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Author: Tinku Tara

find-lim-x-0-ln-sin-xcos-1-x-1-if-it-exits-

Question Number 85057 by kushdasbaghar@gmail.com last updated on 18/Mar/20 $${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}ln}\:\left(\mathrm{sin}\:{x}\mathrm{cos}\:\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\:{if}\:{it}\:{exits}. \\ $$ Commented by abdomathmax last updated on 19/Mar/20 $${we}\:{have}\:{ln}\left(\mathrm{1}+{sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\right)\sim{sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\sim{xcos}\left(\frac{\mathrm{1}}{{x}}\right)\:{and}\:\mid{xcos}\left(\frac{\mathrm{1}}{{x}}\right)\mid\leqslant\mid{x}\mid\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}}…

Let-g-is-the-inverse-function-of-f-and-f-x-x-10-1-x-2-If-g-2-a-then-g-2-

Question Number 150590 by liberty last updated on 13/Aug/21 $$\mathrm{Let}\:\mathrm{g}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{function}\:\mathrm{of} \\ $$$$\mathrm{f}\:\mathrm{and}\:\mathrm{f}\:'\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{10}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\:\mathrm{If}\:\mathrm{g}\left(\mathrm{2}\right)=\:{a}\:\mathrm{then} \\ $$$$\mathrm{g}\:'\left(\mathrm{2}\right)\:=\_\_\: \\ $$ Answered by Olaf_Thorendsen last updated on 13/Aug/21…

Let-Akbar-and-Birbal-together-have-n-marbles-where-n-gt-0-Akbar-says-to-Birbal-If-I-give-you-some-marbles-then-you-will-have-twice-as-many-marbles-as-I-will-have-Birbal-says-to-Akbar-If-I-gi

Question Number 19516 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Let}\:\mathrm{Akbar}\:\mathrm{and}\:\mathrm{Birbal}\:\mathrm{together}\:\mathrm{have}\:{n} \\ $$$$\mathrm{marbles},\:\mathrm{where}\:{n}\:>\:\mathrm{0}. \\ $$$$\mathrm{Akbar}\:\mathrm{says}\:\mathrm{to}\:\mathrm{Birbal},\:“\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$$$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{twice}\:\mathrm{as} \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.''\:\mathrm{Birbal} \\ $$$$\mathrm{says}\:\mathrm{to}\:\mathrm{Akbar},\:“\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$$$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{thrice}\:\mathrm{as} \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.'' \\…

In-the-arrangement-shown-the-wedge-is-smooth-and-has-a-mass-M-The-sphere-has-a-mass-m-The-system-is-released-from-rest-from-the-position-shown-There-is-no-friction-anywhere-Find-the-contact-force

Question Number 19511 by Tinkutara last updated on 12/Aug/17 $$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown},\:\mathrm{the}\:\mathrm{wedge} \\ $$$$\mathrm{is}\:\mathrm{smooth}\:\mathrm{and}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:{M}.\:\mathrm{The}\:\mathrm{sphere} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:{m}.\:\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{released} \\ $$$$\mathrm{from}\:\mathrm{rest}\:\mathrm{from}\:\mathrm{the}\:\mathrm{position}\:\mathrm{shown}. \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{friction}\:\mathrm{anywhere}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{contact}\:\mathrm{force}\:\mathrm{between}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sphere}. \\ $$ Commented…

Blocks-P-and-R-starts-from-rest-and-moves-to-the-right-with-acceleration-a-P-12t-m-s-2-and-a-R-3-m-s-2-Here-t-is-in-seconds-The-time-when-block-Q-again-comes-to-rest-is-

Question Number 19509 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Blocks}\:{P}\:\mathrm{and}\:{R}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{and} \\ $$$$\mathrm{moves}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{with}\:\mathrm{acceleration} \\ $$$${a}_{{P}} \:=\:\mathrm{12}{t}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{and}\:{a}_{{R}} \:=\:\mathrm{3}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{Here}\:{t} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{seconds}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{when}\:\mathrm{block}\:{Q} \\ $$$$\mathrm{again}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{is} \\ $$ Commented…

Prove-that-the-length-of-perpendicular-drawn-from-the-point-z-0-to-the-straight-line-z-z-c-0-is-p-z-0-z-0-c-2-

Question Number 19508 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{perpendicular} \\ $$$$\mathrm{drawn}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:{z}_{\mathrm{0}} \:\mathrm{to}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{line}\:\bar {\alpha}{z}\:+\:\alpha\bar {{z}}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{is} \\ $$$${p}\:=\:\mid\frac{\bar {\alpha}{z}_{\mathrm{0}} \:+\:\alpha\bar {{z}}_{\mathrm{0}} \:+\:{c}}{\mathrm{2}\:\mid\alpha\mid}\mid. \\ $$…

Prove-that-three-points-z-1-z-2-z-3-are-collinear-if-determinant-z-1-z-1-1-z-2-z-2-1-z-3-z-3-1-0-

Question Number 19507 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{three}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{collinear}\:\mathrm{if}\:\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0} \\…

Prove-that-the-equation-of-the-line-joining-the-points-z-1-and-z-2-can-be-put-in-the-form-z-tz-1-1-t-z-2-where-t-is-a-parameter-

Question Number 19506 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{joining}\:\mathrm{the}\:\mathrm{points}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{put} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{z}\:=\:{tz}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:−\:{t}\right){z}_{\mathrm{2}} ,\:\mathrm{where} \\ $$$${t}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parameter}. \\ $$ Answered by ajfour…