VCE Methods Question Thread!

[knightrider]

This may be of some help
(Image removed from quote.)

For the first one we know that 2.442 is less than yet more than
Hence it has to be in the second quadrant
Since sine is positive in the first quadrant and it's asking sin(2.442) the answer will be positive

For the second it's greater than but less than
Therefore it is in the 3rd quadrant
Since only  tan is positive in the 3rd quadrant we end up with a negative answer

I'll let you do the third one

For the last one, it's gone backwards
It would be less than but greater than so it's in the 4th quadrant
Only cos is positive in this quadrant and therefore the answer is negative

I hope this made sense

EDIT: It's meant to be in the picture

Thanks so much paper back really appreciate it

[LiquidPaperz]

answer is C but i got D , because i based it off

AX = C
X = A^-1C

[Phy124]

[LiquidPaperz]

Hey Phy, thanks for the reply. So did you divide both sides by X?, and then in the third step get I the identity matrix, if that is what you are referring to.

[LiquidPaperz]

answer is D, is their a reason its not (C-B)A^-1?

[e^1]

answer is D, is their a reason its not (C-B)A^-1?

Matrix multiplication isn't commutative (ie. , or AB is not always equal to BA). However it is associative, which means that .

If we multiply both sides to the left by A^-1:




If we multiplied both sides by A^-1 on the right side instead:



Since matrix multiplication is not commutative, we cannot write as (which is probably what you would have hoped).

[Phy124]

Hey Phy, thanks for the reply. So did you divide both sides by X?, and then in the third step get I the identity matrix, if that is what you are referring to.

I multiplied both sides by the inverse of X, X^-1.

[soNasty]

How do I do these sorts of questions? I'm having trouble understanding the concept
The answer is 1.

[keltingmeith]

Methods doesn't really teach this, and the only time I've seen that set up is in third party practice exams (I think I saw it in 2013 MAV?). Two ways of going this:

The "recognise" way (and more difficult to get your head around):
Recognise that if we , then this is identical to with some transformations which we will make, getting . Then, realise that since we're integrating by 3x, however just changed this to 3, we must divide by three, giving us

The substitution method (from specialist techniques):
Let , so . This gives us

[soNasty]

Hey thanks! Also, would it be right to think of the question like this?
Since the first equation is f(x) yeah, the second is f(3x+1), meaning it has a dilation of 1/3 parallel to the x axis, meaning the graph is basically condensed by a factor of 1/3. I'm assuming the dilation would affect the value of the area calculated, and hence to account for the dilation we would simply multiply the first area by 1/3. The translation wouldn't have any affect on the area calculated yeah?

[keltingmeith]

Hey thanks! Also, would it be right to think of the question like this?
Since the first equation is f(x) yeah, the second is f(3x+1), meaning it has a dilation of 1/3 parallel to the x axis, meaning the graph is basically condensed by a factor of 1/3. I'm assuming the dilation would affect the value of the area calculated, and hence to account for the dilation we would simply multiply the first area by 1/3. The translation wouldn't have any affect on the area calculated yeah?

This is correct, EXCEPT that a translation would only have no affect on the area IF it is a horizontal translation. A vertical translation will affect the area, compare y=x to y=x+1 (or even y=x+c).

[soNasty]

Ok great, thanks for that

[keltingmeith]

Sorry, let me rephrase:

A horizontal translation won't affect the area IF AND ONLY IF the terminals move along with it. (just realised I wasn't really clear about this, sorry)

For example, if you have f(x+1), the terminals of the definite integral must move along with it in the appropriate direction.

[lzxnl]

Hey thanks! Also, would it be right to think of the question like this?
Since the first equation is f(x) yeah, the second is f(3x+1), meaning it has a dilation of 1/3 parallel to the x axis, meaning the graph is basically condensed by a factor of 1/3. I'm assuming the dilation would affect the value of the area calculated, and hence to account for the dilation we would simply multiply the first area by 1/3. The translation wouldn't have any affect on the area calculated yeah?

I think you're meant to think of it this way in Methods; dilation of factor 1/3 from the y axis (or parallel to the x axis; I don't like that terminology tbh) would shrink the area and translating the entire graph to the left or right while moving the terminals with it would have no effect on the area.

It's easy to see that vertical translations affect area because if you're integrating f(x) + c, you can move the +c to a separate integral and integrating a non-zero constant over any non-trivial interval won't give you zero.

[Brunette15]

Hi everyone!
Can someone please help me with 8b from the following question?