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

Options

A. `(-oo,1]`
B. `[-1,oo)`
C. `[-1,1]`
D. none of these

Explanation

To solve the problem, we need to find the values of \( x \) for which the function \( f(x) \) is increasing, 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, we have:

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

For the right-hand side, we differentiate:

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

Using the Leibniz rule for differentiation under the integral sign, we get:

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

Thus, the right-hand side becomes:

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

Equating both sides gives us:

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

### Step 2: Solve for \( f(x) \)

Rearranging the equation, we have:

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

Factoring out \( f(x) \):

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

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

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

### Step 3: Find the derivative \( f'(x) \)

To determine where \( f(x) \) is increasing, we need to find the derivative \( f'(x) \):

Using the quotient rule:

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

This simplifies to:

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

### Step 4: Determine where \( f'(x) > 0 \)

For \( f(x) \) to be increasing, we need:

\[ f'(x) > 0 \implies \frac{1 - x^2}{(1 + x^2)^2} > 0 \]

Since \( (1 + x^2)^2 > 0 \) for all \( x \), we only need to consider:

\[ 1 - x^2 > 0 \implies x^2 < 1 \]

This gives us:

\[ -1 < x < 1 \]

### Step 5: Conclusion

Thus, the values of \( x \) for which \( f(x) \) is increasing are in the interval:

\[ (-1, 1) \]