Question

Let \(f(x)\) be a continuous function defined in \([0,2] \to \mathbb{R}\) and satisfying the equation
\[ \int_0^2 f(x)[x - f(x)]\,dx = \frac{2}{3} \]
The value of \(f(1)\) is

• A. 1
• B. 2
• C. \(\frac{1}{2}\)
• D. 0

Hint:

Complete the square in \(f(x)\) inside the integrand and compare the leftover term with the value of \(\int_0^2 \frac{x^2}{4}\,dx\).

Answer 1

The Correct Option is C

Step 1: Treat the problem as a pointwise maximization.
For any fixed real number $x$, look at $g(t) = xt - t^2$ as a function of a single variable $t$, standing in for $f(x)$.
$g$ is a downward opening parabola in $t$, so it has one maximum, found by setting $g'(t) = x - 2t = 0$, which gives $t = x/2$.

Step 2: Find the maximum value of g(t).
At $t = x/2$: $g(x/2) = x\cdot\frac{x}{2} - \left(\frac{x}{2}\right)^2 = \frac{x^2}{2} - \frac{x^2}{4} = \frac{x^2}{4}$.
So for every real $t$, $xt - t^2 \le \frac{x^2}{4}$, with equality only when $t = x/2$.

Step 3: Apply this bound pointwise and integrate.
Since $f(x)[x - f(x)] = xf(x) - f(x)^2 \le \frac{x^2}{4}$ for every $x$ in $[0,2]$, integrating both sides over $[0,2]$ gives
\[ \int_0^2 f(x)[x-f(x)]\,dx \le \int_0^2 \frac{x^2}{4}\,dx \]
The right side works out to $\int_0^2 \frac{x^2}{4}\,dx = \frac{1}{4}\cdot\frac{8}{3} = \frac{2}{3}$.

Step 4: Compare with the given value.
The question states $\int_0^2 f(x)[x-f(x)]\,dx = \frac{2}{3}$, which is exactly the upper bound found in Step 3.
Since the integral equals its largest possible value, the pointwise inequality from Step 2 must actually be an equality at every single $x$; if it were strict anywhere, the integral (by continuity of $f$) would come out strictly less than $\frac{2}{3}$.

Step 5: Solve for f(x) and evaluate at x = 1.
Equality in Step 2 happens only when $t = x/2$, so $f(x) = x/2$ for every $x$ in $[0,2]$.
So $f(1) = 1/2 = 0.5$.
This matches option (C), and rules out (A) 1, (B) 2 and (D) 0, since none of those values come from $f(x)=x/2$ at $x=1$.

Final Answer:
\[ f(1) = \frac{1}{2} \]