pi's Specialist Maths Questions

[Mao]

1. More of a query
If we have , and the answer is , do we need to include the line: x is element of R\{0, -2} in the answer? The textbook seems to, and I wonder if its necessary.

Yes

2. More of a query
When do we have to have the modulus in the log when antidifferentiating functions? MQ seems to randomly remove it from some answers (eg. ). Bit confused here...

An antiderivative should be defined on the domain which the integrand is finite and continuous on. So if the integrand is defined for , then the antiderivative should similarly be defined on the same interval, thus the modulus.

[pi]

Thanks Mao!

1. More of a query
If we have , and the answer is , do we need to include the line: x is element of R\{0, -2} in the answer? The textbook seems to, and I wonder if its necessary.

Yes

Still confused here.

Isn't it inferred in the equation that it doesn't exist for those values? We don't write where x = R\{0} ? (or maybe that's my error). Should we ALWAYS be writing the domain?

Confused

[Mao]

Thanks Mao!

1. More of a query
If we have , and the answer is , do we need to include the line: x is element of R\{0, -2} in the answer? The textbook seems to, and I wonder if its necessary.

Yes

Still confused here.

Isn't it inferred in the equation that it doesn't exist for those values? We don't write where x = R\{0} ? (or maybe that's my error). Should we ALWAYS be writing the domain?

Confused

It is the implied domain, but it's always nicer to be explicit.

[pi]

OK, better to be safe than sorry. Thanks again Mao!

[pi]

Couple more questions, this time on DEs

1. Could someone show me how to set out a DE verification problem, to get full marks on the exam? I get the two sides to equal, but I think my working may look a bit clumsy. How about for this simple one:
           Verify the given function is a solution to the DE:

Just a method of working steps would be awesome



2. Is there anyway of doing this other than 'guess and check', and if not, is there an efficient way to guess check (assuming calc-free):

A
B
C
D
E



THANKS 

[TrueTears]

2. Is there anyway of doing this other than 'guess and check', and if not, is there an efficient way to guess check (assuming calc-free):

A
B
C
D
E

DE's are boring but yeah anyways the DE is a homogeneous second-order linear differential equation. If you really want to know how to solve these, then read what's attached.

[pi]

This is actually a methods question that I got wrong from the last SAC

The functions f and g are defined by:


Find the values of x for which (fog)(x) (gof)(x)

I got the solution , but I can't get


Help please

[TrueTears]

This is actually a methods question that I got wrong from the last SAC

The functions f and g are defined by:


Find the values of x for which (fog)(x) (gof)(x)

I got the solution , but I can't get


Help please

Now just split into cases, either and or and

I'm guessing you got because you did this:










My question to you is, think about why that doesn't yield all the solutions


EDIT: WDF? this new latex is fucking gay shit, not only is the font gay as but it doesnt even code "\implies" properly.

[pi]

My question to you is, think about why that doesn't yield all the solutions

Is it due to the fact that  x {-1, 0} from the denominator, and hence x can exist between these asymptotes?

I'm not really sure tbh

[TrueTears]

My question to you is, think about why that doesn't yield all the solutions

Is it due to the fact that  x {-1, 0} from the denominator, and hence x can exist between these asymptotes?

I'm not really sure tbh

Good guess, but it's something simpler!

Think about what you are doing precisely when you "cross-multiplied" to arrive at 4x =>x+1

First you would compute 2(2x) and then you would compute 1(x+1), but who says (2x) is positive and (x+1) is positive? Thus we can not cross multiply in this case since it could change the sign of the inequality!

[pi]

Is it due to the fact that  x {-1, 0} from the denominator, and hence x can exist between these asymptotes?

I'm not really sure tbh

Good guess, but it's something simpler!

Think about what you are doing precisely when you "cross-multiplied" to arrive at 4x =>x+1

First you would compute 2(2x) and then you would compute 1(x+1), but who says (2x) is positive and (x+1) is positive? Thus we can not cross multiply in this case since it could change the sign of the inequality!

Oh. So does that mean I'll have to solve with the equations you outlined earlier instead:

Now just split into cases, either and or and

Simultaneously? Or using quick sketch-graphs for the quadratics and finding the intersections of the values for each pair?

I'm still confused on how the inequality sign changed.

[TrueTears]

Is it due to the fact that  x {-1, 0} from the denominator, and hence x can exist between these asymptotes?

I'm not really sure tbh

Good guess, but it's something simpler!

Think about what you are doing precisely when you "cross-multiplied" to arrive at 4x =>x+1

First you would compute 2(2x) and then you would compute 1(x+1), but who says (2x) is positive and (x+1) is positive? Thus we can not cross multiply in this case since it could change the sign of the inequality!

Oh. So does that mean I'll have to solve with the equations you outlined earlier instead:

Now just split into cases, either and or and

Simultaneously? Or using quick sketch-graphs for the quadratics and finding the intersections of the values for each pair?

I'm still confused on how the inequality sign changed.

U mean how did the inequality sign change in my working? It didn't change, i just split it into cases. It's the exact same concept as solving quadratic inequalities, but here we are solving rational function inequalities, to see how to solve them check here http://vce.atarnotes.com/agora/index.php/topic,39795.msg416590.html#msg416590


With regards to your question, yeah you have to solve my inequalities to yield the correct answer, but besides knowing what inequalities to solve, the more important thing is to realise why you can't solve it the way outlined earlier.

[pi]

What I meant was the sign change in -1<x<0, in the equations you have (and I had, although I did it with the positive assumption, which I see is clearly wrong) they were . I'm not sure on the algebraic way to get this. I got it graphically (kinda), but algebraically, I'm just not sure.



OH, I GET IT NOW! GREAT LINK

(i think I had them on my desktop though, next time I'll d/l + READ )

[pi]

Quick q, hyperbolic trig functions (and their inverses) are definitely NOT on the course right?


(our SACs think otherwise , gotta ask)

[b^3]

As far as I know, no they are not, and we haven't covered them either.