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The value for a for which the equation (1-a^2)x^2+2ax-1=0 has roots belonging to (0,1) is
one month ago

Answers : (1)

Arun
22809 Points
							

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

one month ago
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