A) \[\frac{-3+\sqrt{5+4{{e}^{x}}}}{2}\]
B) \[\frac{-3\pm \sqrt{5+4{{e}^{x}}}}{2}\]
C) \[\frac{-3-\sqrt{5+4{{e}^{x}}}}{2}\]
D) None of the above
Correct Answer: B
Solution :
Given,\[f(x)=\log ({{x}^{2}}+3x+1)\] \[\therefore \] \[f'(x)=\frac{2x+3}{({{x}^{2}}+3x+1)}>0\] \[\forall x\in [1,\,\,3]\]. which is a strictly increasing function. Thus, \[f(x)\] is injective, given that \[f(x)\] is onto. Hence, the given function \[f(x)\] is invertible. Now,\[f\{{{f}^{-1}}(x)\}=x\] \[\Rightarrow \] \[\log {{\{{{f}^{-1}}(x)\}}^{2}}+3\{{{f}^{-1}}(x)+1\}=x\] \[\Rightarrow \] \[{{\{{{f}^{-1}}(x)\}}^{2}}+\{{{f}^{-1}}(x)\}+1-{{e}^{x}}=0\] \[\therefore \] \[{{f}^{-1}}(x)=\frac{-3\pm \sqrt{9-4\cdot 1(1-{{e}^{x}})}}{2}\] \[=\frac{-3\pm \sqrt{5+4{{e}^{x}}}}{2}\] \[\Rightarrow \]\[{{f}^{-1}}(x)=\frac{-3\pm \sqrt{5+4{{e}^{x}}}}{2}\,\,\,\,[\because {{f}^{-1}}(x)\in [1,\,\,3]]\] Hence,\[{{f}^{-1}}(x)=\frac{-3\pm \sqrt{5+4{{e}^{x}}}}{2}\]
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