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Question Number 119657    Answers: 1   Comments: 2

Suppose that the greatest common divisor of the positive integers a,b and c is 1 and ((ab)/(a−b)) = c . Prove that a−b is a perfect square

$${Suppose}\:{that}\:{the}\:{greatest}\:{common}\:{divisor}\:{of} \\ $$$${the}\:{positive}\:{integers}\:{a},{b}\:{and}\:{c}\:{is}\:\mathrm{1}\:{and} \\ $$$$\frac{{ab}}{{a}−{b}}\:=\:{c}\:.\:{Prove}\:{that}\:{a}−{b}\:{is}\:{a} \\ $$$${perfect}\:{square} \\ $$

Question Number 119648    Answers: 1   Comments: 1

Particles of mass m_1 and m_2 (m_2 >m_1 ) are connected by a light inextensible string passing over a smooth fixed pulley. initially both masses hang vertically with mass m_(2 ) at a height X above the floor. if the system is released from rest. with what speed will mass m_2 hit the floor and the mass m_1 will rise a further distance of [(((m_2 −m_1 )x)/(m_1 +m_2 ))] after this occur.

$${Particles}\:{of}\:{mass}\:{m}_{\mathrm{1}} \:{and}\:{m}_{\mathrm{2}} \:\left({m}_{\mathrm{2}} >{m}_{\mathrm{1}} \right) \\ $$$${are}\:{connected}\:{by}\:{a}\:{light}\:{inextensible} \\ $$$${string}\:{passing}\:{over}\:{a}\:{smooth}\:{fixed}\: \\ $$$${pulley}.\:{initially}\:{both}\:{masses}\:{hang} \\ $$$${vertically}\:{with}\:{mass}\:{m}_{\mathrm{2}\:} {at}\:{a}\:{height} \\ $$$${X}\:{above}\:{the}\:{floor}.\:{if}\:{the}\:{system}\:{is}\: \\ $$$${released}\:{from}\:{rest}.\:{with}\:{what}\:{speed} \\ $$$${will}\:{mass}\:{m}_{\mathrm{2}} \:{hit}\:{the}\:{floor}\:{and}\:{the} \\ $$$${mass}\:{m}_{\mathrm{1}} \:{will}\:{rise}\:{a}\:{further}\:{distance}\:{of}\: \\ $$$$\left[\frac{\left({m}_{\mathrm{2}} −{m}_{\mathrm{1}} \right){x}}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }\right]\:{after}\:{this}\:{occur}. \\ $$

Question Number 119647    Answers: 1   Comments: 0

Question Number 119646    Answers: 0   Comments: 1

Question Number 119645    Answers: 1   Comments: 1

A particle of mass m_1 lies on a smooth horizontal table and its connected to a freely hanging particle of mass m_2 by a light inextensible string passing over a smooth fixed pulley situated at the edge of the table. obtain the expression for the time taken for mass m_(1 ) to reach the edge of the table.

$${A}\:{particle}\:{of}\:{mass}\:{m}_{\mathrm{1}} \:{lies}\:{on}\:{a}\:{smooth} \\ $$$${horizontal}\:{table}\:{and}\:{its}\:{connected}\:{to} \\ $$$${a}\:{freely}\:{hanging}\:{particle}\:{of}\:{mass}\:{m}_{\mathrm{2}} \\ $$$${by}\:{a}\:{light}\:{inextensible}\:{string}\:{passing} \\ $$$${over}\:{a}\:{smooth}\:{fixed}\:{pulley}\:{situated} \\ $$$${at}\:{the}\:{edge}\:{of}\:{the}\:{table}.\:{obtain}\:{the} \\ $$$${expression}\:{for}\:{the}\:{time}\:{taken}\:{for} \\ $$$${mass}\:{m}_{\mathrm{1}\:} {to}\:{reach}\:{the}\:{edge}\:{of}\:{the}\:{table}. \\ $$

