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March 29, 2024, 01:10:39 am

Author Topic: VCE Methods Question Thread!  (Read 4802421 times)  Share 

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Jinju-san

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Re: VCE Methods Question Thread!
« Reply #19035 on: December 24, 2020, 11:16:44 am »
0
Heyo again!
Sorry I was also struggling on this other question:

Owlbird83

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Re: VCE Methods Question Thread!
« Reply #19036 on: December 24, 2020, 01:39:36 pm »
+6
Heyo everyone,

So I was doing some practise questions for methods and I was struggling a little on this one.. could someone please help me out?
Q. Find the value of m for which the following simultaneous equations have a unique solution:
-4x + my = -5
-3mx + y =0

I figured out the determinant to be 3m^2 - 4 when I put the coefficients into a matrix and equated it to zero to find the values of m for which they have no unique solutions…
But I’m a little stumped by what ‘a unique solution means’…
Is there another approach I’m supposed to be using?
Thank you!

(Firstly, sorry if this is really wrong I have no confidence in my maths ability.)

A unique solution means that those two lines have one point of intersection, so are orientated somehow like an X (as opposed to parallel with no solutions or on top of each other with infinite solutions).

I don't know how to do it the matrix way, but the way I would do the question would be to:
1. substitute the two equations into eachother like you would to find the point of intersection.
2. then when you get the quadratic with m in it you want to find the discriminant (Δ=b2-4ac)
3. for one solution discriminant=0, so let Δ=0, then solve for m, and the value you get for m should be the value of m when the lines have one unique solution

my (hopefully correct) working out

(If you prefer using matrices you don't have to do it this way)

Also, I cannot view your photo from your second post
« Last Edit: December 24, 2020, 01:44:48 pm by Owlbird83 »
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Jinju-san

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Re: VCE Methods Question Thread!
« Reply #19037 on: December 24, 2020, 02:53:01 pm »
+1
(Firstly, sorry if this is really wrong I have no confidence in my maths ability.)

A unique solution means that those two lines have one point of intersection, so are orientated somehow like an X (as opposed to parallel with no solutions or on top of each other with infinite solutions).

I don't know how to do it the matrix way, but the way I would do the question would be to:
1. substitute the two equations into eachother like you would to find the point of intersection.
2. then when you get the quadratic with m in it you want to find the discriminant (Δ=b2-4ac)
3. for one solution discriminant=0, so let Δ=0, then solve for m, and the value you get for m should be the value of m when the lines have one unique solution

my (hopefully correct) working out

(If you prefer using matrices you don't have to do it this way)

Also, I cannot view your photo from your second post

Ohh okay! Thank you so much! That does make a lot more sense then the method I was using.

I think you should be able to see the picture at the link below (sorry about that):https://drive.google.com/file/d/1ziF94dMbr35zU_O2JFNVWD0xhNeioGHX/view?usp=sharing 

SmartWorker

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Re: VCE Methods Question Thread!
« Reply #19038 on: December 24, 2020, 02:59:13 pm »
+3
Heyo everyone,

So I was doing some practise questions for methods and I was struggling a little on this one.. could someone please help me out?
Q. Find the value of m for which the following simultaneous equations have a unique solution:
-4x + my = -5
-3mx + y =0

I figured out the determinant to be 3m^2 - 4 when I put the coefficients into a matrix and equated it to zero to find the values of m for which they have no unique solutions…
But I’m a little stumped by what ‘a unique solution means’…
Is there another approach I’m supposed to be using?
Thank you!

The condition for a unique solution is that

For

Given this:

solve for m:




For a unique solution



The graph below shows that when m = -(2√3)/3, the lines have no points of intersection, similarly you can do this for m = (2√3)/3
« Last Edit: December 24, 2020, 03:06:35 pm by SmartWorker »
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ArtyDreams

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Re: VCE Methods Question Thread!
« Reply #19039 on: December 24, 2020, 07:12:01 pm »
+3
Heyo everyone,

So I was doing some practise questions for methods and I was struggling a little on this one.. could someone please help me out?
Q. Find the value of m for which the following simultaneous equations have a unique solution:
-4x + my = -5
-3mx + y =0

I figured out the determinant to be 3m^2 - 4 when I put the coefficients into a matrix and equated it to zero to find the values of m for which they have no unique solutions…
But I’m a little stumped by what ‘a unique solution means’…
Is there another approach I’m supposed to be using?
Thank you!

