Question

Let $f(x+y)=f\left(x\right)+f\left(y\right)+2xy-1\forall x,y\in R$. If $f\left(x\right)$ is differentiable and $f^{\prime }\left(0\right)=\sin \varphi$, then

• A. $f\left(x\right)>0\forall x\in R$
• B. $f\left(x\right)<0\forall x\in R$
• C. $f\left(x\right)=\sin \varphi \forall x\in R$
• D. $f\left(x\right)$ is linear

Answer 1

The correct option is A $f\left(x\right)>0\forall x\in R$

Given, $f(x+y)=f\left(x\right)+f\left(y\right)+2xy-1$
Put $x=y=0\Rightarrow f\left(0\right)=2f\left(0\right)-1$
$\Rightarrow f\left(0\right)=1$
$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow {lim}_{h\to 0}\frac{f\left(x\right)+f\left(h\right)+2xh-1-f\left(x\right)}{h}$
$\Rightarrow 2x+{lim}_{h\to 0}\frac{f\left(h\right)-1}{h}$ $\Rightarrow 2x+f^{\prime }\left(0\right)=2x+\sin \varphi$
Integrating, we get $f\left(x\right)=x^2+x\sin \varphi +c$
$\because f\left(0\right)=1\Rightarrow 1=c$
$\therefore f\left(x\right)=x^2+x\sin \varphi +1>0\forall x\in R$