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Question Number 144800 Answers: 1 Comments: 0

Let β be an acute angle such that the equation x^2 +4xcos β+cot β=0 involving variable x has multiple roots. Then the measure of β in radians is __

$$\mathrm{Let}\:\beta\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{angle}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{4xcos}\:\beta+\mathrm{cot}\:\beta=\mathrm{0} \\ $$$$\mathrm{involving}\:\mathrm{variable}\:\mathrm{x}\:\mathrm{has}\:\mathrm{multiple} \\ $$$$\mathrm{roots}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{measure}\:\mathrm{of}\:\beta\:\mathrm{in} \\ $$$$\mathrm{radians}\:\mathrm{is}\:\_\_ \\ $$

Question Number 144799 Answers: 0 Comments: 0

Question Number 144794 Answers: 0 Comments: 0

Question Number 144792 Answers: 1 Comments: 0

∫ ((2x^3 −1)/(x^4 +x)) dx ?

$$\:\int\:\frac{\mathrm{2x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Question Number 144791 Answers: 1 Comments: 0

Express sin 5x as polynomial in terms of sin x.

$$\:\mathrm{Express}\:\mathrm{sin}\:\mathrm{5x}\:\mathrm{as}\:\mathrm{polynomial} \\ $$$$\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{x}.\: \\ $$

Question Number 144813 Answers: 0 Comments: 1

I:=∫_0 ^( 1) ((ln (x))/(1 + x^( 2) )) dx := ∫_0 ^( 1) ln(x ) Σ_(n=0) ^∞ (−1)^( n) x^( 2n) dx := Σ_(n=0) ^∞ ( −1 )^( n) ∫_0 ^( 1) x^( 2n) ln( x )dx : = Σ_(n=0) ^∞ ( −1 )^( n) { [(x^( 2n+1) /(2n +1)) ln ( x )]_0 ^( 1) −(1/((2n +1 )^( 2) )) } : = Σ_(n=1) ^( ∞) ((( −1 )^( n−1) )/(( 2n +1)^( 2) )) = −G (Catalan constant )

$$ \\ $$$$\:\:\:\:\:\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\:\left({x}\right)}{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left({x}\:\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\:{n}} \:{x}^{\:\mathrm{2}{n}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\::=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(\:−\mathrm{1}\:\right)^{\:{n}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{n}} \:\mathrm{ln}\left(\:{x}\:\right){dx} \\ $$$$\:\:\:\:\:\:\:\::\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\:−\mathrm{1}\:\right)^{\:{n}} \left\{\:\left[\frac{{x}^{\:\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}\:+\mathrm{1}}\:\mathrm{ln}\:\left(\:{x}\:\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{\left(\mathrm{2}{n}\:+\mathrm{1}\:\right)^{\:\mathrm{2}} \:}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\::\:=\:\underset{{n}=\mathrm{1}} {\overset{\:\infty} {\sum}}\frac{\left(\:−\mathrm{1}\:\right)^{\:{n}−\mathrm{1}} }{\left(\:\mathrm{2}{n}\:+\mathrm{1}\right)^{\:\mathrm{2}} }\:=\:−\mathrm{G}\:\:\left(\mathrm{Catalan}\:\mathrm{constant}\:\right) \\ $$

Question Number 144789 Answers: 0 Comments: 0

Question Number 144788 Answers: 1 Comments: 0

Question Number 144787 Answers: 1 Comments: 0

Q :: # Calculus # If : 𝛗 ( n ) : = ∫_0 ^( 1) (( x^( 2n) )/(1 + x^( 2) )) dx then find the value of :: S := Σ_(n=1) ^∞ ((( −1 )^( n) 𝛗 ( n ))/n) = ? m.n.july.1970

Question Number 144785 Answers: 0 Comments: 0

Question Number 144783 Answers: 0 Comments: 1

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Question Number 144781 Answers: 1 Comments: 0

Question Number 144780 Answers: 0 Comments: 0

Let x∈[−((5π)/(12)),−(π/3)] , then the maximum value of y=tan (x+((2π)/3))−tan (x+(π/6))+cos (x+(π/6)) is

$$\:\mathrm{Let}\:\mathrm{x}\in\left[−\frac{\mathrm{5}\pi}{\mathrm{12}},−\frac{\pi}{\mathrm{3}}\right]\:,\:\mathrm{then}\:\mathrm{the}\: \\ $$$$\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\mathrm{y}=\mathrm{tan}\:\left(\mathrm{x}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\mathrm{tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{6}}\right)+\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{6}}\right) \\ $$$$\:\mathrm{is}\: \\ $$

Question Number 144777 Answers: 2 Comments: 0

Question Number 144772 Answers: 1 Comments: 0

Question Number 144771 Answers: 0 Comments: 0

A region is enclosed by curves x^2 =4y, x^2 =−4y, x=4 & x=−4 V_1 is the volume of the solid obtained by rotating the above region round the y−axis. Another regions consists of points (x,y) satisfying x^2 +y^2 ≤16, x^2 +(y−2)^2 ≥4 and x^2 +(y+2)^2 ≥4 ,V_2 is the volume of the solid obtained by rotating this region round the y−axis Then V_1 =...

$$\mathrm{A}\:\mathrm{region}\:\mathrm{is}\:\mathrm{enclosed}\:\mathrm{by}\:\mathrm{curves} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4y},\:\mathrm{x}^{\mathrm{2}} =−\mathrm{4y},\:\mathrm{x}=\mathrm{4}\:\&\:\mathrm{x}=−\mathrm{4} \\ $$$$\mathrm{V}_{\mathrm{1}} \mathrm{is}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{obtained} \\ $$$$\mathrm{by}\:\mathrm{rotating}\:\mathrm{the}\:\mathrm{above}\:\mathrm{region}\:\mathrm{round} \\ $$$$\mathrm{the}\:\mathrm{y}−\mathrm{axis}.\:\:\mathrm{Another}\:\mathrm{regions} \\ $$$$\mathrm{consists}\:\mathrm{of}\:\mathrm{points}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{satisfying} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{16},\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{4}\:\mathrm{and} \\ $$$$\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{2}\right)^{\mathrm{2}} \geqslant\mathrm{4}\:,\mathrm{V}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{volume} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{obtained}\:\mathrm{by}\:\mathrm{rotating} \\ $$$$\mathrm{this}\:\mathrm{region}\:\mathrm{round}\:\mathrm{the}\:\mathrm{y}−\mathrm{axis}\: \\ $$$$\mathrm{Then}\:\mathrm{V}_{\mathrm{1}} =... \\ $$$$\: \\ $$

Question Number 144768 Answers: 0 Comments: 0

x^3 −x−c=0 let x=(√(t+p))+(√(t+q)) (t+p)(√(t+p))+(t+q)(√(t+q)) +3(t+p)(√(t+q))+3(t+q)(√(t+p)) −(√(t+p))−(√(t+q))−c=0 let (t+q)+3(t+p)−1=0 ⇒ 4t=1−(p+q) ⇒ 4t+4p=3p−q+1 4t+4q=1−(p−3q) (3p−q+1)^(3/2) +3(1−p+3q)(√(3p−q+1)) −4(√(3p−q+1))−8c=0 2x=(√(3p−q+1))+(√(3q−p+1)) let (√(3p−q+1))=−1 ⇒ q=3p ⇒ −1−3(1−p+3q)+4−8c=0 ⇒ p=−(c/3) 2x=−1+(√(1−((8c)/3))) x=−(1/2)+(√((1/4)−((2c)/3))) for c=(1/4) x=−(1/2)+(1/(2(√3)))

$$\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\sqrt{{t}+{p}}+\sqrt{{t}+{q}} \\ $$$$\left({t}+{p}\right)\sqrt{{t}+{p}}+\left({t}+{q}\right)\sqrt{{t}+{q}} \\ $$$$+\mathrm{3}\left({t}+{p}\right)\sqrt{{t}+{q}}+\mathrm{3}\left({t}+{q}\right)\sqrt{{t}+{p}} \\ $$$$−\sqrt{{t}+{p}}−\sqrt{{t}+{q}}−{c}=\mathrm{0} \\ $$$${let}\:\:\left({t}+{q}\right)+\mathrm{3}\left({t}+{p}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{4}{t}=\mathrm{1}−\left({p}+{q}\right) \\ $$$$\Rightarrow\:\:\mathrm{4}{t}+\mathrm{4}{p}=\mathrm{3}{p}−{q}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{4}{t}+\mathrm{4}{q}=\mathrm{1}−\left({p}−\mathrm{3}{q}\right) \\ $$$$\left(\mathrm{3}{p}−{q}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} +\mathrm{3}\left(\mathrm{1}−{p}+\mathrm{3}{q}\right)\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}} \\ $$$$−\mathrm{4}\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}}−\mathrm{8}{c}=\mathrm{0} \\ $$$$\mathrm{2}{x}=\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}}+\sqrt{\mathrm{3}{q}−{p}+\mathrm{1}} \\ $$$${let}\:\:\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}}=−\mathrm{1} \\ $$$$\Rightarrow\:\:{q}=\mathrm{3}{p} \\ $$$$\Rightarrow\:−\mathrm{1}−\mathrm{3}\left(\mathrm{1}−{p}+\mathrm{3}{q}\right)+\mathrm{4}−\mathrm{8}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{p}=−\frac{{c}}{\mathrm{3}} \\ $$$$\mathrm{2}{x}=−\mathrm{1}+\sqrt{\mathrm{1}−\frac{\mathrm{8}{c}}{\mathrm{3}}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}{c}}{\mathrm{3}}} \\ $$$${for}\:\:{c}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Question Number 144767 Answers: 2 Comments: 0