Free VCE Maths Methods Units 3&4 Practice Exam & Solutions - ATAR Notes

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⭐ Solutions have also been included in a separate attachment.
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⭐ The exam is relevant for the VCE Maths Methods Units 3&4 study design for 2016-2021.

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[redpanda83]

Hey!
Can you double check the methods exam 1 question 2 b solution.
I am getting pi/2.
I think solution missed out, the -1/sqrt(2) part of answer. Just double check pls.
cos(x+pi/4) = +- 1/sqrt(2)
x+pi/4 =pi/4
x=0
x+pi/4 =3pi/4
x=pi/2 --- first positive value of x
x+pi/4 =5pi/4
x=pi ------ 2nd solution
x=3pi/2 -- 3rd solution not first !!
---------------------------------------------------------------------------------
double check question 4 solutions as well pls.
"In a large forest, one quarter of all trees are infected with cinnamon fungus."
p=1/4, not 1/3 
-------------------------------------------------------------------------------
q7 solution needs to revised.  as well

[Joseph41]

Hey!
Can you double check the methods exam 1 question 2 b solution.
I am getting pi/2.
I think solution missed out, the -1/sqrt(2) part of answer. Just double check pls.
cos(x+pi/4) = +- 1/sqrt(2)
x+pi/4 =pi/4
x=0
x+pi/4 =3pi/4
x=pi/2 --- first positive value of x
x+pi/4 =5pi/4
x=pi ------ 2nd solution
x=3pi/2 -- 3rd solution not first !!
---------------------------------------------------------------------------------
double check question 4 solutions as well pls.
"In a large forest, one quarter of all trees are infected with cinnamon fungus."
p=1/4, not 1/3 
-------------------------------------------------------------------------------
q7 solution needs to revised.  as well

Thanks so much - have passed this on for review.

[redpanda83]

Thanks so much - have passed this on for review.

There were a couple of errors in exam 2 solutions as well.
q 20 MCQ - Answer B- working out is correct but just a misprint of option i think
short answers - 3.e.  Prob Eliza will be late to school is 7/23 (calculated in 3c) binomialCdf (5,7/23,3,5)=0.169
4.g. shouldnt (a,2a) but substituted when calculating time for straight road, (a,3a) was substituted in solutions. If i am correct, it would mean a different equation for total time (T) would be true and solution for 4h would also require revision.
Thanks

[dilanka.sam1]

For exam 1, question 3 a) would you still get the mark if you drew the y- asymptote at y=1 ?

[Matthew_Whelan]

Similar to what’s been said, I’ve found a few errors (I think) that are in the solutions.

Because p= 1/4, not 1/3,
4 b) should be 37/64 I believe
and c) should be 1/12, also n =27 not 72, this affects d) too.
Also q 3a) says the endpoint is (-4,5/4) when it should be (-4, 9/8).
7a) should be 3/4 I think, not 11/15, which affects the subsequent question.
9 b) should be (1/2pi) x (2pi - cos(k)) which affects 9 c)

Nice questions though 

[Massimooo123]

Just took the exam, compiling all the mistakes and providing the correct (I think) solutions here:

2a) The answer was correct, but in the notes it says 2sec²(x+pi/4) is an equivalent expression, when it's actually (1/2)sec²(x+pi/4).
2b) As people have said, you have to take the positive AND negative root, which gives you a smaller positive value of x.
sin(x) = +-1/sqrt(2)  -->  x+pi/4 = 3pi/4  -->  x=pi/2

4a) I'm just going to give the answers for question 4 since all errors in the question stem from saying E(P) = 1/3.
E(P) = 1/4
4b) Pr(P≥1/3) = Pr(P=0) = 1-(3/4)^3 = 1-27/64 = 37/64
4c) sqrt( (1/4)(3/4) / (27) ) = sqrt( 3/(16*27) ) = (1/4)*sqrt( 3/27 ) = (1/4)*sqrt(1/9) = 1/(4*3) = 1/12
4d) E(P) = 1/4, SD(P) = 1/12. Pr( P > 1/3 ) = Pr( Z > ((1/3)-(1/4))/(1/12) ) = Pr( Z > (1/12)/(1/12) ) = Pr ( Z > 1 ). By 68-95-99.7 rule, the tail one standard deviation away from the mean has area (1-0.68)/2 = 0.16. Therefore Pr ( P > 1/3 ) = 0.16.

7a) (1-1/2)(1-2/5)(1/6) = 1/20, not 1/30. Substituting this value correctly gives 3/4.
7b) Incorrect usage of the conditional probability formula, should be Pr(Not rejected | Not rejected first), not the other way around.
Pr(Not rejected | Not rejected first) = Pr(Not rejected ∩ Not rejected first) / Pr(Not rejected first).
Pr(Not rejected ∩ Not rejected first) = Pr(Not rejected) = 1 - 3/4 = 1/4, as ALL non-rejected toasters are not rejected first.
Pr(Not rejected first) = 1/2
Therefore Pr(Not rejected | Not rejected first) = (1/4) / (1/2) = 1/2

9b) As Matthew said, the answer you should be getting to is (1/2pi)*(2pi-cos(k)). The incorrect part was the factoring in the very final step.
c) Maximum value of the mean occurs when cos(k) = -1, hence k = pi. Subbing into the correct equation;
(1/2pi)*(2pi-cos(pi))=(1/2pi)*(2pi+1) = 1 + 1/2pi.

