Question and Answers Forum

AllQuestion and Answers: Page 562

Question Number 160638 Answers: 0 Comments: 0

Question Number 160636 Answers: 1 Comments: 0

prove that ∫_(0 ) ^( 1) (((tanh^( −1) ( x ))/x) )^( 2) = ζ ( 2 ) ■ m.n

$$ \\ $$$$\:\:\:\:\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}\:\:} ^{\:\mathrm{1}} \left(\frac{{tanh}^{\:−\mathrm{1}} \left(\:{x}\:\right)}{{x}}\:\right)^{\:\mathrm{2}} =\:\zeta\:\left(\:\mathrm{2}\:\right)\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

Question Number 160630 Answers: 2 Comments: 0

f(x)=x^x^x^x^x . Df(x)=??? svp les baos

$${f}\left({x}\right)={x}^{{x}^{{x}^{{x}^{{x}} } } } . \\ $$$${Df}\left({x}\right)=??? \\ $$$${svp}\:{les}\:{baos} \\ $$

Question Number 160625 Answers: 1 Comments: 0

lim_(x→0+0) (x^x −1)lnx

$$\underset{{x}\rightarrow\mathrm{0}+\mathrm{0}} {\mathrm{lim}}\left(\mathrm{x}^{\mathrm{x}} −\mathrm{1}\right)\mathrm{lnx} \\ $$

Question Number 160624 Answers: 0 Comments: 0

Question Number 160618 Answers: 1 Comments: 0

Question Number 160615 Answers: 0 Comments: 1

Question Number 160609 Answers: 4 Comments: 0

lim_(x→π) (((tan x)/(1+cos x)))=?

$$\:\:\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right)=? \\ $$

Question Number 160608 Answers: 0 Comments: 6

$$ \\ $$$$ \\ $$

Question Number 160606 Answers: 0 Comments: 1

Evaluate (d^2 y/dx^2 ) when x=ln 2 and y=2.

$$\mathrm{Evaluate}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:\mathrm{when}\:{x}=\mathrm{ln}\:\mathrm{2}\:\mathrm{and}\:{y}=\mathrm{2}. \\ $$

Question Number 160605 Answers: 0 Comments: 1

Given that y=(3 sin x−4 cos x+6)^2 , 0≤x≤2π. Find the smallest value of y.

$$\mathrm{Given}\:\mathrm{that}\:{y}=\left(\mathrm{3}\:\mathrm{sin}\:{x}−\mathrm{4}\:\mathrm{cos}\:{x}+\mathrm{6}\right)^{\mathrm{2}} ,\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of}\:{y}. \\ $$

Question Number 160604 Answers: 0 Comments: 0

S_n = ((12)/((4^2 −3^2 )(4^2 −3^2 )))+((12^2 )/((4^2 −3^2 )(4^3 −3^3 )))+((12^3 )/((4^2 −3^2 )(4^4 −3^4 )))+...+((12^n )/((4^2 −3^2 )(4^(n+1) −3^(n+1) ))) lim_(n→∞) S_n = ?

Question Number 160603 Answers: 1 Comments: 0

(x+(√(x^2 +1)))(y+(√(y^2 +1)))=2021 ∀x,y∈R^+ . min (x+y)=?

$$\:\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(\mathrm{y}+\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\right)=\mathrm{2021} \\ $$$$\:\forall\mathrm{x},\mathrm{y}\in\mathbb{R}^{+} \:.\:\mathrm{min}\:\left(\mathrm{x}+\mathrm{y}\right)=? \\ $$

Question Number 160597 Answers: 1 Comments: 0

∫_0 ^( (π/2)) (dx/(2−cos x)) =?

$$\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{2}−\mathrm{cos}\:\mathrm{x}}\:=? \\ $$

Question Number 160596 Answers: 1 Comments: 0

Question Number 160594 Answers: 1 Comments: 0

∫ (dx/(1−tan^2 (x))) =?

$$\:\:\:\:\int\:\frac{\mathrm{dx}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:=? \\ $$

Question Number 160626 Answers: 0 Comments: 1

∫ ((lnz)/(z+lnz)) dz help me sir

$$\int\:\frac{{lnz}}{{z}+{lnz}}\:{dz}\: \\ $$$$ \\ $$$${help}\:{me}\:{sir} \\ $$

Question Number 160591 Answers: 0 Comments: 0

Question Number 160590 Answers: 0 Comments: 0

Ω:=∫_0 ^( (1/2)) (( arcsinh(x))/x) dx =^? (π^2 /(20))

$$ \\ $$$$\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \frac{\:{arcsinh}\left({x}\right)}{{x}}\:{dx}\:\overset{?} {=}\:\frac{\pi^{\mathrm{2}} }{\mathrm{20}} \\ $$

