Question

If the integral $\int\limits_{0}^{10} \frac{[\sin 2 \pi x]}{e^{x-[x]}} d x=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma$, 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 $I=\int\limits_{0}^{10} \frac{[\sin 2 \pi x]}{e^{x-[x]}} d x=\int\limits_{0}^{10} \frac{[\sin 2 \pi x]}{e^{\{x\}}} d x$
Function $f(x)=\frac{[\sin 2 \pi x]}{e^{\{x\}}}$ is periodic with period '1' Therefore
$I=10 \int\limits_{0}^{1} \frac{[\sin 2 \pi x]}{e^{\{x\}}} d x$
$=10 \int\limits_{0}^{1} \frac{[\sin 2 \pi x]}{ e ^{ x }} dx$
$=10\left(\int\limits_{0}^{1 / 2} \frac{[\sin 2 \pi x]}{ e ^{x}} d x+\int\limits_{1 / 2}^{1} \frac{[\sin 2 \pi x]}{ e ^{x}} d x\right)$
$=10\left(0+\int\limits_{1 / 2}^{1} \frac{(-1)}{ e ^{ x }} dx \right)$
$=-10 \int\limits_{1 / 2}^{1} e ^{- x } dx$
$=10\left( e ^{-1}- e ^{-1 / 2}\right)$
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$