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Question Number 149273 Answers: 3 Comments: 0

∫_0 ^(π/4) ((e^(tan x) sin^2 x)/(cos^4 x)) dx =?

$$\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{e}^{\mathrm{tan}\:\mathrm{x}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$

Question Number 149268 Answers: 4 Comments: 0

Question Number 149266 Answers: 1 Comments: 0

Solve for natural numbers: x^2 + y^2 +x + y = 3xy

$${Solve}\:{for}\:{natural}\:{numbers}: \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+{x}\:+\:{y}\:=\:\mathrm{3}{xy} \\ $$

Question Number 149271 Answers: 2 Comments: 0

Question Number 149259 Answers: 1 Comments: 0

cos^(−1) ((√(sin (41°)sin (19°)+(3/4))) )=?

$$\:\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{sin}\:\left(\mathrm{41}°\right)\mathrm{sin}\:\left(\mathrm{19}°\right)+\frac{\mathrm{3}}{\mathrm{4}}}\:\right)=? \\ $$

Question Number 149252 Answers: 4 Comments: 0

e^y =(sin x)(cos x) find (dy/dx)

$$ \\ $$$${e}^{{y}} =\left(\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}\right) \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$

Question Number 149251 Answers: 2 Comments: 0

lim_(x→0^+ ) ((tan x (√(tan x))−sin x (√(sin x)))/(x^3 (√x))) =?

$$\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\:\sqrt{\mathrm{tan}\:\mathrm{x}}−\mathrm{sin}\:\mathrm{x}\:\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{3}} \:\sqrt{\mathrm{x}}}\:=? \\ $$$$ \\ $$

Question Number 149247 Answers: 0 Comments: 3

Question Number 149244 Answers: 0 Comments: 0

Let a,b,c be positive real numbers with sum 3. Prove that (1/a)+(1/b)+(1/c) ≥ (3/(2a^2 +bc))+(3/(2b^2 +ac))+(3/(2c^2 +ab))

$$\:\:{Let}\:{a},{b},{c}\:{be}\:{positive}\:{real}\:{numbers} \\ $$$${with}\:{sum}\:\mathrm{3}.\:{Prove}\:{that}\: \\ $$$$\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}{a}^{\mathrm{2}} +{bc}}+\frac{\mathrm{3}}{\mathrm{2}{b}^{\mathrm{2}} +{ac}}+\frac{\mathrm{3}}{\mathrm{2}{c}^{\mathrm{2}} +{ab}} \\ $$

Question Number 149241 Answers: 2 Comments: 0

Question Number 149240 Answers: 1 Comments: 0

solve for real number x^4 −3x^2 +1 =(√((4/(4−x^2 )) −1))

$$\:\:\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{number}\: \\ $$$$\:\mathrm{x}^{\mathrm{4}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\:=\sqrt{\frac{\mathrm{4}}{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:−\mathrm{1}} \\ $$$$\:\: \\ $$

Question Number 149239 Answers: 1 Comments: 0

lim_(x→a^+ ) (((√x)−(√a) −(√(x−a)))/( (√(x^2 −a^2 )))) =?

$$\:\:\:\underset{{x}\rightarrow\mathrm{a}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:=?\: \\ $$

Question Number 149238 Answers: 0 Comments: 1

z^3 −z=c z^4 −z^2 =cz let z=x+q x^4 +4qx^3 +6q^2 x^2 +4q^3 x+q^4 −x^2 −2qx−q^2 −cx−cq=0 ⇒ x^4 +4qx^3 +(6q^2 −1)x^2 + (4q^3 −2q−c)x+q^4 −q^2 −cq=0 let q^2 =1/6 ⇒ x^4 +4qx^3 −(((4q)/3)+c)x−(5/(36))−cq=0 ≡ (x^2 +mx+k)(x^2 +sx+h)=0 ⇒ m+s=4q h+k+ms=0 mh+ks=−(((4q)/3)+c) kh=−(5/(36))−cq h, k= −((ms)/2)±(√(((m^2 s^2 )/4)+(5/(36))+cq)) m(−((ms)/2)+D)+s(−((ms)/2)−D) +((4q)/3)+c=0 (m−s)D−((ms)/2)(4q)+((4q)/3)+c=0 now let′s assume ms=p & since m+s=4q (m−s)^2 =(8/3)−4p ⇒ ((8/3)−4p)((p^2 /4)+(5/(36))+cq) =(2pq−((4q)/3)−c)^2 p^3 +((5/9)+4cq−(8/9)−4cq)p +(((4q)/3)+c)^2 −(8/3)((5/(36))+cq)=0 ⇒ p^3 −(p/3)+((14)/(27))=0 D_0 =((7/(27)))^2 −((1/9))^3 ....

