VCE Methods Question Thread!

[Angelx001]

(Image removed from quote.)

Thanks for your reply @VanillaRice!! For the solutions, MAV said p hat = 0.8 :/ (image attached)

opps, idk if you can see my photo. i don't think it lets me upload it cause it's a screenshot?

[VanillaRice]

(Image removed from quote.)

Thanks for your reply @VanillaRice!! For the solutions, MAV said p hat = 0.8 :/ (image attached)

I've had a look at the solution for my own copy, and I think it might possibly have been a mistake?.
I might be missing something here, so feel free to correct me, but I'm pretty sure that if anything, it should say E(P hat) = 0.8, since 0.8 is the population proportion. Either way, you should still be able to arrive at the correct answer using the formula.

[mtDNA]

Hey,

Just a quick one - for questions 4di and 4dii from 2007 VCAA Exam 2, I was wondering whether or not you were able to do a proof such as LHS = ..., then RHS= ... then eventually show that both sides are equal, that way fulfilling the proof (hopefully that made sense  )

Thanks in advance!

[Sine]

Hey,

Just a quick one - for questions 4di and 4dii from 2007 VCAA Exam 2, I was wondering whether or not you were able to do a proof such as LHS = ..., then RHS= ... then eventually show that both sides are equal, that way fulfilling the proof (hopefully that made sense  )

Thanks in advance!

yes that's perfectly fine

[Rieko Ioane]

Hi,

For this question https://imgur.com/a/whhb4
They say b E [0,0.2] and say each endpoint gives E(X) min and max respectively. How can we be so confident in saying that b = 0, a=1 means E(X) will be a minimum and when b=0.2 and a=0 it will be a maximum?

[Bri MT]

Hi,

For this question https://imgur.com/a/whhb4
They say b E [0,0.2] and say each endpoint gives E(X) min and max respectively. How can we be so confident in saying that b = 0, a=1 means E(X) will be a minimum and when b=0.2 and a=0 it will be a maximum?

Hope this helps

[Rieko Ioane]

Hope this helps

Thanks!

For finding triangle ABC for this question https://imgur.com/a/ITgWs
I'm not sure what the answer is doing. They have A=1/2(7/6 +3)(2). they're using the area of a triangle formula but 7/6 +3 is the length of the hypotenuse, but don't the lengths have to be the sides that create the 90 degree triangle?

It still gets the same answer as another method that I used similar to theirs but I'm not sure how.

[Eric11267]

Thanks!

For finding triangle ABC for this question https://imgur.com/a/ITgWs
I'm not sure what the answer is doing. They have A=1/2(7/6 +3)(2). they're using the area of a triangle formula but 7/6 +3 is the length of the hypotenuse, but don't the lengths have to be the sides that create the 90 degree triangle?

It still gets the same answer as another method that I used similar to theirs but I'm not sure how.

They are using the most basic formula for calculating a triangle which is 1/2xbasexheight. 7/6+3 is the length of the base and 2 is the height of the triangle

[Bri MT]

Thanks!

For finding triangle ABC for this question https://imgur.com/a/ITgWs
I'm not sure what the answer is doing. They have A=1/2(7/6 +3)(2). they're using the area of a triangle formula but 7/6 +3 is the length of the hypotenuse, but don't the lengths have to be the sides that create the 90 degree triangle?

It still gets the same answer as another method that I used similar to theirs but I'm not sure how.

They are using two length values perpendicular to eachother.
If it helps you understand, draw up a right angled triangle and split it into two triangles so the 90 degrees is now two 45 degrees. Assign letter values for the sides and do the maths for figuring out the area as two right angles triangles. You will see that will always get the same result as treating the hypotenuse as the base (which is what VCAA have done).
Remember that base * height isn't just for right angles triangles
It's not like they're multiplying the hypotenuse by an adjacent side

[DailyInsanity]

Hi, could someone help me out with this question.

Consider the function f:[2,infinity) -->R, f(x) = x^4+2(a-4)x^2-8ax+1.Find the maximal set of values of 'a' for which the inverse function exists.

Thanks in advance!

[Bri MT]

Hi, could someone help me out with this question.

Consider the function f:[2,infinity) -->R, f(x) = x^4+2(a-4)x^2-8ax+1.Find the maximal set of values of 'a' for which the inverse function exists.

