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Question

Let f:RR be a function such that f(x+y)=f(x)+f(y)+x2y+xy2 x,yR. If limx0f(x)x=1, then f(x) is

A
1+x22
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B
xx23
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C
x+x33
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D
1+x23
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Solution

The correct option is C x+x33
We know that,
f(x)=limh0f(x+h)f(x)h =limh0f(x)+f(h)+x2h+xh2f(x)h =limh0(f(h)h+x2+xh) =1+x2+0 (limx0f(x)x=1)f(x)=1+x2

Integrating w.r.t. x, we get
f(x)=x+x33+c

Given f(x+y)=f(x)+f(y)+x2y+xy2
Putting x=y=0, we get
f(0)=2f(0)f(0)=0c=0f(x)=x+x33

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