VCE Specialist 3/4 Question Thread!

[keltingmeith]

How do I Q4, 5 and 11 in the VCAA NHT spesh exam?

Firstly, pro-tip - is definitely helpful if you provide all the questions, please. This time it wasn't too bad, since they're public exams, but they're not always available for us to go looking for (plus it's kind of annoying for us to do so).

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/sm1nht_examrep17.pdf

For Q4, the ans are a=-5, b=1 and a=-2, b=-2. Expanding (z-1-1)(z-1+i)(z+a) then matching coefficient with the given eqn, I only get a=-2, b=-2... how did I NOT get a=-5, b=1?

Interesting - let's give it a try again. So, first we expand:


From here, we can use the fact that constant should be -4 to see that c=2, and get the final polynomial:



Easy. Now, we can equate coefficients of the polynomial to get the two simultaneous equations:



If I add these together, I get:



Solve the quadratic, and you get b=1,-2. Put these into the first equation to get the second a-value you're missing. Not sure what step you've skipped, but I'm guessing you didn't end up at that quadratic?

Unfortunately, I'm a bit short on time, but I'm sure some other lovely person on here can answer your other two questions.

---

Hey guys! Can someone help me with question 5)c) in Section B of Exam 2, 2016 (VCAA)? I understand that I stuffed up by assuming constant acceleration, but I've no clue how to do the question correctly. I can't understand the worked solutions either.

My computer is being an absolute dog and its not letting me attach an image of my working out, but the link is:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2016/2016specmath2-w.pdf

Can someone walk me through this question?

Thanks in advance

Pro-tip: always helps to still give previous answers in a multi-part questions, even if you can't upload your actual working out. By what you said, I'm assuming you just used a constant-acceleration formula? Have you tried finding the integral under the velocity-time curve?

[VanillaRice]

How do I Q4, 5 and 11 in the VCAA NHT spesh exam?

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/sm1nht_examrep17.pdf

For Q4, the ans are a=-5, b=1 and a=-2, b=-2. Expanding (z-1-1)(z-1+i)(z+a) then matching coefficient with the given eqn, I only get a=-2, b=-2... how did I NOT get a=-5, b=1?

For Q5, I get u=3pi/4 for both the upper and lower terminals since u=1-x^2 where x=-1/2 and x=1/2...
The ans is pi squared/ 3 btw

For Q11, I get 4api/ 3 since but the answer is 6a

4) Following on from keltingmeith's post above - since a is already a constant in the question, you should avoid using a factor like (z+a). The only case where (z+a) would be acceptable would be where z = -a was a solution to the equation (in this case, it is not initially clear whether that is the case. Try again with a different constant, and you should be able to get the answer

5) You've made a good start to the question Consider:

Hint: You only need a u-substitution for the first fraction, the second should look familiar 

EDIT: 11) Could you please post your working out or explain what you did? I was able to get 6a.

Hope this helps

[gnaf]

Firstly, pro-tip - is definitely helpful if you provide all the questions, please. This time it wasn't too bad, since they're public exams, but they're not always available for us to go looking for (plus it's kind of annoying for us to do so).

Interesting - let's give it a try again. So, first we expand:


From here, we can use the fact that constant should be -4 to see that c=2, and get the final polynomial:



Easy. Now, we can equate coefficients of the polynomial to get the two simultaneous equations:



If I add these together, I get:



Solve the quadratic, and you get b=1,-2. Put these into the first equation to get the second a-value you're missing. Not sure what step you've skipped, but I'm guessing you didn't end up at that quadratic?

Unfortunately, I'm a bit short on time, but I'm sure some other lovely person on here can answer your other two questions.

---

Pro-tip: always helps to still give previous answers in a multi-part questions, even if you can't upload your actual working out. By what you said, I'm assuming you just used a constant-acceleration formula? Have you tried finding the integral under the velocity-time curve?

