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Question Number 141082    Answers: 3   Comments: 0

........ nice ....... calculus ........ 𝛗:=Ξ£_(n=2) ^∞ ((ΞΆ ( n ))/(n . 4^n ))=?

$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:........\:{nice}\:\:.......\:\:{calculus}\:........ \\ $$$$\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\zeta\:\left(\:{n}\:\right)}{{n}\:.\:\mathrm{4}^{{n}} }=? \\ $$$$ \\ $$

Question Number 141045    Answers: 0   Comments: 4

Question Number 141034    Answers: 0   Comments: 2

I donβ€²t recover my old phone documents. please advise me in briefly how to restore my old phone documents in my new phone. plese help me.

$${I}\:{don}'{t}\:{recover}\:{my}\:{old}\:{phone}\:{documents}. \\ $$$${please}\:{advise}\:{me}\:{in}\:{briefly}\:{how}\:{to}\:{restore} \\ $$$$\:{my}\:{old}\:{phone}\:{documents}\:{in}\:{my}\:{new} \\ $$$$\:{phone}. \\ $$$$ \\ $$$${plese}\:{help}\:{me}. \\ $$

Question Number 141031    Answers: 0   Comments: 0

∫_(Ο€/6) ^( Ο€/3) (√(1+((cos^2 x)/(sin x)))) dx ?

$$\:\:\:\:\:\:\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{3}} \:\sqrt{\mathrm{1}+\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}}\:{dx}\:?\: \\ $$

Question Number 141030    Answers: 1   Comments: 1

β†’βŸ¨Ξ£_(n=1) ^∞ (1/(9n^2 +3n)) =? βŸ©β†

$$\:\:\:\:\:\:\:\:\:\:\rightarrow\langle\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{3}{n}}\:=?\:\rangle\leftarrow \\ $$

Question Number 141024    Answers: 3   Comments: 0

Given X=((√(√x))+(1/( (√(√x)))))^n . What is the coefficient of x^(5βˆ’(n/4) ?) propositions: a. 5((n!)/(10!)) b. ((n!)/(5!)) c. ((n),((10)) ) d. ((n),(5) )

$${Given}\:{X}=\left(\sqrt{\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{{x}}}}\right)^{{n}} . \\ $$$${What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}βˆ’\frac{\mathrm{n}}{\mathrm{4}}\:?} \\ $$$$\mathrm{propositions}: \\ $$$${a}.\:\:\:\:\mathrm{5}\frac{{n}!}{\mathrm{10}!} \\ $$$${b}.\:\:\:\:\frac{{n}!}{\mathrm{5}!} \\ $$$${c}.\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$$${d}.\:\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix} \\ $$

Question Number 141017    Answers: 2   Comments: 1

Question Number 141004    Answers: 0   Comments: 0

Let 0 ≀ a,b < 1. Prove that (1/4)βˆ™(((2βˆ’a)(2βˆ’b))/((1βˆ’a)(1βˆ’b))) β‰₯ ((4+a+b)/(4βˆ’aβˆ’b))

$$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}βˆ’{a}\right)\left(\mathrm{2}βˆ’{b}\right)}{\left(\mathrm{1}βˆ’{a}\right)\left(\mathrm{1}βˆ’{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}βˆ’{a}βˆ’{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140998    Answers: 1   Comments: 0

Question Number 140997    Answers: 0   Comments: 1

Let a,b,c β‰₯ 0. Prove that (1/8)βˆ™(((2+a)(2+b)(2+c))/((1+a)(1+b)(1+c))) β‰₯ ((4βˆ’aβˆ’bβˆ’c)/(4+a+b+c))

$$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}βˆ’{a}βˆ’{b}βˆ’{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140996    Answers: 1   Comments: 0

