Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & {\text{Put x = 1}} \cr & \therefore {\text{1 + }}\frac{1}{{\left( 1 \right)}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisty}}} \right) \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = \left( {1 + 1} \right)\left( {1 + 1} \right) \cr & = 2 \times 2 \cr & = 4 \cr} $$

$$\eqalign{ & {\text{2}}{x^2} - 7x + 12 = 0 \cr & {\text{roots are }}\alpha {\text{ , }}\beta \cr & \therefore \alpha \beta = + \frac{c}{a},\alpha + \beta = \frac{{ - b}}{a} \cr & \therefore \alpha + \beta = + \frac{7}{2},\alpha \beta = \frac{{12}}{2} = 6 \cr & \therefore \frac{\alpha }{\beta } + \frac{\beta }{\alpha } \cr & = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} \cr & = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} \cr & = \frac{{{{\left( {\frac{7}{2}} \right)}^2} - 2 \times 6}}{6} \cr & = \frac{{\frac{{49}}{4} - 12}}{6} \cr & = \frac{{49 - 48}}{{6 \times 4}} \cr & = \frac{1}{{24}} \cr} $$

$$\eqalign{ & {x^3} + \frac{3}{x} = 4\left( {{a^3} + {b^3}} \right)\, . . . . . {\text{(i)}} \cr & 3x + \frac{1}{{{x^3}}} = 4\left( {{a^3} - {b^3}} \right)\, . . . . . (ii) \cr & {\text{Equation (i)}} + {\text{(ii)}} \cr & {\left( {x + \frac{1}{x}} \right)^3} = 8{a^3} \cr & \Leftrightarrow x + \frac{1}{x} = 2a\, . . . . . (iii) \cr & Equation{\text{ }}(i) - (ii) \cr & x - \frac{1}{x} = 2b\, . . . . . (iv) \cr & Equation{\text{ }}(iii) - (iv) \cr & 2\left( {a - b} \right) = \frac{2}{x} \cr & \Rightarrow a - b = \frac{1}{x} \cr & \Rightarrow a + b = x \cr & \because {a^2} - {b^2} = \left( {a + b} \right) \times \left( {a - b} \right) \cr & \Rightarrow {a^2} - {b^2} = x \times \frac{1}{x} \cr & \Rightarrow {a^2} - {b^2} = 1 \cr} $$

$$\eqalign{ & 2x + 1 = 0 \cr & \Leftrightarrow x = \frac{{ - 1}}{2}{\text{ }} \cr & 3y - 9 = 0 \cr & \Leftrightarrow y = \frac{9}{3} = 3 \cr & \therefore \left( {x,y} \right) = \left( {\frac{{ - 1}}{2},3} \right) \cr & {\text{Point of intersection}} \cr} $$

$$\eqalign{ & {\text{121}}{a^2} + 64{b^2} \cr & = {\left( {11a} \right)^2} + {\left( {8b} \right)^2} + 2 \times 11a \times 8b \cr & = {\left( {11a + 8b} \right)^2} \cr} $$
∴ So, term added to make perfect square = 176 ab

$$\eqalign{ & a = 2 + \sqrt 3 \cr & \Rightarrow {a^2} = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow {a^2} = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow {a^2} = 7 + 4\sqrt 3 \cr & \frac{1}{{{a^2}}} = \frac{1}{{7 + 4\sqrt 3 }} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{1} \cr & \Rightarrow \frac{1}{{{a^2}}} = 7 - 4\sqrt 3 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} \cr & = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 \cr & = 14 \cr} $$

$$\eqalign{ & p + \frac{1}{4} + \sqrt p + {k^2} \cr & = p + \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {\left( {\sqrt p } \right)^2} + 2 \times \frac{1}{2} \times \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {{\text{A}}^2} + {\text{2}} \times {\text{A}} \times {\text{B}} + {{\text{B}}^2} \cr & {\text{A}} = \sqrt p \cr & {{\text{B}}^2} = \left( {{k^2} + \frac{1}{4}} \right) \cr & {\text{B}} = \frac{1}{2} \cr & \therefore {k^2} + \frac{1}{4} = {\left( {\frac{1}{2}} \right)^2} \cr & \Rightarrow {k^2} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {k^2} = 0 \cr & \Rightarrow k = 0 \cr} $$

$$\eqalign{ & \frac{{b - c}}{a}{\text{ + }}\frac{{a + c}}{b}{\text{ + }}\frac{{a - b}}{c} = 1 \cr & a - b + c \ne 0 \cr & {\text{Let }}b = c \cr & \therefore \frac{{b - b}}{a}{\text{ + }}\frac{{a + b}}{b}{\text{ + }}\frac{{a - b}}{b} = 1 \cr & \Rightarrow 0 + \frac{a}{b} + 1 + \frac{a}{b} - 1 = 1 \cr & \Rightarrow \frac{a}{b} + \frac{a}{b} = 1 \cr & \Rightarrow \frac{1}{b} + \frac{1}{b} = \frac{1}{a} \cr & {\text{We take }}b = c \cr & \therefore \boxed{\frac{1}{b} + \frac{1}{c} = \frac{1}{a}} \cr} $$

$$\eqalign{ & {\text{Reciprocal of }}\left( {x + \frac{1}{x}} \right){\text{ }} \cr & = \frac{1}{{\left( {x + \frac{1}{x}} \right)}} \cr & = \frac{x}{{{x^2} + 1}} \cr} $$

$$\eqalign{ & {\text{For minimum value of}} \cr & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \cr & a = b = c \cr & a + b + c = 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore a = b = c = \frac{1}{3} \cr & \frac{1}{a} = \frac{1}{b} = \frac{1}{c} = 3 \cr & \therefore {\text{Minimum value of,}} \cr & = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \cr & = 3 + 3 + 3 \cr & = 9 \cr} $$