1.

The value of `a` for which the equation `(1-a^2)x^2+2ax-1=0` has roots belonging to `(0,1)` isA. `a lt (1+ sqrt(5))/(2)`B. `a gt 2`C. `(1+sqrt(5))/(2) lt a lt 2`D. `a gt sqrt(2)`

Answer» Correct Answer - B
Let `f(x) = (1 -a^(2)) x^(2) + 2ax - 1`. Then, f(x) = 0 will have its roots between 0 and 1, if
(i) Discriminant `ge` 0
(ii) 0 and 1 are outside the roots of f(x) = 0
i.e. `(1-a^(2)) f(0) gt 0 and (1-a^(2)) f(1) gt 0`
Now,
(i) Discriminant `ge` 0
`rArr" "4a^(2) + 4(1-a^(2)) gt 0`, which is always true for all `a in R`
(ii) `(1-a^(2)) f(0) gt 0`
`rArr" "-(1-a^(2)) gt 0 rArr a^(2) - 1 gt 0 rArr a lt - 1 or, a gt 1" "...(i)`
and, `(1-a^(2)) f(1) gt 0`
`rArr" "(1-a^(2))(2a - a^(2)) gt 0`
`rArr" "a(a-1)(a+1)(a-2) gt 0`
`rArr" "a lt - 1 or, a gt 2 or, 0 lt a lt 1" "...(ii)`
From (i) and (ii), we get `a lt -1 or, a gt 2`.


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