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Question Number 57389    Answers: 1   Comments: 0

If aεR and the equation : −3{x}^2 +2{x}+a^2 =0 has no integral solution, then all possible value of a lie in the interval : (a)(−1,0)U(0,1) (b)(1,2) (c) (−2,−1) (d)(−∞,−2)U(2,∞)

$${If}\:{a}\epsilon{R}\:{and}\:{the}\:{equation}\:: \\ $$$$−\mathrm{3}\left\{{x}\right\}^{\mathrm{2}} +\mathrm{2}\left\{{x}\right\}+{a}^{\mathrm{2}} =\mathrm{0}\:{has}\:{no}\:{integral} \\ $$$${solution},\:{then}\:{all}\:{possible}\:{value}\:{of}\:{a} \\ $$$${lie}\:{in}\:{the}\:{interval}\:: \\ $$$$\left({a}\right)\left(−\mathrm{1},\mathrm{0}\right)\mathrm{U}\left(\mathrm{0},\mathrm{1}\right)\:\:\:\left({b}\right)\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\left({c}\right)\:\left(−\mathrm{2},−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\left(−\infty,−\mathrm{2}\right)\mathrm{U}\left(\mathrm{2},\infty\right) \\ $$

Question Number 57388    Answers: 0   Comments: 0

Given f(x) = f(x + 2016), ∀x ∈ R If ∫_0 ^3 f(x) = 30, then ∫_3 ^5 f(x + 2016) = ...

$$\mathrm{Given}\:{f}\left({x}\right)\:=\:{f}\left({x}\:+\:\mathrm{2016}\right),\:\:\forall{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{If}\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:{f}\left({x}\right)\:=\:\mathrm{30},\:\mathrm{then}\:\underset{\mathrm{3}} {\overset{\mathrm{5}} {\int}}\:{f}\left({x}\:+\:\mathrm{2016}\right)\:=\:... \\ $$

Question Number 57387    Answers: 0   Comments: 0

Question Number 57385    Answers: 1   Comments: 1

Question Number 57383    Answers: 1   Comments: 1

Question Number 57381    Answers: 0   Comments: 0

Question Number 57377    Answers: 0   Comments: 0

Can i find the sum of a product to infinity ? e.g 1.2.3.4.5 .... infinity

$$\mathrm{Can}\:\mathrm{i}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{a}\:\mathrm{product}\:\mathrm{to}\:\mathrm{infinity}\:? \\ $$$$\:\:\:\mathrm{e}.\mathrm{g}\:\:\:\:\:\:\:\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}\:....\:\:\:\:\mathrm{infinity} \\ $$

Question Number 57373    Answers: 1   Comments: 1

tan 1°+tan 5°+tan 9°+…+tan 177°=...

$$\mathrm{tan}\:\mathrm{1}°+\mathrm{tan}\:\mathrm{5}°+\mathrm{tan}\:\mathrm{9}°+\ldots+\mathrm{tan}\:\mathrm{177}°=... \\ $$

Question Number 57368    Answers: 1   Comments: 0

5^(3x−3) −5^(3x) −5=615 solve for x

$$\mathrm{5}^{\mathrm{3}{x}−\mathrm{3}} −\mathrm{5}^{\mathrm{3}{x}} −\mathrm{5}=\mathrm{615} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$

Question Number 57363    Answers: 0   Comments: 3

Question Number 57357    Answers: 3   Comments: 2

1−2sin(4x)<cos^2 (4x) solve.

$$\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\mathrm{4}\boldsymbol{\mathrm{x}}\right)<\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\mathrm{4}\boldsymbol{\mathrm{x}}\right) \\ $$$$\boldsymbol{\mathrm{solve}}. \\ $$

Question Number 57356    Answers: 0   Comments: 0

S=((𝛑R^2 )/(360°))×𝛂° prove.

$$\boldsymbol{\mathrm{S}}=\frac{\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} }{\mathrm{360}°}×\boldsymbol{\alpha}° \\ $$$$\boldsymbol{\mathrm{prove}}. \\ $$

Question Number 57348    Answers: 1   Comments: 2

Question Number 57345    Answers: 1   Comments: 0

1)if: sinx+tgx=1,then: sin^4 x+tg^4 x=? 2)if: sinx+tgx=2,then: sin4x+tg4x=? 3.if: sinx+tgx=3,then: ((sin4x)/(sin^4 x))+((tg4x)/(tg^4 x))=?

$$\left.\mathrm{1}\right)\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{1},\boldsymbol{\mathrm{then}}:\:\:\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tg}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}=? \\ $$$$\left.\mathrm{2}\right)\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{2},\boldsymbol{\mathrm{then}}:\:\:\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tg}}\mathrm{4}\boldsymbol{\mathrm{x}}=? \\ $$$$\mathrm{3}.\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{3},\boldsymbol{\mathrm{then}}:\:\:\frac{\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}}+\frac{\boldsymbol{\mathrm{tg}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{tg}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}}=? \\ $$

