IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

joyfrancis

**edited...

(assuming the function to be differentiable everywhere)

f'(x) = _{[h]_[0]}(f(x+h) - f(x)) / h

= _{[h]_[0]}(f(x) + f(h) + xh - f(x))/h

= _{[h]_[0]}(f(h)/h) + x........(1)

By putting a=b=0 in the given functional relation we get

f(0)=2f(0)

.: f(0)=0

so f(h)/h is 0/0 form..so by l hosp rule

_{[h]_[0]}(f(h)/h) = _{[h]_[0]}f'(h)

= f'(0)

so f'(x) = f'(0) + x.......(by 1)

.: f(x) = f'(0)x + x²/2 + c

since f(0)=0...c=0

.: f(x) = f'(0)x + x²/2

= kx + x²/2