Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

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

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

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

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

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

a + b + c = 0
$$\frac{1}{{\left( {a + b} \right)\left( {b + c} \right)}} + $$   $$\frac{1}{{\left( {a + c} \right)\left( {b + a} \right)}} + $$   $$\frac{1}{{\left( {c + a} \right)\left( {c + b} \right)}}$$
$$\eqalign{ & \Rightarrow \frac{{\left( {a + c} \right) + \left( {b + c} \right) + \left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right)}} \cr & \Rightarrow \frac{{2\left( {a + b + c} \right)}}{{\left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right)}}{\text{ }}\left( {\because a + b + c = 0} \right) \cr & \Rightarrow 0{\text{ }} \cr} $$

$$\eqalign{ & {x^2} + {y^2} - 4x - 4y + 8 = 0 \cr & \Rightarrow {x^2} + 4 - 4x + {y^2} + 4 - 4y = 0 \cr & \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = 0 \cr & {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2 \cr & {\left( {y - 2} \right)^2} = 0 \Leftrightarrow y = 2 \cr & \therefore x - y \cr & \Rightarrow 2 - 2 \cr & \Rightarrow 0 \cr} $$

x = b + c - 2a
y = c + a - 2b
z = a + b - 2c
⇒ x + y + z = (b + c - 2a) + (c + a - 2b) + (a + b - 2c) = 0
∴ Now,
⇒ x² + y² + 2xy - z²
⇒ (x + y)² - z² [∴ (a² - b²) = (a + b) (a - b)]
⇒ (x + y - z) (x + y + z)
As we know, (x + y + z) = 0
∴ x² + y² - z² + 2xy
= 0 × (x + y - z)
= 0

$$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & \left( {{\text{Take cube on both sides}}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \therefore {x^6} = - 1 \cr & \therefore {x^{18}} + {x^{12}} + {x^6} + 1 \cr & = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 \cr & = - 1 + 1 - 1 + 1 \cr & = 0 \cr} $$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & {\text{Divide by }}x \cr & \frac{{{x^2}}}{x} + \frac{1}{x} = \frac{{3x}}{x} \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 18 \cr} $$