I know this isn't very useful to compile now since I doubt anyone's going to be doing any early morning practice exams, but maybe next year Also, I tallied up the answer key's mark and it got 27/40 = 67.5% lol. Don't mean to be that guy but you should probably have a couple more people review this next time haha.

[AlphaZero]

I'm just going to put this here for the record. I've already reported all the issues I found with the exam 1. If something has already been reported, no point reporting it again. Putting that aside, here was my message from a couple of weeks ago. I think I got everything, but if anyone has anything to add, go right ahead.

When I get the chance, I'll review the exam 2 properly, but just from a quick read, I did find a few errors in Section A.
Exam 1

Question 2
The function \(f\) cannot possibly have the domain \(\mathbb{R}\). For example, \(f(\pi/4)\) is undefined. In Q2b, the solution neglects the possibility \[\cos\left(x+\frac{\pi}{4}\right)=\frac{-1}{\sqrt{2}}.\] The correct answer is \[x_\text{min}=\frac{\pi}{2}.\]
Question 3
In the solution to Q3a, the notes section writes that the endpoint is \((-4,\ 5/4)\), which it isn't. The correct endpoint is \((-4,\ 9/8)\).

Question 4
The question gives \(p=1/4\) yet the solutions take \(p=1/3\) throughout the entire question. So, all the solutions are wrong for this question. Putting this to the side, the notation for the normal distribution is misused. For a normally distributed random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), the correct notation is \[X\sim\mathcal{N}(\mu,\ \sigma^2).\] Notice that the second parameter gives the variance, not the standard deviation. For Q4d, the correct solution yields \(\Pr(\hat{P}\geq 1/3)=\Pr(Z\geq 1)\), which is one of the only ways to give an exact answer without specifying an integral.

Question 5
Infinity is not a 'value' and I think it's a bad idea to treat it as such. In Q5b.i, just ask for the largest subset of \(\mathbb{R}_{\geq 0}\) for which \(h\) is defined. The wording in Q5b.ii can then be fixed by asking for the implied range of \(h\).

Question 6
Nothing wrong with the question other than the solution to Q6a is a bit overkill. The events \(B\) and \(A'\) are also independent, so \[\Pr(B\mid A')=\Pr(B)=p.\]
Question 7
Solution of Q7a has an arithmetic error. Although the method is correct, it's a bit overkill. Where \(R\) denotes 'rejected', we have \[\Pr(R)=1-\Pr(R')=1-\frac12\times \frac35\times \frac56=\frac34.\] If we denote \(R_i\) to mean 'rejected by test \(i\)', then the question asks for \(\Pr(R'\mid R_1')\), not \(\Pr(R_1'\mid R')\). Even if the question did ask for the latter, the solution is still incorrect since a toaster necessarily needs to pass the first test in order to not be rejected. \[\Pr(R'\mid R_1')=\frac{\Pr(R'\cap R_1')}{\Pr(R_1')}=\frac{\Pr(R')}{\Pr(R_1')}=\frac{1/4}{1/2}=\frac12.\]
Question 8
Solution for Q8b makes an algebraic slip despite giving the correct answer. It should have \(-54\), not \(+54\).

Question 9
The wording of Q9b is problematic. The mean value of a function is not the same as the mean of a random variable. The former means average value: \[\overline{f}=\frac{1}{1-0}\int_0^1f(x)\,\text{d}x,\] while the latter means \[\mu=\int_0^1x\,f(x)\,\text{d}x.\] So, one needs to define a random variable (\(X\), say) somewhere in the question. The solution for Q9b is also problematic since the second last line lacks the correct bracketing. The \(x\) needs to multiply the whole function, not just the \(\sin()\) term. That is, \[\mu=\int_0^1 x\big(\!\sin(2\pi x+k)+1\big)\,\text{d}x=\int_0^1 x\sin(2\pi x+k)\,\text{d}x+\int_0^1x\,\text{d}x.\] Solution to Q9c is also really overkill. Maximum \(\mu\) occurs when \(\cos(k)=-1\). Ie. \[\mu_\text{max}=\frac{\pi+1}{2\pi},\quad k=\pi.\]No calculus is required, so the marking scheme needs to be revised.

[yvettek1777]

Are these exams adapted to the changed VCE curriculum?

[ATAR Notes Official]

Are these exams adapted to the changed VCE curriculum?

They were published in 2019, but have not been updated for the amended 2020 study designs.

[Samueliscool223]

amazing exam questions. but goddamn there are a lot of mistakes especially in the first one.

[biology1234]

for functions, could someone explain placing restriction on inverse function on x values. thank you