Login

Welcome, Guest. Please login or register.

March 29, 2024, 12:14:00 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164211 times)  Share 

0 Members and 3 Guests are viewing this topic.

jazzycab

  • Victorian
  • Trendsetter
  • **
  • Posts: 158
  • Respect: +19
  • School: WHS
Re: Specialist 3/4 Question Thread!
« Reply #9150 on: January 06, 2018, 09:37:00 pm »
+2
Further to keltingmeith and brightsky's explanations, you can easily define these cofunction properties by drawing an arbitrary right-triangle.
Consider a right-triangle with angles
The third angle (due to the angle sum of a triangle) must be

ezferns

  • Forum Regular
  • **
  • Posts: 78
  • Respect: 0
Re: Specialist 3/4 Question Thread!
« Reply #9151 on: January 10, 2018, 09:23:15 am »
+1
Hey everyone, I'm having trouble with a question from the Cambridge Textbook 6G questions 16e and 17f.

16e
You're given the function x/sqrt(x-2) and you have to find the equation of the non-vertical asymptote.
I guess you have to know that as x approaches infinity, the -2 becomes irrelevant so the function approaches sqrt(x)
Not sure if this is the right way to go about it.

But then I can't apply this to the next question

17f
Function: (x^2 + x + 7)/sqrt(2x + 1)
and you're supposed to find the non-vertical asymptote. I have absolutely no idea.
The answer is y=sqrt(x)

Help :(

kalopsia

  • Victorian
  • Trailblazer
  • *
  • Posts: 26
  • Respect: +3
Re: Specialist 3/4 Question Thread!
« Reply #9152 on: January 10, 2018, 09:44:04 am »
0
Hey everyone, I'm having trouble with a question from the Cambridge Textbook 6G questions 16e and 17f.

16e
You're given the function x/sqrt(x-2) and you have to find the equation of the non-vertical asymptote.
I guess you have to know that as x approaches infinity, the -2 becomes irrelevant so the function approaches sqrt(x)
Not sure if this is the right way to go about it.

But then I can't apply this to the next question

17f
Function: (x^2 + x + 7)/sqrt(2x + 1)
and you're supposed to find the non-vertical asymptote. I have absolutely no idea.
The answer is y=sqrt(x)

Help :(


I don't think there is a non-vertical asymptote for the second equation.
2016: Methods [45]
2017: English [45] | Chemistry [41] |  Physics  [50] | Specialist Maths [43] | UMEP Maths [4.5] | UMAT [97th]
2018: Bachelor of Medical Science/Doctor of Medicine @ Monash
ATAR: 99.75

PM for tutoring

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Specialist 3/4 Question Thread!
« Reply #9153 on: January 10, 2018, 10:31:52 am »
+1
I don't think there is a non-vertical asymptote for the second equation.
17f
Function: (x^2 + x + 7)/sqrt(2x + 1)
and you're supposed to find the non-vertical asymptote. I have absolutely no idea.
The answer is y=sqrt(x)

Help :(




At first glance, I could immediately tell that \( \to \sqrt{x} \) is a mistake. This is because the leading power on \(x\) in the numerator is \(2\), whereas in the denominator it is \(\frac12\). Hence, as \(x\) gets really large, the behaviour of \(y\) must tend to something, whose highest power on \(x\) is \(2 - \frac12\) i.e. \(\frac32\).
« Last Edit: January 10, 2018, 10:33:33 am by RuiAce »

TheAspiringDoc

  • Guest
Re: Specialist 3/4 Question Thread!
« Reply #9154 on: January 10, 2018, 10:40:25 am »
0
And for the first one, I wouldn't really call it an asymptote, it's just a minimum point at (4,4/sqrt(2))
.. I think

VanillaRice

  • Forum Leader
  • ****
  • Posts: 657
  • Respect: +278
Re: Specialist 3/4 Question Thread!
« Reply #9155 on: January 10, 2018, 10:44:10 am »
+1
Hey everyone, I'm having trouble with a question from the Cambridge Textbook 6G questions 16e and 17f.

