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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

Question Number 144764 Answers: 1 Comments: 0

Question Number 144763 Answers: 0 Comments: 0

Question Number 144760 Answers: 1 Comments: 0

((1+(√x)))^(1/3) + ((1-(√x)))^(1/3) = (5)^(1/3) Find x=?

$$\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{{x}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{1}-\sqrt{{x}}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}} \\ $$$${Find}\:\:\boldsymbol{{x}}=? \\ $$

Question Number 144756 Answers: 1 Comments: 1

Question Number 144754 Answers: 0 Comments: 0

consider a random variable definite by geometric law compute P({X≥4})

$${consider}\:{a}\:{random}\:{variable}\:{definite}\:{by} \\ $$$${geometric}\:{law}\:{compute}\:{P}\left(\left\{{X}\geqslant\mathrm{4}\right\}\right) \\ $$

Question Number 144753 Answers: 1 Comments: 0

Question Number 144744 Answers: 1 Comments: 0

Question Number 144742 Answers: 0 Comments: 0

Let a,b,c>0 and a+b+c = 3. Prove that ((((ab)/(ab+1))+((bc)/(bc+1))+((ca)/(ca+1)))/((1/(ab+1))+(1/(bc+1))+(1/(ca+1)))) ≥ abc (Found by WolframAlpha and inspired by my old problem)

$$\mathrm{Let}\:{a},{b},{c}>\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\frac{{ab}}{{ab}+\mathrm{1}}+\frac{{bc}}{{bc}+\mathrm{1}}+\frac{{ca}}{{ca}+\mathrm{1}}}{\frac{\mathrm{1}}{{ab}+\mathrm{1}}+\frac{\mathrm{1}}{{bc}+\mathrm{1}}+\frac{\mathrm{1}}{{ca}+\mathrm{1}}}\:\geqslant\:{abc} \\ $$$$\left(\mathrm{Found}\:\mathrm{by}\:\mathrm{WolframAlpha}\:\mathrm{and}\:\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{inspired}\:\mathrm{by}\:\mathrm{my}\:\mathrm{old}\:\mathrm{problem}\right) \\ $$

Question Number 144738 Answers: 1 Comments: 0

On souhaite calculer I=∫_0 ^∞ ((sint)/t)dt. (1) On de^ finit la fonction F(x)=∫_0 ^∞ e^(−tx) ((sint)/t)dt. (a) De^ terminer le domaine de de^ finition de f sur R. (b) Montrer que F est de classe C^1 sur R_+ ^∗ et calculer F ′(x). (c) Limite de F en +∞ ? Conse^ quence ? (2) On note Si(t)=∫_0 ^t ((sinu)/u)du pour tout re^ el t. (a) Montrer que G(x)=∫_0 ^∞ e^(−tx) Si(t)dt est de^ finie sur R_+ ^∗ . (b) Montrer que xG(x)→I quand x→0^+ . (c) Au moyen d′une inte^ gration par parties, montrer que F est continue en 0. (3) Calculer I.

$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{On}\:\mathrm{souhaite}\:\mathrm{calculer}\:\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{t}}{{t}}{dt}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{On}\:\mathrm{d}\acute {\mathrm{e}finit}\:\mathrm{la}\:\mathrm{fonction}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} \frac{\mathrm{sin}{t}}{{t}}{dt}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{a}\right)\:\mathrm{D}\acute {\mathrm{e}terminer}\:\mathrm{le}\:\mathrm{domaine}\:\mathrm{de}\:\mathrm{d}\acute {\mathrm{e}finition}\:\mathrm{de}\:{f}\:\mathrm{sur}\:\mathbb{R}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{Montrer}\:\mathrm{que}\:{F}\:\mathrm{est}\:\mathrm{de}\:\mathrm{classe}\:{C}^{\mathrm{1}} \:\mathrm{sur}\:{R}_{+} ^{\ast} \:\mathrm{et}\:\mathrm{calculer}\:{F}\:'\left({x}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{Limite}\:\mathrm{de}\:{F}\:\mathrm{en}\:+\infty\:?\:\mathrm{Cons}\acute {\mathrm{e}quence}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{On}\:\mathrm{note}\:{Si}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{sin}{u}}{{u}}{du}\:\mathrm{pour}\:\mathrm{tout}\:\mathrm{r}\acute {\mathrm{e}el}\:{t}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{a}\right)\:\mathrm{Montrer}\:\mathrm{que}\:{G}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {Si}\left({t}\right){dt}\:\mathrm{est}\:\mathrm{d}\acute {\mathrm{e}finie}\:\mathrm{sur}\:{R}_{+} ^{\ast} . \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{Montrer}\:\mathrm{que}\:{xG}\left({x}\right)\rightarrow{I}\:\mathrm{quand}\:{x}\rightarrow\mathrm{0}^{+} . \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{Au}\:\mathrm{moyen}\:\mathrm{d}'\mathrm{une}\:\mathrm{int}\acute {\mathrm{e}gration}\:\mathrm{par}\:\mathrm{parties},\:\mathrm{montrer}\:\mathrm{que}\:{F}\:\mathrm{est}\:\mathrm{continue}\:\mathrm{en}\:\mathrm{0}. \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Calculer}\:{I}. \\ $$

Question Number 144736 Answers: 3 Comments: 0

P(z)=az^3 +z^2 −(a+6)z+b−6 P(z)=(z^2 +4)∙Q(z) Find a∙b=?

$${P}\left({z}\right)={az}^{\mathrm{3}} +{z}^{\mathrm{2}} −\left({a}+\mathrm{6}\right){z}+{b}−\mathrm{6} \\ $$$${P}\left({z}\right)=\left({z}^{\mathrm{2}} +\mathrm{4}\right)\centerdot{Q}\left({z}\right) \\ $$$${Find}\:\:{a}\centerdot{b}=? \\ $$

Question Number 144734 Answers: 0 Comments: 0