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Let f:R-R be a differential function satisfying f(x+y) + f(x).f(y) = f(xy+1). Also, f(0) = -1, f'(0)= f'(1)= 1. Then, a) f(1) = 0 b) f(2) = f'(2) c) f(1) ≠ 0 d) f(2) ≠ f'(2) Ans is (AB). Please explain how

Let f:R-R be a differential function satisfying
f(x+y) + f(x).f(y) = f(xy+1).
Also,  f(0) = -1, f'(0)= f'(1)= 1. Then, 
 
a) f(1) = 0
b) f(2) = f'(2)
c) f(1) ≠ 0
d) f(2) ≠ f'(2)
 
Ans is (AB). Please explain how  

Grade:12

1 Answers

Aditya Gupta
2081 Points
3 years ago
put x=0 and y=1 in given relation we get
f(1) – f(1)= f(1) so that f(1)= 0.
Now, f(x+y) – f(x) = f(xy+1) – f(x)*[f(y)+1]
or Lt y → 0 [f(x+y) – f(x)]/y = Lt y → 0 f(xy+1)/y – f(x)*Lt y → 0[f(y)+1]/y
or f’(x)= x*Lt y → 0 f(xy+1)/xy – f(x)*Lt y → 0[f(y) – f(0)]/(y – 0)
for Lt y → 0 f(xy+1)/xy, put t= xy, so it becomes Lt t → 0 f(t+1)/t = Lt t → 0 [f(1+t) – f(1)]/t = f’(1)= 1
so, we get  f’(x)= x*1 – f(x)*f’(0)= x – f(x)
let f(x)= z
so dz/dx + z= x
IF= e^x
so d(z.e^x)/dx = xe^x
or z.e^x= e^x.(x – 1) + C
when x=1, z=0, so C=0
So z.e^x= e^x.(x – 1)
or z= f(x)= x – 1.
So, f(2)= 1, f’(x)= 1 so f’(2)= 1.
or f(2) = f'(2) = 1.
So options A, B.
KINDLY APPROVE :D

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