Question Number 96782 by M±th+et+s last updated on 04/Jun/20
$$\left.\mathrm{1}\right)\frac{{cos}^{\mathrm{4}} \left(\theta\right)}{{x}}−\frac{{sin}^{\mathrm{4}} \left(\theta\right)}{{y}}=\frac{\mathrm{1}}{{x}+{y}} \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right){solve}:\mathrm{2}\lfloor{x}−\mathrm{4}+\lfloor{x}\rfloor\rfloor=\mathrm{6}−\mathrm{3}\lfloor{x}\rfloor \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\left({cos}\left({x}\right)\right)^{{x}} −\left({sin}\left({x}\right)\right)^{{x}} −{cos}\left(\mathrm{2}{x}\right)}{\left({x}−\mathrm{4}\right)} \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et+s last updated on 04/Jun/20
$${thanks}\:{for}\:{solutions} \\ $$
Answered by mr W last updated on 04/Jun/20
$$\left(\mathrm{2}\right) \\ $$$$\lfloor{x}\rfloor\:{must}\:{be}\:{even},\:{say}\:{x}=\mathrm{2}{n}+{f} \\ $$$$\mathrm{2}\lfloor\mathrm{2}{n}+{f}−\mathrm{4}+\mathrm{2}{n}\rfloor=\mathrm{6}−\mathrm{3}×\mathrm{2}{n} \\ $$$$\lfloor\mathrm{4}\left({n}−\mathrm{1}\right)+{f}\rfloor=\mathrm{3}\left(\mathrm{1}−{n}\right) \\ $$$$\mathrm{7}{n}=\mathrm{7} \\ $$$$\Rightarrow{n}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\leqslant{x}<\mathrm{3} \\ $$
Answered by Sourav mridha last updated on 04/Jun/20