VCE Methods Question Thread!

[MightyBeh]

any help on this will be appreciated!

Solve the following equation for x, where c is a positive constant:
x^2 - 3cx = 10c^2

i have got it to x = (10c^2)/(x-3c)

but cant extract the the x to the left hand side.........is there a way of doing this??or is this the closest its going to be??

I would use the quadratic formula for this question. Attached my working out. AFAIK there's no easy way to do it the way you've started.

[Perryman]

thanks heaps....sorry i seem to always forget to use tht formula
thanks for the reminder and the help!!
 

[A TART]

This was on a series of revision questions from past exams on Linear Algebra:

The angles between the lines y=-x-2 and y=2x+8 is closest to:

A.45°
B. 52.4°
C.71.6°
D.108.4°
E. 135°

The equations aren't perpendicular so there must be two different angles between them. I got 71.6 and 108.4 and they're both an option. However, I can only have one answer.

I'm not so clear with the "angles between the two lines" part. What is it asking for??? Any help or just discussion will be greatly appreciated! 

EDIT: I can only select one answer right? The paper doesn't have any instructuions.....

[Perryman]

The attached question seems to hav no answer
is tht correct or what am i doing wrong..

as far as I hav worked it out, three of them dont intersect and the other two cross over..

[Perryman]

This was on a series of revision questions from past exams on Linear Algebra:

The angles between the lines y=-x-2 and y=2x+8 is closest to:

A.45°
B. 52.4°
C.71.6°
D.108.4°
E. 135°

The equations aren't perpendicular so there must be two different angles between them. I got 71.6 and 108.4 and they're both an option. However, I can only have one answer.

I'm not so clear with the "angles between the two lines" part. What is it asking for??? Any help or just discussion will be greatly appreciated! 

EDIT: I can only select one answer right? The paper doesn't have any instructuions.....

not exactly sure how to go about this question but if u look at the formula.....π−(tan−1(m1)−tan−1(m2))
u end up with 71.6.....

i understand tht u arrive at 108.4...but the formula has π - at the start... soo presumably u would go with 71.6...
but someone with a bit more knowledge can confirm tht!

[Sine]

The attached question seems to hav no answer
is tht correct or what am i doing wrong..

as far as I hav worked it out, three of them dont intersect and the other two cross over..

yeah I got no answer as well

you could take it a step further and have a go at finding the actual values of k required 

This was on a series of revision questions from past exams on Linear Algebra:

The angles between the lines y=-x-2 and y=2x+8 is closest to:

A.45°
B. 52.4°
C.71.6°
D.108.4°
E. 135°

The equations aren't perpendicular so there must be two different angles between them. I got 71.6 and 108.4 and they're both an option. However, I can only have one answer.

I'm not so clear with the "angles between the two lines" part. What is it asking for??? Any help or just discussion will be greatly appreciated! 

EDIT: I can only select one answer right? The paper doesn't have any instructuions.....

Yes 71.6 is the acute angle and 108.4 is the obtuse angle.

I would probably just go with the acute angle tbh but both are technically correct.

[deStudent]

I'm not sure if I'm missing something obvious but I can't seem to find the intersection between y = -x + 2 and y=1 + ln(x). Apparently x = 1...

Thanks

[Guideme]

Sine, clearly it's good practice to get used to your calculator when doing these sorts of questions, but I think it's extremely important to first have a deep understanding of what the question is asking rather than simply subbing in numbers until you get the correct answer.

Guideme, the fact that you needed help with this tells me that you weren't sure how to do this mathematically (which is obviously completely fine and I'm more than happy to help!). However, before diving into using your calculator for shortcuts, for this question or any others in future, it's important that you know how to solve it mathematically. If you only have a 'rote-learned' surface level of understanding, you'll falter when questions in future are slighly different. That's why I wrote out the proper solutions for you.

Once you know how to do these questions properly with confidence and accuracy, that's when you should start perfecting your calculator skills to save time in exams and SACs

Thank you very much! I like to do every question by hand to help consolidate my foundation in order to do well in tech-free sacs when I don't have a calculator. Furthermore, I believe that when I do more questions by hand I am able to think more critically and effectively to reduce the amount of time answering each question in tech-free sacs.  I would obviously use a calculator to my advantage in tech-active sacs.

[Sine]

Sine, clearly it's good practice to get used to your calculator when doing these sorts of questions, but I think it's extremely important to first have a deep understanding of what the question is asking rather than simply subbing in numbers until you get the correct answer.

Guideme, the fact that you needed help with this tells me that you weren't sure how to do this mathematically (which is obviously completely fine and I'm more than happy to help!). However, before diving into using your calculator for shortcuts, for this question or any others in future, it's important that you know how to solve it mathematically. If you only have a 'rote-learned' surface level of understanding, you'll falter when questions in future are slighly different. That's why I wrote out the proper solutions for you.

Once you know how to do these questions properly with confidence and accuracy, that's when you should start perfecting your calculator skills to save time in exams and SACs

ye Ik. That's why I think stuff should be taught by first principles and not just memorising how to tackle 1 single type of question. (although this helps for the harder differentiator questions because VCAA repeat them)

I make all my students complete multiple choices without CAS (unless CAS is 100% required) and then let them use checking time if they have any in which they can use CAS.

My previous post was merely to show that just because you don't know how to do a question doesn't mean you can't get the answer through various methods.

Pretty sure you posted a full solution before me yeah? I don't see the benefit of two people showing the exact same methods. This sort of thinking doesn't offer extra exposure. You can always gain more benefit seeing different ways to solve stuff and also just to keep in the back of your mind.

