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