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Question Number 143077 Answers: 1 Comments: 0

sin^5 x + cos^5 x = 2 − sin^4 x

$${sin}^{\mathrm{5}} {x}\:+\:{cos}^{\mathrm{5}} {x}\:=\:\mathrm{2}\:−\:{sin}^{\mathrm{4}} {x} \\ $$

Question Number 143072 Answers: 1 Comments: 0

Question Number 143071 Answers: 2 Comments: 0

∫_0 ^(π/4) ((8dx)/(tgx+1))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{8}{dx}}{{tgx}+\mathrm{1}} \\ $$

Question Number 143064 Answers: 4 Comments: 0

lim_(x→0) ((((1+x^2 ))^(1/3) −((1−2x))^(1/4) )/(x+x^2 )) =? lim_(x→1) ((((7+x^2 ))^(1/3) −(√(3+x^2 )))/(x−1)) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:−\sqrt[{\mathrm{4}}]{\mathrm{1}−\mathrm{2x}}}{\mathrm{x}+\mathrm{x}^{\mathrm{2}} }\:=? \\ $$$$\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{7}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{3}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}−\mathrm{1}}\:=? \\ $$

Question Number 143063 Answers: 0 Comments: 0

cos((𝛑/(2n+1)))cos(((2𝛑)/(2n+1)))cos(((3𝛑)/(2n+1))).....cos(((n𝛑)/(2n+1)))=(1/2^n ) prove

$$\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right).....\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{n}\pi}}{\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$

Question Number 143057 Answers: 1 Comments: 0

cos(𝛂)×cos(2α)×cos(4α)×....×cos(2^n 𝛂)=((sin(2^(n+1) 𝛂))/(2^(n+1) sin(α))) prove

$$\mathrm{cos}\left(\boldsymbol{\alpha}\right)×\mathrm{cos}\left(\mathrm{2}\alpha\right)×\mathrm{cos}\left(\mathrm{4}\alpha\right)×....×\mathrm{cos}\left(\mathrm{2}^{\mathrm{n}} \boldsymbol{\alpha}\right)=\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \boldsymbol{\alpha}\right)}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \mathrm{sin}\left(\alpha\right)} \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$

Question Number 143048 Answers: 1 Comments: 0

Question Number 143047 Answers: 2 Comments: 0

Question Number 143046 Answers: 1 Comments: 0

Question Number 143045 Answers: 1 Comments: 0

Question Number 143043 Answers: 1 Comments: 0

Question Number 143042 Answers: 1 Comments: 0

Question Number 143041 Answers: 0 Comments: 2

If f(x)=x^2 +4x+2 then the value of (1−(2/(f(1))))(1−(2/(f(2))))(1−(2/(f(3))))...(1−(2/(f(2021))))=?

$${If}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\:{then}\:{the}\:{value} \\ $$$${of}\:\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{1}\right)}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{2}\right)}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{3}\right)}\right)...\left(\mathrm{1}−\frac{\mathrm{2}}{{f}\left(\mathrm{2021}\right)}\right)=? \\ $$

Question Number 143039 Answers: 0 Comments: 0

cos^(−1) (((x^2 −1)/(x^2 +1)))+(1/2)tan^(−1) (((2x)/(1−x^2 )))=((2π)/3) x=?

$$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${x}=? \\ $$

Question Number 143051 Answers: 1 Comments: 0

_(∗∗∗∗∗) :: Lobachevsky Integral ::_(∗∗∗∗∗) 𝛗:=∫_0 ^( ∞) ((sin^2 ( tan(x)))/x^( 2) )dx=^? (π/2) ..........

$$\:\:\:\:\:\:\:_{\ast\ast\ast\ast\ast} ::\:\:{Lobachevsky}\:{Integral}\:::_{\ast\ast\ast\ast\ast} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{s}{in}^{\mathrm{2}} \left(\:{tan}\left({x}\right)\right)}{{x}^{\:\mathrm{2}} }{dx}\overset{?} {=}\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:.......... \\ $$

Question Number 143036 Answers: 0 Comments: 0

x^(3/2) +x^(1/2) +(x−c)(((3x+1)/(3+x)))^(3/2) =0

$$\:{x}^{\mathrm{3}/\mathrm{2}} +{x}^{\mathrm{1}/\mathrm{2}} +\left({x}−{c}\right)\left(\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{3}+{x}}\right)^{\mathrm{3}/\mathrm{2}} =\mathrm{0} \\ $$

Question Number 143032 Answers: 1 Comments: 0

∫ sin^(−5) x dx =?

