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Question Number 220473 Answers: 0 Comments: 0

lim_(n→∞) (tan((π/4)+(1/n)))^n →^(t=(1/n)) =lim_(t→0) [tan((π/4)+t)]^(1/t) ⇒lim_(t→0) [tan((π/4)+t)−1]×(1/t) =lim_(t→0) (((1+tant)/(1−tant))−1)×(1/t) =lim_(t→0) (((2tant)/(1−tant)))×(1/t)=2 ⇒Ans=e^2 ✓

$${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2} \\ $$$$\Rightarrow{Ans}={e}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$

Question Number 220469 Answers: 0 Comments: 0

∫_0 ^( 1) ((6x(1 − x))/((x + 1)(x^2 + 1) ln (x + 1))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{6}{x}\left(\mathrm{1}\:−\:{x}\right)}{\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{ln}\:\left({x}\:+\:\mathrm{1}\right)}\:{dx} \\ $$$$ \\ $$

Question Number 220468 Answers: 3 Comments: 0

a+b+c=1, a^2 +b^2 +c^2 =1 (a,b,c ∈R) a^(10) +b^(10) +c^(10) =1, a^4 +b^4 +c^4 =?

$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$

Question Number 220459 Answers: 3 Comments: 0

find Σ_(n=1) ^∞ (1/(n^2 +a^2 ))=? (a∈R)

$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=?\:\:\:\:\:\:\:\:\:\:\left({a}\in{R}\right) \\ $$

Question Number 220457 Answers: 0 Comments: 0

Find: 𝛀 = ∫_0 ^( ∞) ∫_0 ^( (𝛑/2)) (((x+1)sin^2 (x)ln(y^3 +1))/(xy(y^2 +1))) dxdy = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{y}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{xy}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{dxdy}\:=\:? \\ $$

Question Number 220456 Answers: 0 Comments: 0

a, b, c, d ≥ 1 ; a + b + c = d show that; ab + bc + ca + (1/a) + (1/b) + (1/c) ≥ 2d − 3 + (9/d)

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{a},\:{b},\:{c},\:{d}\:\geqslant\:\mathrm{1}\:\:\:\:\:\:;\:\:\:{a}\:+\:{b}\:+\:{c}\:=\:{d} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that};\: \\ $$$$\:\:\:{ab}\:+\:{bc}\:+\:{ca}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:\geqslant\:\mathrm{2}{d}\:−\:\mathrm{3}\:+\:\frac{\mathrm{9}}{{d}}\:\:\:\: \\ $$

Question Number 220454 Answers: 0 Comments: 0

In the CO_2 molecule, each oxygen atom forms a double bond with central carbon atom. Given that the Bohr radius (a_o ) is approximately 0.529 A^o and the experimental C=O bond length is about 1.16 A^(o ) , calculate : a) The approximate region (in the terms of Bohr radii) where the shared electrons are most likely to be found between C and O. b) Compare the calculated bond region to the sum of the covalent radii of carbon and oxygen, and comment on the effect of π bonding on the contraction of bond length.

Question Number 220447 Answers: 0 Comments: 1

Question Number 220499 Answers: 3 Comments: 1

Question Number 220405 Answers: 2 Comments: 0

9^x^2 −3^(x+1) =0

$$\mathrm{9}^{{x}^{\mathrm{2}} } −\mathrm{3}^{{x}+\mathrm{1}} =\mathrm{0} \\ $$

Question Number 220403 Answers: 1 Comments: 0

Question Number 220493 Answers: 0 Comments: 1

Let ABC be a triangle such CosA+cosB+cosC=(√2) SinA+sinB+sinC=1+(√2) Find A,B,C

$${Let}\:{ABC}\:{be}\:{a}\:{triangle}\:{such} \\ $$$${CosA}+{cosB}+{cosC}=\sqrt{\mathrm{2}} \\ $$$${SinA}+{sinB}+{sinC}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${Find}\:{A},{B},{C} \\ $$

Question Number 220502 Answers: 1 Comments: 0

each J_ν (z),Y_ν (z) are linear independent....?? W_(Ronskian) {J_ν ^ (z),Y_ν (z)}= determinant (((J_ν (z)),( Y_ν (z))),((J_ν ′(z)),(Y_ν ′(z)))) =J_ν ^((1)) (z)Y_ν (z)−J_ν (z)Y_ν ^((1)) (z) J_ν ^((1)) (z)Y_ν (z)=J_(ν−1) (z)Y_ν (z)−(ν/z)J_ν (z)Y_ν (z) J_ν (z)Y_ν ^((1)) (z)=Y_(ν−1) (z)J_ν (z)−(ν/z)J_ν (z)Y_ν (z) J_(ν−1) (z)Y_ν (z)−J_ν (z)Y_(ν−1) (z).... .....damn..... Result is (2/(πz)) ......

Question Number 220495 Answers: 0 Comments: 6