Answer: Option A. -> -1
$$\eqalign{
& {x^2} + {y^2} + 2x + 1 = 0 \cr
& \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr
& {\text{Let }}{\left( {x + 1} \right)^2} = 0 \cr
& {y^2} = 0 \cr
& x = - 1,y = 0 \cr
& {\text{Now, }}{x^{31}} + {y^{35}} \cr
& = {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr
& = - 1 \cr} $$

Answer: Option C. -> $$\frac{4}{3}$$
$$\eqalign{
& x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& \therefore x = \frac{1}{y} \cr
& \Leftrightarrow xy = 1 \cr
& x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr
& \Rightarrow x + y = \frac{{12}}{4} \cr
& \Rightarrow x + y = 3 \cr
& \Rightarrow x + \frac{1}{x} = 3 \cr
& \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr
& \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr
& {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr
& = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr
& = \frac{{7 + 1}}{{7 - 1}} \cr
& = \frac{8}{6} \cr
& = \frac{4}{3} \cr} $$

Answer: Option B. -> $$\frac{4}{5}$$
$$\eqalign{
& {a^2} + {b^2} + {c^2} = 16,{\text{ }}{x^2} + {y^2} + {z^2} = 25{\text{ }} \cr
& {\text{But }}b = c = 0,{\text{ But }}y = z = 0 \cr
& {\text{}}a = 4{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 5 \cr
& {\text{Now, }} \cr
& ax + by + cz = 20 \cr
& 4 \times 5 + 0 + 0 = 20 \cr
& 20 = 20{\text{ }}\left( {{\text{Satisfy}}} \right) \cr
& {\text{Now, }}\frac{{a + b + c}}{{x + y + z}} \cr
& = \frac{{4 + 0 + 0}}{{5 + 0 + 0}} \cr
& = \frac{4}{5} \cr} $$

Answer: Option B. -> $$\frac{1}{{{x^2} + {y^2}}}$$
$$\left( {1 - \frac{{2xy}}{{{x^2} + {y^2}}}} \right) \div \left( {\frac{{{x^3} - {y^3}}}{{x - y}} - 3xy} \right)$$
$$ = \left( {\frac{{{x^2} + {y^2} - 2xy}}{{{x^2} + {y^2}}}} \right) \div $$    $$\left( {\frac{{{x^3} - {y^3} - 3xy\left( {x - y} \right)}}{{x - y}}} \right)$$
$$\eqalign{
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div \frac{{{{\left( {x - y} \right)}^3}}}{{x - y}} \cr
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div {\left( {x - y} \right)^2} \cr
& = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \times \frac{1}{{{{\left( {x - y} \right)}^2}}} \cr
& = \frac{1}{{{x^2} + {y^2}}} \cr} $$

Answer: Option B. -> a = 6, b = 7
$$\eqalign{
& {x^4} + 2{x^3} + a{x^2} + bx + 9 \cr
& {\text{Put }}x = 1 \cr
& = 1 + 2 \times 1 + a + b + 9 \cr
& = 1 + 2 + a + b + 9 \cr
& = 13 + a + b \cr} $$
To make a perfect square numbers value of a + b must be either 3 or 13
Now, option (B) a = 6, b = 7
$$\eqalign{
& \therefore a + b = 13 \cr
& {\text{make perfect square}} \cr
& \left( {25 = {5^2}} \right) \cr} $$

Answer: Option C. -> 16
$$\eqalign{
& x = a + \frac{1}{a} \cr
& {x^2} = {a^2} + \frac{1}{{{a^2}}} + 2 \cr
& {y^2} = {a^2} + \frac{1}{{{a^2}}} - 2 \cr
& {\text{Now, }} \cr
& {x^4} + {y^4} - 2{x^2}{y^2} \cr
& = {\left( {{x^2} - {y^2}} \right)^2} \cr
& = {\left( {{a^2} + \frac{1}{{{a^2}}} + 2 - {a^2} - \frac{1}{{{a^2}}} + 2} \right)^2} \cr
& = {\left( 4 \right)^2} \cr
& = 16 \cr} $$

Answer: Option A. -> $$\frac{{19}}{4}$$
$$\eqalign{
& 2x - \frac{1}{{2x}} = 5 \cr
& {\text{Divide by 2 both side}} \cr
& x - \frac{1}{{4x}} = \frac{5}{2} \cr
& {\text{Squaring both side}} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - 2 \times x \times \frac{1}{{4x}} = \frac{{25}}{4} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - \frac{1}{2} = \frac{{25}}{4} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{25}}{4} + \frac{1}{2} \cr
& \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{27}}{4} \cr
& {\text{So, }} \cr
& {x^2} + \frac{1}{{16{x^2}}} - 2 \cr
& = \frac{{27}}{4} - 2 \cr
& = \frac{{19}}{4} \cr} $$

Answer: Option C. -> 1
$$\eqalign{
& {\text{Given, }}x + \frac{1}{x} = \sqrt {13} {\text{ }} \cr
& {\text{then ,}}\frac{{3x}}{{\left( {{x^2} - 1} \right)}} \cr
& = \frac{3}{{x - \frac{1}{x}}}\,.............(i) \cr
& {\text{Now, }}x + \frac{1}{x} = \sqrt {13} \cr
& {\text{On squaring both side}} \cr
& = {x^2} + \frac{1}{{{x^2}}} \cr
& = 13 - 2 \cr
& = 11 \cr
& = {x^2} + \frac{1}{{{x^2}}} - 2 \cr
& = 11 - 2 \cr
& = 9 \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = {3^2} \cr
& \Rightarrow x - \frac{1}{x} = 3 \cr
& {\text{Put this value in equation (i)}} \cr
& \Rightarrow \frac{3}{{x - \frac{1}{x}}} \cr
& \Rightarrow \frac{3}{3} \cr
& \Rightarrow 1 \cr} $$

Answer: Option B. -> 0
$$\eqalign{
& x + \frac{1}{x} = \sqrt 3 \cr
& {\text{Cubing both side}} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr} $$

Answer: Option C. -> 0
$$\eqalign{
& {\text{Given that,}} \cr
& {\text{ }}x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{ }} \cr
& {\text{Then, }}{x^2} + 2x - 1 \cr
& = {x^2} + 2x - 1 + 1 - 1 \cr
& = {x^2} + 2x + 1 - 2 \cr
& = {\left( {x + 1} \right)^2} - 2 \cr
& {\text{Now put the value of }}x \cr
& = {\left( {\frac{1}{{\left( {\sqrt 2 + 1} \right)}} + 1} \right)^2} - 2 \cr
& = {\left( {\frac{{1 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr
& = {\left( {\frac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr
& = {\left( {\frac{{\left( {\sqrt 2 + 1} \right) \times \sqrt2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr
& = {\left( {\sqrt 2 } \right)^2} - 2 \cr
& = 2 - 2 \cr
& = 0 \cr} $$