Given $$c < 0 < b$$   and $$\alpha + \beta = - b\,\,\,\,\,\,\,\,\,.....\left( 1 \right)$$
$$\alpha \beta = c\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right)$$
From (2), $$c < 0$$
$$ \Rightarrow \,\,\alpha \beta < 0$$
⇒ either $$\alpha $$ is $$- ve$$  or $$\beta $$ is $$- ve$$  and second ;
quantity is positive.
from (1), $$b > 0$$
⇒ $$- b < 0$$
$$ \Rightarrow \,\,\alpha + \beta < 0$$
⇒ the sum is negative
⇒ modules of negative quantity is > modulus of positive quantity but $$\alpha $$ < $$\beta $$ is given. Therefore, it is clear that $$\alpha $$ is negative and $$\beta $$ is positive and modulus of $$\alpha $$ is greater than modulus of $$\beta $$
$$ \Rightarrow \,\,\alpha < 0 < \beta < \left| \alpha \right|$$

$$\eqalign{ & \alpha ,\beta \,\,{\text{are roots of the equation }}\left( {x - a} \right)\left( {x - b} \right) = c,c \ne 0 \cr & \therefore \,\,\left( {x - a} \right)\left( {x - b} \right) - c = \left( {x - \alpha } \right)\left( {x - \beta } \right) \cr & \Rightarrow \,\,\left( {x - \alpha } \right)\left( {x - \beta } \right) + c = \left( {x - a} \right)\left( {x - b} \right) \cr & \Rightarrow \,\,{\text{roots of }}\left( {x - \alpha } \right)\left( {x - \beta } \right) + c = 0\,\,{\text{are }}\,a\,{\text{and }}b. \cr & \therefore \,\,\left( {\text{C}} \right){\text{is the correct option}}{\text{.}} \cr} $$

Here, $$\alpha + \beta = p,\alpha \beta = q$$
$${\alpha _1} + {\beta _1} = q,{\alpha _1}{\beta _1} = p$$
Sum of given roots
$$\eqalign{ & = \left( {\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}} \right)\,\left( {\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}} \right) \cr & = \frac{{\alpha {\beta _1} + {\alpha _1}\beta + \beta {\beta _1} + \alpha {\alpha _1}}}{{\alpha \beta {\alpha _1}{\beta _1}}} \cr & = \frac{{\left( {\alpha + \beta } \right)\left( {{\alpha _1} + {\beta _1}} \right)}}{{\left( {\alpha \beta } \right)\left( {{\alpha _1}{\beta _1}} \right)}} = \frac{{pq}}{{qp}} = 1 \cr} $$
Product of given roots
$$\eqalign{ & = \left( {\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}} \right)\left( {\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}} \right) \cr & = - \frac{{\left( {\alpha {\beta _1} + {\alpha _1}\beta } \right)\left( {\alpha {\alpha _1} + \beta {\beta _1}} \right)}}{{{\alpha ^2}{\beta ^2}\alpha _1^2\beta _1^2}} \cr & = \frac{{\alpha \beta \left( {\alpha _1^2 + \beta _1^2} \right) + {\alpha _1}{\beta _1}\left( {{\alpha ^2} + {\beta ^2}} \right)}}{{{\alpha ^2}{\beta ^2}\alpha _1^2\beta _1^2}} \cr & = \frac{{\alpha \beta \left[ {{{\left( {{\alpha _1} + {\beta _1}} \right)}^2} - 2{\alpha _1}{\beta _1}} \right] + {\alpha _1}{\beta _1}\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right]}}{{{{\left( {\alpha \beta } \right)}^2}{{\left( {{\alpha _1}{\beta _1}} \right)}^2}}} \cr & = \frac{{q\left( {{q^2} - 2p} \right) + p\left( {{p^2} - 2q} \right)}}{{{q^2}{p^2}}} \cr & = \,\frac{{{p^3} + {q^3} - 4pq}}{{{p^2}{q^2}}} \cr} $$
Hence, the required equation is
$$\left( {{p^2}{q^2}} \right){x^2} - \left( {{p^2}{q^2}} \right)x + {p^3} + {q^3} - 4pq = 0$$

Let, $$a > 0, b > 0, c > 0$$
Given equation $$ax^2 + bx + c = 0$$
we know that $$D = b^2 - 4ac$$   and $$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$
$$\eqalign{ & {\text{Let, }}{b^2} - 4ac > 0,b > 0 \cr & {\text{If, }}a > 0,c > 0{\text{ then }}{b^2} - 4ac < {b^2} \cr} $$
⇒ Roots are negative
$${\text{Let, }}{b^2} - 4ac < 0,{\text{ then }}x = \frac{{ - b \pm i\sqrt {4ac - {b^2}} }}{{2a}}$$
⇒ roots are imaginary and have negative real part. $$\left( {\because b > 0} \right).$$

