VCE Methods Question Thread!

[knightrider]

For this question as follows i got two different sets of answers can you please check if both of these are fine.

Find the general solution to 

my first set of answers

  or    ,

my second set of answers

  or   ,

[IndefatigableLover]

For this question as follows i got two different sets of answers can you please check if both of these are fine.

Find the general solution to 

my first set of answers

  or    ,

my second set of answers

  or   ,

First one is fine and the second one is fine as well

What I did was I said let's find the solutions for the above equation with the domain (0, 2pi) and subbed in n=0, n=1, n=2 etc. into both set of answers and they yielded all solutions within the domain so they're both right

[knightrider]

First one is fine and the second one is fine as well

What I did was I said let's find the solutions for the above equation with the domain (0, 2pi) and subbed in n=0, n=1, n=2 etc. into both set of answers and they yielded all solutions within the domain so they're both right

Thanks so much IndefatigableLover 

[Apink!]

Could someone please explain to me why the min/max x-coordinates is the same in u(t) = -2e^-rt *sin (2t) and in just y= sin(2t)
I thought that min/max x-coordinates would be different because a different function is being multiplied to it. No matter how much i try to understand I don't get it and it's really frustrating me right now ugh

[StupidProdigy]

Pretty simple question but how do I get the median for this. Why can't it be 1.1 or 1.2 haha? cheers

[Adiamond]

Pretty simple question but how do I get the median for this. Why can't it be 1.1 or 1.2 haha? cheers

Hey man, you can't get 1.1 or 1.2 because the question defines it as a discrete variable, meaning that those countable integer values that you see in the probability table (1, 2, 3, 4) are the only numbers that can be taken into account, if it was defined as a continuous variable then 1.1 and 1.2 would be acceptable for a median value.

- Hope this helped

[grannysmith]

If a questions asks for the probability of either A or B or both, assuming they're independent, is this just A_union_B + A_intersection_ B?

Edit: logically, 'both A and B' satisfies  'either A or B', unless an explicit 'but not both' statement is made. So my question is if this is true, then why include the 'or both' tag?

[odeaa]

Doing the 2006 exam 2, and in q3) part e) the question is
Find the positive values of k, where k is a positive real number, for which has one or more solutions for x.

In the assessors report, they say to just solve it for x first, but if i try to do it on my cas it just spits it back out (just gives me what i typed in, no help at all). Am I missing something here? Can you do it by hand? (I tried briefly but to no avail). I understand the rest of the question (solve for k so it is undefined)

[odeaa]

Doing the 2006 exam 2, and in q3) part e) the question is
Find the positive values of k, where k is a positive real number, for which has one or more solutions for x.

In the assessors report, they say to just solve it for x first, but if i try to do it on my cas it just spits it back out (just gives me what i typed in, no help at all). Am I missing something here? Can you do it by hand? (I tried briefly but to no avail). I understand the rest of the question (solve for k so it is undefined)

[IntelxD]

Doing the 2006 exam 2, and in q3) part e) the question is
Find the positive values of k, where k is a positive real number, for which has one or more solutions for x.

In the assessors report, they say to just solve it for x first, but if i try to do it on my cas it just spits it back out (just gives me what i typed in, no help at all). Am I missing something here? Can you do it by hand? (I tried briefly but to no avail). I understand the rest of the question (solve for k so it is undefined)

Try converting the equation into a form where you are able to use the quadratic formula. Look at the spoiler for more help.

Spoiler

Multiply both sides by e^x

[grannysmith]

Doing the 2006 exam 2, and in q3) part e) the question is
Find the positive values of k, where k is a positive real number, for which has one or more solutions for x.

In the assessors report, they say to just solve it for x first, but if i try to do it on my cas it just spits it back out (just gives me what i typed in, no help at all). Am I missing something here? Can you do it by hand? (I tried briefly but to no avail). I understand the rest of the question (solve for k so it is undefined)

Yeah this was a nice question. I just let e^x=a, so:
3-ka-1/a=0
3a-ka^2-1=0
ka^2-3a+1=0
delta=9-4k
one soln: delta=0: k=9/4
more than one: delta>0: 9/4>k
so k<or equal to 9/4. but k>0 so 0<k<=9/4

[odeaa]

Yeah this was a nice question. I just let e^x=a, so:
3-ka-1/a=0
3a-ka^2-1=0
ka^2-3a+1=0
delta=9-4k
one soln: delta=0: k=9/4
more than one: delta>0: 9/4>k
so k<or equal to 9/4. but k>0 so 0<k<=9/4

Oh god how did I not recognise that ahaha

Thanks mate

[odeaa]

When finding the values for x in which f(x) is strictly/decreasing, why do we have to include the turning point (according to insight exam 2014 solutions)? Isn't the gradient 0 there, so not decreasing or increasing?

[keltingmeith]

When finding the values for x in which f(x) is strictly/decreasing, why do we have to include the turning point (according to insight exam 2014 solutions)? Isn't the gradient 0 there, so not decreasing or increasing?

Favourite video eva.

The thing is, strictly increasing/decreasing is NOT defined in terms of the gradient, like you'd expect. Rather, if a<b implies f(a)<f(b) over some interval C, we say that f is strictly increasing over the interval C. Compare to if f is just increasing over C, where a<b implies f(a)<=f(b). This is why the turning point is still included - no matter what a/b you pick, the turning point will always still be above/below it (depending on if the function is increasing or decreasing), and so we include it. However, if you move after the turning point, you can pick a point on that interval that breaks the f(a)<f(b) inequality (in particular, the turning point), hence why we wouldn't choose after the turning point, only up to it. This is a little hard to explain in just words, so feel free to show us the question, and we can use actual points on the graph to help explain it.

It's very easy to see why students (like yourself) get confused, because you're often taught this immediately after you learn about the gradient. However, strictly increasing/decreasing is not defined in terms of gradients (if you continue to do maths at university and do some Real Analysis, you'll learn that strictly increasing/decreasing graphs have some nice properties, hence why we use this definition as opposed to f'(x)>0 or f'(x)<0. For now, you get this nice "black box" explanation)

[odeaa]

Favourite video eva.

The thing is, strictly increasing/decreasing is NOT defined in terms of the gradient, like you'd expect. Rather, if a<b implies f(a)<f(b) over some interval C, we say that f is strictly increasing over the interval C. Compare to if f is just increasing over C, where a<b implies f(a)<=f(b). This is why the turning point is still included - no matter what a/b you pick, the turning point will always still be above/below it (depending on if the function is increasing or decreasing), and so we include it. However, if you move after the turning point, you can pick a point on that interval that breaks the f(a)<f(b) inequality (in particular, the turning point), hence why we wouldn't choose after the turning point, only up to it. This is a little hard to explain in just words, so feel free to show us the question, and we can use actual points on the graph to help explain it.

It's very easy to see why students (like yourself) get confused, because you're often taught this immediately after you learn about the gradient. However, strictly increasing/decreasing is not defined in terms of gradients (if you continue to do maths at university and do some Real Analysis, you'll learn that strictly increasing/decreasing graphs have some nice properties, hence why we use this definition as opposed to f'(x)>0 or f'(x)<0. For now, you get this nice "black box" explanation)

Thanks for the help! Appreciate it