Evaluation of :: Ξ© :=Ξ£_(n=0) ^∞ (((βˆ’1)^n )/(1+n^2 )) solution:: Ξ©:=1+Ξ£_(n=1) ^∞ (((βˆ’1)^n )/(n^2 βˆ’i^2 )) =(1/(2i)){Ξ£_(n=1) ^∞ (((βˆ’1)^n )/(nβˆ’i))βˆ’(((βˆ’1)^n )/(n+i))} :=1+(1/(2i)) (Ξ¦βˆ’Ξ¨) where Ξ¦:=Ξ£_(n=1) ^∞ (((βˆ’1)^n )/(nβˆ’i)) and Ξ¨ :=Ξ£_(n=1) ^∞ (((βˆ’1)^n )/(n+i)) Ξ¦:=Ξ£_(n=1) ^∞ (1/(2nβˆ’k)) βˆ’Ξ£_(n=1) ^∞ (1/(2nβˆ’1βˆ’i)) :=(1/2){Ξ£_(n=1) ^∞ (1/(nβˆ’(i/2)))βˆ’Ξ£_(n=1) ^∞ (1/(nβˆ’((1+i)/2)))} :=(1/2){ψ(1βˆ’((1+i)/2))βˆ’Οˆ(1βˆ’(i/2))} :=(1/2)(ψ(((1βˆ’i)/2))βˆ’Οˆ(1βˆ’(i/2))).... Ξ¨:=Ξ£_(n=1) ^∞ (((βˆ’1)^n )/(n+i)) =Ξ£_(n=1) ^∞ (1/(2n+i))βˆ’Ξ£_(n=1) ^∞ (1/(2nβˆ’1+i)) :=(1/2){Ξ£_(n=1) ^∞ (1/(n+(i/2)))βˆ’Ξ£_(n=1) ^∞ (1/(n+((iβˆ’1)/2)))} :=(1/2)(ψ(((1+i)/2))βˆ’Οˆ(1+(i/2))) .... Ξ¦βˆ’Ξ¨:=(1/2){ψ(((1βˆ’i)/2))βˆ’Οˆ(((1+i)/2))} +(1/2){ψ(1+(i/2))βˆ’Οˆ(1βˆ’(i/2))} :=(1/2)(βˆ’Ο€cotΟ€(((1βˆ’i)/2)))+(1/2)((2/i)βˆ’Ο€cot(Ο€(i/2))) :=βˆ’(Ο€/2)tan(((Ο€i)/2))βˆ’(Ο€/2)cot(((Ο€i)/2))βˆ’i :=βˆ’iβˆ’(Ο€/(sin(Ο€i)))=βˆ’iβˆ’((2iΟ€)/(e^(βˆ’Ο€) βˆ’e^Ο€ )) :=βˆ’i+Ο€icsch(Ο€) .... Ξ© :=1+(1/(2i))(βˆ’i+Ο€icsch(Ο€))=(1/2)+(Ο€/2) csch(Ο€) ... Ξ©:=(1/2)+(Ο€/2) csch(Ο€)....βœ“βœ“βœ“

$$ \\ $$$$\:\:\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Omega\::=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\Omega:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} βˆ’{i}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{{n}βˆ’{i}}βˆ’\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{{n}+{i}}\right\} \\ $$$$\:\:\:\:\:\:\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left(\Phiβˆ’\Psi\right)\:\:\:\:{where}\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{{n}βˆ’{i}} \\ $$$$\:\:\:\:\:\:\:{and}\:\:\:\:\Psi\::=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{{n}+{i}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}βˆ’{k}}\:βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}βˆ’\mathrm{1}βˆ’{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}βˆ’\frac{{i}}{\mathrm{2}}}βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}βˆ’\frac{\mathrm{1}+{i}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}βˆ’\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)βˆ’\psi\left(\mathrm{1}βˆ’\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}βˆ’{i}}{\mathrm{2}}\right)βˆ’\psi\left(\mathrm{1}βˆ’\frac{{i}}{\mathrm{2}}\right)\right).... \\ $$$$\:\:\:\:\Psi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{{n}+{i}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{i}}βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}βˆ’\mathrm{1}+{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}}{\mathrm{2}}}βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}βˆ’\mathrm{1}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)βˆ’\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right)\:.... \\ $$$$\:\:\:\:\:\Phiβˆ’\Psi:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\frac{\mathrm{1}βˆ’{i}}{\mathrm{2}}\right)βˆ’\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)βˆ’\psi\left(\mathrm{1}βˆ’\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(βˆ’\pi{cot}\pi\left(\frac{\mathrm{1}βˆ’{i}}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{i}}βˆ’\pi{cot}\left(\pi\frac{{i}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\::=βˆ’\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi{i}}{\mathrm{2}}\right)βˆ’\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\pi{i}}{\mathrm{2}}\right)βˆ’{i} \\ $$$$\:\:\:\::=βˆ’{i}βˆ’\frac{\pi}{{sin}\left(\pi{i}\right)}=βˆ’{i}βˆ’\frac{\mathrm{2}{i}\pi}{{e}^{βˆ’\pi} βˆ’{e}^{\pi} } \\ $$$$\:\:\:\::=βˆ’{i}+\pi{icsch}\left(\pi\right)\:.... \\ $$$$\:\:\:\:\:\:\Omega\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\left(βˆ’{i}+\pi{icsch}\left(\pi\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)....\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 140988    Answers: 1   Comments: 0

