Question Number 207519 by hardmath last updated on 17/May/24 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\:\frac{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mid\mathrm{x}\:+\:\mathrm{2}\mid}\:\:=\:\:? \\ $$ Answered by Frix last updated on 17/May/24 $${f}\left({x}\right)=\frac{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)}{\mid{x}+\mathrm{2}\mid}=\begin{cases}{−\left({x}+\mathrm{1}\right),\:{x}<−\mathrm{2}}\\{\left({x}+\mathrm{1}\right),\:−\mathrm{2}<{x}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{approach}\:\mathrm{2}\:\mathrm{from}\:\mathrm{the}\:\mathrm{negative}\:\mathrm{side}\:\mathrm{but} \\…
Question Number 207486 by hardmath last updated on 17/May/24 Commented by som(math1967) last updated on 17/May/24 $${DE}\bot{AK}\:? \\ $$ Commented by hardmath last updated on…
Question Number 207487 by hardmath last updated on 17/May/24 $$\mid\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:−\:\mathrm{4}\mid\:=\:\mid\mathrm{x}\:−\:\mathrm{4}\mid \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{min}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{max}}\:\:=\:\:? \\ $$ Answered by A5T last updated on 17/May/24 $$\mid{x}−\mathrm{4}\mid\mid{x}+\mathrm{1}\mid=\mid{x}−\mathrm{4}\mid\Rightarrow\mid{x}−\mathrm{4}\mid=\mathrm{0}\:{or}\:\mid{x}+\mathrm{1}\mid=\mathrm{1} \\ $$$$\Rightarrow{x}−\mathrm{4}=\mathrm{0}\:{or}\:{x}+\mathrm{1}=\mathrm{1}\:{or}\:{x}+\mathrm{1}=−\mathrm{1}…
Question Number 207498 by hardmath last updated on 17/May/24 $$\mathrm{cos2}\boldsymbol{\mathrm{x}}\:+\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:=\:\mathrm{tg}\left(\mathrm{225}°\right)\centerdot\left(\mathrm{0},\mathrm{360}°\right) \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}\:=\:? \\ $$ Answered by MM42 last updated on 17/May/24 $$−\mathrm{2}{sin}^{\mathrm{2}} {x}+{sinx}=\mathrm{0} \\ $$$${sinx}=\mathrm{0}\Rightarrow{x}=\pi…
Question Number 207502 by hardmath last updated on 17/May/24 $$\frac{\mathrm{6}}{\mid\boldsymbol{\mathrm{x}}\:−\:\mathrm{4}\mid\:−\:\mathrm{3}}\:\:\geqslant\:\:\mathrm{1} \\ $$ Answered by efronzo1 last updated on 17/May/24 $$\:\:\:\mathrm{Let}\:\mid\mathrm{x}−\mathrm{4}\mid\:=\:\mathrm{y}\:\geqslant\mathrm{0} \\ $$$$\:\Rightarrow\frac{\mathrm{6}}{\mathrm{y}−\mathrm{3}}\:\geqslant\mathrm{1}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{6}−\left(\mathrm{y}−\mathrm{3}\right)}{\mathrm{y}−\mathrm{3}}\:\geqslant\:\mathrm{0}\: \\…
Question Number 207493 by hardmath last updated on 17/May/24 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{n}\:−\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{3}} \:=\:\:? \\ $$ Answered by A5T last updated on 17/May/24 $$\left(\frac{{n}+\mathrm{2}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\left(\frac{{n}−\mathrm{1}+\mathrm{3}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\left(\mathrm{1}+\frac{\mathrm{3}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} \\…
Question Number 207477 by AliJumaa last updated on 16/May/24 $${is}\:{there}\:{any}\:{generale}\:{form}\:{for}\:{this}\:{sequense}\: \\ $$$$\begin{cases}{{u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}}\\{{u}_{{m}} ={k}}\end{cases} \\ $$$${I}\:{need}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n}\:{i}\:{have}\:{try}\:{to}\:{derrive}\:{it}\:{for}\:{a}\:{long}\:{time}\:{but}\:{i}\:{cant} \\ $$ Commented by Frix last…
Question Number 207462 by hardmath last updated on 16/May/24 $$\mathrm{4}\:\mathrm{sin}\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\:=\:\mathrm{1}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by Frix last updated on 16/May/24 $${x}=\mathrm{4}{n}\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{4}}\:\vee{x}=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi−\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms…
Question Number 207463 by hardmath last updated on 16/May/24 $$\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{x}\:=\:\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 16/May/24 $$\mathrm{cos}^{\mathrm{2}} \:{x}\:=\sqrt{\mathrm{2}}>\mathrm{1}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$ Terms…
Question Number 207450 by York12 last updated on 15/May/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:{m}\:\mathrm{and}\:{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{following}\:\:\mathrm{holds} \\ $$$$\:\frac{{d}\left({y}\right)}{{d}\left({x}\right)}\mid_{{x}={n}} =\left(\frac{{d}\left({x}\right)}{{d}\left({y}\right)}\mid_{{y}={m}} \right)^{−\mathrm{1}} \\ $$ Answered by mr W last updated on 15/May/24 Commented…