Login

Welcome, Guest. Please login or register.

March 29, 2024, 11:05:02 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164440 times)  Share 

0 Members and 7 Guests are viewing this topic.

undefined

  • Forum Obsessive
  • ***
  • Posts: 323
  • Respect: +19
Re: VCE Specialist 3/4 Question Thread!
« Reply #9390 on: January 24, 2019, 05:15:18 pm »
0
2018 Methods
2019 English | Chemistry | Economics | Specialist  | Japanese SL

2020 B.Eng/Comm
2021 - 2025 B.CS/Comm Diplang in Japanese @ Monash

Tatlidil

  • Forum Regular
  • **
  • Posts: 97
  • Math is Life
  • Respect: +1
Re: VCE Specialist 3/4 Question Thread!
« Reply #9391 on: February 13, 2019, 08:22:44 am »
0
Just a general question, what is the best way to improve in trig? Just lots of questions? Mainly prove questions

DBA-144

  • MOTM: APR 19
  • Forum Obsessive
  • ***
  • Posts: 211
  • Respect: +35
Re: VCE Specialist 3/4 Question Thread!
« Reply #9392 on: February 13, 2019, 01:26:10 pm »
0
Just a general question, what is the best way to improve in trig? Just lots of questions? Mainly prove questions

Give an example of a question and I might be able to provide some more specific help?
PM me for Methods (raw 46) and Chemistry (raw 48) resources (notes, practice SACs, etc.)

I also offer tutoring for these subjects, units 1-4 :)

jollyboat

  • Trailblazer
  • *
  • Posts: 26
  • Respect: 0
Re: VCE Specialist 3/4 Question Thread!
« Reply #9393 on: February 16, 2019, 01:55:03 pm »
0
Hi,

I was wondering if trig proof questions are in the spec 3/4 course? There was one in my textbook, but only one, it wasn't gone over any more than that. We did them last year in spec 1/2, so I roughly know how to do them, I was just wondering if I should be practicing how to do them since I haven't done any since last year and have got pretty rusty?

I also had another question about trig, does anyone know if VCAA wants us to label asymptotes with their equations (eg. x=pi/4) when drawing tan, cot, sec and cosec graphs?

Thanks :)
« Last Edit: February 16, 2019, 02:00:13 pm by jollyboat »

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: VCE Specialist 3/4 Question Thread!
« Reply #9394 on: February 16, 2019, 02:26:21 pm »
+1
Hi,

I was wondering if trig proof questions are in the spec 3/4 course? There was one in my textbook, but only one, it wasn't gone over any more than that. We did them last year in spec 1/2, so I roughly know how to do them, I was just wondering if I should be practicing how to do them since I haven't done any since last year and have got pretty rusty?

I also had another question about trig, does anyone know if VCAA wants us to label asymptotes with their equations (eg. x=pi/4) when drawing tan, cot, sec and cosec graphs?

Thanks :)

Q1) Yes, proofs involving trigonometric functions are in the course, but emphasis isn't really placed on them. If anywhere, they are usually in exam 1 and are around 3 marks. However, most problems don't plain ask you to prove that two expressions are equal. VCAA generally likes to present students with questions where using trigonometric identities and manipulating expressions is part of a broader problem. Some great examples of these questions are Q3 and Q9 from the 2018 NHT Exam 1. I think it's beneficial to practice proofs to become more familiar and comfortable with using trigonometric functions.

Q2) VCAA will be very clear in their instructions with how they want you to answer it. It general, they use the phrasing "Show any asymptotes and label them with their equations". That is, show them with a dashed line and label it with its equation (eg:  \(y=0,\ \ x=\pi/4\)).
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne


jollyboat

  • Trailblazer
  • *
  • Posts: 26
  • Respect: 0
Re: VCE Specialist 3/4 Question Thread!
« Reply #9395 on: February 16, 2019, 02:49:29 pm »
0
Thanks so much!

Ansaki

  • Adventurer
  • *
  • Posts: 20
  • With hardship come ease.
  • Respect: 0
Re: VCE Specialist 3/4 Question Thread!
« Reply #9396 on: February 21, 2019, 05:09:54 pm »
0
when i put solve(0=|x-4|+1,x) into ti nspire cas it comes up with false. i have solved it by hand and got x intercepts, im pretty confused, is their a property im missing out on? where am i going wrong? by the way its a question from exercise 1E question 3 of the cambridge textbook for spesh.
thank you

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: VCE Specialist 3/4 Question Thread!
« Reply #9397 on: February 21, 2019, 05:50:32 pm »
+1
when i put solve(0=|x-4|+1,x) into ti nspire cas it comes up with false. i have solved it by hand and got x intercepts, im pretty confused, is their a property im missing out on? where am i going wrong? by the way its a question from exercise 1E question 3 of the cambridge textbook for spesh.
thank you

Hey there.

Your CAS calculator is correct. There are no solutions. A simple graph shows that \(y=|x-4|+1\) will never touch the \(x\)-axis.



This is because  \(|x-4|\geq0\)  for all  \(x\in\mathbb{R}\),  and so  \(|x-4|=-1<0\)  will have no solutions.

A common mistake that students make is to drop the absolute value brackets and consider the positive and negative cases without any justification.

Recall that the definition of the absolute value function is  \[|x|=\begin{cases}x, & x\geq 0\\ -x, & x<0\end{cases},\]and so we have \[|x-4|+1=\begin{cases}x-3, & x\geq 4\\ 5-x, & x<4\end{cases}.\] Let's consider the first case. Are there any solutions when \(x\geq 4\)?  Well, \(x-3=0\implies x=3\),  but  \(3<4\), so we reject this solution.

