Maths, Addmaths and Statistics: Post your doubts here!

[Ram97]

just asking..did ur comp pprs got cancelled or wht?

i dont have comp

 

[M.Omar]

no the equation

2^x(2^x-1)=0
was the equation frm which we wud later on tell why no sol...wht did u write for tht

 

[Dark Destination]

gf(x) kia aaraha tha???

2e^2x + 2e^x

 

[M.Omar]

What was it?

I wrote plus infinity to minus infinity, as the range.
And x> 1 as the domain.

as transposed
range was f^-1>2
and domain x>3
just opposite values of fx

 

[Dark Destination]

as transposed
range was f^-1>2
and domain x>3
just opposite values of fx

I don't think so. It was a logarithmic function, doesn't work that way.

 

[Ram97]

2e^2x + 2e^x

ye e kahan se aya????

 

[M.Omar]

I don't think so. It was a logarithmic function, doesn't work that way.

simply sketch its graph...mind map and whtever and see tht infact values r transposed..check the past pprs do mind map check the book even u will see tht they r transposed values..

 

[M.Omar]

ye e kahan se aya????

tht,s wht I was thinking and yeah abt the equation
2^x(2^x-1)=0
wht did u write abt being no solutions...for me have already told u

 

[Dark Destination]

What does this mean? What meeting?

 

[Dark Destination]

simply sketch its graph...mind map and whtever and see tht infact values r transposed..check the past pprs do mind map check the book even u will see tht they r transposed values..

You should google ''how to find domain and range of logarithmic functions''..

 

[M.Omar]

What does this meaning? What meeting?

View attachment 45314

tht,s wht I am saying abt xtra sol which u put up tht they penalize as it violates accuracy mark..smtimes during checking they might see a general trend or sth or doubts of their own regarding xtra sol and so they hold meetings e.t.c

 

[Dark Destination]

tht,s wht I am saying abt xtra sol which u put up tht they penalize as it violates accuracy mark..smtimes during checking they might see a general trend or sth or doubts of their own regarding xtra sol and so they hold meetings e.t.c

In a trignometry question, the range is there to make sure than ALL ANSWERS in that specific range are present.
So if an extra answer is there, out of the range, they will deduct a whole mark?

 

[M.Omar]

You should google ''how to find domain and range of logarithmic functions''..

Oh ho I know wht ur talking abt ...have read it in my add maths book ...but for this graph u had to check its initial domain for the orginal fx to be defined and thus for its inverse to be defined as well I know..these were the extra restrictions
lg(x+1)/lg2 so x>-1 is think wht u wrote and this wrong due to the initial domain and range of org fx

 

[Ram97]

tht,s wht I was thinking and yeah abt the equation
2^x(2^x-1)=0
wht did u write abt being no solutions...for me have already told u

i solved the equation further so it became 2^2x - 2^x
Maybe i could be wrong
So i wrote the bases cannot get cancelled as they are not being multiplied or divided
I guess i was wrong :/

 

[Dark Destination]

Oh ho I know wht ur talking abt ...have read it in my add maths book ...but for this graph u had to check its initial domain for the orginal fx to be defined and thus for its inverse to be defined as well I know..these were the extra restrictions
lg(x+1)/lg2 so x>-1 is think wht u wrote and this wrong due to the initial domain and range of org fx

So the domain of f. That was your range of f inverse?

And the range of f that was >3 was your domain of f inverse?

 

[M.Omar]

i solved the equation further so it became 2^2x - 2^x
Maybe i could be wrong
So i wrote the bases cannot get cancelled as they are not being multiplied or divided
I guess i was wrong :/

since they were already sort of factorized then it wud simply be 2^x=0 not possible and 2^x-1=0 i.e x=o which was not in domain so hence no real sol

 

[M.Omar]

So the domain of f. That was your range of f inverse?

And the range of f that was >3 was your domain of f inverse?

x>3 and f^-1>2
tht,s it

 

[Dark Destination]

since they were already sort of factorized then it wud simply be 2^x=0 not possible and 2^x-1=0 i.e x=o which was not in domain so hence no real sol

But that wasn't even factorized.
It was 2e^2x + 2e^x

 

[Dark Destination]

x>3 and f^-1>2
tht,s it

But Oct-Nov 2007 paper 2 last question? :/

 

[Dark Destination]

since they were already sort of factorized then it wud simply be 2^x=0 not possible and 2^x-1=0 i.e x=o which was not in domain so hence no real sol

f(x) = 2^x -1

g(x) = x(x+1)

Make an expression for gf(x)...

It's not what you are saying.