Question Number 119644    Answers: 5   Comments: 0

(1) ∫ x^5 (√(x^3 +1)) dx (2) (dy/dx) = (3x−2y+1)^2

$$\:\left(\mathrm{1}\right)\:\int\:{x}^{\mathrm{5}} \:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\:\frac{{dy}}{{dx}}\:=\:\left(\mathrm{3}{x}−\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Question Number 119636    Answers: 1   Comments: 0

Test wheather convergent Σ_(n = 1) ^∞ ((3n − 1)/(7^n + 2n))

$$\mathrm{Test}\:\mathrm{wheather}\:\mathrm{convergent} \\ $$$$\:\:\:\underset{\mathrm{n}\:\:=\:\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{3n}\:\:−\:\:\mathrm{1}}{\mathrm{7}^{\mathrm{n}} \:\:+\:\:\mathrm{2n}} \\ $$

Question Number 119635    Answers: 1   Comments: 0

If a function f:R→R satisfies the relation f(x+1)+f(x−1)=(√3)f(x) for all x∈R then a period of f is (A) 10 (B) 12 (C) 6 (D) 4

$$\mathrm{If}\:\mathrm{a}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:{f}\left(\mathrm{x}+\mathrm{1}\right)+{f}\left(\mathrm{x}−\mathrm{1}\right)=\sqrt{\mathrm{3}}{f}\left(\mathrm{x}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Question Number 119634    Answers: 2   Comments: 0

Q1 If f:R→R is defined by f(x)=[x]+[x+(1/2)]+[x+(2/3)]−3x+5 where [x] is the integral part of x, then a period of f is (A) 1 (B) 2/3 (C) 1/2 (D) 1/3 Q2 Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$$\mathrm{Q1} \\ $$$$\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:\:\:\:\:\:\:\:{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]+\left[\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]+\left[\mathrm{x}+\frac{\mathrm{2}}{\mathrm{3}}\right]−\mathrm{3x}+\mathrm{5} \\ $$$$\mathrm{where}\:\left[\mathrm{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:\mathrm{x},\:\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}/\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{1}/\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Q2} \\ $$$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$

Question Number 119629    Answers: 1   Comments: 0

Question Number 119618    Answers: 1   Comments: 0

Question Number 119611    Answers: 0   Comments: 0

Question Number 119600    Answers: 2   Comments: 0

Find gcd of x^4 +x^3 −4x^2 +x+5 and x^3 +x^2 −9x−9

$${Find}\:{gcd}\:{of}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{5}\: \\ $$$${and}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9} \\ $$

Question Number 119588    Answers: 1   Comments: 0

Question Number 119580    Answers: 3   Comments: 0

Given k ∈ N. 1) justify these relations: 3^(2k) +1≡2[8] and 3^(2k+1) +1≡4[8]. 2) Given (E): 2^n −3^m =1. n and m are unknowed. • Show that if m is even , (E) does not have solution. ■ Deduct from the first question 1) that the couple (2;1) is the only solution of (E).

$$\mathrm{Given}\:\mathrm{k}\:\in\:\mathbb{N}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{justify}\:\mathrm{these}\:\mathrm{relations}: \\ $$$$\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\equiv\mathrm{2}\left[\mathrm{8}\right]\:\mathrm{and}\:\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}\equiv\mathrm{4}\left[\mathrm{8}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\left(\mathrm{E}\right):\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1}.\:\mathrm{n}\:\mathrm{and}\:\mathrm{m}\:\mathrm{are}\:\mathrm{unknowed}. \\ $$$$\bullet\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:,\:\left(\mathrm{E}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\: \\ $$$$\mathrm{solution}. \\ $$$$\left.\blacksquare\:\mathrm{Deduct}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{question}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{couple}\:\left(\mathrm{2};\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{of}\:\left(\mathrm{E}\right). \\ $$

