VCE Methods Questions Thread

VividReverie

PizzaMaster
I think you have to consider the cases slightly differently.

Case 1: Ends with zero
Since 0 must be at the end, there is only one possibility for the last digit. We can fill this in:
_ _ _ 1
After this, we are left with 7 numbers that could still come at the start, as 0 has been used.
7 _ _ 1
Continue filling this in to get 7 X 6 X 5 X 1 = 210 ways

Case 2: Ends with 2, 4, or 6
The last digit can be either 2, 4, or 6, so that’s three possibilities for the last position:
_ _ _ 3
After this, there are 6 options for the first digit (excluding 0 and whichever number we just used)
6 _ _ 3
Then 6 again for the next digit (excluding the two we already used, but including 0)
6 6 _ 3
Which is 6 X 6 X 5 X 3 = 540 ways

As you already calculated, the total is 1470, so the probability that the number is even is (210+540)/1470 = 25/49

Hope this helps!

PizzaMaster

Oh so you have to consider zero as 1 in the blank and as a separate case.

I get it now!!!
Thank you so much.

PizzaMaster

Hello everyone

Just a quick question:
Is radians only activated on CAS when dealing with circular functions/ trig
OR
does it effect your other day-to-day calculations and graphing as well?

Thank you!

God

If you look in the top right-hand corner, it should either say 'rad' or 'deg' next to the battery symbol.

If you click on it with the mouse - it will switch to the other mode. (You will still need to re-do calculations though)
Just be careful - if you press the up/down arrows while the mouse is still over it - it may switch back.

Hope this helps!

God

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Edit: It seems if u get 100%, you can only get a 45..... Lol

PizzaMaster

Hey guys,

How do you find the domain of:
√x-1/x+2 ?

We know that x-1/ x+2 has to be >= 0

I tried doing it in a way that I brought x+2 to the other side and then I had:
x-1 >= 0

But the answer says:
x <-2 U x >=1

How...
Is there a method to do this?

Please help.

Thanks.

Revan

PizzaMaster

Function:
√(x-1)/( x+2)

√(x-1) implies that x≥1 as you cannot have a negative value inside the root (to get a real value). x+2 is the denominator and cannot equal zero. So the domain of x+2 would be any real number except -2. But since the domain of √(x-1) is x≥1, we can disregard the -2 and be left with the final answer of x≥1. I'm not sure if I interpreted your function correctly tho, my answer is correct tho according to my interpretation. Let me know if the function is listed above is the one you are talking about, I CAS verified it as well.

PizzaMaster

Thank you for your help.
However, I meant that the (x-1) and (x+2) are both under the square root sign.
Sorry for the confusion.

Revan

No worries, in that case we have a whole new situation. The same principle applies in that x+2 cannot equal zero and everything inside the bracket must be 0 or greater. To be honest, I would convert (x-1)/(x+2) into a partial fraction which would yield 1 - 3/(x+2) which can be obtained through long division. If you graph that you get a hyperbola (I think it's called that). You just gotta find when the graph is greater than or equal to zero. I would strongly suggest graphing it and you'll come to realise that from -2<x<1 the graph is negative. Thus the domain of that function would be the direct opposite, aka your aforementioned answer.

You'll find graphing stuff is always gonna help you and you need to learn and get used to do doing it. Hope that Helps.

ILovePizzaJay

PizzaMaster Nice to meet a fellow pizza lover

PizzaMaster

Can a one-to-one function have a many-to one inverse function?
For example:
f(x) = 2x^ 3/4 + 1

has an inverse:
f-1(x) = [(x-1) / 2]^ 4/3

How come?

Billzene

PizzaMaster
If you have f(x) = x⁰.5

Then f-1(x) = x², but restricted to the domain R+ u {0}, which means the inverse is also 1-1. The negative branch of x² isn't a reflection of y = x⁰.5 in the line y = x, so it's not part of the inverse function

PizzaMaster

Ohh so you mean that the range of the original is the domain of the inverse?

And so that's why
f(x) = 2x^ 3/4 + 1

does have an inverse:
f-1(x) = [(x-1) / 2]^ 4/3
BUT with domain [1, infinity)

Billzene

PizzaMaster

Correct that's why you need to restrict the domain of the original function

snowflake

How do we know whether the question wants us to give the dilation from x axis or y axis when given both the original and transformed function.

For example:
f(x)= 1/x²
f1(x)= 5/x²

I got dilation by a factor of 1/5 from the y axis but the answers say factor of 5 from the x axis. Is my answer still valid?

Billzene

snowflake Your answer wouldn't be considered correct, if the coefficient of x2 were 5 ie y = 1 / (5x2) then it'd be correct since that would be a dilation from the y-axis

snowflake

Billzene

Ohh ok but how do we whether to work out it’s dilation from the X or Y axis if it doesn’t say in the question?

Billzene

snowflake by recognition of the form y = a/(1/b(x-h))ⁿ + k

a = dilation factor from x axis
b = dilation factor from y axis
h = translation in +ve direction of x axis
n = power
k = translation in the positive direction of y axis

This is similar to turning point form of parabolas

snowflake

Billzene

thank youuu

chemistry1111

find values of a and b such that a(x+2)+b(x+3)=18x+8 for all values of x
if someone could help that would be great

beep_boop_booop

chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28