Answer: Option C. -> 2
$$\eqalign{
& x = \sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } \cr
& \Rightarrow {x^2} = 2 + \sqrt {2 + \sqrt {2 + .......} } \cr
& \Rightarrow {x^2} = 2 + x \cr
& \Rightarrow {x^2} - x - 2 = 0 \cr
& \Rightarrow x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0 \cr
& \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right) = 0 \cr
& \Rightarrow x = 2 \cr} $$

Answer: Option A. -> 0
$$\eqalign{
& {\text{2 + }}\sqrt {0.09} - \root 3 \of {0.008} - 75\% \,{\text{of }}2.80 \cr
& = 2 + 0.3 - 0.2 - \left( {\frac{3}{4} \times 2.8} \right) \cr
& = 2 + 0.3 - 0.2 - 2.10 \cr
& = 2.3 - 2.3 \cr
& = 0 \cr} $$

Answer: Option D. -> 198
$$\eqalign{
& {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr
& = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$
  $$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$      $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$
$$\eqalign{
& = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr
& = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr
& a = 3 - 2\sqrt 2 \cr
& b = 3 + 2\sqrt 2 \cr
& \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr
& = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr
& = \left( 6 \right)\left( {33} \right) \cr
& = 198 \cr} $$

Answer: Option D. -> 25
$$\eqalign{
& \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 2y} \right)}} = \frac{5}{6} \cr
& \Rightarrow 18x - 12y = 10x + 15y \cr
& \Rightarrow 8x = 27y \cr
& \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr
& \Rightarrow {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr
& \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr
& \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr
& \Rightarrow {\left( 5 \right)^2} \cr
& \Rightarrow 25 \cr} $$

Answer: Option C. -> 5.9
$$\eqalign{
& {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr
& \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr
& \Rightarrow ? + 6.5 = 12.4 \cr
& \Rightarrow ? = 12.4 - 6.5 \cr
& \Rightarrow ? = 5.9 \cr} $$

Answer: Option D. -> 275 × 1025
$$\eqalign{
& {\text{Expression,}} \cr
& {\text{ = }}\frac{{{{\left( {10} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr
& = \frac{{{{\left( {2 \times 5} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr
& = \frac{{{{\left( 2 \right)}^{100}} \times {{\left( 5 \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr
& = {2^{100}} \times \frac{{{5^{100}}}}{{{5^{75}}}} \cr
& = {2^{100}} \times {5^{\left( {100 - 75} \right)}}.....\left[ {\because \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right] \cr
& = {2^{100}} \times {5^{25}} \cr
& = {2^{25}} \times {5^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr
& = {\left( {10} \right)^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {b^m} = a{b^m}} \right] \cr} $$

Answer: Option C. -> $${{\text{6}}^{\frac{1}{4}}}$$
$$\eqalign{
& {\text{The exponential form of }} \cr
& \sqrt {\sqrt 2 \times \sqrt 3 } \cr
& = \sqrt {{6^{\frac{1}{2}}}} \cr
& = {\left( {{6^{\frac{1}{2}}}} \right)^{\frac{1}{2}}} \cr
& = {6^{\frac{1}{4}}} \cr} $$

Answer: Option C. -> 1
$$\eqalign{
& \frac{1}{{1 + \sqrt 2 + \sqrt 3 }} + \frac{1}{{1 - \sqrt 2 + \sqrt 3 }} \cr
& = \frac{1}{{1 + \sqrt 3 + \sqrt 2 }} + \frac{1}{{1 + \sqrt 3 - \sqrt 2 }} \cr
& = \frac{{1 + \sqrt 3 - \sqrt 2 + 1 + \sqrt 3 + \sqrt 2 }}{{{{\left( {1 + \sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \frac{{2 + 2\sqrt 3 }}{{4 + 2\sqrt 3 - 2}} \cr
& = \frac{{2 + 2\sqrt 3 }}{{2 + 2\sqrt 3 }} \cr
& = 1 \cr} $$

Answer: Option A. -> $${\text{2}}\sqrt 2 $$
$$\eqalign{
& {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr
& = x + \frac{1}{x} - 2 \cr
& = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr
& = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr
& = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr
& = 10 - 2 \cr
& = 8 \cr
& \therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$

Answer: Option A. -> $$ - \frac{1}{{36}}$$
$$\eqalign{
& \frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}} \cr
& = \frac{{{2^{n - 1}}\left( {1 - 2} \right)}}{{{2^{n + 1}}\left( {{2^3} + 1} \right)}} \cr
& = \left( { - \frac{1}{9}} \right){.2^{\left( {n - 1} \right) - \left( {n + 1} \right)}} \cr
& = \left( { - \frac{1}{9}} \right){.2^{ - 2}} \cr
& = \left( { - \frac{1}{9}} \right).\frac{1}{{{2^2}}} \cr
& = \left( { - \frac{1}{9}} \right) \times \frac{1}{4} \cr
& = - \frac{1}{{36}} \cr} $$