A) \[\log ({{x}^{2}}+px+q)\]
B) \[\log ({{x}^{2}}-px+q)\]
C) \[\log (1+px+q{{x}^{2}})\]
D) \[\log (1-px+q{{x}^{2}})\]
E) \[\log ({{x}^{2}}+qx+p)\]
Correct Answer: D
Solution :
\[\therefore \]\[\alpha \]and\[\beta \]are the roots of the equation\[{{x}^{2}}\text{+}px+q=0,\]then \[\alpha +\beta =-p\] and\[\alpha \beta =q\] Now, \[(\alpha +\beta )x-\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{2}{{x}^{2}}\] \[+\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{3}{{x}^{3}}-.....\] \[=\left( \alpha x-\frac{\alpha {{x}^{2}}}{2}+\frac{{{\alpha }^{3}}{{x}^{3}}}{3}-..... \right)\] \[+\left( \beta x-\frac{{{\beta }^{2}}{{x}^{2}}}{2}+\frac{{{\beta }^{2}}{{x}^{3}}}{3}-.... \right)\] \[=\log (1+\alpha x)+\log (1+\beta x)\] \[=\log \{1+(\alpha +\beta )x+\alpha \beta {{x}^{2}}\}\] \[=\log \{1+(-p)x+q{{x}^{2}}\}\] \[=\log (1-px+q{{x}^{2}})\]
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