Thanks in advance!

For the inverse to exist the function must be one-to-one. (Pass horizontal and vertical line test )
The easiest way to approach this question is to sketch the graph and see where the turning points are.

[VanillaRice]

Hi, could someone help me out with this question.

Consider the function f:[2,infinity) -->R, f(x) = x^4+2(a-4)x^2-8ax+1.Find the maximal set of values of 'a' for which the inverse function exists.

Thanks in advance!

Nice answer by miniturtle
If you're having trouble sketching the graph initially (since there is an unknown, a), the turning point (in terms of a) can be found by finding the value of x for which the derivative is equal to 0.

[DailyInsanity]

Thanks for the responses guys!

I've tried finding the general form for the turning points in terms of 'a'. I solved f'(x)=0 and got T.Ps at x=2, x=(1-a)^1/2 -1 and a=-(1-a)^1/2 -1. For the function to be one to one, the T.P at 2 must be the last turning point on the right side of the graph, since the domain is [2, infinity), and if there is another T.P past x=2 then it will not pass the horizontal line test. So I solved for (1-a)^1/2 -1 to be less than or equal to 2, since this T.P x co-ordinate must be to the left of the x=2 T.P, and i got   -8<(or equal to) a < 1 (or equal to). However the answer is a is from -8 (inclusive) to infinity.

I'm thinking that a>1 is included because the x co-ordinates of the T.Ps are undefined for a>1(in second line from solving f'(x)=0) and hence the only T.P is at x=2, and therefore has inverse for a>1. Therefore combine these values (a>1) with -8<a<1 to get, -8<a<infinity. I'm not sure if this thinking is correct or where I've gone wrong - it's still quite confusing. Do you guys have a better way of doing this (assuming that what I said makes any sense  )?

[VanillaRice]

Thanks for the responses guys!

I've tried finding the general form for the turning points in terms of 'a'. I solved f'(x)=0 and got T.Ps at x=2, x=(1-a)^1/2 -1 and a=-(1-a)^1/2 -1. For the function to be one to one, the T.P at 2 must be the last turning point on the right side of the graph, since the domain is [2, infinity), and if there is another T.P past x=2 then it will not pass the horizontal line test. So I solved for (1-a)^1/2 -1 to be less than or equal to 2, since this T.P x co-ordinate must be to the left of the x=2 T.P, and i got   -8<(or equal to) a < 1 (or equal to). However the answer is a is from -8 (inclusive) to infinity.

I'm thinking that a>1 is included because the x co-ordinates of the T.Ps are undefined for a>1(in second line from solving f'(x)=0) and hence the only T.P is at x=2, and therefore has inverse for a>1. Therefore combine these values (a>1) with -8<a<1 to get, -8<a<infinity. I'm not sure if this thinking is correct or where I've gone wrong - it's still quite confusing. Do you guys have a better way of doing this (assuming that what I said makes any sense  )?

That's some good logic there!

Another way of putting it is that only 1 stationary point exists when a > 1. However, the equation still has a valid inverse for a > 1, so that's why it's included. If you sketch the graph in your calculator and plug in any value a > 1, you'll find that x=2 is the only stationary point, but the curve will still have a valid inverse.

Note: I will add that I think that this question is quite difficult in terms of the amount of thought required to get the entire answer. I don't think I would have been able to get the full answer if this came up in an exam 

[DailyInsanity]

That's some good logic there!

Another way of putting it is that only 1 stationary point exists when a > 1. However, the equation still has a valid inverse for a > 1, so that's why it's included. If you sketch the graph in your calculator and plug in any value a > 1, you'll find that x=2 is the only stationary point, but the curve will still have a valid inverse.

Note: I will add that I think that this question is quite difficult in terms of the amount of thought required to get the entire answer. I don't think I would have been able to get the full answer if this came up in an exam 

Ah yes! Only one turning point exists for a>1, which is at x=2, hence inverse exists for a,1. Much more nicely said.

Just so you know, this question is actually Qn 20 from the multiple choice section of the 2017 NHT Methods exam 2. So you didn't have to work from scratch, that being said i'm not sure how much more easy it is in M.C form - trial and error ? Thanks for the help