The equation $x^{2} + a x - a^{2}$ - 1 = 0 will have roots of opposite signs if :

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a [-1 , 1 ]

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None of these

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Solution

The correct option is A
a

$△$ $\geq$ and ac < 0

$△$ = $a^{2}$ + $4 a^{2}$ + 4 is always greater than 0

- $(a^{2} + 1)$ Is always negative

a $\in$ R

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