Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & a\left( {2 + \sqrt 3 } \right) = b\left( {2 - \sqrt 3 } \right) = 1 \cr & a = \frac{1}{{\left( {2 + \sqrt 3 } \right)}} \cr & b = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \Rightarrow a = \frac{1}{b} \cr & \Rightarrow \frac{1}{{{a^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{1}{{{b^2}}} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{{1 + {b^2}}}{{{b^2}}}}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2}}}{{{b^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2} + 1}}{{{b^2} + 1}} \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & \left( {2 + \sqrt 3 } \right)a = \left( {2 - \sqrt 3 } \right)b = 1 \cr & \Rightarrow \frac{1}{a} = \left( {2 + \sqrt 3 } \right) \cr & {\text{By rationals}} \cr & \Rightarrow \frac{1}{b} = \left( {2 - \sqrt 3 } \right) \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} = 4 \cr} $$

$$a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$$
To save your time assume values of a, b, c according to the question.
$$\eqalign{ & {\text{Let }}a = 2,{\text{ }}b = - 1\& c = \frac{1}{2} \cr & 2 + \frac{1}{{ - 1}} = - 1 + \frac{1}{{\frac{1}{2}}} = \frac{1}{2} + \frac{1}{2} \cr & \therefore abc = 2 \times - 1 \times \frac{1}{2} = - 1 \cr} $$

$$\eqalign{ & \frac{x}{y} = \frac{4}{5} \cr & \frac{4}{7} + \frac{{2y - x}}{{2y + x}} \cr & = \frac{4}{7} + \frac{{y\left( {2 - \frac{x}{y}} \right)}}{{y\left( {2 + \frac{x}{y}} \right)}} \cr & = \frac{4}{7} + \frac{{\left( {2 - \frac{4}{5}} \right)}}{{\left( {2 + \frac{4}{5}} \right)}} \cr & = \frac{4}{7} + \frac{{10 - 4}}{{10 + 4}} \cr & = \frac{4}{7} + \frac{6}{{14}} \cr & {\text{ = }}\frac{{8 + 6}}{{14}} \cr & {\text{ = }}\frac{{14}}{{14}} \cr & {\text{ = 1}} \cr} $$

$$\eqalign{ & a + b = 12\,.....{\text{(i)}} \cr & ab = 22\,.....{\text{(ii)}} \cr} $$
Squaring both sides of equation (i)
$$\eqalign{ & \Rightarrow {a^2} + {b^2} + 2ab = 144 \cr & \Rightarrow {a^2} + {b^2} + 2 \times 22 = 144 \cr & \Rightarrow {a^2} + {b^2} = 144 - 44 \cr & \Rightarrow {a^2} + {b^2} = 100 \cr} $$

$$\eqalign{ & x = \sqrt 3 - \frac{1}{{\sqrt 3 }}{\text{ and }}y = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{{{x^2}}}{y} + \frac{{{y^2}}}{x} \cr & = \frac{{{x^3} + {y^3}}}{{xy}} \cr & = \frac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy}} \cr & \therefore x + y \cr & = \sqrt 3 - \frac{1}{{\sqrt 3 }} + \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & = 2\sqrt 3 \cr & \therefore xy \cr & = \sqrt 3 - \frac{1}{{\sqrt 3 }} \times \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & = 3 - \frac{1}{3} \cr & = \frac{8}{3} \cr & \Rightarrow \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2} + 2xy - 2xy - xy} \right)}}{{xy}} \cr & \Rightarrow \frac{{\left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 3xy} \right)}}{{xy}} \cr & \Rightarrow \frac{{2\sqrt 3 \left( {{{\left( {2\sqrt 3 } \right)}^2} - 3 \times \frac{8}{3}} \right)}}{{\frac{8}{3}}} \cr & \Rightarrow \frac{{2\sqrt 3 \left( {12 - 8} \right)}}{{\frac{8}{3}}} \cr & \Rightarrow \frac{{2 \times 3\sqrt 3 \left( 4 \right)}}{8} \cr & \Rightarrow 3\sqrt 3 \cr} $$

$$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & {\text{Put }} \cr & a = 0 \cr & b = 0 \cr & c = 2 \cr & d = 2 \cr & \therefore a + b + c + d = 4 \cr & \Rightarrow 0 + 0 + 2 + 2 = 4 \cr & \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr & \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr & = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr & = - 1 + 1 + 1 - 1 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0 \cr & \Rightarrow {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} = - {c^{\frac{1}{3}}} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)^3} = {\left( { - {c^{\frac{1}{3}}}} \right)^3} \cr & \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right) = - c \cr & \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }} - {c^{\frac{1}{3}}}} \right) = - c \cr & \Rightarrow a + b + c = 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again taking cube }} \cr & \Rightarrow {\left( {a + b + c} \right)^3} = 27abc \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 1{\text{ }} \cr & \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr & {\text{Divide and multiply by }}x \cr & \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr & \Rightarrow \frac{4}{8} \cr & \Rightarrow \frac{1}{2} \cr} $$

$$\eqalign{ & x + y = 15 \cr & \Rightarrow x - 10 = 5 - y \cr & \Rightarrow x - 10 = - \left( {y - 5} \right) \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - 10} \right)^3} = - {\left( {y - 5} \right)^3} \cr & \Rightarrow {\left( {x - 10} \right)^3} + {\left( {y - 5} \right)^3} = 0 \cr} $$