Question Number 191926 by Rupesh123 last updated on 04/May/23 Answered by AST last updated on 04/May/23 $$\left[{x}\right]+\left\{{x}\right\}={x} \\ $$$$\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\}<\left[{x}\right]+\left[{y}\right]+\mathrm{2} \\ $$$$\Rightarrow{Case}\:{I}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]\:{or}\:{II}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\mathrm{1} \\ $$$${I}\Rightarrow\left[{x}\right]\left(\left[{y}\right]−\mathrm{1}\right)=\left[{y}\right]\Rightarrow\left[{x}\right]=\frac{\left[{y}\right]}{\left[{y}\right]−\mathrm{1}}=\mathrm{1}+\frac{\mathrm{1}}{\left[{y}\right]−\mathrm{1}} \\ $$$$\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\mathrm{1}\Rightarrow\left[{y}\right]−\mathrm{1}=\mathrm{1}\:{or}\:\left[{y}\right]−\mathrm{1}=−\mathrm{1}…
Question Number 126389 by shaker last updated on 20/Dec/20 Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\:\left[\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\right]}{\:\left[\frac{\mathrm{2cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}}\right]}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}×\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2cos}\:\mathrm{2}{x}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\:=\:\mathrm{1} \\ $$ Answered…
Question Number 191921 by Rupesh123 last updated on 03/May/23 Answered by AST last updated on 03/May/23 $$\int\left(\mathrm{3}{x}+\mathrm{5}\right){dx}=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{5}{x}+{c} \\ $$$$\int_{−\mathrm{4}} ^{\mathrm{2}} \left(\mathrm{3}{x}+\mathrm{5}\right){dx}=\left(\mathrm{16}+{c}\right)−\left(\mathrm{4}+{c}\right)=\mathrm{12} \\ $$$$\Rightarrow\left(\mathrm{4}+{log}_{\mathrm{3}} {x}\right)\left({log}_{\mathrm{3}}…
Question Number 60849 by Kunal12588 last updated on 26/May/19 $${For}\:{all}\:\theta\:{in}\:\left[\mathrm{0},\:\pi/\mathrm{2}\right]\:{show}\:{that}\:{cos}\left({sin}\theta\right)\geqslant{sin}\left({cos}\theta\right). \\ $$ Commented by Prithwish sen last updated on 26/May/19 $$\mathrm{when}\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right] \\ $$$$\mathrm{sin}\theta\:\mathrm{and}\:\mathrm{cos}\theta\:\mathrm{lies}\:\mathrm{between}\:\mathrm{0}\:\mathrm{to}\:\mathrm{1} \\ $$$$\therefore\:\mathrm{For}\:\alpha\in\left[\mathrm{0},\mathrm{1}\right]…
Question Number 126383 by Lordose last updated on 20/Dec/20 $$\mathrm{Evaluate}\:\Omega\:\mathrm{if} \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{cos}\left(\mathrm{mx}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$ Answered by mathmax by abdo last updated…
Question Number 126381 by liberty last updated on 20/Dec/20 $$\:\mathrm{arctan}\:\mathrm{1}\:+\:\mathrm{arctan}\:\mathrm{2}\:+\:\mathrm{arctan}\:\mathrm{3}\:=? \\ $$ Answered by benjo_mathlover last updated on 20/Dec/20 $$\left(\bullet\right)\:\mathrm{arctan}\:\mathrm{1}+\mathrm{arctan}\:\mathrm{2}=\mathrm{arctan}\:\left(\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}−\mathrm{1}.\mathrm{2}}\right) \\ $$$$\:\mathrm{arctan}\:\left(−\mathrm{3}\right) \\ $$$$\left(\bullet\bullet\right)\:\mathrm{arctan}\:\left(−\mathrm{3}\right)+\mathrm{arctan}\:\left(\mathrm{3}\right)=\mathrm{arctan}\:\left(\frac{−\mathrm{3}+\mathrm{3}}{\mathrm{1}−\left(−\mathrm{3}\right)\left(\mathrm{3}\right)}\right) \\…
Question Number 126378 by benjo_mathlover last updated on 20/Dec/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}+{x}^{\mathrm{3}} \mathrm{cos}\:\left(\frac{\pi}{{x}}\right)}{{x}^{\mathrm{2}} }? \\ $$ Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} }+\underset{{x}\rightarrow\mathrm{0}}…
Question Number 126376 by benjo_mathlover last updated on 20/Dec/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\mathrm{3}{x}\right)^{−\frac{\mathrm{5}}{{x}}} \:? \\ $$ Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{3}{x}\right)^{−\frac{\mathrm{5}}{{x}}} =\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{3}{x}−\mathrm{1}\right)\left(−\frac{\mathrm{5}}{{x}}\right)}…
Question Number 126369 by I want to learn more last updated on 19/Dec/20 $$\int\:\frac{\mathrm{tan}\left(\mathrm{x}\right)}{\mathrm{x}}\:\:\mathrm{dx} \\ $$ Answered by Olaf last updated on 20/Dec/20 $$\mathrm{I}\:\mathrm{believe}\:\mathrm{the}\:\mathrm{only}\:\mathrm{way}\:\mathrm{to}\:\mathrm{handle}\:\mathrm{this} \\…
Question Number 191901 by mehdee42 last updated on 03/May/23 $${if}\:\:\mathrm{0}<\theta<\mathrm{45}^{\mathrm{0}} \:\:{which}\:{is}\:{bigger}\:? \\ $$$$\mathrm{2}{tan}\theta\:\:{or}\:\:{tan}\mathrm{2}\theta \\ $$ Answered by mr W last updated on 03/May/23 Commented by…