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(C) Let $n = [a]$,where $n$ is a non-negative integer. Then $a = n + \{a\}$,where $0 \le \{a\} < 1$.The integral can be split as:$\int_{0}^{a} e^{x-[x

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$10 + \log_{e} 2$

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(C) Let $n = [a]$,where $n$ is a non-negative integer. Then $a = n + \{a\}$,where $0 \le \{a\} The integral can be split as:$\int_{0}^{a} e^{x-[x]} dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} e^{x-k} dx + \int_{n}^{a} e^{x-n} dx = 10e - 9$Evaluating the sum:$\sum_{k=0}^{n-1} [e^{x-k}]_{k}^{k+1} = \sum_{k=0}^{n-1} (e^1 - e^0) = \sum_{k=0}^{n-1} (e - 1) = n(e - 1)$Evaluating the remaining part:$\int_{n}^{a} e^{x-n} dx = [e^{x-n}]_{n}^{a} = e^{a-n} - e^0 = e^{\{a\}} - 1$Combining these:$n(e - 1) + e^{\{a\}} - 1 = 10e - 9$$ne - n + e^{\{a\}} - 1 = 10e - 10$Comparing the terms,we get $n = 10$ and $e^{\{a\}} - 1 = -1 + 10e - 10e = 0$ is incorrect. Let's re-evaluate:$ne - n + e^{\{a\}} - 1 = 10e - 10$$ne + e^{\{a\}} - (n + 1) = 10e - 10$If $n = 10$,then $10e + e^{\{a\}} - 11 = 10e - 9 \Rightarrow e^{\{a\}} = 2 \Rightarrow \{a\} = \log_{e} 2$.Thus,$a = n + \{a\} = 10 + \log_{e} 2$.

Let $a$ be a positive real number such that $\int_{0}^{a} e^{x-[x]} dx = 10e - 9$,where $[x]$ is the greatest integer less than or equal to $x$. Then $a$ is equal to:

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