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Author: Tinku Tara

If-f-x-2a-1-x-1-x-a-and-f-x-f-1-x-Find-a-2-3-

Question Number 208453 by hardmath last updated on 16/Jun/24 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\left(\mathrm{2a}\:+\:\mathrm{1}\right)\centerdot\mathrm{x}\:+\:\mathrm{1}}{\mathrm{x}\:−\:\mathrm{a}}\:\:\:\mathrm{and}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{3}\:=\:? \\ $$ Answered by efronzo1 last updated on 16/Jun/24 $$\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{ax}+\mathrm{1}}{\mathrm{x}−\left(\mathrm{2a}+\mathrm{1}\right)}\:=\:\frac{\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{a}}…

Question-208412

Question Number 208412 by efronzo1 last updated on 15/Jun/24 Answered by A5T last updated on 15/Jun/24 $$\mathrm{8}=\mathrm{24}^{\frac{\mathrm{1}}{{a}}} ,\mathrm{27}=\mathrm{24}^{\frac{\mathrm{1}}{{b}}} ,\mathrm{64}=\mathrm{24}^{\frac{\mathrm{1}}{{c}}} \\ $$$$\Rightarrow\mathrm{24}^{\mathrm{3}} =\mathrm{8}×\mathrm{27}×\mathrm{64}=\mathrm{24}^{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{{ab}+{bc}+{ca}}{{abc}}\right)} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}}{{abc}}=\mathrm{3}\Rightarrow\frac{\mathrm{2022}{abc}}{{ab}+{bc}+{ca}}=\mathrm{2022}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{674} \\…

Find-0-2-1-x-dx-

Question Number 208409 by hardmath last updated on 15/Jun/24 $$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\mid\mathrm{1}\:−\:\mathrm{x}\mid\:\mathrm{dx}\:=\:? \\ $$ Answered by mr W last updated on 15/Jun/24 $$=\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}−\mathrm{1}\right){dx}…

resoudre-dans-R-3-x-y-3-y-z-5-x-z-4-

Question Number 208420 by lepuissantcedricjunior last updated on 15/Jun/24 $$\boldsymbol{{resoudre}}\:\boldsymbol{{dans}}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{3}}\\{\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{5}}\end{cases}\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{4} \\ $$ Commented by A5T last updated on 15/Jun/24 $${You}\:{should}\:{learn}\:{to}\:{signify}\:{that}\:{you}\:{changed} \\ $$$${a}\:{question}. \\…

Question-208421

Question Number 208421 by lepuissantcedricjunior last updated on 15/Jun/24 Commented by Frix last updated on 15/Jun/24 $$\Psi=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{tan}\:\mathrm{2}{x}}}\:\overset{{t}=\sqrt{\mathrm{tan}\:\mathrm{2}{x}}} {=}\:\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{2}} \\ $$…