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Question Number 110395    Answers: 0   Comments: 5

prove to ealier problem of ∫_0 ^1 ∫_0 ^1 ((tanh^(−1) ((x)^(1/4) )tanh^(−1) ((y)^(1/4) ))/(x(√y)))dxdy=π^2 solution let I=∫_0 ^1 ((tanh^(−1) ((x)^(1/4) ))/x)dx∫_0 ^1 ((tanh^(−1) ((y)^(1/4) ))/(√y))dy=A.B let m=x^(1/4) and dx=4m^3 A=∫_0 ^1 ((tanh^(−1) (m))/m^4 )×4m^3 dm=4∫_0 ^1 ((tanh^(−1) (m))/m)dm but series of tanh^(−1) (m)=Σ_(k=0) ^∞ (m^(2k+1) /(2k+1)) A=4Σ_(k=0) ^∞ (1/(2k+1))∫_0 ^1 (m^(2k+1) /m)dm=4Σ_(k=o) ^∞ (1/(2k+1))∫_0 ^1 m^(2k) dm A=4Σ_(k=0) ^∞ (1/((2k+1)^2 ))=4×(1/2)Σ_(k=−∞) ^∞ (1/((2k+1)^2 )) recall that Σ_(n=−∞) ^∞ (1/((an+1)^m ))=−(π/((a)^m ))lim_(z→−(1/a) ) (1/((m−1)!))(d^((m−1)) /dz^((m−1)) )[cot(πz)] ∵A=2[−(π/((2)^2 ))lim_(z→−(1/2)) (1/((2−1)!))(d/dz)[cot(πz)] A=−(π/2)lim_(z→−(1/2)) [−πcosec^2 (−(π/2))]=(π^2 /2).....(1) then B=∫_0 ^1 ((tanh^(−1) ((y)^(1/4) ))/(√y))dy let n=y^(1/4) and dy=4n^3 B=∫_0 ^1 ((tanh^(−1) (n))/n^2 )×4n^3 dn=4∫_0 ^1 ntanh^(−1) (n)dn B=4Σ_(k=0) ^∞ (1/(2k+1))∫_0 ^1 n^(2k+2) dn=4Σ_(k=0) ^∞ (1/((2k+1)(2k+3))) B=4×(1/2)Σ_(k=0) ^∞ ((1/(2k+1))−(1/(2k+3))) B=2lim_(k→∞) (1−(1/3)+(1/3)−(1/5)+(1/5).......+(1/(2k+1)))=2.....(2) the series is telescoping but I=A×B=(π^2 /2)×2=π^2 ..............(3) ∫_0 ^1 ∫_0 ^1 ((tanh^(−1) ((x)^(1/4) )tanh^(−1) ((y)^(1/4) ))/(x(√y)))dxdy=π^2 Q.E.D by mathdave(28/08/2020)

