The sum of all non-integer roots of the equation $x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0$ is

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-11

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-5

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Solution

The correct option is D 3
Putting $x = 1$
$\Rightarrow 1 - 6 + 11 - 5 - 3 + 2 = 0$
On putting $x = 2$
$\Rightarrow 32 - 96 + 88 - 20 - 6 + 2 = 0$
$∴ (x - 1) (x - 2) (x^{3} - 3 x^{2} + 1) = 0$
So, sum of non integer roots $= 3$

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