The second degree polynomial f(x), satisfying f(0)=o,
`f(1)=1,f'(x)gt0AAx in (0,1)`

Options

A. `f(x)=phi`
B. `f(x)=ax+(1-a)x^(2),AAa in (0,oo)`
C. `f(x)=ax+(1-a)x^(2), a in (0,2)`
D. No such polynomial

Answer

To solve the problem, we need to find a second-degree polynomial \( f(x) = ax^2 + bx + c \) that satisfies the conditions:

1. \( f(0) = 0 \)
2. \( f(1) = 1 \)
3. \( f'(x) > 0 \) for \( x \in (0, 1) \)

### Step 1: Use the condition \( f(0) = 0 \)

Since \( f(0) = 0 \), we can substitute \( x = 0 \) into the polynomial:

\[ f(0) = a(0)^2 + b(0) + c = c = 0 \]

Thus, the polynomial simplifies to:

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

### Step 2: Use the condition \( f(1) = 1 \)

Now we substitute \( x = 1 \):

\[ f(1) = a(1)^2 + b(1) = a + b = 1 \]

This gives us our first equation:

\[ a + b = 1 \quad \text{(1)} \]

### Step 3: Find the derivative and apply the condition \( f'(x) > 0 \)

Next, we find the derivative of \( f(x) \):

\[ f'(x) = 2ax + b \]

We need this derivative to be positive for all \( x \in (0, 1) \).

1. At \( x = 0 \):
\[ f'(0) = b > 0 \quad \text{(2)} \]

2. At \( x = 1 \):
\[ f'(1) = 2a + b > 0 \quad \text{(3)} \]

### Step 4: Solve the inequalities

From equation (2), we have \( b > 0 \).

From equation (1), we can express \( b \) in terms of \( a \):

\[ b = 1 - a \]

Substituting this into inequality (2):

\[ 1 - a > 0 \implies a < 1 \]

Now substituting \( b = 1 - a \) into inequality (3):

\[ 2a + (1 - a) > 0 \]
\[ 2a + 1 - a > 0 \implies a + 1 > 0 \implies a > -1 \]

### Step 5: Combine the conditions

We have two conditions for \( a \):

1. \( a < 1 \)
2. \( a > -1 \)

Since \( a \) must also be non-negative (as it is a coefficient of \( x^2 \) in a polynomial that opens upwards), we conclude:

\[ 0 \leq a < 1 \]

### Step 6: Summary of the polynomial

Thus, the polynomial can be expressed as:

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

where \( a \) is in the range \( [0, 1) \).