Surds and Indices Questions and Answers Part-9

1. The value of $$\sqrt {40 + \sqrt {9\sqrt {81} } }$$    is = ?
a) $$\sqrt {111} $$
b) 9
c) 7
d) 11

Answer: c
Explanation:
$$\eqalign{ & \sqrt {40 + \sqrt {9\sqrt {81} } } \cr & \sqrt {40 + \sqrt {9 \times 9} } \cr & \sqrt {40 + 9} \cr & \sqrt {49} \cr & 7 \cr} $$

2. $$\sqrt {8 - 2\sqrt {15} } $$   is equal to = ?
a) $${\text{3}} - \sqrt 5 $$
b) $$\sqrt 5 - \sqrt 3 $$
c) $${\text{5}} - \sqrt 3 $$
d) $$\sqrt 5 + \sqrt 3 $$

Answer: b
Explanation:
$$\eqalign{ & \sqrt {8 - 2\sqrt {15} } \cr & = \sqrt {5 + 3 - 2 \times \sqrt 5 \times \sqrt 3 } \cr & = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 3 } \cr & = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \cr & = \left( {\sqrt 5 - \sqrt 3 } \right) \cr} $$

3. $$\sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } $$       is equal to = ?
a) $${\text{1}} - \sqrt 3 $$
b) $$\sqrt 3 - 1$$
c) $${\text{2}}\left( {2 - \sqrt 3 } \right)$$
d) $${\text{2}}\left( {2 + \sqrt 3 } \right)$$

Answer: b
Explanation:
$$\eqalign{ & \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr & = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr & = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr & = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr & = \sqrt 3 - 1 \cr} $$

4. If N = $$\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - $$     $$\sqrt {3 - 2\sqrt 2 } {\text{,}}$$   then the value of N is = ?
a) $${\text{2}}\sqrt 2 - 1$$
b) 3
c) 1
d) 2

Answer: c
Explanation:
$$\eqalign{ & {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr & {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
  $${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{ & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & {{\text{X}}^2} = 2 \cr & {\text{X = }}\sqrt 2 \cr & {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr & {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr & {\text{N}} = 1 \cr} $$

5. If $$\frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}$$      = 1 then the value of x is = ?
a) $$\sqrt 5 $$
b) 5
c) $${\text{2}}\sqrt 5 $$
d) $${\text{3}}\sqrt 5 $$

Answer: b
Explanation:
$$\eqalign{ & \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}{\text{ = 1 }} \cr & \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }}{\text{ }} \cr & \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr & \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr & x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr & x = 5{\text{ }} \cr} $$

6.Find the simplest value of $${\text{2}}\sqrt {50} $$  + $$\sqrt {18} $$  - $$\sqrt {72} $$ = ?(given $$\sqrt 2 $$ = 1.414)
a) 4.242
b) 9.898
c) 10.6312
d) 8.484

Answer: b
Explanation:
$$\eqalign{ & {\text{2}}\sqrt {50} {\text{ + }}\sqrt {18} - \sqrt {72} \cr & {\text{2}} \times {\text{5}}\sqrt 2 {\text{ + 3}}\sqrt 2 - 6\sqrt 2 \cr & 13\sqrt 2 - 6\sqrt 2 \cr & 7\sqrt 2 \cr & 7 \times 1.414 \cr & 9.898 \cr} $$

7. 553 + 173 - 723 + 201960 is equal to = ?
a) -1
b) 0
c) 1
d) 17

Answer: b
Explanation:
$$\eqalign{ & {\text{Let }}a = 55, \cr & \,\,\,\,\,\,\,\,\,\,\,b = 17, \cr & \,\,\,\,\,\,\,\,\,\,\,c = - 72 \cr & a + b + c \cr & = 55 + 17 - 72 \cr & = 0 \cr & {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \left( {a + b + c} \right) = 0 \cr & {\text{Answer is }}0. \cr} $$

8. The simplification value of $$\left( {\sqrt 3 + 1} \right)$$  $$\left( {10 + \sqrt {12} } \right)$$  $$\left( {\sqrt {12} - 2} \right)$$  $$\left( {5 - \sqrt 3 } \right)$$  is = ?
a) 16
b) 88
c) 176
d) 132

Answer: c
Explanation:
$$\eqalign{ & \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr} $$
$$ \Rightarrow \left( {\sqrt 3 + 1} \right) \times $$   $$2\left( {5 + \sqrt 3 } \right) \times $$   $$2\left( {\sqrt 3 - 1} \right)$$  $$\left( {5 - \sqrt 3 } \right)$$
$$\eqalign{ & \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 + \sqrt 3 } \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow 4\left[ {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} \right]\left[ {{{\left( 5 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \cr & \Rightarrow 4 \times 2 \times 22 \cr & \Rightarrow 176 \cr} $$

9. $$\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} = ?$$
a) 0
b) $$\frac{1}{2}$$
c) 1
d) am+n

Answer: c
Explanation:
$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{1 + \frac{{{a^n}}}{{{a^m}}}}} + \frac{1}{{1 + \frac{{{a^m}}}{{{a^n}}}}} \cr & = \frac{{{a^m}}}{{{a^m} + {a^n}}} + \frac{{{a^n}}}{{{a^m} + {a^n}}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$

10. $$\frac{1}{{1 + {x^{\left( {b - a} \right)}} + {x^{\left( {c - a} \right)}}}} \,+ $$    $$\frac{1}{{1 + {x^{\left( {a - b} \right)}} + {x^{\left( {c - b} \right)}}}} \,+ $$    $$\frac{1}{{1 + {x^{\left( {b - c} \right)}} + {x^{\left( {a - c} \right)}}}} = ?$$
a) 0
b) 1
c) xa-b-c
d) None of these

Answer: b
Explanation:
$$\eqalign{ & \frac{1}{{1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}}} + \frac{1}{{1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}}} + \frac{1}{{1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}}} \cr} $$
  $$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$    $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^{b}} + {x^c}} \right)}}$$
$$\eqalign{ & = \frac{{\left( {{x^a} + {x^b} + {x^c}} \right)}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr & = 1 \cr} $$