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If $p^{2} + q^{2} = 7 p q,$ ${p^2} + {q^2} = 7pq{\text{,}}$ then the value of $\frac{p}{q} + \frac{q}{p}$ $\frac{p}{q}{\text{ + }}\frac{q}{p}$ is equal to?

19
25
37
43

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} p^{2} + q^{2} = 7 p q \\ \frac{p}{q} + \frac{q}{p} = 7 \end{aligned}

$\eqalign{ & {p^2} + {q^2} = 7pq \cr & \frac{p}{q}{\text{ + }}\frac{q}{p} = 7 \cr}$
(Divide whole equation by pq)

If $x = {(0.25)}^{\frac{1}{2}},$ $x = {\left( {0.25} \right)^{\frac{1}{2}}},$ $y = {(0.4)}^{2},$ $y = {\left( {0.4} \right)^2},$ $z = {(0.216)}^{\frac{1}{2}}$ $z = {\left( {0.216} \right)^{\frac{1}{2}}}$ then-

1y > x > z
2x > y > z
3z > x > y
4x > z > y

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = {(0.25)}^{\frac{1}{2}} \\ x = \sqrt[2]{0.25} = 0.5 \\ y = {(0.4)}^{2} \\ y = 0.16 \\ z = {(0.216)}^{\frac{1}{2}} \\ z = \sqrt[2]{0.216} = 0.464 \\ ∴ x > z > y \end{aligned}

$\eqalign{ & x = {\left( {0.25} \right)^{\frac{1}{2}}} \cr & x = \root 2 \of {0.25} = 0.5{\text{ }} \cr & y = {\left( {0.4} \right)^2} \cr & y = 0.16 \cr & z = {\left( {0.216} \right)^{\frac{1}{2}}} \cr & z = \root 2 \of {0.216} = 0.464 \cr & \therefore x > z > y \cr}$

03. If p³ - q³ = (p - q){(p + q)² - xpq}, then the value of x is?

11
2-1
32
4-2

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} p^{3} - q^{3} = (p - q) {p^{2} + q^{2} + p q} \\ \Rightarrow (p - q) {{(p + q)}^{2} - x p q} = (p - q) (p^{2} + q^{2} + p q) \\ \Rightarrow p^{2} + q^{2} + 2 p q - x p q = p^{2} + q^{2} + p q \\ \Rightarrow 2 p q - p q = x p q \\ \Rightarrow p q = x p q \\ \Rightarrow x = 1 \end{aligned}

$\eqalign{ & {p^3} - {q^3} = \left( {p - q} \right)\left\{ {{p^2} + {q^2} + pq} \right\} \cr & \Rightarrow \left( {p - q} \right)\left\{ {{{\left( {p + q} \right)}^2} - xpq} \right\} = \left( {p - q} \right)\left( {{p^2} + {q^2} + pq} \right) \cr & \Rightarrow {p^2} + {q^2} + 2pq - xpq = {p^2} + {q^2} + pq \cr & \Rightarrow 2pq - pq = xpq \cr & \Rightarrow pq = xpq \cr & \Rightarrow x = 1 \cr}$

If pq(p + q) = 1, then the value of $\frac{1}{p^{3} q^{3}}$ $\frac{1}{{{p^3}{q^3}}}$ - $p^{3}$ ${p^3}$ - $q^{3},$ ${q^3}{\text{,}}$ is equal to?

11
22
33
44

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} p q (p + q) = 1 \\ \Rightarrow p + q = \frac{1}{p q} \\ \Rightarrow p^{3} + q^{3} + p q (p + q) = \frac{1}{p^{3} q^{3}} \\ \Rightarrow \frac{1}{p^{3} q^{3}} - p^{3} - q^{3} = (\frac{13}{p + q}) (p + q) \\ \Rightarrow \frac{1}{p^{3} q^{3}} - p^{3} - q^{3} = 3 \end{aligned}

$\eqalign{ & pq\left( {p + q} \right) = 1 \cr & \Rightarrow p + q = \frac{1}{{pq}} \cr & \Rightarrow {p^3} + {q^3} + pq\left( {p + q} \right) = \frac{1}{{{p^3}{q^3}}} \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = \left( {\frac{{13}}{{p + q}}} \right)\left( {p + q} \right) \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = 3 \cr}$

If $\frac{x^{2} + 1}{x^{2}} = 2,$ $\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$ then the value of $\frac{x - 1}{x}$ $\frac{{x - 1}}{x}$ is?

