Let `f:RtoR` be a differentiable function such that `f(x)=x^(2)+int_(0)^(x)e^(-t)f(x-t)dt`.
`y=f(x)` is

Options

A. (a) injective but not surjective
B. (b) surjective but not injective
C. (c) bijective
D. (d) neither injective nor surjective

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined by the equation:

\[ f(x) = x^2 + \int_0^x e^{-t} f(x-t) dt \]

### Step 1: Differentiate \( f(x) \)

We start by differentiating \( f(x) \) with respect to \( x \). Using the Leibniz rule for differentiation under the integral sign, we have:

\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\int_0^x e^{-t} f(x-t) dt\right) \]

The derivative of \( x^2 \) is \( 2x \). For the integral, we apply the Leibniz rule:

\[ \frac{d}{dx}\left(\int_0^x e^{-t} f(x-t) dt\right) = e^{-x} f(0) + \int_0^x e^{-t} \frac{d}{dx} f(x-t) dt \]

### Step 2: Substitute into the derivative

Thus, we have:

\[ f'(x) = 2x + e^{-x} f(0) \]

### Step 3: Find \( f(0) \)

To find \( f(0) \), we substitute \( x = 0 \) into the original equation:

\[ f(0) = 0^2 + \int_0^0 e^{-t} f(0-t) dt = 0 + 0 = 0 \]

So, \( f(0) = 0 \).

### Step 4: Simplify \( f'(x) \)

Substituting \( f(0) = 0 \) back into the derivative:

\[ f'(x) = 2x + e^{-x} \cdot 0 = 2x \]

### Step 5: Integrate \( f'(x) \)

Now we integrate \( f'(x) \):

\[ f(x) = \int 2x \, dx = x^2 + C \]

### Step 6: Determine the constant \( C \)

Using the value of \( f(0) \):

\[ f(0) = 0^2 + C = 0 \implies C = 0 \]

Thus, we have:

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

### Step 7: Analyze the function \( f(x) \)

Now we need to determine if \( f(x) = x^2 \) is injective, surjective, or bijective.

1. **Injective (One-to-One)**: A function is injective if \( f(a) = f(b) \) implies \( a = b \). For \( f(x) = x^2 \), \( f(1) = f(-1) = 1 \), hence it is not injective.

2. **Surjective (Onto)**: A function is surjective if for every \( y \) in the codomain, there exists an \( x \) such that \( f(x) = y \). The range of \( f(x) = x^2 \) is \( [0, \infty) \), which does not cover all real numbers \( \mathbb{R} \), hence it is not surjective.

3. **Bijective**: A function is bijective if it is both injective and surjective. Since \( f(x) \) is neither, it is not bijective.

### Conclusion

The function \( f(x) = x^2 \) is neither injective nor surjective.

### Final Answer

The function \( y = f(x) \) is neither injective nor surjective.