Question

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

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

Answer 1

Answer: (B)

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