131.
If $$a = 2 + \sqrt 3 {\text{,}}$$ ÃÂÃÂ then the value of $$\left( {{a^2} + \frac{1}{{{a^2}}}} \right) = \,?$$
(A) 12
(B) 14
(C) 16
(D) 10
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Solution:
$$\eqalign{ & a = 2 + \sqrt 3 \cr & \Rightarrow {a^2} = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow {a^2} = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow {a^2} = 7 + 4\sqrt 3 \cr & \frac{1}{{{a^2}}} = \frac{1}{{7 + 4\sqrt 3 }} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{1} \cr & \Rightarrow \frac{1}{{{a^2}}} = 7 - 4\sqrt 3 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} \cr & = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 \cr & = 14 \cr} $$
132.
If a3 - b3 = 56 and a - b = 2, then the value of a2 + b2 will be?
(A) 48
(B) 20
(C) 22
(D) 5
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Solution:
$$\eqalign{ & {a^3} - {b^3} = 56 \cr & \Rightarrow a - b = 2 \cr & \,\,\,\,\left( {{\text{By cubing}}} \right) \cr & \Rightarrow {a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( 2 \right)^2} \cr & \Rightarrow 56 - 3ab \times 2 = 8 \cr & \Rightarrow - 6ab = 8 - 56 \cr & \Rightarrow 6ab = 48 \cr & \Rightarrow ab = 8 \cr & \left( {a - b} \right) = 2 \cr & \,\,{\text{ }}\left( {{\text{By squaring}}} \right) \cr & \Rightarrow {\left( {a - b} \right)^2} = {\left( 2 \right)^2} \cr & \Rightarrow {a^2} + {b^2} - 2ab = 4 \cr & \Rightarrow {a^2} + {b^2} = 4 + 2ab \cr & \Rightarrow {a^2} + {b^2} = 4 + 2 \times 8 \cr & \Rightarrow {a^2} + {b^2} = 20 \cr} $$
133.
If p = 101, then the value of $$\root 3 \of {p\left( {{p^2} - 3p + 3} \right) - 1} $$ ÃÂÃÂ ÃÂÃÂ is?
(A) 100
(B) 101
(C) 102
(D) 1000
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Solution:
$$\eqalign{ & p = 101 \cr & \root 3 \of {p\left( {{p^2} - 3p + 3} \right) - 1} \cr & = \root 3 \of {{p^3} - 3{p^2} + 3p - 1} \cr & \therefore \left[ {{{\left( {p - 1} \right)}^3} = {p^3} - {{\left( 1 \right)}^3} - 3p\left( {p - 1} \right)} \right] \cr & = \root 3 \of {{{\left( {p - 1} \right)}^3}} \cr & = p - 1 \cr & = 101 - 1 \cr & = 100{\text{ }} \cr} $$
134.
If x = 255, y = 256, z = 257, then find the value of x3 + y3 + z3 - 3xyz.
(A) 1984
(B) 2304
(C) 1876
(D) 1378
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Solution:
$$\eqalign{ & {x^3} + {y^3} + {z^3} - 3xyz \cr & = \frac{{\left( {x + y + z} \right)}}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{{255 + 256 + 257}}{2}\left[ {{1^2} + {1^2} + {2^2}} \right] \cr & = \frac{{768 \times 6}}{2} \cr & = \frac{{4608}}{2} \cr & = 2304 \cr} $$
135.
The simplified value of following is: $$\left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right)$$ ÃÂàÃÂà$$ ÃÂÃÂÃÂ÷ $$ $$\frac{{10}}{{27}}{a^2}b{c^3}$$
(A) $$\frac{3}{{10}}a{b^4}{c^3}$$
(B) $$\frac{9}{{10}}{a^2}b{c^4}$$
(C) $$\frac{3}{{10}}{a^4}{b^{10}}{c^4}$$
(D) $$\frac{1}{{10}}{a^4}{b^4}{c^{10}}$$
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Solution:
$$\eqalign{ & \left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right) \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{1}{9}{a^6}{b^{11}}{c^7} \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{{\frac{1}{9}{a^6}{b^{11}}{c^7}}}{{\frac{{10}}{{27}}{a^2}b{c^3}}} \cr & \Rightarrow \frac{3}{{10}}{a^4}{b^{10}}{c^4} \cr} $$
136.
