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Question Number 110715    Answers: 1   Comments: 2

a,b,c,d are unit digits whose pairwise sums form an arithmetic progression. Given that a+b+c+d is even, find the common positive difference of the arithmetic progression.

$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{are}\:\mathrm{unit}\:\mathrm{digits}\:\mathrm{whose} \\ $$$$\mathrm{pairwise}\:\mathrm{sums}\:\mathrm{form}\:\mathrm{an}\:\mathrm{arithmetic} \\ $$$$\mathrm{progression}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\:\mathrm{is} \\ $$$$\mathrm{even},\:\mathrm{find}\:\mathrm{the}\:\mathrm{common}\:\mathrm{positive} \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arithmetic} \\ $$$$\mathrm{progression}. \\ $$

Question Number 110498    Answers: 1   Comments: 3

How many ways can the letters in the word MATHEMATICS be rearranged such that the word formed either starts or ends with a vowel, and any three consecutive letters must contain a vowel?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{word}\:\mathrm{MATHEMATICS}\:\mathrm{be} \\ $$$$\mathrm{rearranged}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{word}\:\mathrm{formed} \\ $$$$\mathrm{either}\:\mathrm{starts}\:\mathrm{or}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{a}\:\mathrm{vowel},\:\mathrm{and} \\ $$$$\mathrm{any}\:\mathrm{three}\:\mathrm{consecutive}\:\mathrm{letters}\:\mathrm{must} \\ $$$$\mathrm{contain}\:\mathrm{a}\:\mathrm{vowel}? \\ $$

Question Number 110490    Answers: 1   Comments: 0

e^(iπ) = −1

$$ \\ $$$${e}^{\mathrm{i}\pi} \:=\:−\mathrm{1} \\ $$

Question Number 110480    Answers: 0   Comments: 0

Question Number 110471    Answers: 1   Comments: 0

the sum of all x values (x^2 −6x+9)^((x−3)/(x−1)) = (x−3)^((x−4))

$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{x}\:\mathrm{values} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{\frac{{x}−\mathrm{3}}{{x}−\mathrm{1}}} \:=\:\left({x}−\mathrm{3}\right)^{\left({x}−\mathrm{4}\right)} \\ $$

Question Number 110463    Answers: 0   Comments: 2

Find the gcd(n−1,n+1)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{gcd}\left({n}−\mathrm{1},{n}+\mathrm{1}\right) \\ $$

Question Number 110467    Answers: 2   Comments: 1

Question Number 110460    Answers: 0   Comments: 0

Find all continuous real function f such that f(ln(x+y))=f(sin(xy))+f(cos(y/x)) for all x,y∈R^+ .

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{continuous}\:\mathrm{real}\:\mathrm{function}\:{f}\: \\ $$$$\mathrm{such}\:\mathrm{that}\: \\ $$$${f}\left(\mathrm{ln}\left({x}+{y}\right)\right)={f}\left(\mathrm{sin}\left({xy}\right)\right)+{f}\left(\mathrm{cos}\left({y}/{x}\right)\right) \\ $$$$\mathrm{for}\:\mathrm{all}\:{x},{y}\in\mathbb{R}^{+} . \\ $$

Question Number 110456    Answers: 1   Comments: 0

If (f(x).g(x))′ = f(x)′ . g(x)′ find the function of f(x) .

$$\:\:\:{If}\:\left({f}\left({x}\right).{g}\left({x}\right)\right)'\:=\:{f}\left({x}\right)'\:.\:{g}\left({x}\right)'\: \\ $$$$\:{find}\:{the}\:{function}\:{of}\:{f}\left({x}\right)\:. \\ $$

Question Number 110451    Answers: 1   Comments: 0

calculate U_n =∫_([(1/n),n[^2 ) (x^2 −y^2 )e^(−x^2 −y^2 ) dxdy and lim_(n→+∞) U_n

$$\mathrm{calculate}\:\mathrm{U}_{\mathrm{n}} =\int_{\left[\frac{\mathrm{1}}{\mathrm{n}},\mathrm{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } \mathrm{dxdy} \\ $$$$\mathrm{and}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{U}_{\mathrm{n}} \\ $$

