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 $$x\, = $$

Let a, b, c be real numbers, $$a \ne 0$$. If $$\alpha \,$$ is a root of $${a^2}{x^2} + bx + c = 0$$. $$\beta \,$$ is the root of $${a^2}{x^2} - bx - c = 0$$ and $$0 < \alpha \, < \,\beta $$, then the equation $${a^2}{x^2} + 2bx + 2c = 0$$ has a root $$\gamma $$ that always satisfies