If f(x+y) = f(x) f(y) for all real x and y and f(0) is not equal to 0 , then the function g(x) = f(x) /[1+{f(x)}^2] is
a) odd function
b) even function
c) oddiff (x) >0
d) none of these
kindly explain step by step
let z =f(x)
then denominator always +
so look at numerator z
y = 0, f(y) is not 0, say it is -1
then f(x+0) = f(x)*-1 = -f(x)
but if when y = 0 ,f(y) = +2
then f(x+0) = f(x)*2
so the sign depends on what f(0) sign is
can u complete it please. I have a problem in this sum
i request ... because these type of questions are hardly to understand for me
consider f(z) = e^z. then e^(x+y) = e^x e^y
then g(x) = e^x/(1+e^2x)
is that even or odd, or neither?
Dont know its even or odd
i request
even if f(-x) = f(x)
odd if f(-x) = -f(x)
its even
let z =f(x)
then denominator is always positive.
Now, Numerator z
y = 0
f(y) is not 0, say it is -1
then f(x+0) = f(x)*-1 = -f(x)
but if when y = 0 ,f(y) = +2
then f(x+0) = f(x)*2
[even if f(-x) = f(x)
odd if f(-x) = -f(x)]
therefore, answer is even function
I have done kindly cross check please
let z =f(x)
then denominator is always positive.
Now, Numerator z
[consider f(z) = e^z. then e^(x+y) = e^x e^y
then g(x) = e^x/(1+e^2x)]
y = 0
f(y) is not 0, say it is -1
then f(x+0) = f(x)*-1 = -f(x)
but if when y = 0 ,f(y) = +2
then f(x+0) = f(x)*2
[even if f(-x) = f(x)
odd if f(-x) = -f(x)]
therefore, answer is even function
Have I done wrong ?
..because no one is answering