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Mathematics · Definite Integration

JEE Advanced 2018 — Paper 1 — Question 37

For each positive integer n , let yn=1n((n+1)(n+2)…(n+n))1/n.\mathrm{y}_{\mathrm{n}}=\frac{1}{\mathrm{n}}((\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{n}))^{1 / \mathrm{n}} . For x∈Rx \in R, let [x][x] be the greatest integer less than or equal to xx. If lim⁡n→∞yn=L\lim _{n \rightarrow \infty} y_{n}=L, then the value of [L] is ____\_\_\_\_.

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

ln⁡(L)=lim⁡n→∞1n∑r=1nln⁡(n+rn)=∫01ln⁡(1+x)dx=∣xln⁡(1+x)∣01−∣x−ln⁡(1+x)∣01=ln⁡2−(1−ln⁡2)=ln⁡(4/e)⇒L=4/e⇒[ L]=1\begin{aligned} \ln (\mathrm{L}) & =\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}} \sum_{\mathrm{r}=1}^{\mathrm{n}} \ln \left(\frac{\mathrm{n}+\mathrm{r}}{\mathrm{n}}\right) \\& =\int_{0}^{1} \ln (1+\mathrm{x}) \mathrm{dx} \\& =|\mathrm{x} \ln (1+\mathrm{x})|_{0}^{1}-|\mathrm{x}-\ln (1+\mathrm{x})|_{0}^{1} \\& =\ln 2-(1-\ln 2) \\& =\ln (4 / \mathrm{e}) \\& \Rightarrow \quad \mathrm{L} =4 / \mathrm{e} \Rightarrow[\mathrm{~L}]=1 \end{aligned}

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Mathematics
Chapter
Definite Integration
Topic
Integration as a limit of sum

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