Answer: Option B. -> $$\frac{b}{a}$$
$$\eqalign{
& a + b = 1 \cr
& c + d = 1 \cr
& a - b = \frac{d}{c} \cr
& or\,\,\frac{1}{{a - b}} = \frac{c}{d} \cr
& \Rightarrow \frac{{a + b}}{{a - b}} = \frac{c}{d}\left( {\therefore a + b = 1} \right) \cr
& {\text{By C & D rule }} \cr
& \Rightarrow \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} = \frac{{c + d}}{{c - d}} \cr
& \Rightarrow \frac{{2a}}{{2b}} = \frac{{c + d}}{{c - d}} \cr
& \Rightarrow \frac{a}{b} = \frac{{c + d}}{{c - d}} \cr} $$
Now multiply & divide by (c + d)
$$\eqalign{
& \frac{a}{b} = \frac{{\left( {c + d} \right)}}{{\left( {c - d} \right)}} \times \frac{{\left( {c + d} \right)}}{{\left( {c + d} \right)}} = \frac{{{{\left( {c + d} \right)}^2}}}{{{c^2} - {d^2}}} \cr
& \Rightarrow \frac{a}{b} = \frac{{{{\left( {c + d} \right)}^2}}}{{\left( {{c^2} - {d^2}} \right)}} \cr
& \Rightarrow c + d = 1 \cr
& \Rightarrow \frac{a}{b} = \frac{1}{{{c^2} - {d^2}}} \cr
& \Rightarrow {c^2} - {d^2} = \frac{b}{a} \cr} $$

Answer: Option B. -> $$\frac{1}{2}$$
$$\eqalign{
& x = 3t\,......(i) \cr
& y = \frac{1}{2}\left( {t + 1} \right) \cr
& x = 2y \cr
& \Rightarrow x = 2 \times \frac{1}{2}\left( {t + 1} \right) \cr
& \Rightarrow x = t + 1\,......(ii) \cr
& \therefore 3t = t + 1 \cr
& \left( {{\text{From equation (i) and (ii)}}} \right) \cr
& \Rightarrow 2t = 1 \cr
& \Rightarrow t = \frac{1}{2} \cr} $$

Answer: Option C. -> $$\frac{1}{{10}}$$
$$\eqalign{
& {x^2} + \frac{1}{5}x + {a^2} \cr
& {{\text{A}}^2} + {\text{2}} \times {\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} + {\text{B}}} \right)^2} \cr
& {x^2} + 2 \times \frac{1}{{10}} \times x + {a^2} = {\left( {x + \frac{1}{{10}}} \right)^2} \cr
& {\text{A}} = x \cr
& {\text{B}} = \frac{1}{{10}} \cr
& {\text{B}} = a = \frac{1}{{10}} \cr} $$

Answer: Option D. -> 5
$$\eqalign{
& \frac{{5x}}{{2{x^2} + 5x + 1}} = \frac{1}{3} \cr
& \Rightarrow \frac{5}{{\frac{{2{x^2}}}{x} + \frac{{5x}}{x} + \frac{1}{x}}} = \frac{1}{3} \cr
& \Rightarrow \frac{5}{{2x + \frac{1}{x} + 5}} = \frac{1}{3} \cr
& \Rightarrow 2x + \frac{1}{x} + 5 = 15 \cr
& \Rightarrow 2x + \frac{1}{x} = 10 \cr
& {\text{Divide by 2 both sides }} \cr
& \Rightarrow x + \frac{1}{{2x}} = \frac{{10}}{2} \cr
& \Rightarrow 2x + \frac{1}{x} = 5 \cr} $$