1-2
20
31
4-1

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} + \frac{1}{x^{2}} = 2 \\ \Rightarrow {(x - \frac{1}{x})}^{2} + 2. x . \frac{1}{x} = 2 \\ \Rightarrow x - \frac{1}{x} = 2 - 2 \\ \Rightarrow x - \frac{1}{x} = 0 \end{aligned}

$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr}$

If $\frac{a^{2} + b^{2}}{c^{2}}$ $\frac{{{a^2} + {b^2}}}{{{c^2}}}$ = $\frac{b^{2} + c^{2}}{a^{2}}$ $\frac{{{b^2} + {c^2}}}{{{a^2}}}$ = $\frac{c^{2} + a^{2}}{b^{2}}$ $\frac{{{c^2} + {a^2}}}{{{b^2}}}$ = $\frac{1}{k},$ $\frac{1}{k}{\text{,}}$ $(k \neq 0)$ $\left( {k \ne 0} \right)$ then k = ?

12
21
30
4 $\frac{1}{2}$ $\frac{1}{2}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a^{2} + b^{2}}{c^{2}} = \frac{b^{2} + c^{2}}{a^{2}} = \frac{c^{2} + a^{2}}{b^{2}} = \frac{1}{k} \\ Put a = b = c = 1 \\ \Rightarrow \frac{1 + 1}{1} + \frac{1 + 1}{1} + \frac{1 + 1}{1} = \frac{1}{k} \\ \Rightarrow 2 = 2 = 2 = \frac{1}{k} \\ \Rightarrow k = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{{{a^2} + {b^2}}}{{{c^2}}} = \frac{{{b^2} + {c^2}}}{{{a^2}}} = \frac{{{c^2} + {a^2}}}{{{b^2}}} = \frac{1}{k} \cr & {\text{Put }}a = b = c = 1 \cr & \Rightarrow \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} = \frac{1}{k} \cr & \Rightarrow 2 = 2 = 2 = \frac{1}{k} \cr & \Rightarrow k = \frac{1}{2} \cr}$

If ${(a + \frac{1}{a})}^{2} = 3,$ ${\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}}$ the value of $a^{3} + \frac{1}{a^{3}}$ ${a^3} + \frac{1}{{{a^3}}}$ = ?

10
2 $3 (a + \frac{1}{a})$ $3\left( {a + \frac{1}{a}} \right)$
3 $3 (a^{2} + \frac{1}{a^{2}})$ $3\left( {{a^2} + \frac{1}{{{a^2}}}} \right)$
41

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(a + \frac{1}{a})}^{2} = 3 \\ \Rightarrow a + \frac{1}{a} = \sqrt{3} \\ Taking cube on both sides \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (\sqrt{3}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 0 \end{aligned}

$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr}$

If $\frac{a}{q - r}$ $\frac{a}{{q - r}}$ = $\frac{b}{r - p}$ $\frac{b}{{r - p}}$ = $\frac{c}{p - q},$ $\frac{c}{{p - q}}{\text{,}}$ find the value of pa + qb + rc is?

10
221
32
4-6

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let \frac{a}{q - r} = \frac{b}{r - p} = \frac{c}{p - q} = k \\ \frac{a}{q - r} = k (On multiplying by p) \\ p a = k (p q - p r) . . . . . . (i) \\ In the same way we can write \\ qb = k (q r - q p) . . . . . . . . (i i) \\ And r c = k (r p - r p) . . . . . i i i) \\ On adding equation (i), (ii) and (iii) \\ p a + q b + r c \\ = k (p q - p r + q r - q p + r p - r q) \\ = 0 \end{aligned}

$\eqalign{ & {\text{Let }}\frac{a}{{q - r}} = \frac{b}{{r - p}} = \frac{c}{{p - q}} = k \cr & \frac{a}{{q - r}} = k{\text{ }}\left( {{\text{On multiplying by }}p} \right) \cr & pa = k\left( {pq - pr} \right)\,......(i) \cr & {\text{In the same way we can write}} \cr & {\text{qb = k}}\left( {qr - qp} \right)\,........(ii) \cr & {\text{And }}rc = k\left( {rp - rp} \right)\,.....iii) \cr & {\text{On adding equation (i), (ii) and (iii) }} \cr & pa + qb + rc \cr & = k\left( {pq - pr + qr - qp + rp - rq} \right) \cr & = 0 \cr}$

If $a^{3} + \frac{1}{a^{3}} = 2,$ ${a^3} + \frac{1}{{{a^3}}} = 2{\text{,}}$ then the value of $\frac{a^{2} + 1}{a}$ $\frac{{{a^2} + 1}}{a}$ is (a positive number) ?

