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Category: Differentiation

dx-1-x-x-2-

Question Number 104927 by bobhans last updated on 24/Jul/20 $$\int\frac{{dx}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated on 24/Jul/20 $$\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C}…

1-aT-1-bT-1-yT-1-zT-

Question Number 104860 by ~blr237~ last updated on 24/Jul/20 $$\left(\mathrm{1}−{aT}\right)\left(\mathrm{1}−{bT}\right)…..\left(\mathrm{1}−{yT}\right)\left(\mathrm{1}−{zT}\right)\:\:\:\:???? \\ $$ Answered by ~blr237~ last updated on 24/Jul/20 $${Just}\:\:{to}\:{smile}\::\:{if}\:{we}\:{remember}\:{the}\:{formula}\:{of}\:{the}\:{frequency}\:\:\:{f}=\frac{\mathrm{1}}{{T}}\:\:{then}\:\:\mathrm{1}−{fT}=\mathrm{0}\: \\ $$$${And}\:{that}\:{product}\:{will}\:{be}\:{null} \\ $$ Commented…

lim-z-0-z-z-lim-z-i-z-4-z-4-lim-z-0-sinz-z-

Question Number 104856 by ~blr237~ last updated on 24/Jul/20 $$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\overset{−} {{z}}}{{z}}\:\:\:\:,\:\:\:\:\underset{{z}\rightarrow{i}} {\mathrm{lim}}\:\frac{\left(\overset{−} {{z}}\right)^{\mathrm{4}} }{{z}^{\mathrm{4}} }\:\:,\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinz}}{{z}}\: \\ $$ Answered by abdomathmax last updated on…

1-a-3-1-b-3-1-x-3-1-y-3-1-z-3-

Question Number 104858 by ~blr237~ last updated on 24/Jul/20 $$\:\:\left(\mathrm{1}−{a}^{\mathrm{3}} \right)\left(\mathrm{1}−{b}^{\mathrm{3}} \right)……\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\left(\mathrm{1}−{y}^{\mathrm{3}} \right)\left(\mathrm{1}−{z}^{\mathrm{3}} \right)\:\:\:\:\:\:\:\:???? \\ $$ Commented by 1549442205PVT last updated on 24/Jul/20 $$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{your}}\:\:\boldsymbol{\mathrm{question}}?…

y-y-x-y-x-y-0-0-

Question Number 104857 by ~blr237~ last updated on 24/Jul/20 $$\:\:{y}'=\frac{{y}−{x}}{{y}+{x}}\:\:\:\:\:\:\:{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Answered by bemath last updated on 24/Jul/20 $${y}={px}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{p}+{x}\frac{{dp}}{{dx}} \\ $$$$\Leftrightarrow{p}+{x}\frac{{dp}}{{dx}}\:=\:\frac{{x}\left({p}−\mathrm{1}\right)}{{x}\left({p}+\mathrm{1}\right)} \\ $$$${x}\frac{{dp}}{{dx}}\:=\:\frac{{p}−\mathrm{1}}{{p}+\mathrm{1}}−\frac{{p}\left({p}+\mathrm{1}\right)}{{p}+\mathrm{1}} \\…

if-g-C-R-R-and-0-1-g-x-dx-1-3-0-1-g-2-x-2-dx-then-0-1-g-x-dx-2-3-and-0-1-g-2-x-dx-1-2-

Question Number 104852 by ~blr237~ last updated on 24/Jul/20 $$\:\:{if}\:\:{g}\in{C}\left(\mathbb{R},\mathbb{R}\right)\:{and}.\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{g}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}+\int_{\mathrm{0}} ^{\mathrm{1}} {g}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right){dx}\: \\ $$$${then}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({x}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}}\:{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}^{\mathrm{2}} \left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\:\: \\ $$…

F-x-x-3-9x-2-24x-c-0-has-three-real-and-distinct-roots-amp-Q-1-Possible-value-of-c-is-Q-2-If-8-then-c-is-Q-3-If-7-then-c-is-Options-for-the-above

Question Number 39142 by rahul 19 last updated on 03/Jul/18 $$\mathrm{F}\left({x}\right)\:=\:{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{24}{x}+{c}=\mathrm{0}\:{has}\:\mathrm{three} \\ $$$$\mathrm{real}\:\mathrm{and}\:\mathrm{distinct}\:\mathrm{roots}\:\alpha\:,\:\beta\:\&\:\gamma\:. \\ $$$$\mathrm{Q}.\mathrm{1}\:\rightarrow\:\mathrm{Possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{c}\:\mathrm{is}\:: \\ $$$$\mathrm{Q}.\mathrm{2}\:\rightarrow\:\mathrm{If}\:\left[\alpha\right]+\left[\beta\right]+\left[\gamma\right]=\:\mathrm{8}\:\mathrm{then}\:\mathrm{c}\:\mathrm{is}\:: \\ $$$$\mathrm{Q}.\mathrm{3}\:\rightarrow\:\mathrm{If}\:\left[\alpha\right]+\left[\beta\right]+\left[\gamma\right]=\mathrm{7}\:\mathrm{then}\:\mathrm{c}\:\mathrm{is}\:: \\ $$$$ \\ $$$$\mathrm{Options}\:\mathrm{for}\:\mathrm{the}\:\mathrm{above}\:\mathrm{3}\:\mathrm{Q}.\:\rightarrow…