If the integral \(\int_{0}^{10} \frac{[\sin 2 \pi x]}{e^{x-[x]}} d x=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma\) 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:

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Definite Integrals

If the integral

\int_{0}^{10} \frac{[ s i n 2 π x ]}{e ^{x - [x]}} d x = α e^{- 1} + β e^{- \frac{1}{2}} + γ

where

α, β, γ

are integers and

[x]

denotes the greatest integer less than or equal to

x

, then the value of

α + β + γ

is equal to:

Answer: (D)

Both

[sin 2 π x]

and

e^{x - [x]}

have period

1

, so

\int_{0}^{10} \frac{[ sin 2 π x ]}{e ^{x - [x]}} d x = 10 \int_{0}^{1} \frac{[ sin 2 π x ]}{e ^{x}} d x

(0, \frac{1}{2})

the sine lies in

(0, 1]

, so its greatest integer part is

0

apart from the single point where the sine equals

1

; on

(\frac{1}{2}, 1)

the sine lies in

[- 1, 0)

, so the greatest integer part is

- 1

. Therefore

\int_{0}^{1} \frac{[ sin 2 π x ]}{e ^{x}} d x = - \int_{1/2}^{1} e^{- x} d x = - (e^{- 1/2} - e^{- 1}) = e^{- 1} - e^{- \frac{1}{2}}

and the whole integral is

10 e^{- 1} - 10 e^{- \frac{1}{2}}

. Comparing with

α e^{- 1} + β e^{- \frac{1}{2}} + γ

gives

α = 10

β = - 10

γ = 0

, so

α + β + γ = 0

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option