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Question Number 141457 by gajmer last updated on 19/May/21

Answered by mathmax by abdo last updated on 19/May/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{{2\mathrm{x}}^2+4}-\sqrt{{\mathrm{x}}^2+5}}{\mathrm{x}-1}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mid \mathrm{x}\mid \sqrt{2}\sqrt{1+\frac{2}{{\mathrm{x}}^2}}-\mid \mathrm{x}\mid \sqrt{1+\frac{5}{{\mathrm{x}}^2}}}{\mathrm{x}(1-\frac{1}{\mathrm{x}})}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{\mid \mathrm{x}\mid }{\mathrm{x}}\times \frac{\sqrt{2}+\frac{1}{{\mathrm{x}}^2}-1-\frac{5}{{2\mathrm{x}}^2}}{1-\frac{1}{\mathrm{x}}}=\frac{\mid \mathrm{x}\mid }{\mathrm{x}}.\frac{\sqrt{2}-1-\frac{3}{{2\mathrm{x}}^2}}{1-\frac{1}{\mathrm{x}}}$$$$\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{2}-1\mathrm{and}{\lim }_{\mathrm{x}\to -\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=1-\sqrt{2}$$

Answered by bramlexs22 last updated on 19/May/21

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{2x^2+4}-\sqrt{x^2+5}}{x-1}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{2+\frac{4}{x^2}}-\sqrt{1+\frac{5}{x^2}}}{1-\frac{1}{x}}$$$$=\frac{\sqrt{2}-1}{1}=\sqrt{2}-1$$

Answered by bramlexs22 last updated on 19/May/21

$${\tan }^{-1}x+{\tan }^{-1}y=u$$$$\Rightarrow \tan ({\tan }^{-1}x+{\tan }^{-1}y)=\tan u$$$$\Rightarrow \frac{\tan \left({\tan }^{-1}x\right)+\tan \left({\tan }^{-1}y\right)}{1-\tan \left({\tan }^{-1}x\right)\tan \left({\tan }^{-1}y\right)}=\tan u$$$$\Rightarrow \frac{x+y}{1-xy}=\tan u$$$$\Rightarrow u={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$$

Answered by mathmax by abdo last updated on 19/May/21

$$\mathrm{b})\mathrm{let}\mathrm{arctanx}=\mathrm{a}\mathrm{and}\mathrm{arctany}=\mathrm{b}\Rightarrow \mathrm{x}=\mathrm{tana}\mathrm{and}\mathrm{y}=\mathrm{tanb}$$$$\Rightarrow \frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}=\frac{\mathrm{tana}+\mathrm{tanb}}{1-\mathrm{tana}.\mathrm{tanb}}=\tan (\mathrm{a}+\mathrm{b})\Rightarrow$$$$\mathrm{a}+\mathrm{b}=\arctan \left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)=\mathrm{arctanx}+\mathrm{arctany}$$

Answered by bramlexs22 last updated on 19/May/21

$$\sec x+\tan x=\sqrt{2}$$$$\frac{1+\sin x}{\cos x}=\sqrt{2};\cos x>0$$$$\Rightarrow 1+\sin x=\sqrt{2(1-\sin {}^{2}x)}$$$$\Rightarrow 1+2\sin x+\sin {}^{2}x=2-2\sin {}^{2}x$$$$\Rightarrow 3\sin {}^{2}x+2\sin x-1=0$$$$\Rightarrow (3\sin x-1)(\sin x+1)=0$$$$\Rightarrow \sin x=\frac{1}{3};\sin x=-1\left(rejected\right)$$$$\Rightarrow x=\arcsin \left(\frac{1}{3}\right)+2k\pi ,k\in \mathbb{Z}$$

Answered by mathmax by abdo last updated on 19/May/21

$$1)\mathrm{use}\mathrm{scalar}\mathrm{product}{\mathrm{a}}^2=\mathrm{B}\stackrel{\to 2}{\mathrm{C}}={(\mathrm{A}\overrightarrow{\mathrm{C}}-\mathrm{A}\overrightarrow{\mathrm{B}})}^2$$$$=\mathrm{A}\stackrel{{\to }_2}{\mathrm{C}}-2\mathrm{A}\overrightarrow{\mathrm{B}}.\mathrm{A}\stackrel{{\to }^2}{\mathrm{C}}+\mathrm{A}\stackrel{{\to }^2}{\mathrm{B}}={\mathrm{b}}^2+{\mathrm{c}}^2-2\mathrm{bc}\mathrm{cosA}\Rightarrow$$$$\mathrm{cosA}=\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{bc}}$$

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