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pi's Specialist Maths Questions

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TrueTears:

--- Quote from: Rohitpi on July 21, 2011, 08:37:39 pm ---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

--- End quote ---
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:

--- Quote from: Rohitpi on July 25, 2011, 09:04:03 am ---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


--- End quote ---




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:

--- Quote from: TrueTears on July 25, 2011, 03:15:06 pm ---My question to you is, think about why that doesn't yield all the solutions ;)

--- End quote ---

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:

--- Quote from: Rohitpi on July 25, 2011, 07:07:42 pm ---
--- Quote from: TrueTears on July 25, 2011, 03:15:06 pm ---My question to you is, think about why that doesn't yield all the solutions ;)

--- End quote ---

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 :(

--- End quote ---
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!

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