IIT JEE , AIEEE Forum - Mathematics , Algebra

deedee

f(x+y)=f(x)+f(y)

then f(x)=1+x^n

then

f(x*y)=f(x)+f(y)

then f(x)=??

and

wen f(x+y)=f(x)*f(y)

then f(x)=??

plezz help soon!!

akhil_o

1) f(x)=log x
2)
g(x)=a^x

kishore.subramanian.b

If

f(x+y)=f(x)+f(y) then f(x)=mx

but u have written f(x)=1+x^n

Which of these is right????

sboosy

f(x+y) = f(x) + f(y)

if this is true for all x and y ..then

f(0+0) = f(0)+f(0)

f(0) = 2f(0)

=> f(0) = 0

also

f(2) = f(1+1) = f(1)+f(1) = 2f(1)

f(3) = f(1+2) = f(1)+f(2) = f(1)+f(1)+f(1) = 3f(1)

....

for integers we find

f(n) = nf(1) ............1

f(2.5) = f(1+1.5) = f(1) + f(1.5) = f(1)+f(1) + f(0.5)

thus it can be shown that for non integers and in general also

f(x) = [x] f(1) +f( {x} )

where [ ] represents greatest integer function and { } represents fractional part of x

and the answer is not 1+xⁿ as written by the author

now consider

f(x) *f(1/x) = f(x) +f(1/x)

then in this case f(x) = 1+(or)- xⁿ

konichiwa2x

1.

Putting

in

we get,

.

Assuming,

.

Thus,

Thus by induction,

[

]

The answer you provided is wrong.

I think you probably got confused with

as sboosy pointed out.

edit : in red

konichiwa2x

2.

Put y = 0 to obtain,

=>

.

Putting

in

,

konichiwa2x

3.

Putting y = 1, we get,

=>

(1)

Putting

we get,

Similarly,

Thus,

(2) (since f(1) = 0 from (1))

From

and

,

.

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