Question

# Consider a continuous function and differentiable function $$'f'$$ satisfies the functional equation    $$f(x)+f(y)=f(x+y+xy)+xy$$ for $$x,y>-1$$ and $$f'(0)=0$$,thenThe value of $$f(e-1)$$is

A
e2
B
e1
C
1e
D
2e

Solution

## The correct option is D $$2-e$$ $$f(x)+f(y)=f(x+y+xy)+xy$$  $$f(x)+f(y)=f(x+y(1+x))+xy$$ Put x=0 gives f(0) = 0also given $$f'(0)=0$$hence $$\lim_{x \rightarrow 0} \cfrac{f(x)}{x} = 0$$$$\cfrac{f(x +y(1+x)) - f(x)}{y(1+x)} = \cfrac{f(y) -xy}{y(1+x)}$$Putting limit $$y \rightarrow 0$$$$\lim_{y \rightarrow 0} \cfrac{f(x +y(1+x)) - f(x)}{y(1+x)} = \cfrac{\lim_{y \rightarrow 0} \cfrac{f(y)}{y} -x}{(1+x)}$$$$f'(x) = \cfrac{f'(0) -x}{(1+x)}$$$$f'(x) = \cfrac{-x}{(1+x)}$$$$\cfrac{df(x)}{dx} = \cfrac{f'(0) -x}{(1+x)}$$Integrating to get$$f(x) = - x +ln(1+x)$$Hence $$f(e-1)=2-e$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More