If the integral $\int_{0}^{10} \frac{\left[\sin 2 \pi_{\mathrm{x}}\right]}{\mathrm{e}^{\mathrm{x}-[\mathrm{x}]}} \mathrm{d} \mathrm{x}=\alpha \mathrm{e}^{-1}+\beta \mathrm{e}^{-\frac{1}{2}}+\gamma$, where $\alpha, \beta, \gamma$ are integers and $[x]$ denotes the greatest integer less than or equal to $\mathrm{x}$, then the value of $\alpha+\beta+\gamma$ is equal to :
Correct Option: 1,
Let $I=\int_{0}^{10} \frac{[\sin 2 \pi x]}{e^{x-[x]}} d x=\int_{0}^{10} \frac{[\sin 2 \pi x]}{e^{(x)}} d x$
Function $\mathrm{f}(\mathrm{x})=\frac{[\sin 2 \pi \mathrm{x}]}{\mathrm{e}^{|x|}}$ is periodic with
period '1' Therefore
$\mathrm{I}=10 \int_{0}^{1} \frac{[\sin 2 \pi \mathrm{x}]}{\mathrm{e}^{|\mathrm{x}|}} \mathrm{dx}$
$=10 \int_{0}^{1} \frac{\left[\sin 2 \pi_{\mathrm{x}}\right]}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}$
$=10\left(\int_{0}^{1 / 2} \frac{[\sin 2 \pi \mathrm{x}]}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}+\int_{1 / 2}^{1} \frac{[\sin 2 \pi \mathrm{x}]}{\mathrm{e}^{\mathrm{x}}} \mathrm{d} \mathrm{x}\right)$
$=10\left(0+\int_{1 / 2}^{1} \frac{(-1)}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\right)$
$=-10 \int_{1 / 2}^{1} e^{-x} d x$
$=10\left(e^{-1}-e^{-1 / 2}\right)$
Now,
$10 \cdot \mathrm{e}^{-1}-10 \cdot \mathrm{e}^{-1 / 2}=\alpha \mathrm{e}^{-1}+\beta \mathrm{e}^{-1 / 2}+\gamma($ given $)$
$\Rightarrow \alpha=10, \beta=-10, \gamma=0$
$\Rightarrow \alpha+\beta+\gamma=0$