Q. If $\alpha$ and $\beta$ are the roots of the equation $x^{2} + px + q = 0$ and if the sum $(\alpha + \beta)x - \frac{\alpha^{2} + \beta^{2}}{2}.x^{2} + \frac{\alpha^{3} + \beta^{3}}{3}.x^{3} - ...$ exists then it is equal to
A) $\log{(x^{2} + px + q)}$
B) $\log{(x^{2} - px + q)}$
C) $\log{(1 + px + qx^{2})}$
D) $\log{(1 - px + qx^{2})}$
Explanation
(d)
Here, $\alpha + \beta = - p$ and $\alpha\beta = q$
Now, $(\alpha + \beta)x - \frac{\alpha^{2}{+ \beta}^{2}}{2}x^{2} + \frac{\alpha^{3} + \beta^{3}}{3}x^{3} - \ldots$
$= \left( \alpha x - \frac{\alpha^{2}x^{2}}{2} + \frac{\alpha^{3}x^{3}}{3} - ... \right) + \left( \beta x - \frac{\beta^{2}x^{2}}{2} + \frac{\beta^{3}x^{3}}{3} - ... \right)$
$= \log{(1 + \alpha x)} + \log{(1 + \beta x)}$
$= \log{\{ 1 + (\alpha + \beta)x + \alpha\beta x^{2}\}}$
$= \log{(1 - px + qx^{2})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.