Let `f prime(x)=(192x^3)/(2+sin^4 pix)` for all `x in RR` with `f(1/2)=0.` If `mlt=int_(1/2)^1f(x)dxlt=M` then the possible values of m and M are (i) `m=13,M= 24` (ii)` m=1/4,M=1/2` (iii) `m=-11,M = 0` (iv) `m=1,M=12`

Options

A. `m=13,M=24`
B. `m=1/4,M=1/2`
C. `m=-11,M=0`
D. `m=1,M=12`

Explanation

To solve the problem, we start with the given derivative of the function:

\[ f'(x) = \frac{192x^3}{2 + \sin^4(\pi x)} \]

We also know that \( f\left(\frac{1}{2}\right) = 0 \).

We need to find the integral \( m = \int_{\frac{1}{2}}^{1} f(x) \, dx \) and \( M \) such that \( m \leq \int_{\frac{1}{2}}^{1} f(x) \, dx \leq M \).

### Step 1: Establish bounds for \( f'(x) \)

To find the bounds for \( f'(x) \), we evaluate it at the endpoints of the interval \( \left[\frac{1}{2}, 1\right] \).

1. **At \( x = \frac{1}{2} \)**:
\[ f'\left(\frac{1}{2}\right) = \frac{192 \left(\frac{1}{2}\right)^3}{2 + \sin^4\left(\frac{\pi}{2}\right)} = \frac{192 \cdot \frac{1}{8}}{2 + 1} = \frac{24}{3} = 8 \]

2. **At \( x = 1 \)**:
\[ f'(1) = \frac{192 \cdot 1^3}{2 + \sin^4(\pi)} = \frac{192}{2 + 0} = 96 \]

Thus, we have:
\[ 8 \leq f'(x) \leq 96 \quad \text{for } x \in \left[\frac{1}{2}, 1\right] \]

### Step 2: Integrate to find bounds for \( f(x) \)

Now we integrate \( f'(x) \) over the interval \( \left[\frac{1}{2}, 1\right] \):

1. **Lower bound**:
\[ \int_{\frac{1}{2}}^{1} f'(x) \, dx \geq \int_{\frac{1}{2}}^{1} 8 \, dx = 8 \left(1 - \frac{1}{2}\right) = 4 \]

2. **Upper bound**:
\[ \int_{\frac{1}{2}}^{1} f'(x) \, dx \leq \int_{\frac{1}{2}}^{1} 96 \, dx = 96 \left(1 - \frac{1}{2}\right) = 48 \]

### Step 3: Relate \( f(x) \) to the bounds

Using the Fundamental Theorem of Calculus, we have:
\[ f(1) - f\left(\frac{1}{2}\right) = \int_{\frac{1}{2}}^{1} f'(x) \, dx \]
Since \( f\left(\frac{1}{2}\right) = 0 \):
\[ f(1) = \int_{\frac{1}{2}}^{1} f'(x) \, dx \]
Thus, we have:
\[ 4 \leq f(1) \leq 48 \]

### Step 4: Determine possible values of \( m \) and \( M \)

From our calculations:
- The lower bound \( m \) is \( 4 \).
- The upper bound \( M \) is \( 48 \).

### Conclusion

Now we compare these values with the options given in the question:

- (i) \( m = 13, M = 24 \)
- (ii) \( m = \frac{1}{4}, M = \frac{1}{2} \)
- (iii) \( m = -11, M = 0 \)
- (iv) \( m = 1, M = 12 \)

None of the provided options match our calculated bounds of \( m = 4 \) and \( M = 48 \).

### Final Answer

The possible values of \( m \) and \( M \) based on the calculations are \( m = 4 \) and \( M = 48 \).