16e
You're given the function x/sqrt(x-2) and you have to find the equation of the non-vertical asymptote.
I guess you have to know that as x approaches infinity, the -2 becomes irrelevant so the function approaches sqrt(x)
Not sure if this is the right way to go about it.

But then I can't apply this to the next question

17f
Function: (x^2 + x + 7)/sqrt(2x + 1)
and you're supposed to find the non-vertical asymptote. I have absolutely no idea.
The answer is y=sqrt(x)

Help :(


16e) I don't think this is a question that would come up in a VCAA exam (it's not a rational function), but I believe this is the method to find the asymptote:

As x approaches infinity, the 2/x in the denominator tends to zero, and the numerator dominates.

17f) I checked the answers, and it looks like the author just copied the answer from the last question (but forgot to change it to something?), and the worked solutions even go on to say there is no other asymptote. If you sketch the function, you can see that it doesn't approach y=sqrt(x)
VCE 2015-16
2017-20: BSc (Stats)/BBiomedSc [Monash]

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Specialist 3/4 Question Thread!
« Reply #9156 on: January 10, 2018, 10:46:15 am »
+1
And for the first one, I wouldn't really call it an asymptote, it's just a minimum point at (4,4/sqrt(2))
.. I think
The minimum point doesn't really influence the existence of an asymptote. Functions like \( f(x)=x+\frac{1}{x} \) have a minimum but also a non-vertical asymptote.

A TART

  • Forum Regular
  • **
  • Posts: 86
  • "Dont ever look back"~Spesh Teacher
  • Respect: +32
Re: Specialist 3/4 Question Thread!
« Reply #9157 on: January 10, 2018, 10:53:47 pm »
0
I've been scratching my head over this question for a while now. I've been able to produce an answer but I'm not really satisfied with it.

For the question, I tried working backwards:

a+c=b+d

OA+OC=OB+OD

OA-OD=OB-OC

DO+OA=CO+OB

DA=CB

Now that's not true for most parallelograms, however, if the parallelogram is a special kind with 90-degree angles (a square), then it would satisfy that.

(And I really shouldn't be starting with a+c=b+d, since it's asking me to prove it)

So idk. (The answer is probably just there and I can't see it)

EDIT: Forgot to add "n't" to "shouldn't"



« Last Edit: January 10, 2018, 11:04:10 pm by A TART »
2018-English, Chinese SL, Chemistry, Physics, Maths Methods and Specialist

2019- Hug trees and hopefully do something related to Environmental science @ UniMelb

exit

  • Forum Obsessive
  • ***
  • Posts: 433
  • COALESCE
  • Respect: +38
Re: Specialist 3/4 Question Thread!
« Reply #9158 on: January 10, 2018, 11:07:44 pm »
+1
I've been scratching my head over this question for a while now. I've been able to produce an answer but I'm not really satisfied with it.

For the question, I tried working backwards:

a+c=b+d

OA+OC=OB+OD

OA-OD=OB-OC

DO+OA=CO+OB

DA=CB





Now that's not true for most parallelograms, however, if the parallelogram is a special kind with 90-degree angles (a square), then it would satisfy that.

(And I really should be starting with a+c=b+d, since it's asking me to prove it)

So idk. (The answer is probably just there and I can't see it)





The key to these sort of proofs is to not overcomplicate them. To prove that something is a parallelogram, two sides must equal the same vector essentially. (forcing the other two vectors to also be the same)

 Thus, if ABCD is a parallelogram AB=DC. (draw this out. always draw a vector question out if possible - regardless of simplicity)

Now let's solve.

1. AB=DC
2. b-a=c-d (convert to vectors)
3. [rearrange] b+d=a+c

As you can see, we already got our answer. No need to confuse yourself.

