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Question-200004

Question Number 200004 by Mingma last updated on 12/Nov/23 Answered by aleks041103 last updated on 12/Nov/23 $$−\mathrm{1}<{x}<\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{{x}+\mathrm{2}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{{x}}{\mathrm{2}^{{k}+\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }<{f}\left({x}\right)<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} } \\…

Question-200005

Question Number 200005 by Mingma last updated on 12/Nov/23 Answered by AST last updated on 12/Nov/23 $$\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\Rightarrow{a}+{b}+{c}\leqslant\mathrm{3} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{2}{a}+{b}}\geqslant\frac{\mathrm{9}}{\mathrm{3}\left({a}+{b}+{c}\right)}\geqslant\frac{\mathrm{9}}{\mathrm{3}×\mathrm{3}}=\mathrm{1} \\ $$$${Equality}\:{holds}\:{when}\:{a}={b}={c}=\mathrm{1}…

f-x-x-g-x-2x-1-fog-x-f-2x-1-fog-x-2x-1-gof-x-g-x-gof-x-2-x-

Question Number 200007 by Hummayoun last updated on 12/Nov/23 $${f}\left({x}\right)=\sqrt{{x}} \\ $$$${g}\left({x}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${fog}\left({x}\right)={f}\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${fog}\left({x}\right)=\sqrt{\mathrm{2}{x}+\mathrm{1}} \\ $$$${gof}\left({x}\right)={g}\left(\sqrt{\left.{x}\right)}\right. \\ $$$${gof}\left({x}\right)=\mathrm{2}\sqrt{{x}} \\ $$ Terms of Service…

1-If-3-ab-bc-115-Find-max-a-b-c-2-a-b-c-N-If-a-2-b-3-c-4-Find-min-a-b-c-

Question Number 199932 by hardmath last updated on 11/Nov/23 $$\mathrm{1}. \\ $$$$\mathrm{If}\:\:\:\mathrm{3}\:\centerdot\:\overline {\mathrm{ab}}\:+\:\overline {\mathrm{bc}}\:=\:\mathrm{115} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{max}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=? \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{If}\:\:\:\frac{\mathrm{a}}{\mathrm{2}}\:\:+\:\:\frac{\mathrm{b}}{\mathrm{3}}\:\:=\:\:\frac{\mathrm{c}}{\mathrm{4}} \\…

ABCD-trapezium-EF-middle-line-EG-2-GF-4-DC-y-and-AB-x-Find-x-y-Here-point-G-is-the-point-of-intersection-of-the-extension-of-sides-AB-and-CD-

Question Number 199988 by hardmath last updated on 11/Nov/23 $$\mathrm{ABCD}\:-\:\mathrm{trapezium} \\ $$$$\mathrm{EF}\:-\:\mathrm{middle}\:\mathrm{line} \\ $$$$\mathrm{EG}\:=\:\mathrm{2}\:\:,\:\:\mathrm{GF}\:=\:\mathrm{4}\:\:,\:\:\mathrm{DC}\:=\:\mathrm{y}\:\:\mathrm{and}\:\:\mathrm{AB}\:=\:\mathrm{x} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}\:-\:\mathrm{y}\:=\:? \\ $$$$\left(\mathrm{Here}\:\mathrm{point}\:\mathrm{G}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\right. \\ $$$$\left.\mathrm{of}\:\mathrm{the}\:\mathrm{extension}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{CD}\right) \\ $$ Terms of Service…

Question-199921

Question Number 199921 by Mingma last updated on 11/Nov/23 Answered by des_ last updated on 12/Nov/23 $$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:{I}; \\ $$$${I}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx},\:{a}\:>\mathrm{0};…