Mathematics - Functions Question with Solution | TestHub

Consider $\mathrm{f}(\mathrm{x})=1\pm {\mathrm{x}}^{\mathrm{n}}\Rightarrow (\mathrm{f}(\mathrm{x})-1)(\mathrm{f}(1/\mathrm{x})-1)=\left(\pm {\mathrm{x}}^{\mathrm{n}}\right)\left(\pm \frac{1}{{\mathrm{x}}^{\mathrm{n}}}\right)=1\Rightarrow f(x)\cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$. Consider $\mathrm{f}(\mathrm{x})=\frac{2}{1+\mathrm{k}\ln |\mathrm{x}|}$.

$$f(x)\cdot f\left(\frac{1}{x}\right)=\frac{2}{1+k\ln |x|}\times \frac{2}{1-k\ln |x|}=\frac{4}{1-k^2{\ln }^2|x|}$$

Consider $\mathrm{f}(\mathrm{x})=\frac{\pi }{2{\tan }^{-1}|\mathrm{x}|}$.

$$f(x)\cdot f\left(\frac{1}{x}\right)=\frac{\pi }{2{\tan }^{-1}|x|}\cdot \frac{\pi }{2{\tan }^{-1}\left|\frac{1}{x}\right|}=\frac{{\pi }^2}{4{\tan }^{-1}|x|\cdot {\cot }^{-1}|x|}$$

$$f(x)+f\left(\frac{1}{x}\right)=\frac{\pi }{2{\tan }^{-1}|x|}+\frac{\pi }{2{\tan }^{-1}\frac{1}{|x|}}=\frac{\pi }{2}\left(\frac{1}{{\tan }^{-1}|x|}+\frac{1}{{\cot }^{-1}|x|}\right)$$