Let a be a positive real number

$a>0$

Let $n \leq a

Here $[a]=n$

Now, $\int_{0}^{a} e^{x-[x]} d x=10 e-9$

$\Rightarrow \int_{0}^{n} e^{\{x\}} d x+\int_{n}^{a} e^{x-[x]} d x=10 e-9$

$\therefore \quad n \int_{0}^{1} e^{x} d x+\int_{n}^{a} e^{x-n} d x=10 e-9$

$\Rightarrow \mathrm{n}(\mathrm{e}-1)+\left(\mathrm{e}^{\mathrm{a}-\mathrm{n}}-1\right)=10 \mathrm{e}-9$

$\therefore \quad \mathrm{n}=0 \quad$ and $\{\mathrm{a}\}=\log _{\mathrm{e}} 2$

So, $a=[a]+\{a\}=\left(10+\log _{e} 2\right)$

$\Rightarrow$ Option (2) is correct