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  • Suppose that is a differentiable function with the property that <img alt="\mathrm{f(x+y)=f(x)+f(y)+x

Suppose that \mathrm{f}is a differentiable function with the property that \mathrm{f(x+y)=f(x)+f(y)+x y\: and \: \lim _{h \rightarrow 0} \frac{1}{h} f(h)=3}, then

Option: 1

\mathrm{f} is a linear function


Option: 2

\mathrm{f(x)=3 x+x^2}


Option: 3

\mathrm{f(x)=3 x+\frac{x^2}{2}}


Option: 4

none of these

 


Answers (1)

best_answer

\mathrm{f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x)+f(h)+x h-f(x)}{h}}
                                                         \mathrm{=\lim _{h \rightarrow 0} \frac{1}{h} f(h)+x=3+x},

Hence \mathrm{f(x)=3 x+\frac{x^2}{2}+c}

Putting \mathrm{x=y=0} in the given equation, we have
 \mathrm{f(0)=f(0+0)=f(0)+f(0)+0 \Rightarrow f(0)=0}

Thus \mathrm{c=0} and \mathrm{f(x)=3 x+\frac{x^2}{2}}

Posted by

sudhir.kumar

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