Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & 2 + \frac{6}{{\sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 - 2}} \cr & \Rightarrow 2 + \frac{{2 \times 3\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} - \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow 2 + 2\sqrt 3 + \left( {\frac{{\left( {2 - \sqrt 3 } \right) - \left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}} \right) \cr & \Rightarrow 2 + 2\sqrt 3 + \left( {\frac{{2 - \sqrt 3 - 2 - \sqrt 3 }}{{4 - 3}}} \right) \cr & \Rightarrow 2 + 2\sqrt 3 - 2\sqrt 3 \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt {11 - 2\sqrt {30} } }} \cr & = \frac{1}{{\sqrt {6 + 5 - 2 \times \sqrt 6 \times \sqrt 5 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2 \times \sqrt 6 \times \sqrt 5 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}} }} \cr & = \frac{1}{{\sqrt 6 - \sqrt 5 }} \cr & = \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 - \sqrt 5 } \right)\left( {\sqrt 6 + \sqrt 5 } \right)}} \cr & = \sqrt 6 + \sqrt 5 \cr & \frac{3}{{\sqrt {7 - 2\sqrt {10} } }} \cr & = \frac{3}{{\sqrt {5 + 2 - 2 \times \sqrt 5 \times \sqrt 2 } }} \cr & = \frac{3}{{\sqrt 5 - \sqrt 2 }} \cr & = \frac{{3 \times \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}} \cr & = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}} \cr & = \sqrt 5 + \sqrt 2 \cr & \frac{4}{{\sqrt {8 + 4\sqrt 3 } }} \cr & = \frac{4}{{\sqrt {8 + 2\sqrt {12} } }} \cr & = \frac{4}{{\sqrt {6 + 2 + 2 \times \sqrt 6 \times \sqrt 2 } }} \cr & = \frac{4}{{\sqrt {{{\left( {\sqrt 6 + \sqrt 2 } \right)}^2}} }} \cr & = \frac{{4 \times \left( {\sqrt 6 - \sqrt 2 } \right)}}{{\left( {\sqrt 6 + \sqrt 2 } \right)\left( {\sqrt 6 - \sqrt 2 } \right)}} \cr & = \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{{6 - 2}} \cr & = \sqrt 6 - \sqrt 2 \cr & \therefore {\text{Expression}} \cr & = \left( {\sqrt 6 + \sqrt 5 } \right) - \left( {\sqrt 5 + \sqrt 2 } \right) - \left( {\sqrt 6 - \sqrt 2 } \right) \cr & = \sqrt 6 + \sqrt 5 - \sqrt 5 - \sqrt 2 - \sqrt 6 + \sqrt 2 \cr & = 0 \cr} $$

Given,
$$\root 4 \of 7 ,\,\root 3 \of {11} $$   and $$\root {12} \of {1257} $$
Take LCM of 4, 3 and 12 = 12
$${\left( 7 \right)^{\frac{1}{4}}},\,{\left( {11} \right)^{\frac{1}{3}}},\,{\left( {1257} \right)^{\frac{1}{{12}}}}$$
Multiplying the power by 12 = 7³, 11⁴, 1257
i.e., 343, 14641, 1257
Therefore, from above greater one is $$\root 3 \of {11} $$

$$\eqalign{ & \frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }} = {\text{A}} + \sqrt {\text{B}} \cr & \Rightarrow \sqrt {7 + 4\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( 2 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2 \times 2\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \cr & \Rightarrow \left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} = {\text{A}} + \sqrt {\text{B}} \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = {\text{A}} + \sqrt {\text{B}} \cr & \Rightarrow \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = {\text{A}} + \sqrt {\text{B}} \cr & \Rightarrow 8 - 4\sqrt 3 + 6\sqrt 3 - 9 = {\text{A}} + \sqrt {\text{B}} \cr & \Rightarrow 2\sqrt 3 - 1 = {\text{A}} + \sqrt {\text{B}} \cr & {\text{A}} = - 1{\text{ and }}\sqrt {\text{B}} = 2\sqrt 3 \cr & {\text{B}} = 2\sqrt 3 \times 2\sqrt 3 = 12 \cr & {\text{B }} - {\text{ A}} = 12 - \left( { - 1} \right) = 13 \cr} $$

