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  • A differentiable function f(x) satisfies the condition <img alt="f(x+y)=f(x)+f(y)+x y" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x&plus;y%29%3Df%28

A differentiable function f(x) satisfies the condition f(x+y)=f(x)+f(y)+x y and \ln _{h \rightarrow 0} \frac{1}{h} f(h)=3 then f is

Option: 1

Linear


Option: 2

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


Option: 3

f(x)=3 x+x^2


Option: 4

none of these


Answers (1)

best_answer

Since f is differentiable

\therefore \quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x)+f(h)+h x-f(x)}{h} \\
=\lim _{h \rightarrow 0} \frac{f(h)}{h}+x=3+x \\

Integrating  f(x)=3 x+\frac{x^2}{2}+k           ............(i)

Putting \mathrm{x}=0, \mathrm{y}=0 in the given relation

\mathrm{f}(0)=\mathrm{f}(0)+\mathrm{f}(0)+0 \quad \therefore f(0)=0

Now from (i) we have        \mathrm{f}(0)=0+\mathrm{k} \quad \therefore \mathrm{f}(0)=0

New from (i) we have \mathrm{f}(0)=0+\mathrm{k} \quad \therefore \mathrm{k}=0

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

Posted by

chirag

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