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Given that: both the roots £ & β , of the equation (1-a²)x² + 2ax - 1 = 0 , lie between 0 & 1
To find: the values of ‘a’
£ + β = (-2a) / (1-a²)
Here, LHS £ + β is positive as both the roots lie between 0 & 1
=> RHS should be positive
=> a > 1 ( as denominator has to be negative)
But, values of ‘a’ between 1 & 2 do not satisfy the condition that both the roots lie between 0 & 1
So, a > 2
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