Both roots of $(a^{2} - 1) x^{2} + 2 a x + 1 = 0$ belong to the interval (0,1) then exhaustive set of values of 'a' is :

( -

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( -

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None of these

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Solution

The correct option is B
( -

$L e t f (x) = (a^{2} - 1) x^{2} + 2 a x + 1$
Case 1: $a^{2} - 1$ > 0
Given both roots of f(x)=0 are in (0,1)
$\Rightarrow$ f(0)>0,f(1)>0
f(1) > 0 $\Rightarrow a^{2} + 2 a$ > 0 $\Rightarrow$ a < -2 or a > 0
$∴ a ϵ (- (\infty, - 2) \cup (1, \infty)$
Case 2 : $(a^{2} - 1)$ <0, f(1) < 0 & f(0) < 0 which is not possible.

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