VCE Methods Question Thread!

[Lear]

The touch must occur on the line y=x.
Therefore the coordinate will be (a,a)

If it touches y=x the gradient must be 1 at point a.
Solve the derivative of f for 1 to find the x coordinate in terms of k (our a value)
Now solve f(a)=a for k.

[S_R_K]

Mutually exclusive
Pr(a and b) is 0
Independent events
Pr(a and b) is equal to Pr(a) * Pr(b)
Are these correct?

Yes. This is something that is easily checked in your textbook or any online source on basic definitions for probability.

I have no idea how to do MCQ Q20 of VCAA 2018 exam 2 (nht)
The solution makes no sense either. Can someone please explain what’s going on in that question?

Sorry I couldn’t upload an attachment. I’m on my phone and it’s saying the jpeg file format is not supported.

Thanks.

Yes, VCAA's solution to that one is garbled nonsense.

Easiest way to do this one is to realise that if two lines are reflections of each other about the line y = x, then their gradients will be reciprocals of eachother.

So given that f(3) = 7, then we have f–1(7) = 3 (by definition). Now given that f'(3) = 2, this says that the tangent to f at x = 3 has gradient 2. Hence, the tangent to f–1 at x = 7 will have a gradient that is the reciprocal of that, hence 1/2.

[S_R_K]

i also dont understand Q19 of mcq 2017 nht

This is a standard conditional probability question that can be answered using the formula: Pr(A|B) = Pr(A and B)/Pr(B).

Just find the intersection of the two events, and use the rule for the probability distribution. You'll need to use some log laws to get the final answer.

[Lear]

This is a standard conditional probability question that can be answered using the formula: Pr(A|B) = Pr(A and B)/Pr(B).

Just find the intersection of the two events, and use the rule for the probability distribution. You'll need to use some log laws to get the final answer.

Lol the thing that troubled most of my peers was to recognise you could only have integer values so you can't simply integrate between 8 and 9. You just have to sub in d=9

[minhalgill]

In question 9, do we have to give our final answer in radians or can it be in degrees? i got 9 x squareroot 2. is that correct? or when finding the area do we have to convert it to radians?

[S_R_K]

understood the probability got a little confused with the log laws, thanks man

No worries. If you're still having trouble, post your working and someone here can help you out further.

[Lear]

In question 9, do we have to give our final answer in radians or can it be in degrees? i got 9 x squareroot 2. is that correct? or when finding the area do we have to convert it to radians?

This is no longer relevant to Methods I think

[VinnyD]

This is no longer relevant to Methods I think

Yep, related rates were taken out of the study design for Methods

[minhalgill]

Yep, related rates were taken out of the study design for Methods

oh okay, but if for another question where we're asked to find the area, should we convert to radians, or can we find the answer using degrees?

[Sine]

oh okay, but if for another question where we're asked to find the area, should we convert to radians, or can we find the answer using degrees?

Differentiation and Integration techniques only work with radians if that is what you are asking.

Also although related rates is out of the study design similar questions could come up just as an application of the chain rule (which is on the study design).

[am123321]

Could someone help me out with this question from the 2017 exam 2. How would i use convert this to a quadratic equation to use the discriminant?

[Lear]

Could someone help me out with this question from the 2017 exam 2. How would i use convert this to a quadratic equation to use the discriminant?

I don’t think this is a question where you can transform it into a quadratic and then use discriminant. That does not work for every function.

Have a read of my reply to Darth Vader’s question 4 hours ago here —> https://atarnotes.com/forum/index.php?topic=128232.msg1074051#msg1074051

It’s essentially the same question and I would approach it the same way.

[MathMethodsGuru]

.

[Unsplash]

so does no one no how to do it? ^^

As both the function gk(x) and its inverse both go through the origin (0,0) and we also know that the inverse of a function is a reflection across the line y=x. That implies that the gradient of both the tangent of the function and its inverse are reflections across the line y=x. As y=x has makes an angle of 45 degrees with the positive direction of the x-axis, we have two cases. Either the gradient of the function makes a 60 degree angle with the positive direction with the x-axis and therefore the inverse makes 30 degrees, or the gradient of the inverse make an angle of 60 degrees with the positive direction and therefore the function makes an angle of 30 degrees. The derivative of gk at 0 is 2k. Therefore solve it equal to both tan(60 degrees) and tan(30 degrees).

For the next part both the graphs must have the same gradient at the origin (ie touch once) therefore solve 2k=1/2k to get k=1/2 therefore p=1/2.

Finally, for i.ii consider the area formed by the intersection of the two graphs. May help to use the diagram given in part d. It can be seen that the area will approach 4 (area of the rectangle), but will never be 4 due to the asymptotes present. Therefore the smallest value of b is 4.

[Mattjbr2]

Hi guys, this has been troubling me.

If we have f: [0,6]->R, f(x)= sin(pi*x/3) and g: [-3,3]->R, g(x)=e^x

Shouldn't the domain of g(f(x)) be the domain of f(x), [0,6]?
So then, shouldn't the range of g(f(x)) be [1/e, e]?

The answers claim that the minimum value for f is 0 and that the maximum value for f is 1, so the range of g(f(x)) is [0,e].
But I claim that the minimum value for f is -1 and the maximum value is 1, so the range of g(f(x)) should be [1/e, e]

If you define both functions (including their x restrictions) then graph g(f(x)), it'll display a function with a domain of [0,6] and range [1/e,e].

Am I wrong, or are the answers wrong?