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Question Number 71235 by Rio Michael last updated on 13/Oct/19

sinh[ln (x + (√(1 + x^2 ))) ] ≡ A. 2x B. (1/x) C. x^2 D. x

$${sinh}\left[{ln}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:\right]\:\equiv\: \\ $$$$ \\ $$$${A}.\:\:\mathrm{2}{x} \\ $$$${B}.\:\:\frac{\mathrm{1}}{{x}} \\ $$$${C}.\:\:{x}^{\mathrm{2}} \\ $$$${D}.\:\:{x} \\ $$

Commented by mathmax by abdo last updated on 13/Oct/19

sh{ln(x+(√(1+x^2 )))} =((e^(ln(x+(√(1+x^2 )))) −e^(−ln(x+(√(1+x^2 )))) )/2) =((x+(√(1+x^2 ))−(1/(x+(√(1+x^2 )))))/2) =(((x+(√(1+x^2 )))^2 −1)/(2(x+(√(1+x^2 ))))) =((x^2 +2x(√(1+x^2 )) +1+x^2 −1)/(2(x+(√(1+x^2 ))))) =((2x^2 +2x(√(1+x^2 )))/(2(x+(√(1+x^2 ))))) =((x(x+(√(1+x^2 )))/(x+(√(1+x^2 )))) =x also we can use that ln(x+(√(1+x^2 )))=argsh(x)=sh^(−1) (x)

$${sh}\left\{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right\}\:=\frac{{e}^{{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} −{e}^{−{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.} }{\mathrm{2}} \\ $$$$=\frac{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}}{\mathrm{2}}\:=\frac{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:+\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\right.}\:=\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{x}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right.}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:={x} \\ $$$${also}\:{we}\:{can}\:{use}\:{that}\:{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)={argsh}\left({x}\right)={sh}^{−\mathrm{1}} \left({x}\right) \\ $$

Answered by MJS last updated on 13/Oct/19

sinh α =(1/2)(e^α −e^(−α) ) ⇒ answer is D. x

$$\mathrm{sinh}\:\alpha\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\alpha} −\mathrm{e}^{−\alpha} \right) \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{D}.\:{x} \\ $$

Commented by Rio Michael last updated on 13/Oct/19

i don′t understand the working sir

$${i}\:{don}'{t}\:{understand}\:{the}\:{working}\:{sir} \\ $$

Commented by MJS last updated on 13/Oct/19

first, sorry I made a typo in the formula sinh ln (x+(√(x^2 +1))) =(1/2)(x+(√(x^2 +1))−(1/(x+(√(x^2 +1)))))= =(1/2)×(((x+(√(x^2 +1)))^2 −1)/(x+(√(x^2 +1))))=(1/2)×((2x^2 +2x(√(x^2 +1)))/(x+(√(x^2 +1))))=x

$$\mathrm{first},\:\mathrm{sorry}\:\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{the}\:\mathrm{formula} \\ $$$$ \\ $$$$\mathrm{sinh}\:\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}={x} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 13/Oct/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by mr W last updated on 13/Oct/19

let t=ln (x+(√(1+x^2 ))) e^t =x+(√(1+x^2 )) ...(i) e^(−t) =(1/(x+(√(1+x^2 ))))=−x+(√(1+x^2 )) −e^(−t) =x−(√(1+x^2 )) ...(ii) e^t −e^(−t) =2x ((e^t −e^(−t) )/2)=x ⇒sinh t=x i.e. sinh [ln (x+(√(1+x^2 )))]=x ⇒answer D

$${let}\:{t}=\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$${e}^{{t}} ={x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:...\left({i}\right) \\ $$$${e}^{−{t}} =\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=−{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$−{e}^{−{t}} ={x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:...\left({ii}\right) \\ $$$${e}^{{t}} −{e}^{−{t}} =\mathrm{2}{x} \\ $$$$\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}={x} \\ $$$$\Rightarrow\mathrm{sinh}\:{t}={x} \\ $$$${i}.{e}.\:\mathrm{sinh}\:\left[\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]={x} \\ $$$$\Rightarrow{answer}\:{D} \\ $$

Commented by Rio Michael last updated on 13/Oct/19

thanks sirs

$${thanks}\:{sirs} \\ $$

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