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No!!!But that wasn't even factorized.
It was 2e^2x + 2e^x
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No!!!But that wasn't even factorized.
It was 2e^2x + 2e^x
Shukar!!!Lol, sorry. I put the e there by mistake.
It was 2^2x + 2^x
it wud bef(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?
last one:What were the answers to differentiation qs????
how does it become 2x=2?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
so exactly proves my point since in this question f^-1>0 tht is the original domainWhat do you think of this:
View attachment 45315
Marking scheme (oct-nov2007):
View attachment 45316
wht if x=0Yes, 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
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 rememberwhat was the integration answer where we had to integrate y then apply limits from 1.5 to 0
i got 12.375 but ppl were saying it was 3.75wht did u get? for me it was in fraction I guess...I,ll try to remember
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
there were 2 inegrations in the whole question right?part I and 2i got 12.375 but ppl were saying it was 3.75
because firstly it cud not be neg as range and domain of org function said soI don't understand :/
Why not that infinity thing?
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