11
22
33
44

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} But a = 1 \\ a^{3} + \frac{1}{a^{3}} = 2 \\ \Rightarrow 1^{3} + \frac{1}{1^{3}} = 2 \\ \Rightarrow 2 = 2 (Satisfy) \\ So, \frac{a^{2} + 1}{a} \\ = \frac{1^{2} + 1}{1} \\ = 2 \end{aligned}

$\eqalign{ & {\text{But }}a = 1 \cr & {a^3} + \frac{1}{{{a^3}}} = 2 \cr & \Rightarrow {1^3} + \frac{1}{{{1^3}}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{So, }}\frac{{{a^2} + 1}}{a} \cr & = \frac{{{1^2} + 1}}{1} \cr & = 2 \cr}$

If $x + \frac{1}{9 x} = 4,$ $x + \frac{1}{{9x}} = 4{\text{,}}$ then $9 x^{2} + \frac{1}{9 x^{2}}$ ${\text{9}}{x^2} + \frac{1}{{9{x^2}}}$ is?

1140
2142
3144
4146

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x + \frac{1}{9 x} = 4 \\ Multiply by 3 both side \\ \Rightarrow 3 x + \frac{1}{3 x} = 12 \\ Squaring both sides \\ \Rightarrow 9 x^{2} + \frac{1}{9 x^{2}} + 2 \times 3 x \times \frac{1}{3 x} = 144 \\ \Rightarrow 9 x^{2} + \frac{1}{9 x^{2}} + 2 = 144 \\ \Rightarrow 9 x^{2} + \frac{1}{9 x^{2}} = 142 \end{aligned}

$\eqalign{ & {\text{ }}x + \frac{1}{{9x}} = 4 \cr & {\text{Multiply by 3 both side}} \cr & \Rightarrow {\text{3}}x + \frac{1}{{3x}} = 12 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 \times 3x \times \frac{1}{{3x}} = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} = 142 \cr}$

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algebra - 02

If $p^{2} + q^{2} = 7 p q,$ ${p^2} + {q^2} = 7pq{\text{,}}$ then the value of $\frac{p}{q} + \frac{q}{p}$ $\frac{p}{q}{\text{ + }}\frac{q}{p}$ is equal to?

If $x = {(0.25)}^{\frac{1}{2}},$ $x = {\left( {0.25} \right)^{\frac{1}{2}}},$ $y = {(0.4)}^{2},$ $y = {\left( {0.4} \right)^2},$ $z = {(0.216)}^{\frac{1}{2}}$ $z = {\left( {0.216} \right)^{\frac{1}{2}}}$ then-

03. If p³ - q³ = (p - q){(p + q)² - xpq}, then the value of x is?

If pq(p + q) = 1, then the value of $\frac{1}{p^{3} q^{3}}$ $\frac{1}{{{p^3}{q^3}}}$ - $p^{3}$ ${p^3}$ - $q^{3},$ ${q^3}{\text{,}}$ is equal to?

If $\frac{x^{2} + 1}{x^{2}} = 2,$ $\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$ then the value of $\frac{x - 1}{x}$ $\frac{{x - 1}}{x}$ is?

If $\frac{a^{2} + b^{2}}{c^{2}}$ $\frac{{{a^2} + {b^2}}}{{{c^2}}}$ = $\frac{b^{2} + c^{2}}{a^{2}}$ $\frac{{{b^2} + {c^2}}}{{{a^2}}}$ = $\frac{c^{2} + a^{2}}{b^{2}}$ $\frac{{{c^2} + {a^2}}}{{{b^2}}}$ = $\frac{1}{k},$ $\frac{1}{k}{\text{,}}$ $(k \neq 0)$ $\left( {k \ne 0} \right)$ then k = ?

If ${(a + \frac{1}{a})}^{2} = 3,$ ${\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}}$ the value of $a^{3} + \frac{1}{a^{3}}$ ${a^3} + \frac{1}{{{a^3}}}$ = ?

If $\frac{a}{q - r}$ $\frac{a}{{q - r}}$ = $\frac{b}{r - p}$ $\frac{b}{{r - p}}$ = $\frac{c}{p - q},$ $\frac{c}{{p - q}}{\text{,}}$ find the value of pa + qb + rc is?

If $a^{3} + \frac{1}{a^{3}} = 2,$ ${a^3} + \frac{1}{{{a^3}}} = 2{\text{,}}$ then the value of $\frac{a^{2} + 1}{a}$ $\frac{{{a^2} + 1}}{a}$ is (a positive number) ?

If $x + \frac{1}{9 x} = 4,$ $x + \frac{1}{{9x}} = 4{\text{,}}$ then $9 x^{2} + \frac{1}{9 x^{2}}$ ${\text{9}}{x^2} + \frac{1}{{9{x^2}}}$ is?

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algebra - 02

01. If p2+q2=7pq,p2+q2=7pq,{p^2} + {q^2} = 7pq{\text{,}} then the value of pq + qppq + qp\frac{p}{q}{\text{ + }}\frac{q}{p} is equal to?

