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Grade: 11

                        

Please solve it. ____________________________________________

7 months ago

Answers : (3)

Arun
25333 Points
							
 
Dear student
 
 
put y = 0
then you will get 
 
f(0) ( 1 + f(x)) = 0
 
In case of any difficulty, Please feel free to ask again
 
7 months ago
Saurabh Koranglekar
askIITians Faculty
10233 Points
							Dear student

Assume y = 0 to simplify the given equation and determine required

Regards
7 months ago
Aditya Gupta
2069 Points
							
Dear student, above answer is wrong. Correct answer is:
On putting y=0, we see that f(0)(1+f(x))= 0
Since f(x) is non constant, f(0)= 0
Now, we can write
f(x+y)-f(x)= f(y)(1+f(x))
Or Lt y tends to 0 [f(x+y)-f(x)]/y = (1+f(x))*Lt y tends to 0 [f(y)-f(0)]/(y-0)
Or f'(x)= f'(0)(1+f(x))= k(1+f(x))
This differential equation can be easily solved by integrating as
∫f'(x)dx/(1+f(x))= ∫kdx
Or ln(1+f(x))= kx+c
Or f(x)= Ae^kx-1
Put x=0 to get f(0)= A-1
Or A=1
So, f(x)= e^kx - 1
KINDLY APPROVE :))
7 months ago
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