VCE Specialist 3/4 Question Thread!

[zvezda]

is undefined over the domain .

As the function has a minimum value of and a maximum value of this means the minimum value of the can be put into the function is and the maximum value is (simplicticly, the range of is therefore the domain of its inverse ,, will be

Therefore the domain is represented as





Meaning the domain of the function   is:

and

wow, just realised how stupid I was. But even still, if you did for example
1: -1<3/x<1  (take it as greater than or equal to etc. I dont like using latex)
2: -1<x/3<1
3: -3<x<3

Unless I'm am missing something outrageously stupid, isnt this correct mathematical working?

[b^3]

When is negative, you are multiplying by a negative number, meaning that you would need to flip the direction of the signs, so its not mathematically correct to just flip it like that. You need to split it up into each case, then look what happens.

[Homer]

hey guys i'm having trouble with Euler's method for numerical solutions of differential equations, like i never get the right answer.

here's an example of  such question:

Use Euler's method with h=0.5 to find the approximate value of the solution of the following differential equations at the given value

y'=2x+1 , y(0)=0 at x=2

[polar]

hey guys i'm having trouble with Euler's method for numerical solutions of differential equations, like i never get the right answer.

here's an example of  such question:

Use Euler's method with h=0.5 to find the approximate value of the solution of the following differential equations at the given value

y'=2x+1 , y(0)=0 at x=2

So the formula given is , we want to find y(2) which is since .
First, we find :

Next, we find : but we already know that so that means


If we keep going, we find a bit of shortcut for finding :


See if this method works better

[saba.ay]

Could someone please explain how I would do the following?

If z= x + yi, determine the values of x such that z=(3 + 4i)^(0.5)

Thank you

[sin0001]

(x+yi)^2 = 3 + 4i
x^2 + 2xyi - y^2 = 3 + 4i
(x^2 - y^2) + 2xyi = 3 + 4i
By equating coefficients, we get the following equations:
x^2 - y^2 = 3 (1)
2xy = 4, therefore:
x=2/y (2)
Subbing (2) into (1):
(2/y)^2 - y^2 = 3
4 - y^4 = 3y^2
Letting y^2 = a, we get a quadratic:
a^2 + 3a - 4 = 0
(a+4)(a-1)=0
a = y^2 = -4, 1
Since y^2 > 0
y = +/- 1
Now sub. this in (2), to get values for 'x':
Let y = 1, x = 2/1 = 2
Let y = -1, x = 2/-1 = -2
Therefore the following complex numbers are square roots of z=3 + 4i
2 + i  and -2-i
Hope I didn't stuff up

[Conic]

2 + i  and -2-i
Hope I didn't stuff up

You should reject -2-i since it's in the 4th quadrant. In the question it says z=(3 + 4i)^(0.5) so it should be in the 1st quadrant (ie 2+i).

[Henreezy]

If f: [0, pi] -> R, f(x) = tan (x), find the coordinates of the points on the graph where the gradient is equal to:

a) 1
b) 4/3
c) 4
d) 10

I'm unsure about how to approach gradient questions with circular functions, I know that you would first need to diff. f(x) = tan (x) which gives you gradient fn f'(x) = sec^2(x), but I'm unsure where to go from there.

[saba.ay]

(x+yi)^2 = 3 + 4i
x^2 + 2xyi - y^2 = 3 + 4i
(x^2 - y^2) + 2xyi = 3 + 4i
By equating coefficients, we get the following equations:
x^2 - y^2 = 3 (1)
2xy = 4, therefore:
x=2/y (2)
Subbing (2) into (1):
(2/y)^2 - y^2 = 3
4 - y^4 = 3y^2
Letting y^2 = a, we get a quadratic:
a^2 + 3a - 4 = 0
(a+4)(a-1)=0
a = y^2 = -4, 1
Since y^2 > 0
y = +/- 1
Now sub. this in (2), to get values for 'x':
Let y = 1, x = 2/1 = 2
Let y = -1, x = 2/-1 = -2
Therefore the following complex numbers are square roots of z=3 + 4i
2 + i  and -2-i
Hope I didn't stuff up

Thank you so much. Can't believe I didn't think to expand the brackets in step 1. Thanks again.

[Phy124]

If f: [0, pi] -> R, f(x) = tan (x), find the coordinates of the points on the graph where the gradient is equal to:

a) 1
b) 4/3
c) 4
d) 10

I'm unsure about how to approach gradient questions with circular functions, I know that you would first need to diff. f(x) = tan (x) which gives you gradient fn f'(x) = sec^2(x), but I'm unsure where to go from there.

a.











b.











I think you get the pattern, same for the other two.

[Homer]

[ashoni]

i'm having trouble finding the implied domain and range of inverse trigonometric functions... could someone please explain this to me?

[Daenerys Targaryen]

You find the original graph (without the inverse), find it's domain and range. Then for the inverse, the domain and range interchange. i.e. Dom->Ran of inverse. and Ran-> Dom of Inverse

[Homer]

i'm having trouble finding the implied domain and range of inverse trigonometric functions... could someone please explain this to me?

any specific question?

[polar]