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Author: Tinku Tara

Given-m-gt-0-n-gt-0-m-n-a-Find-the-range-of-a-such-that-m-1-m-n-1-n-gets-its-minimum-iff-m-n-

Question Number 215883 by CrispyXYZ last updated on 20/Jan/25 $$\mathrm{Given}\:{m}>\mathrm{0},\:{n}>\mathrm{0},\:{m}+{n}=\sqrt{{a}}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{a}\:\mathrm{such}\:\mathrm{that} \\ $$$$“\left({m}+\frac{\mathrm{1}}{{m}}\right)\left({n}+\frac{\mathrm{1}}{{n}}\right)\:\mathrm{gets}\:\mathrm{its}\:\mathrm{minimum}\:\mathrm{iff}\:{m}={n}''. \\ $$ Commented by MathematicalUser2357 last updated on 21/Jan/25 $$“{iff}''?\:{does}\:{it}\:{mean}\:{ifh}.{cc}? \\…

If-x-2-3x-2-y-2-5y-8-Prove-that-x-3-4y-2-20y-33-2-

Question Number 215874 by MathematicalUser2357 last updated on 20/Jan/25 $$\mathrm{If}\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}={y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{8}, \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+\mathrm{33}}}{\mathrm{2}}. \\ $$ Answered by MrGaster last updated on 20/Jan/25 $${x}^{\mathrm{2}}…

If-b-3-a-2-c-ac-2-3abc-then-prove-that-one-root-of-ax-2-bx-c-0-is-the-square-of-the-other-one-

Question Number 215837 by MATHEMATICSAM last updated on 19/Jan/25 $$\mathrm{If}\:{b}^{\mathrm{3}} \:+\:{a}^{\mathrm{2}} {c}\:+\:{ac}^{\mathrm{2}} \:=\:\mathrm{3}{abc}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{the}\:\mathrm{square}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}. \\ $$ Answered by MrGaster last updated…

Let-be-a-hyperbola-with-foci-F-1-and-F-2-eccentricity-e-M-is-an-arbitrary-point-on-Let-x-MF-1-F-2-y-MF-2-F-1-Prove-that-cos-x-cos-y-1-cos-x-cos-y-2e-e-2-1-

Question Number 215828 by CrispyXYZ last updated on 19/Jan/25 $$\mathrm{Let}\:\Gamma\:\mathrm{be}\:\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{with}\:\mathrm{foci}\:{F}_{\mathrm{1}} \:\mathrm{and}\:{F}_{\mathrm{2}} ,\: \\ $$$$\mathrm{eccentricity}\:{e}.\:{M}\:\mathrm{is}\:\mathrm{an}\:\mathrm{arbitrary}\:\mathrm{point}\:\mathrm{on}\:\Gamma. \\ $$$$\mathrm{Let}\:{x}=\angle{MF}_{\mathrm{1}} {F}_{\mathrm{2}} ,\:{y}=\angle{MF}_{\mathrm{2}} {F}_{\mathrm{1}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\frac{\mid\:\mathrm{cos}\:{x}\:−\:\mathrm{cos}\:{y}\:\mid}{\mathrm{1}\:−\:\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}}\:=\:\frac{\mathrm{2}{e}}{{e}^{\mathrm{2}} +\mathrm{1}}. \\ $$ Answered…

lim-x-x-1-3-x-1-3-x-

Question Number 215831 by golsendro last updated on 19/Jan/25 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }−\sqrt{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }}{\:\sqrt{\mathrm{x}}}\:=? \\ $$ Answered by efronzo1 last updated on 19/Jan/25 $$\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}}…

1-3-1-3-7-21-25-Next-three-terms-

Question Number 215845 by Tawa11 last updated on 19/Jan/25 $$\mathrm{1},\:\mathrm{3},\:−\:\mathrm{1},\:−\:\mathrm{3},\:−\:\mathrm{7},\:\:−\:\mathrm{21},\:\:−\:\mathrm{25},\:\:\:\:\:\_\_\_,\:\:\:\:\:\_\_\_,\:\:\:\:\:\_\_\_ \\ $$$$ \\ $$$$\mathrm{Next}\:\mathrm{three}\:\mathrm{terms}?? \\ $$ Commented by mr W last updated on 20/Jan/25 $${this}\:{is}\:{a}\:{game},\:{not}\:{mathematics}!!!…

log-24-3-a-and-log-24-6-b-6-log-8-b-4a-

Question Number 215840 by golsendro last updated on 19/Jan/25 $$\:\:\:\mathrm{log}\:_{\mathrm{24}} \:\mathrm{3}=\:{a}\:\mathrm{and}\:\mathrm{log}\:_{\mathrm{24}} \:\mathrm{6}\:=\:\frac{{b}}{\mathrm{6}} \\ $$$$\:\:\:\mathrm{log}\:_{\sqrt{\mathrm{8}}} \:\left({b}−\mathrm{4}{a}\right)=\:? \\ $$ Commented by oubiji last updated on 19/Jan/25 $$\:\:\:\:{b}−\mathrm{4}{a}=\mathrm{log}_{\mathrm{24}}…

Question-215820

Question Number 215820 by Jubr last updated on 18/Jan/25 Commented by A5T last updated on 18/Jan/25 $$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{generally}\:\mathrm{true},\:\mathrm{it}\:\mathrm{fails}\:\mathrm{when}\:\mathrm{c}=\mathrm{0}. \\ $$ Answered by Rasheed.Sindhi last updated on…