∫((βˆ’csc^2 x)/((cscx+cotx)^3 ))dx

$$\int\frac{βˆ’{csc}^{\mathrm{2}} {x}}{\left({cscx}+{cotx}\right)^{\mathrm{3}} }{dx} \\ $$

Question Number 141012    Answers: 0   Comments: 0

s^2 (s+1)^2 +(c/2)s(s+1)^2 βˆ’(c^2 /2)(s+1) +(c^3 /2) = 0 solve for s in terms of 0<c<(2/(3(√3))) .

$$\:{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} βˆ’\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left({s}+\mathrm{1}\right) \\ $$$$\:\:+\frac{{c}^{\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{0}\:\:\:\:\:{solve}\:{for}\:{s}\:{in}\:{terms} \\ $$$${of}\:\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

Question Number 140982    Answers: 0   Comments: 3

convergence and value of Σ_(n=1) ^∞ (n^n /((n!)^2 ))

$${convergence}\:{and}\:{value}\:{of} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{{n}} }{\left({n}!\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Question Number 140981    Answers: 1   Comments: 0

Let Ξ±β‰ 1 and Ξ±^(13) =1. If a=Ξ±+Ξ±^3 +Ξ±^4 +Ξ±^(βˆ’4) +Ξ±^(βˆ’3) + Ξ±^(βˆ’1) and b=Ξ±^2 +Ξ±^5 +Ξ±^6 +Ξ±^(βˆ’6) +Ξ±^(βˆ’5) +Ξ±^(βˆ’2) then the quadratic equation whose roots are a and b is (A) x^2 +x+3=0 (B) x^2 +x+4=0 (C) x^2 +xβˆ’3=0 (D) x^2 +xβˆ’4=0

$$\mathrm{Let}\:\alpha\neq\mathrm{1}\:\mathrm{and}\:\alpha^{\mathrm{13}} =\mathrm{1}.\:\mathrm{If}\:{a}=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{βˆ’\mathrm{4}} +\alpha^{βˆ’\mathrm{3}} + \\ $$$$\alpha^{βˆ’\mathrm{1}} \:\mathrm{and}\:{b}=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{βˆ’\mathrm{6}} +\alpha^{βˆ’\mathrm{5}} +\alpha^{βˆ’\mathrm{2}} \:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:{a}\:\mathrm{and}\:{b}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}βˆ’\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}βˆ’\mathrm{4}=\mathrm{0} \\ $$

Question Number 140978    Answers: 1   Comments: 0

calculate Σ_(n=0) ^∞ (1/((n!)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}!\right)^{\mathrm{2}} } \\ $$

Question Number 140977    Answers: 2   Comments: 0

find ∫_0 ^∞ (e^(βˆ’t(1+x^2 )) /(1+x^2 ))dx with tβ‰₯0

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{βˆ’{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$

Question Number 140976    Answers: 1   Comments: 0

find ∫_0 ^Ο€ (dx/((2βˆ’cosxβˆ’sinx)^2 ))

$${find}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\left(\mathrm{2}βˆ’{cosx}βˆ’{sinx}\right)^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 140974    Answers: 2   Comments: 3