This is how I would have done it - hope it helped!

DA=CB is true for all parallelograms. Not only when it is a square. If you still don't understand, help me clarify this for you.
« Last Edit: January 10, 2018, 11:11:07 pm by exit »
VCE [ATAR: 99.25]: Physics 1/2, English 1/2, EngLang,Methods, Spesh, Accounting, Chem, German

2018-2021: Bachelor Of Commerce @ University of Melbourne
VCE English Language: A+ Short Answer Guide[pm for extra guidance!]

A TART

  • Forum Regular
  • **
  • Posts: 86
  • "Dont ever look back"~Spesh Teacher
  • Respect: +32
Re: Specialist 3/4 Question Thread!
« Reply #9159 on: January 10, 2018, 11:11:51 pm »
0
The key to these sort of proofs is to not overcomplicate them. To prove that something is a parallelogram, two sides must equal the same vector essentially. (forcing the other two vectors to also be the same)

 Thus, if ABCD is a parallelogram AB=DC. (draw this out. always draw a vector question out if possible - regardless of simplicity)

Now let's solve.

1. AB=DC
2. b-a=c-d (convert to vectors)
3. [rearrange] b+d=a+c

As you can see, we already got our answer. No need to confuse yourself.

This is how I would have done it - hope it helped!


I know why now. I didn't draw my parallelogram correctly XD

Thanks
« Last Edit: January 10, 2018, 11:16:40 pm by A TART »
2018-English, Chinese SL, Chemistry, Physics, Maths Methods and Specialist

2019- Hug trees and hopefully do something related to Environmental science @ UniMelb

ezferns

  • Forum Regular
  • **
  • Posts: 78
  • Respect: 0
Re: Specialist 3/4 Question Thread!
« Reply #9160 on: January 14, 2018, 08:07:25 pm »
0
Hey everyone, can I please have some help making a program for Euler's Method? I know how to make a spreadsheet but in the Cambridge textbook Chapter 9H it asks for 100+ iterations. I saw in the worked solutions it says 'use the program as outlined in the textbook' however I could only find the spreadsheet in the textbook.

Sine

  • Werewolf
  • National Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 5135
  • Respect: +2103
Re: Specialist 3/4 Question Thread!
« Reply #9161 on: January 14, 2018, 08:11:10 pm »
0
Hey everyone, can I please have some help making a program for Euler's Method? I know how to make a spreadsheet but in the Cambridge textbook Chapter 9H it asks for 100+ iterations. I saw in the worked solutions it says 'use the program as outlined in the textbook' however I could only find the spreadsheet in the textbook.
check the solutions for the sample questions iirc that is how I ended up doing some of that. For the exam you definitely don't need to know how to create a program just understand eulers methods and doing a couple iterations by hand.+ learn the CAS commands as a shortcut

noregret

  • Forum Regular
  • **
  • Posts: 76
  • Respect: 0
  • School: somewhere
  • School Grad Year: 2018
Re: Specialist 3/4 Question Thread!
« Reply #9162 on: January 24, 2018, 12:04:18 pm »
0
Hello, can anyone help with this question please by using integration by parts Method?

7x times (6x^2+2)^1\4

Sine

  • Werewolf
  • National Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 5135
  • Respect: +2103
Re: Specialist 3/4 Question Thread!
« Reply #9163 on: January 24, 2018, 12:07:56 pm »
+3
Hello, can anyone help with this question please by using integration by parts Method?

7x times (6x^2+2)^1\4
integration by parts isn't in the spesh course. Try using a u substitution

noregret

  • Forum Regular
  • **
  • Posts: 76
  • Respect: 0
  • School: somewhere
  • School Grad Year: 2018
Re: Specialist 3/4 Question Thread!
« Reply #9164 on: January 24, 2018, 01:34:22 pm »
0
integration by parts isn't in the spesh course. Try using a u substitution
thanks it worked.