Suppose a differentiable function


Suppose a differentiable function $\mathrm{f}(\mathrm{x})$ satisfies the identity $f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y$,

for all real $x$ and $y$. If $\operatorname{Lim}_{x \rightarrow 0} \frac{f(x)}{x}=1$, then $f^{\prime}(3)$

is equal to




Since, $\lim _{x \rightarrow 0} \frac{f(x)}{x}$ exist $\Rightarrow f(0)=0$

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

$=\lim _{h \rightarrow 0} \frac{f(h)+x h^{2}+x^{2} h}{h}($ take $y=h)$

$=\lim _{h \rightarrow 0} \frac{f(h)}{h}+\lim _{h \rightarrow 0}(x h)+x^{2}$

$\Rightarrow f^{\prime}(x)=1+0+x^{2}$

$\Rightarrow f^{\prime}(3)=10$


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