Question Number 57336    Answers: 2   Comments: 0

If n be even, show that the expression ((n(n + 2)(n + 4) ... (2n − 2))/(1.3.5 ... (n − 1))) simplify to 2^(n − 1)

$$\mathrm{If}\:\:\mathrm{n}\:\mathrm{be}\:\mathrm{even},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{expression}\:\:\:\:\frac{\mathrm{n}\left(\mathrm{n}\:+\:\mathrm{2}\right)\left(\mathrm{n}\:+\:\mathrm{4}\right)\:...\:\left(\mathrm{2n}\:−\:\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}\:...\:\left(\mathrm{n}\:−\:\mathrm{1}\right)} \\ $$$$\mathrm{simplify}\:\mathrm{to}\:\:\mathrm{2}^{\mathrm{n}\:−\:\mathrm{1}} \\ $$

Question Number 57332    Answers: 0   Comments: 1

find the value of k such that k(x^2 +y^2 )+(y−2x+1)(y+2x+3)=0 is a circle hence obtain the centre and radius of the resulting circle.

$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{such}\:\mathrm{that} \\ $$$${k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\left({y}−\mathrm{2}{x}+\mathrm{1}\right)\left({y}+\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{hence}\:\mathrm{obtain}\: \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{resulting}\:\mathrm{circle}. \\ $$

Question Number 57330    Answers: 0   Comments: 0

Question Number 57329    Answers: 1   Comments: 1

∫_(−1) ^2 ∣x∣ ⌊x⌋ dx = ?

$$\underset{−\mathrm{1}} {\int}\overset{\mathrm{2}} {\:}\:\mid{x}\mid\:\lfloor{x}\rfloor\:{dx}\:\:=\:\:\:? \\ $$

Question Number 57328    Answers: 1   Comments: 0

Question Number 57325    Answers: 0   Comments: 1

calculate ∫_0 ^(π/2) ((ln(1+sinx))/(sinx))dx

$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+{sinx}\right)}{{sinx}}{dx} \\ $$

Question Number 57324    Answers: 0   Comments: 0

we want to find the vslue of I =∫_0 ^1 ((ln(1+x))/(1+x^2 )) dx let A=∫∫_W (x/((1+x^2 )(1+xy)))dxdy with W=[0,1]^2 calculate A by two method and conclude the value of I .

$${we}\:{want}\:{to}\:{find}\:{the}\:{vslue}\:{of} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:{let} \\ $$$${A}=\int\int_{{W}} \frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}{dxdy} \\ $$$${with}\:{W}=\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} \\ $$$${calculate}\:{A}\:{by}\:{two}\:{method}\:{and} \\ $$$${conclude}\:{the}\:{value}\:{of}\:{I}\:. \\ $$

Question Number 57323    Answers: 0   Comments: 1

calculate ∫∫_D ((x+y)/(3+(√(x^2 +y^2 ))))dxdy with D={(x,y)∈R^2 /x^2 +y^2 ≤2 and x≥0 ,y≥0}

$${calculate}\:\int\int_{{D}} \:\:\frac{{x}+{y}}{\mathrm{3}+\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}\right. \\ $$$$\left.{and}\:{x}\geqslant\mathrm{0}\:,{y}\geqslant\mathrm{0}\right\} \\ $$

Question Number 57321    Answers: 1   Comments: 1

calculate ∫∫_D (x−y)(√(x^2 +y^2 ))dxdy with D ={ (x,y)∈R^2 /x^2 +y^2 ≤2 and x≥0}

$${calculate}\:\int\int_{{D}} \left({x}−{y}\right)\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy} \\ $$$${with}\:{D}\:=\left\{\:\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{2}\:{and}\:{x}\geqslant\mathrm{0}\right\} \\ $$

Question Number 57320    Answers: 1   Comments: 1

calculate ∫∫_D xy e^(−x^2 −y^2 ) dxdy with D={(x,y)∈R^2 / 0≤x≤2 and 1≤y≤3}

$${calculate}\:\int\int_{{D}} {xy}\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \:{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\:{and}\right. \\ $$$$\left.\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\right\} \\ $$

Question Number 57319    Answers: 1   Comments: 1

calculate ∫∫_D e^(x−y) dxdy with D={(x,y)∈R^2 /∣x∣<1 and 0≤y≤1}

$${calculate}\:\int\int_{{D}} \:{e}^{{x}−{y}} \:{dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\mid{x}\mid<\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\right\} \\ $$

Question Number 57310    Answers: 0   Comments: 1

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