Statement-1: Let $f$ be a differentiable function satisfying $f(x+y)=f(x)+f(y)+2 x y-1$ for all $x, y \in R$ and $f'(0)=a$, where $0<a<1$ then, $f(x)>0$ for all x.

Statement-2: $f(x)$ in statement -1 is of the form $x^2+a x+1$.

Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for Statement-1.

We have,

$f(x+y)=f(x)+f(y)+2 x y-1$  for all  $x, y \in R$

On replacing y by 0 , we get

$f(x)=f(x)+f(0)-1 \Rightarrow f(0)=1$

Now,

$f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} \frac{f(x)+f(h)+2 h x-1-f(x)}{h}$

$\Rightarrow f'(x)=\lim\limits_{h \rightarrow 0} 2 x+\lim\limits_{h \rightarrow 0} \frac{f(h)-1}{h}$

$\Rightarrow f'(x)=2 x+\lim\limits_{h \rightarrow 0} \frac{f(h)-f(0)}{h}$

$\Rightarrow f'(x)=2 x+f'(0) \Rightarrow f'(x)=2 x+a \Rightarrow f(x)=x^2+a x+b $

∴  $f(0)=1 \Rightarrow b=1$

Hence, f(x) = x² + ax + 1

So, statement - 2 is true.

Again, $f(x)=x^2+a x+1=\left(x+\frac{a}{2}\right)^2+\left(1-\frac{a^2}{4}\right)>0$ for all x

So, statement - 1 is true.