Hi! I learnt how to do these types of questions using the matrix method, so I thought I’d attach a solution here. A unique solution is when the graphs have exactly one point of intersection, and thus, the determinant is not equal to zero.
(I’m terrible at typing maths on here so I thought I’d just attach my writing  :D)
My diagrams might help with your next question, just let us know if you need more help!

my working out

Also.....my maths is rusty lol.....someone please correct me if this is wrong!




p0kem0n21

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Re: VCE Methods Question Thread!
« Reply #19040 on: January 09, 2021, 02:36:39 pm »
+2
Hey is anyone able to help me out with this question! A little stumped :(

Couple ways to solve it. One thing you could directly do is just substitute f(x) for 2x-4. Then you get 2x-4 ≥ 2, and you solve for x. You could also use a graphical method to determine the values as well (i.e. sketch the graph of y=2x-4, and see for what values it is greater than/intersects with the graph of y=2). Hope that helps  :)
« Last Edit: January 09, 2021, 03:42:57 pm by p0kem0n21 »

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Re: VCE Methods Question Thread!
« Reply #19041 on: January 11, 2021, 05:48:36 pm »
0
For q 13 I have no clue how to do it

13. show that the equation (k+1)x^2-2x -k =0 has a solution for all values of k

also, how do you find y-int for y= 1/2(9-x^2). I got x int as -3 and 3 but unsure of how to find y int-

tia

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Re: VCE Methods Question Thread!
« Reply #19042 on: January 11, 2021, 06:11:43 pm »
+6
For q 13 I have no clue how to do it

13. show that the equation (k+1)x^2-2x -k =0 has a solution for all values of k

also, how do you find y-int for y= 1/2(9-x^2). I got x int as -3 and 3 but unsure of how to find y int-

tia

For first question, find discriminant and complete the square. You will find that the resultant quadratic is above the x-axis (or alternatively you could find the discriminant of the discriminant function and show that its less than 0). Hence, the discriminant of the original function is always greater than zero so the equation always has two solutions for all values of k.

For the second question, just set x to be zero, so the y-int is (0,4.5)
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miyukiaura

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Re: VCE Methods Question Thread!
« Reply #19043 on: January 14, 2021, 01:19:41 pm »
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Hey guys, are we expected to use general solutions for trig equations on the exam?
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Sine

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Re: VCE Methods Question Thread!
« Reply #19044 on: January 14, 2021, 01:38:38 pm »
+7
Hey guys, are we expected to use general solutions for trig equations on the exam?
If there is no domain yes. Even if there is a domain it can really help you get solutions quite easily. Although commonly it comes up in mcqs.

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Re: VCE Methods Question Thread!
« Reply #19045 on: January 14, 2021, 03:12:47 pm »
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Hey guys, are we expected to use general solutions for trig equations on the exam?

I just want to add on to Sine - note that general solutions don't have to be as hard as the textbooks make them out to be. Here is a great guide by an old member, TrueTears, which details an intuitive way of solving circular functions generally without having to memorise weird formula and is almost as simple as solving circular functions for a given domain.

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Re: VCE Methods Question Thread!
« Reply #19046 on: January 16, 2021, 02:57:45 pm »
+2
Hey can someone help me solve this bit confused!


Having a unique solution means the lines intercept each other once / there is one x value for which the equations have the same y value.

When looking at 2 linear equations, this means that they can't be parallel i.e. they need to have a different gradient.

Rearrange the equations to be in the standard form (y = [gradient]x + c) then see what values of m mean that the gradients are different

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Re: VCE Methods Question Thread!
« Reply #19047 on: January 16, 2021, 08:18:11 pm »
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No worries!

I'm just doing the maths in my head so I might have made a mistake but B looks right. Keep in mind that B is everything except -5 and 5 since we want all the situations where the lines are not parallel.

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Re: VCE Methods Question Thread!
« Reply #19048 on: January 17, 2021, 09:23:15 am »
+1
Ok thanks! I'm having some second thoughts now do you think it would be D then [-5,5]? sorry to be a pain!

Nah it's not D because D suggests that any value x from -5 to 5 would mean that the lines are parallel / have the same gradient

My point was that B is R \ {-5, 5}  not  {-5,5}, which is correct because {-5,5} is when the lines have the same gradient and we want every value where they do not have the same gradient.

I hope this clears things up :)

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Re: VCE Methods Question Thread!
« Reply #19049 on: January 17, 2021, 05:20:22 pm »
+3
Thanks so much makes heaps more sense!
Theres 2 more questions i've tried to work out but can't get to an answer.

1. Find the value of m for which the following simultaneous equations have infinitely many solutions
mx−4y=6
3x−(m−1)y=2m

2. A function has rule y=ae^kt. Given that y=3 when t=2 and that y=6 when t=3,  find the values of a and  k.




For the first question, the simultaneous equations will have infinitely many solutions if they can be graphed on top of each which would mean they are the same line. To solve this equation, you would rearrange the equations and make y the subject. Then you would make the two lines equal each other and solve for m. Alternatively, you could make the gradients = each other and then solve for m but you will have to test if they're truly the same line by substituting the solutions of m back into the equations as they could simply be parallel and not the same equation.

For the second question, you would simply create a simultaneous equation by substituting in the numbers stated. So you would end up having two equations and then you would be able to solve for the required variables.