Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Cube both sides}} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} + 1 = 0 \cr & \therefore {\text{ }}{a^{30}} + {a^{24}} + {a^{18}} + {a^{12}} + {a^6} + 1 = ? \cr & \Rightarrow {\text{ }}{a^{24}}\left( {{a^6} + 1} \right) + {a^{12}}\left( {{a^6} + 1} \right) + {a^6} + 1 \cr & \Rightarrow {a^{24}}\left( 0 \right) + {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & x:y = 3:5{\text{ }} \cr & x - y = - 2 \cr & \frac{x}{y} = \frac{3}{5} \cr & x - y = 3 - 5 \cr & \Leftrightarrow - 2 = - 2 \cr & x = 3,{\text{ }}y = 5 \cr & x + y = 3 + 5 \cr & \Leftrightarrow x + y = 8 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & x = 1 + \sqrt 2 + \sqrt 3 \,.....(i) \cr & y = 1 + \sqrt 2 - \sqrt 3 \,.....(ii) \cr & \Rightarrow \frac{{{x^2} + 4xy + {y^2}}}{{x + y}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & {\text{From equation (i)}} + {\text{(ii)}} \cr & \Rightarrow x + y = 2 + 2\sqrt 2 \cr & xy = {\left( {1 + \sqrt 2 } \right)^2} - {\left( {\sqrt 3 } \right)^2} \cr & \Rightarrow xy = 3 + 2\sqrt 2 - 3 \cr & \Rightarrow xy = 2\sqrt 2 \cr & {\text{So, }}\frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & = \frac{{{{\left( {2 + 2\sqrt 2 } \right)}^2} + 2 \times 2\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{4 + 8 + 8\sqrt 2 + 4\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12 + 12\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12\left( {1 + \sqrt 2 } \right)}}{{2\left( {1 + \sqrt 2 } \right)}} \cr & = 6 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + 1 = 3x\,.....(i) \cr & \Rightarrow {\left( {{x^2} + 1} \right)^2} = {\left( {3x} \right)^2} \cr & \Rightarrow {x^4} + 1 + 2{x^2} = 9{x^2} \cr & \Rightarrow {x^4} + 1 = 7{x^2}\,.....(ii) \cr & \therefore \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{7{x^2} + 3{x^3} + 5{x^2} + 3x}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{12{x^2} + 3{x^3} + 3x}}{{7{x^2}}} \cr & {\text{From equation (i)}} \cr & \Rightarrow \frac{{12x + 3\left( {{x^2} + 1} \right)}}{{7x}} \cr & \Rightarrow \frac{{12x + 3 \times 3x}}{{7x}} \cr & \Rightarrow \frac{{21x}}{{7x}} \cr & \Rightarrow 3 \cr} $$

$$\eqalign{ & p{\left( {x + y} \right)^2} = 5{\text{ and }}q{\left( {x - y} \right)^2} = 3 \cr & {\text{Put the value of }}x = 2{\text{ and }}y = 1 \cr & p{\left( {2 + 1} \right)^2} = 5 \cr & \Leftrightarrow p = \frac{5}{9} \cr & q{\left( {2 - 1} \right)^2} = 3 \cr & \Leftrightarrow q = 3 \cr & {p^2}{\left( {x + y} \right)^2} + 4pqxy - {q^2}{\left( {x - y} \right)^2} \cr} $$
  $$ = {\left( {\frac{5}{9}} \right)^2}{\left( {2 + 1} \right)^2} + 4 \times \frac{5}{9} \times 3 \times $$      $$2 \times $$ $$1 - $$ $${\left( 3 \right)^2}$$ $${\left( {2 - 1} \right)^2}$$
$$\eqalign{ & = \frac{{25}}{{81}} \times 9 + \frac{{40}}{3} - 9 \cr & = \frac{{25}}{9} + \frac{{40}}{3} - 9 \cr & = \frac{{25 + 120 - 81}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Put the value of p and q in option (A)}} \cr & {\text{Option 1}} \to 2\left( {p + q} \right) \cr & = 2\left( {\frac{5}{9} + 3} \right) \cr & = 2 \times \frac{{32}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Option A is satisfied}} \cr & {\text{So, }}2\left( {p + q} \right){\text{ is answer}} \cr} $$

$$\eqalign{ & {\text{ }}{x^2}{\text{ + }}\alpha x + \beta = 0 \cr & {\text{Sum of root}} \cr & \alpha + \beta = \frac{{ - \alpha }}{1}\,......(i) \cr & \alpha \beta = \beta \,......(ii) \cr & {\text{From (i) and (ii)}} \cr & {\text{Then, }}\alpha = 1 \cr & {\text{Then, }}\beta = - 2 \cr & {\text{Then value of }} \cr & \Leftrightarrow {\alpha ^3} + {\beta ^3} \cr & = 1 + {\left( { - 2} \right)^3} \cr & = - 7 \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 1 \cr & {\text{Then, }}{\left( {x + \frac{1}{x}} \right)^2} = 1 + 2 \cr & \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\sqrt 3 } \right)^3} - 3\sqrt 3 \cr & = 3\sqrt 3 - 3\sqrt 3 \cr & = 0 \cr & {\text{Then,}} \cr & {x^{102}} + {x^{96}} + {x^{90}} + {x^{84}} + {x^{78}} + {x^{72}} + 5 \cr} $$
  $$ = {x^{96}}\left( {{x^6} + 1} \right) + $$    $${x^{84}}\left( {{x^6} + 1} \right) + $$   $${x^{72}}\left( {{x^6} + 1} \right) + $$   $$5$$
  $$ = 5$$

If (x - 1) and (x + 1) are the factors y equation then,
$$\eqalign{ & x - 1 = 0 \cr & x = 1 \cr & \Rightarrow {\text{Put }}x = 1{\text{, we get }} \cr & 1 + a - 3 + 2 + b = 0 \cr & a + b = 0\,.....(i) \cr & \Rightarrow x + 1 = 0 \cr & \Rightarrow x = - 1 \cr & {\text{Put }}x = - 1,{\text{we get }} \cr & 1 - a - 3 - 2 + b = 0 \cr & b - a = 4\,.....(ii) \cr & {\text{After solving (i) & (ii),}} \cr & {\text{We get }} \cr & a = - 2,{\text{ }}b = 2 \cr} $$

$$\eqalign{ & {x^3} - 7{x^2} + 11x - 5 \geqslant 0 \cr & \Rightarrow {x^3} - 5{x^2} - 2{x^2} + 10x + x - 5 \geqslant 0 \cr & \Rightarrow {x^2}\left( {x - 5} \right) - 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 2x + 1} \right) \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right){\left( {x - 1} \right)^2} \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x - 1} \right) \geqslant 0 \cr & {\text{So, }}x = 1\& 5 \cr} $$
Equation satisfies at both the values, but the minimum value of these two
x = 1

$$\eqalign{ & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {\left( {26} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {109} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 676 - 218 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 458 \cr} $$