If $$x = 3 + 2\sqrt 2 {\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of $$\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right){\text{ is?}}$$
(A) 1
(B) 2
(C) $${\text{2}}\sqrt 2 $$
(D) $${\text{3}}\sqrt 3 $$
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Solution:
$$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \Rightarrow x = 2 + 1 + 2\sqrt 2 \cr & \Rightarrow x = {\left( {\sqrt 2 + 1} \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 2 + 1 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 2 - 1 \cr & \therefore \sqrt x - \frac{1}{{\sqrt x }} \cr & = \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right) \cr & = \sqrt 2 + 1 - \sqrt 2 + 1 \cr & = 2 \cr} $$
137.
If $${\text{2}}x - \frac{1}{{2x}} = 5{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ $${\text{x}} \ne {\text{0,}}$$ ÃÂÃÂ then find the value of $${x^2} + \frac{1}{{16{x^2}}} - 2$$ ÃÂÃÂ ÃÂÃÂ = ?
(A) $$\frac{{19}}{4}$$
(B) $$\frac{{23}}{4}$$
(C) $$\frac{{27}}{4}$$
(D) $$\frac{{31}}{4}$$
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Solution:
$$\eqalign{ & 2x - \frac{1}{{2x}} = 5 \cr & {\text{Divide by 2 both side}} \cr & x - \frac{1}{{4x}} = \frac{5}{2} \cr & {\text{Squaring both side}} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - 2 \times x \times \frac{1}{{4x}} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - \frac{1}{2} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{25}}{4} + \frac{1}{2} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{27}}{4} \cr & {\text{So, }} \cr & {x^2} + \frac{1}{{16{x^2}}} - 2 \cr & = \frac{{27}}{4} - 2 \cr & = \frac{{19}}{4} \cr} $$
138.
If $$x = 3 + 2\sqrt 2 {\text{,}}$$ ÃÂÃÂ then the value of $${x^2}{\text{ + }}\frac{1}{{{x^2}}}{\text{ is?}}$$
(A) 36
(B) 30
(C) 32
(D) 34
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Solution:
$$\eqalign{ & {\text{ }}x = 3 + 2\sqrt 2 \cr & \Rightarrow {x^2} = {\left( {3 + 2\sqrt 2 } \right)^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} = 9 + 8 + 12\sqrt 2 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr & \Rightarrow \frac{1}{{{x^2}}} = \frac{1}{{17 + 12\sqrt 2 }} \times \frac{{17 - 12\sqrt 2 }}{{17 - 12\sqrt 2 }} \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 12\sqrt 2 \cr & \therefore {\text{ }}{x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 17 + 12\sqrt 2 + 17 - 12\sqrt 2 \cr & = 34 \cr} $$
139.
If x is real, $$x + \frac{1}{x} \ne 0$$ ÃÂÃÂ and $${x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{,}}$$ ÃÂÃÂ then the value of $${\left( {x + \frac{1}{x}} \right)^4}\,{\text{is?}}$$
(A) 4
(B) 9
(C) 16
(D) 25
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Solution:
$$\eqalign{ & {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{ }} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = 3\left( {x + \frac{1}{x}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \,\,\,\,\left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\left[ {{{\left( {x + \frac{1}{x}} \right)}^2}} \right]^2} = {\left( 3 \right)^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$
140.
If $$x + \frac{1}{x} = 5{\text{,}}$$ ÃÂÃÂ then $$\frac{{2x}}{{3{x^2} - 5x + 3}}$$ ÃÂÃÂ is equal to?
(A) 5
(B) $$\frac{1}{5}$$
(C) 3
(D) $$\frac{1}{3}$$
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Solution:
$$\eqalign{ & x + \frac{1}{x} = 5 \cr & \therefore \frac{{2x}}{{3{x^2} - 5x + 3}}\,\left( {{\text{Divide by }}x} \right) \cr & = \frac{{\frac{{2x}}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{5x}}{x} + \frac{3}{x}}} \cr & = \frac{2}{{3x + \frac{3}{x} - 5}} \cr & = \frac{2}{{3\left( {x + \frac{1}{x}} \right) - 5}} \cr & = \frac{2}{{3 \times 5 - 5}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$