If P(x) is polynomial satisfying `P(x^(2))=x^(2)P(x)andP(0)=-2, P'(3//2)=0andP(1)=0.`
The maximum value of P(x) is

Options

A. (a) `-(1)/(3)`
B. (b) `(1)/(4)`
C. (c) `-(1)/(2)`
D. (d) none of the above

Answer

To solve the problem, we need to find the polynomial \( P(x) \) that satisfies the given conditions and then determine its maximum value.

### Step-by-step Solution:

1. **Understanding the Polynomial Condition**:
We have the condition:
\[ P(x^2) = x^2 P(x) \]
This implies that if \( P(x) \) is a polynomial of degree \( n \), then \( P(x^2) \) will be of degree \( 2n \) and \( x^2 P(x) \) will also be of degree \( n + 2 \). Therefore, we can equate the degrees:
\[ 2n = n + 2 \]
Solving this gives:
\[ n = 2 \]
Thus, \( P(x) \) is a quadratic polynomial.

2. **Assuming the Form of the Polynomial**:
Let’s assume:
\[ P(x) = ax^2 + bx + c \]

3. **Using Given Conditions**:
- From \( P(0) = -2 \):
\[ P(0) = c = -2 \]
- From \( P(1) = 0 \):
\[ P(1) = a(1)^2 + b(1) + c = a + b - 2 = 0 \implies a + b = 2 \quad \text{(Equation 1)} \]
- From \( P'(3/2) = 0 \):
The derivative \( P'(x) = 2ax + b \).
Setting \( P'(3/2) = 0 \):
\[ 2a(3/2) + b = 0 \implies 3a + b = 0 \quad \text{(Equation 2)} \]

4. **Solving the Equations**:
Now we have two equations:
- Equation 1: \( a + b = 2 \)
- Equation 2: \( 3a + b = 0 \)

We can solve these equations simultaneously. From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = -3a \]
Substituting this into Equation 1:
\[ a - 3a = 2 \implies -2a = 2 \implies a = -1 \]
Now substituting \( a = -1 \) back into Equation 1:
\[ -1 + b = 2 \implies b = 3 \]

5. **Finding the Polynomial**:
Now we have:
\[ a = -1, \quad b = 3, \quad c = -2 \]
Thus, the polynomial is:
\[ P(x) = -x^2 + 3x - 2 \]

6. **Finding the Maximum Value**:
To find the maximum value, we differentiate \( P(x) \):
\[ P'(x) = -2x + 3 \]
Setting \( P'(x) = 0 \):
\[ -2x + 3 = 0 \implies x = \frac{3}{2} \]
To confirm that this is a maximum, we check the second derivative:
\[ P''(x) = -2 \]
Since \( P''(x) < 0 \), \( P(x) \) has a maximum at \( x = \frac{3}{2} \).

7. **Calculating the Maximum Value**:
Now we calculate \( P\left(\frac{3}{2}\right) \):
\[ P\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) - 2 \]
Simplifying:
\[ = -\frac{9}{4} + \frac{9}{2} - 2 = -\frac{9}{4} + \frac{18}{4} - \frac{8}{4} = \frac{1}{4} \]

### Final Answer:
The maximum value of \( P(x) \) is:
\[ \boxed{\frac{1}{4}} \]