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Question Number 164364    Answers: 0   Comments: 0

Lim_(x→∞) (e^(tan(x)) ) {Z.A}

$$\boldsymbol{{Lim}}_{\boldsymbol{{x}}\rightarrow\infty} \:\left(\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \right) \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{\mathrm{A}}\right\} \\ $$

Question Number 164347    Answers: 0   Comments: 4

Question Number 164345    Answers: 2   Comments: 1

Question Number 164341    Answers: 0   Comments: 1

x^4 =x^2 +cx (x^2 −(1/2))^2 =cx+(1/4) let x^2 −(1/2)=t ⇒ (t^2 −(1/4))^2 =c^2 (t+(1/2)) now let t^2 −(1/4)=z ⇒ (z^2 −(c^2 /2))^2 =c^4 (z+(1/4)) z^2 −(c^2 /2)=p ⇒ (p^2 −(c^4 /4))^2 =c^8 (p+(c^2 /2)) ⇒ p^4 −((c^4 p^2 )/2)−c^8 p+(c^8 /(16))−(c^(10) /2)=0 p^4 −Ap^2 −Bp−λ=0 (p^2 +sp+h)(p^2 −sp−(λ/h))=0 h−(λ/h)=s^2 −A s(h+(λ/h))=B (B^2 /s^2 )−(s^2 −A)^2 =4λ (s^2 −A)^3 +A(s^2 −A)^2 +4λ(s^2 −A) +4λA−B^2 =0 m^3 +Am^2 +4λm+(4λA−B^2 )=0 let m=w−(A/3) ⇒ w^3 +(4λ−(A^2 /3))w+((2A^3 )/(27))+((8λA)/3)−B^2 =0 4λ−(A^2 /3)=2c^(10) −(c^8 /4)−(c^8 /(12)) =2c^8 (c^2 −(1/6))

$$\:\:{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +{cx} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={cx}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}={t}\:\:\:\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${now}\:\:{let}\:\:\:{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}={z}\:\:\:\Rightarrow \\ $$$$\left({z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{4}} \left({z}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}={p}\:\:\Rightarrow\:\:\left({p}^{\mathrm{2}} −\frac{{c}^{\mathrm{4}} }{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{8}} \left({p}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:{p}^{\mathrm{4}} −\frac{{c}^{\mathrm{4}} {p}^{\mathrm{2}} }{\mathrm{2}}−{c}^{\mathrm{8}} {p}+\frac{{c}^{\mathrm{8}} }{\mathrm{16}}−\frac{{c}^{\mathrm{10}} }{\mathrm{2}}=\mathrm{0} \\ $$$${p}^{\mathrm{4}} −{Ap}^{\mathrm{2}} −{Bp}−\lambda=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{sp}+{h}\right)\left({p}^{\mathrm{2}} −{sp}−\frac{\lambda}{{h}}\right)=\mathrm{0} \\ $$$${h}−\frac{\lambda}{{h}}={s}^{\mathrm{2}} −{A} \\ $$$${s}\left({h}+\frac{\lambda}{{h}}\right)={B} \\ $$$$\frac{{B}^{\mathrm{2}} }{{s}^{\mathrm{2}} }−\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} =\mathrm{4}\lambda \\ $$$$\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{3}} +{A}\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} +\mathrm{4}\lambda\left({s}^{\mathrm{2}} −{A}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}^{\mathrm{3}} +{Am}^{\mathrm{2}} +\mathrm{4}\lambda{m}+\left(\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\:\:{m}={w}−\frac{{A}}{\mathrm{3}}\:\:\:\Rightarrow \\ $$$${w}^{\mathrm{3}} +\left(\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}\right){w}+\frac{\mathrm{2}{A}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}\lambda{A}}{\mathrm{3}}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}{c}^{\mathrm{10}} −\frac{{c}^{\mathrm{8}} }{\mathrm{4}}−\frac{{c}^{\mathrm{8}} }{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{c}^{\mathrm{8}} \left({c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$

Question Number 164339    Answers: 1   Comments: 0

In AB^Δ C : cos^( 2) (A )+ cos^( 2) (B )+ cos^( 2) ( C )=1 . Prove that AB^Δ C is right angled. −−−−−−−−

$$ \\ $$$$\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\:\:\::\:\:\:{cos}^{\:\mathrm{2}} \left({A}\:\right)+\:{cos}^{\:\mathrm{2}} \left({B}\:\right)+\:{cos}^{\:\mathrm{2}} \left(\:{C}\:\right)=\mathrm{1}\:\:. \\ $$$$\:\:\:\:\:\:\:\:{Prove}\:{that}\:\:{A}\overset{\Delta} {{B}C}\:\:\:{is}\:\:\:{right}\:{angled}. \\ $$$$\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 164331    Answers: 0   Comments: 2

