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Question Number 225047    Answers: 2   Comments: 0

∫ x^x dx

$$\int\:{x}^{{x}} \:{dx} \\ $$

Question Number 223735    Answers: 0   Comments: 0

for all n ∈ Z , Show that τ ( ϕ ( n )) ≥ ϕ (τ (n ))

$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\:{n}\:\in\:\mathbb{Z}\:, \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{Show}}\:\boldsymbol{\mathrm{that}}\:\tau\:\left(\:\varphi\:\left(\:{n}\:\right)\right)\:\geqslant\:\varphi\:\left(\tau\:\left({n}\:\right)\right) \\ $$$$ \\ $$

Question Number 223006    Answers: 1   Comments: 0

L { tsin((√t) )}=?

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathscr{L}\:\:\left\{\:{tsin}\left(\sqrt{{t}}\:\right)\right\}=? \\ $$

Question Number 222778    Answers: 1   Comments: 0

lim_(x→0) ((2log(1+x)−((x(3x+2))/((x+1)^2 )))/x^3 )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2log}\left(\mathrm{1}+\mathrm{x}\right)−\frac{\mathrm{x}\left(\mathrm{3x}+\mathrm{2}\right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{3}} } \\ $$

Question Number 222601    Answers: 0   Comments: 1

L 60 1. y=x^2 +5x Find the equation of a line with the slope of 7 that touches y=x^2 +5x. [Sol.] Let f(x)=x^2 +5x Then f′(x)=2x+5 Since 2x+5=7⇒x=1 then the point is (1, 1^2 +5∙1)=(1, 6) So the equation of a line is y−6=7(x−1)⇒y=7x−1 2. y=ax^2 +bx (2, 2) a, b Find the values of constants a, b that the slope of the line that touches (2, 2) and y=ax^2 +bx is 5. [Sol.] Let f(x)=ax^2 +bx Then f′(x)=2ax+b and build two equations to solve for a and b { ((f(2)=a∙2^2 +b∙2=4a+2b=2)),((f′(2)=2a∙2+b=4a+b=5)) :} Solving for a, b gives a=2, b=−3 3. y=x^3 −3x^2 −1 Find the equation of a line that is drawn, touches y=x^3 −3x^2 −1. [Sol.] The line of the equation is y−(a^3 −3a^2 −1)=(3a^2 −6a)(x−a) Calculating gives y=(3a^2 −6a)x−(3a^2 −6a)a+(a^3 −3a^2 −1) y=(3a^2 −6a)x+(−3a^3 +6a^2 )+(a^3 −3a^2 −1) y=(3a^2 −6a)x+(−2a^3 +3a^2 −1) −2a^3 +3a^2 −1=0 a=−(1/2) or a=2 ...a=−3x, a=((15)/4)x

