Let $$\alpha $$ and $$\beta $$ be the roots of$${x^2} - x - 1 = 0$$, with $$\alpha $$ > $$\beta $$. For all positive integers n, define

$${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$$

$${b_1} = 1\,and\,{b_n} = {a_{n - 1}} + {a_{n + 1}},\,n \ge 2$$

Then which of the following options is/are correct?

Let $$S$$ be the set of all non-zero real numbers $$\alpha $$ such that the quadratic equation $$\alpha {x^2} - x + \alpha = 0$$ has two distinct real roots $${x_1}$$ and $${x_2}$$ satisfying the inequality $$\left| {{x_1} - {x_2}} \right| < 1.$$ Which of the following intervals is (are) $$a$$ subset(s) os $$S$$?

If $${3^x}\, = \,{4^{x - 1}},$$ then $$x\, = $$

If $$\alpha $$ and $$\beta $$ are the roots of $${x^2}$$+ px + q = 0 and $${\alpha ^4},{\beta ^4}$$ are the roots of $$\,{x^2} - rx + s = 0$$, then the equation $${x^2} - 4qx + 2{q^2} - r = 0$$ has always