Similarly, for the case \(x<4\),  we get  \(x=5\).  But,  \(5>4\),  so we must reject this too.

Thus, there are no solutions for \(x\).
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne


DBA-144

  • MOTM: APR 19
  • Forum Obsessive
  • ***
  • Posts: 211
  • Respect: +35
Re: VCE Specialist 3/4 Question Thread!
« Reply #9398 on: February 21, 2019, 05:53:48 pm »
0
when i put solve(0=|x-4|+1,x) into ti nspire cas it comes up with false. i have solved it by hand and got x intercepts, im pretty confused, is their a property im missing out on? where am i going wrong? by the way its a question from exercise 1E question 3 of the cambridge textbook for spesh.
thank you


wait so how did you solve it by hand? My understanding is that the modulus function is |sqrt (x^2)|=x. That is, that we only take the positive square root. So, since you can't take the square root of -1, at least not in the real numbers, then how did you solve it...? Sorry if I'm missing something but I think you are struggling with the notion that we only take the positive square root and thus we cannot have -1 and cannot solve the equation? At least that is what I think of it.

Beaten by AlphaZero and I think I was right but will post this nonetheless.
PM me for Methods (raw 46) and Chemistry (raw 48) resources (notes, practice SACs, etc.)

I also offer tutoring for these subjects, units 1-4 :)

Ansaki

  • Adventurer
  • *
  • Posts: 20
  • With hardship come ease.
  • Respect: 0
Re: VCE Specialist 3/4 Question Thread!
« Reply #9399 on: February 21, 2019, 06:26:47 pm »
0
AlphaZero and DBA-144 thank you!

S_R_K

  • MOTM: Feb '21
  • Forum Obsessive
  • ***
  • Posts: 487
  • Respect: +58
Re: VCE Specialist 3/4 Question Thread!
« Reply #9400 on: February 22, 2019, 02:31:42 pm »
0
wait so how did you solve it by hand? My understanding is that the modulus function is |sqrt (x^2)|=x. That is, that we only take the positive square root. So, since you can't take the square root of -1, at least not in the real numbers, then how did you solve it...? Sorry if I'm missing something but I think you are struggling with the notion that we only take the positive square root and thus we cannot have -1 and cannot solve the equation? At least that is what I think of it.

Yes, you are right, but just to clarify, the issue is not that with this equation you'll need to evaluate sqrt(–1) over the real numbers. Rather, the problem is that the equation you end up with is sqrt((x-4)^2) = –1, and the LHS is a positive number whereas the RHS is negative, hence no solution. I think that is what you meant, but just clarifying for other people who may be following this discussion.

Ansaki

  • Adventurer
  • *
  • Posts: 20
  • With hardship come ease.
  • Respect: 0
Re: VCE Specialist 3/4 Question Thread!
« Reply #9401 on: February 22, 2019, 07:33:21 pm »
0
hey guys! I've got another question!
how do i find the x intercepts, by hand, if f(x)=|x|^2-4|x|
i know you make f(x)=0 and then solve but i just dont know the process.

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: VCE Specialist 3/4 Question Thread!
« Reply #9402 on: February 22, 2019, 09:20:38 pm »
0
hey guys! I've got another question!
how do i find the x intercepts, by hand, if f(x)=|x|^2-4|x|
i know you make f(x)=0 and then solve but i just dont know the process.

Edit: I way over-complicated the problem. There's clearly a much better method... but I'll leave this here to be laughed at lol

\[f(x)=x^2-4|x|\] (Note is that  \(|x|^2=x^2\). The absolute value brackets are redundant).

To proceed with solving equations involving the absolute value function, we generally want to break it up into cases. In some cases like this one, you can actually make a few clever observations to arrive at the answer quicker, but I'll show that process at the end.

Recall that the definition of the absolute value function is: \[|x|=\begin{cases}x, & x\geq 0\\
-x, & x<0\end{cases}.\] In our case, we would like to break up the problem into the cases  \(x\geq0\)  and  \(x<0\).

Case 1: \(x\geq 0\)
\[x^2-4x=0\implies x(x-4)=0\implies \ \ x=0,\ 4\] Case 2: \(x<0\) \[x^2-4(-x)=0\implies x(x+4)=0\implies \ \ x=-4\]
Putting our results together, we see that the solutions to  \(f(x)=0\)  are  \(x=-4,\ 0,\ 4\).

Some clever observations we can make to make our lives easier

Note that in this question,  \(f(x)\)  is an even function  (that is,  \(f(-x)=f(x)\) ).

So, we must have  \(f(a)=0\iff f(-a)=0\).

This means, we could've just considered Case 1, and realise that since  \(x=4\)  is a solution,  then so must  \(x=-4\).
« Last Edit: February 22, 2019, 10:01:42 pm by AlphaZero »
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne


RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: VCE Specialist 3/4 Question Thread!
« Reply #9403 on: February 22, 2019, 09:43:57 pm »
+3
Can't you just do this?
\begin{align*} |x|^2 - 4|x| &= 0\\ |x| \big( |x|-4 \big)&=0\\ |x|&=0, 4 \end{align*}
\[ |x| = 0 \implies x=0\\ |x|=4\implies x=\pm 4 \]

schoolstudent115

  • Trendsetter
  • **
  • Posts: 117
  • Respect: +6
Re: VCE Specialist 3/4 Question Thread!
« Reply #9404 on: February 23, 2019, 09:21:31 pm »
0
Why is the body in the first diagram accelerating upwards, but in the second diagram, downwards? Shouldn’t it be the other way around as the friction is moving with the pulley in the second diagram?
https://m.imgur.com/CAAH3du
2021: ATAR: 99.95
2022-2024: University of Melbourne, BSci (Major in Mathematics and Statistics)