Thank you for reporting, we will resolve it shortly
Q.
Let $f$ be a differentiable function such that $f ( x + y )= f ( x )+ f ( y )+2 xy -1$, for all real $x$ and $y$.
If $f ^{\prime}(0)=\cos \alpha$, then $\forall x \in R$.
Integrals
Solution:
$f(x+y)=f(x)+f(y)+2 x y-1$,
$[$ Put $x =0, y =0, $
$ \therefore f (0)=1]$
Partial differentiation w. r. to ' $x$ '
$f ^{\prime}( x + y )= f ^{\prime}( x )+2( y )$
Replace $y \rightarrow x, x \rightarrow 0$
$f ^{\prime}( x )= f ^{\prime}(0)+2 x $
$f ^{\prime}( x )=2 x +\cos \alpha$
Integral, $f(x)=x^2+x \cos \alpha+c$
$f (0)=1 \Rightarrow 1= c $
$f ( x )= x ^2+ x \cos x +1$
$D =\cos ^2 \alpha-4<0,\left(\text { coefficient of } x ^2>0\right. \text { ) } $
$\therefore f ( x )>0 .$