Question Number 160589 Answers: 0 Comments: 0

Question Number 160577 Answers: 1 Comments: 0

Ω=∫_0 ^( 1) tan^( −1) (x).ln(x) = ? −−−−solution−−−− f(a)=∫_0 ^( 1) tan^( −1) ( x) .x^( a) dx = Σ_(n=1) ^∞ (( (−1)^( n−1) )/(2n−1)) ∫_0 ^( 1) x^( 2n+a−1) dx = Σ_(n=0) ^∞ (((−1)^( n−1) )/(2n−1)) ((1/(2n+ a)) ) Ω= f ′ (a )∣_(a=0) = Σ_(n=1) ^∞ (((−1)^n )/((2n−1)( 2n+ a)^( 2) )) Ω= f ′ (0 )=(1/4)Σ(( (−1 )^(n−1) (2n−1−2n ) )/(( 2n−1)n^( 2) )) =(1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^( 2) ) − Σ_(n=1) ^∞ (((−1)^( n−1) )/((2n−1)(2n))) = (π^( 2) /(48)) −{ Σ_(n=1) ^∞ {(((−1)^(n−1) )/(2n−1)) −(((−1)^( n−1) )/(2n))}} ∴ Ω = (π^( 2) /(48)) − (π/4) +(1/2) ln(2)

$$ \\ $$$$\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} {tan}^{\:−\mathrm{1}} \:\left({x}\right).{ln}\left({x}\right)\:=\:? \\ $$$$\:\:\:\:\:−−−−{solution}−−−− \\ $$$$\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} {tan}^{\:−\mathrm{1}} \left(\:{x}\right)\:.{x}^{\:{a}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{n}+{a}−\mathrm{1}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\:{a}}\:\right) \\ $$$$\:\:\:\:\:\Omega=\:{f}\:'\:\left({a}\:\right)\mid_{{a}=\mathrm{0}} =\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\:\mathrm{2}{n}+\:{a}\right)^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\Omega=\:{f}\:'\:\left(\mathrm{0}\:\right)=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\:\left(−\mathrm{1}\:\right)^{{n}−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}−\mathrm{2}{n}\:\right)\:}{\left(\:\mathrm{2}{n}−\mathrm{1}\right){n}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\:\mathrm{2}} }\:−\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)} \\ $$$$\:\:\:\:\:\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{48}}\:−\left\{\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}\:−\frac{\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} }{\mathrm{2}{n}}\right\}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\Omega\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{48}}\:−\:\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

Question Number 160576 Answers: 1 Comments: 2

Question Number 160580 Answers: 0 Comments: 0

Calculate 1. lim_(x→+∞) ((x(√(ln (x^2 +1))))/(1+e^(x−3) )) 2. lim_(x→+∞) (((x^3 +5)/(x^2 +2)))^((x+1)/(x^2 +1))

$${Calculate}\: \\ $$$$\mathrm{1}.\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{1}+{e}^{{x}−\mathrm{3}} } \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{3}} +\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$

Question Number 160569 Answers: 1 Comments: 0

Calculate 1. lim_(x→0) [2e^(x/(x+1)) −1]^((x^2 +1)/x) 2. lim_(x→0) (((1+x×2^x )/(1+x×3^x )))^(1/x^2 )

$${Calculate}\: \\ $$$$\mathrm{1}.\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{2}{e}^{\frac{{x}}{{x}+\mathrm{1}}} −\mathrm{1}\right]^{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}} \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{x}×\mathrm{2}^{{x}} }{\mathrm{1}+{x}×\mathrm{3}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$ \\ $$

Question Number 160564 Answers: 3 Comments: 2

if the roots of the equation ax^2 +bx+c=0 are in the ratio 3:4,then show that 12b^2 =49ac.

$${if}\:{the}\:{roots}\:{of}\:{th}\mathrm{e}\:{equation}\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0} \\ $$$${are}\:{in}\:{the}\:{ratio}\:\mathrm{3}:\mathrm{4},{then}\:{show}\:{that}\: \\ $$$$\mathrm{12b}^{\mathrm{2}} =\mathrm{49ac}. \\ $$

Question Number 160563 Answers: 1 Comments: 0

∫x{x}[x]dx=?

$$\int\mathrm{x}\left\{\mathrm{x}\right\}\left[\mathrm{x}\right]\mathrm{dx}=? \\ $$

Pg 557 Pg 558 Pg 559 Pg 560 Pg 561 Pg 562 Pg 563 Pg 564 Pg 565 Pg 566

Contact: info@tinkutara.com