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

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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