Question Number 119576    Answers: 2   Comments: 0

Question Number 119575    Answers: 1   Comments: 0

a ∈ N. a is not a multiple of 3. 1) Show that a^3 ≡−1[9] or a^3 ≡1[9]. 2) Given a; b; c ∈ Z. Deduct from 1) that if a^3 +b^3 +c^3 ≡0[9] , then one of integers a; b; c is divisible by 3.

$$\mathrm{a}\:\in\:\mathbb{N}.\:\mathrm{a}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{a}^{\mathrm{3}} \equiv−\mathrm{1}\left[\mathrm{9}\right]\:\mathrm{or}\:\mathrm{a}^{\mathrm{3}} \equiv\mathrm{1}\left[\mathrm{9}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\mathrm{a};\:\mathrm{b};\:\mathrm{c}\:\in\:\mathbb{Z}. \\ $$$$\left.\mathrm{Deduct}\:\mathrm{from}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \equiv\mathrm{0}\left[\mathrm{9}\right]\:,\:\mathrm{then}\:\: \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{integers}\:\mathrm{a};\:\mathrm{b};\:\mathrm{c}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}. \\ $$

Question Number 119572    Answers: 0   Comments: 3

Question Number 119570    Answers: 3   Comments: 0

... ♣_♣ ^♣ nice calculus♣_♣ ^♣ ... prove that :: Ω=∫_0 ^( ∞) e^(−2x) ln(((1+e^(−x) )/(1−e^(−x) )))=1 ...★ M.N.1970★...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}{nice}\:\:{calculus}\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}... \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {e}^{−\mathrm{2}{x}} {ln}\left(\frac{\mathrm{1}+{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\bigstar\:\mathscr{M}.\mathscr{N}.\mathrm{1970}\bigstar... \\ $$

Question Number 119567    Answers: 1   Comments: 0

(1 −x)(d^2 y/dx^(2 ) ) + x(dy/dx) − xy = (1/(1 − x)) , x≠1 has power series solution for ∣x∣<1

$$ \\ $$$$\left(\mathrm{1}\:−{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }\:+\:{x}\frac{{dy}}{{dx}}\:−\:{xy}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:,\:{x}\neq\mathrm{1} \\ $$$${has}\:{power}\:{series}\:{solution}\:{for}\:\mid{x}\mid<\mathrm{1}\: \\ $$

Question Number 119591    Answers: 0   Comments: 6

...nice calculus... prove that :: ∫_0 ^( (π/2)) (√(((2^x −1)sin^3 (x))/((2^x +1)(sin^3 (x)+cos^3 (x))))) dx<(π/8) ...m.n.1970...

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\: \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\left(\mathrm{2}^{{x}} −\mathrm{1}\right){sin}^{\mathrm{3}} \left({x}\right)}{\left(\mathrm{2}^{{x}} +\mathrm{1}\right)\left({sin}^{\mathrm{3}} \left({x}\right)+{cos}^{\mathrm{3}} \left({x}\right)\right)}}\:\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\mathrm{1970}... \\ $$$$ \\ $$

Question Number 119565    Answers: 2   Comments: 1

Question Number 119564    Answers: 0   Comments: 0

((3/x) − ((15)/(2y)) ) ÷ (6/(xy)) = ((6y − 15x)/(2xy)) × ((xy)/6).

$$\left(\frac{\mathrm{3}}{{x}}\:−\:\frac{\mathrm{15}}{\mathrm{2}{y}}\:\right)\:\boldsymbol{\div}\:\frac{\mathrm{6}}{{xy}}\:=\:\frac{\mathrm{6}{y}\:−\:\mathrm{15}{x}}{\mathrm{2}{xy}}\:×\:\frac{{xy}}{\mathrm{6}}. \\ $$

Question Number 119548    Answers: 1   Comments: 2

Question Number 119540    Answers: 1   Comments: 0

Question Number 119538    Answers: 1   Comments: 1

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