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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
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B
a+2b+3c+4d
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C
a+b2+c3+d4
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D
0
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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, $$\displaystyle \dfrac{1}{2}\left( a+b\alpha +c{ \alpha  }^{ 2 }+d\alpha ^{ 3 } \right)+\displaystyle \frac{1}{2}(a+b\beta+c\beta^{2}+d\beta^{3})$$

$$ = \displaystyle  \frac{1}{2}(a+a) +\frac{b}{2}(\alpha+\beta) + \frac{c}{2}(\alpha^{2}+\beta^{2}) + \frac{d}{2}(\alpha^{3}+\beta^{3}) $$

$$ =  \displaystyle a+ \frac{b}{2} \times 1 + \frac{c}{2}\times\frac{2}{3}+\frac{d}{2}\times\frac{1}{2}$$

$$  \displaystyle = a+\frac{b}{2}+ \frac{c}{3} +\frac{d}{4}$$


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