Using, $$\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$$
$$ \Rightarrow {{\sqrt 3 } \over 2} = {{{{({x^2} + x + 1)}^2} + {{({x^2} - 1)}^2} - {{(2x + 1)}^2}} \over {2({x^2} + x + 1)({x^2} - 1)}}$$
$$ \Rightarrow (x + 2)(x + 1)(x - 1)x + {({x^2} - 1)^2} = \sqrt 3 ({x^2} + x + 1)({x^2} - 1)$$
$$ \Rightarrow {x^2} + 2x + ({x^2} - 1) = \sqrt 3 ({x^2} + x + 1)$$
$$ \Rightarrow (2 - \sqrt 3 ){x^2} + (2 - \sqrt 3 )x - (\sqrt 3 + 1) = 0$$
$$ \Rightarrow x = - (2 + \sqrt 3 )$$ and $$x = 1 + \sqrt 3 $$
But, $$x = - (2 + \sqrt 3 ) \Rightarrow c$$ is negative.
$$\therefore$$ $$x = 1 + \sqrt 3 $$ is the only solution.
Hence, (b) is the correct option.
Since, A, B, C are in AP
$$\Rightarrow$$ 2B = A + C i.e., $$\angle$$B = 60$$^\circ$$
$$\therefore$$ $${a \over c}$$(2 sin C cos C) + $${c \over a}$$ (2 sin A cos A)
= 2k (a cos C + c cos A)
[using, $${a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {1 \over k}$$]
= 2k (b)
= 2 sin B
[using, b = a cos C + c cos A]
= $$\sqrt3$$
