Answer: Option D. -> 2
$$\eqalign{
& {\text{ }}x = \frac{{4ab}}{{a + b}}{\text{ }} \cr
& \Rightarrow \frac{x}{{2a}} = {\text{ }}\frac{{2b}}{{a + b}}{\text{ }} \cr
& \frac{{x + 2a}}{{x - 2a}} \cr
& = \frac{{2b + a + b}}{{2b - a - b}} \cr
& = \frac{{3b + a}}{{b - a}} \cr} $$
( By Componendo and Dividendo rule )
$$\eqalign{
& \Rightarrow {\text{again}}\frac{x}{{2b}} = \frac{{2a}}{{a + b}} \cr
& \frac{{x + 2b}}{{x - 2b}} \cr
& = \frac{{2a + a + b}}{{2a - a - b}} \cr
& = \frac{{3a + b}}{{a - b}} \cr
& \Rightarrow \frac{{x + 2a}}{{x - 2a}}{\text{ + }}\frac{{x + 2b}}{{x - 2b}} \cr
& \Rightarrow \frac{{3b + a}}{{b - a}} - \frac{{3a + b}}{{a - b}} \cr
& \Rightarrow \frac{{3b + a - 3a - b}}{{b - a}} \cr
& \Rightarrow \frac{{2b - 2a}}{{b - a}} \cr
& \Rightarrow \frac{{2\left( {b - a} \right)}}{{\left( {b - a} \right)}} \cr
& \Rightarrow 2 \cr} $$

Answer: Option B. -> 1331000
$$\eqalign{
& {\text{ }}{a^{\frac{1}{3}}} = 11 \cr
& \Leftrightarrow a = {11^3} = 1331 \cr
& {a^2} - 331a \cr
& = a\left( {a - 331} \right) \cr
& = 1331\left( {1331 - 331} \right) \cr
& = 1331\left( {1000} \right) \cr
& = 1331000 \cr} $$

Answer: Option D. -> 2
$$\eqalign{
& {a^2} + {b^2} = 2{\text{ }} \cr
& {c^2} + {d^2} = 1 \cr
& {\text{Put values of }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \cr
& {\text{Take ,}}a = b = 1 \cr
& c = 1 \cr
& d = 0 \cr
& \Rightarrow {\left( {ad - bc} \right)^2}{\text{ + }}{\left( {ac - bd} \right)^2} \cr
& \Rightarrow {\left( {0 - 1} \right)^2} + {\left( {1 + 0} \right)^2} \cr
& \Rightarrow {\left( { - 1} \right)^2} + {\left( 1 \right)^2} \cr
& \Rightarrow 2 \cr} $$

Answer: Option C. -> 2
$$\eqalign{
& m + \frac{1}{{m - 2}} = 4 \cr
& \Rightarrow m - 2 + \frac{1}{{m - 2}} = 2 \cr
& \left( {{\text{Squaring the both sides}}} \right) \cr} $$
$$ \Rightarrow {\left( {m - 2} \right)^2}{\text{ + }}\frac{1}{{{{\left( {m - 2} \right)}^2}}} + 2 \times $$      $$\left( {m - 2} \right) \times $$  $$\frac{1}{{\left( {m - 2} \right)}}$$   $$ = 4$$
$$ \Rightarrow {\left( {m - 2} \right)^2}{\text{ + }}\frac{1}{{{{\left( {m - 2} \right)}^2}}} = 2$$

Answer: Option C. -> $$\frac{1}{3}$$
$$\eqalign{
& {a^2} + {b^2} + 2b + 4a + 5 = 0 \cr
& \Rightarrow {a^2} + {b^2} + 2b + 4a + 4 + 1 = 0 \cr
& \Rightarrow {a^2} + 4a + 4 + {b^2} + 2b + 1 = 0 \cr
& \Rightarrow {\left( {a + 2} \right)^2} + {\left( {b + 1} \right)^2} = 0 \cr
& a + 2 = 0{\text{ }} \Rightarrow {\text{ a}} = - 2 \cr
& b + 1 = 0\,\,\,\, \Rightarrow \,\,\,\,b = - 1 \cr
& \frac{{a - b}}{{a + b}} \Rightarrow \frac{{ - 2 + 1}}{{ - 2 - 1}} \cr
& \Rightarrow \frac{{ - 1}}{{ - 3}} = \frac{1}{3} \cr} $$

