Let the roots of given equation be $$\alpha $$ and $$2\alpha $$  then
$$\eqalign{ & \alpha + 2\alpha = 3\alpha = \frac{{1 - 3a}}{{{a^2} - 5a + 3}} \cr & \& \,\,\alpha .2\alpha = 2{\alpha ^2} = \frac{2}{{{a^2} - 5a + 3}} \cr & \Rightarrow \,\,\alpha = \frac{{1 - 3a}}{{3\left( {{a^2} - 5a + 3} \right)}} \cr & \therefore \,\,2\left[ {\frac{1}{9}\frac{{{{\left( {1 - 3a} \right)}^2}}}{{{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right] = \frac{2}{{{a^2} - 5a + 3}} \cr & \frac{{{{\left( {1 - 3a} \right)}^2}}}{{\left( {{a^2} - 5a + 3} \right)}} = 9\,\,{\text{or}}\,\,9{a^2} - 6a + 1 = 9{a^2} - 45a + 27 \cr & {\text{or}}\,\,39a = 26\,\,\,{\text{or }}a = \frac{2}{3} \cr} $$

$${x^2} - \left( {k - 2} \right)x + {k^2} = 0\,\,{\text{and }}{x^2} + kx + 2k - 1 = 0$$          should have both roots common or each should have equal roots.
$$\eqalign{ & \therefore \,\,\left( {\text{i}} \right)\frac{1}{1} = \frac{{ - \left( {k - 2} \right)}}{k} = \frac{{{k^2}}}{{2k - 1}} \cr & {\text{or, }}\left( {{\text{ii}}} \right){\left( {k - 2} \right)^2} - 4{k^2} = 0\,\,{\text{and }}{k^2} - 4\left( {2k - 1} \right) = 0. \cr & \left( {\text{i}} \right) \Rightarrow \,\,k = - k + 2,2k - 1 = {k^2}.\,{\text{So, }}k = 1 \cr & \left( {{\text{ii}}} \right)\, \Rightarrow \,\,\left( {3k - 2} \right)\left( { - k - 2} \right) = 0,{k^2} - 8k + 4 = 0.\,\,{\text{So, no value of }}k{\text{ is possible}}{\text{.}} \cr} $$

Let $$\alpha $$ be the common root of given equations, then
$$\eqalign{ & {\alpha ^2} + b\alpha - 1 = 0\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}{\alpha ^2} + \alpha + b = 0\,\,\,\,\,.....\left( 2 \right) \cr} $$
Subtracting (2) from (1), we get
$$\left( {b - 1} \right)\alpha - \left( {b + 1} \right) = 0\,\,\,{\text{or }}\alpha = \frac{{b + 1}}{{b - 1}}$$
Substituting this value of $$\alpha $$ in equation (1), we get
$$\eqalign{ & {\left( {\frac{{b + 1}}{{b - 1}}} \right)^2} + b\left( {\frac{{b + 1}}{{b - 1}}} \right) - 1 = 0 \cr & {\text{or, }}{b^3} + 3b = 0 \cr & \Rightarrow b = 0,i\sqrt 3 , - i\sqrt 3 \cr} $$

Clearly $$\alpha ,\beta $$  are the roots of the equation $${x^2} - 5x + 3 = 0.$$    Use $$\alpha + \beta = 5,\alpha \beta = 3.$$

$$f\left( x \right) = \frac{{{x^2} + 1 - 2}}{{{x^2} + 1}} = 1 - \frac{2}{{{x^2} + 1}};f\left( x \right)\,\,{\text{is minium when }}\frac{2}{{{x^2} + 1}}\,{\text{is maximum, i}}{\text{.e}}{\text{., 2}}{\text{.}}$$

When $$x < 0,\left| x \right| = - x$$
$$\therefore $$ Equation is $$x^2 - x - 6 = 0$$
⇒ $$x = - 2, 3$$
$$\because $$ $$x < 0,$$  $$\therefore $$ $$x = - 2$$  is the solution.
When $$x \geqslant 0,\left| x \right| = x$$
$$\therefore $$ Equation is $$x^2 + x - 6 = 0$$
⇒ $$x = 2, - 3$$
$$\because x \geqslant 0,\therefore x = 2$$    is the solution,
Hence, $$x = 2, - 2$$   are the solutions and their sum is zero.

For $$x < - 2,$$
$$\left| {x + 2} \right| = - \left( {x + 2} \right)$$     and the inequality becomes
$$\eqalign{ & {x^2} + x + 2 + x > 0 \cr & \Rightarrow \,{\left( {x + 1} \right)^2} + 1 > 0 \cr & {\text{which is valid }}\forall \,\,x \in R\,\,{\text{but }}x < - 2 \cr & \therefore \,\,x \in \left( { - \infty , - 2} \right)\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{For }}x \geqslant 2,\left| {x + 2} \right| = x + 2\,\,{\text{and the inequality becomes}} \cr & {x^2} - x - 2 + x > 0 \cr & \Rightarrow \,\,{x^2} > 2 \cr & \Rightarrow \,\,x > \sqrt 2 \,\,{\text{or }}x < - \sqrt 2 \cr & {\text{i}}{\text{.e}}{\text{., }}x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr & {\text{but }}x \geqslant - 2 \cr & \Rightarrow \,\,x \in \left[ { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{From}}\left( 1 \right){\text{and}}\left( 2 \right) \cr & x \in \left( { - \infty , - 2} \right) \cup \left[ { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr & \Rightarrow \,\,x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr} $$

$$\eqalign{ & {\text{if }}p = 5,q = 3,r = 2 \cr & {\text{max }}\left( {p,q} \right) = 5\,;\,\,\,\max \left( {p,q,r} \right) = 5 \cr & \Rightarrow \,\,{\text{max}}\left( {p,q} \right) = \max \,\left( {p,q,r} \right) \cr} $$
∴ (A) is not true. Similarly we can show that (C) is not true.
$$\eqalign{ & {\text{Also min}}\left( {p,q} \right) = \frac{1}{2}\left( {p + q - \left| {p - q} \right|} \right) \cr & {\text{Let }}p < q\,\,{\text{then L}}{\text{.H}}{\text{.S}} = p \cr & {\text{and R}}{\text{.H}}{\text{.S = }}\frac{1}{2}\left( {p + q - q + p} \right) = p \cr} $$
Similarly, we can prove that $$(B)$$ is true for $$q < p$$  too.

$$\eqalign{ & \alpha - 3 < 0,\beta - 3 < 0\,\,{\text{and }}D \geqslant 0 \cr & \Rightarrow \,\,\alpha + \beta - 6 < 0,\alpha \beta - 3\left( {\alpha + \beta } \right) + 9 > 0\,\,{\text{and }}4{a^2} - 4\left( {{a^2} + a - 3} \right) \geqslant 0 \cr & \Rightarrow \,\,2a - 6 < 0,{a^2} + a - 3 - 3 \cdot 2a + 9 > 0\,\,{\text{and }}3 - a \geqslant 0. \cr} $$
Solve these and find the common value.

$$\eqalign{ & {\log _{\frac{1}{3}}}\left( {{x^2} + x + 1} \right) > - 1 = {\log _{\frac{1}{3}}}{\left( {\frac{1}{3}} \right)^3} \cr & \Rightarrow \,\,{x^2} + x + 1 < {\left( {\frac{1}{3}} \right)^{ - 1}} \cr & \Rightarrow \,\,{x^2} + x - 2 < 0. \cr} $$
Use sign scheme.