A) \[\frac{1}{4}(a+b+c+d)\]
B) \[\frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}\]
C) \[\frac{a}{2}-\frac{b}{2}+\frac{c}{3}-\frac{d}{4}\]
D) None of these
Correct Answer: B
Solution :
\[\alpha ,\beta \] are the roots of the equation \[6{{x}^{2}}-6x+1=0\] \[\Rightarrow \alpha +\beta =1,\,\,\alpha \beta =1/6\] \[\therefore \,\,\frac{1}{2}\left[ \,a+b\alpha +c{{\alpha }^{2}}+d{{\alpha }^{3}} \right]+\frac{1}{2}\left[ \,a+b\beta +c{{\beta }^{2}}+d{{\beta }^{3}} \right]\] = \[a+\frac{1}{2}b(\alpha +\beta )+\frac{1}{2}c\,({{\alpha }^{2}}+{{\beta }^{2}})+\frac{1}{2}d\,({{\alpha }^{3}}+{{\beta }^{3}})\] = \[a+\frac{1}{2}b+\frac{1}{2}c[\,{{(\alpha +\beta )}^{2}}-2\alpha \beta ]+\frac{1}{2}d\,[{{(\alpha +\beta )}^{3}}\]\[-3\alpha \beta (\alpha +\beta )]\] = \[a+\frac{b}{2}+\frac{1}{2}c\,\left[ {{(1)}^{2}}-2.\,\frac{1}{6} \right]+\frac{1}{2}d\,\left[ {{(1)}^{3}}-3.\frac{1}{6} \right]\] = \[\frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}\].
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