Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {a^2} + {b^2} = ab\,........(i) \cr & {a^2} + {b^2} - ab = 0 \cr & \because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( 0 \right) \cr & \Rightarrow {a^3} + {b^3} = 0 \cr} $$

$$\eqalign{ & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr} $$
(On adding and subtracting one both side)
$$\eqalign{ & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {100^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr} $$

$$\eqalign{ & {a^3} - {b^3} = 61 \cr & \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = 61 \cr & a - b = 1{\text{ }}\left( {{\text{ }}Given} \right) \cr & 1 \times {a^2} + ab + {b^2} = 61\,......(i) \cr & {\text{Now,}}a - b = 1 \cr & {\text{On squaring both sides}} \cr & {a^2} + {b^2} - 2ab = 1 \cr & {a^2} + {b^2} + ab - 3ab = 1 \cr & {\text{From equation (i)}} \cr & \Rightarrow 61 - 3ab = 1 \cr & \Rightarrow 3ab = 60 \cr & \Rightarrow ab = \frac{{60}}{3} \cr & \Rightarrow ab = 20 \cr} $$

$$\eqalign{ & {\text{If }}{x^2} + 5x + 6 = 0 \cr & {\text{then, }}{x^2} + 6 = - 5x \cr & {\text{So,}}\frac{{2x}}{{{x^2} + 6 - 7x}} \cr & = \frac{{2x}}{{ - 5x - 7x}} \cr & = \frac{{2x}}{{ - 12x}} \cr & = - \frac{1}{6}{\text{ }} \cr} $$

$$\eqalign{ & a + b = 5{\text{ }} \cr & \Leftrightarrow {a^2} + {b^2} + 2ab = 25\,.....(i) \cr & a - b = 3 \cr & \Leftrightarrow {a^2} + {b^2} - 2ab = 9\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Leftrightarrow {\text{2}}\left( {{a^2} + {b^2}} \right) = 34 \cr & \Leftrightarrow {a^2} + {b^2} = 17 \cr} $$

$$\eqalign{ & x = a + \frac{1}{a} \cr & {x^2} = {a^2} + \frac{1}{{{a^2}}} + 2 \cr & {y^2} = {a^2} + \frac{1}{{{a^2}}} - 2 \cr & {\text{Now, }} \cr & {x^4} + {y^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {{a^2} + \frac{1}{{{a^2}}} + 2 - {a^2} - \frac{1}{{{a^2}}} + 2} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & {\text{Cubing both side}} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr} $$

$$\eqalign{ & 2x - \frac{1}{{2x}} = 5 \cr & {\text{Divide by 2 both side}} \cr & x - \frac{1}{{4x}} = \frac{5}{2} \cr & {\text{Squaring both side}} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - 2 \times x \times \frac{1}{{4x}} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - \frac{1}{2} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{25}}{4} + \frac{1}{2} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{27}}{4} \cr & {\text{So, }} \cr & {x^2} + \frac{1}{{16{x^2}}} - 2 \cr & = \frac{{27}}{4} - 2 \cr & = \frac{{19}}{4} \cr} $$

$$\eqalign{ & {\text{Given that,}} \cr & {\text{ }}x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{ }} \cr & {\text{Then, }}{x^2} + 2x - 1 \cr & = {x^2} + 2x - 1 + 1 - 1 \cr & = {x^2} + 2x + 1 - 2 \cr & = {\left( {x + 1} \right)^2} - 2 \cr & {\text{Now put the value of }}x \cr & = {\left( {\frac{1}{{\left( {\sqrt 2 + 1} \right)}} + 1} \right)^2} - 2 \cr & = {\left( {\frac{{1 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\left( {\sqrt 2 + 1} \right) \times \sqrt2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\sqrt 2 } \right)^2} - 2 \cr & = 2 - 2 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = \sqrt {13} {\text{ }} \cr & {\text{then ,}}\frac{{3x}}{{\left( {{x^2} - 1} \right)}} \cr & = \frac{3}{{x - \frac{1}{x}}}\,.............(i) \cr & {\text{Now, }}x + \frac{1}{x} = \sqrt {13} \cr & {\text{On squaring both side}} \cr & = {x^2} + \frac{1}{{{x^2}}} \cr & = 13 - 2 \cr & = 11 \cr & = {x^2} + \frac{1}{{{x^2}}} - 2 \cr & = 11 - 2 \cr & = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = {3^2} \cr & \Rightarrow x - \frac{1}{x} = 3 \cr & {\text{Put this value in equation (i)}} \cr & \Rightarrow \frac{3}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{3}{3} \cr & \Rightarrow 1 \cr} $$