Mathematics - Functions Question with Solution | TestHub

Differentiating both sides partially with respect to x , (keepiring y as constant), we get

$$f^{\prime }\left(\frac{x^2+y^2}{2}\right)\times \frac{1}{2}(2x)=\frac{1}{2}\left[2f(x)f^{\prime }(x)+0\right]\Rightarrow x{f}^{\prime }\left(\frac{x^2+y^2}{2}\right)=f(x)f^{\prime }(x)..\left(1\right)$$

Similarly differentiating both sides partially with respect to $y$ (keeping $x$ as constant) or interchanging $x$ and $y$, we get

$$y{f}^{\prime }\left(\frac{x^2+y^2}{2}\right)=f(y)f^{\prime }(y)b..\dots \left(2\right)$$

Now, dividing (2) by (1), we have $\frac{x}{y}=\frac{f(x)f^{\prime }(x)}{f(y)f^{\prime }(y)}$ $\Rightarrow \frac{\mathrm{f}(\mathrm{x}){\mathrm{f}}^{\prime }(\mathrm{x})}{\mathrm{x}}=\frac{\mathrm{f}(\mathrm{y}){\mathrm{f}}^{\prime }(\mathrm{y})}{\mathrm{y}}=\mathrm{k}\left(=\mathrm{f}(1){\mathrm{f}}^{\prime }(1)\right)$ say $\Rightarrow \mathrm{f}(\mathrm{x}){\mathrm{f}}^{\prime }(\mathrm{x})=\mathrm{kx}\Rightarrow \int \mathrm{f}(\mathrm{x}){\mathrm{f}}^{\prime }(\mathrm{x})\mathrm{dx}=\mathrm{k}\int \mathrm{xdx}$ $\Rightarrow \frac{\{\mathrm{f}(\mathrm{x}){\}}^2}{2}=\mathrm{k}\frac{{\mathrm{x}}^2}{2}+\mathrm{c}\Rightarrow \{\mathrm{f}(\mathrm{x}){\}}^2={\mathrm{kx}}^2+2\mathrm{c}$.

When $\mathrm{x}=0,\mathrm{y}=0,\mathrm{f}(0)=0$ or 1 .

$\Rightarrow \{\mathrm{f}(0){\}}^2=0+2\mathrm{c}\Rightarrow$ Either $0=2\mathrm{c}$ or $1=2\mathrm{c}$.

$\therefore \mathrm{c}=0$ or $\mathrm{c}=\frac{1}{2}$.

Hence $\mathrm{f}(\mathrm{x})=\pm {\mathrm{kx}}^2$ or $\pm \sqrt{{\mathrm{kx}}^2+1}$.