$$\eqalign{ & \frac{{\sqrt 6 + 2}}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} - \frac{{\sqrt 6 - 2}}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }} - \frac{{2\sqrt 2 }}{{2 + \sqrt 2 }} \cr & \Rightarrow \frac{{\sqrt 6 + 2}}{{\sqrt 2 + \frac{{\sqrt 3 + 1}}{{\sqrt 2 }}}} - \frac{{\sqrt 6 - 2}}{{\sqrt 2 - \frac{{\sqrt 3 - 1}}{{\sqrt 2 }}}} - \frac{2}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{{\left( {\sqrt 6 + 2} \right)\sqrt 2 }}{{2 + \sqrt 3 + 1}} - \frac{{\left( {\sqrt 6 - 2} \right)\sqrt 2 }}{{2 - \sqrt 3 + 1}} - \frac{2}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\frac{{\sqrt 6 + 2}}{{\left( {\sqrt 3 + 1} \right)}} - \frac{{\sqrt 6 - 2}}{{\left( {\sqrt 3 - 1} \right)}}} \right] - \frac{2}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\frac{{\sqrt {18} + 2\sqrt 3 - \sqrt 6 - 2 - \sqrt {18} + 2\sqrt 3 - \sqrt 6 + 2}}{{3 - 1}}} \right] - \frac{{2\left( {\sqrt 2 - 1} \right)}}{{2 - 1}} \cr & \Rightarrow \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\frac{{2\left( { - \sqrt 6 + 2\sqrt 3 } \right)}}{2}} \right] - 2\left( {\sqrt 2 - 1} \right) \cr & \Rightarrow - \sqrt 3 \times \sqrt 2 \times \frac{{\sqrt 2 }}{{\sqrt 3 }} + 2\sqrt 3 \times \frac{{\sqrt 2 }}{{\sqrt 3 }} - 2\left( {\sqrt 2 - 1} \right) \cr & \Rightarrow - 2 + 2\sqrt 2 - 2\sqrt 2 + 2 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \frac{1}{{0.2}} + \frac{1}{{0.02}} + \frac{1}{{0.002}} + \,.....\,{9^{{\text{th}}}}{\text{ term}} \cr & {\text{ = }}\frac{1}{2}\left[ {10 + 100 + 1000 + \,.....\,{9^{{\text{th}}}}{\text{ term}}} \right] \cr & = \frac{1}{2}\left[ {1111111110} \right] \cr & = 555555555 \cr} $$

$$\eqalign{ & {\text{Let m}} = \sqrt {5 + \sqrt {5 + \sqrt 5 } } \cr & {\text{Factor}} = \left( a \right) \times \left( {a + 1} \right) \cr & {\text{Here m}} = a + 1 \cr & {\text{or m}} - 1 = a\,.\,.\,.\,.\,.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{Let n}} = \sqrt {5 - \sqrt {5 - \sqrt 5 } } \cr & {\text{Factor}} = \left( a \right) \times \left( {a + 1} \right) \cr & {\text{Here n}} = a\,.\,.\,.\,.\,.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & {\text{From equation}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & {\text{m}} - 1 = {\text{n}} \cr & {\text{or m}} - {\text{n}} - 1 = 0 \cr} $$

$$\eqalign{ & \sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} = \frac{1}{{\sqrt 6 }} \cr & {\text{On squaring both sides,}} \cr & \Rightarrow {\left( {\sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} } \right)^2} = \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} - 2 = \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} = 2 + \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} = \frac{{13}}{6} \cr & \Rightarrow \frac{{{{\left( {1 + x} \right)}^2} + {x^2}}}{{x\left( {1 + x} \right)}} = \frac{{13}}{6} \cr & \Rightarrow \frac{{1 + {x^2} + 2x + {x^2}}}{{x + {x^2}}} = \frac{{13}}{6} \cr & \Rightarrow \left( {2{x^2} + 2x + 1} \right)6 = 13\left( {x + {x^2}} \right) \cr & \Rightarrow 12{x^2} + 12x + 6 = 13x + 13{x^2} \cr & \Rightarrow {x^2} + x - 6 = 0 \cr & \Rightarrow {x^2} + 3x - 2x - 6 = 0 \cr & \Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0 \cr & \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2{\text{ as }}x \ne - 3 \cr & {\text{Or, of the given options, when }}x = 2 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} \cr & = \sqrt {\frac{{1 + 2}}{2}} - \sqrt {\frac{2}{{1 + 2}}} \cr & = \frac{{\sqrt 3 }}{{\sqrt 2 }} - \frac{{\sqrt 2 }}{{\sqrt 3 }} \cr & = \frac{{3 - 2}}{{\sqrt 6 }} \cr & = \frac{1}{{\sqrt 6 }} \cr} $$

$$\eqalign{ & \left( {\sqrt 6 + \sqrt {10} - \sqrt {21} - \sqrt {35} } \right)\left( {\sqrt 6 - \sqrt {10} + \sqrt {21} - \sqrt {35} } \right) \cr & = \left\{ {\left( {\sqrt 6 - \sqrt {35} } \right) + \left( {\sqrt {10} - \sqrt {21} } \right)} \right\}\left\{ {\left( {\sqrt 6 - \sqrt {35} } \right) - \left( {\sqrt {10} - \sqrt {21} } \right)} \right\} \cr & = {\left( {\sqrt 6 - \sqrt {35} } \right)^2} - {\left( {\sqrt {10} - \sqrt {21} } \right)^2} \cr & = \left( {6 - 35 - 2\sqrt {210} } \right) - \left( {10 + 21 - 2\sqrt {210} } \right) \cr & = 41 - 2\sqrt {210} - 31 + 2\sqrt {210} \cr & = 41 - 31 \cr & = 10 \cr} $$

$$\eqalign{ & \frac{1}{{1 + \sqrt 2 + \sqrt 3 }} + \frac{1}{{1 - \sqrt 2 + \sqrt 3 }} \cr & \Rightarrow \frac{1}{{1 + \sqrt 3 + \sqrt 2 }} + \frac{1}{{1 + \sqrt 3 - \sqrt 2 }} \cr & \Rightarrow \frac{{1 + \sqrt 3 - \sqrt 2 + 1 + \sqrt 3 + \sqrt 2 }}{{{{\left( {1 + \sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & \Rightarrow \frac{{2 + 2\sqrt 3 }}{{4 + 2\sqrt 3 - 2}} \cr & \Rightarrow \frac{{2 + 2\sqrt 3 }}{{2 + 2\sqrt 3 }} \cr & \Rightarrow 1 \cr} $$