Let f:R→R be a function such that f(x+y)=f(x)+f(y)+x2y+xy2∀x,y∈R. If limx→0f(x)x=1, then f(x) is
A
1+x22
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B
x−x23
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C
x+x33
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D
1+x23
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Solution
The correct option is Cx+x33 We know that, f′(x)=limh→0f(x+h)−f(x)h=limh→0f(x)+f(h)+x2h+xh2−f(x)h=limh→0(f(h)h+x2+xh)=1+x2+0(∵limx→0f(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)=0⇒c=0∴f(x)=x+x33