Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\text{ }}x + \frac{1}{x} = 2{\text{, }}\,\,\,x \ne 0 \cr & {\text{Put }}x = 1 \cr & 1 + 1 = 2 \cr & \therefore {x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \Rightarrow {x^6} + 1 = 0 \cr & \Rightarrow {x^6} = - 1 \cr & \Rightarrow {x^{72}} + {x^{66}} + {x^{54}} + {x^{24}} + {x^6} + 1 \cr & \Rightarrow {\left( {{x^6}} \right)^{12}} + {\left( {{x^6}} \right)^{11}} + {\left( {{x^6}} \right)^9} + {\left( {{x^6}} \right)^4} + {x^6} + 1 \cr & \Rightarrow {\left( { - 1} \right)^{12}} + {\left( { - 1} \right)^{11}} + {\left( { - 1} \right)^9} + {\left( { - 1} \right)^4} - 1 + 1 \cr & \Rightarrow 1 - 1 - 1 + 1 - 1 + 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {a^3} - {b^3} = 56 \cr & \Rightarrow a - b = 2 \cr & \,\,\,\,\left( {{\text{By cubing}}} \right) \cr & \Rightarrow {a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( 2 \right)^2} \cr & \Rightarrow 56 - 3ab \times 2 = 8 \cr & \Rightarrow - 6ab = 8 - 56 \cr & \Rightarrow 6ab = 48 \cr & \Rightarrow ab = 8 \cr & \left( {a - b} \right) = 2 \cr & \,\,{\text{ }}\left( {{\text{By squaring}}} \right) \cr & \Rightarrow {\left( {a - b} \right)^2} = {\left( 2 \right)^2} \cr & \Rightarrow {a^2} + {b^2} - 2ab = 4 \cr & \Rightarrow {a^2} + {b^2} = 4 + 2ab \cr & \Rightarrow {a^2} + {b^2} = 4 + 2 \times 8 \cr & \Rightarrow {a^2} + {b^2} = 20 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 5 \cr & \left( {{\text{By squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = {\left( 5 \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = 23 \cr & {\text{Now,}} \cr & \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & {\text{Divided by }}{x^2}, \cr & \Rightarrow \frac{{\frac{{{x^4}}}{{{x^2}}} + \frac{{3{x^3}}}{{{x^2}}} + \frac{{5{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}}}{{\frac{{{x^4}}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + 3x + 5 + \frac{3}{x} + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}} + 3\left( {x + \frac{1}{x}} \right) + 5}}{{{x^2} + \frac{1}{{{x^2}}}}} \cr & \Rightarrow \frac{{23 + 3\left( 5 \right) + 5}}{{23}} \cr & \Rightarrow \frac{{43}}{{23}} \cr} $$

$$\eqalign{ & {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{ }} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = 3\left( {x + \frac{1}{x}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \,\,\,\,\left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\left[ {{{\left( {x + \frac{1}{x}} \right)}^2}} \right]^2} = {\left( 3 \right)^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \left( {{\text{Squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{On cubing both sides}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 18 \times 7 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + \left( {x + \frac{1}{x}} \right) = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + 3 = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) = 123 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 3 \cr & \left( {{\text{By cubing both sides}}} \right) \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.\left( 3 \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 36 \cr} $$

$$\eqalign{ & x + y + z = 6 \cr & {\left( {x - 1} \right)^3}{\text{ + }}{\left( {y - 2} \right)^3}{\text{ + }}{\left( {z - 3} \right)^3} \cr & \therefore {\text{As, }}x + y + z = 6 \cr & {\text{Take values}} \cr & x = 1 \cr & y = 2 \cr & z = 3 \cr & \left( {1 + 2 + 3} \right) = 6 \cr & \therefore {\left( {1 - 1} \right)^3}{\text{ + }}{\left( {2 - 2} \right)^3}{\text{ + }}{\left( {3 - 3} \right)^3} \cr & = 0 \cr} $$
Now assume values in options.
Option 'D' satisfies the given relation.
Hence 'D' is correct.

$$\eqalign{ & a + b = 1 \cr & {\text{By cubing,}} \cr & {a^3} + {b^3} + 3ab\left( {a + b} \right) = {1^3} \cr & \Rightarrow {a^3} + {b^3} + 3ab = 1\left[ {\because a + b = 1} \right] \cr & \Rightarrow {a^3} + {b^3} + 3ab = k \cr & {\text{From above both equations,}} \cr & k = 1 \cr} $$

$$\eqalign{ & 2x + y = 5..............(i) \cr & x + 2y = 4..............(ii) \cr & {\text{Multiply equation (ii) by 2}} \cr & 2x + 4y = 8............(iii) \cr} $$
Now subtracting equation (i) from (iii)
$$\eqalign{ & {\text{ }}2x + 4y = 8 \cr & \mathop {}\limits_ - 2x\mathop + \limits_ - \,\,y\, = 5 \cr & \overline {\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3y = 3{\text{ }}} } \cr & y = 1 \cr & x = 2 \cr & \therefore {\text{Intersection point = }}\left( {2,1} \right) \cr} $$