Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\frac{2}{{\sqrt 7 + \sqrt 5 }} + \frac{7}{{\sqrt {12} - \sqrt 5 }} - \frac{5}{{\sqrt {12} - \sqrt 7 }}$$ $$ = \frac{2}{{\sqrt 7 + \sqrt 5 }} \times $$  $$\frac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} + $$  $$\frac{7}{{\sqrt {12} - \sqrt 5 }} \times $$  $$\frac{{\sqrt {12} + \sqrt 5 }}{{\sqrt {12} + \sqrt 5 }} - $$  $$\left( {\frac{5}{{\sqrt {12} - \sqrt 7 }} \times \frac{{\sqrt {12} + \sqrt 7 }}{{\sqrt {12} + \sqrt 7 }}} \right)$$ $$ = \frac{{2\left( {\sqrt 7 - \sqrt 5 } \right)}}{2} + $$   $$\frac{{7\left( {\sqrt {12} + \sqrt 5 } \right)}}{7} - $$   $$\frac{{5\left( {\sqrt {12} + \sqrt 7 } \right)}}{5}$$ $$\eqalign{ &= \sqrt 7 - \sqrt 5 + \sqrt {12} + \sqrt 5 - \sqrt {12} - \sqrt 7 \cr &= 0 \cr} $$

Solution:

 $$\eqalign{ & \left( {4 + \sqrt 7 } \right) \cr & = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$

Solution:

 $$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$

Solution:

 $$\eqalign{ & a = \frac{1}{{2 - \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} \times \frac{{3 + \sqrt 8 }}{{3 + \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \times \frac{{4 + \sqrt {15} }}{{4 + \sqrt {15} }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} + \frac{{3 + \sqrt 8 }}{{9 - 8}} + \frac{{4 + \sqrt {15} }}{{16 - 15}} \cr & \Rightarrow 2 + \sqrt 3 + 3 + \sqrt 8 + 4 + \sqrt {15} \cr & \Rightarrow 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} \cr & a = 9 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} 18 \cr & \sqrt 3 = 1.73,\,\sqrt 2 = 1.41,\,\sqrt {15} = 3.9 \cr & \Rightarrow 9 9 + 1.73 + \left( {2 \times 1.41} \right) + 3.9 = 17.4 18 \cr} $$

Solution:

Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr & {\text{Then,}} \cr & {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr & \Rightarrow {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr & \Rightarrow x = 210 \cr} $$

Solution:

 $$\eqalign{ & \sqrt {121} + \sqrt {12321} + \sqrt {1234321} + \sqrt {123454321} \cr & = 11 + 111 + 1111 + 11111 \cr & = 12344 \cr} $$

Solution:

 $$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$

Solution:

 $$\eqalign{ & {\text{1}}{{\text{0}}^{ - 8x}} \cr & = {\left[ {{{10}^x}} \right]^{ - 8}} \cr & = {\left[ {\frac{1}{2}} \right]^{ - 8}} \cr & = {2^8} \cr & = 256 \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} \cr & = \frac{{2.236 - 1.732}}{2} \cr & = \frac{{0.504}}{2} \cr & = 0.252 \cr} $$

Solution:

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