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AllQuestion and Answers: Page 1084

Question Number 109715    Answers: 1   Comments: 0

{ ((sin 2x+sin 2y=(5/4))),((cos (x−y)=2sin (x+y))) :}where 0<x,y<(π/2) cos^2 (x+y) = ? △((♭o♭)/(hans))▽

$$\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{sin}\:\mathrm{2}{y}=\frac{\mathrm{5}}{\mathrm{4}}}\\{\mathrm{cos}\:\left({x}−{y}\right)=\mathrm{2sin}\:\left({x}+{y}\right)}\end{cases}{where}\:\mathrm{0}<{x},{y}<\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left({x}+{y}\right)\:=\:? \\ $$$$\:\:\:\:\:\:\bigtriangleup\frac{\flat{o}\flat}{{hans}}\bigtriangledown \\ $$

Question Number 109734    Answers: 1   Comments: 0

solve the following integral 1)∫_3 ^7 4(√((x−3)(7−x)))dx 2)∫_0 ^∞ ((xln^2 (1+x))/((1+x)^3 ))dx 3)∫_0 ^(π/4) [ln(1−tanx)]^2 dx=(π/2)ln2−2G 4)∫_0 ^(π/4) ln(1+cotx)dx=(π/8)ln2+G 5)∫_0 ^(π/2) ln(2+cosx)dx

$${solve}\:{the}\:{following}\:{integral} \\ $$$$\left.\mathrm{1}\right)\int_{\mathrm{3}} ^{\mathrm{7}} \mathrm{4}\sqrt{\left({x}−\mathrm{3}\right)\left(\mathrm{7}−{x}\right)}{dx} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \frac{{x}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{dx} \\ $$$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left[\mathrm{ln}\left(\mathrm{1}−\mathrm{tan}{x}\right)\right]^{\mathrm{2}} {dx}=\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\mathrm{2}{G} \\ $$$$\left.\mathrm{4}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{cot}{x}\right){dx}=\frac{\pi}{\mathrm{8}}\mathrm{ln2}+{G} \\ $$$$\left.\mathrm{5}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}+\mathrm{cos}{x}\right){dx} \\ $$

Question Number 109733    Answers: 2   Comments: 0

lim_(x→0) ((x−tan x)/(sin x−x)) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}−{x}}\:? \\ $$

Question Number 109709    Answers: 2   Comments: 0

∫_(0 ) ^(π/4) ln(tanx+1)dx

$$\:\:\:\:\:\int_{\mathrm{0}\:} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tanx}+\mathrm{1}\right)\mathrm{dx} \\ $$

Question Number 109729    Answers: 3   Comments: 0

Question Number 109728    Answers: 4   Comments: 1

△((♭ε)/(math))▽ lim_(x→π) ((sin^2 x)/(cos 3x+1))

$$\:\:\:\bigtriangleup\frac{\flat\epsilon}{{math}}\bigtriangledown \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:\mathrm{3}{x}+\mathrm{1}} \\ $$

Question Number 109699    Answers: 1   Comments: 0

x(x−1)^2 ≥ 12(x−1)

$$\:\:\:\:\:{x}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{12}\left({x}−\mathrm{1}\right) \\ $$

Question Number 109674    Answers: 2   Comments: 0

Find the area of the region enclose by the curve y = 1 − x^2 , and the line y = 1 + 3x, x = 1.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclose}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:+\:\:\mathrm{3x},\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1}. \\ $$

Question Number 109658    Answers: 1   Comments: 0

Question Number 109848    Answers: 0   Comments: 0

Question Number 109655    Answers: 0   Comments: 0

Question Number 109642    Answers: 1   Comments: 0

Question Number 109640    Answers: 0   Comments: 2

Question Number 109639    Answers: 2   Comments: 0

Question Number 109626    Answers: 1   Comments: 1

Question Number 109625    Answers: 0   Comments: 2

((sin 2𝛂+2sin 𝛂∙cos 2𝛂)/(1+cos 𝛂+cos2 𝛂+cos3 𝛂))

$$\frac{\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}+\mathrm{2sin}\:\boldsymbol{\alpha}\centerdot\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\alpha}+\mathrm{cos2}\:\boldsymbol{\alpha}+\mathrm{cos3}\:\boldsymbol{\alpha}} \\ $$

Question Number 109619    Answers: 0   Comments: 2

Σ_(n=1) ^∞ ((n!)/n^n )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} } \\ $$

