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Question Number 137848 Answers: 1 Comments: 0

Solve the equation (x^2 −x+1)^2 +(y^2 +y+1)^2 =2021 x,y∈Z

$$\boldsymbol{{Solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}} \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2021} \\ $$$$\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{Z}} \\ $$

Question Number 137873 Answers: 0 Comments: 3

.......nice ............calculus....... 𝛗=Σ_(n=1) ^∞ ((sin(nx))/n) =(π/2)−(x/2) 𝛗=−Im(1−e^(ix) )=−Imln{(1−cos(x)−isin(x))} =−Im{ln((√((1−cos(x))^2 +sin^2 (x))) +itan^(−1) (((−sin(x))/(1−cos(x))))} =−Im{(√(2−2cos(x)_ )) −itan^(−1) (tan((π/2)−(x/2)))} =tan^(−1) (tan((π/2)−(x/2))) ........ 𝛗=(π/2)−(x/2) ...✓.....⟨∗⟩ ^(′′) ∫ ^(′′) both sides of ⟨∗⟩ −Σ_(n=1) ^∞ ((cos(nx))/n^2 )dx+C=((πx)/2)−(x^2 /4) x=0 ⇒ C=(π^2 /6) ...✓ Σ_(n=1) ^∞ ((cos(nx))/n^2 )=−((πx)/2)+(x^2 /4)+(π^2 /6) ....✓ x=π(1+(√(1/3)) ) .... Σ((cos(nπ(1+(√(1/3)) )))/n^2 )=(1/2)(−π^2 (1+(√(1/3)) )+((π^2 (1+(√(1/3)) )^2 )/2)+(π^2 /3)) =(π^2 /2)(−1−((√3)/3) +((1+2.((√3)/3) +(1/3))/2)+(1/3)) =(π^2 /2)(−1−(((√3) )/3)+(4/6)+(((√3) )/3)+(1/3))=0..✓✓ ......prepared by mr (Dwaipayan)..... m.n#

$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:.......{nice}\:\:............{calculus}....... \\ $$$$\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\phi}=−{Im}\left(\mathrm{1}−{e}^{{ix}} \right)=−{Imln}\left\{\left(\mathrm{1}−{cos}\left({x}\right)−{isin}\left({x}\right)\right)\right\} \\ $$$$\:\:\:\:=−{Im}\left\{{ln}\left(\sqrt{\left(\mathrm{1}−{cos}\left({x}\right)\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \left({x}\right)}\:+{itan}^{−\mathrm{1}} \left(\frac{−{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\right)\right\}\right. \\ $$$$=−{Im}\left\{\sqrt{\mathrm{2}−\mathrm{2}{cos}\left({x}\right)_{\:} }\:−{itan}^{−\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\right)\right\} \\ $$$$\:\:={tan}^{−\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\right)\:\: \\ $$$$\:\:\:\:\:\:\:........\:\:\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\:\:...\checkmark.....\langle\ast\rangle \\ $$$$\:\:\:\:\:^{''} \:\:\int\:\:^{''} \:\:{both}\:{sides}\:{of}\:\langle\ast\rangle \\ $$$$\:\:\:\:\:\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }{dx}+{C}=\frac{\pi{x}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\:\Rightarrow\:{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:\:...\checkmark \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=−\frac{\pi{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:....\checkmark \\ $$$$\:\:\:\:\:\:\:\:{x}=\pi\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)\:.... \\ $$$$\:\:\:\:\:\:\:\:\Sigma\frac{{cos}\left({n}\pi\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)\right)}{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(−\pi^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)+\frac{\pi^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\left(−\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}+\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\left(−\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{6}}+\frac{\sqrt{\mathrm{3}}\:}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0}..\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:......{prepared}\:{by}\:{mr}\:\left({Dwaipayan}\right)..... \\ $$$$\:\:{m}.{n}# \\ $$$$\:\:\:\: \\ $$

Question Number 137841 Answers: 0 Comments: 1

Question Number 137838 Answers: 1 Comments: 1

∫log_x edx=??

$$\int{log}_{{x}} {edx}=?? \\ $$

Question Number 137837 Answers: 4 Comments: 0

∫_0 ^∞ ((sin x−sin x^2 )/x)=(π/4) ∫_0 ^∞ ((cos x−cos x^2 )/x)dx=−(γ/2)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{x}−\mathrm{sin}\:{x}^{\mathrm{2}} }{{x}}=\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:{x}−\mathrm{cos}\:{x}^{\mathrm{2}} }{{x}}{dx}=−\frac{\gamma}{\mathrm{2}} \\ $$

Question Number 137836 Answers: 0 Comments: 0

Question Number 137829 Answers: 1 Comments: 0

.......nice ... ... .... calculus..... prove that :::: 𝛗=∫_0 ^( 1) (((log(1−x))/x))^2 dx=2ζ(2)....

$$\:\:\:\:\:.......{nice}\:\:...\:...\:....\:{calculus}..... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:{that}\::::: \\ $$$$\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{log}\left(\mathrm{1}−{x}\right)}{{x}}\right)^{\mathrm{2}} {dx}=\mathrm{2}\zeta\left(\mathrm{2}\right).... \\ $$$$ \\ $$

Question Number 137820 Answers: 2 Comments: 0

lim_(x→π/2) (((1−tan (x/2))(1−sin x))/((1+tan (x/2))(π−2x)^3 )) ?

