Mathematics - Functions Question with Solution | TestHub

$\because \underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=1 \Rightarrow f\text{'}\left(0\right)=1$

$f\left(x+y\right)=f\left(x\right)+f\left(y\right)+x{y}^2+x^{2}y$

Differentiate w.r.t. $x$ keeping $y$ constant

$f\text{'}\left(x+y\right)=f\text{'}\left(x\right)+0+y^2+2xy$

put $y=-x$

$f\text{'}\left(0\right)=f\text{'}\left(x\right)+x^2-2x^2$

$1=f\text{'}\left(x\right)-x^2$

$f\text{'}\left(x\right)=1+x^2$

$f\text{'}\left(3\right)=10$.