Answer: Option B. -> 2
$$\eqalign{
& {\text{ }}x + \frac{1}{x} = 2{\text{, }}\,\,\,x \ne 0 \cr
& {\text{Put }}x = 1 \cr
& 1 + 1 = 2 \cr
& \therefore {x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr
& = 1 + 1 \cr
& = 2 \cr} $$

Answer: Option A. -> 0
$$\eqalign{
& {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr
& \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr
& \Rightarrow {x^6} + 1 = 0 \cr
& \Rightarrow {x^6} = - 1 \cr
& \Rightarrow {x^{72}} + {x^{66}} + {x^{54}} + {x^{24}} + {x^6} + 1 \cr
& \Rightarrow {\left( {{x^6}} \right)^{12}} + {\left( {{x^6}} \right)^{11}} + {\left( {{x^6}} \right)^9} + {\left( {{x^6}} \right)^4} + {x^6} + 1 \cr
& \Rightarrow {\left( { - 1} \right)^{12}} + {\left( { - 1} \right)^{11}} + {\left( { - 1} \right)^9} + {\left( { - 1} \right)^4} - 1 + 1 \cr
& \Rightarrow 1 - 1 - 1 + 1 - 1 + 1 \cr
& \Rightarrow 0 \cr} $$

Answer: Option B. -> 9
$$\eqalign{
& {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{ }} \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 0 \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) = 0 \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = 3\left( {x + \frac{1}{x}} \right) \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr
& \,\,\,\,\left( {{\text{Squaring both sides}}} \right) \cr
& \Rightarrow {\left[ {{{\left( {x + \frac{1}{x}} \right)}^2}} \right]^2} = {\left( 3 \right)^2} \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$

Answer: Option A. -> $$\frac{{43}}{{23}}$$
$$\eqalign{
& x + \frac{1}{x} = 5 \cr
& \left( {{\text{By squaring both sides}}} \right) \cr
& {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = {\left( 5 \right)^2} \cr
& {x^2} + \frac{1}{{{x^2}}} = 23 \cr
& {\text{Now,}} \cr
& \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr
& {\text{Divided by }}{x^2}, \cr
& \Rightarrow \frac{{\frac{{{x^4}}}{{{x^2}}} + \frac{{3{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr
& \Rightarrow \frac{{{x^2} + 3x + 5 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr
& \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}} + 3\left( {x + \frac{1}{x}} \right) + 5}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr
& \Rightarrow \frac{{23 + 3\left( 5 \right) + 5}}{{23}} \cr
& \Rightarrow \frac{{43}}{{23}} \cr} $$

Answer: Option B. -> 20
$$\eqalign{
& {a^3} - {b^3} = 56 \cr
& \Rightarrow a - b = 2 \cr
& \,\,\,\,\left( {{\text{By cubing}}} \right) \cr
& \Rightarrow {a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( 2 \right)^2} \cr
& \Rightarrow 56 - 3ab \times 2 = 8 \cr
& \Rightarrow - 6ab = 8 - 56 \cr
& \Rightarrow 6ab = 48 \cr
& \Rightarrow ab = 8 \cr
& \left( {a - b} \right) = 2 \cr
& \,\,{\text{ }}\left( {{\text{By squaring}}} \right) \cr
& \Rightarrow {\left( {a - b} \right)^2} = {\left( 2 \right)^2} \cr
& \Rightarrow {a^2} + {b^2} - 2ab = 4 \cr
& \Rightarrow {a^2} + {b^2} = 4 + 2ab \cr
& \Rightarrow {a^2} + {b^2} = 4 + 2 \times 8 \cr
& \Rightarrow {a^2} + {b^2} = 20 \cr} $$