$$\:\:\:\:\:\int\:\mathrm{sin}^{−\mathrm{5}} {x}\:{dx}\:=? \\ $$

Question Number 143031 Answers: 2 Comments: 0

Question Number 143027 Answers: 1 Comments: 1

Question Number 143024 Answers: 1 Comments: 0

Question Number 143013 Answers: 0 Comments: 3

lim_(x→∞) (((x+1))^(1/3) /(x+1))=?

$${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$

Question Number 143012 Answers: 1 Comments: 1

lim_(α→∞) (e^x /α^(60!) )=?

$${li}\underset{\alpha\rightarrow\infty} {{m}}\frac{{e}^{{x}} }{\alpha^{\mathrm{60}!} }=? \\ $$

Question Number 143018 Answers: 1 Comments: 0

Question Number 143011 Answers: 0 Comments: 0

x^4 +bx^2 +cx=s let x^2 =px+t ⇒ p^2 x^2 +2ptx+t^2 +bx^2 +cx=s ⇒ (p^2 +b)x^2 +(2pt+c)x =s−t^2 ⇒ (p^2 +b)(px+t)+(2pt+c)x =s−t^2 ⇒ p(p^2 +b)+2pt+c=0 and (p^2 +b)t=s−t^2 ((s/t)−t−b)((s/t)+t)^2 =c^2 ⇒ (A−b)(A^2 +4s)=c^2 ⇒ A^3 −bA^2 +4sA−4bs−c^2 =0 let A=z+(b/3) ⇒ z^3 +(4s−(b^2 /3))z−(((2b^3 )/(27))+((8bs)/3)+c^2 )=0 D=((b^3 /(27))+((4bs)/3)+(c^2 /2))^2 −((b^2 /9)−((4s)/3))^3 If s=0, b=−1, c→−c then D=(−(1/(27))+(c^2 /2))^2 −((1/9))^3 ...

$$\:\:{x}^{\mathrm{4}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$${let}\:\:{x}^{\mathrm{2}} ={px}+{t} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{ptx}+{t}^{\mathrm{2}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{pt}+{c}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:\:={s}−{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right)\left({px}+{t}\right)+\left(\mathrm{2}{pt}+{c}\right){x} \\ $$$$\:\:\:\:\:\:\:={s}−{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{p}\left({p}^{\mathrm{2}} +{b}\right)+\mathrm{2}{pt}+{c}=\mathrm{0} \\ $$$${and}\:\:\left({p}^{\mathrm{2}} +{b}\right){t}={s}−{t}^{\mathrm{2}} \\ $$$$\left(\frac{{s}}{{t}}−{t}−{b}\right)\left(\frac{{s}}{{t}}+{t}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({A}−{b}\right)\left({A}^{\mathrm{2}} +\mathrm{4}{s}\right)={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:{A}^{\mathrm{3}} −{bA}^{\mathrm{2}} +\mathrm{4}{sA}−\mathrm{4}{bs}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:{A}={z}+\frac{{b}}{\mathrm{3}}\:\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}{s}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\right){z}−\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}{bs}}{\mathrm{3}}+{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${D}=\left(\frac{{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{4}{bs}}{\mathrm{3}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{4}{s}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$${If}\:\:{s}=\mathrm{0},\:{b}=−\mathrm{1},\:{c}\rightarrow−{c}\:\: \\ $$$${then}\:{D}=\left(−\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{3}} \\ $$$$... \\ $$

Question Number 143008 Answers: 1 Comments: 1

Solve : x=p^3 −p+2 , y^′ =p

$${Solve}\::\:\:{x}={p}^{\mathrm{3}} −{p}+\mathrm{2}\:\:,\:{y}^{'} ={p} \\ $$

Question Number 143006 Answers: 0 Comments: 0

1. y(∂z/∂x) + z(∂z/∂y) = (y/x) 2. x^2 (∂z/∂x) − xy(∂z/∂y) + y^2 = 0 3. { (((∂z/∂x) = (z/x))),(((∂z/∂y) = ((2z)/y))) :}

$$\mathrm{1}.\:{y}\frac{\partial{z}}{\partial{x}}\:+\:{z}\frac{\partial{z}}{\partial{y}}\:=\:\frac{{y}}{{x}} \\ $$$$\mathrm{2}.\:{x}^{\mathrm{2}} \frac{\partial{z}}{\partial{x}}\:−\:{xy}\frac{\partial{z}}{\partial{y}}\:+\:{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{3}.\:\begin{cases}{\frac{\partial{z}}{\partial{x}}\:=\:\frac{{z}}{{x}}}\\{\frac{\partial{z}}{\partial{y}}\:=\:\frac{\mathrm{2}{z}}{{y}}}\end{cases} \\ $$

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