$${z}^{\mathrm{3}} −{z}={c} \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} ={cz} \\ $$$${let}\:\:{z}={x}+{q} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} +\mathrm{6}{q}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{3}} {x}+{q}^{\mathrm{4}} \\ $$$$−{x}^{\mathrm{2}} −\mathrm{2}{qx}−{q}^{\mathrm{2}} −{cx}−{cq}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} +\left(\mathrm{6}{q}^{\mathrm{2}} −\mathrm{1}\right){x}^{\mathrm{2}} + \\ $$$$\:\:\:\:\left(\mathrm{4}{q}^{\mathrm{3}} −\mathrm{2}{q}−{c}\right){x}+{q}^{\mathrm{4}} −{q}^{\mathrm{2}} −{cq}=\mathrm{0} \\ $$$${let}\:\:{q}^{\mathrm{2}} =\mathrm{1}/\mathrm{6}\:\:\Rightarrow \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{qx}^{\mathrm{3}} −\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right){x}−\frac{\mathrm{5}}{\mathrm{36}}−{cq}=\mathrm{0} \\ $$$$\equiv\:\left({x}^{\mathrm{2}} +{mx}+{k}\right)\left({x}^{\mathrm{2}} +{sx}+{h}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{m}+{s}=\mathrm{4}{q} \\ $$$$\:\:\:\:\:\:{h}+{k}+{ms}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{mh}+{ks}=−\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right) \\ $$$$\:\:\:\:\:\:{kh}=−\frac{\mathrm{5}}{\mathrm{36}}−{cq} \\ $$$${h},\:{k}=\:−\frac{{ms}}{\mathrm{2}}\pm\sqrt{\frac{{m}^{\mathrm{2}} {s}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+{cq}} \\ $$$${m}\left(−\frac{{ms}}{\mathrm{2}}+{D}\right)+{s}\left(−\frac{{ms}}{\mathrm{2}}−{D}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}=\mathrm{0} \\ $$$$\left({m}−{s}\right){D}−\frac{{ms}}{\mathrm{2}}\left(\mathrm{4}{q}\right)+\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}=\mathrm{0} \\ $$$${now}\:{let}'{s}\:{assume}\:\:{ms}={p} \\ $$$$\&\:{since}\:\:{m}+{s}=\mathrm{4}{q} \\ $$$$\left({m}−{s}\right)^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{4}{p} \\ $$$$\Rightarrow\:\:\left(\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{4}{p}\right)\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+{cq}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{2}{pq}−\frac{\mathrm{4}{q}}{\mathrm{3}}−{c}\right)^{\mathrm{2}} \\ $$$$\:\:{p}^{\mathrm{3}} +\left(\frac{\mathrm{5}}{\mathrm{9}}+\mathrm{4}{cq}−\frac{\mathrm{8}}{\mathrm{9}}−\mathrm{4}{cq}\right){p} \\ $$$$\:\:\:\:\:\:\:+\left(\frac{\mathrm{4}{q}}{\mathrm{3}}+{c}\right)^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}\left(\frac{\mathrm{5}}{\mathrm{36}}+{cq}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{3}} −\frac{{p}}{\mathrm{3}}+\frac{\mathrm{14}}{\mathrm{27}}=\mathrm{0} \\ $$$${D}_{\mathrm{0}} =\left(\frac{\mathrm{7}}{\mathrm{27}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{3}} \\ $$$$.... \\ $$

Question Number 149237 Answers: 1 Comments: 0