The minimum value of the twice differentiable function f(x)=∫₀ˣ eˣ⁻ᵗ f′(t) dt-(x²-x+1) eˣ, x ∈ R, is :

Solution

Maths sir · 4 steps

Pull eˣ out of the integral, divide it across, and one differentiation hands you f(x) itself.

1. Separate the variables. Inside the integral eˣ⁻ᵗ = eˣ e⁻ᵗ, and eˣ does not depend on t, so it comes outside. Divide the whole equation by eˣ.

   f(x) e⁻ˣ = ∫₀ˣ e⁻ᵗ f'(t) dt - (x² - x + 1)

2. Differentiate once. The left side gives f'(x)e⁻ˣ - f(x)e⁻ˣ. The right side gives e⁻ˣf'(x) - (2x-1). The f'(x)e⁻ˣ terms cancel, leaving f(x)e⁻ˣ = 2x - 1, so f(x) = (2x-1)eˣ. Check: f(0) = -1 from both forms.
3. Find the critical point. f'(x) = 2eˣ + (2x-1)eˣ = (2x+1)eˣ. Since eˣ > 0, f'(x) = 0 only at x = -1/2. The sign of f' goes from negative to positive there, so it is the minimum.
4. So. f(-1/2) = (-1-1)e^(-1/2) = -2/√e, which is option (a).

The idea to keep. When x sits inside an integral with t, split it off first, as in eˣ⁻ᵗ = eˣ e⁻ᵗ. Then differentiate using the Leibniz rule, and the unknown function usually falls out.

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