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Author: Tinku Tara

lim-x-sin-x-1-sin-x-

Question Number 200103 by cortano12 last updated on 14/Nov/23 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{sin}\:\sqrt{\mathrm{x}}\:=? \\ $$ Answered by Frix last updated on 14/Nov/23 $$\mathrm{sin}\:\alpha\:−\mathrm{sin}\:\beta\:=\mathrm{2cos}\:\frac{\alpha+\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\alpha−\beta}{\mathrm{2}} \\ $$$$\mathrm{For}\:\mathrm{large}\:{x}:\:\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}{\mathrm{2}}\sim\sqrt{{x}}\:\wedge\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}}{\mathrm{2}}\sim\mathrm{0} \\ $$$$\Rightarrow…

Question-200092

Question Number 200092 by ajfour last updated on 13/Nov/23 Commented by ajfour last updated on 13/Nov/23 $${If}\:{semicircle}\:{has}\:{radius}\:\mathrm{1},\:{find} \\ $$$${radii}\:{of}\:{blue}\:{and}\:{green}\:{circles}. \\ $$$${One}\:{end}\:{of}\:{each}\:{semicircle}\:{is}\:{at} \\ $$$${center}\:{of}\:{other}. \\ $$…

Question-200060

Question Number 200060 by sonukgindia last updated on 13/Nov/23 Answered by cortano12 last updated on 13/Nov/23 $$\:\mathrm{L}=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\mathrm{2}{ab}−{a}\sqrt{{ab}}−{ab}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ab}−{a}\sqrt{{ab}}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\sqrt{{ab}}\:\left(\sqrt{{ab}}−{a}\right)}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)}\: \\…

if-1-is-a-root-of-unity-aand-z-is-a-complex-number-such-that-z-1-then-2-3-4z-2-4-3-2-z-2z-

Question Number 200087 by universe last updated on 13/Nov/23 $$\:\:\mathrm{if}\:\omega\:\neq\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{aand}\:\mathrm{z}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mid{z}\mid\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\:\:\mid\frac{\mathrm{2}+\mathrm{3}\omega+\mathrm{4}{z}\omega^{\mathrm{2}} }{\mathrm{4}\omega+\mathrm{3}\omega^{\mathrm{2}} {z}+\mathrm{2}{z}}\mid=\:? \\ $$ Commented by Frix last updated on 13/Nov/23…