If the integral
∫₀¹⁰ [sin 2π x]/e^(x - [x]) dx = α e⁻¹ + β e^(- 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 :

Solution

Maths sir · 4 steps

Both [sin 2π x] and x-[x] repeat every 1 unit, so solve one period and multiply by 10.

1. Use the period. sin 2π x has period 1, and so does the fractional part x-[x]. The integrand repeats every unit, so ∫₀¹⁰ = 10∫₀¹.
2. Split one period. On (0, 1/2), 0 < sin 2π x ≤ 1, so [sin 2π x] = 0 (it equals 1 only at the single point x = 1/4, which does not affect the integral). On (1/2, 1), -1 ≤ sin 2π x < 0, so [sin 2π x] = -1. Also x - [x] = x on (0,1).

   ∫₀¹ [sin 2π x]/eˣ dx = -∫_(1/2)¹ e⁻ˣ dx = [e⁻ˣ]_(1/2)¹ = e⁻¹ - e^(-1/2)

3. Multiply by 10. The full integral is 10e⁻¹ - 10e^(-1/2) + 0. Matching terms gives α = 10, β = -10, γ = 0.
4. So. α + β + γ = 10 - 10 + 0 = 0, which is option (a).

The idea to keep. When a greatest integer function sits on a periodic function, first look for the common period, then find where the bracket is constant inside one period.

Answer from the NTA final answer key. Solution by PrepWiser.