If P (x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that `P (0) =0, P (1) =1, and P'(x) gt 0 AA x in [0.1],` then

Options

A. `S = phi`
B. `S = ax + (1- a )x ^(2) AA a in (0,2)`
C. `S = ax + (1-a) x ^(2) AA a in (0,oo)`
D. `S = ax + (1-a) x ^(2) AA a in (0,1)`

Answer

To solve the problem, we need to find the set of all polynomials \( P(x) \) of degree less than or equal to 2 that satisfy the conditions given in the problem statement. Let's break down the solution step by step.

### Step 1: Define the Polynomial
Since \( P(x) \) is a polynomial of degree less than or equal to 2, we can express it in the standard form:
\[ P(x) = ax^2 + bx + c \]
where \( a, b, c \) are constants.

### Step 2: Apply the Condition \( P(0) = 0 \)
We know that \( P(0) = 0 \). Substituting \( x = 0 \) into the polynomial gives:
\[ P(0) = a(0)^2 + b(0) + c = c \]
Thus, we have:
\[ c = 0 \]
So now our polynomial simplifies to:
\[ P(x) = ax^2 + bx \]

### Step 3: Apply the Condition \( P(1) = 1 \)
Next, we apply the condition \( P(1) = 1 \):
\[ P(1) = a(1)^2 + b(1) = a + b \]
Setting this equal to 1 gives us:
\[ a + b = 1 \]
From this, we can express \( a \) in terms of \( b \):
\[ a = 1 - b \]

### Step 4: Differentiate the Polynomial
Now we need to find the derivative \( P'(x) \):
\[ P'(x) = \frac{d}{dx}(ax^2 + bx) = 2ax + b \]

### Step 5: Apply the Condition \( P'(x) > 0 \) for \( x \in [0, 1] \)
We require that \( P'(x) > 0 \) for all \( x \) in the interval \([0, 1]\). This means:
\[ 2ax + b > 0 \quad \text{for all } x \in [0, 1] \]

### Step 6: Evaluate the Derivative at the Endpoints
1. At \( x = 0 \):
\[ P'(0) = 2a(0) + b = b > 0 \]
Thus, we have:
\[ b > 0 \]

2. At \( x = 1 \):
\[ P'(1) = 2a(1) + b = 2a + b > 0 \]
Substituting \( a = 1 - b \) into this inequality:
\[ 2(1 - b) + b > 0 \implies 2 - 2b + b > 0 \implies 2 - b > 0 \implies b < 2 \]

### Step 7: Combine the Conditions
From the above steps, we have two conditions:
1. \( b > 0 \)
2. \( b < 2 \)

Thus, we can conclude that:
\[ 0 < b < 2 \]

### Step 8: Write the Final Form of the Polynomial
Substituting back for \( a \):
\[ a = 1 - b \]
Since \( b \) is in the interval \( (0, 2) \), \( a \) will be in the interval:
\[ 1 - 2 < a < 1 \implies -1 < a < 1 \]

### Conclusion
The set of all such polynomials \( P(x) \) can be expressed as:
\[ P(x) = (1 - b)x^2 + bx \quad \text{for } 0 < b < 2 \]