Question Number 133934 by liberty last updated on 25/Feb/21 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}\right)^{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{4}} \:\leqslant\:\mathrm{1} \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\left(\mathrm{i}\right)\:\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}\:>\:\mathrm{0}\:;\:\begin{cases}{\mathrm{x}<\mathrm{1}}\\{\mathrm{x}>\mathrm{3}}\end{cases} \\…
Question Number 68397 by Faradtimmy last updated on 10/Sep/19 Answered by $@ty@m123 last updated on 10/Sep/19 $$\left({a}\right)\:{LHS}=\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$$=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$$=\mathrm{2}\left(\mathrm{cos}\:\mathrm{30cos}\:\theta−\mathrm{sin}\:\mathrm{30sin}\:\theta\right) \\ $$$$=\mathrm{2cos}\:\left(\mathrm{30}+\theta\right) \\ $$$${Pl}.\:{check}\:{the}\:{question}.…
Question Number 133929 by bemath last updated on 25/Feb/21 $$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°=? \\ $$ Answered by liberty last updated on 25/Feb/21 $$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}…
Question Number 2859 by Syaka last updated on 29/Nov/15 $${if}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{cx}\:−\:\mathrm{5}{d}\:=\:\mathrm{0}\:{has}\:{root}\:{a}\:{and}\:{b}, \\ $$$${also}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{ax}\:−\:\mathrm{5}{b}\:=\:\mathrm{0}\:{has}\:{root}\:{c}\:{and}\:{d} \\ $$$${then}\:{a},\:{b},\:{c}\:{and}\:{d}\:{are}\:{different}\:{real}\:{number}. \\ $$$${What}\:{the}\:{value}\:{of}\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:=\:? \\ $$ Commented by Syaka last updated…
Question Number 133928 by mnjuly1970 last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 25/Feb/21 $$\left.\mathrm{2}\right)\:\mathrm{u}_{\mathrm{n}} =\left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \:=\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}}…
Question Number 2857 by Syaka last updated on 29/Nov/15 $${By}\:{Induction}\:{Prove}\:{that}\:: \\ $$$$ \\ $$$$\mathrm{12}\mid\left({n}^{\mathrm{4}} \:−\:{n}^{\mathrm{2}} \right) \\ $$ Commented by Filup last updated on 29/Nov/15…
Question Number 133925 by bemath last updated on 25/Feb/21 $$\:\mathrm{Given}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:=\:\hat {\mathrm{i}}−\mathrm{2}\hat {\mathrm{j}}+\hat {\mathrm{k}}\:,\: \\ $$$$\overset{\rightarrow} {{b}}=\:\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}−\mathrm{2}\hat {\mathrm{k}}\:,\:\overset{\rightarrow} {{c}}=−\hat {\mathrm{i}}+\mathrm{3}\hat {\mathrm{j}}−\hat {\mathrm{k}} \\…
Question Number 68390 by TawaTawa last updated on 10/Sep/19 Answered by som(math1967) last updated on 10/Sep/19 $${tan}\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}\:\right)+{tan}\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}}…
Question Number 2851 by 123456 last updated on 28/Nov/15 $${x}^{\mathrm{2}} ={x}+{x}+{x}+\centerdot\centerdot\centerdot+{x}\:\left({x}\:\mathrm{times}\right) \\ $$$$\mathrm{taking}\:\mathrm{derivate} \\ $$$$\mathrm{2}{x}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\centerdot\centerdot\centerdot+\mathrm{1}\:\left({x}\:\mathrm{times}\right) \\ $$$$\mathrm{2}{x}={x}\:\left({x}\neq\mathrm{0}\right) \\ $$$$\mathrm{2}=\mathrm{1} \\ $$$$\mathrm{where}\:\mathrm{the}\:\mathrm{problem}? \\ $$ Commented by…
Question Number 133923 by bemath last updated on 25/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{x}^{\mathrm{2}} \:=\:\mid\mathrm{x}\mid^{\mathrm{2}} \:\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mid\mathrm{x}\mid}\:.\frac{\mathrm{1}}{\mid\mathrm{x}\mid}=\:\underset{{x}\rightarrow\mathrm{0}}…