Question

# lf $$\alpha$$ and $$\beta$$ are the roots of $$6x^{2}-6x+1=0$$, then $$\displaystyle \frac{1}{2}\left( a+b\alpha +c{ \alpha }^{ 2 }+d\alpha ^{ 3 } \right)+\displaystyle \frac{1}{2}(a+b\beta+c\beta^{2}+d\beta^{3})=$$

A
a+b+c+d
B
a+2b+3c+4d
C
a+b2+c3+d4
D
0

Solution

## The correct option is B $$a+\dfrac b2+\dfrac c3+\dfrac d4$$If $$\alpha$$ and $$\beta$$ are the roots of $$6x^{2}-6x+1$$, then $$\alpha + \beta = -\dfrac{-6}{6} = 1$$$$\alpha \beta =\dfrac{1}{6}$$$$\alpha^{2} + \beta^{2} = (\alpha+\beta)^{2}-2\alpha \beta = 1-\dfrac{1}{3} = \dfrac{2}{3}$$$$\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3\alpha \beta(\alpha+\beta) = 1-3\times\dfrac{1}{6} = \dfrac{1}{2}$$Now, $$\displaystyle \dfrac{1}{2}\left( a+b\alpha +c{ \alpha }^{ 2 }+d\alpha ^{ 3 } \right)+\displaystyle \frac{1}{2}(a+b\beta+c\beta^{2}+d\beta^{3})$$$$= \displaystyle \frac{1}{2}(a+a) +\frac{b}{2}(\alpha+\beta) + \frac{c}{2}(\alpha^{2}+\beta^{2}) + \frac{d}{2}(\alpha^{3}+\beta^{3})$$$$= \displaystyle a+ \frac{b}{2} \times 1 + \frac{c}{2}\times\frac{2}{3}+\frac{d}{2}\times\frac{1}{2}$$$$\displaystyle = a+\frac{b}{2}+ \frac{c}{3} +\frac{d}{4}$$Maths

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