Come exam time these shortcuts is what allows you to finish early and have enough time to check answers.

thanks 

P.S Great work on the forums lately 

[evandowsett]

Thank you very much! I like to do every question by hand to help consolidate my foundation in order to do well in tech-free sacs when I don't have a calculator. Furthermore, I believe that when I do more questions by hand I am able to think more critically and effectively to reduce the amount of time answering each question in tech-free sacs.  I would obviously use a calculator to my advantage in tech-active sacs.

Sounds good Guideme, I think that's a great approach!

ye Ik. That's why I think stuff should be taught by first principles and not just memorising how to tackle 1 single type of question. (although this helps for the harder differentiator questions because VCAA repeat them)

I make all my students complete multiple choices without CAS (unless CAS is 100% required) and then let them use checking time if they have any in which they can use CAS.

My previous post was merely to show that just because you don't know how to do a question doesn't mean you can't get the answer through various methods.

Pretty sure you posted a full solution before me yeah? I don't see the benefit of two people showing the exact same methods. This sort of thinking doesn't offer extra exposure. You can always gain more benefit seeing different ways to solve stuff and also just to keep in the back of your mind.

Come exam time these shortcuts is what allows you to finish early and have enough time to check answers.

thanks 

P.S Great work on the forums lately 

Yeah for sure Sine, I agree with you I definitely think that's a great way to do it as a tutor and I understand why you wrote the answers you did in terms of using CAS - it's good to get extra exposure especially with methods of using your calculator! I just wanted to make it clear to Guideme that he should try and fundamentally understand the questions before jumping into CAS shortcuts

Thank you, just trying to help out!

[evandowsett]

I'm not sure if I'm missing something obvious but I can't seem to find the intersection between y = -x + 2 and y=1 + ln(x). Apparently x = 1...

Thanks

Hey deStudent, here's the worked solution. I'm pretty sure that there is no arithmetical way to prove the last step, but it's logical that if xe^x = e, then x must equal 1. If x was anything other than 1, the statement wouldn't be true because of the index on e. Make sense?

[deStudent]

Hey deStudent, here's the worked solution. I'm pretty sure that there is no arithmetical way to prove the last step, but it's logical that if xe^x = e, then x must equal 1. If x was anything other than 1, the statement wouldn't be true because of the index on e. Make sense?

Thanks. I think I sometimes fall in the trap of trying to fully work things out 'algebraically' without thinking about them logically..

---
Edit:
Some small problems here:
http://m.imgur.com/a/2eXOA

Q10ii) Is the domain in the blue wrong? Isn't the domain of gof = domain of f?

Iv) I'm having trouble finding the left half of the maximal domain?

Q8c) I'm having trouble understanding what the reverse operation is for a "reflection in the x-axis". My brain thinks the reverse of this is no reflection at all, but it gets me the wrong answer.

Thanks

[lzxnl]

This was on a series of revision questions from past exams on Linear Algebra:

The angles between the lines y=-x-2 and y=2x+8 is closest to:

A.45°
B. 52.4°
C.71.6°
D.108.4°
E. 135°

The equations aren't perpendicular so there must be two different angles between them. I got 71.6 and 108.4 and they're both an option. However, I can only have one answer.

I'm not so clear with the "angles between the two lines" part. What is it asking for??? Any help or just discussion will be greatly appreciated! 

EDIT: I can only select one answer right? The paper doesn't have any instructuions.....

By convention, you pick the acute angle as the angle between two lines. Of course, vectors are a different story where you may need an obtuse angle.

[Shadowxo]

Thanks. I think I sometimes fall in the trap of trying to fully work things out 'algebraically' without thinking about them logically..

---
Edit:
Some small problems here:
http://m.imgur.com/a/2eXOA

Q10ii) Is the domain in the blue wrong? Isn't the domain of gof = domain of f?

Iv) I'm having trouble finding the left half of the maximal domain?

Q8c) I'm having trouble understanding what the reverse operation is for a "reflection in the x-axis". My brain thinks the reverse of this is no reflection at all, but it gets me the wrong answer.

Thanks

For that solution (xe^x=e, or x=e/e^x), I don't think there's any way to solve it directly, just that when x =1, x=e/e^x, and as x becomes greater than 1, x>e/e^x and as x becomes less than 1, x<e/e^x so therefore x=1 is the only solution.

For your questions:
for 10 ii) I think they may be wrong. Let me know if I've made an error but:
g(f(x)) exists if the range of f is a subset of the domain of g.
Range of f = (0,infinity)
Domain of g = [0,infinity)
Therefore it exists.
The range of x values that gives the range of f to be (0,infinity) is R\{0} therefore
Domain of g(f(x)) should be R\{0}, not (0,infinity). Can anyone confirm?

iv) f o h is defined if range of h is a subset of domain of f.
Domain of f = R \ {0}
Range of g = [-1,infinity)
Therefore the g must be restricted such that the Range of g ≠ 0, ie such that range of g = [-1,0) u (0,infinity).
g(x)=0 where √2x - 1 =0
2x=1
x=1/2
So the domain is [0,infinity) without x=1/2, aka
[0,1/2) u (1/2,infinity)

8c) When you reflect in the x  axis, y goes to -y, so to reverse it you divide y by -1 (or multiply). With the reverse, y goes to negative y just like when doing the initial reflection.

Hope this helps

[clarke54321]

Could someone please guide me with these questions. I get really close to the answer, but not quite there.

Thanks!