11.
If x1 x2 x3 = 4(4 + x1 + x2 + x3 ), then what is the value of $$\left[ {\frac{1}{{2 + {x_1}}}} \right] + \left[ {\frac{1}{{2 + {x_2}}}} \right] + \left[ {\frac{1}{{2 + {x_3}}}} \right]?$$
(A) 1
(B) $$\frac{1}{2}$$
(C) 2
(D) $$\frac{1}{3}$$
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Solution:
$$\eqalign{ & {x_1}{x_2}{x_3} = 4\left( {4 + {x_1} + {x_2} + {x_3}} \right) \cr & \left[ {\frac{1}{{2 + {x_1}}}} \right] + \left[ {\frac{1}{{2 + {x_2}}}} \right] + \left[ {\frac{1}{{2 + {x_3}}}} \right] = ? \cr & {\text{Assume value of }}{x_1},\,{x_2}\,\& \,{x_3} \cr & {x_1} = 4,\,{x_2} = 4,\,{x_3} = 4 \cr & 4 \times 4 \times 4 = 4\left( {4 + 4 + 4 + 4} \right) \cr & 64 = 64{\text{ value satisfied}} \cr & \therefore \left[ {\frac{1}{{2 + 4}} + \frac{1}{{2 + 4}} + \frac{1}{{2 + 4}}} \right] \cr & = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \cr & = \frac{3}{6} \cr & = \frac{1}{2} \cr} $$
12.
If x + y + z = 19, xy + yz + zx = 144, then the value of $$\sqrt {{x^3} + {y^3} + {z^3} - 3xyz} $$ ÃÂÃÂ ÃÂÃÂ is:
(A) 21
(B) 17
(C) 19
(D) 13
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Solution:
$$\eqalign{ & x + y + z = 19, \cr & xy + yz + zx = 144, \cr & \sqrt {{x^3} + {y^3} + {z^3} - 3xyz} \cr & {\text{Let }}z = 0 \cr & x + y = 19,\,xy = 144,\,\sqrt {{x^3} + {y^3}} = ? \cr & \sqrt {{x^3} + {y^3}} \cr & = \sqrt {\left( {x + y} \right)\left[ {{{\left( {x + y} \right)}^2} - 3xy} \right]} \cr & = \sqrt {19\left( {{{19}^2} - 3 \times 144} \right)} \cr & = \sqrt {19 \times 19} \cr & = 19 \cr} $$
13.
If x = 1 + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ2 + ÃÂÃÂÃÂâÃÂÃÂÃÂÃÂ3, then find the value of x - 2x + 4.2
(A) 2(7 + √6)
(B) 2(4 + √6)
(C) 2(3 + √7)
(D) (4 + √6)
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Solution:
x = 1 + √2 + √3 ⇒ x - 1 = √2 + √3 On squaring both sides, (x - 1)2 = (√2 + √3)2 ⇒ x2 - 2x + 1 = 2 + 3 + 2√6 Add 3 both side ⇒ x2 - 2x + 4 = 5 + 2√6 + 3 ⇒ x2 - 2x + 4 = 8 + 2√6 ⇒ x2 - 2x + 4 = 2(4 + √6)
14.
The value of $$\frac{{{{\left( {243} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{9^n}{{.3}^{n - 1}}}}{\text{ is?}}$$
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Solution:
$$\eqalign{ & \frac{{{{\left( {243} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{9^n}{{.3}^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{3^{2n}}{{.3}^{n - 1}}}} \cr & = \frac{{{3^{n + 2n + 1}}}}{{{3^{2n + n - 1}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{3n + 1 - 3n + 1}} \cr & = {3^2} \cr & = 9 \cr} $$
15.
If $$\frac{4}{3}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 110\frac{2}{3},$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ find $$\frac{1}{9}\left( {{x^3} - \frac{1}{{{x^3}}}} \right),$$ ÃÂÃÂ where x > 0.
