VCE Methods Question Thread!

[Phy124]

what is the logic behind VCAA 2011 exam 1 question 4b?

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas1-w.pdf

is this right? ::

step 1: A square root functions contents must be positive
step 2: therefore (X+8) (X+2) must be both positive? so x<-8 ad x>-2? is that all?

The square contents must be greater than or equal to zero.

So x is less than or equal to -8 and greater than or equal to -2.

(Your reasoning was right, except you didn't include zero)

[sh00my]

Hi

I was just wondering what impact adding modulas symbols to... Lets say sin(x) has on the period of the function? Does it halve the period? Or does the period remain constant regardless?

Thanks for any help.

[Homer]

I think unless there are any vertical translations on the original graph, the period would be halved

[supermanflyaway]

2009 Exam 2 Extended Response Q 1e)

With f(g(x)), VCAA have given the domain as R. Shouldn't it be [0, ∞), the domain of f(x)?

[Phy124]

Hi everyone, just need help on Question 10 b of VCAA 2009 Exam 1. I've stumbled across these type of questions for many practice exams and I still don't understand the concept. Any help appreciated! Cheers.

The value you are trying to approximate will be in a function. You use this function as . You then differentiate that to find . You'll then have a known value for that function e.g. in this case we know that the cube root of 8 is 2, so we use that as x, then h becomes 0.06. Put these values into the equation and we're good to go.

[supermanflyaway]

I don't get the last question of 2011 Exam 2 part f)... why is dT/dx ≤0?

[BubbleWrapMan]

I don't even know actually, but this is what I PM'd Moko (too lazy to type again haha)

"The previous parts of the question demonstrated that the larger k is, the closer to the plant he will have to run to achieve minimum time. This is because the time taken to swim increases as k increases, so running would be more beneficial than swimming if k is bigger.

From this I gathered that for every point on the lake near the plant, there was an associated value of k for which he should run there to achieve minimum time. So I figured that there was a value of k for which minimum time would be achieved if he ran straight to the plant, i.e. k was too large for him to bother swimming because it would take too long. So I solved dT/dx = 0 for k, when x was equal to the x-coordinate of the desalination plant.

If k was any larger, however, it still made sense that he should run straight to the plant, since the swimming process would take even longer. So any value of k bigger than this would require him to run straight there."

[nina_rox]

Also sorry to be annoying:

How do you find the general solution of:

cos(x) + cos(3x) = 1/2

I found this equation in the study design. Thanks!

cos(x) + cos(2x+x)=1/2
expand them all, then use double angle formula then solve for cos(x)

Hold on, are we supposed to know double angle formulae?

My teacher told us to put it in our bound references and she said it would be very very very unlikely to come up in exam 1

Yeah, the exact question is in the study design so maybe we do need to know them?

[supermanflyaway]

I don't even know actually, but this is what I PM'd Moko (too lazy to type again haha)

"The previous parts of the question demonstrated that the larger k is, the closer to the plant he will have to run to achieve minimum time. This is because the time taken to swim increases as k increases, so running would be more beneficial than swimming if k is bigger.

From this I gathered that for every point on the lake near the plant, there was an associated value of k for which he should run there to achieve minimum time. So I figured that there was a value of k for which minimum time would be achieved if he ran straight to the plant, i.e. k was too large for him to bother swimming because it would take too long. So I solved dT/dx = 0 for k, when x was equal to the x-coordinate of the desalination plant.

If k was any larger, however, it still made sense that he should run straight to the plant, since the swimming process would take even longer. So any value of k bigger than this would require him to run straight there."

Great, thanks a lot for that! Appreciated

[VCEstudentguy]

For anyone that has done the MAV 2012 exam 2 can you explain how you found Var(C) for Q2dii? I don't quite understand in their solutions how they found Var(C) by 1.25^2 * 49.

[polar]

since the standard deviation is 7, the variance is 49, using and writing the event as

[VCEstudentguy]

Ah, I thought there was a formula I was missing, thanks for that.

[captainamy]

Hi everyone, just need help on Question 10 b of VCAA 2009 Exam 1. I've stumbled across these type of questions for many practice exams and I still don't understand the concept. Any help appreciated! Cheers.

The value you are trying to approximate will be in a function. You use this function as . You then differentiate that to find . You'll then have a known value for that function e.g. in this case we know that the cube root of 8 is 2, so we use that as x, then h becomes 0.06. Put these values into the equation and we're good to go.

Thanks Rangaaaa, I needed question 10 b though, "Explain why this approximate value is greater than the exact value of cubed root 8.06." Sorry!

[Phy124]

Thanks Rangaaaa, I needed question 10 b though, "Explain why this approximate value is greater than the exact value of cubed root 8.06." Sorry!

Oh my mistake,

have a look at this diagram I've made:



As you can see, it is merely or in this case rearranged



As the slope of the curve decreases as x increases at the estimated value will be above that of the real.

Additionally, for graphs such as the gradient will be increasing as increases, therefore the approximated value will be lower than the actual.

I'm pretty sure all VCAA would be looking for is that that the approximation for is reliant on the gradient of , what the gradient type is (increasing/decreasing) and hence whether the value is higher or lower than the actual.

[Homer]

if , then what would