Question

Let $f$ be a differentiable function such that $f(x+y)=f\left(x\right)+f\left(y\right)+2xy-1$ for all real $x$ and $y$. If $f^{\prime }\left(0\right)=\cos \alpha$, then $\forall x\in R$

• A. $f\left(x\right)=0$
• B. $f\left(x\right)<0$
• C. $f\left(x\right)>0$
• D. $f\left(x\right)=x$

Answer 1

The correct option is C $f\left(x\right)>0$

We have, $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+2hx-1-1f\left(x\right)}{h}=\underset{h\to 0}{lim}(2x+\frac{f\left(h\right)-1}{h})$

Now substituting $x=y=0$ in the given functional relation, we get,

$f\left(0\right)=f\left(0\right)+f\left(0\right)+0-1\Rightarrow f\left(0\right)=1$

$\therefore f^{\prime }\left(x\right)=2x+\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=2x+f^{\prime }\left(0\right)\Rightarrow f^{\prime }\left(x\right)=2x+\cos \alpha$

Integrating, $f\left(x\right)=x^2+x\cos \alpha +C$

Here, $x=0$ and $f\left(0\right)=1$

$\therefore 1=C\Rightarrow f\left(x\right)=x^2+x\cos \alpha +1$

It is quadratic in $x$ with discriminant

$D={\cos }^2\alpha -4<0$

and coefficients of $x^2=1>0$

$\therefore f\left(x\right)>0\forall x\in R$