact, if you look at the graph for this relationship, you will see that BOTH of those solutions are present in the graph. In fact, if you look at this desmos graph, you can see both of those functions map onto the relationship.

The truth is, whether you pick one or the other is going to depend on the value of y. Is there maybe more information to this question that you haven't given us?

Yes sorry, it also tells us that y(0) = -1 (from a differential equation), meaning it should be negative, but, what would the steps be in order to turn it into a negative? (is it just a negative sign in front?) but again I don't think that works 
Thanks

[keltingmeith]

Yes sorry, it also tells us that y(0) = -1 (from a differential equation), meaning it should be negative, but, what would the steps be in order to turn it into a negative? (is it just a negative sign in front?) but again I don't think that works 
Thanks

Well, let's try plugging this into the LHS of our equation! We have:



Looking at this, we had to flip the sign in the modulus - this means that the branch we choose must be the negative branch. Remember the definition of |x|:



Since we have 2y+1<0, then we must take -f(x). So, the next step of solving for y is:



Alternative: if you look at what's in the modulus, you know it will be equal to the positive branch if it's greater than 0. So, solve for when y is greater than or equal to 0:



Since y=-1 is NOT greater than or equal it -1/2, then you know we must take the negative branch.

[aspiringantelope]

Alright!
Thanks for the explanation!

[justinator2]

Vector Functions Notation

I have always been writing out in full "make vertical component of v(t) = 0" for questions where I need to find the maximum height, and let horizontal component of r(t) = 0 to then find the time it takes to reach the ground again etc...

Are there any accpetable short-hand ways of writing this out instead of spending an extras 5-10 seconds writing out a full senctence

such as v_y(t) = 0      to mean that the vertical component of the velocity vector = 0

Do examiners or teachers who have been past examiners care about/understand you if you write this

[S_R_K]

What's wrong with just writing out the equation you get by setting the vertical component equal to 0?

Eg. given \(\textbf{r}(t)=10t\textbf{i} + \left(10t-\frac{g}{2}t^2\right)\textbf{j}\), just write: \(10t-\frac{g}{2}t^2 = 0\).

[cutiepie30]

Hey Guys,

I need help with question 8C and have linked it below, could someone please help

Thanks so much

[S_R_K]

Hey Guys,

I need help with question 8C and have linked it below, could someone please help

Thanks so much

Draw a line from A intersecting with OB, call this point of intersection X. The shortest distance between A and OB is when AX is perpendicular to OB. Draw a diagram illustrating the right-angled triangle OAX and consider how the lengths of the line segments AX and OX relate to parts a) and b) of the question.

Hopefully that helps, can provide more hints if required.

[cutiepie30]

Draw a line from A intersecting with OB, call this point of intersection X. The shortest distance between A and OB is when AX is perpendicular to OB. Draw a diagram illustrating the right-angled triangle OAX and consider how the lengths of the line segments AX and OX relate to parts a) and b) of the question.

Hopefully that helps, can provide more hints if required.

Hey S_R_K,

Thanks so much for the help! Hope you have a great day

[amyzzwq]

Anyone knows how I can format my TI such that it gives answers like the Classpad?

[S_R_K]

For the TI, if you input expressions using decimal notation, it gives answers using decimal notation. So for exact answers, convert those decimals to fractions of integers.