VCE Methods Question Thread!

[Ahmad_A_1999]

Edit: started writing this but paused for a bit so Zhen beat me to it
27.
so for any log, eg log_ab, b≠0, b must be positive aka b>0. This is because a (positive) number to a power can never equal zero or go negative, so what's inside the log therefore cannot be less than zero either. So we know (x-2)²>0 and therefore x≠2 but can be anything else. So, maximal or implied domain is x belongs to R \{2}. And (x-2)² can be any number between 0 and infinity not inclusive (can't be zero) so the log range is R (you can see this using a standard log graph). So answer is B

Also, 2logx is different to logx² only because you can take the negative values of x when it's squared as it'll end up as a positive number (squaring things can be annoying in maths)

28.
Not sure if you were taught this (don't think you have to know this, just a bit of additional info), but if you line a ruler up to the right side of a unit circle, the value of the tan of an angle is just the the ruler measurement where the line extends out to the ruler. That's always how I remembered it (image attached). So tan(0)=0. From this you can also tell that the period of tan x is π without having to memorise it, just imagine the angles changing (worked for me and just a little trivia - you don't need to know this but it may help)
From the graph you're given you know that the period is π/2, half of what a regular tan graph would be so you know as period = π/n for tan, n=2. First thing you do is dilate it accordingly, to tan(2x). You also know it's translated π/4 (or negative π/4 or 3π/4 etc, you can just read off the graph for various possible translations) because you know tan(0)=0 and the x intercept/s have moved along by π/4. So, you replace x with (x-π/4), which makes the equation tan(2(x-π/4)), C

Thanks for the help man!

Yeah I was mainly worried about the squared in the logarithm, I'll just accept it 

[LPadlan]

Factorise each of the following:
8a^3+27b^3

[Shadowxo]

Factorise each of the following:
8a^3+27b^3

You know that the expression is equal to
(2a)³ + (3b)³, just use the formula a³ + b³ = (a+b)(a²-ab+b²) to factorise it into (2a+3b)(4a²-6ab+9b²)

[LPadlan]

Hey how do you know a sign diagram is negative or positive? eg.(3-x)(x-1)(x-6) has ints 3,1,6. In the book it shows it as negative, i dont understand why?

[QueenSmarty]

Hey how do you know a sign diagram is negative or positive? eg.(3-x)(x-1)(x-6) has ints 3,1,6. In the book it shows it as negative, i dont understand why?

If you rearrange the first bracket, it becomes (-x+3). Since there's a negative in front of the x, the graph must also be negative

[zxcvbnm18]

Hi guys,
I was wondering whether we need to memorise complementary relationships between angles (e.g sin (pi + theta) = cos theta ) for the end of year exams. If so, can someone tell me how to memorise these relationships easily?
Thnx

[zhen]

Hi guys,
I was wondering whether we need to memorise complementary relationships between angles (e.g sin (pi + theta) = cos theta ) for the end of year exams. If so, can someone tell me how to memorise these relationships easily?
Thnx

Yes, you will have to remember them. For me I remember them by looking at from what axis of the unit circle it makes. For example sin(π+θ) makes an angle θ from the x-axis, so it remains as sine. But sin(π/2+θ) makes an angle θ from the y-axis, so it turning into cosine. Also, since sin(π/2+θ) is in the second quadrant it remains positive. But if it was something like sin(3π/2+θ) it would be -cos(θ), since it's in the fourth quadrant, where sine is negative. Basically if sine or cosine makes and angle θ from the y-axis, they change into the other one and make sure to be careful of whether they are positive or negative. Hope that helps and someone should correct me if I made a mistake, since I'm not too sure about this.

[zxcvbnm18]

Yes, you will have to remember them. For me I remember them by looking at from what axis of the unit circle it makes. For example sin(π+θ) makes an angle θ from the x-axis, so it remains as sine. But sin(π/2+θ) makes an angle θ from the y-axis, so it turning into cosine. Also, since sin(π/2+θ) is in the second quadrant it remains positive. But if it was something like sin(3π/2+θ) it would be -cos(θ), since it's in the fourth quadrant, where sine is negative. Basically if sine or cosine makes and angle θ from the y-axis, they change into the other one and make sure to be careful of whether they are positive or negative. Hope that helps and someone should correct me if I made a mistake, since I'm not too sure about this.

Thank you!

[Max Kawasakii]

Can someone please explain how the calculator is doing this?

In my cas calculator I put 1/(-1+5)^1/2 and the answer is 1/2.

Whilst if I put 1/(-4)^1/2, the answer is as expected, a non real result.

Same equation, why the difference???

Super simple don't even know if this qualifies as a MM question but will appreciate the clarification.

[zhen]

Can someone please explain how the calculator is doing this?

In my cas calculator I put 1/(-1+5)^1/2 and the answer is 1/2.

Whilst if I put 1/(-4)^1/2, the answer is as expected, a non real result.

Same equation, why the difference???

Super simple don't even know if this qualifies as a MM question but will appreciate the clarification.

The square root of -4 is an imaginary number for the second 1. The first one is different to the second because it's the square root of positive 4, as 5-1=4, which equals 2. So the difference is one is 1/sqrt(-4) and one is 1/sqrt(4)

[Max Kawasakii]

The square root of -4 is an imaginary number for the second 1. The first one is different to the second because it's the square root of positive 4, as 5-1=4, which equals 2. So the difference is one is 1/sqrt(-4) and one is 1/sqrt(4)

Well probably the easiest up vote you have ever gotten, my brain must be cooked to think -1 + 5 =-4

... embarrassing to say the least, but thankyou

[Perryman]

Hey

Any help with this question will be greatly appreciated!

Thanks!

[Hungry4Apples]

Hey

Any help with this question will be greatly appreciated!

Thanks!

Hey so the best way to do this question is to sub in values of n. However be aware that n is an element of z so it must be integer values. start with n=0 and work from there untill you find all the solutions within the domain!!

[Max Kawasakii]

Find the inverse f^-1, of the function f: R\{-3/5}->R, f(x)= (2x+1)/(5x+3). State the domain and range of the inverse function.

Just having trouble transposing the inverse for y. Thanks in advance!

[zhen]

Find the inverse f^-1, of the function f: R\{-3/5}->R, f(x)= (2x+1)/(5x+3). State the domain and range of the inverse function.

Just having trouble transposing the inverse for y. Thanks in advance!

I just simplified the fraction so that there's only one x, which makes it easier to find the inverse. My working is in the image attached below. From here you should be able to find the inverse by swapping x and y.