Let a be a positive real number such that ∫₀ᵃ e^(x - [x]) dx = 10e - 9 where [ x ] is the greatest integer less than or equal to x. Then a is equal to:

Solution

Maths sir · 5 steps

The function repeats every 1 unit, so count the full periods first and then find the leftover piece.

1. Spot the period. x - [x] is the fractional part of x, so e^(x-[x]) repeats with period 1. On [0, 1) it is just eˣ.
2. One period. ∫₀¹ eˣ dx = e - 1. So n full periods contribute n(e-1).
3. Split the target. Write 10e - 9 = 10(e-1) + 1. That is 10 full periods plus a leftover integral equal to 1. So a = 10 + t with 0 ≤ t < 1.

   ∫₁₀¹⁰⁺ᵗ e^(x-[x]) dx = ∫₀ᵗ eˣ dx = eᵗ - 1 = 1

4. Solve for t. eᵗ = 2, so t = logₑ 2 ≈ 0.693. This lies in [0, 1), so it is a valid leftover.
5. So. a = 10 + logₑ 2, which is option (b).

The idea to keep. For a periodic integrand, split the integral into whole periods plus one leftover piece. The leftover length must be less than one period, so always check it.

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