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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

Question Number 119529 Answers: 3 Comments: 7

Question Number 119524 Answers: 1 Comments: 0

Question Number 119517 Answers: 2 Comments: 0

lim_(x→∞) [ sin (x+(1/x))−sin x ] =?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{sin}\:{x}\:\right]\:=? \\ $$

Question Number 119527 Answers: 2 Comments: 2

Question Number 119510 Answers: 0 Comments: 0

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

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

Question Number 119506 Answers: 1 Comments: 0

cos (π/5)cos x+sin (π/5)sin x ≤ ((√2)/2)

$$\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{5}}\mathrm{cos}\:{x}+\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\mathrm{sin}\:{x}\:\leqslant\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Question Number 119503 Answers: 2 Comments: 0

solve ⌊ (√x) ⌋ = ⌊ (x)^(1/(3 )) ⌋

$${solve}\:\lfloor\:\sqrt{{x}}\:\rfloor\:=\:\lfloor\:\sqrt[{\mathrm{3}\:}]{{x}}\:\rfloor\: \\ $$

Question Number 119495 Answers: 0 Comments: 0

Question Number 119494 Answers: 3 Comments: 0

Russian olympiad find real solution of the system { ((sin x+2sin (x+y+z)=0)),((sin y+3sin (x+y+z)=0)),((sin z+4sin (x+y+z)=0)) :}

$${Russian}\:{olympiad}\: \\ $$$${find}\:{real}\:{solution}\:{of}\:{the}\:{system}\: \\ $$$$\begin{cases}{\mathrm{sin}\:{x}+\mathrm{2sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{y}+\mathrm{3sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{z}+\mathrm{4sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\end{cases} \\ $$$$ \\ $$

Question Number 119490 Answers: 0 Comments: 0

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)

$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$

Question Number 119487 Answers: 2 Comments: 0

find element of set S = { ((x^3 −3x^2 +2)/(2x+1)) ∈ Z for x∈Z }

$${find}\:{element}\:{of}\:{set}\:{S}\:=\:\left\{\:\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\:\in\:\mathbb{Z}\:{for}\:{x}\in\mathbb{Z}\:\right\} \\ $$

Question Number 119483 Answers: 1 Comments: 0

Let a>0 and f:R→R a function satisfying f(x+a)=1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3) for all x∈R. Then a period of f(x) is ka where k is a positive integer whose value is (A)1 (B)2 (C)3 (D)4

$$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{x}+{a}\right)=\mathrm{1}+\left[\mathrm{2}−\mathrm{3}{f}\left(\mathrm{x}\right)+\mathrm{3}{f}\left(\mathrm{x}\right)^{\mathrm{2}} −{f}\left(\mathrm{x}\right)^{\mathrm{3}} \right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}.\:\mathrm{Then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\left(\mathrm{x}\right)\:\mathrm{is}\:{ka}\:\mathrm{where}\:{k}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{whose}\:\mathrm{value}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\mathrm{4} \\ $$

Question Number 119482 Answers: 1 Comments: 0

lim_(x→0) x^2 cos ((1/x)) ?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}}\right)\:? \\ $$

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