$$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }\:? \\ $$

Question Number 137866 Answers: 0 Comments: 0

Question Number 137817 Answers: 1 Comments: 0

Let f(x)=x^2 −2x−3; x≥1 & g(x)=1+(√(x+4)) ; x≥−4 then the number of real solutions of equation f(x)=g(x) is ...

$${Let}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3};\:{x}\geqslant\mathrm{1}\:\& \\ $$$${g}\left({x}\right)=\mathrm{1}+\sqrt{{x}+\mathrm{4}}\:;\:{x}\geqslant−\mathrm{4}\:{then} \\ $$$${the}\:{number}\:{of}\:{real}\:{solutions} \\ $$$${of}\:{equation}\:{f}\left({x}\right)={g}\left({x}\right)\:{is}\:... \\ $$

Question Number 137813 Answers: 1 Comments: 0

cos (arctan (((21)/(60))))=?

$$\mathrm{cos}\:\left(\mathrm{arctan}\:\left(\frac{\mathrm{21}}{\mathrm{60}}\right)\right)=? \\ $$

Question Number 137811 Answers: 2 Comments: 0

Given { ((cos (x−y)=−1+(1/2)cos x)),((cos (x+y)= 1+(1/3)cos x)) :} where 270° < y< 360° . find sin 2y .

$${Given}\:\begin{cases}{\mathrm{cos}\:\left({x}−{y}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{x}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:{x}}\end{cases} \\ $$$${where}\:\mathrm{270}°\:<\:{y}<\:\mathrm{360}°\:.\:{find} \\ $$$$\mathrm{sin}\:\mathrm{2}{y}\:. \\ $$

Question Number 137799 Answers: 2 Comments: 0

..... mathematical .. ... ... analysis.... evaluate :: 𝛗=∫_0 ^( 1) (((ln^2 (1−x^2 ))/x))=?

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.....\:{mathematical}\:..\:...\:...\:{analysis}.... \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\right)=? \\ $$$$ \\ $$

Question Number 137787 Answers: 1 Comments: 0

A string AB of lenght 2l has a particle attached to its midpoint C. The ends A and B of the string are fastened to two fixed points with A distance l vertically above B.With both parts of the string taunt,the particle describes a horizontal circle about the line AB with constant angular speed ω. If the tension in CA is three times that in CB, prove that the angular velocity is 2(√(g/l)) .

$$\mathrm{A}\:\mathrm{string}\:{AB}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{2}{l}\:\mathrm{has}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{its}\:\mathrm{midpoint}\:\mathrm{C}.\: \\ $$$$\mathrm{The}\:\mathrm{ends}\:{A}\:\mathrm{and}\:{B}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:\mathrm{are}\:\mathrm{fastened}\:\mathrm{to}\:\mathrm{two}\:\mathrm{fixed}\:\mathrm{points} \\ $$$$\mathrm{with}\:{A}\:\mathrm{distance}\:{l}\:\mathrm{vertically}\:\mathrm{above}\:\mathrm{B}.\mathrm{With}\:\mathrm{both}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string} \\ $$$$\mathrm{taunt},\mathrm{the}\:\mathrm{particle}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:{AB}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{angular}\:\mathrm{speed}\:\omega.\:\mathrm{If}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:{CA}\:\mathrm{is}\:\mathrm{three}\:\mathrm{times} \\ $$$$\mathrm{that}\:\mathrm{in}\:{CB},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angular}\:\mathrm{velocity}\:\mathrm{is}\:\mathrm{2}\sqrt{\frac{\mathrm{g}}{{l}}}\:. \\ $$

Question Number 137784 Answers: 2 Comments: 0

∫_0 ^( (π/6)) (dx/(sinx))=?

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{6}}} \frac{{dx}}{{sinx}}=? \\ $$$$ \\ $$

Question Number 137783 Answers: 1 Comments: 0

Some birds were flying and met a bird, the bird greeted them and said how are you hundred. The birds said we are not hundred, we need half of us plus you to make hundred. How many birds were flying

$$\mathrm{Some}\:\mathrm{birds}\:\mathrm{were}\:\mathrm{flying}\:\mathrm{and}\:\mathrm{met}\:\mathrm{a}\: \\ $$$$\mathrm{bird},\:\mathrm{the}\:\mathrm{bird}\:\mathrm{greeted}\:\mathrm{them}\:\mathrm{and}\:\mathrm{said} \\ $$$$\mathrm{how}\:\mathrm{are}\:\mathrm{you}\:\mathrm{hundred}.\:\mathrm{The}\:\mathrm{birds}\:\mathrm{said} \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{not}\:\mathrm{hundred},\:\mathrm{we}\:\mathrm{need}\:\mathrm{half}\:\mathrm{of}\:\mathrm{us} \\ $$$$\mathrm{plus}\:\mathrm{you}\:\mathrm{to}\:\mathrm{make}\:\mathrm{hundred}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{birds}\:\mathrm{were}\:\mathrm{flying} \\ $$

Question Number 137902 Answers: 2 Comments: 0

lim_(x→0) ((x−sin x)/(x^2 (√x))) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:=? \\ $$

Question Number 137773 Answers: 2 Comments: 0

lim_(x→0) ((cos px−cos qx)/x^2 )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{px}−\mathrm{cos}\:{qx}}{{x}^{\mathrm{2}} } \\ $$

Question Number 137772 Answers: 2 Comments: 0