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Author: Tinku Tara

Question-211029

Question Number 211029 by BaliramKumar last updated on 26/Aug/24 Answered by Ar Brandon last updated on 26/Aug/24 $${f}\left(\mathrm{2}\right)=\mathrm{5}\:\mathrm{therefore}\:{f}\left({x}\right)\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{polynomial}. \\ $$$$\mathrm{Let}\:{f}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{a}_{{n}} {x}^{{n}}…

Question-211020

Question Number 211020 by peter frank last updated on 26/Aug/24 Answered by mahdipoor last updated on 26/Aug/24 $${dV}=\left(\pi{y}^{\mathrm{2}} \right)\left({dx}\right)=\pi\frac{\left({x}−{a}\right)^{\mathrm{2}} \left({x}−{b}\right)^{\mathrm{2}} }{{c}^{\mathrm{2}} }{dx} \\ $$$${V}=\int{dV}=\frac{\pi}{{c}^{\mathrm{2}} }\int_{{a}}…

Question-211019

Question Number 211019 by peter frank last updated on 26/Aug/24 Answered by som(math1967) last updated on 26/Aug/24 $$\:{let}\:{x}={sin}\alpha\:\:{y}={sin}\beta \\ $$$$\:{cos}\alpha+{cos}\beta={a}\left({sin}\alpha−{sin}\beta\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{cos}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}}}{\mathrm{2}{sin}\frac{\alpha−\beta}{\mathrm{2}}{cos}\frac{\alpha+\beta}{\mathrm{2}}}={a} \\ $$$$\Rightarrow{cot}\frac{\alpha−\beta}{\mathrm{2}}={a} \\…

Question-211043

Question Number 211043 by MATHEMATICSAM last updated on 26/Aug/24 Answered by Ghisom last updated on 26/Aug/24 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9}= \\ $$$$=\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{5}=\mathrm{2024} \\ $$ Answered…