As $$a, b, c > 0, a, b, c$$    should be real (note that order
relation is not defined in the set of complex numbers)
∴ Roots of equation are either real or complex conjugate.
Let $$\alpha {\text{,}}\beta $$  be the roots of $$a{x^2} + bx + c = 0,$$    then
$$\alpha + \beta = - \frac{b}{a} = - ve,\,\,\,\,\,\,\alpha \beta = \frac{c}{a} = + ve$$
⇒ Either both $$\alpha {\text{,}}\beta $$  are $$- ve$$  (if roots are real) or both $$\alpha {\text{,}}\beta $$  have $$- ve$$  real parts (if roots are complex conjugate)

$$\eqalign{ & {\text{Given expression }}{x^{12}} - {x^9} + {x^4} - x + 1 = f\left( x \right)\,\left( {{\text{say}}} \right) \cr & {\text{For }}x < 0\,{\text{ put}}\,\,x = - y\,\,{\text{where}}\,y > 0{\text{ }} \cr & {\text{then we get }}\,f\left( x \right) = {y^{12}} + {y^9} + {y^4} + y + 1 > 0\,\,{\text{for }}\,y > 0 \cr & {\text{For }}0 < x < 1,\,\,{x^9} < {x^4} \cr & \Rightarrow \, - {x^9} + {x^4} > 0 \cr & {\text{Also }}1 - x > 0\,\,{\text{and }}{x^{12}} > 0 \cr & \Rightarrow \,\,{x^{12}} - {x^9} + {x^4} + 1 - x > 0 \cr & \Rightarrow \,\,f\left( x \right) > 0 \cr & {\text{For }}x > 1 \cr & f\left( x \right) = x\left( {{x^3} - 1} \right)\left( {{x^8} + 1} \right) + 1 > 0 \cr & {\text{So}}\,\,f\left( x \right) > 0\,\,{\text{for}}\,\, - \infty < x < \infty . \cr} $$

$$\left[ {\frac{x}{2}} \right] + \left[ {\frac{x}{3}} \right] + \left[ {\frac{x}{5}} \right] = \frac{{31}}{{30}}x = \frac{x}{2} + \frac{x}{3} + \frac{x}{5}$$
$$ \Rightarrow \,\,\frac{x}{2},\frac{x}{3},\frac{x}{5}$$   are all integers. So, $$x =$$  multiple of the LCM of 2,3,5
$$\therefore \,\,x = 30 \times 1,30 \times 2,30 \times 3,.....,30 \times 33.$$

$$\eqalign{ & {\left| x \right|^2} - 3\left| x \right| + 2 = 0 \cr & {\bf{Case I :}}\,\,x < 0{\text{ then }}\left| x \right| = - x \cr & \Rightarrow \,{x^2} + 3x + 2 = 0 \cr & \Rightarrow \,\left( {x + 1} \right)\left( {x + 2} \right) = 0 \cr & \,\,\,\,\,\,\,x = - 1, - 2\left( {{\text{both acceptable as}} < 0} \right) \cr & {\bf{Case II :}}\,\,x > 0{\text{ then }}\left| x \right| = x \cr & \Rightarrow \,{x^2} - 3x + 2 = 0 \cr & \Rightarrow \,\left( {x - 1} \right)\left( {x - 2} \right) = 0 \cr & \,\,\,\,\,\,\,x = 1,2\left( {{\text{both acceptable as}} > 0} \right) \cr & \therefore \,\,{\text{There are 4 real solutions}}{\text{.}} \cr} $$

$$\eqalign{ & \ell ,m,n\,\,{\text{are real, }}\ell \ne m \cr & {\text{Given equation is }} \cr & \left( {\ell - m} \right){x^2} - 5\left( {\ell + m} \right)x - 2\left( {\ell - m} \right) = 0 \cr & D = 25{\left( {\ell + m} \right)^2} + 8{\left( {\ell - m} \right)^2} > 0,\ell ,m \in R \cr & \therefore \,\,\,{\text{Roots are real and unequal}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Equation }}{x^2} - 2mx + {m^2} - 1 = 0 \cr & {\left( {x - m} \right)^2} - 1 = 0\,\,{\text{or }}\left( {x - m + 1} \right)\left( {x - m - 1} \right) = 0 \cr & x = m - 1,m + 1 \cr & m - 1 > - 2\,\,{\text{and }}m + 1 < 4 \cr & \Rightarrow \,\,m > - 1\,\,{\text{and }}m < 3\,\,{\text{or, }} - 1 < m < 3 \cr} $$

