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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,thenThe value of 1f′′(2) is

A
9
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B
9
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C
4
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D
4
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Solution

## The correct option is A −9 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)=0hence limx→0f(x)x=0f(x+y(1+x))−f(x)y(1+x)=f(y)−xyy(1+x)Putting limit y→0limy→0f(x+y(1+x))−f(x)y(1+x)=limy→0f(y)y−x(1+x)f′(x)=f′(0)−x(1+x)f′(x)=−x(1+x)Differentiate using product rulef′′(x)=−1(1+x)2f′′(2)=−19

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