Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\text{Subtracting 2 from both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} = 83 - 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & \Rightarrow x - \frac{1}{x} = 9 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3{\text{ }}\left( {x - \frac{1}{x}} \right) = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times {\text{9}} = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 729 + 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$

Solution:

 $$\eqalign{ & \because p = 99 \cr & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {\left( {100} \right)^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & x = - 2k{\text{ and }}y = 1 - 3k \cr & \therefore {\text{For }}x = y \cr & \Rightarrow - 2k = 1 - 3k \cr & \Rightarrow k = 1 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 2{\text{ }} \cr & {\text{But }}x = 1 \cr & {\text{So, }}{x^2} + \frac{1}{{{x^2}}} = {1^2} + \frac{1}{{{1^2}}} = 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & \because a = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}{\text{, }}b = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{Find }}\frac{{{a^2}}}{b} + \frac{{{b^2}}}{a} = \,? \cr & \Rightarrow \frac{{{a^3} + {b^3}}}{{ab}} = \,? \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - 3ab\left( {a + b} \right)}}{{ab}} = \,? \cr & \Rightarrow a + b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} + \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \Rightarrow \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2} + {{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\sqrt 3 }^2} - {{\sqrt 2 }^2}}} \cr & \Rightarrow \frac{{2\left( {{{\sqrt 3 }^2} + {{\sqrt 2 }^2}} \right)}}{{3 - 2}} \cr & \Rightarrow \frac{{2 \times \left( 5 \right)}}{1} \cr & \Rightarrow a + b = 10 \cr & {\text{Again, }} \cr & \Rightarrow a \times b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \Rightarrow ab = 1 \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^3} - 3ab\left( {a + b} \right)}}{{ab}} \cr & \Rightarrow \frac{{{{10}^3} - 3 \times 1 \times 10}}{1} \cr & \Rightarrow 1000 - 30 \cr & \Rightarrow 970 \cr} $$

Solution:

 $$\eqalign{ & a + b + c = 0 \cr & {\text{Let, }}{a^3} + {b^3} + {c^3} = T \cr} $$ $$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \frac{1}{2}\left( {a + b + c} \right)$$      $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$ $$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)$$     $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$ $$\eqalign{ & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \Rightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$

Solution:

 $$\eqalign{ & \frac{{\left( {a + b} \right)}}{c} = \frac{6}{5} \cr & \frac{{\left( {b + c} \right)}}{a} = \frac{9}{2} \cr & \frac{{\left( {a + c} \right)}}{b} = ? \cr & c = 5,\,\frac{{\left( {b + c} \right)}}{a} = \frac{9}{2} \cr & \Rightarrow \frac{{b + 5}}{2} = \frac{9}{2} \cr & b = 4,\,a = 2 \cr & \Rightarrow \frac{{\left( {a + c} \right)}}{b} = \frac{{2 + 5}}{4} \cr & \Rightarrow \frac{{\left( {a + c} \right)}}{b} = \frac{7}{4} \cr} $$

Solution:

 $$\eqalign{ & \left( {x - 2} \right)\left( {x - 9} \right) \cr & = {x^2} - 9x - 2x + 18 \cr & = {x^2} - 11x + 18 \cr & = a{x^2} + bx + c = 0 \cr & {\text{For minimum value}} \cr & = \frac{{4ac - {b^2}}}{{4a}} \cr & = \frac{{4 \times 1 \times 18 - {{\left( { - 11} \right)}^2}}}{{4 \times 1}} \cr & = \frac{{72 - 121}}{4} \cr & = \frac{{ - 49}}{4} \cr & = - \frac{{49}}{4} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given, }}{x^2} + \frac{1}{{{x^2}}} = \frac{7}{4} \cr & {\text{Adding 2 to both sides}} \cr & {x^2} + \frac{1}{{{x^2}}} + 2 = \frac{7}{4} + 2 \cr & {\left( {x + \frac{1}{x}} \right)^2} = \frac{{15}}{4} \cr & x + \frac{1}{x} = \frac{{\sqrt {15} }}{2} \cr & \because \,{x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) \cr & \therefore \,{x^3} + \frac{1}{{{x^3}}} = \frac{{15 \times \sqrt {15} }}{8} - 3 \times \frac{{\sqrt {15} }}{2} \cr & = \frac{{15\sqrt {15} - 12\sqrt {15} }}{8} \cr & = \frac{{3\sqrt {15} }}{8} \cr} $$

Solution:

 $$\because $$ x = 11 x5 - 12x4 + 12x3 - 12x2 + 12x - 1 = x5 - 11x4 - x4 + 11x3 + x3 - 11x2 - x2 + 11x + x - 1 = 115 - 11.114 - 114 + 11.113 + 113 - 11.112 - 112 + 11.11 + 11 - 1 = 0 - 0 + 0 + 0 + 11 - 1 = 10