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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

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one year ago

Arun
25285 Points
```							Given that: both the roots £ & β , of the equation (1-a²)x² + 2ax - 1 = 0 , lie between 0 & 1To 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 & 1So, a > 2
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one year ago
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Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions