Question Number 171762 by pete last updated on 20/Jun/22 $$\mathrm{The}\:\mathrm{curve}\:\mathrm{for}\:\mathrm{which}\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}\left(\mathrm{x}−\mathrm{p}\right)\left(\mathrm{x}−\mathrm{q}\right), \\ $$$$\mathrm{where}\:\mathrm{a},\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{constants},\:\mathrm{has}\:\mathrm{turning} \\ $$$$\mathrm{points}\:\mathrm{at}\:\left(\mathrm{2},\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{1},\mathrm{1}\right). \\ $$$$\left.\mathrm{i}\right)\:\mathrm{state}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}. \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{using}\:\mathrm{these}\:\mathrm{values},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{a} \\ $$ Answered by mr…
Question Number 40662 by Tinkutara last updated on 25/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18 $${pls}\:{check}\:{the}\:{question}…{it}\:{seems}\:{wrong}… \\ $$ Commented by maxmathsup by imad last…
Question Number 40479 by Raj Singh last updated on 22/Jul/18 Answered by MJS last updated on 22/Jul/18 $${a}=\mathrm{8}+\mathrm{2}{x} \\ $$$${b}={c}={d}=\mathrm{8} \\ $$$${h}=\sqrt{\mathrm{64}−{x}^{\mathrm{2}} } \\ $$$${A}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right){h}=\left(\mathrm{8}+{x}\right)\sqrt{\mathrm{64}−{x}^{\mathrm{2}}…
Question Number 171529 by cortano1 last updated on 17/Jun/22 Commented by kapoorshah last updated on 17/Jun/22 $$\mathrm{25} \\ $$ Answered by FelipeLz last updated on…
Question Number 171519 by JumanneM last updated on 16/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 40319 by ajfour last updated on 19/Jul/18 $${y}=\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{a}} \\ $$$${Find}\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:. \\ $$ Commented by math khazana by abdo…
Question Number 171379 by mathlove last updated on 14/Jun/22 $${f}\left({x}\right)=\begin{cases}{\mathrm{2}{x}+\mathrm{3}\:\:\:;{x}>\mathrm{0}}\\{\mathrm{3}{x}−\mathrm{5}\:\:;{x}\leqslant\mathrm{0}}\end{cases} \\ $$$$\frac{{df}\left({x}\right)}{{dx}}=? \\ $$ Commented by MJS_new last updated on 14/Jun/22 $${f}\left({x}\right)\:=\begin{cases}{\mathrm{3}{x}−\mathrm{5};\:{x}\leqslant\mathrm{0}}\\{\mathrm{not}\:\mathrm{defined};\:\mathrm{0}<{x}\leqslant\mathrm{2}}\\{\mathrm{2}{x}+\mathrm{3};\:{x}>\mathrm{2}}\end{cases} \\ $$$${f}\:'\left({x}\right)=\begin{cases}{\mathrm{3};\:{x}\leqslant\mathrm{0}}\\{\mathrm{not}\:\mathrm{defined};\:\mathrm{0}<{x}\leqslant\mathrm{2}}\\{\mathrm{2};\:{x}>\mathrm{2}}\end{cases} \\…
Question Number 40277 by Raj Singh last updated on 18/Jul/18 Answered by MJS last updated on 18/Jul/18 $$\frac{{d}}{{d}\theta}\left[\mathrm{sin}^{{p}} \:\theta\:\mathrm{cos}^{{q}} \:\theta\right]=\mathrm{sin}^{{p}−\mathrm{1}} \:\theta\:\mathrm{cos}^{{q}−\mathrm{1}} \:\theta\:\left({p}\mathrm{cos}^{\mathrm{2}} \:\theta\:−{q}\mathrm{sin}^{\mathrm{2}} \:\theta\right) \\…
Question Number 40276 by Raj Singh last updated on 18/Jul/18 Answered by MJS last updated on 18/Jul/18 $$\frac{{d}}{{dx}}\left[\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}\right]=\frac{\mathrm{cos}^{\mathrm{3}} \:{x}\:−\mathrm{sin}^{\mathrm{3}} \:{x}}{\mathrm{1}+\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}} \\ $$$$\mathrm{zeros}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{4}{k}+\mathrm{1}\right)\:\wedge\:{k}\in\mathbb{Z} \\ $$$$\mathrm{max}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{4}{k}+\mathrm{1}\right)\:\wedge\:{k}=\mathrm{2}{n}\:\wedge\:{n}\in\mathbb{Z}\:\Rightarrow \\…
Question Number 171315 by pete last updated on 12/Jun/22 $$\mathrm{The}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{2} \\ $$$$\mathrm{at}\:\left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{10}\:\mathrm{at}\:\left(−\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}. \\ $$ Answered by som(math1967) last updated…