91.
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:
$$\eqalign{ & \left( {{x^4} + 64} \right) \cr & = {x^4} + {8^2} + 2.{x^2}.8 - 2.{x^2}.8 \cr & = {\left( {{x^2} + 8} \right)^2} - \left( {16{x^2}} \right) \cr & = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2} \cr & = \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right) \cr & = \left( {{x^2} + 4x + 8} \right)\left( {{x^2} - 4x + 8} \right) \cr} $$
92.
If p = 999, then the value of $$\root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} {\text{ is?}}$$
(A) 1000
(B) 999
(C) 998
(D) 1002
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Solution:
$$\eqalign{ & \because p = 999 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {999 + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {1000} \right)}^3}} \cr & = 1000 \cr} $$
93.
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:
$$\eqalign{ & {a^3} - {b^3} = 56 \cr & \Rightarrow a - b = 2 \cr & \,\,\,\,\left( {{\text{By cubing}}} \right) \cr & \Rightarrow {a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( 2 \right)^2} \cr & \Rightarrow 56 - 3ab \times 2 = 8 \cr & \Rightarrow - 6ab = 8 - 56 \cr & \Rightarrow 6ab = 48 \cr & \Rightarrow ab = 8 \cr & \left( {a - b} \right) = 2 \cr & \,\,{\text{ }}\left( {{\text{By squaring}}} \right) \cr & \Rightarrow {\left( {a - b} \right)^2} = {\left( 2 \right)^2} \cr & \Rightarrow {a^2} + {b^2} - 2ab = 4 \cr & \Rightarrow {a^2} + {b^2} = 4 + 2ab \cr & \Rightarrow {a^2} + {b^2} = 4 + 2 \times 8 \cr & \Rightarrow {a^2} + {b^2} = 20 \cr} $$
94.
If $${a^{\frac{1}{3}}} = 11,$$ ÃÂÃÂ then the value of a2 - 331a is?
(A) 1331331
(B) 1331000
(C) 1334331
(D) 1330030
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Solution:
$$\eqalign{ & {\text{ }}{a^{\frac{1}{3}}} = 11 \cr & \Leftrightarrow a = {11^3} = 1331 \cr & {a^2} - 331a \cr & = a\left( {a - 331} \right) \cr & = 1331\left( {1331 - 331} \right) \cr & = 1331\left( {1000} \right) \cr & = 1331000 \cr} $$
95.
If $${p^2} + \frac{1}{{{p^2}}} = 47{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of $$p + \frac{1}{p}$$ ÃÂÃÂ is?
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Solution:
$$\eqalign{ & {p^2} + \frac{1}{{{p^2}}} = 47 \cr & {\text{On adding 2 both side}} \cr & {p^2} + \frac{1}{{{p^2}}} + 2 = 47 + 2 \cr & \Rightarrow {\left( {p + \frac{1}{p}} \right)^2} = 49 \cr & \Rightarrow \left( {p + \frac{1}{p}} \right) = 7 \cr & \Rightarrow p + \frac{1}{p} = 7 \cr} $$
96.
If $$x + \frac{1}{x} = 2{\text{,}}$$ ÃÂÃÂ then the value of $${x^{12}} - \frac{1}{{{x^{12}}}}$$ ÃÂÃÂ is?
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Solution:
$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 2\ , . . . . . . (i) \cr & {\text{The value of }}{x^{12}} - \frac{1}{{{x^{12}}}} =\, ? \cr & \Rightarrow {\text{If }}x = 1 \cr & \Rightarrow x + \frac{1}{x} = 2 \cr & \Rightarrow 1 + 1 = 2 \cr & {\text{Then, }}{x^{12}} - \frac{1}{{{x^{12}}}} \cr & \Rightarrow {1^{12}} - \frac{1}{{{1^{12}}}} \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$
97.
If a2 + b2 = 4b + 6a - 13, then what is the value of a + b?
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Solution:
a2 + b2 = 4b + 6a - 13 a2 + 9 - 3a + b2 + 4 - 4b = 0 (a - 3)2 + (b - 2)2 = 0 a = 3, b = 2 a + b = 5
98.
If [8(x + y)3 - 27(x - y)3 ] ÃÂÃÂÃÂ÷ (5y - x) = Ax2 + Bxy + Cy2 , then the value of (A + B + C) is:
(A) 26
(B) 19
(C) 16
(D) 13
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Solution:
[8(x + y)3 - 27(x - y)3] ÷ (5y - x) = Ax2 + Bxy + Cy2 Let, x = 1, y = 1 [8(1 + 1)3 - 27 × 0] ÷ (5 - 1) = (A + B + C) $$\frac{{8 \times 8}}{4}$$ = A + B + C 16 = A + B + C
99.
If $$x - \frac{1}{x} = 1{\text{,}}$$ ÃÂÃÂ then the value of $$\frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}}$$ ÃÂÃÂ = ?
(A) $$\frac{1}{4}$$
(B) $$\frac{1}{2}$$
(C) $$\frac{3}{4}$$
(D) 0
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Solution:
$$\eqalign{ & x - \frac{1}{x} = 1{\text{ }} \cr & \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr & {\text{Divide and multiply by }}x \cr & \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr & \Rightarrow \frac{4}{8} \cr & \Rightarrow \frac{1}{2} \cr} $$
100.
For what value of k the expression $$p + \frac{1}{4} + \sqrt p + {k^2}$$ ÃÂÃÂ ÃÂÃÂ is perfect square?
(A) 0
(B) $$ \pm \frac{1}{4}$$
(C) $$ \pm \frac{1}{8}$$
(D) $$ \pm \frac{1}{2}$$
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Solution:
$$\eqalign{ & p + \frac{1}{4} + \sqrt p + {k^2} \cr & = p + \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {\left( {\sqrt p } \right)^2} + 2 \times \frac{1}{2} \times \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {{\text{A}}^2} + {\text{2}} \times {\text{A}} \times {\text{B}} + {{\text{B}}^2} \cr & {\text{A}} = \sqrt p \cr & {{\text{B}}^2} = \left( {{k^2} + \frac{1}{4}} \right) \cr & {\text{B}} = \frac{1}{2} \cr & \therefore {k^2} + \frac{1}{4} = {\left( {\frac{1}{2}} \right)^2} \cr & \Rightarrow {k^2} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {k^2} = 0 \cr & \Rightarrow k = 0 \cr} $$