If the roots of the equation $m x^{2} + (2 m - 1) x + m - 2 = 0$ are rational, then if $m \in I$ it will be

odd integer

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even integer

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zero only

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none of these

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Solution

The correct option is A odd integer
$m x^{2} + (2 m - 1) x + (m - 2) = 0$
$D = b^{2} - 4 a c$
$D = (2 m - 1)^{2} - 4 m (m - 2)$
$= 4 m^{2} + 1 - 4 - 4 m^{2} + 8 m$
$\Rightarrow 4 m + 1$
$4$ m will always be even, so $(4 m + 1)$ will be odd integer.

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