Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & {a^3} - {b^3} - {c^3} - 3abc = 0 \cr & \therefore a - b - c = 0 \cr & \Rightarrow a = b + c \cr} $$

Solution:

 $$\eqalign{ & a = \sqrt 6 - \sqrt 5 \cr & b = \sqrt 5 - 2 \cr & c = 2 - \sqrt 3 \cr & {\text{Rationalize all the terms}} \cr & a = \frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} \times \sqrt 6 + \sqrt 5 \cr & \,\,\,\,\, = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \times \sqrt 5 + 2 \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + 2}} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} \times 2 + \sqrt 3 \cr & \,\,\,\,\,\, = \frac{1}{{2 + \sqrt 3 }} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr & {\text{Now, }} \cr & a = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr} $$ Now the term whose denominator is largest is the smallest term. So, a b c

Solution:

 $$\eqalign{ & \frac{{3\left( {{x^2} + 1} \right) - 7x}}{{3x}} = 6 \cr & 3{x^2} + 3 - 7x = 18x \cr & 3{x^2} + 3 = 25x \cr & 3\left( {{x^2} + 1} \right) = 25x \cr & {\text{divided by }}'x' \cr & 3\left( {x + \frac{1}{x}} \right) = 25 \cr & x + \frac{1}{x} = 25 \cr & x + \frac{1}{x} + 2 = \frac{{25}}{3} \cr & {\left( {x + \frac{1}{x}} \right)^2} = \frac{{25}}{3} + 2 \cr & {\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = \frac{{31}}{3} \cr & \sqrt x + \frac{1}{{\sqrt x }} = \sqrt {\frac{{31}}{3}} \cr} $$

Solution:

 $$\eqalign{ & 5\sqrt 5 {x^3} + 2\sqrt 2 {y^3} = \left( {Ax + \sqrt 2 y} \right)\left( {B{x^2} + 2{y^2} + Cxy} \right) \cr & {\left( {\sqrt 5 x} \right)^3} + {\left( {\sqrt 2 y} \right)^3} = \left( {\sqrt 5 x + \sqrt 3 y} \right)\left( {5{x^2} + 9{y^2} - \sqrt {10} xy} \right) \cr & \left( {\sqrt 5 x + \sqrt 3 y} \right)\left( {5{x^2} + 9{y^2} - \sqrt {10} xy} \right) = \left( {Ax + \sqrt 2 y} \right)\left( {B{x^2} + 2{y^2} + Cxy} \right) \cr & {\text{Comparison both side:}} \cr & A = \sqrt 5 \cr & B = 5 \cr & C = - \sqrt {10} \cr & {A^2} + {B^2} + {C^2} = 5 + 25 - 10 = 20 \cr} $$

Solution:

 $$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$

Solution:

 $$\eqalign{ & {\text{Put }}\left( {x + y + z} \right) = 10 \cr & x = 2 \cr & y = 3 \cr & z = 5 \cr & x\left( {x + y + z} \right) = 20 \cr & \Leftrightarrow 2\left( {10} \right) = 20 \cr & \Leftrightarrow 20 = 20 \cr & {\text{Similarly other will satisfied,}} \cr & {\text{So, value of 2}}\left( {x + y + z} \right) \cr & = 2\left( {10} \right) \cr & = 20 \cr} $$

Solution:

 $$\eqalign{ & p{\left( {x + y} \right)^2} = 5{\text{ and }}q{\left( {x - y} \right)^2} = 3 \cr & {\text{Put the value of }}x = 2{\text{ and }}y = 1 \cr & p{\left( {2 + 1} \right)^2} = 5 \cr & \Leftrightarrow p = \frac{5}{9} \cr & q{\left( {2 - 1} \right)^2} = 3 \cr & \Leftrightarrow q = 3 \cr & {p^2}{\left( {x + y} \right)^2} + 4pqxy - {q^2}{\left( {x - y} \right)^2} \cr} $$   $$ = {\left( {\frac{5}{9}} \right)^2}{\left( {2 + 1} \right)^2} + 4 \times \frac{5}{9} \times 3 \times $$      $$2 \times $$ $$1 - $$ $${\left( 3 \right)^2}$$ $${\left( {2 - 1} \right)^2}$$ $$\eqalign{ & = \frac{{25}}{{81}} \times 9 + \frac{{40}}{3} - 9 \cr & = \frac{{25}}{9} + \frac{{40}}{3} - 9 \cr & = \frac{{25 + 120 - 81}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Put the value of p and q in option (A)}} \cr & {\text{Option 1}} \to 2\left( {p + q} \right) \cr & = 2\left( {\frac{5}{9} + 3} \right) \cr & = 2 \times \frac{{32}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Option A is satisfied}} \cr & {\text{So, }}2\left( {p + q} \right){\text{ is answer}} \cr} $$

Solution:

 $$\eqalign{ & {a^2} - 4a - 1 = 0 \cr & {a^2} - 1 = 4a \cr & a - \frac{1}{a} = 4 \cr & {\text{Squring both sides}} \cr & {a^2} + \frac{1}{{{a^2}}} - 2 = 16 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = 18 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} + 3a - \frac{3}{a} \cr & = {a^2} + \frac{1}{{{a^2}}} + 3\left( {a - \frac{1}{a}} \right) \cr & = 18 + 3 \times 4 \cr & = 18 + 12 \cr & = 30 \cr} $$

Solution:

 $$\eqalign{ & {a^2} + {b^2} + {c^2} + 96 = 8\left( {a + b - 2c} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 96 = 8a + 8b - 16c \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 96 - 8a - 8b + 16c = 0 \cr & \Rightarrow {a^2} - 8a + 16 + {b^2} - 8b + 16 + {c^2} + 16c + 64 = 0 \cr & \Rightarrow {a^2} - 2 \times a \times 4 + {4^2} + {b^2} - 2 \times b \times 4 + {4^2} + {c^2} + 2 \times c \times 8 + {8^2} = 0 \cr & \Rightarrow {\left( {a - 4} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c + 8} \right)^2} = 0 \cr & {\text{Now we can say,}} \cr & \Rightarrow {\left( {a - 4} \right)^2} = 0 \cr & \Rightarrow a - 4 = 0 \cr & \Rightarrow a = 4 \cr & {\text{Similarly,}} \cr & \Rightarrow {\left( {b - 4} \right)^2} = 0 \cr & \Rightarrow b - 4 = 0 \cr & \Rightarrow b = 4 \cr & {\text{Similarly,}} \cr & \Rightarrow {\left( {c + 8} \right)^2} = 0 \cr & \Rightarrow c + 8 = 0 \cr & \Rightarrow c = - 8 \cr & {\text{Now, put }}a,{\text{ }}b{\text{ and }}c{\text{ value in }}\sqrt {ab - bc - ca} \cr & \Rightarrow \sqrt {4 \times 4 - 4 \times \left( { - 8} \right) + \left( { - 8} \right) \times 4} \cr & \Rightarrow \sqrt {16 + 32 - 32} \cr & \Rightarrow \sqrt {16} \cr & \Rightarrow 4 \cr} $$

Solution:

 $$\eqalign{ & \sqrt {1 - \frac{{{x^3}}}{{100}}} = \frac{3}{5} \cr & \Rightarrow 1 - \frac{{{x^3}}}{{100}} = {\left( {\frac{3}{5}} \right)^2} \cr & \Rightarrow 1 - \frac{9}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow \frac{{16}}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow {x^3} = \frac{{16 \times 100}}{{25}} \cr & \Rightarrow {x^3} = 16 \times 4 \cr & \Rightarrow {x^3} = 64 \cr & \Rightarrow \boxed{x = 4} \cr} $$