Question Number 109616    Answers: 1   Comments: 1

find ∫_0 ^1 (√(1+x^4 ))dx

$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$

Question Number 109611    Answers: 4   Comments: 1

Question Number 109606    Answers: 0   Comments: 15

Question Number 109605    Answers: 0   Comments: 3

Question Number 109595    Answers: 1   Comments: 1

For any Real numbers x,y and z, if (x+y+z)=2, then prove that xyz≥8(1−x)(1−y)(1−z)

$${For}\:{any}\:{Real}\:{numbers}\:{x},{y}\:{and}\:{z}, \\ $$$$\:{if}\:\:\left({x}+{y}+{z}\right)=\mathrm{2},\:{then}\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\:\:{xyz}\geqslant\mathrm{8}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right) \\ $$

Question Number 109591    Answers: 1   Comments: 3

An isosceles AEF is inscribed into a square ABCD such that pointE is on side BC,point F is on side CDand AE=EF. Knownthat tanAEF^() =2.Find tanFEC^()

$$\mathrm{An}\:\mathrm{isosceles}\:\mathrm{AEF}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{into}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{ABCD}\:\mathrm{such}\:\mathrm{that}\:\mathrm{pointE}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{side}\:\mathrm{BC},\mathrm{point}\:\mathrm{F}\:\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CDand}\:\mathrm{AE}=\mathrm{EF}. \\ $$$$\mathrm{Knownthat}\:\mathrm{tan}\widehat {\mathrm{AEF}}=\mathrm{2}.\mathrm{Find}\:\mathrm{tan}\widehat {\mathrm{FEC}} \\ $$

Question Number 109585    Answers: 1   Comments: 0

If (xy+yz+zx)=1, then prove that (x/(1−x^2 ))+(y/(1−y^2 ))+(z/(1−z^2 ))=((4xyz)/((1−x^2 )(1−y^2 )(1−z^2 )))

$${If}\:\left({xy}+{yz}+{zx}\right)=\mathrm{1},\:{then}\:{prove}\:{that} \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{y}}{\mathrm{1}−{y}^{\mathrm{2}} }+\frac{{z}}{\mathrm{1}−{z}^{\mathrm{2}} }=\frac{\mathrm{4}{xyz}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\left(\mathrm{1}−{z}^{\mathrm{2}} \right)} \\ $$

Question Number 109584    Answers: 3   Comments: 1

((−♭o♭−)/(hans)) (1)(√(x−(√(x−(1/4))))) ≥ (1/4) (2)∣a^→ ∣ = 1, ∣b^→ ∣ = 2 , ∣c^→ ∣=3 , ∠(a^→ ,b^→ )=90° ∠(b^→ ,c^→ )=60° , ∠(a^→ ,c^→ )=120° , then ∣a^→ +b^→ −c^→ ∣=? (3) { ((x^(log _3 (y)) +y^(log _3 (x)) =18)),((log _3 (x)+log _3 (y)=3)) :} . Find the solution

$$\:\:\:\frac{−\flat{o}\flat−}{{hans}} \\ $$$$\left(\mathrm{1}\right)\sqrt{{x}−\sqrt{{x}−\frac{\mathrm{1}}{\mathrm{4}}}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\mid\overset{\rightarrow} {{a}}\mid\:=\:\mathrm{1},\:\mid\overset{\rightarrow} {{b}}\mid\:=\:\mathrm{2}\:,\:\mid\overset{\rightarrow} {{c}}\mid=\mathrm{3}\:,\:\angle\left(\overset{\rightarrow} {{a}},\overset{\rightarrow} {{b}}\right)=\mathrm{90}° \\ $$$$\:\:\:\:\:\:\:\:\angle\left(\overset{\rightarrow} {{b}},\overset{\rightarrow} {{c}}\right)=\mathrm{60}°\:,\:\angle\left(\overset{\rightarrow} {{a}},\overset{\rightarrow} {{c}}\right)=\mathrm{120}°\:,\:{then}\: \\ $$$$\:\:\:\:\:\:\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}−\overset{\rightarrow} {{c}}\mid=? \\ $$$$\left(\mathrm{3}\right)\begin{cases}{{x}^{\mathrm{log}\:_{\mathrm{3}} \left({y}\right)} +{y}^{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)} =\mathrm{18}}\\{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)+\mathrm{log}\:_{\mathrm{3}} \left({y}\right)=\mathrm{3}}\end{cases}\:.\:{Find}\:{the}\:{solution} \\ $$$$\:\:\: \\ $$

Question Number 109577    Answers: 4   Comments: 1

Given x^4 +x^2 y^2 +y^4 =133 and x^2 −xy+y^2 =7 then what is the value of xy ?

$$\:\:\:\mathrm{G}{iven}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} =\mathrm{133} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\:\:{then}\:{what}\:{is}\:{the}\:{value}\:{of}\:{xy}\:? \\ $$

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