A) ke
B) \[{{k}^{-1}}e\]
C) \[k{{e}^{-1}}\]
D) \[{{k}^{-1}}{{e}^{-1}}\]
Correct Answer: D
Solution :
| [d] Let \[P=\underset{n\to \infty }{\mathop{Lim}}\,{{\left\{ \frac{n!}{{{(kn)}^{n}}} \right\}}^{\frac{1}{n}}}\] |
| Taking log of both the sides at the base e |
| \[{{\log }_{e}}P=\underset{n\to \infty }{\mathop{Lim}}\,\frac{1}{n}{{\log }_{e}}\left\{ \frac{n!}{{{(kn)}^{n}}} \right\}\] |
| \[=\underset{n\to \infty }{\mathop{Lim}}\,\frac{1}{n}{{\log }_{e}}\left\{ \frac{1}{kn}.\frac{2}{kn}.\frac{3}{kn}.........\,\,.........\frac{n}{kn} \right\}\] |
| \[=\underset{n\to \infty }{\mathop{Lim}}\,\frac{1}{n}\left[ \log \left( \frac{1}{kn} \right)+\log \left( \frac{2}{kn} \right)+......+\log \left( \frac{n}{kn} \right) \right]\] |
| \[=\underset{n\to \infty }{\mathop{Lim}}\,\frac{1}{n}\sum\limits_{r=1}^{n}{\log \left( \frac{r}{kn} \right)}\] |
| \[=\int\limits_{0}^{1}{\log \left( \frac{x}{k} \right)dx=\int\limits_{0}^{1}{(\log \,\,x-\log \,\,k)dx}}\] |
| \[=\int\limits_{0}^{1}{\log x\,dx-\int\limits_{0}^{1}{\log \,\,k\,\,dx}}\] |
| \[=\left[ x\log x-x \right]_{0}^{1}-\log k\left[ x \right]_{0}^{1}\] |
| \[=\left[ 0-1-0+0 \right]-\log k=-1-\log \,\,k\] |
| \[=-(\log \,\,e+\log \,\,k)=-\log (ek)=\log \frac{1}{ek}\] |
| [Value of \[x\,\,\log x\,\,at\,\,x=0\] is \[\underset{x\to {{0}^{+}}}{\mathop{Lim}}\,x\log x=0\]] |
| \[\therefore \,\,\,\,\,P=\frac{1}{ek}={{k}^{-1}}{{e}^{-1}}\] |
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