Question Number 66599 by aliesam last updated on 17/Aug/19 Commented by kaivan.ahmadi last updated on 17/Aug/19 $$\left({a}−{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \Rightarrow \\…
Question Number 132135 by benjo_mathlover last updated on 11/Feb/21 Answered by Olaf last updated on 11/Feb/21 $${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}+{x}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{x}\:=\:\frac{{u}−\mathrm{1}}{{u}} \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\frac{{u}−\mathrm{1}}{{u}}−\mathrm{1}}{\frac{{u}−\mathrm{1}}{{u}}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}} \\ $$$${f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\::…
Question Number 132131 by Ar Brandon last updated on 11/Feb/21 Commented by Ar Brandon last updated on 11/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\: \\ $$$$\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{and}\:\mathrm{triangle}\:\mathrm{A}'\mathrm{B}'\mathrm{C}' \\ $$ Commented by…
Question Number 1057 by 123456 last updated on 25/May/15 $${f}:\mathbb{R}_{+} \rightarrow\mathbb{R} \\ $$$${x}={i}+{j} \\ $$$${x}\in\mathbb{R}_{+} \\ $$$${i}\in\mathbb{N} \\ $$$${j}\in\left[\mathrm{0},\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\begin{cases}{{f}\left({i}−\mathrm{1}\right)+\left({i}+\mathrm{1}\right)\left({j}+\mathrm{1}\right)}&{{x}\geqslant\mathrm{1}}\\{{j}}&{\mathrm{0}\leqslant{x}<\mathrm{1}}\\{{x}}&{{x}<\mathrm{0}}\end{cases} \\ $$$${f}\left(\mathrm{9}.\mathrm{5}\right)=? \\ $$…
Question Number 1056 by 123456 last updated on 25/May/15 $${y}''+{y}=\mathrm{8}{cos}\left({t}\right) \\ $$$${y}\left(\mathrm{0}\right)=+\mathrm{1} \\ $$$${y}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$ Commented by prakash jain last updated on 26/May/15 $$\mathrm{characterstic}\:\mathrm{equation}…
Question Number 1055 by 123456 last updated on 24/May/15 $$\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R} \\ $$$${g}:\left[\mathrm{0},\mathrm{1}\right]×\mathbb{N}\rightarrow\mathbb{R} \\ $$$${g}_{{n}} \left({x}\right)={f}\left[{g}_{{n}−\mathrm{1}} \left({x}\right)\right] \\ $$$${g}_{\mathrm{0}} \left({x}\right)={x} \\ $$$$\mathscr{A}\left\{{f}\right\}\left({n}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({t}\right){g}_{{n}} \left({t}\right){dt} \\…
Question Number 132124 by Chhing last updated on 11/Feb/21 $$ \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{tan}\left(\mathrm{x}\right)+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{tan}\left(\mathrm{x}\right)−\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}}\:\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\: \\ $$ Commented by liberty…
Question Number 1054 by 123456 last updated on 24/May/15 $$\mathrm{L}\frac{{di}}{{dt}}+\mathrm{R}{i}+\frac{\mathrm{1}}{{C}}\underset{\mathrm{0}} {\overset{{t}} {\int}}{idt}=\mathrm{E} \\ $$$${i}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 66589 by Tanmay chaudhury last updated on 17/Aug/19 $$\int\frac{{sinx}}{\mathrm{1}+{sinx}+{sin}\mathrm{2}{x}}{dx} \\ $$ Commented by MJS last updated on 17/Aug/19 $$\mathrm{Weierstrass}\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right]\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{4}\int\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}}…
Question Number 132121 by abdullahquwatan last updated on 11/Feb/21 $$\int_{\mathrm{2}} ^{\mathrm{8}} \frac{\sqrt{{x}}}{\:\sqrt{\mathrm{10}−{x}}\:+\sqrt{{x}}}\:\mathrm{dx} \\ $$ Commented by Dwaipayan Shikari last updated on 11/Feb/21 $$\mathrm{3} \\ $$…