The minimum value of the function f(x) = ∫₀² e^(|x - t|)dt is :

Solution

Maths sir · 4 steps

Split the integral at t = x so the modulus opens cleanly, then minimise using AM-GM.

1. Open the modulus. For x in [0, 2], |x - t| = x - t when t ≤ x and t - x when t ≥ x. So split the integral at t = x.

   f(x) = ∫₀ˣ eˣ⁻ᵗ dt + ∫ₓ² eᵗ⁻ˣ dt = (eˣ - 1) + (e²⁻ˣ - 1)

2. Apply AM-GM. So f(x) = eˣ + e²⁻ˣ - 2. Since eˣ · e²⁻ˣ = e², AM-GM gives eˣ + e²⁻ˣ ≥ 2√e² = 2e. Equality holds when eˣ = e²⁻ˣ, that is, at x = 1.
3. Outside the interval. For x ≥ 2, f(x) = eˣ⁻²(e² - 1), which only increases. By symmetry about x = 1, f also increases as x goes below 0. So the minimum lies inside [0, 2].
4. So. The minimum is f(1) = 2e - 2 = 2(e - 1), which is option (b).

The idea to keep. When the modulus contains both the variable of integration and a parameter, split the limits at the point where the inside becomes zero. Then treat what is left as an ordinary function of the parameter.

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