Mathematics - Quadratic Equation Question with Solution | TestHub

Let $\mathrm{f}(\mathrm{x})={\mathrm{a}}_0{\mathrm{x}}^{\mathrm{n}}+{\mathrm{a}}_1{\mathrm{x}}^{\mathrm{n}-1}+\dots .+{\mathrm{a}}_{\mathrm{n}}$

Then $x^{n}f\left(\frac{1}{x}\right)=x^n\left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\dots .+a_n\right)$

$=a_0+a_{1}x+\dots .+a_n{x}^n$

Since, $\mathrm{f}(\mathrm{x})={\mathrm{x}}^{\mathrm{n}}\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)$,

$\therefore {\mathrm{a}}_0={\mathrm{a}}_{\mathrm{n}},{\mathrm{a}}_1={\mathrm{a}}_{\mathrm{n}-1},\dots ,{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_0$

$\therefore \mathrm{f}(\mathrm{x})=0$ is a reciprocal equation of first type.