1. The value of $$\frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}}{\text{is = ?}}$$
a) $$ - \frac{1}{{36}}$$
b) $$\frac{2}{3}$$
c) $$\frac{1}{{13}}$$
d) $$\frac{5}{{13}}$$
Explanation:
$$\eqalign{ & \frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}} \cr & = \frac{{{2^{n - 1}}\left( {1 - 2} \right)}}{{{2^{n + 1}}\left( {{2^3} + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{\left( {n - 1} \right) - \left( {n + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{ - 2}} \cr & = \left( { - \frac{1}{9}} \right).\frac{1}{{{2^2}}} \cr & = \left( { - \frac{1}{9}} \right) \times \frac{1}{4} \cr & = - \frac{1}{{36}} \cr} $$
2. If $$x = 5 + 2\sqrt 6 {\text{,}}$$ then $$\sqrt x - \frac{1}{{\sqrt x }}$$ = is?
a) $${\text{2}}\sqrt 2 $$
b) $${\text{2}}\sqrt 3 $$
c) $$\sqrt 3 + \sqrt 2 $$
d) $$\sqrt 3 - \sqrt 2 $$
Explanation:
$$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr & = 10 - 2 \cr & = 8 \cr & \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$
3. $$\left( {4 + \sqrt 7 } \right),$$ expressed as a perfect square, is equal to = ?
a) $${\left( {2 + \sqrt 7 } \right)^2}$$
b) $${\left( {\frac{{\sqrt 7 }}{2} + \frac{1}{2}} \right)^2}$$
c) $$\left\{ {\frac{1}{2}{{\left( {\sqrt 7 + 1} \right)}^2}} \right\}$$
d) $$\left( {\sqrt 3 + \sqrt 4 } \right)$$
Explanation:
$$\eqalign{ & \left( {4 + \sqrt 7 } \right) \cr & = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$
4. If 3x+y = 81 and 81x-y = 3, then the value of $$\frac{x}{y}$$ is = ?
a) $$\frac{{15}}{{17}}$$
b) $$\frac{{17}}{{30}}$$
c) $$\frac{{15}}{{34}}$$
d) $$\frac{{17}}{{15}}$$
Explanation:
$$\eqalign{ & {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr & x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr & x + y = 4......{\text{(i)}} \cr & {\text{ }}x - y = \frac{1}{4}.....(ii) \cr & {\text{Solve the equation of (i) and (ii)}} \cr & x = \frac{{17}}{8}, \cr & y = \frac{{15}}{8}, \cr & \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$
5. $$\left( {\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}} \right)$$ simplifies to = ?
a) $$\sqrt 5 + \sqrt 6 $$
b) $${\text{2}}\sqrt 5 + \sqrt 6 $$
c) $$\sqrt 5 - \sqrt 6 $$
d) $${\text{2}}\sqrt 5 - 3\sqrt 6 $$
Explanation:
$$\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}$$
$$ \frac{{\left( {1 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) + \left( {1 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}$$
$$ \frac{{\sqrt 5 - \sqrt 3 + \sqrt {10} - \sqrt 6 + \sqrt 5 + \sqrt 3 - \sqrt {10} - \sqrt 6 }}{{5 - 3}}$$
$$\eqalign{ & \frac{{2\sqrt 5 - 2\sqrt 6 }}{2} \cr & \frac{{2\left( {\sqrt 5 - \sqrt 6 } \right)}}{2} \cr & \sqrt 5 - \sqrt 6 \cr} $$
6. $${6^{1.2}} \times {36^?} \times {30^{2.4}} \times {25^{1.3}} = {30^5}$$
a) 0.1
b) 0.7
c) 1.4
d) 2.6
Explanation:
$$\eqalign{ & {\text{Let }}\,{6^{1.2}} \times {36^x} \times {30^{2.4}} \times {25^{1.3}} = {30^5} \cr & {\text{Then,}}\,{6^{1.2}} \times {({6^2})^x} \times {(6 \times 5)^{2.4}} \times {({5^2})^{1.3}} = {30^5} \cr & {6^{1.2}} \times {6^{2x}} \times {6^{2.4}} \times {5^{2.4}} \times {5^{2.6}} = {(6 \times 5)^5} \cr & {6^{\left( {1.2 + 2x + 2.4} \right)}} \times {5^{\left( {2.4 + 2.6} \right)}} = {6^5} \times {5^5} \cr & {6^{\left( {3.6 + 2x} \right)}} \times {5^5} = {6^5} \times {5^5} \cr & 3.6 + 2x = 5 \cr & 2x = 1.4 \cr & x = 0.7 \cr} $$
7.$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$ simplifies to = ?
a) $${\text{2}}\sqrt 6 $$
b) $${\text{4}}\sqrt 6 $$
c) $${\text{2}}\sqrt 3 $$
d) $${\text{3}}\sqrt 2 $$
Explanation:
$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$
$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$
8. If $$\sqrt 3 $$ = 1.732 is given, then the value of $$\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}$$ is = ?
a) 11.732
b) 13.928
c) 12.928
d) 13.925
Explanation:
$$\eqalign{ & \sqrt 3 = 1.732 \cr & \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{4 - 3}} \cr & 4 + 3 + 4\sqrt 3 \cr & 7 + 4 \times 1.732 \cr & 7 + 6.928 \cr & 13.928 \cr} $$
9. Evaluate: $$16\sqrt {\frac{3}{4}} - 9\sqrt {\frac{4}{3}} $$ if $$\sqrt {12} $$ = 3.46
a) 3.46
b) 10.38
c) 13.84
d) 24.22
Explanation:
$$\eqalign{ & 16\sqrt {\frac{3}{4}} - 9\sqrt {\frac{4}{3}} \cr & 16\sqrt {\frac{{3 \times 4}}{{4 \times 4}}} - 9\sqrt {\frac{{3 \times 4}}{{3 \times 3}}} \cr & 16 \times \frac{{\sqrt {12} }}{4} - \frac{{9\sqrt {12} }}{3} \cr & 4\sqrt {12} - 3\sqrt {12} \cr & \sqrt {12} \cr & 3.46 \cr} $$
10. $${2^{3.6}} \times {4^{3.6}} \times {4^{3.6}} \times {(32)^{2.3}} = $$ $${\left( {32} \right)^?}$$
a) 5.9
b) 7.7
c) 9.5
d) 13.1
Explanation:
$$\eqalign{ & {\text{Let }}{2^{3.6}} \times {4^{3.6}} \times {4^{3.6}} \times {(32)^{2.3}} = {\left( {32} \right)^x} \cr & {\text{Then,}}{2^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {2^{3.6}} \times {2^{\left( {2 \times 3.6} \right)}} \times {2^{\left( {2 \times 3.6} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {2^{\left( {3.6 + 7.2 + 7.2} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {2^{18}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {\left( {{2^5}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & {\left( {{2^5}} \right)^{\left( {3.6 + 2.3} \right)}} = {\left( {{2^5}} \right)^x} \cr & {\left( {{2^5}} \right)^{5.9}} = {\left( {{2^5}} \right)^x} \cr & x = 5.9 \cr} $$