A) \[\ln 2-1+\lambda \]
B) \[\ln 2-3+3\lambda \]
C) \[2\ln 2-\lambda \]
D) \[\ln 4-3+3\lambda \]
Correct Answer: B
Solution :
\[P=\underset{n\to \infty }{\mathop{\text{Lim}}}\,{{\frac{\left( \prod\limits_{r=1}^{n}{({{n}^{3}}+{{r}^{3}})} \right)}{{{n}^{3}}}}^{1/n}}\] \[=\underset{n\to \infty }{\mathop{\text{Lim}}}\,{{\left( \prod\limits_{r=1}^{n}{\left( 1+\frac{{{r}^{3}}}{{{n}^{3}}} \right)} \right)}^{1/n}}\] \[\Rightarrow \]\[\ln p=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\prod\limits_{r=1}^{n}{\ln }\left( 1+\frac{{{r}^{3}}}{{{n}^{3}}} \right)\] \[=\int\limits_{0}^{1}{\ln (1+{{x}^{3}})dx}\] \[\left. =x.\ln (1+{{x}^{3}}) \right]_{0}^{1}-\int\limits_{0}^{1}{\frac{1}{1+{{x}^{3}}},3{{x}^{2}}\times x\,dx}\] \[=\ln 2-3\int\limits_{0}^{1}{\frac{1+{{x}^{3}}-1}{1+{{x}^{3}}}dx}\] \[=\ln 2-3\int\limits_{0}^{1}{\frac{1}{1+{{x}^{3}}}dx}\] \[=\ln 2-3+3\int\limits_{0}^{1}{\frac{1}{1+{{x}^{3}}}dx}\]
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