P(x) is a non-zero polynomial such that P(0)=0 and `P(x^3)=x^4P(x),P'(1)=7` and `int_0^1P(x)=1.5` then `int_0^1P(x) P'(x) dx =`

Options

A. 6
B. 8
C. `7.5`
D. None of these

Answer

To solve the problem step by step, we will analyze the given conditions and use them to find the required integral.

### Step 1: Understand the Given Conditions
We have a polynomial \( P(x) \) such that:
1. \( P(0) = 0 \)
2. \( P(x^3) = x^4 P(x) \)
3. \( P'(1) = 7 \)
4. \( \int_0^1 P(x) \, dx = 1.5 \)

We need to find \( \int_0^1 P(x) P'(x) \, dx \).

### Step 2: Set Up the Integral
Let:
\[ I = \int_0^1 P(x) P'(x) \, dx \]

### Step 3: Use Integration by Parts
Using the property of integration by parts, we can rewrite the integral:
\[ I = \int_0^1 P(x) P'(x) \, dx = \left[ \frac{P(x)^2}{2} \right]_0^1 \]
This means we will evaluate \( \frac{P(x)^2}{2} \) at the limits \( 0 \) and \( 1 \).

### Step 4: Evaluate the Limits
Now we need to calculate \( P(1) \) and \( P(0) \):
- Since \( P(0) = 0 \), we have \( \frac{P(0)^2}{2} = \frac{0^2}{2} = 0 \).
- We need to find \( P(1) \).

### Step 5: Use the Given Condition \( P(x^3) = x^4 P(x) \)
Substituting \( x = 1 \) in the equation \( P(x^3) = x^4 P(x) \):
\[ P(1^3) = 1^4 P(1) \implies P(1) = P(1) \]
This doesn't provide new information. We need to differentiate the equation.

### Step 6: Differentiate \( P(x^3) = x^4 P(x) \)
Differentiating both sides with respect to \( x \):
\[ P'(x^3) \cdot 3x^2 = 4x^3 P(x) + x^4 P'(x) \]

### Step 7: Substitute \( x = 1 \)
Substituting \( x = 1 \):
\[ P'(1) \cdot 3 = 4 P(1) + P'(1) \]
Given \( P'(1) = 7 \):
\[ 3 \cdot 7 = 4 P(1) + 7 \]
\[ 21 = 4 P(1) + 7 \]
\[ 4 P(1) = 21 - 7 = 14 \implies P(1) = \frac{14}{4} = \frac{7}{2} \]

### Step 8: Substitute Back into the Integral
Now substituting \( P(1) \) back into the integral:
\[ I = \left[ \frac{P(x)^2}{2} \right]_0^1 = \frac{P(1)^2}{2} - \frac{P(0)^2}{2} = \frac{(\frac{7}{2})^2}{2} - 0 = \frac{\frac{49}{4}}{2} = \frac{49}{8} \]

### Final Answer
Thus, the value of \( \int_0^1 P(x) P'(x) \, dx \) is:
\[ \boxed{\frac{49}{8}} \]