JEE Main 2021 (Online) 17th March Evening ShiftQuestion 1
If the integral $\int_{0}^{10}\frac{[\sin2\pi x]}{e^{x-[x]}}dx=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma$, 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 :
Definite Integration
Solution
Concept. Both $[\sin2\pi x]$ and $e^{x-[x]}$ have period $1$. In one period, $[\sin2\pi x]$ is $0$ on the first half and $-1$ on the second half. Formulas used. (1) On $[0,1)$: $[\sin2\pi x]=0$ for $0\le x\le\frac12$ (value $1$ only at $x=\frac14$), $=-1$ for $\frac12<x<1$(2) $\int_{1/2}^1e^{-x}dx=e^{-1/2}-e^{-1}$Step 1 - One period. By (1), (2):$$\int_0^1\frac{[\sin2\pi x]}{e^{x}}dx=-\big(e^{-1/2}-e^{-1}\big).$$ Step 2 - Ten periods. $10e^{-1}-10e^{-1/2}$. So $\alpha=10$, $\beta=-10$, $\gamma=0$. Step 3 - Value. $\alpha+\beta+\gamma=0$. Answer: (A)