Question

In the intergal $\underset{0}{\overset{10}{\int }}\frac{\left[\sin 2\pi x\right]}{e^{x-\left[x\right]}}dx=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma ,$ where $\alpha ,\beta ,\gamma$ are integers and $\left[x\right]$ denotes the greatest integer less than or equal to $x,$ then the value of $\alpha +\beta +\gamma$ is equal to:

• A. $20$
• B. $0$
• C. $25$
• D. $10$

Answer 1

The correct option is B $0$

$\frac{\left[\sin 2\pi x\right]}{e^{x-\left[x\right]}}$ is periodic function with period $1.$
So,
$\underset{0}{\overset{10}{\int }}\frac{\left[\sin 2\pi x\right]}{e^{x-\left[x\right]}}dx=10\underset{0}{\overset{1}{\int }}\frac{\left[\sin 2\pi x\right]}{e^{\text{{}x\text{}}}}dx$
$=10\left[\underset{0}{\overset{\frac{1}{2}}{\int }}0\cdot dx+\underset{\frac{1}{2}}{\overset{1}{\int }}\frac{-1}{e^x}dx\right]$
$=-10{\left[\frac{e^{-x}}{-1}\right]}_{\frac{1}{2}}^1$
$=-10[-e^{-1}+e^{-\frac{1}{2}}]$
$=10e^{-1}-10e^{-\frac{1}{2}}$
Comparing with the given relation,
$\Rightarrow \alpha =10,\beta =-10,\gamma =0$
$\therefore \alpha +\beta +\gamma =0.$