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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 $$f'(0)=0$$,then
The value of $$f(e-1)$$is


A
e2
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B
e1
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C
1e
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D
2e
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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) = 0
also 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

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