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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

f(x)=(((√(2x^2 +4))−(√(x^2 +5)))/(x−1)) ⇒ f(x)=((∣x∣(√2)(√(1+(2/x^2 )))−∣x∣(√(1+(5/x^2 ))))/(x(1−(1/x)))) ⇒ f(x)∼((∣x∣)/x)×(((√2)+(1/x^2 )−1−(5/(2x^2 )))/(1−(1/x))) =((∣x∣)/x).(((√2)−1−(3/(2x^2 )))/(1−(1/x))) ⇒lim_(x→+∞) f(x) =(√2)−1 and lim_(x→−∞) f(x)=1−(√2)

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

Answered by bramlexs22 last updated on 19/May/21

lim_(x→∞) (((√(2x^2 +4))−(√(x^2 +5)))/(x−1)) = lim_(x→∞) (((√(2+(4/x^2 )))−(√(1+(5/x^2 ))))/(1−(1/x))) = (((√2)−1)/1) = (√2) −1

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}{{x}−\mathrm{1}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}−\sqrt{\mathrm{1}+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\: \\ $$$$=\:\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{1}}\:=\:\sqrt{\mathrm{2}}\:−\mathrm{1} \\ $$

Answered by bramlexs22 last updated on 19/May/21

tan^(−1) x + tan^(−1) y = u ⇒tan (tan^(−1) x+tan^(−1) y)=tan u ⇒((tan (tan^(−1) x)+tan (tan^(−1) y))/(1−tan (tan^(−1) x)tan (tan^(−1) y)))=tan u ⇒ ((x+y)/(1−xy)) = tan u ⇒u = tan^(−1) (((x+y)/(1−xy)))

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

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

b)let arctanx =a and arctany =b ⇒x=tana and y =tanb ⇒((x+y)/(1−xy)) =((tana +tanb)/(1−tana .tanb)) =tan(a+b) ⇒ a+b =arctan(((x+y)/(1−xy)))=arctanx +arctany

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

Answered by bramlexs22 last updated on 19/May/21

sec x + tan x =(√2) ((1+sin x)/(cos x)) = (√2) ; cos x>0 ⇒1+sin x =(√(2(1−sin^2 x))) ⇒1+2sin x+sin^2 x=2−2sin^2 x ⇒3sin^2 x+2sin x−1=0 ⇒(3sin x−1)(sin x+1)=0 ⇒sin x=(1/3) ; sin x=−1(rejected) ⇒x = arcsin ((1/3))+2kπ , k∈Z

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

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

1) use scalar product a^2 =BC^(→2) =(AC^→ −AB^→ )^2 =AC^→_2 −2AB^→ .AC^→^2 +AB^→^2 =b^2 +c^2 −2bc cosA ⇒ cosA =((b^2 +c^2 −a^2 )/(2bc))

$$\left.\mathrm{1}\right)\:\mathrm{use}\:\mathrm{scalar}\:\mathrm{product}\:\:\:\mathrm{a}^{\mathrm{2}} =\mathrm{B}\overset{\rightarrow\mathrm{2}} {\mathrm{C}}=\left(\mathrm{A}\overset{\rightarrow} {\mathrm{C}}−\mathrm{A}\overset{\rightarrow} {\mathrm{B}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{A}\overset{\rightarrow_{\mathrm{2}} } {\mathrm{C}}−\mathrm{2A}\overset{\rightarrow} {\mathrm{B}}.\mathrm{A}\overset{\rightarrow^{\mathrm{2}} } {\mathrm{C}}\:+\mathrm{A}\overset{\rightarrow^{\mathrm{2}} } {\mathrm{B}}\:=\mathrm{b}^{\mathrm{2}} \:+\mathrm{c}^{\mathrm{2}} −\mathrm{2bc}\:\mathrm{cosA}\:\Rightarrow \\ $$$$\mathrm{cosA}\:=\frac{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }{\mathrm{2bc}} \\ $$

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