$${prove}\:{to}\:{ealier}\:{problem}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{{x}\sqrt{{y}}}{dxdy}=\pi^{\mathrm{2}} \\ $$$${solution}\: \\ $$$${let} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)}{{x}}{dx}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{\sqrt{{y}}}{dy}={A}.{B} \\ $$$${let}\:{m}={x}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{and}\:\:{dx}=\mathrm{4}{m}^{\mathrm{3}} \:\: \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left({m}\right)}{{m}^{\mathrm{4}} }×\mathrm{4}{m}^{\mathrm{3}} {dm}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left({m}\right)}{{m}}{dm} \\ $$$${but}\:{series}\:{of}\:\mathrm{tanh}^{−\mathrm{1}} \left({m}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}^{\mathrm{2}{k}+\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$${A}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{m}^{\mathrm{2}{k}+\mathrm{1}} }{{m}}{dm}=\mathrm{4}\underset{{k}={o}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {m}^{\mathrm{2}{k}} {dm} \\ $$$${A}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${recall}\:{that} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({an}+\mathrm{1}\right)^{{m}} }=−\frac{\pi}{\left({a}\right)^{{m}} }\underset{{z}\rightarrow−\frac{\mathrm{1}}{{a}}\:} {\mathrm{lim}} \\ $$$$\frac{\mathrm{1}}{\left({m}−\mathrm{1}\right)!}\frac{{d}^{\left({m}−\mathrm{1}\right)} }{{dz}^{\left({m}−\mathrm{1}\right)} }\left[\mathrm{cot}\left(\pi{z}\right)\right] \\ $$$$\because{A}=\mathrm{2}\left[−\frac{\pi}{\left(\mathrm{2}\right)^{\mathrm{2}} }\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\frac{{d}}{{dz}}\left[\mathrm{cot}\left(\pi{z}\right)\right]\right. \\ $$$${A}=−\frac{\pi}{\mathrm{2}}\underset{{z}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\left[−\pi\mathrm{cosec}^{\mathrm{2}} \left(−\frac{\pi}{\mathrm{2}}\right)\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}.....\left(\mathrm{1}\right) \\ $$$${then}\:\:{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{\sqrt{{y}}}{dy} \\ $$$${let}\:\:{n}={y}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:\:{and}\:\:\:{dy}=\mathrm{4}{n}^{\mathrm{3}} \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left({n}\right)}{{n}^{\mathrm{2}} }×\mathrm{4}{n}^{\mathrm{3}} {dn}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {n}\mathrm{tanh}^{−\mathrm{1}} \left({n}\right){dn} \\ $$$${B}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {n}^{\mathrm{2}{k}+\mathrm{2}} {dn}=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)} \\ $$$${B}=\mathrm{4}×\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\right) \\ $$$${B}=\mathrm{2}\underset{{k}\rightarrow\infty} {\mathrm{li}{m}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}.......+\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right)=\mathrm{2}.....\left(\mathrm{2}\right) \\ $$$${the}\:{series}\:{is}\:{telescoping} \\ $$$${but}\:{I}={A}×{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}×\mathrm{2}=\pi^{\mathrm{2}} ..............\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{{x}\sqrt{{y}}}{dxdy}=\pi^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{Q}.{E}.{D}\:\:\:\: \\ $$$$\:{by}\:{mathdave}\left(\mathrm{28}/\mathrm{08}/\mathrm{2020}\right) \\ $$

Question Number 110385    Answers: 0   Comments: 3

Question Number 110374    Answers: 2   Comments: 0

If P(x) is a polynomial whose sum of coefficients is 3 and P(x) can be factorised into two polynomials Q(x),R(x) with integer coefficients, the sum of the coefficients Q(x)^2 +R(x)^2 is

$$\mathrm{If}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{whose}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{coefficients}\:\mathrm{is}\:\mathrm{3}\:\mathrm{and}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{factorised}\:\mathrm{into}\:\mathrm{two}\:\mathrm{polynomials} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right),\mathrm{R}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{integer}\:\mathrm{coefficients}, \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{coefficients} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{x}\right)^{\mathrm{2}} \:\mathrm{is} \\ $$

Question Number 110367    Answers: 2   Comments: 0

Question Number 110365    Answers: 3   Comments: 0

Question Number 110359    Answers: 1   Comments: 0

Let f(x)=∣x−2∣+∣x−4∣−∣2x−6∣, for 2≤x≤8. The sum of the largest and smallest values of f(x) is

$$\mathrm{Let} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{x}−\mathrm{2}\mid+\mid\mathrm{x}−\mathrm{4}\mid−\mid\mathrm{2x}−\mathrm{6}\mid, \\ $$$$\mathrm{for}\:\mathrm{2}\leqslant\mathrm{x}\leqslant\mathrm{8}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{largest}\:\mathrm{and} \\ $$$$\mathrm{smallest}\:\mathrm{values}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is} \\ $$

Question Number 110358    Answers: 1   Comments: 0

if positive integer x satisfies x^2 −4x+56 ≡14 (mod 17) , what is the minimum value of x.

$${if}\:{positive}\:{integer}\:{x}\:{satisfies}\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{56}\:\equiv\mathrm{14}\:\left({mod}\:\mathrm{17}\right)\: \\ $$$$,\:{what}\:{is}\:{the}\:{minimum}\:{value}\:{of}\:{x}. \\ $$

Question Number 110357    Answers: 2   Comments: 1

Given that p,q are primes and pq divides p^2 +q^2 −4. How many possible values does ∣p−q∣ have?