Thanks so much keltinmeithg Just realised I used (z-a) instead of (z-c)... note to myself, don't use the same variable as in the question...
Sorry about not posing pics, they weren't loading for some reason but I'll keep that in mind!

[gnaf]

4) Following on from keltingmeith's post above - since a is already a constant in the question, you should avoid using a factor like (z+a). The only case where (z+a) would be acceptable would be where z = -a was a solution to the equation (in this case, it is not initially clear whether that is the case. Try again with a different constant, and you should be able to get the answer

5) You've made a good start to the question Consider:

Hint: You only need a u-substitution for the first fraction, the second should look familiar 

Ahh thank you VanillaRice, silly me! 

I got that fraction , but when I let u be the denom of the first fraction (u=1-x squared) where x=-1/2 and x=1/2, the u value is the same?

[VanillaRice]

Ahh thank you VanillaRice, silly me! 

I got that fraction , but when I let u be the denom of the first fraction (u=1-x squared) where x=-1/2 and x=1/2, the u value is the same?

Ah I see the problem now. It's due to that particular function being symmetric about the y-axis. The next best option is the evaluate the indefinite integral first, then evaluate your definite integral based on that information

Also, I made an edit to my post above RE: Q11 - would you be able to explain or post your workings?

[gnaf]

Thanks VanillaRice, I'll give it a go 

For Q11, I got dx/d theta= -acos(theta)
dy/d theta= asin(theta)
So the length is the integral of square root a^2(cos(theta))^2 + a^2(sin(theta))^2 (from 2pi/3 to 2pi)
So it simplifies to the integral of a from 2pi/3 to 2 pi
I antidiff that to get a*theta from 2pi/3 to 2pi which gives me 4api/3

I have two more questions but mcq this time (attached):
Q3- I tried subbing in random values, which got me to B and E but I don't know how to choose between the two (ans is B)
Q8- I got it down to C or D by seeing when the gradient was negative or positive but once again, not sure how to choose between the two? I graphed C and D in my CAS, but it looks completely different to the diagram in the question?

[Willba99]

Have you tried finding the integral under the velocity-time curve?

Yeah I did that and got the right answer (I think??)

Thanks

[VanillaRice]

Thanks VanillaRice, I'll give it a go 

For Q11, I got dx/d theta= -acos(theta)
dy/d theta= asin(theta)
So the length is the integral of square root a^2(cos(theta))^2 + a^2(sin(theta))^2 (from 2pi/3 to 2pi)
So it simplifies to the integral of a from 2pi/3 to 2 pi
I antidiff that to get a*theta from 2pi/3 to 2pi which gives me 4api/3

I have two more questions but mcq this time (attached):
Q3- I tried subbing in random values, which got me to B and E but I don't know how to choose between the two (ans is B)
Q8- I got it down to C or D by seeing when the gradient was negative or positive but once again, not sure how to choose between the two? I graphed C and D in my CAS, but it looks completely different to the diagram in the question?

11) Have a second look at the derivative of x - it should be a - a*cos
The rest of the method looks fine though.

3) Have a think about the y value of the lower asymptote of a standard graph y = arctan(x), which is -Pi/2. Now apply the transformations (only worrying about the ones that change y values i.e. c and k) which should change the y-value to -k(pi)/2 + c. We want this to be greater than 0, so -k(pi)/2 + c > 0. Solving that for c gives the answer.

8 ) Are you on the Ti-nspire? Try changing the window settings (i.e. the max and min values for x and y) so that they match the ones in the question/image, that usually works for me.

Hope this helps

[shiba woof woof]

Hi Guys

Just a question, when they ask for the new variance of 2 object As with Var(X) and 2 object Bs with Var(Y) is it

Var(X) + Var(x) + Var(Y) + Var (Y) or Var(2X+2Y)?

And why? How do i know when the question is asking for either of these?