Question Number 110450    Answers: 1   Comments: 0

find ∫∫_([0,1]^2 ) ln(x^2 +3y^2 ) e^(−x^2 −3y^2 ) dxdy

$$\mathrm{find}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \right)\:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{3y}^{\mathrm{2}} } \:\mathrm{dxdy} \\ $$

Question Number 110449    Answers: 2   Comments: 0

find lim_(x→0) ((arctan(x−sinx)−arctan(1−cosx))/x^2 )

$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{arctan}\left(\mathrm{x}−\mathrm{sinx}\right)−\mathrm{arctan}\left(\mathrm{1}−\mathrm{cosx}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 110448    Answers: 1   Comments: 0

calculate ∫_(−∞) ^(+∞) ((cos(2x))/((x^2 −4i)^3 ))dx (i=(√(−1)))

$$\mathrm{calculate}\:\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4i}\right)^{\mathrm{3}} }\mathrm{dx}\:\:\:\:\:\left(\mathrm{i}=\sqrt{−\mathrm{1}}\right) \\ $$

Question Number 110447    Answers: 1   Comments: 0

calculate ∫_(−∞) ^(+∞) (dx/((x^2 −ix +1)^2 )) (i=(√(−1)))

$$\mathrm{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{ix}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\left(\mathrm{i}=\sqrt{−\mathrm{1}}\right) \\ $$

Question Number 110440    Answers: 1   Comments: 0

Given lim_(x→0) (f(x)+(1/(f(x)))) = 2 , find the value of lim_(x→0) f(x).

$$\mathrm{Given}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{f}\left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{f}\left(\mathrm{x}\right)}\right)\:=\:\mathrm{2}\:,\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right). \\ $$

Question Number 110436    Answers: 2   Comments: 0

If 2f(x) + f((1/x)) = 3x what is f(x)

$$\:\:\:\:\:\mathrm{If}\:\mathrm{2f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\mathrm{3x} \\ $$$$\:\:\:\:\:\mathrm{what}\:\mathrm{is}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$

Question Number 110434    Answers: 1   Comments: 1

The product of the four terms of an increasing arithmetic progression is a, their sum is b, and the sum of their reciprocal is c. Suppose that a,b,c form a geometric progression whose product is 8000, find the sum of the first and fourth term.

$$\mathrm{The}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{increasing}\:\mathrm{arithmetic}\:\mathrm{progression}\:\mathrm{is}\:\mathrm{a}, \\ $$$$\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{b},\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their} \\ $$$$\mathrm{reciprocal}\:\mathrm{is}\:\mathrm{c}.\:\mathrm{Suppose}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{form} \\ $$$$\mathrm{a}\:\mathrm{geometric}\:\mathrm{progression}\:\mathrm{whose} \\ $$$$\mathrm{product}\:\mathrm{is}\:\mathrm{8000},\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{and}\:\mathrm{fourth}\:\mathrm{term}. \\ $$

Question Number 110425    Answers: 2   Comments: 3

Which of the following is not a factor of x^6 −56x+55 A. x−1 B. x^2 −x+5 C. x^3 +2x^2 −2x−11 D. x^4 +x^3 +4x^2 −9x+11 E. x^5 +x^4 +x^3 +x^2 +x−55 Please show all workings clearly. Thanks.

$$ \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{factor} \\ $$$$\mathrm{of}\:\mathrm{x}^{\mathrm{6}} −\mathrm{56x}+\mathrm{55} \\ $$$$\mathrm{A}.\:\mathrm{x}−\mathrm{1}\:\mathrm{B}.\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{5}\:\mathrm{C}. \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{11}\:\mathrm{D}. \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{4x}^{\mathrm{2}} −\mathrm{9x}+\mathrm{11}\:\mathrm{E}. \\ $$$$\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{55} \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{show}\:\mathrm{all}\:\mathrm{workings}\:\mathrm{clearly}. \\ $$$$\mathrm{Thanks}. \\ $$

Question Number 110420    Answers: 1   Comments: 0

Prove that x^5 −3x^4 −17x^3 −x^2 −3x+17 cannot be factorized completely over the set of polynomials with integral coefficients.

