Values of $^{'} m^{'}$ such that the roots of the equation $2 x^{2} - x - 1 = 0$ lie inside the roots of the equation $x^{2} + (2 m - m^{2}) x - 2 m^{3} = 0,$ is

2

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\frac{1}{8}

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\frac{3}{4}

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\frac{2}{3}

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Solution

The correct option is A $2$
$2 x^{2} - x - 1 = 0 \Rightarrow x = - \frac{1}{2}, 1$
For the equation, $x^{2} + (2 m - m^{2}) x - 2 m^{3} = 0,$
$D = (2 m - m^{2})^{2} + 8 m^{3} = (m^{2} + 2 m)^{2} > 0$
As $D > 0 \Rightarrow f (- \frac{1}{2}) < 0, f (1) < 0$
So $m > \frac{1}{4}$
$f (1) < 0 \Rightarrow m \in (- 1, - \frac{1}{2}) \cup (1, \infty)$
After intersection we get, $m \in (1, \infty)$

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