Question

Let $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{y}\right)+2\mathrm{xy}-1$ for all $\mathrm{x},\mathrm{y}\in \mathrm{R}$. If $\mathrm{f}\left(\mathrm{x}\right)$ is differentiable and $\mathrm{f}’\left(0\right)=\sqrt{3+\mathrm{a}-{\mathrm{a}}^2}$, then

• A. $\mathrm{f}\left(\mathrm{x}\right)>0\forall \mathrm{x}\in \mathrm{R}$
• B. $\mathrm{f}\left(\mathrm{x}\right)<0\forall \mathrm{x}\in \mathrm{R}$
• C. $\mathrm{f}\left(\mathrm{x}\right)=0\forall \mathrm{x}\in \mathrm{R}$
• D. $\mathrm{f}\left(\mathrm{x}\right)<{\mathrm{x}}^2\forall \mathrm{x}\in \mathrm{R}$

Answer 1

The correct option is A

$\mathrm{f}\left(\mathrm{x}\right)>0\forall \mathrm{x}\in \mathrm{R}$

Determining the correct option

Step 1: Determine the value of $f\left(0\right)$

Given, $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{y}\right)+2\mathrm{xy}-1$ and $\mathrm{f}’\left(0\right)=\sqrt{3+\mathrm{a}-{\mathrm{a}}^2}$

Put $\mathrm{x}=\mathrm{y}=0$ in given function $\mathrm{f}(\mathrm{x}+\mathrm{y})$ we get,

$$\begin{gathered} \mathrm{f}\left(0\right)=\mathrm{f}\left(0\right)-\mathrm{f}\left(0\right)+0-1 \\ \Rightarrow \mathrm{f}\left(0\right)=-1 \end{gathered}$$

Step 2: differentiation of function

$$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}\left(1\right) \\ =\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{h}\right)+2\mathrm{xh}-1-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}} \\ =\underset{\mathrm{h}\to 0}{\lim }\frac{-\mathrm{f}\left(\mathrm{h}\right)+2\mathrm{xh}-1}{\mathrm{h}} \\ =2\mathrm{x}-\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}\left(\mathrm{h}\right)-1}{\mathrm{h}} \\ =2\mathrm{x}-\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(0+\mathrm{h})+\mathrm{f}\left(0\right)}{\mathrm{h}}\left(2\right) \end{gathered}$$

Now compare the negative term of the equation $\left(2\right)$ with equation $\left(1\right)$ we get,

$$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=2\mathrm{x}-\mathrm{f}\text{'}\left(0\right) \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=2\mathrm{x}-(\sqrt{3+\mathrm{a}-{\mathrm{a}}^2} \end{gathered}$$

Step 3: Integrate on both sides

On integrating we get,

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-\left(\sqrt{3+\mathrm{a}-{\mathrm{a}}^2}\right)\mathrm{x}+\mathrm{C}$

Now $\mathrm{f}\left(0\right)=-1$, hence:

$$\begin{gathered} -1=0-0+\mathrm{C} \\ \Rightarrow \mathrm{C}=-1(\mathrm{C}=\text{integration constant)} \end{gathered}$$

Hence, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-\left(\sqrt{3+\mathrm{a}-{\mathrm{a}}^2}\right)\mathrm{x}-1$. The discriminant of this quadratic function is given below,

$$\begin{gathered} \mathrm{D}={\mathrm{b}}^2-4\mathrm{ac} \\ \Rightarrow \mathrm{D}=\left(3+\mathrm{a}-{\mathrm{a}}^2\right)-4\times 1\times (-1) \\ \Rightarrow \mathrm{D}=7+\mathrm{a}-{\mathrm{a}}^2 \\ \Rightarrow \mathrm{D}=-\left[{\left(\mathrm{a}-\frac{1}{2}\right)}^2-\frac{29}{4}\right] \\ \Rightarrow \mathrm{D}<0 \end{gathered}$$

Therefore, $\mathrm{f}\left(\mathrm{x}\right)>0\forall \mathrm{x}\in \mathrm{R}$.

Hence, option A is the correct answer.