Question

if f be a decreasing continuous function satisfying f(x+y) = f(x) + f(y) - f(x)f(y)
f'(0)=-1

Answer 1

then \[\int\limits_{0}^{1}f(x)dx = ?\]

Answer 2

@hartnn @ganeshie8

Answer 3

@wio

Answer 4

any ideas?

Answer 5

hmm im trying
but im getting nothing :o

Answer 6

I got
\[f(x) = 1 - e^x\]

Answer 7

how !

Answer 8

First lemme confirm the answer.
@random231 is the final answer (-e) ??

Answer 9

no

Answer 10

:(

Answer 11

but u did it right
ur integration came wrong
watch the signs

Answer 12

ans is 2-e

Answer 13

Auh..okay..my bad XD

Answer 14

Partially differentiate the given equation w.r.t x
f'(x+y) = f'(x) [ 1 - f(y)]

put x = 0 and f'(x) = -1
f'(y) = f(y) - 1

Then it is a simple diff equation.. :)

Answer 15

Solution -
\[f(x) = Ce^x + 1\]
implies
f'(0) = C = -1

Hence,
\[f(x) = 1 - e^x\]

Answer 16

oh ok the tricky part was the partial differentiation!

Answer 17

Yep :)

Answer 18

but on wat basis did u hav the idea of using partial diff

Answer 19

& I need to work on my arithmetic.. :(

Answer 20

Experience..I took a drop..so had seen such Q's before ;)

Answer 21

@random231 They gave you the value of the derivative at \(0\), that that is the basis for finding a derivative.

Answer 22

hmm getting it, thanks wio!

Answer 23

In addition, this property f(x+y) = f(x) + f(y) - f(x)f(y) is inherently to some sort of exponential functions.

Answer 24

Since \[
e^{x+y}=e^xe^y
\]after all.

Answer 25

oh dear i didnt notice that thanks wio!!! and lastday!!!

Answer 26

Yea..that makes sense..
I never saw the incentive to use Partial differentiation..just knew I had to use it XD

Answer 27

okay