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Question Number 201629 Answers: 0 Comments: 3

Question Number 201627 Answers: 1 Comments: 0

Question Number 201657 Answers: 1 Comments: 0

if f(x)= { ((((sin (1+[x]))/([x])) for [x]≠0)),((0 for [x]=0)) :} where [x] represents an integer x greatest ≤ x Find lim_(x→0^− ) f(x).

$${if}\:\:{f}\left({x}\right)=\begin{cases}{\frac{\mathrm{sin}\:\left(\mathrm{1}+\left[{x}\right]\right)}{\left[{x}\right]}\:\:{for}\:\left[{x}\right]\neq\mathrm{0}}\\{\mathrm{0}\:\:{for}\:\left[{x}\right]=\mathrm{0}}\end{cases} \\ $$$${where}\:\left[{x}\right]\:{represents}\:{an}\:{integer}\:\boldsymbol{{x}}\:{greatest}\:\leqslant\:\boldsymbol{{x}} \\ $$$${Find}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right). \\ $$

Question Number 201615 Answers: 1 Comments: 0

x,y,z ∈ R a,b,c>0 prove that: (x^2 /a) + (y^2 /b) + (z^2 /c) ≥ (((x + y + z)^2 )/(a + b + c))

$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{c}}\:\geqslant\:\frac{\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)^{\mathrm{2}} }{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}} \\ $$

Question Number 201613 Answers: 2 Comments: 0

x,y,z ∈ R { ((xy + yz + zx = 3)),((x + y + z = 5)) :} → max(z) = ?

$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}\:=\:\mathrm{3}}\\{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{5}}\end{cases}\:\:\:\:\:\rightarrow\:\:\:\:\mathrm{max}\left(\boldsymbol{\mathrm{z}}\right)\:=\:? \\ $$

Question Number 201604 Answers: 2 Comments: 1

Question Number 201599 Answers: 1 Comments: 2

Question Number 201595 Answers: 3 Comments: 0

1) ∣((3+2x)/(3x))∣ ≤1 2) 1≤ ∣ ((x−3)/(1−2x))∣≤ 2 3) ((x^2 +2x−35)/(x+2)) > 0 4) −1 ≤ ((x+1)/(x−2)) ≤2

$$\left.\mathrm{1}\right)\:\:\mid\frac{\mathrm{3}+\mathrm{2}{x}}{\mathrm{3}{x}}\mid\:\leq\mathrm{1} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{1}\leq\:\mid\:\frac{{x}−\mathrm{3}}{\mathrm{1}−\mathrm{2}{x}}\mid\leq\:\mathrm{2} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{35}}{{x}+\mathrm{2}}\:>\:\mathrm{0} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\:−\mathrm{1}\:\leq\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}\:\leq\mathrm{2} \\ $$

Question Number 201582 Answers: 0 Comments: 0

Solve.... y′′(t)−sin(t)y(t)=0 , y^((2)) (0)=0 , y^((1)) (0)=−1 , y(0)=0 L{y′′(t)−sin(t)y(t)}=0 s^2 F(s)−sy(0)−y′(0)−L{sin(t)y(t)}=0 Holy...×uck I already know y′′(t)−ty(t)=0 solution C_1 Ai(t)+C_2 Bi(t) But I Can′t Solve y′′(t)−sin(t)y(t)=0....

$$\mathrm{Solve}.... \\ $$$${y}''\left({t}\right)−\mathrm{sin}\left({t}\right){y}\left({t}\right)=\mathrm{0}\:,\: \\ $$$${y}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{0}\:,\:{y}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)=−\mathrm{1}\:,\:{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{\mathcal{L}}\left\{{y}''\left({t}\right)−\mathrm{sin}\left({t}\right){y}\left({t}\right)\right\}=\mathrm{0} \\ $$$${s}^{\mathrm{2}} \boldsymbol{\mathrm{F}}\left({s}\right)−{sy}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)−\boldsymbol{\mathcal{L}}\left\{\mathrm{sin}\left({t}\right){y}\left({t}\right)\right\}=\mathrm{0} \\ $$$$\mathrm{Holy}...×\mathrm{uck} \\ $$$$\mathrm{I}\:\mathrm{already}\:\mathrm{know}\:{y}''\left({t}\right)−{ty}\left({t}\right)=\mathrm{0}\:\:\:\mathrm{solution} \\ $$$$\mathrm{C}_{\mathrm{1}} \mathrm{Ai}\left({t}\right)+{C}_{\mathrm{2}} \mathrm{Bi}\left({t}\right) \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{Can}'\mathrm{t}\:\mathrm{Solve}\:{y}''\left({t}\right)−\mathrm{sin}\left({t}\right){y}\left({t}\right)=\mathrm{0}....\: \\ $$

Question Number 201581 Answers: 1 Comments: 0