The number of polynomials f(x) with non-negative integer coefficients of degree `le 2`, satisfying f(0) = 0 and `int_(0)^(1) f(x) dx = 1`, is

Options

A. 8
B. 2
C. 4
D. 0

Explanation

To solve the problem step by step, we will analyze the conditions given for the polynomial \( f(x) \).

### Step 1: Define the polynomial
Since we need a polynomial \( f(x) \) of degree less than or equal to 2 with non-negative integer coefficients, we can express it in the general form:
\[ f(x) = ax^2 + bx + c \]
where \( a, b, c \) are non-negative integers.

### Step 2: Apply the first condition
The first condition given is \( f(0) = 0 \). This means:
\[ f(0) = a(0)^2 + b(0) + c = c = 0 \]
Thus, we have \( c = 0 \). Therefore, the polynomial simplifies to:
\[ f(x) = ax^2 + bx \]

### Step 3: Apply the second condition
The second condition is given by the integral:
\[ \int_0^1 f(x) \, dx = 1 \]
Substituting \( f(x) \) into the integral, we get:
\[ \int_0^1 (ax^2 + bx) \, dx = 1 \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ \int_0^1 ax^2 \, dx + \int_0^1 bx \, dx \]
Calculating each part:
\[ \int_0^1 ax^2 \, dx = a \left[ \frac{x^3}{3} \right]_0^1 = \frac{a}{3} \]
\[ \int_0^1 bx \, dx = b \left[ \frac{x^2}{2} \right]_0^1 = \frac{b}{2} \]
Thus, we have:
\[ \frac{a}{3} + \frac{b}{2} = 1 \]

### Step 5: Clear the fractions
To eliminate the fractions, we can multiply the entire equation by 6 (the least common multiple of 3 and 2):
\[ 6 \left( \frac{a}{3} + \frac{b}{2} \right) = 6 \cdot 1 \]
This simplifies to:
\[ 2a + 3b = 6 \]

### Step 6: Find non-negative integer solutions
Now we need to find non-negative integer solutions for the equation \( 2a + 3b = 6 \).

1. **If \( b = 0 \)**:
\[ 2a = 6 \implies a = 3 \]
So one solution is \( (a, b) = (3, 0) \).

2. **If \( b = 1 \)**:
\[ 2a + 3(1) = 6 \implies 2a = 3 \implies a = 1.5 \quad \text{(not an integer)} \]

3. **If \( b = 2 \)**:
\[ 2a + 3(2) = 6 \implies 2a = 0 \implies a = 0 \]
So another solution is \( (a, b) = (0, 2) \).

4. **If \( b \geq 3 \)**:
The left-hand side \( 2a + 3b \) would exceed 6, so no further solutions exist.

### Step 7: List the valid polynomials
From the valid pairs \( (a, b) \):
- For \( (3, 0) \): \( f(x) = 3x^2 \)
- For \( (0, 2) \): \( f(x) = 2x \)

### Conclusion
Thus, the total number of polynomials \( f(x) \) satisfying the given conditions is **2**.