161.
If x = 5 + 2ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ6, then what is the value of $$\sqrt x + \frac{1}{{\sqrt x }}?$$
(A) 2√3
(B) 3√2
(C) 2√6
(D) 6√2
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Solution:
$$\eqalign{ & x = 5 + 2\sqrt 6 \cr & \Rightarrow x = {\left( {\sqrt 3 + \sqrt 2 } \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 3 + \sqrt 2 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 3 - \sqrt 2 \cr & \Rightarrow \sqrt x + \frac{1}{{\sqrt x }} = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & \Rightarrow \sqrt x + \frac{1}{{\sqrt x }} = 2\sqrt 3 \cr} $$
162.
If $$3\sqrt {\frac{{1 - a}}{a}} + 9 = 19 - 3\sqrt {\frac{a}{{1 - a}}} ,$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ then what is the value of a?
(A) $$\frac{3}{{10}},\,\frac{7}{{10}}$$
(B) $$\frac{1}{{10}},\,\frac{9}{{10}}$$
(C) $$\frac{2}{5},\,\frac{3}{5}$$
(D) $$\frac{1}{5},\,\frac{4}{5}$$
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Solution:
$$\eqalign{ & {\text{Put }}\sqrt {\frac{{1 - a}}{a}} = x \cr & \Rightarrow 3x + 9 = 19 - \frac{3}{x} \cr & \Rightarrow 3x + \frac{3}{x} = 10 \cr & \Rightarrow x = 3,\,\frac{1}{3} \cr & {\text{Now, }}\sqrt {\frac{{1 - a}}{a}} = 3 \cr & \Rightarrow \frac{{1 - a}}{a} = 9 \cr & \Rightarrow 10a = 1 \cr & \Rightarrow a = \frac{1}{{10}} \cr & {\text{and }}\sqrt {\frac{{1 - a}}{a}} = \frac{1}{3} \cr & \Rightarrow \frac{{1 - a}}{a} = \frac{1}{9} \cr & \Rightarrow 10a = 9 \cr & \Rightarrow a = \frac{9}{{10}} \cr & \therefore \,a = \frac{1}{{10}}\,{\text{and }}\frac{9}{{10}} \cr} $$
163.
If $$\frac{{4x - 3}}{x}$$ ÃÂÃÂ + $$\frac{{4y - 3}}{y}$$ ÃÂÃÂ + $$\frac{{4z - 3}}{z} = 0{\text{,}}$$ ÃÂÃÂ then the value of $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$ ÃÂÃÂ is?
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Solution:
$$\eqalign{ & \frac{{4x - 3}}{x} + \frac{{4y - 3}}{y} + \frac{{4z - 3}}{z} = 0 \cr & \Rightarrow \frac{{4x}}{x} - \frac{3}{x} + \frac{{4y}}{y} - \frac{3}{y} + \frac{{4z}}{z} - \frac{3}{z} = 0 \cr & \Rightarrow 4 - \frac{3}{x} + 4 - \frac{3}{y} + 4 - \frac{3}{z} = 0 \cr & \Rightarrow 12 - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = 0 \cr & \Rightarrow - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = - 12 \cr & \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 \cr} $$
164.
If $$x + \frac{1}{{16x}} = 3,$$ ÃÂÃÂ then the value of $$16{x^3} + \frac{1}{{256{x^3}}}$$ ÃÂÃÂ is:
(A) 423
(B) 441
(C) 432
(D) 414
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Solution:
$$\eqalign{ & x + \frac{1}{{16x}} = 3 \cr & 2x + \frac{1}{{8x}} = 6 \cr & {\text{Cube both side}} \cr & 8{x^3} + \frac{1}{{512{x^3}}} + 3 \times 2 \times \frac{1}{8} \times 6 = 216 \cr & 8{x^3} + \frac{1}{{512{x^3}}} = 216 - \frac{9}{2} \cr & {\text{Multiply by '2' both side}} \cr & 16{x^3} + \frac{1}{{256{x^3}}} = 432 - 9 = 423 \cr} $$
165.
If x(x + y + z) = 20, y = (x + y + z) = 30 & z(x + y + z) = 50, then the value of 2(x + y + z) is?
(A) 20
(B) 10
(C) 15
(D) 18
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Solution:
$$\eqalign{ & {\text{Put }}\left( {x + y + z} \right) = 10 \cr & x = 2 \cr & y = 3 \cr & z = 5 \cr & x\left( {x + y + z} \right) = 20 \cr & \Leftrightarrow 2\left( {10} \right) = 20 \cr & \Leftrightarrow 20 = 20 \cr & {\text{Similarly other will satisfied,}} \cr & {\text{So, value of 2}}\left( {x + y + z} \right) \cr & = 2\left( {10} \right) \cr & = 20 \cr} $$
166.
If x = $$2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}},$$ ÃÂÃÂ then find the value of x3 - 6x2 + 18x.
