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Question Number 143063    Answers: 0   Comments: 0

cos((𝛑/(2n+1)))cos(((2𝛑)/(2n+1)))cos(((3𝛑)/(2n+1))).....cos(((n𝛑)/(2n+1)))=(1/2^n ) prove

$$\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right).....\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{n}\pi}}{\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$

Question Number 143057    Answers: 1   Comments: 0

cos(𝛂)Γ—cos(2Ξ±)Γ—cos(4Ξ±)Γ—....Γ—cos(2^n 𝛂)=((sin(2^(n+1) 𝛂))/(2^(n+1) sin(Ξ±))) prove

$$\mathrm{cos}\left(\boldsymbol{\alpha}\right)Γ—\mathrm{cos}\left(\mathrm{2}\alpha\right)Γ—\mathrm{cos}\left(\mathrm{4}\alpha\right)Γ—....Γ—\mathrm{cos}\left(\mathrm{2}^{\mathrm{n}} \boldsymbol{\alpha}\right)=\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \boldsymbol{\alpha}\right)}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \mathrm{sin}\left(\alpha\right)} \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$

Question Number 143048    Answers: 1   Comments: 0

Question Number 143047    Answers: 2   Comments: 0

Question Number 143046    Answers: 1   Comments: 0

Question Number 143045    Answers: 1   Comments: 0

Question Number 143043    Answers: 1   Comments: 0

Question Number 143042    Answers: 1   Comments: 0

Question Number 143041    Answers: 0   Comments: 2

If f(x)=x^2 +4x+2 then the value of (1βˆ’(2/(f(1))))(1βˆ’(2/(f(2))))(1βˆ’(2/(f(3))))...(1βˆ’(2/(f(2021))))=?

$${If}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\:{then}\:{the}\:{value} \\ $$$${of}\:\left(\mathrm{1}βˆ’\frac{\mathrm{2}}{{f}\left(\mathrm{1}\right)}\right)\left(\mathrm{1}βˆ’\frac{\mathrm{2}}{{f}\left(\mathrm{2}\right)}\right)\left(\mathrm{1}βˆ’\frac{\mathrm{2}}{{f}\left(\mathrm{3}\right)}\right)...\left(\mathrm{1}βˆ’\frac{\mathrm{2}}{{f}\left(\mathrm{2021}\right)}\right)=? \\ $$

Question Number 143039    Answers: 0   Comments: 0

cos^(βˆ’1) (((x^2 βˆ’1)/(x^2 +1)))+(1/2)tan^(βˆ’1) (((2x)/(1βˆ’x^2 )))=((2Ο€)/3) x=?

$$\:\mathrm{cos}^{βˆ’\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} βˆ’\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{βˆ’\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${x}=? \\ $$

Question Number 143051    Answers: 1   Comments: 0

_(βˆ—βˆ—βˆ—βˆ—βˆ—) :: Lobachevsky Integral ::_(βˆ—βˆ—βˆ—βˆ—βˆ—) 𝛗:=∫_0 ^( ∞) ((sin^2 ( tan(x)))/x^( 2) )dx=^? (Ο€/2) ..........

$$\:\:\:\:\:\:\:_{\ast\ast\ast\ast\ast} ::\:\:{Lobachevsky}\:{Integral}\:::_{\ast\ast\ast\ast\ast} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{s}{in}^{\mathrm{2}} \left(\:{tan}\left({x}\right)\right)}{{x}^{\:\mathrm{2}} }{dx}\overset{?} {=}\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:.......... \\ $$

Question Number 143036    Answers: 0   Comments: 0

x^(3/2) +x^(1/2) +(xβˆ’c)(((3x+1)/(3+x)))^(3/2) =0

$$\:{x}^{\mathrm{3}/\mathrm{2}} +{x}^{\mathrm{1}/\mathrm{2}} +\left({x}βˆ’{c}\right)\left(\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{3}+{x}}\right)^{\mathrm{3}/\mathrm{2}} =\mathrm{0} \\ $$

Question Number 143032    Answers: 1   Comments: 0

∫ sin^(βˆ’5) x dx =?

$$\:\:\:\:\:\int\:\mathrm{sin}^{βˆ’\mathrm{5}} {x}\:{dx}\:=? \\ $$

Question Number 143031    Answers: 2   Comments: 0

Question Number 143027    Answers: 1   Comments: 1

Question Number 143024    Answers: 1   Comments: 0

Question Number 143013    Answers: 0   Comments: 3

lim_(xβ†’βˆž) (((x+1))^(1/3) /(x+1))=?

$${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$

Question Number 143012    Answers: 1   Comments: 1

lim_(Ξ±β†’βˆž) (e^x /Ξ±^(60!) )=?

$${li}\underset{\alpha\rightarrow\infty} {{m}}\frac{{e}^{{x}} }{\alpha^{\mathrm{60}!} }=? \\ $$

Question Number 143018    Answers: 1   Comments: 0

Question Number 143011    Answers: 0   Comments: 0

x^4 +bx^2 +cx=s let x^2 =px+t β‡’ p^2 x^2 +2ptx+t^2 +bx^2 +cx=s β‡’ (p^2 +b)x^2 +(2pt+c)x =sβˆ’t^2 β‡’ (p^2 +b)(px+t)+(2pt+c)x =sβˆ’t^2 β‡’ p(p^2 +b)+2pt+c=0 and (p^2 +b)t=sβˆ’t^2 ((s/t)βˆ’tβˆ’b)((s/t)+t)^2 =c^2 β‡’ (Aβˆ’b)(A^2 +4s)=c^2 β‡’ A^3 βˆ’bA^2 +4sAβˆ’4bsβˆ’c^2 =0 let A=z+(b/3) β‡’ z^3 +(4sβˆ’(b^2 /3))zβˆ’(((2b^3 )/(27))+((8bs)/3)+c^2 )=0 D=((b^3 /(27))+((4bs)/3)+(c^2 /2))^2 βˆ’((b^2 /9)βˆ’((4s)/3))^3 If s=0, b=βˆ’1, cβ†’βˆ’c then D=(βˆ’(1/(27))+(c^2 /2))^2 βˆ’((1/9))^3 ...

$$\:\:{x}^{\mathrm{4}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$${let}\:\:{x}^{\mathrm{2}} ={px}+{t} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{ptx}+{t}^{\mathrm{2}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{pt}+{c}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:\:={s}βˆ’{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right)\left({px}+{t}\right)+\left(\mathrm{2}{pt}+{c}\right){x} \\ $$$$\:\:\:\:\:\:\:={s}βˆ’{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{p}\left({p}^{\mathrm{2}} +{b}\right)+\mathrm{2}{pt}+{c}=\mathrm{0} \\ $$$${and}\:\:\left({p}^{\mathrm{2}} +{b}\right){t}={s}βˆ’{t}^{\mathrm{2}} \\ $$$$\left(\frac{{s}}{{t}}βˆ’{t}βˆ’{b}\right)\left(\frac{{s}}{{t}}+{t}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({A}βˆ’{b}\right)\left({A}^{\mathrm{2}} +\mathrm{4}{s}\right)={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:{A}^{\mathrm{3}} βˆ’{bA}^{\mathrm{2}} +\mathrm{4}{sA}βˆ’\mathrm{4}{bs}βˆ’{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:{A}={z}+\frac{{b}}{\mathrm{3}}\:\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}{s}βˆ’\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\right){z}βˆ’\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}{bs}}{\mathrm{3}}+{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${D}=\left(\frac{{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{4}{bs}}{\mathrm{3}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} βˆ’\left(\frac{{b}^{\mathrm{2}} }{\mathrm{9}}βˆ’\frac{\mathrm{4}{s}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$${If}\:\:{s}=\mathrm{0},\:{b}=βˆ’\mathrm{1},\:{c}\rightarrowβˆ’{c}\:\: \\ $$$${then}\:{D}=\left(βˆ’\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} βˆ’\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{3}} \\ $$$$... \\ $$

Question Number 143008    Answers: 1   Comments: 1

Solve : x=p^3 βˆ’p+2 , y^β€² =p

$${Solve}\::\:\:{x}={p}^{\mathrm{3}} βˆ’{p}+\mathrm{2}\:\:,\:{y}^{'} ={p} \\ $$

Question Number 143006    Answers: 0   Comments: 0

1. y(βˆ‚z/βˆ‚x) + z(βˆ‚z/βˆ‚y) = (y/x) 2. x^2 (βˆ‚z/βˆ‚x) βˆ’ xy(βˆ‚z/βˆ‚y) + y^2 = 0 3. { (((βˆ‚z/βˆ‚x) = (z/x))),(((βˆ‚z/βˆ‚y) = ((2z)/y))) :}

$$\mathrm{1}.\:{y}\frac{\partial{z}}{\partial{x}}\:+\:{z}\frac{\partial{z}}{\partial{y}}\:=\:\frac{{y}}{{x}} \\ $$$$\mathrm{2}.\:{x}^{\mathrm{2}} \frac{\partial{z}}{\partial{x}}\:βˆ’\:{xy}\frac{\partial{z}}{\partial{y}}\:+\:{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{3}.\:\begin{cases}{\frac{\partial{z}}{\partial{x}}\:=\:\frac{{z}}{{x}}}\\{\frac{\partial{z}}{\partial{y}}\:=\:\frac{\mathrm{2}{z}}{{y}}}\end{cases} \\ $$

Question Number 143003    Answers: 1   Comments: 0

Question Number 142992    Answers: 0   Comments: 3

Question Number 142990    Answers: 2   Comments: 0

find ∫_0 ^∞ (e^(βˆ’x^2 ) /((3+x^2 )^2 ))dx

$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{3}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 142989    Answers: 2   Comments: 0

calculate ∫_0 ^∞ (e^(βˆ’3x^2 ) /(1+x^2 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{βˆ’\mathrm{3x}^{\mathrm{2}} } }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$

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