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{p},\mathrm{q}\:\mathrm{are}\:\mathrm{primes}\:\mathrm{and}\:\mathrm{pq} \\ $$$$\mathrm{divides}\:\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} −\mathrm{4}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{possible}\:\mathrm{values}\:\mathrm{does}\:\mid\mathrm{p}−\mathrm{q}\mid\:\mathrm{have}? \\ $$

Question Number 110354    Answers: 0   Comments: 1

The Diophantine equation x^2 +y^2 +1 =N(xy+1) has infinitely many integer solutions if N equals?

$$\mathrm{The}\:\mathrm{Diophantine}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{1}\:=\mathrm{N}\left(\mathrm{xy}+\mathrm{1}\right)\:\mathrm{has} \\ $$$$\mathrm{infinitely}\:\mathrm{many}\:\mathrm{integer} \\ $$$$\mathrm{solutions}\:\mathrm{if}\:\mathrm{N}\:\mathrm{equals}? \\ $$

Question Number 110320    Answers: 3   Comments: 0

find the point of intersection of the line r^→ =(1−2t,3+4t,t) and the plane 3x−2y+5z=15

$${find}\:{the}\:{point}\:{of}\:{intersection} \\ $$$${of}\:{the}\:{line}\:\overset{\rightarrow} {{r}}=\left(\mathrm{1}−\mathrm{2}{t},\mathrm{3}+\mathrm{4}{t},{t}\right) \\ $$$${and}\:{the}\:{plane}\:\mathrm{3}{x}−\mathrm{2}{y}+\mathrm{5}{z}=\mathrm{15}\: \\ $$

Question Number 110318    Answers: 1   Comments: 0

(a+b−c)^2 =??

$$\left({a}+{b}−{c}\right)^{\mathrm{2}} =?? \\ $$

Question Number 110307    Answers: 6   Comments: 0

(1)lim_(x→−∞) ((3−3x)/( (√(x^2 −4x+1)))) ? (2) ∫_0 ^1 arctan (((2x−1)/(1+x−x^2 ))) dx (3)how many integer solution sets exist for the equation x^2 +y^2 =2

$$\left(\mathrm{1}\right)\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{arctan}\:\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)\:{dx} \\ $$$$\left(\mathrm{3}\right){how}\:{many}\:{integer}\:{solution}\:{sets} \\ $$$${exist}\:{for}\:{the}\:{equation}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$

Question Number 110306    Answers: 1   Comments: 2

Question Number 110920    Answers: 1   Comments: 1

Question Number 110919    Answers: 1   Comments: 0

lim_(n→∞) (Σ_(r=1) ^n (1/(3^r r!))(Π_(k=1) ^r (2k−1)))

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{r}} {r}!}\left(\underset{{k}=\mathrm{1}} {\overset{{r}} {\prod}}\left(\mathrm{2}{k}−\mathrm{1}\right)\right)\right) \\ $$

Question Number 110301    Answers: 0   Comments: 1

Question Number 110299    Answers: 3   Comments: 0

Question Number 110294    Answers: 1   Comments: 0

∣2x+1∣−∣x−2∣ < 4 find the solution set

$$\mid\mathrm{2}{x}+\mathrm{1}\mid−\mid{x}−\mathrm{2}\mid\:<\:\mathrm{4}\: \\ $$$${find}\:{the}\:{solution}\:{set} \\ $$

Question Number 110293    Answers: 0   Comments: 0

Simplify: ((tan((3π)/7) − 4sin(π/7))/(tan((6π)/7) + 4sin((2π)/7)))

$$\mathrm{Simplify}:\:\:\:\frac{\mathrm{tan}\frac{\mathrm{3}\pi}{\mathrm{7}}\:\:\:−\:\:\mathrm{4sin}\frac{\pi}{\mathrm{7}}}{\mathrm{tan}\frac{\mathrm{6}\pi}{\mathrm{7}}\:\:+\:\:\mathrm{4sin}\frac{\mathrm{2}\pi}{\mathrm{7}}} \\ $$

Question Number 110288    Answers: 1   Comments: 0

Question Number 110287    Answers: 3   Comments: 2

Question Number 110286    Answers: 1   Comments: 0

Question Number 110285    Answers: 1   Comments: 0

Question Number 110281    Answers: 1   Comments: 0

Question Number 110280    Answers: 0   Comments: 0

Question Number 110269    Answers: 1   Comments: 4

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