Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+px+q=0$and the sum$(\alpha +\beta )x-\frac{({\alpha }^2+{\beta }^2)}{2}{x}^2+\frac{({\alpha }^3+{\beta }^3)}{3}{x}^3-\dots ....$ exists, then it is

• A. $\log (x^2+px+q)$
• B. $\log (x^2–px+q)$
• C. $\log (1+px+q{x}^2)$
• D. $\log (1–px+q{x}^2)$

Answer 1

The correct option is D

$\log (1–px+q{x}^2)$

Explanation for the correct option:

Find the sum $\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{\beta }\mathbf{)}\mathit{x}\mathbf{}\mathbf{-}\frac{\mathbf{(}{\mathbf{\alpha }}^{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}^{\mathbf{2}}\mathbf{)}}{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{(}{\mathbf{\alpha }}^{\mathbf{3}}\mathbf{+}{\mathbf{\beta }}^{\mathbf{3}}\mathbf{)}}{\mathbf{3}}{\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$ :

Given that $\alpha$ and $\beta$ are the roots of the equation $x^2+px+q=0$.

$\Rightarrow (\alpha +\beta )=-p;\alpha .\beta =q$

and also given the sum $(\alpha +\beta )x-\frac{({\alpha }^2+{\beta }^2)}{2}{x}^2+\frac{({\alpha }^3+{\beta }^3)}{3}{x}^3-\dots ....$exists.

$$\begin{gathered} (\alpha +\beta )x-\frac{({\alpha }^2+{\beta }^2)}{2}{x}^2+\frac{({\alpha }^3+{\beta }^3)}{3}{x}^3-\dots .... \\ =\left(\alpha x-\frac{{\alpha }^2}{2}{x}^2+\frac{{\alpha }^3}{3}{x}^3-.........\right)+\left(\beta x-\frac{{\beta }^2}{2}{x}^2+\frac{{\beta }^3}{3}{x}^3-.........\right) \\ =\log (1+\alpha x)+\log (1+\beta x) \\ =\log (1+x(\alpha +\beta )+\alpha \beta x^2) \\ =\log (1-px+q{x}^2) \end{gathered}$$

Hence, the correct option is D.