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Question Number 149885    Answers: 1   Comments: 0

I_n =∫_0 ^(π/4) (dx/(cos^(2n+1) x)) to show that : ∀ n∈N^∗ , 2nI_n =(2n−1)I_(n−1) +(2^n /( (√2))) (I_n =∫_0 ^(π/4) ((1/(cos^(2n−1) x))×(1/(cos^2 x)))dx)...

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{{cos}^{\mathrm{2}{n}+\mathrm{1}} {x}} \\ $$$${to}\:{show}\:{that}\:: \\ $$$$\forall\:{n}\in\mathbb{N}^{\ast} ,\:\mathrm{2}{nI}_{{n}} =\left(\mathrm{2}{n}−\mathrm{1}\right){I}_{{n}−\mathrm{1}} +\frac{\mathrm{2}^{{n}} }{\:\sqrt{\mathrm{2}}} \\ $$$$\left({I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}{n}−\mathrm{1}} {x}}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}\right){dx}\right)... \\ $$

Question Number 149883    Answers: 1   Comments: 0

Prove that ((2+(√5)))^(1/3) +((2−(√5)))^(1/3) is a rational number

$$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rational}\:\mathrm{number} \\ $$

Question Number 149876    Answers: 2   Comments: 0

Question Number 149871    Answers: 1   Comments: 0

lim_(x→2) ((3^(x!) −9)/(x−2))

$${lim}_{{x}\rightarrow\mathrm{2}} \frac{\mathrm{3}^{{x}!} −\mathrm{9}}{{x}−\mathrm{2}} \\ $$

Question Number 149870    Answers: 0   Comments: 3

if x;y;z;m;n∈R^+ then: Σ_(cyc) (b^(−1) /((m(√x) + n(√y))^2 )) ≥ (3/((m + n)^2 ))

$$\mathrm{if}\:\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{m};\mathrm{n}\in\mathbb{R}^{+} \:\:\mathrm{then}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{b}^{−\mathrm{1}} }{\left(\mathrm{m}\sqrt{\mathrm{x}}\:+\:\mathrm{n}\sqrt{\mathrm{y}}\right)^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\left(\mathrm{m}\:+\:\mathrm{n}\right)^{\mathrm{2}} } \\ $$

Question Number 149852    Answers: 0   Comments: 0

What angle is subtended at the centre of the Earth by an arc of the equator of length 1) 2002km 2) 30030km

What angle is subtended at the centre of the Earth by an arc of the equator of length 1) 2002km 2) 30030km

Question Number 149868    Answers: 1   Comments: 0

if a;b and c are the dimensions of a cuboid with the diagonal d then prove d ≤ (√((a^3 /b) + (b^3 /c) + (c^3 /a)))

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{and}\:\:\mathrm{c}\:\:\mathrm{are}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of}\:\:\mathrm{a} \\ $$$$\mathrm{cuboid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{diagonal}\:\boldsymbol{\mathrm{d}}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{d}\:\leqslant\:\sqrt{\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{c}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{a}}} \\ $$

Question Number 150356    Answers: 3   Comments: 0

solve... I:= ∫_0 ^( 1) (( Arcsin ((√x) ))/(1−x + x^( 2) )) dx=?

$$\:{solve}... \\ $$$$\:\:\:\:\:\mathrm{I}:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{Arcsin}\:\left(\sqrt{{x}}\:\right)}{\mathrm{1}−{x}\:+\:{x}^{\:\mathrm{2}} }\:{dx}=? \\ $$

Question Number 149865    Answers: 1   Comments: 0

Question Number 149830    Answers: 1   Comments: 0

lim_(x→0) ((√(x−(√(x−(√(x−...))))))/(sin (sin (sin (....x)))=?

$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}−\sqrt{{x}−\sqrt{{x}−...}}}}{{sin}\:\left({sin}\:\left({sin}\:\left(....{x}\right)\right.\right.}=? \\ $$

Question Number 149955    Answers: 1   Comments: 0

If a group consist of 8 men and 6 women, in how many ways can a committee of 5 be selected if: i) the committee is to consist of 3 men and 3 women. ii) there are no restrictions on the number of men and women on the committee. iii) at least one man

$$\mathrm{If}\:\mathrm{a}\:\mathrm{group}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{8}\:\mathrm{men}\:\mathrm{and}\:\mathrm{6}\:\mathrm{women}, \\ $$$$\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{a}\:\mathrm{committee}\:\mathrm{of} \\ $$$$\mathrm{5}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{if}: \\ $$$$\left.\:\:\:\:\:\:\mathrm{i}\right)\:\mathrm{the}\:\mathrm{committee}\:\mathrm{is}\:\mathrm{to}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{3}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{3}\:\mathrm{women}. \\ $$$$\left.\:\:\:\:\:\:\mathrm{ii}\right)\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{restrictions}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{men}\:\mathrm{and}\:\mathrm{women}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{committee}. \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{iii}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{man} \\ $$

