VCE Methods Question Thread!

[SS1314]

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)

[miyukiaura]

Hey guys, are we expected to use general solutions for trig equations on the exam?

[Sine]

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.

[keltingmeith]

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.

[Bri MT]

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

[Bri MT]

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.

[Bri MT]

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

[Danzorr]

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.

[SmartWorker]

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.

[bluebird]

Hi, I hope this is the right place but I'm kinda stumped on these distance speed time questions. Any help is appreciated.
Thanks
-bluebird

[SmartWorker]

Hi, I hope this is the right place but I'm kinda stumped on these distance speed time questions. Any help is appreciated.
Thanks
-bluebird

Hey Bluebird!

To do these type of questions i would recommend making diagrams and logically drawing conclusions.

Qu 4:


Qu 5:

[fun_jirachi]

Hey is someone able to help me out with these q's.
1. The volume, Vcm3, of water in a tank at time t seconds is given by V(t)=5t^3+5t−8. What is the average rate of change of volume between t=1 and t=4.
-I first differentiated the equation to be 5t^2+5.
-Then I tried to sub t=1 into the equation which gave me 20. Then I subbed 4 into the equation and I got 245. So I subtracted 20 from 245 and it gave me 225cm3. But its wrong? Where've I gone wrong?

A golf ball is hit so that its height h(t) metres above the ground t seconds after it is hit is given by h(t)=10t(4−t). What is the maximum height reached.
I graphed the equation and found the max height to be 40 on the y-axis? idk if thats right?

For the first question, you're pretty damn close - the question asks for the average rate of change, which you haven't done. I'm not going to give anything away because you're literally that close.

For the second question, you're definitely right

[alexandra.vo]

Is someone able to check my answers for me plz!

[fun_jirachi]

Last two seem correct, first one is wrong.

When checking things like these, there are a few checks you can do to definitively check if your answer is wrong.
- Test a few points (not boundary points, points either clearly outside or inside the zone defined by the equation) and check if they satisfy the equation. Common points you might like to try are points on the x-axis, y-axis or most commonly the origin.
- In the case of multiple equations defining a composite area, do a bit of a simple sanity check and check that your answer actually defines an enclosed area that makes enough sense

Hope this helps

[alexandra.vo]

Last two seem correct, first one is wrong.

When checking things like these, there are a few checks you can do to definitively check if your answer is wrong.
- Test a few points (not boundary points, points either clearly outside or inside the zone defined by the equation) and check if they satisfy the equation. Common points you might like to try are points on the x-axis, y-axis or most commonly the origin.
- In the case of multiple equations defining a composite area, do a bit of a simple sanity check and check that your answer actually defines an enclosed area that makes enough sense

Hope this helps

Thanks so much! Makes more sense, I've re-looked at the graph and I'm thinking B now for the first graph?