$$\mathrm{L}\:\mathrm{60} \\ $$$$ \\ $$$$\mathrm{1}.\: {y}={x}^{\mathrm{2}} +\mathrm{5}{x} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{with}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{7}\:\mathrm{that}\:\mathrm{touches}\:{y}={x}^{\mathrm{2}} +\mathrm{5}{x}. \\ $$$$\left[\mathrm{Sol}.\right]\:\mathrm{Let}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{5}{x}\:\mathrm{Then}\:{f}'\left({x}\right)=\mathrm{2}{x}+\mathrm{5} \\ $$$$\mathrm{Since}\:\mathrm{2}{x}+\mathrm{5}=\mathrm{7}\Rightarrow{x}=\mathrm{1}\:\mathrm{then}\:\mathrm{the}\:\mathrm{point}\:\mathrm{is}\:\left(\mathrm{1},\:\mathrm{1}^{\mathrm{2}} +\mathrm{5}\centerdot\mathrm{1}\right)=\left(\mathrm{1},\:\mathrm{6}\right) \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{is}\:{y}−\mathrm{6}=\mathrm{7}\left({x}−\mathrm{1}\right)\Rightarrow{y}=\mathrm{7}{x}−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{2}.\: \:{y}={ax}^{\mathrm{2}} +{bx}\: \:\left(\mathrm{2},\:\mathrm{2}\right) \:{a},\:{b} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{constants}\:{a},\:{b}\:\mathrm{that}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{that}\:\mathrm{touches}\:\left(\mathrm{2},\:\mathrm{2}\right)\:\mathrm{and}\:{y}={ax}^{\mathrm{2}} +{bx}\:\mathrm{is}\:\mathrm{5}. \\ $$$$\left[\mathrm{Sol}.\right]\:\mathrm{Let}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}\:\mathrm{Then}\:{f}'\left({x}\right)=\mathrm{2}{ax}+{b}\:\mathrm{and}\:\mathrm{build}\:\mathrm{two}\:\mathrm{equations}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{a}\:\mathrm{and}\:{b} \\ $$$$\begin{cases}{{f}\left(\mathrm{2}\right)={a}\centerdot\mathrm{2}^{\mathrm{2}} +{b}\centerdot\mathrm{2}=\mathrm{4}{a}+\mathrm{2}{b}=\mathrm{2}}\\{{f}'\left(\mathrm{2}\right)=\mathrm{2}{a}\centerdot\mathrm{2}+{b}=\mathrm{4}{a}+{b}=\mathrm{5}}\end{cases} \\ $$$$\mathrm{Solving}\:\mathrm{for}\:{a},\:{b}\:\mathrm{gives}\:{a}=\mathrm{2},\:{b}=−\mathrm{3} \\ $$$$ \\ $$$$\mathrm{3}.\: \:{y}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{that}\:\mathrm{is}\:\mathrm{drawn},\:\mathrm{touches}\:{y}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}. \\ $$$$\left[\mathrm{Sol}.\right]\:\mathrm{The}\:\mathrm{line}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:{y}−\left({a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right)=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right)\left({x}−{a}\right) \\ $$$$\mathrm{Calculating}\:\mathrm{gives}\:{y}=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){x}−\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){a}+\left({a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${y}=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){x}+\left(−\mathrm{3}{a}^{\mathrm{3}} +\mathrm{6}{a}^{\mathrm{2}} \right)+\left({a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${y}=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){x}+\left(−\mathrm{2}{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$−\mathrm{2}{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:{a}=\mathrm{2} \\ $$$$...{a}=−\mathrm{3}{x},\:{a}=\frac{\mathrm{15}}{\mathrm{4}}{x} \\ $$

Question Number 222427    Answers: 1   Comments: 0

if lim_(x→0) (((sin2x)/x^3 )+(a/x^2 )+b)=1 find a and b without using LHopial rule

$$\:\:\boldsymbol{{if}}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{3}} }+\frac{\boldsymbol{{a}}}{\boldsymbol{{x}}^{\mathrm{2}} }+\boldsymbol{{b}}\right)=\mathrm{1}\: \\ $$$$\:\:\:\:\boldsymbol{{find}}\:\boldsymbol{{a}}\:\boldsymbol{{and}}\:\boldsymbol{{b}}\:\:\boldsymbol{{without}} \\ $$$$\:\:\:\:\:\:\boldsymbol{{using}}\:\boldsymbol{{LH}}{opial}\:{rule} \\ $$

Question Number 222389    Answers: 0   Comments: 0

Prove: ∫_0 ^∞ ((cos(nx)cos(p arctan x))/((1+x^2 )^(p/2) ))=(π/2) ((n^(p−1) e^(−n) )/(Γ(p))) (p>0)

$$\mathrm{Prove}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left({nx}\right)\mathrm{cos}\left({p}\:\mathrm{arctan}\:{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{{p}}{\mathrm{2}}} }=\frac{\pi}{\mathrm{2}}\:\frac{{n}^{{p}−\mathrm{1}} {e}^{−{n}} }{\Gamma\left({p}\right)}\:\left({p}>\mathrm{0}\right) \\ $$

Question Number 222329    Answers: 0   Comments: 1

lim_(x→∞) 4x+(√(16x^2 −3x))

$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\infty} \:\mathrm{4}\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$

Question Number 222284    Answers: 1   Comments: 0

y=(8^x /((in8)^3 )) find (d^6 y/dx^6 )

$$\boldsymbol{\mathrm{y}}=\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} }{\left(\boldsymbol{\mathrm{in}}\mathrm{8}\right)^{\mathrm{3}} } \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{6}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{6}} } \\ $$

