VCE Methods Question Thread!

[Lear]

2013 methods exam 1
Question 10c
Is it possible to integrate a log in the methods study design?

You don’t need to integrate a log for that question.

[sailinginwater]

You don’t need to integrate a log for that question.

But how else do you do the question apart from area between curves?

[Lear]

I’ll give you a hint. Note ST and the y axis form a triangle.

[S_R_K]

dont we obtain 1 tangent from the inverse and 1 tangent from the graph of g? and use the gradient of the inverse with 30 and the gradient of g with 60?

You are told that the angle between the tangents is 30°. There are two ways this can occur.

Tangent of g has an angle of 30° with x-axis, tangent of g^–1 has an angle of 60° with x-axis.

OR

Tangent of g has an angle of 60° with x-axis, tangent of g^-1 has an angle of 30° with x-axis.

This gives two equations in terms of k.

[Azim.m]

Could someone help me out?
Thanks in advance

[S_R_K]

Could someone help me out?
Thanks in advance

There's some key information in the question - you are working with the sampling distribution of the proportion.

What have you tried?

[minhalgill]

im having trouble solving for general solutions. i understand the base angle is pi/6 but im confused as to how to progress after ive gotten the base angle? ive also attatched the formula for the general solution below.

thanks in advance,

[S200]

Well, it has to be negative...

Find the places that cos is negative at that value and plus/minus \(n\times 2\pi\)

[minhalgill]

can soemone please help me with part (c)? i think we're supposed to use the previous working to find the domain but i cant figuire out how?

2) in question 3, i dont understand how we get the upper restriction of the discriminant? like, how do we know its supposed to be less than 25?

[S_R_K]

can soemone please help me with part (c)? i think we're supposed to use the previous working to find the domain but i cant figuire out how?

The maximal domain of the square root function is the set of all non-negative real numbers. So you can use part (b) to help you out by finding all values of x where the graph of that hyperbola has non-negative y coordinates.

2) in question 3, i dont understand how we get the upper restriction of the discriminant? like, how do we know its supposed to be less than 25?

If the discriminant is greater than 25, then one of the solutions to a^2 – 5a – k = 0 will be negative. However, a = 2^x, which is always positive, so we discard any values of the discriminant that give negative solutions to the quadratic.

[Yertle the Turtle]

can soemone please help me with part (c)? i think we're supposed to use the previous working to find the domain but i cant figuire out how?

2) in question 3, i dont understand how we get the upper restriction of the discriminant? like, how do we know its supposed to be less than 25?

1) Part c is separate from part b, so you just look at the maximum and minimum values of f(x) rather than g(x). Therefore x is between -ve infinity and -2 and between 3 and infinity.
2) You rearrange the formula to create a quadratic function, and then use the quadratic formula to find the discriminant of the equation. Since 5 is the coefficient of the second term, then the discriminant is 5²-(4 times the coefficient of the first term times the coefficient of the third term). That's where the 25 comes from.

Hope this helps!

[minhalgill]

The maximal domain of the square root function is the set of all non-negative real numbers.
does this apply for any/all cases?

[S_R_K]

Yes, of course. The square root of any negative number is undefined; the square root of zero is zero; and every positive number has one positive square root.

[nottelling12345ftw]

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

[DrDin213]

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.