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Let $\alpha $ and $\beta $ are roots of $5{x^2} - 3x - 1 = 0$ , then $\left[ {\left( {\alpha + \beta } \right)x - \left( {\frac{{{\alpha ^2} + {\beta ^2}}}{2}} \right){x^2} + \left( {\frac{{{\alpha ^3} + {\beta ^3}}}{3}} \right){x^3} -......} \right]$ is
$x^2 + 3x -5$
$x^2 -3x -5$
$-x^2 + 3x + 5$
none of these
Solution
$\alpha + \beta = \frac{3}{5}$
${\alpha \beta} {=-\frac{1}{5}}$
$\left[ {(\alpha + \beta )x – \left( {\frac{{{\alpha ^2} + {\beta ^2}}}{2}} \right){x^2} + \left( {\frac{{{\alpha ^3} + {\beta ^3}}}{3}} \right){x^3} \cdots \cdots \cdots } \right]$
$=\left(\alpha \mathrm{x}-\frac{(\alpha \mathrm{x})^{2}}{2}+\frac{(\alpha \mathrm{x})^{3}}{3} \ldots \ldots \ldots\right)+\left(\beta \mathrm{x}-\frac{(\beta \mathrm{x})^{2}}{2}+\frac{(\beta \mathrm{x})^{3}}{3} \ldots \ldots\right)$
$=\ln (1-\alpha x)+ln(1-\beta x)$
$=\ln ((1-\alpha x)(1-\beta x))$
$ = l{\rm{n}}\left( {1 – {\rm{x}}(\alpha + \beta ) + \alpha \beta {{\rm{x}}^2}} \right)$
$=\ln \left(1-x \frac{3}{5}-\frac{1}{5} x^{2}\right)$