Question Number 222113    Answers: 0   Comments: 2

name the following compound

$$\mathrm{name}\:\mathrm{the}\:\mathrm{following}\:\mathrm{compound} \\ $$

Question Number 221044    Answers: 0   Comments: 1

If V be a function of x and y, prove that (∂^2 V/∂x^2 )+(∂^2 V/∂y^2 )=(∂^2 V/∂r^2 )+(1/r) (∂V/∂r)+(1/r^2 ) (∂^2 V/∂θ^2 ), where x=r cos θ , y=rsin θ

$${If}\:{V}\:{be}\:{a}\:{function}\:{of}\:{x}\:{and}\:{y},\:{prove}\:{that} \\ $$$$\frac{\partial^{\mathrm{2}} {V}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {V}}{\partial{y}^{\mathrm{2}} }=\frac{\partial^{\mathrm{2}} {V}}{\partial{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\:\frac{\partial{V}}{\partial{r}}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial^{\mathrm{2}} {V}}{\partial\theta^{\mathrm{2}} }, \\ $$$${where}\:{x}={r}\:\mathrm{cos}\:\theta\:,\:{y}={r}\mathrm{sin}\:\theta \\ $$

Question Number 220964    Answers: 0   Comments: 0

Find the general solution of the differential equation x^2 (d^3 y/dx^3 ) + x(d^2 y/dx^2 )−6(dy/dx)+6(y/x)=((x ln x+1)/x^2 ),[x>0]

$${Find}\:{the}\:{general}\:{solution}\:{of}\:{the}\:{differential}\:{equation} \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:{x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{6}\frac{{dy}}{{dx}}+\mathrm{6}\frac{{y}}{{x}}=\frac{{x}\:\mathrm{ln}\:{x}+\mathrm{1}}{{x}^{\mathrm{2}} },\left[{x}>\mathrm{0}\right] \\ $$

Question Number 220863    Answers: 1   Comments: 0

(211) Find the derivative of Δx, where Δx= determinant (((f_1 (x)),(φ_1 (x)),(Ψ_1 (x))),((f_2 (x)),(φ_2 (x)),(Ψ_2 (x))),((f_3 (x)),(φ_3 (x)),(Ψ_3 (x)))) and f_1 (x) ,f_2 (x), f_3 (x),φ_1 (x), etc. are different functions of x.

$$\left(\mathrm{211}\right) \\ $$$$\:\: \\ $$$${Find}\:{the}\:{derivative}\:{of}\:\Delta{x},\:{where} \\ $$$$\Delta{x}=\begin{vmatrix}{{f}_{\mathrm{1}} \left({x}\right)}&{\phi_{\mathrm{1}} \left({x}\right)}&{\Psi_{\mathrm{1}} \left({x}\right)}\\{{f}_{\mathrm{2}} \left({x}\right)}&{\phi_{\mathrm{2}} \left({x}\right)}&{\Psi_{\mathrm{2}} \left({x}\right)}\\{{f}_{\mathrm{3}} \left({x}\right)}&{\phi_{\mathrm{3}} \left({x}\right)}&{\Psi_{\mathrm{3}} \left({x}\right)}\end{vmatrix} \\ $$$${and}\:{f}_{\mathrm{1}} \left({x}\right)\:,{f}_{\mathrm{2}} \left({x}\right),\:{f}_{\mathrm{3}} \left({x}\right),\phi_{\mathrm{1}} \left({x}\right),\:{etc}.\:{are}\:{different}\:{functions}\:{of}\:{x}. \\ $$

Question Number 220232    Answers: 1   Comments: 0

prove that (π/(16)) < ∫_0 ^( 1 ) (√((x(1−x))/(sin(πx)+cos(πx)+2))) dx<(π/8)

$$ \\ $$$$\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:\:\frac{\pi}{\mathrm{16}}\:<\:\int_{\mathrm{0}} ^{\:\mathrm{1}\:} \sqrt{\frac{{x}\left(\mathrm{1}−{x}\right)}{{sin}\left(\pi{x}\right)+{cos}\left(\pi{x}\right)+\mathrm{2}}}\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\: \\ $$

Question Number 220131    Answers: 2   Comments: 0

If f(x,y)=(((x^2 +y^2 )^n )/(2n(2n−1)))+xφ((y/x))+Ψ((y/x)), then using Euler′s theorem on homogenous functions,show that x^2 ((δ^2 f)/(δx^2 ))+2xy((δ^2 f)/(δxδy))+y^2 ((δ^2 f)/(δy^2 ))=(x^2 +y^2 )^n

$${If}\:\:\:{f}\left({x},{y}\right)=\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} }{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}+{x}\phi\left(\frac{{y}}{{x}}\right)+\Psi\left(\frac{{y}}{{x}}\right), \\ $$$${then}\:{using}\:{Euler}'{s}\:{theorem}\:{on}\:{homogenous}\:{functions},{show}\:{that} \\ $$$${x}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{x}^{\mathrm{2}} }+\mathrm{2}{xy}\frac{\delta^{\mathrm{2}} {f}}{\delta{x}\delta{y}}+{y}^{\mathrm{2}} \frac{\delta^{\mathrm{2}} {f}}{\delta{y}^{\mathrm{2}} }=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{{n}} \\ $$