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} −\mathrm{3x}^{\mathrm{4}} −\mathrm{17x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{17}\:\mathrm{cannot}\:\mathrm{be} \\ $$$$\mathrm{factorized}\:\mathrm{completely}\:\mathrm{over}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of} \\ $$$$\mathrm{polynomials}\:\mathrm{with}\:\mathrm{integral}\:\mathrm{coefficients}. \\ $$

Question Number 110419    Answers: 1   Comments: 2

Let t be a root of x^3 −3x+1=0, if ((t^2 +pt+1)/(t^2 −t+1)) can be written as t+c for some p,c ∈ Z, then p−c equals?

$$\mathrm{Let}\:\mathrm{t}\:\mathrm{be}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}+\mathrm{1}=\mathrm{0},\:\mathrm{if}\: \\ $$$$\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{pt}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{t}+\mathrm{c}\:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{p},\mathrm{c}\:\in\:\mathbb{Z},\:\mathrm{then}\:\mathrm{p}−\mathrm{c}\:\mathrm{equals}? \\ $$

Question Number 110410    Answers: 0   Comments: 0

Question Number 110403    Answers: 0   Comments: 1

if W represents the Runesky determinant of the two independent solutions linearly (y_1 ,y_2 )of the equation y^(′′) +p(x)y^′ +Q(x)=0 then demonstrate that W satisfies the differential equation (W^( ′) +p(x)W=0) and solve this equation to qet W ? help me sir please

$${if}\:{W}\:{represents}\:{the}\:{Runesky}\:{determinant}\:{of}\:{the}\:{two} \\ $$$${independent}\:{solutions}\:{linearly}\:\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right){of}\:{the}\:{equation}\:{y}^{''} +{p}\left({x}\right){y}^{'} +{Q}\left({x}\right)=\mathrm{0}\:{then}\:{demonstrate}\:{that}\:{W}\:{satisfies}\:{the}\:{differential}\:{equation}\:\left({W}^{\:'} +{p}\left({x}\right){W}=\mathrm{0}\right)\:{and}\:{solve}\:{this}\:{equation}\:{to}\:{qet}\:{W}\:? \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$

Question Number 110399    Answers: 0   Comments: 5

The identity 2[16a^4 +81b^4 +c^4 ]=[4a^2 +9b^2 +c^2 ]^2 cannot result from which of the following equations? A. 6b=4a+2c B. 6a=9b+3c C. 6b=−4a+2c D.c= −2a−3b E. 6c=2b+3a

$$\mathrm{The}\:\mathrm{identity} \\ $$$$\mathrm{2}\left[\mathrm{16a}^{\mathrm{4}} +\mathrm{81b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}} \right]=\left[\mathrm{4a}^{\mathrm{2}} +\mathrm{9b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right]^{\mathrm{2}} \\ $$$$\mathrm{cannot}\:\mathrm{result}\:\mathrm{from}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{equations}?\: \\ $$$$\mathrm{A}.\:\mathrm{6b}=\mathrm{4a}+\mathrm{2c}\:\mathrm{B}.\:\mathrm{6a}=\mathrm{9b}+\mathrm{3c}\:\mathrm{C}. \\ $$$$\mathrm{6b}=−\mathrm{4a}+\mathrm{2c}\:\mathrm{D}.\mathrm{c}=\:−\mathrm{2a}−\mathrm{3b}\:\mathrm{E}. \\ $$$$\mathrm{6c}=\mathrm{2b}+\mathrm{3a} \\ $$

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

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

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