Practice MCQ Questions and Answer on Algebra

Solution:

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

Solution:

 $$\eqalign{ & {t^2} - 4t + 1 = 0 \cr & \Rightarrow {t^2} + 1 = 4t \cr & \Rightarrow \frac{{{t^2} + 1}}{t} = \frac{{4t}}{t} \cr & \Rightarrow t + \frac{1}{t} = 4 \cr & \left[ {{\text{Take cube both sides}}} \right] \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3.t.\frac{1}{t}\left( {t + \frac{1}{t}} \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3\left( 4 \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 64 - 12 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 52 \cr} $$

Solution:

 $$\eqalign{ & \frac{{3a + 5b}}{{3a - 5b}} = 5 \cr & \Rightarrow 3a + 5b = 15a - 25b \cr & \Rightarrow 12a = 30b \cr & \Rightarrow 2a = 5b \cr & a:b \cr & 5:2 \cr} $$

Solution:

 $${a^2} + a + 1 = 0$$ \[\left[ \begin{array}{l} {a^3} + {1^3} = \left( {a + 1} \right)\left( {{a^2} + a + 1} \right)\\ {a^3} - {1^3} = \left( {a - 1} \right)\left( {{a^2} + a + 1} \right) \end{array} \right]\] $$\eqalign{ & \therefore \left( {{a^3} - 1} \right) = \left( {a - 1} \right) \times 0 \cr & \Rightarrow {a^3} - 1 = 0 \cr & \Rightarrow {a^3} = 1 \cr & \Rightarrow {\left( {{a^3}} \right)^3} = {1^3} \cr & \Rightarrow {a^9} = 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{We know that }} \cr & {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & {x^3} - p = \left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right) \cr & \Rightarrow {x^3} - p = \left( {{x^3} - {4^3}} \right) \cr & \Rightarrow p = {4^3}{\text{ }}\left( {{\text{By comparison}}} \right) \cr & {\text{So, }}p = 64 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ , }}y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & {\text{Find, }}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? \cr & \Rightarrow {\text{ }}\frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} \cr & \Rightarrow {\text{ }}\frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} = ? \cr & {\text{Now,}} \cr & {\text{ }}x + y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow x + y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} \cr & \Rightarrow {\text{ }}x + y = \frac{{2\left( {{{\sqrt 5 }^2} + {{\sqrt 3 }^2}} \right)}}{{5 - 3}} \cr & \Rightarrow x + y = 8\,.......(i) \cr & Again, \cr & x - y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}x - y = \frac{{4 \times \sqrt 5 \times \sqrt 3 }}{2} \cr & \Rightarrow x - y = 2\sqrt {15} ..............(ii) \cr & {\text{And, }}xy = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow {\text{ }}xy = 1 \cr & {\text{Substitutes values in the question}}{\text{.}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} \cr & \Rightarrow \frac{{{8^2} - 1}}{{{{\left( {2\sqrt {15} } \right)}^2} + 1}} \cr & \Rightarrow \frac{{63}}{{61}} \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 12x + 33 = 0 \cr & {x^2} - 12x + 36 - 3 = 0 \cr & {\left( {x - 6} \right)^2} - 3 = 0 \cr & x = \sqrt 3 + 6 \cr & {\left( {x - 4} \right)^2} + \frac{1}{{{{\left( {x - 4} \right)}^2}}} \cr & \therefore {\left( {\sqrt 3 + 2} \right)^2} + \frac{1}{{{{\left( {\sqrt 3 + 2} \right)}^2}}} \cr & = {\left( {\sqrt 3 + 2} \right)^2} + {\left( {\sqrt 3 - 2} \right)^2} \cr & = 2\left( {{{\sqrt 3 }^2} + {2^2}} \right) \cr & = 2 \times 7 \cr & = 14 \cr} $$

Solution:

 $$\eqalign{ & a*b = a + b + ab \cr & 3*4 \cr & = 3 + 4 + 3 \times 4 \cr & = 19 \cr & 2*3 \cr & = 2 + 3 + 2 \times 3 \cr & = 11 \cr & \therefore 3*4 - 2*3{\text{ }} \cr & = 19 - 11 \cr & = 8 \cr} $$

Solution:

 x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2 (x - z) = ? (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) 361 = 133 + 2(xy + yz + y2) 228 = 2(x + y + z)y $$y = \frac{{114}}{{19}} = 6$$ x + z = 13 xz = 36 x - z = ? (x + z)2 - (x - z)2 = 4xz 169 - (x - z)2 = 144 x - z = 5

Solution:

 $$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & {\text{Put }} \cr & a = 0 \cr & b = 0 \cr & c = 2 \cr & d = 2 \cr & \therefore a + b + c + d = 4 \cr & \Rightarrow 0 + 0 + 2 + 2 = 4 \cr & \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr & \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr & = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr & = - 1 + 1 + 1 - 1 \cr & = 0 \cr} $$