VCE Methods Question Thread!

[HERculina]

I get yaaaa, I also kept thinking that if I'd done
y=-4e^(1/2) the asymptote here would be y=4
And then y=-4e^(1/2) + 1 would make asymptote be y=4+1=5.
But the asymptote in the first step is y=0 not y=4!!!!! Totally confused myself.
Thanks for the help

[Phy124]

y=6x^2+4kx=k=3

find the yint in terms of K

and find the discriminant ?

Did you put one too many equals signs in your equation?

For whatever the equation is, just let x equal zero and solve for y as you would normally to find the y-intercept of a graph.

To find the discriminant put the equation in the form

Then

[sin0001]

y=6x^2+4kx=k=3

Did you mean:
?

[Hutchoo]

Can someone help me understand questions 12, 13 and 22  (MCQ) from VCAA's 2010 exam 2?

Thank you.

Link:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010mmcas2-w.pdf

[catwoman101]

y=6x^2+4kx=k=3

Did you mean:
?

yes

[Phy124]

y=6x^2+4kx=k=3

Did you mean:
?

yes

Following on from what I said before;

[charmanderp]

Can someone help me understand questions 12, 13 and 22  (MCQ) from VCAA's 2010 exam 2?

Thank you.

Link:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010mmcas2-w.pdf

Question 12 is just binomial CDF - upper bound 6, lower bound 0, n=15, p=0.6. Answer is B.

With Question 13, it asks you to find the probability that Z between two SDs below the mean and one SD above the mean. This is equal to the probability that X is between two SDs below the mean and one SD above the mean or, thanks to symmetry, between one SD below the mean and two SDs above the mean.

Hence we want either 8<X<26 or 14<X<32. Looking through the available options, the latter is manifest in B.

I have no idea how to answer Question 22 properly. I just rationalise it based on the fact that of the available options only b represents an expression where ab=a+b in some way, thanks to the log law ln(ab)=ln(a)+ln(b).

[Homer]

With question 22, anti derivative of f(x) would be F(X). When you simplify the expression you'd end up with something like

F(ab+2) = F(a+b) + F(b+2) - 2F(3)

By subing in the a's b's and 3's, The only option which would seem right would be B.

[polar]

but that's the property of the antiderivative, which means being the derivative needs to be in the form

[soccerboi]

With the very last question of VCAA 2011 exam 2, can someone please explain to me why there's an inequality sign there? I really have no idea.

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/mmcas2_assessrep_11.pdf

Thanks 

[Phy124]

With the very last question of VCAA 2011 exam 2, can someone please explain to me why there's an inequality sign there? I really have no idea.

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/mmcas2_assessrep_11.pdf

Thanks 

if he goes directly from his camp at to the plant then since the coordinates of the plant is

the function that describes how long he takes to go from his camp to the plant is

the question requires that T is as small as possible, and since x is kept as a constant, only k can be varied. so, differentiating T and substituting gives thus, solving for k gives

Reference: 2011 VCAA exam 2 last question

[Homer]

I always have trouble sketching trig graphs with a horizontal translation. For instance question 4 on 2006 exam 1 took me ages to draw. I get all the x intercepts right but the end point and the shapes are different. Plus I get confused with the 2 after the cos. (5cos(2(x+ (pi/3)) . does it expand to become 2pi/3 or does it just stay there? Plus does anyone have a quicker way of sketching such graphs? 

[nina_rox]

Hey, could someone possibly explain the signed area and total area? When a graph is above and under the x axis do you add the integrals that represents the area above and under for the total area? and then for a signed area, would you subtract them? I'm so confused. Any help is really appreciated!

[b^3]

An integral will give you the area of something, counting everything that is below the x-axis as negative, and above it positive. This means that:
 - for Signed Area, you just integrate from to , as it will count the areas below the x-axis as negative.
 - for Total Area, you split the integral up at x-intercepts, and then add the areas above the x-axis, and minus those below the x-axis (as minusing a minus will 'add' the area) to give the total area.

[TrueTears]

signed area by definition means that you attach the corresponding "sign" to the area depending on whether the area is above or below the x axis.

If the area is below x axis, then its sign is "-"

If above then it is "+"

The total signed area is the sum of all the areas but with their corresponding signs, does it represent the 'actual' area? No, because it is signed. Why do we have this definition? Well sometimes things like displacement requires the addition of signed area but not the actual area.