Question

If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + c =0$, then $log (a - bx + cx^2)$ is equal to

• A. $log a + \left(\alpha+\beta\right)x +\frac{\alpha^{2} +\beta^{2}}{2}x^{2} +\frac{\alpha^{3}+\beta^{3}}{3} x^{3}+...$
• B. $log a + \left(\alpha+\beta\right)x -\left(\frac{\alpha^{2} +\beta^{2}}{2}\right)\cdot x^{2} +\left(\frac{\alpha^{3}+\beta^{3}}{3}\right) x^{3}-...$
• C. $log a - \left(\alpha+\beta\right)x -\left(\frac{\alpha^{2} +\beta^{2}}{2}\right)x^{2} -\left(\frac{\alpha^{3}+\beta^{3}}{3}\right) x^{3}-...$
• D. None of the above

Answer 1

Answer: (B)

Since, $\alpha, \beta$ are roots of the equation $a x^{2}+b x+c=0$, we have
$\therefore \alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
$\therefore a-b x+c x^{2}=a\left(1-\frac{b}{a} x+\frac{c}{a} x^{2}\right)$
$=a\left\{1+(\alpha+\beta) x+\alpha \beta x^{2}\right\}=a\{(1+\alpha x)(1+\beta x)\}$
Hence, $\log \left(a-b x+c x^{2}\right)$
$=\log \{a(1+\alpha x)(1+\beta x)\}$
$=\log a+\log (1+\alpha x)+\log (1+\beta x)$
$=\log a+\left(\alpha x-\frac{(\alpha x)^{2}}{2}+\frac{(\alpha x)^{3}}{3}-\ldots\right)$
$+\left(\beta x-\frac{(\beta x)^{2}}{2}+\frac{(\beta x)^{3}}{3}-\ldots\right)$
$=\log a+(\alpha+\beta) x-\left(\frac{\alpha^{2}+\beta^{2}}{2}\right) x^{2}$
$+\left(\frac{\alpha^{3}+\beta^{3}}{3}\right) x^{3}-\ldots$