Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & \root 3 \of {11} ,\,\sqrt 7 ,\,\root 4 \of {45} \cr & {11^{\frac{1}{3}}},\,{7^{\frac{1}{2}}},\,{45^{\frac{1}{4}}} \cr & {\text{Take LCM of 3, 2, 4 is 12}} \cr & {11^4},\,{7^6},\,{45^3} \cr & {\text{14641,}}\,{\text{117649,}}\,{\text{91125}} \cr & 117649 > 91125 > 14641 \cr & {\text{Option C is true}} \cr & \sqrt 7 > \root 4 \of {45} > \root 3 \of {11} \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}} \cr & = \frac{{y - 1 + 2y + 1}}{{{y^2} - 1}} \cr & = \frac{{3y}}{{{y^2} - 1}} \cr & = \frac{{3\left( {\frac{{2 - x}}{{1 + x}}} \right)}}{{{{\left( {\frac{{2 - x}}{{1 + x}}} \right)}^2} - 1}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{4 + {x^2} - 4x - 1 - {x^2} - 2x}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{ - 6x + 3}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{3\left( {1 - 2x} \right)}} \cr & = \frac{{\left( {2 - x} \right)\left( {1 + x} \right)}}{{\left( {1 - 2x} \right)}} \cr & \cr & {\bf{Alternate}}\,{\bf{solution:}} \cr & y = \frac{{2 - x}}{{1 + x}} \cr & {\text{Put }}x = 0, \cr & y = 2 \cr & \frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}} \cr & = \frac{1}{3} + \frac{5}{3} \cr & = \frac{6}{3} \cr & = 2 \cr & {\text{Go through option}} \cr & \frac{{\left( {1 + x} \right)\left( {2 - x} \right)}}{{\left( {1 - 2x} \right)}} \cr & = \frac{{1 \times 2}}{1} \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & \frac{{343 \times 49}}{{216 \times 16 \times 81}} \cr & = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}} \cr & = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}} \cr & = \frac{{{7^5}}}{{{6^3} \times {6^4}}} \cr & = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}} \cr & = \frac{{{7^5}}}{{{6^7}}} \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{\sqrt 3 + \sqrt 4 }} \cr & = \frac{1}{{\sqrt 4 + \sqrt 3 }} \times \frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} \cr & = \frac{{\sqrt 4 - \sqrt 3 }}{1} \cr & = \sqrt 4 - \sqrt 3 \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt 4 + \sqrt 5 }} = \sqrt 5 - \sqrt 4 \cr & \Rightarrow \frac{1}{{\sqrt 5 + \sqrt 6 }} = \sqrt 6 - \sqrt 5 \cr & \Rightarrow \frac{1}{{\sqrt 6 + \sqrt 7 }} = \sqrt 7 - \sqrt 6 \cr & \Rightarrow \frac{1}{{\sqrt 7 + \sqrt 8 }} = \sqrt 8 - \sqrt 7 \cr & \Rightarrow \frac{1}{{\sqrt 8 + \sqrt 9 }} = \sqrt 9 - \sqrt 8 \cr & {\text{Now put values}} \cr & \Rightarrow \sqrt 4 - \sqrt 3 + \sqrt 5 - \sqrt 4 + \sqrt 6 - \sqrt 5 + \sqrt 7 - \sqrt 6 + \sqrt 8 - \sqrt 7 + \sqrt 9 - \sqrt 8 \cr & \Rightarrow \sqrt 9 - \sqrt 3 \cr & \Rightarrow 3 - \sqrt 3 \cr} $$

Solution:

 $$\eqalign{ & f\left( 1 \right) = 1 - \frac{1}{2} \cr & f\left( 2 \right) = \frac{1}{2} - \frac{1}{3} \cr & f\left( 3 \right) = \frac{1}{3} - \frac{1}{4} \cr & f\left( 4 \right) = \frac{1}{4} - \frac{1}{5} \cr & 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}\,.\,.\,.\,.\,.\,\frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{11}} \cr & = 1 - \frac{1}{{11}} \cr & = \frac{{10}}{{11}} \cr} $$

Solution:

 $$\eqalign{ & \sqrt 2 = 1.414 \cr & \Rightarrow \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} \cr & \Rightarrow 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 \cr & \Rightarrow 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & \Rightarrow 6\sqrt 2 \cr & \Rightarrow 6 \times 1.414 \cr & \Rightarrow 8.484{\text{ }} \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr & = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr & = \frac{{{3^1}}}{{{7^1}}} \cr & = \frac{3}{7} \cr} $$

Solution:

Answer & Solution Answer: Option A No explanation is given for this question Let's Discuss on Board

Solution:

 $$\eqalign{ & {a^1} \cr & = {c^z} \cr & = {\left( {{b^y}} \right)^z} \cr & = {b^{yz}} \cr & = {\left( {{a^x}} \right)^{yz}} \cr & = {a^{xyz}} \cr & \Rightarrow xyz = 1 \cr} $$

Solution:

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