Answer: Option B. -> k1 = 2, k2 = -3
$$\eqalign{
& \therefore {\text{When}}\left( {x - 1} \right) = 0 \cr
& x = 1 \cr
& {x^2} + {k_1}x + {k_2}{\text{ }} = 0 \cr
& \Rightarrow 1 + {k_1} + {k_2}{\text{ }} = 0 \cr
& \Rightarrow {k_1} + {k_2} = - 1\,.......(i) \cr
& {\text{When}}\left( {x + 3} \right) = 0 \cr
& x = - 3 \cr
& 9 - 3{k_1} + {k_2}{\text{ }} = 0 \cr
& \Rightarrow - 3{k_1} + {k_2}{\text{ }} = - 9\,.......(ii) \cr
& {\text{From equation (i) and (ii)}} \cr
& {k_1} = 2,\,\,\,\,\,\,{\text{ }}{k_2} = - 3 \cr} $$

Answer: Option C. -> $$\frac{7}{8}$$
$$\eqalign{
& {\text{ }}2x + \frac{2}{x} = 3 \cr
& \Rightarrow {\text{ }}x + \frac{1}{x} = \frac{3}{2} \cr
& {\text{Taking cube on both sides}} \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\frac{3}{2}} \right)^3}{\text{ }} \cr
& x + \frac{1}{x} = a \cr
& {x^3} + \frac{1}{{{x^3}}} = {a^3} - 3a \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right) = \frac{{27}}{8} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times \frac{3}{2} = \frac{{27}}{8} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{27}}{8} - \frac{9}{2} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{ - 9}}{8} \cr
& \therefore {x^3} + \frac{1}{{{x^3}}} + 2 \cr
& = \frac{{ - 9}}{8} + 2 \cr
& = \frac{{ - 9 + 16}}{8} \cr
& = \frac{7}{8} \cr} $$

Answer: Option B. -> 2
$$\eqalign{
& x + \frac{1}{{x + 1}} = 1 \cr
& {\text{Adding both 1 sides}} \cr
& \Rightarrow x + 1 + \frac{1}{{x + 1}} = 1 + 1 \cr
& \Rightarrow \left( {x + 1} \right) + \frac{1}{{\left( {x + 1} \right)}} = 2 \cr
& {\text{Put }}x + 1 = 1 \cr
& {\text{And }}\frac{1}{{x + 1}} = 1 \cr
& \therefore {\left( {x + 1} \right)^5}{\text{ + }}\frac{1}{{{{\left( {x + 1} \right)}^5}}} \cr
& = 1 + 1 \cr
& = 2 \cr} $$

Answer: Option C. -> 1000
x = y = 333,     z = 334
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2]
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (333 + 333 + 334) (333 - 333)2 + (333 - 334)2 + (334 - 333)2
⇒ x3 + y3 + z3 - 3xyz = $$\frac{1}{2}$$ (1000) (0 + 1 + 1)
⇒ x3 + y3 + z3 - 3xyz = 1000

Answer: Option B. -> 180
$$\eqalign{
& a + b + c = 15{\text{ }} \cr
& {a^2} + {b^2} + {c^2} = 83{\text{ }}\left( {{\text{Given}}} \right) \cr
& \therefore a + b + c = 15 \cr
& \left( {{\text{Squaring both sides}}} \right){\text{ }} \cr
& \Rightarrow {\left( {a + b + c} \right)^2} = {\left( {15} \right)^2}{\text{ }} \cr
& \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 225 \cr
& \Rightarrow 83 + 2\left( {ab + bc + ca} \right) = 225 \cr
& \Rightarrow 2\left( {ab + bc + ca} \right) = 142 \cr
& \Rightarrow ab + bc + ca = 71 \cr} $$
  $$\therefore {a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$  $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$\eqalign{
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15\left( {83 - 71} \right) \cr
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15 \times 12 \cr
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 180 \cr} $$

Answer: Option D. -> 3abc
$$\eqalign{
& {\text{If }}a + b + c = 0 \cr
& {\text{then,}} \cr
& {a^3} + {b^3} + {c^3} - 3abc = 0 \cr
& \Leftrightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$