(A) 40
(B) 33
(C) 45
(D) 22
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Solution:
$$\eqalign{ & x = 2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}} \cr & x - 2 = {2^{\frac{2}{3}}} - {2^{\frac{1}{3}}} \cr & {\left( {x - 2} \right)^3} = {\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right)^3} \cr & {x^3} - 8 - 3 \times 2x\left( {x - 2} \right) = 4 - 2 - 3 \times {2^{\frac{2}{3}}} \times {2^{\frac{1}{3}}}\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right) \cr & {x^3} - 8 - 6{x^2} + 12x = 4 - 2 - 6\left( {x - 2} \right) \cr & {x^3} - 8 - 6{x^2} + 12x = 2 - 6x + 12 \cr & {x^3} - 6{x^2} + 12x + 6x = 2 + 12 + 8 \cr & {x^3} - 6{x^2} + 18x = 22 \cr} $$
167.
If $$x = \frac{{\sqrt 3 }}{2}{\text{,}}$$ ÃÂÃÂ then $$\frac{{\sqrt {1 + x} }}{{1 + \sqrt {1 + x} }}{\text{ + }}$$ ÃÂÃÂ $$\frac{{\sqrt {1 - x} }}{{1 - \sqrt {1 - x} }}$$ ÃÂÃÂ is equal to?
(A) 1
(B) $$\frac{2}{{\sqrt 3 }}$$
(C) $${\text{2}} - \sqrt 3 $$
(D) 2
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Solution:
$$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & {\text{or }}1 + x = 1 + \frac{{\sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2 + \sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2\left( {2 + \sqrt 3 } \right)}}{{2 \times 2}} \cr & \left( {{\text{Divided and multiply by 2}}} \right) \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{1 + 3 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.1.\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{4} \cr & \therefore \sqrt {1 + x} = \frac{{1 + \sqrt 3 }}{2} \cr & \cr & {\bf{Similarly:}} \cr & \sqrt {1 - x} \cr & = \frac{{\sqrt 3 - 1}}{2} \cr & \therefore \frac{{\sqrt {1 + x} }}{{1 + \sqrt {1 + x} }}{\text{ + }}\frac{{\sqrt {1 - x} }}{{1 - \sqrt {1 - x} }} \cr & = \frac{{\frac{{1 + \sqrt 3 }}{2}}}{{1 + \frac{{1 + \sqrt 3 }}{2}}} + \frac{{\frac{{\sqrt 3 - 1}}{2}}}{{1 - \frac{{\sqrt 3 - 1}}{2}}} \cr & = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{3 - \sqrt 3 }} \cr & = \frac{{1 + \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}} + \frac{{1 - \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}} \cr & = \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$
168.
If a3 = 117 + b3 and a = 3 + b, then the value of a + b is?
(A) 7
(B) 13
(C) 49
(D) 0
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Solution:
$$\eqalign{ & {a^3} = 117 + {b^3}{\text{ , }}a = 3 + b \cr & {a^3} - {b^3}{\text{ = 117}}\,......{\text{(i)}} \cr & a - b = 3\,........(ii) \cr & {\text{Put }}a = 5,{\text{ }}b = 2 \cr & {\text{Both equation satisfy}} \cr & {\text{Now, }}a + b = 5 + 2 = 7 \cr} $$
169.
If $$x = p + \frac{1}{p}$$ ÃÂÃÂ and $$y = p - \frac{1}{p}$$ ÃÂÃÂ then the value of x4 - 2x2 y2 + y4 = ?
(A) 24
(B) 4
(C) 16
(D) 8
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Solution:
$$\eqalign{ & x = p + \frac{1}{p}{\text{ }} \cr & y = p - \frac{1}{p} \cr & \therefore x + y = p + \frac{1}{p} + p - \frac{1}{p} \cr & \Leftrightarrow x + y = 2p \cr & \therefore x - y = p + \frac{1}{p} - p + \frac{1}{p} \cr & \Leftrightarrow x - y = \frac{2}{p} \cr & \therefore {x^4} - 2{x^2}{y^2} + {y^4} \cr & = {x^4} + {y^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left[ {\left( {x + y} \right)\left( {x - y} \right)} \right]^2} \cr & = {\left( {2p \times \frac{2}{p}} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$
170.
If $$A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}}$$ ÃÂÃÂ ÃÂÃÂ and $$B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}},$$ ÃÂÃÂ ÃÂÃÂ then what is the value of (A2 + B2 )2 ?
(A) 0.8
(B) 1
(C) 1.4
(D) 2.2
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Solution:
$$\eqalign{ & A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}} \cr & = \frac{{{{\left( {0.6} \right)}^3} + {{\left( {0.2} \right)}^3}}}{{{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)}} \cr & = \frac{{\left( {0.6 + 0.2} \right)\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}}{{\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}} \cr & = 0.8 \cr & B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}} \cr & = \frac{{{{\left( {0.9} \right)}^3} - {{\left( {0.3} \right)}^3}}}{{{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)}} \cr & = \frac{{\left( {0.9 - 0.3} \right)\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}}{{\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}} \cr & = 0.6 \cr & \therefore \,{\left( {{A^2} + {B^2}} \right)^2} \cr & = {\left[ {{{\left( {0.8} \right)}^2} + {{\left( {0.6} \right)}^2}} \right]^2} \cr & = {\left( {0.64 + 0.36} \right)^2} \cr & = {\left( 1 \right)^2} \cr & = 1 \cr} $$