`y=f(x)` satisfies the relation `int_(2)^(x)f(t)dt=(x^(2))/2+int_(x)^(2)t^(2)f(t)dt`
The range of `y=f(x)` is

Options

A. `[0,oo)`
B. `R`
C. `(-oo,0]`
D. `[-1/2,1/2]`

Explanation

To solve the problem, we need to find the range of the function \( f(x) \) given the relation:

\[ \int_{2}^{x} f(t) dt = \frac{x^2}{2} + \int_{x}^{2} t^2 f(t) dt \]

### Step 1: Differentiate both sides with respect to \( x \)

We start by differentiating both sides of the equation with respect to \( x \).

Using the Fundamental Theorem of Calculus on the left-hand side, we have:

\[ \frac{d}{dx} \left( \int_{2}^{x} f(t) dt \right) = f(x) \]

For the right-hand side, we differentiate each term:

1. The derivative of \( \frac{x^2}{2} \) is \( x \).
2. For the integral \( \int_{x}^{2} t^2 f(t) dt \), we apply the Leibniz rule:

\[ \frac{d}{dx} \left( \int_{x}^{2} t^2 f(t) dt \right) = -x^2 f(x) \]

Putting it all together, we get:

\[ f(x) = x - x^2 f(x) \]

### Step 2: Rearranging the equation

Now, we rearrange the equation to isolate \( f(x) \):

\[ f(x) + x^2 f(x) = x \]

Factoring out \( f(x) \):

\[ f(x)(1 + x^2) = x \]

### Step 3: Solve for \( f(x) \)

Now, we can solve for \( f(x) \):

\[ f(x) = \frac{x}{1 + x^2} \]

### Step 4: Find the range of \( f(x) \)

To find the range of \( f(x) = \frac{x}{1 + x^2} \), we set \( y = f(x) \):

\[ y = \frac{x}{1 + x^2} \]

Rearranging gives us:

\[ y(1 + x^2) = x \implies yx^2 - x + y = 0 \]

This is a quadratic equation in \( x \). For \( x \) to have real solutions, the discriminant must be non-negative:

\[ D = (-1)^2 - 4(y)(y) \geq 0 \]

Calculating the discriminant:

\[ 1 - 4y^2 \geq 0 \]

### Step 5: Solve the inequality

Solving the inequality:

\[ 1 \geq 4y^2 \implies y^2 \leq \frac{1}{4} \]

Taking the square root gives:

\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

### Conclusion

Thus, the range of the function \( f(x) \) is:

\[ \boxed{[-\frac{1}{2}, \frac{1}{2}]} \]