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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)) ?

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

Question Number 119479    Answers: 3   Comments: 0

Π_(k=1) ^∞ cos((x/2^k )) = ?

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:=\:? \\ $$

Question Number 119474    Answers: 0   Comments: 0

Please help create an algorithm when the userviews in website around 0− 11:59 it will alert “Good Morning” and from 12− 15:59 it will alert “Goodafternoon” and lastly from 16− 23:59 it will alert “Good evening”.

$$ \\ $$$$\boldsymbol{{P}\mathrm{lease}}\:\boldsymbol{\mathrm{help}} \\ $$$$\mathrm{create}\:\mathrm{an}\:\mathrm{algorithm}\:\mathrm{when}\:\mathrm{the}\: \\ $$$$\mathrm{userviews}\:\mathrm{in}\:\mathrm{website}\:\mathrm{around}\:\mathrm{0}− \\ $$$$\mathrm{11}:\mathrm{59}\:\mathrm{it}\:\mathrm{will}\:\mathrm{alert}\:``\mathrm{Good}\:\mathrm{Morning}''\: \\ $$$$\mathrm{and}\:\mathrm{from}\:\mathrm{12}−\:\mathrm{15}:\mathrm{59}\:\mathrm{it}\:\mathrm{will}\:\mathrm{alert}\: \\ $$$$``\mathrm{Goodafternoon}''\:\mathrm{and}\:\mathrm{lastly}\:\mathrm{from}\: \\ $$$$\mathrm{16}−\:\mathrm{23}:\mathrm{59}\:\mathrm{it}\:\mathrm{will}\:\mathrm{alert}\:``\mathrm{Good}\:\mathrm{evening}''. \\ $$

Question Number 119473    Answers: 4   Comments: 1

Question Number 119472    Answers: 1   Comments: 0

Question Number 122799    Answers: 1   Comments: 0

Question Number 122798    Answers: 2   Comments: 1

Question Number 119464    Answers: 2   Comments: 0

Expand as far as the term in x^3 (1) (3−2x−x^2 )^9 (2) (1−x+x^2 )^8

$$\mathrm{Expand}\:\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{8}} \\ $$

Question Number 119463    Answers: 1   Comments: 2

one cube of side length 3cm is painted by three colour say (red blue and black) opposte face same colour Now cube is cut into 27 small cube i)How many cube has three face coloured ii)How many cube two face coloured iii)How many cube has no colour in its face

$${one}\:{cube}\:{of}\:{side}\:{length}\:\mathrm{3}{cm}\:{is}\:{painted}\:{by}\:{three}\:{colour} \\ $$$${say}\:\left({red}\:{blue}\:{and}\:{black}\right)\:{opposte}\:{face}\:{same}\:{colour} \\ $$$${Now}\:{cube}\:{is}\:{cut}\:{into}\:\mathrm{27}\:{small}\:{cube} \\ $$$$\left.{i}\right){How}\:{many}\:{cube}\:{has}\:{three}\:{face}\:{coloured} \\ $$$$\left.{ii}\right){How}\:{many}\:{cube}\:{two}\:{face}\:{coloured} \\ $$$$\left.{iii}\right){How}\:{many}\:{cube}\:\:{has}\:{no}\:{colour}\:{in}\:{its}\:{face} \\ $$

Question Number 119462    Answers: 2   Comments: 0

... advanced calculus... evaluate :: Ω=∫_0 ^( ∞) ((tan^(−1) (x))/(e^(2πx) −1))dx =? m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119452    Answers: 1   Comments: 1

Question Number 119451    Answers: 1   Comments: 0

Question Number 119446    Answers: 0   Comments: 0

calculste ∫_0 ^∞ ((ln(2+x^2 ))/(1+x^3 ))dx

$${calculste}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$

Question Number 119445    Answers: 0   Comments: 0

...nice calculus... prove that:: Σ_(n=1) ^∞ (((−1)^(n−1) )/(n^3 (((2n)),(n) ))) =^(???) ζ(3) m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119442    Answers: 0   Comments: 0

... advanced calculus... prove that : Σ_(n=1) ^∞ (1/(n^2 (((2n)),(n) ))) =^(???) ((ζ(2))/3) solution:: Σ_(n=1) ^∞ (1/(n^2 ∗(((2n)!)/((n!)^2 ))))=Σ_(n=1) ^∞ ((n!∗n!)/(n^2 ∗(2n)!)) =Σ_(n=1) ^∞ ((Γ(n)Γ(n+1))/(nΓ(2n+1)))=Σ_(n=1) ^∞ ((β(n,n+1))/n) =Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n−1) (1−x)^n =∫_0 ^( 1) (1/x)Σ(((x−x^2 )^n )/n)dx =−∫_0 ^( 1) ((ln(1−x+x^2 ))/x)dx =−∫_0 ^( 1) ((ln(1+x^3 )−ln(1+x))/x)dx =∫_0 ^( 1) ((ln(1+x))/x)dx −∫_0 ^( 1) ((ln(1+x^3 ))/x)dx =−li_2 (−1) −∫_0 ^( 1) ((Σ_(n=1) (((−1)^(n+1) x^(3n) )/n))/x) dx =(π^2 /(12))+Σ_(n=1) ^∞ (((−1)^n )/n)∫_0 ^( 1) x^(3n−1) dx =(π^2 /(12)) +(1/3)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(12))−(π^2 /(36)) =(π^2 /(18)) =((ζ(2))/3) ✓✓ m.n.july.1970..

$$\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${solution}::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \ast\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!\ast{n}!}{{n}^{\mathrm{2}} \ast\left(\mathrm{2}{n}\right)!}\: \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\Gamma\left({n}\right)\Gamma\left({n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta\left({n},{n}+\mathrm{1}\right)}{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}}\Sigma\frac{\left({x}−{x}^{\mathrm{2}} \right)^{{n}} }{{n}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}}{dx} \\ $$$$\:=−{li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\underset{{n}=\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{3}{n}} }{{n}}}{{x}}\:{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}−\mathrm{1}} {dx}\:\:\:\:\: \\ $$$$ \\ $$$$\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}.. \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 119440    Answers: 1   Comments: 1

Determine the integer which can be written: N=xyz^(−) ^7 =zyx^(−) ^(11)

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{which}\:\mathrm{can}\: \\ $$$$\mathrm{be}\:\mathrm{written}:\:\mathrm{N}=\overline {\mathrm{xyz}}\:^{\mathrm{7}} =\overline {\mathrm{zyx}}\:^{\mathrm{11}} \\ $$

Question Number 119425    Answers: 1   Comments: 0

decompose F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3))) and calculate ∫_(√2) ^(+∞) F(x)dx

$$\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{2x}−\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)} \\ $$$$\mathrm{and}\:\mathrm{calculate}\:\int_{\sqrt{\mathrm{2}}} ^{+\infty} \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

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