lf $\mathrm{f}(\mathrm{x})$ is a quadratic expression which is positive for all real vaues of $\mathrm{x}$ and $\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}'(\mathrm{x})+\mathrm{f}''(\mathrm{x})$ then for any real value of $\mathrm{x}$

Let $f(x) =ax^2+bx+c$

According to the given condition $a>0, b^2-4ac <0$.........(i)

$\therefore g(x) = ax^2+bx+c+2ax+b+2a=ax^2+(b+2a)x+(b+c+2a)$

Now discriminant of $g(x)$ is $D=(b+2a)^2-4a(2a+c+b)=b^2+4a^2+4ab-4ab-4ac-8a^2=(b^2-4ac)-4a^2< 0$ using (i)

Hence $g(x)> 0$ for any $x \in R$