Equal First Three Moments on [0,1]

Detailed Solution

Since \[ \int_0^1 f(x)\,dx = \int_0^1 f(x)^2\,dx, \] we have \[ \int_0^1\bigl(f(x)-f(x)^2\bigr)\,dx=0. \] Factoring, \[ \int_0^1 f(x)\bigl(1-f(x)\bigr)\,dx=0. \] Because \(0\le f(x)\le1\) for all \(x\), the integrand \[ f(x)(1-f(x)) \] is nonnegative everywhere on \([0,1]\). A nonnegative continuous function whose integral is zero must vanish identically. Hence \[ f(x)(1-f(x))=0 \qquad\text{for all }x\in[0,1]. \] Therefore, \[ f(x)\in\{0,1\} \qquad\text{for all }x\in[0,1]. \] Since \(f\) is continuous and \([0,1]\) is connected, the image \(f([0,1])\) must also be connected. The only connected subsets of \(\{0,1\}\) are \(\{0\}\) and \(\{1\}\). Consequently, \(f\) is constant, and therefore \[ f\equiv0 \quad\text{or}\quad f\equiv1. \] Thus the correct option is \[ \boxed{\mathrm{A}}. \] Option (B) is false because continuity forces a \(\{0,1\}\)-valued function on a connected interval to be constant. Option (C) is false since \[ \int_0^1 x\,dx=\frac12 \neq \frac13=\int_0^1 x^2\,dx. \] Option (D) is false because \(f\equiv1\) also satisfies all the given conditions.