A) \[\log (x+\sqrt{{{x}^{2}}+1})\]
B) \[\log (x\pm \sqrt{{{x}^{2}}-1})\]
C) \[\log (x\mp \sqrt{{{x}^{2}}+1})\]
D) \[\log (x-\sqrt{{{x}^{2}}-1})\]
Correct Answer: B
Solution :
Let \[{{\cosh }^{-1}}x=y\] \[\Rightarrow \] \[x=cosh\text{ }y\] \[\Rightarrow \] \[x=\frac{{{e}^{y}}+{{e}^{-y}}}{2}\] \[\Rightarrow \] \[{{e}^{y}}+{{e}^{-y}}=2x\] \[\Rightarrow \] \[{{e}^{2y}}-2{{e}^{y}}x+1=0\] \[\Rightarrow \] \[{{e}^{y}}=\frac{2x\pm \sqrt{4{{x}^{2}}-4}}{2}\] \[\Rightarrow \] \[{{e}^{y}}=x\pm \sqrt{{{x}^{2}}-1}\] \[\Rightarrow \] \[y=\log (x\pm \sqrt{{{x}^{2}}-1})\] \[\Rightarrow \] \[{{\cosh }^{-1}}x=\log (x\pm \sqrt{{{x}^{2}}-1})\]
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