Question Number 216077 by MATHEMATICSAM last updated on 27/Jan/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{non}\:\mathrm{negative} \\ $$$$\mathrm{integer}\:{p}\:\mathrm{for}\:\mathrm{which}\: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left\{\frac{−\:{px}\:+\:\mathrm{sin}\left({x}\:−\:\mathrm{1}\right)\:+\:{p}}{{x}\:+\:\mathrm{sin}\left({x}\:−\:\mathrm{1}\right)\:−\:\mathrm{1}}\right\}^{\frac{\mathrm{1}\:−\:{x}}{\mathrm{1}\:−\:\sqrt{{x}}}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$ Answered by mahdipoor last updated on 27/Jan/25…
Question Number 216110 by Tawa11 last updated on 27/Jan/25 Commented by mr W last updated on 28/Jan/25 $${this}\:{workings}\:{is}\:{ok},\:{if}\:{we}\:{don}'{t}\:{think} \\ $$$${about}\:{how}\:{the}\:{particle}\:{moves}\:{from} \\ $$$${P}\:\:{to}\:{Q}. \\ $$ Commented…
Question Number 216078 by MathematicalUser2357 last updated on 27/Jan/25 $${see}\:{comments} \\ $$ Commented by Tinku Tara last updated on 27/Jan/25 Commented by Tinku Tara last…
Question Number 216105 by mr W last updated on 27/Jan/25 Answered by A5T last updated on 27/Jan/25 $$\mathrm{x}^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \Rightarrow\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{2}\underset{−}…
Question Number 216106 by mr W last updated on 27/Jan/25 Commented by mr W last updated on 27/Jan/25 $${three}\:{squares}\:{with}\:{side}\:{lengthes} \\ $$$${a},{b},{c}\:{respectively}. \\ $$$${find}\:{the}\:{area}\:{of}\:{the}\:{red}\:{triangle}. \\ $$…
Question Number 216074 by CrispyXYZ last updated on 27/Jan/25 $$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{y}=\mid\mathrm{sin}\:{x}\mid+\mid\mathrm{sin}\:\mathrm{2}{x}\mid. \\ $$ Answered by efronzo1 last updated on 27/Jan/25 $$\:\mathrm{y}\:=\:\mid\mathrm{sin}\:\mathrm{x}\mid\:+\:\mathrm{2}\mid\mathrm{sin}\:\mathrm{x}\mid\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\:\mathrm{y}=\:\mid\mathrm{sin}\:\mathrm{x}\mid\:\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:\right) \\…
Question Number 216060 by mr W last updated on 26/Jan/25 Commented by mr W last updated on 26/Jan/25 $${find}\:{radius}\:{of}\:{inscribed}\:{circle}\:{r}=? \\ $$ Commented by Tawa11 last…
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Question Number 216058 by mokys last updated on 26/Jan/25 Commented by mokys last updated on 26/Jan/25 $${graph}\:{the}\:{function} \\ $$ Commented by mokys last updated on…
Question Number 216055 by efronzo1 last updated on 26/Jan/25 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{x}}}\:=? \\ $$ Answered by golsendro last updated on 27/Jan/25 $$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{x}}}…