02. If x=(0.25)12,x=(0.25)12,x = {\left( {0.25} \right)^{\frac{1}{2}}}, y=(0.4)2,y=(0.4)2,y = {\left( {0.4} \right)^2}, z=(0.216)12z=(0.216)12z = {\left( {0.216} \right)^{\frac{1}{2}}} then-

03. If p3 - q3 = (p - q){(p + q)2 - xpq}, then the value of x is?

04. If pq(p + q) = 1, then the value of 1p3q31p3q3\frac{1}{{{p^3}{q^3}}} - p3p3{p^3} - q3,q3,{q^3}{\text{,}} is equal to?

05. If x2+1x2=2, x2+1x2=2,\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}} then the value of x−1xx−1x\frac{{x - 1}}{x} is?

06. If a2+b2c2a2+b2c2\frac{{{a^2} + {b^2}}}{{{c^2}}} = b2+c2a2b2+c2a2\frac{{{b^2} + {c^2}}}{{{a^2}}} = c2+a2b2c2+a2b2\frac{{{c^2} + {a^2}}}{{{b^2}}} = 1k,1k,\frac{1}{k}{\text{,}} (k≠0)(k≠0)\left( {k \ne 0} \right) then k = ?

07. If (a+1a)2=3,(a+1a)2=3,{\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}} the value of a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} = ?

08. If aq−raq−r\frac{a}{{q - r}} = br−pbr−p\frac{b}{{r - p}} = cp−q,cp−q,\frac{c}{{p - q}}{\text{,}} find the value of pa + qb + rc is?

09. If a3+1a3=2,a3+1a3=2,{a^3} + \frac{1}{{{a^3}}} = 2{\text{,}} then the value of a2+1aa2+1a\frac{{{a^2} + 1}}{a} is (a positive number) ?

10. If x+19x=4,x+19x=4,x + \frac{1}{{9x}} = 4{\text{,}} then 9x2+19x29x2+19x2{\text{9}}{x^2} + \frac{1}{{9{x^2}}} is?

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If $p^{2} + q^{2} = 7 p q,$ ${p^2} + {q^2} = 7pq{\text{,}}$ then the value of $\frac{p}{q} + \frac{q}{p}$ $\frac{p}{q}{\text{ + }}\frac{q}{p}$ is equal to?

If $x = {(0.25)}^{\frac{1}{2}},$ $x = {\left( {0.25} \right)^{\frac{1}{2}}},$ $y = {(0.4)}^{2},$ $y = {\left( {0.4} \right)^2},$ $z = {(0.216)}^{\frac{1}{2}}$ $z = {\left( {0.216} \right)^{\frac{1}{2}}}$ then-

03. If p³ - q³ = (p - q){(p + q)² - xpq}, then the value of x is?

If pq(p + q) = 1, then the value of $\frac{1}{p^{3} q^{3}}$ $\frac{1}{{{p^3}{q^3}}}$ - $p^{3}$ ${p^3}$ - $q^{3},$ ${q^3}{\text{,}}$ is equal to?

If $\frac{x^{2} + 1}{x^{2}} = 2,$ $\frac{{{\text{ }}{x^2} + 1}}{{{x^2}}} = 2{\text{,}}$ then the value of $\frac{x - 1}{x}$ $\frac{{x - 1}}{x}$ is?

If $\frac{a^{2} + b^{2}}{c^{2}}$ $\frac{{{a^2} + {b^2}}}{{{c^2}}}$ = $\frac{b^{2} + c^{2}}{a^{2}}$ $\frac{{{b^2} + {c^2}}}{{{a^2}}}$ = $\frac{c^{2} + a^{2}}{b^{2}}$ $\frac{{{c^2} + {a^2}}}{{{b^2}}}$ = $\frac{1}{k},$ $\frac{1}{k}{\text{,}}$ $(k \neq 0)$ $\left( {k \ne 0} \right)$ then k = ?

If ${(a + \frac{1}{a})}^{2} = 3,$ ${\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}}$ the value of $a^{3} + \frac{1}{a^{3}}$ ${a^3} + \frac{1}{{{a^3}}}$ = ?

If $\frac{a}{q - r}$ $\frac{a}{{q - r}}$ = $\frac{b}{r - p}$ $\frac{b}{{r - p}}$ = $\frac{c}{p - q},$ $\frac{c}{{p - q}}{\text{,}}$ find the value of pa + qb + rc is?

If $a^{3} + \frac{1}{a^{3}} = 2,$ ${a^3} + \frac{1}{{{a^3}}} = 2{\text{,}}$ then the value of $\frac{a^{2} + 1}{a}$ $\frac{{{a^2} + 1}}{a}$ is (a positive number) ?

If $x + \frac{1}{9 x} = 4,$ $x + \frac{1}{{9x}} = 4{\text{,}}$ then $9 x^{2} + \frac{1}{9 x^{2}}$ ${\text{9}}{x^2} + \frac{1}{{9{x^2}}}$ is?

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6 to 12