Question

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

• A. 0
• B. 20
• C. 25
• D. 10

Answer 1

Answer: (A)

Let $I=\underset{0}{\overset{10}{\int }}\frac{[\sin 2\pi x]}{e^{x-[x]}}dx=\underset{0}{\overset{10}{\int }}\frac{[\sin 2\pi x]}{e^{\{x\}}}dx$
Function $f(x)=\frac{[\sin 2\pi x]}{e^{\{x\}}}$ is periodic with period '1' Therefore
$I=10\underset{0}{\overset{1}{\int }}\frac{[\sin 2\pi x]}{e^{\{x\}}}dx$
$=10\underset{0}{\overset{1}{\int }}\frac{[\sin 2\pi x]}{e^x}dx$
$=10\left(\underset{0}{\overset{1/2}{\int }}\frac{[\sin 2\pi x]}{e^x}dx+\underset{1/2}{\overset{1}{\int }}\frac{[\sin 2\pi x]}{e^x}dx\right)$
$=10\left(0+\underset{1/2}{\overset{1}{\int }}\frac{(-1)}{e^x}dx\right)$
$=-10\underset{1/2}{\overset{1}{\int }}{e}^{-x}dx$
$=10\left(e^{-1}-e^{-1/2}\right)$
Now,
$10\cdot e^{-1}-10\cdot e^{-1/2}=\alpha e^{-1}+\beta e^{-1/2}+\gamma$ (given)
$\Rightarrow \alpha =10,\beta =-10,\gamma =0$
$\Rightarrow \alpha +\beta +\gamma =0$