Aditya Gupta
Last Activity: 4 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
