A) \[0\]
B) \[{{\log }_{e}}6\]
C) \[{{\log }_{e}}7\]
D) \[{{\log }_{e}}42\]
Correct Answer: C
Solution :
[c] \[I=\int\limits_{5}^{41}{\frac{dx}{{{({{f}^{-1}}(x))}^{5}}+5{{f}^{-1}}(x)}}\] Put \[{{f}^{-1}}(x)=t\] Now \[f(x)={{x}^{5}}+5x-1\] \[\Rightarrow \,\,\,f(1)=5\Rightarrow \,\,{{f}^{-1}}(5)=1\] and \[f(2)=41\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,{{f}^{-1}}(41)=2\] \[\therefore \,\,\,\,I=\int\limits_{1}^{2}{\frac{f'(t)}{{{t}^{5}}+5t}}dt=\int\limits_{1}^{2}{\frac{5{{t}^{4}}+5}{{{t}^{5}}+5t}}dt\] \[=\left[ {{\log }_{e}}({{t}^{5}}+5t) \right]_{1}^{2}={{\log }_{e}}42-{{\log }_{e}}6={{\log }_{e}}7\]
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