VCAA EXAMS-Easiest to Hardest?

[psyxwar]

I think the way that question was created was meant to separate students. The exam panel creates separate marking schemes for each exams - so they don't always have to give you method marks, all the marks can go to the answer only -  and so if you didn't get the gradient right, the remaining questions remained inaccessible according to my teacher, who marked last year's exam.

It stops sneaky people making the gradient be something easy like "1" or "a" and do the rest of the question hoping to get conseqs when if you did it using the actual correct answer, it would be a lot harder.

Wait wot, this doesnt make any sense for tech active where your CAS does everything for you

[lzxnl]

Yeah but from memory the gradient function was something like "square root m -4". Then you had to find the area function (and the terminals for the integral were messed up as well I think) and that would have been really convoluted because you had to integrate with unknowns.

THEN you had to derive area function to find max/min which would have been harder with square root m-4 than a variable "a" or a random positive constant.

If you made it a simple constant like 'a" or a integer value like "1", you make it really easy for yourself in an unfair way because you don't have m or square roots (or whatever) in your area function since your equation is much more simplified.

Debatable. If you REALLY wanted to, you could have just stated 'let gradient = m' and then at the end m for the gradient back into what you got.
If anything, having an explicit gradient potentially makes it easier as there might be a possibility of some simplification involved by virtue of what the gradient is. For instance, if you have derivative of (x^x * f(x)), that's a pain to do normally, but if f(x) happens to equal 2x^-x, the derivative is trivial. See what I mean by possible simplification?