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Question

If $$\displaystyle f:R\rightarrow R$$ be a differentiable function, such that $$\displaystyle f(x+2y)=f(x)+f(2y)+4xy\:\forall\:x,y\epsilon R,$$ then


A
f(1)=f(0)+1
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B
f(1)=f(0)1
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C
f(0)=f(1)+2
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D
f(0)=f(1)2
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Solution

The correct option is A $$\displaystyle f'(0)=f'(1)-2$$
$$\displaystyle f(x+2y)=f(x)+f(2y)+4xy\:\forall\:x,y\epsilon R,$$
$$ \cfrac{f(x+2y) -f(x)}{2y} = \cfrac{f(2y)}{2y} + 2x $$
$$ \lim _{2y \rightarrow 0} \cfrac{f(x+2y) -f(x)}{2y} = \lim _{2y \rightarrow 0} \cfrac{f(2y)}{2y} + 2x $$ 
If limit exits for all values of x and y, then $$ \lim _{2y \rightarrow 0} \cfrac{f(2y)}{2y} = k $$
Thus $$ f'(x) = k +2x $$
$$f'(1) -f'(0) = 2$$

Mathematics

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