Question Number 164328    Answers: 2   Comments: 0

solve Σ_(n=0) ^∞ (( cos^(2n) (x))/(n!)) = (e)^(1/4) ■ −−−−−−−

$$ \\ $$$$\:\:\:\:\:{solve} \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:{cos}^{\mathrm{2}{n}} \left({x}\right)}{{n}!}\:=\:\sqrt[{\mathrm{4}}]{{e}}\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$\:\:\:\:\:\:\:\:−−−−−−− \\ $$$$\:\: \\ $$

Question Number 164315    Answers: 0   Comments: 3

Question Number 164312    Answers: 0   Comments: 3

Question Number 164311    Answers: 0   Comments: 0

Question Number 164310    Answers: 1   Comments: 0

Question Number 164305    Answers: 2   Comments: 1

Question Number 164299    Answers: 1   Comments: 0

if f(f(f(...f(x)...)))_(n times) =2x+1, find f(1)=?

$${if}\:\underset{{n}\:{times}} {{f}\left({f}\left({f}\left(...{f}\left({x}\right)...\right)\right)\right)}=\mathrm{2}{x}+\mathrm{1},\: \\ $$$${find}\:{f}\left(\mathrm{1}\right)=? \\ $$

Question Number 164300    Answers: 1   Comments: 0

Question a.which numbers have an odd number of divisors b. Is there a number with exactly 13 divisors c. Generalize

$$\mathrm{Question} \\ $$$$\mathrm{a}.\mathrm{which}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{number} \\ $$$$\:\:\:\:\:\mathrm{of}\:\mathrm{divisors} \\ $$$$\mathrm{b}.\:\:\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{number}\:\mathrm{with}\:\mathrm{exactly}\:\mathrm{13} \\ $$$$\:\:\:\:\:\:\mathrm{divisors} \\ $$$$\mathrm{c}.\:\:\mathrm{Generalize} \\ $$

Question Number 164290    Answers: 2   Comments: 0

Question Number 164279    Answers: 3   Comments: 0

x^n =n^x x=?

$${x}^{{n}} ={n}^{{x}} \:\:\:{x}=? \\ $$

Question Number 164272    Answers: 1   Comments: 0

Question Number 164269    Answers: 1   Comments: 0

Solve the inequality; (x+3)[(x−1)^2 −x(x+(3/4))]+6+((7x^2 +29x)/4)>0 {Z.A}

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Solve}}\:\boldsymbol{{the}}\:\boldsymbol{{inequality}}; \\ $$$$\:\:\left(\boldsymbol{{x}}+\mathrm{3}\right)\left[\left(\boldsymbol{{x}}−\mathrm{1}\right)^{\mathrm{2}} −\boldsymbol{{x}}\left(\boldsymbol{{x}}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right]+\mathrm{6}+\frac{\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{29}\boldsymbol{{x}}}{\mathrm{4}}>\mathrm{0} \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$

Question Number 164264    Answers: 0   Comments: 0

Question Number 164266    Answers: 2   Comments: 0

lim_(x→∞) (x−2) − (√(x^2 +2x−5)) = ?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\left({x}−\mathrm{2}\right)\:−\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{5}}\:\:=\:\:? \\ $$

Question Number 164256    Answers: 1   Comments: 0

Question Number 164242    Answers: 0   Comments: 0

Question Number 164240    Answers: 2   Comments: 0

Find: Σ_(n=1) ^∞ ln (n) = ?

$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{ln}\:\left(\mathrm{n}\right)\:=\:? \\ $$

Question Number 164237    Answers: 1   Comments: 0

Question Number 164236    Answers: 2   Comments: 1

Question Number 164231    Answers: 0   Comments: 0

∫_(−∞) ^∞ −(1/3) ∫_0 ^1 ((((^4 log 3.^4 log 6)/( ^4 log 9.^8 log 2+^4 log 9.^8 log 3))/((^9 log 4.^2 log27+^3 log 81)/(^2 log 64−^2 log 8))))dxdy {z.}

$$\int_{−\infty} ^{\infty} \:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\frac{\frac{\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{3}.^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{6}}{\:\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{9}.^{\mathrm{8}} \boldsymbol{\mathrm{log}}\:\mathrm{2}+\:^{\mathrm{4}} \boldsymbol{\mathrm{log}}\:\mathrm{9}.^{\mathrm{8}} \boldsymbol{\mathrm{log}}\:\mathrm{3}}}{\frac{\:^{\mathrm{9}} \boldsymbol{\mathrm{log}}\:\mathrm{4}.^{\mathrm{2}} \boldsymbol{\mathrm{log}}\mathrm{27}+^{\mathrm{3}} \boldsymbol{\mathrm{log}}\:\mathrm{81}}{\:^{\mathrm{2}} \boldsymbol{\mathrm{log}}\:\mathrm{64}−^{\mathrm{2}} \boldsymbol{\mathrm{log}}\:\mathrm{8}}}\right)\boldsymbol{\mathrm{dxdy}} \\ $$$$\left\{\boldsymbol{\mathrm{z}}.\right\} \\ $$

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