Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & 3{a^2} = {b^2}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{ }}\frac{{{{\left( {a + b} \right)}^3} - {{\left( {a - b} \right)}^3}}}{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}} \cr} $$   $$ = \frac{{{a^3} + {b^3} + 3ab\left( {a + b} \right)\, - \,\left( {{a^3} - {b^3} - 3ab\left( {a - b} \right)} \right){\text{ }}}}{{{a^2} + {b^2} + 2ab + {\text{ }}{a^2} + {b^2} - 2ab}}$$ $$\eqalign{ & = \frac{{2{b^3}{\text{ + 6}}{{\text{a}}^2}{\text{b }}}}{{2{a^2} + 2{b^2}{\text{ }}}} \cr & = \frac{{{b^3}{\text{ + 3}}{{\text{a}}^2}{\text{b }}}}{{{a^2} + {b^2}{\text{ }}}} \cr & = \frac{{{b^3} + {b^3}{\text{ }}}}{{\frac{{{b^2}}}{3} + {b^2}{\text{ }}}} \cr & = \frac{{2{b^3}}}{{{b^2}\left( {\frac{1}{3} + 1} \right)}} \cr & = \frac{{2b}}{{\frac{4}{3}}} \cr & = \frac{{3b}}{2} \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 5x + 1 = 0 \cr & \left( {{x^4} + \frac{1}{{{x^2}}}} \right) \div \left( {{x^2} + 1} \right) = ? \cr & {x^2} - 5x + 1 = 0 \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 110 \cr & \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{x\left( {x + \frac{1}{x}} \right)}} = \frac{{110}}{5} = 22 \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & \therefore {x^2} + x + \frac{1}{x} + \frac{1}{{{x^2}}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + x + \frac{1}{x} \cr & \Rightarrow 7 + 3 \cr & \Rightarrow 10 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 5 \cr & \therefore \frac{{2x}}{{3{x^2} - 5x + 3}}\,\left( {{\text{Divide by }}x} \right) \cr & = \frac{{\frac{{2x}}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{5x}}{x} + \frac{3}{x}}} \cr & = \frac{2}{{3x + \frac{3}{x} - 5}} \cr & = \frac{2}{{3\left( {x + \frac{1}{x}} \right) - 5}} \cr & = \frac{2}{{3 \times 5 - 5}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{4}\sqrt x + {a^2} \cr & = {\left( {\sqrt x } \right)^2} + 2 \times \frac{1}{8} \times \sqrt x + {a^2} \cr & \left[ {\left( {{{\text{A}}^2} + {\text{2AB}} + {{\text{B}}^2}} \right) = {{\left( {{\text{A}} + {\text{B}}} \right)}^2}} \right] \cr & {\text{Here, A}} = \sqrt x {\text{ and }} \cr & {\text{B}} = a \cr & {\text{B}} = \frac{1}{8} \cr & \therefore a = \frac{1}{8} \cr} $$

Solution:

 Here, Base = 3 units Height = 4 units Area of Δ = $$\frac{1}{2}$$ × b × h = $$\frac{1}{2}$$ × 3 × 4 = 6 sq. units

Solution:

 $$\eqalign{ & a\left( {2 + \sqrt 3 } \right) = b\left( {2 - \sqrt 3 } \right) = 1 \cr & a = \frac{1}{{\left( {2 + \sqrt 3 } \right)}} \cr & b = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \Rightarrow a = \frac{1}{b} \cr & \Rightarrow \frac{1}{{{a^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{1}{{{b^2}}} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{{1 + {b^2}}}{{{b^2}}}}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2}}}{{{b^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2} + 1}}{{{b^2} + 1}} \cr & \Rightarrow 1 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\text{Given, }}x + y = 2a \cr & {\text{Find, }}\frac{a}{{x - a}} + \frac{a}{{y - a}}{\text{ = ?}} \cr & \mathop {\mathop x\limits_ \downarrow \,}\limits_3 + \mathop {\mathop y\limits_ \downarrow }\limits_1 = \mathop {\mathop {2a}\limits_ \downarrow }\limits_2 \cr & {\text{Let }}x = 3,{\text{ }}y = 1,{\text{ }}a = 2 \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{2}{{\left( {3 - 2} \right)}} + \frac{2}{{\left( {1 - 2} \right)}} \cr & = \frac{2}{1} + \frac{2}{{ - 1}} \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & {\text{or }}1 + x = 1 + \frac{{\sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2 + \sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2\left( {2 + \sqrt 3 } \right)}}{{2 \times 2}} \cr & \left( {{\text{Divided and multiply by 2}}} \right) \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{1 + 3 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.1.\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{4} \cr & \therefore \sqrt {1 + x} = \frac{{1 + \sqrt 3 }}{2} \cr & \cr & {\bf{Similarly:}} \cr & \sqrt {1 - x} \cr & = \frac{{\sqrt 3 - 1}}{2} \cr & \therefore \frac{{\sqrt {1 + x} }}{{1 + \sqrt {1 + x} }}{\text{ + }}\frac{{\sqrt {1 - x} }}{{1 - \sqrt {1 - x} }} \cr & = \frac{{\frac{{1 + \sqrt 3 }}{2}}}{{1 + \frac{{1 + \sqrt 3 }}{2}}} + \frac{{\frac{{\sqrt 3 - 1}}{2}}}{{1 - \frac{{\sqrt 3 - 1}}{2}}} \cr & = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{3 - \sqrt 3 }} \cr & = \frac{{1 + \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}} + \frac{{1 - \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}} \cr & = \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$