VCE Methods Question Thread!

[Hancock]

For there to be exactly one solution, the discriminant of the quadratic must equal 0.









Stupid error.

[e^1]

firstly, what are two applications of the process of "completing the square" for a quadratic expression?

Check me if I'm wrong.

1. To get it in turning point form . You may also use this to assist you in graphing.

2. Can be used to derive the quadratic formula from 

3. Solve quadratic equations.

You can actually find the answer here: http://en.wikipedia.org/wiki/Completing_the_square

[darklight]

For pi < a < 3pi/2, with cos a = - sin b, where 0 < b < pi/2, find a in terms of b.

[Limista]

For pi < a < 3pi/2, with cos a = - sin b, where 0 < b < pi/2, find a in terms of b.

cosa= -sinb
-sin (3pi/2 - a) = -sinb
3pi/2 - a = b
a = 3pi/2 - b

[darklight]

How do you work out that cos a  = - sin (3pi/2 - a) ?

[chisel]

For there to be exactly one solution, the discriminant of the quadratic must equal 0.









Stupid error.

Thanks mate, nice working, clear and concise

[chisel]

Check me if I'm wrong.

1. To get it in turning point form . You may also use this to assist you in graphing.

2. Can be used to derive the quadratic formula from 

I know the first one is right, and the second one sounds good too, thanks man!

[Limista]

How do you work out that cos a  = - sin (3pi/2 - a) ?

sin(3pi/2 - a) is in the 3rd quadrant, where cosine and sine are negative
therefore      sin(3pi/2 - a) = -cos a
according to the rules of trignometry, you should know that the 3pi/2 after the sin changes the sin to cos. In this instance I have just applied these rules.

putting minus sign in front of sin(3pi/2 - a) = cos a (this is what we want)

[polar]

For pi < a < 3pi/2, with cos a = - sin b, where 0 < b < pi/2, find a in terms of b.

one is in quadrant 3, the other is in quadrant 1 - so lets put them in the same quadrant first (think about , but 30 isn't equal to 150)
lets choose to put them both in quadrant 1 (add to get to quadrant 3 from quadrant 1, so subtract (awks, forgot to put a slash) to get from quadrant 3 to quadrant 1)

you remove the negative in front of because they're now both in quadrant 1 (both positive), whereas before you needed to make negative since was negative

now, write them both as a function of cosine

[Jaswinder]

how would i find the domain of without the aid of calculator?

[b^3]

There are a few things to note when finding maximal domains.
1. The bottom of a fraction cannot equal zero, since we cannot divide by zero. i.e. we end up with the function being undefined there.
2. Anything under a square root has to be equal to or greater than zero, as in methods we cannot square root a negative number.
3. Anything inside a log has to be greater than zero, as we cannot log a zero or a negative number.

So in our case we have 1 and 2, so we take the intersection of the two restrictions sets, that is

So we will need to find when . So to do this we can solve for when that is zero, and then do a little sketch and find when the graph is above the x axis.

We can do a little sketch of the graph of and as the coefficient on the is positive, we know that it will be a U shaped graph, with intercepts and . So the curve will be above the x axis for . I.e. . Hope that helps

EDIT: beaten, shitty net keeps dc'ing -.-
EDIT2: fixed the missing square roots.

[Will T]

Consider where g(x) := x^2-5 > 0.

Also, I believe the answer to Exercise 3J question 2f. of the Essentials is incorrect. Verification appreciated.

[b^3]

Consider where g(x) := x^2-5 > 0.

Also, I believe the answer to Exercise 3J question 2f. of the Essentials is incorrect. Verification appreciated.

Why do you believe that it is incorrect? It looks fine to me.

EDIT: In that exercise you are using the mapping to find your inverse functions, i.e. it the inverse will be the original flipped in the line . If you graph them then you can see the arm of the curve for would know have come from anything under the mapping. Plus since Dom Ran we get rid of the other arm anyway.

[pi]

how would i find the domain of without the aid of calculator?

Quick summary:

Implied domains of certain scenarios

#1

Take and solve for
Domain = \{}

#2

Take and solve (use a "quick-sketch" if needed)
Domain = solution of inequation

#3

Take and solve (use a "quick-sketch if needed)
Domain = solution of inequation

#4

Take and solve
Domain = solution

[Will T]

Why do you believe that it is incorrect? It looks fine to me.

EDIT: In that exercise you are using the mapping to find your inverse functions, i.e. it the inverse will be the original flipped in the line . If you graph them then you can see the arm of the curve for would know have come from anything under the mapping. Plus since Dom Ran we get rid of the other arm anyway.

Thanks, what on Earth was I on about....