(A) 84
(B) 85
(C) 74
(D) 76
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Solution:
$$\eqalign{ & \frac{4}{3}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 110\frac{2}{3} \cr & {x^2} + \frac{1}{{{x^2}}} = \frac{{332}}{3} \times \frac{3}{4} \cr & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} - 2 \times x \times \frac{1}{x} \cr & {\left( {x + \frac{1}{x}} \right)^2} = 83 - 2 \cr & {\left( {x + \frac{1}{x}} \right)^2} = 81 \cr & x + \frac{1}{x} = 9 \cr & {\text{Hence,}} \cr & \frac{1}{9}\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = \frac{1}{9}\left[ {{{\left( {x - \frac{1}{x}} \right)}^3} + 3 \times \left( {x - \frac{1}{x}} \right)} \right] \cr & = \frac{1}{9}\left[ {729 + 3 \times 9} \right] \cr & = 84 \cr} $$
16.
A complete factorisation of x4 + 64 is?
(A) (x2 + 8)2
(B) (x2 + 8)(x2 - 8)
(C) (x2 - 4x + 8)(x2 - 4x - 8)
(D) (x2 + 4x + 8)(x2 - 4x + 8)
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Solution:
$$\eqalign{ & \left( {{x^4} + 64} \right) \cr & = {x^4} + {8^2} + 2.{x^2}.8 - 2.{x^2}.8 \cr & = {\left( {{x^2} + 8} \right)^2} - \left( {16{x^2}} \right) \cr & = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2} \cr & = \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right) \cr & = \left( {{x^2} + 4x + 8} \right)\left( {{x^2} - 4x + 8} \right) \cr} $$
17.
If $$\frac{p}{a}$$ + $$\frac{q}{b}$$ + $$\frac{r}{c}$$ = 1 and $$\frac{a}{p}$$ + $$\frac{b}{q}$$ + $$\frac{c}{r}$$ = 0 where p, q, r and a, b, c are non - zero, then value of $$\frac{{{p^2}}}{{{a^2}}}$$ + $$\frac{{{q^2}}}{{{b^2}}}$$ + $$\frac{{{r^2}}}{{{c^2}}}$$ = ?
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Solution:
$$\eqalign{ & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 \cr & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0{\text{ }} \cr & \Rightarrow \frac{p}{a} = x,{\text{ }}\frac{q}{b} = y,{\text{ }}\frac{r}{c} = z \cr & \Rightarrow \left( {x + y + z} \right) = 1 \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 1 \cr & {\text{and }}\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \cr & \Rightarrow \frac{{xy + yz + zx}}{{xyz}} = 0 \cr & \Rightarrow xy + yz + zx = 0 \cr & \therefore {x^2} + {y^2} + {z^2} = 1 \cr & {\text{So, }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1 \cr} $$
18.
The minimum value of (x - 2)(x - 9) is?
(A) $$ - \frac{{11}}{4}$$
(B) $$\frac{{49}}{4}$$
(C) 0
(D) $$ - \frac{{49}}{4}$$
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Solution:
$$\eqalign{ & \left( {x - 2} \right)\left( {x - 9} \right) \cr & = {x^2} - 9x - 2x + 18 \cr & = {x^2} - 11x + 18 \cr & = a{x^2} + bx + c = 0 \cr & {\text{For minimum value}} \cr & = \frac{{4ac - {b^2}}}{{4a}} \cr & = \frac{{4 \times 1 \times 18 - {{\left( { - 11} \right)}^2}}}{{4 \times 1}} \cr & = \frac{{72 - 121}}{4} \cr & = \frac{{ - 49}}{4} \cr & = - \frac{{49}}{4} \cr} $$
19.
If $$\frac{a}{{1 - 2a}}$$ÃÂÃÂ $$+$$ $$\frac{b}{{1 - 2b}}$$ÃÂÃÂ $$+$$ $$\frac{c}{{1 - 2c}}$$ÃÂÃÂ = $$\frac{1}{2}{\text{,}}$$ ÃÂÃÂ then the value of $$\frac{1}{{1 - 2a}}$$ÃÂÃÂ $$+$$ $$\frac{1}{{1 - 2b}}$$ÃÂÃÂ $$+$$ $$\frac{1}{{1 - 2c}}$$ÃÂÃÂ is?
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Solution:
$$\eqalign{ & \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr & {\text{Adding 3 both side}} \cr} $$ $$ \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + $$ $$\frac{{2c}}{{1 - 2c}} = $$ $$1 + 3$$ $$ \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4$$
20.
If $$\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of $$\frac{{{a^6} + 1}}{{{a^3}}}$$ ÃÂÃÂ = ?
(A) 9
(B) 18
(C) 27
(D) 1
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Solution:
$$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr} $$