Let the distance of school from $$A$$ = $$x$$
∴ The distance of the school form $$B$$ = $$60 - x$$
Total distance covered by 200 students
$$= 2[150 x +50(60 - x)] = 2[100 x + 3000]$$
This is min., when $$x$$ = 0
∴ school should be built at town $$A.$$

Let $$\alpha $$ and $$\beta $$ are roots of the equation $${x^2} + ax + 1 = 0\,$$
So, $$\alpha \, + \beta = - a\,\,{\text{and }}\alpha \beta = 1$$
$$\eqalign{ & {\text{given}}\,\,\left| {\alpha - \beta } \right| < \sqrt 5 \cr & \Rightarrow \,\,\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 \,\,\,\,\,\,\,\,\,\left( {\because {{\left( {\alpha - \beta } \right)}^2} = {{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } \right) \cr & \Rightarrow \,\,\sqrt {{a^2} - 4} < \sqrt 5 \cr & \Rightarrow \,\,{a^2} - 4 < 5 \cr & \Rightarrow \,\,{a^2} - 9 < 0 \cr & \Rightarrow \,\,{a^2} < 9 \cr & \Rightarrow \,\, - 3 < a < 3 \cr & \Rightarrow \,\,a \in \left( { - 3,3} \right) \cr} $$

We have
$$\eqalign{ & y = 5{x^2} + 2x + 3\, \cr & \,\,\,\, = 5{\left( {x + \frac{1}{5}} \right)^2} + \frac{{14}}{5} > 2,\,\forall \,x \in R \cr & {\text{while }}y = 2\sin x \leqslant 2,\,\forall \,x \in R \cr & \Rightarrow \,\,{\text{The two curves do not meet at all}}{\text{.}} \cr} $$

Let the vertices be $${z_0},{z_1},.....,{z_5}$$   w.r.t centre $$O$$ as origin $$\left| {{z_0}} \right| = 1,$$
$$\eqalign{ & {A_0}{A_1} = \left| {{z_1} - {z_0}} \right| = \left| {{z_0}{e^{i\theta }} - {z_0}} \right| \cr & \therefore {A_0}{A_1} = \left| {{z_0}} \right|\left| {\cos \theta + i\sin \theta - 1} \right| \cr & = 1 \cdot \sqrt {{{\left( {\cos \theta - 1} \right)}^2} + {{\sin }^2}\theta } = \sqrt {2\left( {1 - \cos \theta } \right)} \cr & \therefore {A_0}{A_1} = \sqrt {2.2{{\sin }^2}\frac{\theta }{2}} = 2\sin \frac{\theta }{2} \cr} $$
Where $$\theta = \frac{{2\pi }}{6} = \frac{\pi }{3}.$$   Replacing $$\theta $$ by $$2\theta $$ and $$4\theta ,$$
we get, $${A_0}{A_2} = 2\sin \frac{{2\theta }}{2} = 2\sin \theta \,\,\& \,\,{A_0}{A_4} = 2\sin \frac{{4\theta }}{2} = 2\sin 2\theta $$
$$\eqalign{ & \therefore \left( {{A_0}{A_1}} \right)\left( {{A_0}{A_2}} \right)\left( {{A_0}{A_4}} \right) \cr & = 8\sin \frac{\pi }{6}\sin \frac{\pi }{3}\sin \frac{{2\pi }}{3} \cr & = 8\left( {\frac{1}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) = 3 \cr} $$