$$\tan \left( {\frac{P}{2}} \right),\tan \left( {\frac{Q}{2}} \right)$$    are the roots of $$a{x^2} + bx + c = 0$$
$$\eqalign{ & \tan \left( {\frac{P}{2}} \right) + \tan \left( {\frac{Q}{2}} \right) = - \frac{b}{a},\tan \left( {\frac{P}{2}} \right) \cdot \tan \left( {\frac{Q}{2}} \right) = \frac{c}{a} \cr & \frac{{\tan \left( {\frac{P}{2}} \right) + \tan \left( {\frac{Q}{2}} \right)}}{{1 - \tan \left( {\frac{P}{2}} \right)\tan \left( {\frac{Q}{2}} \right)}} = \tan \left( {\frac{P}{2} + \frac{Q}{2}} \right) = 1 \cr & \Rightarrow \,\,\frac{{ - \frac{b}{a}}}{{1 - \frac{c}{a}}} = 1 \cr & \Rightarrow \,\, - \frac{b}{a} = \frac{a}{a} - \frac{c}{a} \cr & \Rightarrow \,\, - b = a - c\,\,\,{\text{or }}c = a + b. \cr} $$

$$\eqalign{ & \frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2} \cr & \Rightarrow \,\,\frac{{ - {x^2} + 2x + 3}}{{{x^2} + 1}} > 0 \cr & \Rightarrow \,\, - {x^2} + 2x + 3 > 0\left\{ {\because \,\,{x^2} + 1\,\,{\text{is always positive}}} \right\}. \cr} $$
By sign scheme, $$ - 1 < x < 3.\,{\text{As }}x\,{\text{is integer,}}\,x = 0,1,2.$$

$$\eqalign{ & {\text{Given equation - }} \cr & {x^2} + 2ax + \left( {10 - 3a} \right) > 0{\text{ for all }}x \in R \cr & {\text{Here, }}A = 1,\,B = 2a,\,C = \left( {10 - 3a} \right) \cr & {\text{As we know that,}} \cr & A{x^2} + Bx + C > 0{\text{ for all }}x \in R{\text{ if}} \cr & A > 0{\text{ and }}D < 0 \cr & A = 1 > 0 \cr & {\text{Now, }}D < 0 \cr & {B^2} - 4AC < 0 \cr & {\left( {2a} \right)^2} - 4\left( 1 \right)\left( {10 - 3a} \right) < 0 \cr & 4{a^2} - 40 + 12a < 0 \cr & {a^2} + 3a - 10 < 0 \cr & \left( {a + 5} \right)\left( {a - 2} \right) < 0 \cr & a \in \left( { - 5,\,2} \right) \cr & \therefore \, - 5 < a < 2 \cr} $$

$$\eqalign{ & {4^{{{\log }_2}\lambda }} = 8 \cr & \Rightarrow \,\,{2^{2{{\log }_2}\lambda }} = {2^3} \cr & \Rightarrow \,\,2{\log _2}\lambda = 3 \cr & \Rightarrow \,\,{\lambda ^2} = {2^3} \cr & \Rightarrow \,\,\lambda = 2\sqrt 2 \,\,\left( {\because \,\,\lambda > 0} \right).2 \cr} $$

$$\eqalign{ & {\cos ^4}x - {\cos ^2}x + 1 - p = 0;\,{\text{as }}0 \leqslant {\cos ^2}x \leqslant 1,\,{\text{the roots of }}{y^2} - y + 1 - p = 0\,\,{\text{lie between 0 and 1}}{\text{.}} \cr & \therefore \,\,\alpha \geqslant 0,\beta \geqslant 0,\alpha - 1 \leqslant 0,\beta - 1 \leqslant 0\,\,{\text{and }}D \geqslant 0 \cr & \Rightarrow \,\,\alpha + \beta \geqslant 0,\alpha \beta \geqslant 0,\alpha + \beta - 2 \leqslant 0,\alpha \beta - \left( {\alpha + \beta } \right) + 1 \geqslant 0\,\,{\text{and }}D \geqslant 0 \cr & \therefore \,\,1 \geqslant 0,1 - p \geqslant 0,1 - 2 \leqslant 0,1 - p - 1 + 1 \geqslant 0\,\,{\text{and }}1 - 4\left( {1 - p} \right) \geqslant 0. \cr} $$

$$\eqalign{ & {\left( {\frac{5}{{13}}} \right)^x} + {\left( {\frac{{12}}{{13}}} \right)^x} \geqslant 1 \cr & \therefore \,\,{\cos ^x}\alpha + {\sin ^x}\alpha \geqslant 1,\,\,{\text{where }}\cos \alpha = \frac{5}{{13}}. \cr} $$
Equality holds for $$x = 2.$$  If $$x < 2,$$  both $$\cos\alpha $$  and $$\sin\alpha $$  increase (being positive fractions). So, $${\cos ^x}\alpha + {\sin ^x}\alpha \geqslant 1\,\,{\text{if }}x < 2.\,{\text{Thus }}x \leqslant 2.$$