Let `S` denote the set of all polynomials `P(x)` of degree `lt=2` such that `P(1)=1,P(0)=0a n dP^(prime)(x)>0AAx in [0,1]` , then `S=varphi` b. `S={(1-a)x^2+a x ;0

Options

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

Explanation

To solve the problem, we need to find the set \( S \) of all polynomials \( P(x) \) of degree less than or equal to 2 that satisfy the conditions \( P(1) = 1 \), \( P(0) = 0 \), and \( P'(x) > 0 \) for all \( x \in [0, 1] \).

### Step-by-Step Solution:

1. **Assume the form of the polynomial**:
Since \( P(x) \) is a polynomial of degree less than or equal to 2, we can write it in the form:
\[ P(x) = ax^2 + bx + c \]

2. **Use the condition \( P(0) = 0 \)**:
From the condition \( P(0) = 0 \), we substitute \( x = 0 \):
\[ P(0) = c = 0 \]
Thus, we can simplify our polynomial to:
\[ P(x) = ax^2 + bx \]

3. **Use the condition \( P(1) = 1 \)**:
Now, we apply the condition \( P(1) = 1 \):
\[ P(1) = a(1)^2 + b(1) = a + b = 1 \]
Therefore, we have:
\[ b = 1 - a \]
Substituting this back into our polynomial gives:
\[ P(x) = ax^2 + (1 - a)x \]

4. **Find the derivative \( P'(x) \)**:
Next, we need to find the derivative of \( P(x) \):
\[ P'(x) = 2ax + (1 - a) \]

5. **Set the condition \( P'(x) > 0 \) for \( x \in [0, 1] \)**:
We need \( P'(x) > 0 \) for all \( x \) in the interval \([0, 1]\).
- At \( x = 0 \):
\[ P'(0) = 1 - a > 0 \implies a < 1 \]
- At \( x = 1 \):
\[ P'(1) = 2a + (1 - a) = a + 1 > 0 \implies a > -1 \]
Since \( a \) must be positive (as \( P'(x) \) must be greater than 0 throughout the interval), we combine these conditions:
\[ 0 < a < 1 \]

6. **Conclusion**:
The set \( S \) of all such polynomials is:
\[ S = \{ (1-a)x^2 + ax \; | \; 0 < a < 1 \} \]

### Final Answer:
The correct option is:
**c. \( S = \{ (1-a)x^2 + ax \; | \; 0 < a < 1 \} \)**