Question:
Let a be a positive real number such that $\int_{0}^{\mathrm{a}} \mathrm{e}^{\mathrm{x}-[\mathrm{x}]} \mathrm{dx}=10 \mathrm{e}-9$ where $[\mathrm{x}]$ is the greatest integer less than or equal to $x$. Then a is equal to :
Correct Option: , 2
Solution:
$a>0$
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