VCE Methods Question Thread!

[snowisawesome]

Methods exam 2 2013
Multiple choice question 20
A transformation T: R^2 → R^2, T [

• ,[y]]=[[1,0],[0,-1]]*[
• ,[y]]+[[5],[0]]

maps the graph of a function f to the graph of y = x^2, x ∈ R.


The rule of f is
a. f(x) = -(x+5)^2
b. f(x) = (5-x)^2
c. f(x) = -(x-5)^2
d. f(x) = -x^2 + 5
e. f(x) = x^2 - 5

Would trial and error be a good way to solve this question?
There's meant to be an x next to both the y values, it's meant to be
[

• ,[y]] for the first one and [
• ,[y]] for the second one aswell

Don't know why x won't show up, but before both the y values in my question, there's meant to be an x

[Sine]

The general rule is, if you apply an increasing function to both sides, the inequality sign stays the same. For instance, x^3 < 1 -> x < 1 is valid because cube roots are monotonically increasing.

If you apply a decreasing function to both sides, the inequality sign flips. For instance, -x < 1 -> x > -1 because applying the function f(x) = -x (i.e. multiplication by -1) is a decreasing function.

yes it's definitely possible to do so. However I found personally that graphing and just visualising it is much less error prone.

[snowisawesome]

is it possible for the equation 2^x < 0.3 to show up in exam 1?

[zhen]

is it possible for the equation 2^x < 0.3 to show up in exam 1?

Yup, it’s definitely possible. I would just graph it and then solve for the x value for when 2^x=0.3. So, I’d get x=log₂(0.3). And from the graph, we’d know that x<log₂(0.3)

[snowisawesome]

Yup, it’s definitely possible. I would just graph it and then solve for the x value for when 2^x=0.3. So, I’d get x=log₂(0.3). And from the graph, we’d know that x<log₂(0.3)

But how would we know the shape of y = 2^x without a cas?

[zhen]

But how would we know the shape of y = 2^x without a cas?

You should remember the shape of exponential graphs. Even if you don’t, all you need to know for this one is that when x increases, y increases. When x decreases, y decreases. You can see this by subbing in random values. For 2^x<0.3, you just need to realise that you want to make x smaller and smaller to get it less than 0.3.

[snowisawesome]

You should remember the shape of exponential graphs. Even if you don’t, all you need to know for this one is that when x increases, y increases. When x decreases, y decreases. You can see this by subbing in random values. For 2^x<0.3, you just need to realise that you want to make x smaller and smaller to get it less than 0.3.

But wouldn't it be difficult to find the exact value that it starts to be less than 0.3 for without a cas?

[RuiAce]

But wouldn't it be difficult to find the exact value that it starts to be less than 0.3 for without a cas?

[snowisawesome]

But how would we work out the decimal equivalent of log₂(0.3) without a cas?

[Sine]

But how would we work out the decimal equivalent of log₂(0.3) without a cas?

you don't

always leave answers as exact values

[snowisawesome]

you don't

always leave answers as exact values

But how do you know whether the decimal value is negative or positive so you know which side of the graph it's on, say there was
log₂(0.3) = ~-1.7
and there was log₄(9) = 1.6

[RuiAce]

But how do you know whether the decimal value is negative or positive so you know which side of the graph it's on, say there was
log₂(0.3) = ~-1.7
and there was log₄(9) = 1.6

[snowisawesome]

Is it possible for you to provide an example of a graph please?

[pha0015]

How would you use integration by recognition to solve anti differentiation for log?

bump

[VanillaRice]

Methods exam 2 2013
Multiple choice question 20
A transformation T: R^2 → R^2, T [

• ,[y]]=[[1,0],[0,-1]]*[
• ,[y]]+[[5],[0]]

maps the graph of a function f to the graph of y = x^2, x ∈ R.


The rule of f is
a. f(x) = -(x+5)^2
b. f(x) = (5-x)^2
c. f(x) = -(x-5)^2
d. f(x) = -x^2 + 5
e. f(x) = x^2 - 5

Would trial and error be a good way to solve this question?
There's meant to be an x next to both the y values, it's meant to be
[

• ,[y]] for the first one and [
• ,[y]] for the second one aswell

Don't know why x won't show up, but before both the y values in my question, there's meant to be an x

In my opinion, trial and error probably isn't the most efficient way to solve this type of problem. You should start by multiplying and adding the matrices together, to give you your x' and y' expressions, and then work from there (as per a standard transformation by matrices problem).

How would you use integration by recognition to solve anti differentiation for log?

A differential equation will be given to you. Using this equation, you'll have to rearrange the equation to get the log term as the subject, and then integrate both sides, hence giving you an antiderivative of the log.

Is it possible for you to provide an example of a graph please?

Rui is referring to the standard log, of form y = log_a(x). Have a graph of this on your CAS (subbing in any positive number for a) to observe the features mentioned (you should be able to even Google it).

Hope this helps