If the equations x^2 βˆ’3x+a=0 and x^2 +axβˆ’3=0 have a common root, then a possible value of a is (A) 3 (B) 1 (C) βˆ’2 (D) 2

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{x}^{\mathrm{2}} βˆ’\mathrm{3x}+{a}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} +{a}\mathrm{x}βˆ’\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{then}\:\mathrm{a}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:βˆ’\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{2} \\ $$

Question Number 140973    Answers: 1   Comments: 0

If Ξ± and Ξ² are roots of the equation 2x^2 +ax+b=0, then one of the roots of the equation 2(Ξ±x+Ξ²)^2 + a(Ξ±x+Ξ²)+b=0 is (A) 0 (B) ((Ξ±+2b)/Ξ±^2 ) (C) ((aΞ±+b)/(2Ξ±^2 )) (D) ((aΞ±βˆ’2b)/(2Ξ±^2 ))

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2x}^{\mathrm{2}} +{a}\mathrm{x}+{b}=\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}\left(\alpha\mathrm{x}+\beta\right)^{\mathrm{2}} + \\ $$$${a}\left(\alpha\mathrm{x}+\beta\right)+{b}=\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\alpha+\mathrm{2}{b}}{\alpha^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:\frac{{a}\alpha+{b}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{a}\alphaβˆ’\mathrm{2}{b}}{\mathrm{2}\alpha^{\mathrm{2}} } \\ $$

Question Number 140971    Answers: 0   Comments: 1

Question Number 140969    Answers: 0   Comments: 0

The roots of the equation x^2 βˆ’(mβˆ’3)x+m=0 are such that exactly one of them lies in the interval (1, 2). Then (A) 5<m<7 (B) m<10 (C) 2<m<5 (D) m>10

$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} βˆ’\left({m}βˆ’\mathrm{3}\right)\mathrm{x}+{m}=\mathrm{0}\:\mathrm{are} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right).\:\mathrm{Then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{5}<{m}<\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{m}<\mathrm{10} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{2}<{m}<\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{m}>\mathrm{10} \\ $$

Question Number 140965    Answers: 0   Comments: 0

If a≠0 and a(l+m)^2 +2blm+c=0 and a(l+n)^2 + 2bln+c=0, then (A) mn=l^2 +c/a (B) lm=n^2 +c/a (C) ln=m^2 +c/a (D) mn=l^2 +bc/a

$$\mathrm{If}\:{a}\neq\mathrm{0}\:\mathrm{and}\:{a}\left({l}+{m}\right)^{\mathrm{2}} +\mathrm{2}{blm}+{c}=\mathrm{0}\:\mathrm{and}\:{a}\left({l}+{n}\right)^{\mathrm{2}} + \\ $$$$\mathrm{2}{bln}+{c}=\mathrm{0},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:{mn}={l}^{\mathrm{2}} +{c}/{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{lm}={n}^{\mathrm{2}} +{c}/{a} \\ $$$$\left(\mathrm{C}\right)\:{ln}={m}^{\mathrm{2}} +{c}/{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{mn}={l}^{\mathrm{2}} +{bc}/{a} \\ $$

Question Number 140961    Answers: 2   Comments: 0

∫_0 ^( 1) ((ln (x+(√(1βˆ’x^2 ))))/x) dx =?

$$\:\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\:\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}βˆ’{x}^{\mathrm{2}} }\right)}{{x}}\:{dx}\:=?\: \\ $$

Question Number 140959    Answers: 1   Comments: 0

Let a,b β‰₯ 0. Prove that (1/4)βˆ™(((2+a)(2+b))/((1+a)(1+b))) β‰₯ ((4βˆ’aβˆ’b)/(4+a+b))

$$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}\:\geqslant\:\frac{\mathrm{4}βˆ’{a}βˆ’{b}}{\mathrm{4}+{a}+{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140958    Answers: 0   Comments: 0

find e^ (((βˆ’1 1)),((2 βˆ’1)) )

$$\mathrm{find}\:\mathrm{e}^{\begin{pmatrix}{βˆ’\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:βˆ’\mathrm{1}}\end{pmatrix}} \\ $$

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