Given, `f(x)= sin^(3) x and P(x) ` is a quadratic polynomial with coefficient unity.
Statement I ` int_(0)^(2pi) P(x).f''(x) dx ` vanishes.
Statement II `int _(0)^(2pi) f (x) dx` vanishes.

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement 1.
B. Statement I is true, Statement II is also true , Statement II is not the correct explanation of Statement II.
C. Statement I is true, Statement II is false
D. Statement I is false , Statement II is true

Answer

To solve the problem, we need to analyze the two statements given and verify their validity.

### Step 1: Analyze Statement I

We need to evaluate the integral:
\[ I = \int_{0}^{2\pi} P(x) f''(x) \, dx \]
where \( f(x) = \sin^3 x \) and \( P(x) \) is a quadratic polynomial with coefficient unity. We can express \( P(x) \) as:
\[ P(x) = x^2 + bx + c \]
for some constants \( b \) and \( c \).

### Step 2: Calculate \( f''(x) \)

First, we need to find the second derivative \( f''(x) \) of \( f(x) = \sin^3 x \).

Using the chain rule:
1. First derivative:
\[ f'(x) = 3\sin^2 x \cos x \]
2. Second derivative:
\[ f''(x) = 3(2\sin x \cos^2 x - \sin^3 x) = 3\sin x(2\cos^2 x - \sin^2 x) \]

### Step 3: Evaluate the Integral

Now we substitute \( f''(x) \) into the integral:
\[ I = \int_{0}^{2\pi} P(x) \cdot 3\sin x(2\cos^2 x - \sin^2 x) \, dx \]
Since \( P(x) \) is a polynomial and \( \sin x \) is an odd function, the product \( P(x) \sin x \) will also be an odd function over the interval \( [0, 2\pi] \).

### Step 4: Conclusion for Statement I

The integral of an odd function over a symmetric interval around zero (or any complete period of \( 2\pi \)) is zero. Thus:
\[ I = 0 \]
This means Statement I is true.

### Step 5: Analyze Statement II

Now we need to evaluate:
\[ J = \int_{0}^{2\pi} f(x) \, dx = \int_{0}^{2\pi} \sin^3 x \, dx \]

### Step 6: Use the Identity for \( \sin^3 x \)

Using the identity:
\[ \sin^3 x = \frac{3\sin x - \sin(3x)}{4} \]
we can rewrite the integral:
\[ J = \int_{0}^{2\pi} \frac{3\sin x - \sin(3x)}{4} \, dx \]
This can be split into two integrals:
\[ J = \frac{1}{4} \left( 3 \int_{0}^{2\pi} \sin x \, dx - \int_{0}^{2\pi} \sin(3x) \, dx \right) \]

### Step 7: Evaluate Each Integral

Both integrals \( \int_{0}^{2\pi} \sin x \, dx \) and \( \int_{0}^{2\pi} \sin(3x) \, dx \) evaluate to zero:
\[ \int_{0}^{2\pi} \sin x \, dx = 0 \]
\[ \int_{0}^{2\pi} \sin(3x) \, dx = 0 \]

Thus:
\[ J = \frac{1}{4} (3 \cdot 0 - 0) = 0 \]

### Step 8: Conclusion for Statement II

Since \( J = 0 \), Statement II is also true.

### Final Conclusion

Both statements are true:
- Statement I: \( \int_{0}^{2\pi} P(x) f''(x) \, dx \) vanishes.
- Statement II: \( \int_{0}^{2\pi} f(x) \, dx \) vanishes.