Question Number 219868    Answers: 1   Comments: 0

Prove that; (d/dx) (((sin^( 2) x)/(1+cot x)) + ((cos^( 2) x)/(1+tan x))) = −cos 2x

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{{d}}{{dx}}\:\left(\frac{\mathrm{sin}^{\:\mathrm{2}} {x}}{\mathrm{1}+\mathrm{cot}\:{x}}\:+\:\frac{\mathrm{cos}^{\:\mathrm{2}} {x}}{\mathrm{1}+\mathrm{tan}\:{x}}\right)\:=\:−\mathrm{cos}\:\mathrm{2}{x}\:\:\:\: \\ $$$$ \\ $$

Question Number 219451    Answers: 0   Comments: 1

Question Number 219243    Answers: 3   Comments: 2

Question Number 219098    Answers: 2   Comments: 0

ζ(α)=Σ_(n=1) ^(+∞) (1/n^α )

$$\zeta\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }\:\: \\ $$

Question Number 216800    Answers: 1   Comments: 0

Question Number 216694    Answers: 4   Comments: 0

Prove that ^3 (√((√5)+2)) −^3 (√((√5)−2)) =1

$${Prove}\:{that}\:\:^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:−^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\:=\mathrm{1} \\ $$

Question Number 216638    Answers: 1   Comments: 1

without using LHopital rule evalute lim_(x→0) ((ln(1−x)−sin(x) )/(1−cox^2 (x)))

$$\:\:\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{LHopital}} \\ $$$$\:\:\:\boldsymbol{{rule}}\:\boldsymbol{{evalute}}\: \\ $$$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}−{x}\right)−\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{cox}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)} \\ $$

Question Number 216411    Answers: 1   Comments: 0

(dx/dx)

$$\frac{{dx}}{{dx}} \\ $$

Question Number 215979    Answers: 0   Comments: 0

u_n = Σ_(k=n+1) ^(2n) (1/k) and v_n = Σ_(k=n) ^(2n−1) (1/k) • show that u_n and v_n are adjacent use ln(x+1) ≤ x and x≤−ln(1−x) and • show that u_n ≤ Σ_(k=n+1) ^(2n) (ln(k)−ln(k−1)) hence deduce that u_n ≤ ln2 • show that v_n ≥ Σ_(k=n) ^(2n−1) (ln(k+1)−ln(k)) hence deduce that v_n ≥ln2

$${u}_{{n}} \:=\:\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:{and}\:{v}_{{n}} \:=\:\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\bullet\:{show}\:{that}\:{u}_{{n}} \:{and}\:{v}_{{n}} \:{are}\:{adjacent} \\ $$$${use}\:{ln}\left({x}+\mathrm{1}\right)\:\leqslant\:{x}\:{and}\:{x}\leqslant−{ln}\left(\mathrm{1}−{x}\right)\:{and} \\ $$$$\bullet\:{show}\:{that}\:{u}_{{n}} \:\leqslant\:\underset{{k}={n}+\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left({ln}\left({k}\right)−{ln}\left({k}−\mathrm{1}\right)\right) \\ $$$${hence}\:{deduce}\:{that}\:{u}_{{n}} \:\leqslant\:{ln}\mathrm{2} \\ $$$$\bullet\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\underset{{k}={n}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left({ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right) \\ $$$${hence}\:{deduce}\:{that}\:{v}_{{n}} \geqslant{ln}\mathrm{2} \\ $$

Question Number 215958    Answers: 2   Comments: 0

Find the only function that satisfy the expression below: ((dy/dx))^2 = (d^2 y/dx^2 )

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{only}\:\mathrm{function}\:\mathrm{that}\:\mathrm{satisfy} \\ $$$$\mathrm{the}\:\mathrm{expression}\:\mathrm{below}: \\ $$$$\:\:\:\:\:\:\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:\:=\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} } \\ $$

Question Number 215528    Answers: 0   Comments: 2

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