$$\eqalign{ & {\text{Given that }}\,{a^2} + {b^2} + {c^2} = 1\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{We know }}{\left( {a + b + c} \right)^2} \geqslant 0 \cr & \Rightarrow \,{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca \geqslant 0 \cr & \Rightarrow \,2\left( {ab + bc + ca} \right) \geqslant - 1\,\,\,\,\,\,\,\,\,\left[ {{\text{Using}}\left( 1 \right)} \right] \cr & \Rightarrow \,ab + bc + ca \geqslant - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{Also we know that}} \cr & \frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \geqslant 0 \cr & \Rightarrow \,{a^2} + {b^2} + {c^2} - ab - bc - ca \geqslant 0 \cr & \Rightarrow \,ab + bc + ca \leqslant 1\,\,\,\,\,\,\,\,\,\left[ {{\text{Using}}\left( 1 \right)} \right]\,\,\,\,\,.....\left( 3 \right) \cr & \,\,\,\,\,{\text{Combining}}\left( 2 \right){\text{and}}\left( 3 \right){\text{,we get}} \cr & \,\, - \frac{1}{2} \leqslant ab + bc + ca \leqslant 1 \cr & \therefore \,\,ab + bc + ca \in \left[ { - \frac{1}{2},1} \right] \cr & \therefore \,\,\left( {\text{C}} \right){\text{is the correct answer}}{\text{.}} \cr} $$

$$\eqalign{ & {x^3} + 27 = 0 \cr & \Rightarrow \,\,x = {\left( { - 27} \right)^{\frac{1}{3}}} = - 3, - 3\omega , - 3{\omega ^2} \cr & \therefore \,\,\frac{\gamma }{\alpha } = {\omega ^2},\frac{\beta }{\alpha } = \omega \,\,\,{\text{or, }}\frac{\gamma }{\alpha } = \frac{1}{\omega },\frac{\beta }{\alpha } = \omega \,\,{\text{or, }}\frac{\gamma }{\alpha } = \frac{1}{{{\omega ^2}}},\frac{\beta }{\alpha } = \frac{1}{\omega }. \cr} $$
In all the cases, $$\frac{\gamma }{\alpha },\frac{\beta }{\alpha }$$  are $$\omega \,\,{\text{or }}{\omega ^2}.$$
∴ the equation is $${x^2} - \left( {\omega + {\omega ^2}} \right)x + \omega \cdot {\omega ^2} = 0.$$

$$xy + {y^2} + x = \frac{y}{2},{x^2} + xy = - \frac{y}{2}.$$
Adding, $${\left( {x + y} \right)^2} + x = 0,$$    and subtracting $${y^2} - {x^2} = y - x.$$
Solving these equations, $$x = - 1, y = 2$$    and $$x = - \frac{1}{4} = y.$$

Statement 2 is
$$\eqalign{ & \sqrt {n\left( {n + 1} \right)} < n + 1,n \geqslant 2 \cr & \Rightarrow \,\,\sqrt n < \sqrt {n + 1} ,n \geqslant 2{\text{ which is true}} \cr & \Rightarrow \,\,\sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - < \sqrt n \cr & {\text{Now}}\,\,\sqrt 2 < \sqrt n \cr & \Rightarrow \,\,\frac{1}{{\sqrt 2 }} > \frac{1}{{\sqrt n }} \cr & \sqrt 3 < \sqrt n \Rightarrow \,\,\frac{1}{{\sqrt 3 }} > \frac{1}{{\sqrt n }}; \cr & \sqrt n \leqslant \sqrt n \Rightarrow \,\,\frac{1}{{\sqrt n }} \geqslant \frac{1}{{\sqrt n }} \cr & {\text{Also }}\frac{1}{{\sqrt 1 }} > \frac{1}{{\sqrt n }}\,\,\,\,\,\,\,\,\,\,\therefore \,{\text{Adding all,we get}} \cr & \frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + ...... + \frac{1}{n} > \frac{n}{{\sqrt n }} = \sqrt n \cr} $$
Hence both the statements are correct and statement - 2 is a correct explanation of statement - 1.

