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 for $\forall x\in R$ 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. None of these

Answer 1

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

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