Let $f$ be a function such that $f(x) \cdot f(y)+2=f(x)+f(y)+f(x y), \forall x, y \in R-\{0\}$ and $

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Let $f$ be a function such that $f(x) \cdot f(y)+2=f(x)+f(y)+f(x y), \forall x, y \in R-\{0\}$ and $f(0)=1, f^{\prime}(1)=2$, then:

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We have, $f(x)\cdot f(y)+2=f(x)+f(y)+f(xy)...(i)$ On replacing $x,y\rightarrow1$ we get $(f(1))^{2}+2=3f(1)$ $\Rightarrow f^{2}(1)-3f(1)+2=0\Rightarrow f(1)=2$ or 1 But $f(1)$ cannot be equal to one as $f(0)=1\Rightarrow f(1)=2...(ii)$ On replacing y by $1/x$ in Eq. (i) we get $f(x)\cdot f(1/x)+2=f(x)+f(1/x)+f(1)$ $\Rightarrow f(x)\cdot f(1/x)+2=f(x)+f(1/x)+2$ f(1/x) ? f(x)= and f (1/x)= f(x) ...(iii) f (1/x)-1 f(x)-1 [using Eq. (ii) Now, $f^{\prime}(x)=lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=lim_{h\rightarrow0}\frac{f(x+h)+\frac{f(1/x)}{1-f(1/x)}}{h}$ $=lim_{h\rightarrow0}\frac{f(x+h)-f(x+h)\cdot f(1/x)+f(1/x)}{h\{1-f(1/x)\}}$ $=lim_{h\rightarrow0}\frac{f(x+h)-f(x+h)-f(1/x)-f(\frac{x+h}{x})+2+f(1/x)}{h[1-f(1/x)\}}[usingEq.(i)]$ $=lim_{h\rightarrow0}\frac{f(1+\frac{h}{x})-2}{h(f(1/x)-1)}=lim_{h\rightarrow0}\frac{f(1+h/x)-f(1)}{\frac{h}{x}x(f(1/x)-1)}$ [usingEq.(ii)] $=\frac{f^{\prime}(0)}{x(f(1/x)-1)}[:fom~eq,(iii),f(x):f(\frac{1}{x})=\frac{f(x)f(\frac{1}{x})}{[f(\frac{1}{x})-1\}(f(x)-1)}]$ $f^{\prime}(x)=\frac{2\{f(x)-1\}}{x}\Rightarrow xf^{\prime}(x)=2\{f(x)-1\}$ On integrating both the sides of the above expression, we get $x\hat{f}(x)-\int f(x)dx=2\int f(x)dx-2x+\lambda$ Where is the constant of integral $\Rightarrow3\int f(x)dx=x\{2+f(x)\}-\lambda$ Hence, $3\int f(x)dx-x\{2+f(x)\}=\lambda$ (constant)