Question Number 165054 by mathlove last updated on 25/Jan/22 Answered by alephzero last updated on 25/Jan/22 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\sqrt{\mathrm{sin}\:\frac{\pi}{{x}}}\::\:\mathrm{cos}\:\frac{\pi}{{x}+\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \:= \\ $$$$=\:\sqrt{\sqrt{\mathrm{sin}\:\frac{\pi}{\mathrm{2}}}\::\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}}\:= \\ $$$$=\:\sqrt{\mathrm{1}\::\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\:\sqrt{\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\sqrt{\mathrm{2}}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\:=\: \\ $$$$=\:\frac{\sqrt{\mathrm{2}}\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{3}}…
Question Number 165044 by alephzero last updated on 25/Jan/22 $$\mathrm{If}\:{n}\:=\:{j}\:+\:{k},\:\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove} \\ $$$$\frac{{d}^{{n}} }{{dx}^{{n}} }{f}\left({x}\right)\:=\:\frac{{d}^{{j}} }{{dx}^{{j}} }\left(\frac{{d}^{{k}} }{{dx}^{{k}} }{f}\left({x}\right)\right)\:=\:\frac{{d}^{{k}} }{{dx}^{{k}} }\left(\frac{{d}^{{j}} }{{dx}^{{j}} }{f}\left({x}\right)\right) \\ $$ Commented…
Question Number 165021 by mathlove last updated on 25/Jan/22 $${fog}=\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3} \\ $$$${gof}=\mathrm{12}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)=? \\ $$ Answered by mr W last updated on…
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Question Number 164970 by Zaynal last updated on 24/Jan/22 $$\:\:\lfloor\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{1}} }{\frac{\boldsymbol{{x}}}{\mathrm{10}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{2}} }{\frac{\boldsymbol{{x}}}{\mathrm{9}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{3}} }{\frac{\boldsymbol{{x}}}{\mathrm{8}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\boldsymbol{{x}}\:\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{4}} }{\frac{\boldsymbol{{x}}}{\mathrm{7}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{5}} }{\frac{\boldsymbol{{x}}}{\mathrm{6}−\mathrm{3}^{\boldsymbol{{x}}} }}\:\geqslant\:\frac{\mathrm{1}}{\frac{\mathrm{25}}{\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}\:\left(\frac{\mathrm{1}}{\mathrm{25}}\right)} } }}\rfloor \\ $$$$\:\:\:\left\{\mathrm{Z}.\mathrm{A}\right\} \\…
Question Number 99430 by I want to learn more last updated on 20/Jun/20 Answered by mr W last updated on 20/Jun/20 $${f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} −\frac{\mathrm{44}}{\mathrm{9}}{x}−\frac{\mathrm{40}}{\mathrm{9}}=\mathrm{0} \\…
Question Number 164966 by Zaynal last updated on 24/Jan/22 $$\:\:\:\:\:\lfloor\left(\frac{\mathrm{5}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{3}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{2}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\:\bullet\:\left(\frac{\boldsymbol{{x}}}{\mathrm{1}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{2}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{3}}\:\:−\:\frac{\boldsymbol{{x}}}{\mathrm{4}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{5}}\:\right)^{\mathrm{2}} >\:\frac{\mathrm{1}}{\mathrm{15}}\rfloor \\ $$$$\:\:\:\left\{\mathrm{za}\right\} \\ $$ Answered by alephzero last updated on 24/Jan/22 $$\frac{\mathrm{15}}{{x}}\:×\:\frac{\mathrm{17}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{60}^{\mathrm{2}}…