Maths, Addmaths and Statistics: Post your doubts here!

[Ram97]

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

No!!!

 

[Dark Destination]

No!!!

Lol, sorry. I put the e there by mistake.

It was 2^2x + 2^x

 

[Ram97]

Lol, sorry. I put the e there by mistake.

It was 2^2x + 2^x

Shukar!!!

 

[M.Omar]

f(x) = 2^x -1

g(x) = x(x+1)

Make an expression for gf(x)...

It's not what you are saying.

it wud be
(2^x-1)(2^x-1+1)=0
ab khud hi nikal lo ka kya eq
2^x-1=0 or 2^x=0
nahi banti?

 

[Ram97]

What were the answers to differentiation qs????

 

[Dark Destination]

it wud be
(2^x-1)(2^x-1+1)=0
ab khud hi nikal lo ka kya eq
2^x-1=0 or 2^x=0
nahi banti?

2^2x - 2^x = 0

2^2x = 2^x

2x = 2

Not possible. Since, x>2 so no solution

 

[M.Omar]

What were the answers to differentiation qs????

last one:
(cosx(1+x^1/2)x^1/2-sinx)/((1+x^1/2)^2*x^1/2) like tht

 

[M.Omar]

2^2x - 2^x = 0

2^2x = 2^x

2x = 2

Not possible. Since, x>2 so no solution

how does it become 2x=2?
shud n,t it be
2x=x and u cud work out further

 

[Dark Destination]

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

What do you think of this:



Marking scheme (oct-nov2007):

 

[Dark Destination]

how does it become 2x=2?
shud n,t it be
2x=x and u cud work out further

Yes, 2x =x which is not possible. I wrote this. Is this correct?

 

[M.Omar]

What do you think of this:

View attachment 45315

Marking scheme (oct-nov2007):

View attachment 45316

so exactly proves my point since in this question f^-1>0 tht is the original domain
for tht question x>2 not x>0
and hence range of f^-1>2 and so on and so tht...I think it wud be pretty clear now

 

[M.Omar]

Yes, 2x =x which is not possible. I wrote this. Is this correct?

wht if x=0
u shud have wrote abt x>2 and all tht
and also I think they wud reserve mark for the fact tht 2^x=0 is not possible because it was quite conspicuous when we made the equation

 

[Dark Destination]

wht if x=0
u shud have wrote abt x>2 and all tht
and also I think they wud reserve mark for the fact tht 2^x=0 is not possible because it was quite conspicuous when we made the equation

I Did write that. That it doesn't fit in the domain.. and x>2

2x = x is possible if x is zero. But zero isn't in the domain.. so.. no solution

 

[Ram97]

what was the integration answer where we had to integrate y then apply limits from 1.5 to 0

 

[M.Omar]

2x = x is possible if x is zero. But zero isn't in the domain.. so.. no solution[/quote]
then more probably 2 marks can be achieved

 

[M.Omar]

what was the integration answer where we had to integrate y then apply limits from 1.5 to 0

what was the integration answer where we had to integrate y then apply limits from 1.5 to 0

wht did u get? for me it was in fraction I guess...I,ll try to remember

 

[Ram97]

wht did u get? for me it was in fraction I guess...I,ll try to remember

i got 12.375 but ppl were saying it was 3.75

 

[Dark Destination]

so exactly proves my point since in this question f^-1>0 tht is the original domain
for tht question x>2 not x>0
and hence range of f^-1>2 and so on and so tht...I think it wud be pretty clear now

I don't understand :/

Why not that infinity thing?

 

[M.Omar]

i got 12.375 but ppl were saying it was 3.75

there were 2 inegrations in the whole question right?part I and 2
wht was ur integrated form

 

[M.Omar]

I don't understand :/

Why not that infinity thing?

because firstly it cud not be neg as range and domain of org function said so
secondly it was to be greater than 2 i.e domain of org f
and thus u cud write 2<f<infinity or simply f>2