Question Number 149817    Answers: 0   Comments: 0

x^3 −x=c let x=t+h t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c=0 let t(t^2 +3h^2 −1)=p 3ht^2 +h^3 −h−c=q ⇒ t^2 =((q+c+h−h^3 )/(3h)) p+q=0 (((q+c+h−h^3 )/(3h)))(((q+c+h−h^3 )/(3h))+3h^2 −1)^2 =q^2 ⇒ (q+c+h−h^3 )(q+c−2h+8h^3 )^2 = 27h^3 q^2 let q+c−2h=0 ⇒ 64h^4 (3−h^2 )=27(2h−c)^2 ⇒ 8h^2 (√(3−h^2 ))=3(√3)(2h−c) let h=(√3)sin θ ⇒ 8sin^2 θcos θ=2(√3)sin θ−c 4sin θsin 2θ=2(√3)sin θ−c 2sin θ((√3)−2sin 2θ)=c ......

$$\:\:\:{x}^{\mathrm{3}} −{x}={c} \\ $$$$\:\:{let}\:\:{x}={t}+{h} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{t}\left({t}^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={p} \\ $$$$\mathrm{3}{ht}^{\mathrm{2}} +{h}^{\mathrm{3}} −{h}−{c}={q} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} =\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}} \\ $$$${p}+{q}=\mathrm{0} \\ $$$$\left(\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}}\right)\left(\frac{{q}+{c}+{h}−{h}^{\mathrm{3}} }{\mathrm{3}{h}}+\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:={q}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({q}+{c}+{h}−{h}^{\mathrm{3}} \right)\left({q}+{c}−\mathrm{2}{h}+\mathrm{8}{h}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{27}{h}^{\mathrm{3}} {q}^{\mathrm{2}} \\ $$$${let}\:\:{q}+{c}−\mathrm{2}{h}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{64}{h}^{\mathrm{4}} \left(\mathrm{3}−{h}^{\mathrm{2}} \right)=\mathrm{27}\left(\mathrm{2}{h}−{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{8}{h}^{\mathrm{2}} \sqrt{\mathrm{3}−{h}^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{3}}\left(\mathrm{2}{h}−{c}\right) \\ $$$${let}\:\:{h}=\sqrt{\mathrm{3}}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{8sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{c} \\ $$$$\mathrm{4sin}\:\theta\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{c} \\ $$$$\mathrm{2sin}\:\theta\left(\sqrt{\mathrm{3}}−\mathrm{2sin}\:\mathrm{2}\theta\right)={c} \\ $$$$...... \\ $$

Question Number 149813    Answers: 1   Comments: 1

∫(−1)^([x]) dx = ?

$$\int\left(−\mathrm{1}\right)^{\left[\boldsymbol{\mathrm{x}}\right]} \:\mathrm{dx}\:=\:? \\ $$

Question Number 149809    Answers: 2   Comments: 0

Question Number 149854    Answers: 0   Comments: 11

Question Number 149853    Answers: 1   Comments: 0

Question Number 149805    Answers: 1   Comments: 0

Question Number 149795    Answers: 1   Comments: 0

lim_(n→∞) ((1∙3∙5∙7∙ ... ∙(2n-1))/(2∙4∙6∙ ... ∙2n)) = ?

$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}\centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{7}\centerdot\:...\:\centerdot\left(\mathrm{2n}-\mathrm{1}\right)}{\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}\centerdot\:...\:\centerdot\mathrm{2n}}\:=\:? \\ $$

Question Number 149793    Answers: 0   Comments: 1

Compare: tan(11°) and (1/5)

$$\mathrm{Compare}: \\ $$$$\mathrm{tan}\left(\mathrm{11}°\right)\:\:\:\mathrm{and}\:\:\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$

Question Number 149789    Answers: 0   Comments: 0

form

$${form} \\ $$

Question Number 149782    Answers: 2   Comments: 0

Question Number 149781    Answers: 1   Comments: 2

a − (√((20)/a)) = 7 ⇒ (√(5a)) − a = ?

$$\mathrm{a}\:\:−\:\:\sqrt{\frac{\mathrm{20}}{\mathrm{a}}}\:=\:\mathrm{7}\:\:\Rightarrow\:\:\sqrt{\mathrm{5a}}\:−\:\mathrm{a}\:=\:? \\ $$

Question Number 149778    Answers: 1   Comments: 2

Simplify: (√(8(√(24(√(8(√(24...)))))))) = ?

$$\mathrm{Simplify}: \\ $$$$\sqrt{\mathrm{8}\sqrt{\mathrm{24}\sqrt{\mathrm{8}\sqrt{\mathrm{24}...}}}}\:\:=\:? \\ $$

Question Number 149775    Answers: 0   Comments: 2

Solve the equation: 5^x = (3^x − 2^x )^2

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:=\:\left(\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \right)^{\mathrm{2}} \\ $$

Question Number 149773    Answers: 1   Comments: 0

Question Number 149772    Answers: 0   Comments: 0

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