VCE Methods Question Thread!

[S_R_K]

the ruling out part

We can rule out the function being strictly decreasing for x ≥ 2, since as x approaches positive infinity, f(x) approaches positive infinity (the x^4 term becomes overwhelmingly large compared to the other terms). So because the function eventually becomes strictly increasing, the only way it can have an inverse for x ≥ 2 is if its strictly increasing for all x ≥ 2.

[Freddie Hg]

We can rule out the function being strictly decreasing for x ≥ 2, since as x approaches positive infinity, f(x) approaches positive infinity (the x^4 term becomes overwhelmingly large compared to the other terms). So because the function eventually becomes strictly increasing, the only way it can have an inverse for x ≥ 2 is if its strictly increasing for all x ≥ 2.

f'(x) = 4(x – 2)(x^2 + 2x + a)
interesting, would an explanation also be true if you were to say the part in the derivative 4(x – 2) will always be positive for x>2, therefore the second part(x^2 + 2x + a) must be always negative for x>2 for it to satisfy a strictly decreasing , however no posibile value of A can achieve this considering the parabola is a positive one and for x>2 will eventually become positive, thus changing the gradient of the overall function making it have no inverse defined for x>2?  f'(x) = 4(x – 2)(x^2 + 2x + a)

[S_R_K]

f'(x) = 4(x – 2)(x^2 + 2x + a)
interesting, would an explanation also be true if you were to say the part in the derivative 4(x – 2) will always be positive for x>2, therefore the second part(x^2 + 2x + a) must be always negative for x>2 for it to satisfy a strictly decreasing , however no posibile value of A can achieve this considering the parabola is a positive one and for x>2 will eventually become positive, thus changing the gradient of the overall function making it have no inverse defined for x>2?  f'(x) = 4(x – 2)(x^2 + 2x + a)

Yes, it is correct to reason that, for any constant A, the derivative will eventually be positive for x > 2, hence the function can not be strictly decreasing for all x > 2.

[Freddie Hg]

Yes, it is correct to reason that, for any constant A, the derivative will eventually be positive for x > 2, hence the function can not be strictly decreasing for all x > 2.

Thanks  alot for explaining this question, for questions like this on the exam which require the student to unpack its basic elements, how would i need to approach it?

[S_R_K]

Thanks  alot for explaining this question, for questions like this on the exam which require the student to unpack its basic elements, how would i need to approach it?

Hard to give a general answer to this. These sorts of questions can take a lot of time, and are generally best left until the end and only if you are confident that you (i) have answered everything else correctly, and (ii) have time to spare.

One thing that can be useful for questions involving families of functions (or functions where one of the coefficients / constants is a parameter), try sketching a graph and using the slider function on CAS to get a feel for how changing the value of the coefficient / constant changes the key features of the function. This approach won't necessarily tell you what to write as your answer, but it can suggest the approach to take (ie. start by finding coordinates of stationary points, or whatever).

[Melodyt]

LOL did you find the answer to this uestion??

[jasontran11]

Hey, for exam 1 do we need to memorise general solutions of trigonometry?

General solutions as in the solutions for sin, cas and tan. Not exact values

[Lear]

Yes

[Azim.m]

Could we use limits instead, to answer question 4fii?

And if not, what are they saying when they’re talking about the region bounded by the triangle?

2016 exam 2

[Sine]

Could we use limits instead, to answer question 4fii?

And if not, what are they saying when they’re talking about the region bounded by the triangle?

2016 exam 2

yes, a friend of mine and I both used limits for this question in the actual 2016 exam and we got marks.

[smamsmo22]

I am terrible at general solutions.
How would you approach this?

Thanks

[Sine]

I am terrible at general solutions.
How would you approach this?

Thanks

sin(2x) - cos(2x) = 0
sin(2x) = cos(2x)
tan(2x) = 1

then just continue as you would with a normal general solution quesiton

[fruitbowl34]

Can someone please help me with how to approach questions that require to find the derivative then afterwards its 'hence find the antiderivative of . . .' These questions always confuse me and I have no idea how to approach or structure an answer!

[itsaishah]

Can someone help me with Question 10 from 2015 Exam 1?

[hums_student]

I know that diagrams aren't generally drawn to scale in VCAA exams, but what do we do when they're RIDICULOUSLY off?
Example: VCAA 2017 Exam 1

As you can see there is a VERY OBVIOUS gap between the x-intercept of the curve at (1,0) and point B, yet after doing the calculations it's found that they're actually the same point. I ended up rejecting x=1 for point B because I thought the diagram made it very clear that B>1.

If this comes up on the final exam do we just trust our own calculations and assume that the diagrams are wrong? Thanks.