VCE Methods Question Thread!

[b^3]

Now since the is part of the power on the , we want to try and get on it's own.

Now we need to get the out of the power. If we turn it into some kind of log we can then bring the power down to the front, but we want to pick a log that will give us the simpliest expression. , so we can take the of both sides.

Now, the answer might express it using log with base , if so you can just swap it to that using


Although you could go straight to that by taking log with base of both sides instead of base .
e.g.

[M-D]

Thanks guys for all your help.

I've got another question from the same textbook (Maths Quest 11 Mathematical Methods CAS 3rd edition). I actually think some info is missing in the question. Here's the question:

Question 10:

A golf ball is hit from the bottom of a bunker as shown at right. The height, h metres, of the ball above the ground is given by h = 5.2t − 1.5t² − 1, where t seconds is the time the ball has been in flight.

    (a) How deep is the bunker?
    (b) When is the ball first level with the top of the bunker?
    (c) At which times is the ball at a height of 3 metres?

I have attached the diagram pertaining to this question.

The answer at the back of the book is:
Question 10
    (a) 1 metre

    (b) At t = 0.2 seconds

    (c) At t = 1.2 and 2.3 seconds

Thanks in advance for your help
 

[M_BONG]

Our teacher taught us crazy-hard absolute functions today and I didn't really fully understand it.

Can someone explain how to sketch:
y= f(|x^2-7x|)

Thanks in advance!

[b^3]

What function is ? Normally there's two types of modulus composite functions that you'll come across.
The first being , where you flip all the negative parts of the curve, that is the parts that are below the axis, in and across the axis to become positive.
The second being , where you take the curve for , that is the part of the curve to the right of the axis, and copy and flip it in the axis.

You normally need to know to be able to do it though. With the above, if you have then you can just look at it as a composite function of and (the latter which you can draw with the information above).

[M_BONG]

What function is ? Normally there's two types of modulus composite functions that you'll come across.
The first being , where you flip all the negative parts of the curve, that is the parts that are below the axis, in and across the axis to become positive.
The second being , where you take the curve for , that is the part of the curve to the right of the axis, and copy and flip it in the axis.

You normally need to know to be able to do it though. With the above, if you have then you can just look at it as a composite function of and (the latter which you can draw with the information above).

Oh sorry,

f(x) = x^2-7x

[b^3]

Ok, going to assume where .

Let's look at what happens as you plug a number into our function. It first has to go through the , if it's positive, then it'll be unchanged and you'll get the same output as . If it's negative, then the modulus will change it into a positive, which is then feed into . So say we had where is positive, then , but if we have then we get as well, as the modulus will turn our negative input into a positive one. So if you put or in, you'll get the same value, same goes for and . What this effectively does is restrict us to putting positive (and zero) numbers into .

This means that for , and you can just draw . But for , (note that is negative so is positive). So we have the curve flipped in the axis.

So in the end, you draw the curve of on the right hand side of the axis, then copy this curve and flip it in the axis to get the curve for .



If you were to write this out as a hybrid function, then we'd have

[Phy124]

The height, h metres, of the ball above the ground is given by h = ...

So our vertical axis represents the height of the ball above (or below) the ground surface (the top of the bunker). This means that when h=0 then ball is at surface height, when h<0 it is below surface height and either in the air or sitting in the bunker and finally when h>0 the ball is in the air above surface height.

As our function is dependent on time after contact with the ball, we know that when t = 0 the ball has not yet been hit and hence we can use this to find out the depth of the bunker (the "height" of ball at this point will be the distance between the ground level and the sand).

Using this information you should be able to do each of the questions, but I will supply solutions below:

Spoiler

(a)


Hence the bunker is 1 metre below the surface.

(b)




Hence it is first at level with the top of the bunker after 0.2s

(c)



Hence the ball is at a height of 3 metres, 1.2 seconds after contact and 2.3 seconds after contact.

[ETTH96]

Hey guys, how do I go about doing these questions?

1. The simultaneous linear equations (m-2)x+3y=6 and 2x+(m-3)y=m-1 have no solution for:
a) m (belonging to) R\{0,5}
b) m (belonging to) R\{0}
c) m (belonging to) R\{6}
d) m=5
e) m=0


2. The simultaneous linear equations mx+12y=24 and 3x+my=m have a unique solution only for:
a) m=6 or m=-6
b) m=12 or m=3
c) m (belonging to) R\{-6,6}
d) m=2 or m=1
e) m (belonging to) R\ {-12,-3}

How should I approach these types of questions?
Thanks.

[b^3]

There are two ways to do this (well technically three but the latter is CAS-ing things but it won't be reliable), the first is to rearrange the equations into form and then compare the gradients and intercept, the second is to use the determinate of a matrix from the matrix system of equations. I'll go through each, the second is probably quicker though.

If you rearrange both equations we get

Now there are a couple of rules.
1. For no solutions, the two lines need to be parallel, but not the same line. So that means the gradients need to be the same but the y-ints need to be different.
2. For infinite solutions, the two lines need to be the same line, so both the gradient and y-int need to be the same in both equations.
3. For a unique solution, the two lines need to be not parallel, that is the gradients must be different.

So in the above we want no solutions, so that is we want the gradients to be the same but have different y-ints.
Equating gradients.

Having different y-ints.


So the intersection of the two will give us what satifies both conditions that we want, which is .
Now for the matrix method. The system of equations can be written in matrix form as the following.

Now to solve this system of equations we would normally do the following.

Now for the inverse of to exist, we need the determinate of to be non zero.
- If the determinate is zero, then the is no unique solution, that is there is either infinite solutions or no solutions
- If the determinate is zero, then there is a unique solution.
In the first case you would then substitute the values into the equations and see if you get the same two equations or not.

So for the above our det of is
So we want that to be equal to zero for no solutions, so

Now we substitute the values back into the equations to get

So we get the same two equations, which relates to the infinite solutions case. So we know that is not what we're looking for, for no solutions.

Which are two different equations, so for no solutions we have .
I guess you could also attempt to solve the system on your calc for and and then look at the denominators of the solutions to see which values gives you solutions and which don't but it won't always be reliable, and you should know the theory behind it and how to do it by hand anyways.

See if you can do the second question on your own. Hope that helps

[alchemy]

^^^ Why do we even need a textbook when we've got you?

[lzxnl]

Because you can't bring b^3 into the exam or SACs?

[alchemy]

Because you can't bring b^3 into the exam or SACs?

Hehe, fair enough. On another note: has anyone found the textbook helpful during an exam (if they brought it in for reference)?

[soNasty]

can someone show me a step by step solution in differentiating:



using the product rule, thanks

[Thu Thu Train]

can someone show me a step by step solution in differentiating:



using the product rule, thanks

Let

Let and

The product rule states that





By the chain rule




I think.... I'm so unsure of my answer and don't want to post it because it's potentially wrong :/

[lzxnl]

Now, the first derivative is just 2x. As for the derivative of , a chain rule application yields

Putting this altogether, you should get

As mentioned above.