I keep getting this confused and I'm never sure which one it is asking for. Thanks a lot

[LifeisaConstantStruggle]

it's normally specified in the question, if they want to calculate independent events or not.If the events are independent then it should be Var(X1+X2)=Var(X1)+Var(X2), not Var(2X)

[shiba woof woof]

it's normally specified in the question, if they want to calculate independent events or not.If the events are independent then it should be Var(X1+X2)=Var(X1)+Var(X2), not Var(2X)

ohhh I see,
thanks so much!

[gnaf]

Thanks so much VanillaRice, I'll try the questions again and let you know how I go 
1) General question- in complex numbers, are the roots supposed to be spread around evenly, like even angles, aroujnd the Argand diagram? One questions asked me to plot z=0, z=root 3 + i and z= root 3-i on the argand diagram but their definitely not evenly distributed e.g. 360/3= 120 degrees apart?

Sorry for not posting pics for 2) and 3), it's extended response so if I took screenshots, there'd probably be heaps of pics! If it's actually easier though, lmk!!

2) I'm not sure how to do Q5 e) from http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/2017SM2-nht-w.pdf
I tried using the T1 which I found in c), which I realised doesn't work since I get R=0...

3) I think the answers are wrong for Q6d)? The ans is C = $409 870 but we're solving for the sample mean? I got n=16?

[VanillaRice]

Thanks so much VanillaRice, I'll try the questions again and let you know how I go 
1) General question- in complex numbers, are the roots supposed to be spread around evenly, like even angles, aroujnd the Argand diagram? One questions asked me to plot z=0, z=root 3 + i and z= root 3-i on the argand diagram but their definitely not evenly distributed e.g. 360/3= 120 degrees apart?

Sorry for not posting pics for 2) and 3), it's extended response so if I took screenshots, there'd probably be heaps of pics! If it's actually easier though, lmk!!

2) I'm not sure how to do Q5 e) from http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2017/nht/2017SM2-nht-w.pdf
I tried using the T1 which I found in c), which I realised doesn't work since I get R=0...

3) I think the answers are wrong for Q6d)? The ans is C = $409 870 but we're solving for the sample mean? I got n=16?

1) That is true only if you are solving for a root of a complex number i.e. zⁿ = k, where k is a complex number, then the solutions will be 2pi/n from each other.

2) You aren't required (or should) use the answers from the previous questions, since they are referring to a different situation (where R does not exist, hence why you got R = 0). In this case, you are setting up an equation
R + 5g*sin(30) = 3g + 2g
and solving for R.

3) You're solving for the sample mean, so retain the mean = 400 000, n = 25 and sd = 30000/5 = 6000, but want a new sample mean value.
You're looking for k such that Pr(sample mean > k|mu = 400 000) = 0.05
Using the inverse normal function on our calc, we find Pr(Z > 1.64485) = 0.05
So we can use the equation for z-scores

and solve for k, which is your answer.

Hope this helps

[shiba woof woof]

it's normally specified in the question, if they want to calculate independent events or not.If the events are independent then it should be Var(X1+X2)=Var(X1)+Var(X2), not Var(2X)

Hey i'm a bit confused actually now that i'm trying to put this into work. Because in a multiple choice question, it states that the two events are independent, and therefore to find variance it's Var(X1 + X2)
But then in a short answer question, it states that the two events are independent, and to find the Variance it's Var(4X1+4X2)

So i'm a bit confused on what you said. I attached the questions below.

Thanks in advance

[LifeisaConstantStruggle]

Okay I might be a bit sloppy in my explanation just now, but for the SA question, I'm pretty sure the "independent" in this case refers to R and C, as in R is independent to C.
But for the total weight of R alone, we have to find Var(4R) instead of Var(R) alone, the terminologies might be confusing but you can sorta get the difference when you do more questions.
Edit: if they somehow add the variables together (the SA question does this, as the total weight is measured instead of measuring the weight of one tyre and then another), instead of calculating it independently like the MCQ (as they measure the mass of chocolate respectively, instead of combining their masses) the variables would be independent to one another, and you'd have to use Var(X1+X2)