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

Answer 1

The correct option is B $\log a+(\alpha +\beta )x-\left(\frac{{\alpha }^2+{\beta }^2}{2}\right)\cdot x^2+\left(\frac{{\alpha }^3+{\beta }^3}{3}\right)x^3$

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