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

Options

A. `agt0`
B. `alt0`
C. `agt2`
D. none of these

Answer

C

Explanation

`:'(1-a^(2))x^(2)+2ax-1=0`
`1-a^(2)!=0`
`x^(2)-((2a)/(1-a^(2)))x-(1/(1-a^(2)))0`
Let `f(x)=x^(2)+((2a)/(1-a^(2)))x-(1/((1-a^(2))))`

The following cases arise:
Case I `Dge0` ltbr. `((2a)/(1-a^(2)))^(2)-4.1.((-1)/(1-a^(2)))ge0`
`implies(4a^(2))/((1-a^(2))^(2))+4/((1-a^(2)))ge0`
`implies(4a^(2)+4-4a^(2))/((1-a^(2))^(2))ge0`
`implies4/((1-a^(2))^(2))ge0` [always true]
CaseII `f(0)gt0`
`implies(-1)/((1-a^(2)))gt0implies1/(1-a^(2))lt0`
`implies1-a^(2)lt0`
`:.a epsilon(-oo,o-1)uu(1,oo)`
Case III `f(1)gt0`
`implies1+(2a)/((1-a^(2)))-1/((1-a^(2)))gt0`
`implies(1-a^(2)+2a-1)/((1-a^(2)))gt0implies(a^(2)-2a)/(1-a^(2))lg0`

`implies(a(a-2))/((a+1)(a-1))gt0`
`:.a epsilon(-oo,-10uu(0,1)uu(2,oo)`
Case Iv `0ltx` coordiane of vertex `lt1`
`implies0lt-(2a)/(2(1-a^(2)))lt1implies0lta/(a^(2)-1)lt1`
`implies0lta/((a+1)(a-1))` an `1=a/(a^(2)-1)gt0`
`impliesa/((a+1)(a-1))gt0`

`impliesa epsilon(-1,0)uu(1,oo)`
and `((a-(1+sqrt(5))/2)(a-(1-sqrt(5))/2))/((a+1)(a-1))gt0`

and `a epsilon(-oo,-1)uu((1-sqrt(5))/2,1)uu((1+sqrt(5))/2,oo)`
`:.a epsilon((1-sqrt(5))/2,0)uu((1+sqrt(5))/2,oo)`
Combining all cases, we get