For $$a{x^2} - bxy - a{y^2},D = {b^2} + 4{a^2} > 0.$$

$$\eqalign{ & \sum\limits_{r = 1}^n {{s_r}} = \left( {\alpha + \beta } \right) + \left( {{\alpha ^2} + {\beta ^2}} \right) + ..... + \left( {{\alpha ^n} + {\beta ^n}} \right) \cr & \sum\limits_{r = 1}^n {{s_r}} = \left( {\alpha + {\alpha ^2} + ..... + {\alpha ^n}} \right) + \left( {\beta + {\beta ^2} + ..... + {\beta ^n}} \right) \cr & \therefore \,\,\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{s_r}} = \left( {\alpha + {\alpha ^2} + {\alpha ^3} + .....\,{\text{to }}\infty } \right) + \left( {\beta + {\beta ^2} + {\beta ^3} + .....\,{\text{to }}\infty } \right) \cr & \therefore \,\,\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{s_r}} = \frac{\alpha }{{1 - \alpha }} + \frac{\beta }{{1 - \beta }}\left( {{\text{clearly }}\left| \alpha \right| < 1,\left| \beta \right| < 1} \right). \cr} $$

$$\eqalign{ & \therefore \,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0 \cr & \Rightarrow \,\,2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0 \cr & \Rightarrow \,\,2 > 2\left( {ab + bc + ca} \right) \cr & \Rightarrow \,\,ab + bc + ca < 1 \cr} $$

$$\eqalign{ & {\text{Given}}\,{\text{that}}\,\,\,\,\left| {z - \frac{4}{z}} \right| = 2, \cr & {\text{Now}}\left| z \right|\,\,\, = \,\,\,\left| {z - \frac{4}{z} + \frac{4}{{ - z}}} \right|\,\,\,\, \leqslant \,\,\,\,\left| {z - \frac{4}{z}} \right| + \frac{4}{{\left| z \right|}} \cr & \Rightarrow \,\,\left| z \right|\,\,\, \leqslant 2 + \frac{4}{{\left| z \right|}} \cr & \Rightarrow \,\,{\left| z \right|^2} - 2\left| z \right| - 4 \leqslant 0 \cr & \Rightarrow \,\,\left( {\left| z \right| - \frac{{2 + \sqrt {20} }}{2}} \right)\,\left( {\left| z \right| - \frac{{2 - \sqrt {20} }}{2}} \right) \leqslant 0 \cr & \Rightarrow \left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\,\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \leqslant 0 \cr & \Rightarrow \,\left( { - \sqrt 5 + 1} \right)\, \leqslant \,\,\left| z \right|\, \leqslant \,\left( {\sqrt 5 + 1} \right) \cr & \Rightarrow \,{\left| z \right|_{\max }} = \sqrt 5 + 1 \cr} $$

The roots of $$x^2 + x + 1 = 0$$    are $$\omega $$ and $$\omega ^2$$
[see the cube roots of unity in complex numbers]
Let, $$\alpha = \omega ,\beta = {\omega ^2}$$
Now, $${\alpha ^{229}} = {\omega ^{229}} = {\omega ^{228}}.\omega = {\left( {{\omega ^3}} \right)^{76}}.\omega = \omega = \alpha \left( {\because {\omega ^3} = 1} \right)$$
$${\alpha ^{1004}} = {\omega ^{1002}}.{\omega ^2} = {\left( {{\omega ^3}} \right)^{334}}.{\omega ^2} = {\omega ^2} = \beta $$
∴ equation with roots $${\alpha ^{229}}$$  and $${\alpha ^{1004}}$$  is same as the equation with roots $$\alpha$$ and $$\beta\,$$ i.e., the original equation.

As all the coefficients are not real, one common root does not imply that the other root is also common.
Let $$\alpha $$ be the common root. Then $${\alpha ^2} + i\alpha + a = 0,{\alpha ^2} - 2\alpha + ia = 0$$
$$\eqalign{ & \Rightarrow \,\,\frac{{{\alpha ^2}}}{a} = \frac{\alpha }{{a\left( {1 - i} \right)}} = \frac{1}{{ - 2 - i}} \cr & \Rightarrow \,\,{a^2}{\left( {1 - i} \right)^2} = a\left( { - 2 - i} \right). \cr} $$