Correct : c

The given condition is ∫₀² f(x)[x − f(x)]dx = 2/3, which expands to ∫₀² [xf(x) − f(x)²]dx = 2/3.
The key trick is to consider the integral of the perfect square [f(x) − x/2]²:
∫₀² [f(x) − x/2]² dx = ∫₀² f(x)²dx − ∫₀² xf(x)dx + ∫₀² (x²/4)dx
From the given condition, ∫₀² xf(x)dx − ∫₀² f(x)²dx = 2/3, so ∫₀² f(x)²dx − ∫₀² xf(x)dx = −2/3.
Computing the remaining integral: ∫₀² (x²/4)dx = [x³/12]₀² = 8/12 = 2/3.
Adding everything: ∫₀² [f(x) − x/2]² dx = −2/3 + 2/3 = 0.
Since [f(x) − x/2]² is non-negative everywhere on [0,2] and continuous, and its integral equals zero, the function itself must be identically zero throughout the interval. This forces f(x) = x/2 for all x in [0,2].
At x = 1: f(1) = 1/2.
Correct answer: C) 1/2.