For `-1<=p<=1` , the equation `4x^3-3x-p = 0` has 'n' distinct real roots equation in the interval `[1/2,1]` and one its root is `cos(kcos^-1p)`, then the value of `n+1/k` is :

Answer

To solve the problem, we need to analyze the equation \(4x^3 - 3x - p = 0\) for \(p\) in the interval \([-1, 1]\) and find the number of distinct real roots \(n\) in the interval \([\frac{1}{2}, 1]\). We also need to determine the value of \(k\) such that one of the roots is \( \cos(k \cos^{-1}(p)) \).

### Step-by-Step Solution:

1. **Define the function**:
Let \(y = 4x^3 - 3x - p\).

2. **Differentiate the function**:
We find the derivative:
\[ \frac{dy}{dx} = 12x^2 - 3. \]

3. **Set the derivative to zero**:
To find the critical points, set the derivative equal to zero:
\[ 12x^2 - 3 = 0. \]
Solving for \(x\):
\[ 12x^2 = 3 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2}. \]

4. **Analyze the critical points**:
The critical points are \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\). Since we are interested in the interval \([\frac{1}{2}, 1]\), we will evaluate the function at these points.

5. **Evaluate the function at the critical points**:
- At \(x = \frac{1}{2}\):
\[ y\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) - p = 4 \cdot \frac{1}{8} - \frac{3}{2} - p = \frac{1}{2} - \frac{3}{2} - p = -1 - p. \]
- At \(x = 1\):
\[ y(1) = 4(1)^3 - 3(1) - p = 4 - 3 - p = 1 - p. \]

6. **Determine the number of roots**:
For the function to have distinct real roots in the interval \([\frac{1}{2}, 1]\), we need to check the signs of \(y\) at the endpoints:
- At \(x = \frac{1}{2}\), \(y\left(\frac{1}{2}\right) = -1 - p\) which is negative for \(p \in [-1, 1]\).
- At \(x = 1\), \(y(1) = 1 - p\) which is positive for \(p \in [-1, 1]\).

Since \(y\left(\frac{1}{2}\right) < 0\) and \(y(1) > 0\), by the Intermediate Value Theorem, there is at least one root in the interval \([\frac{1}{2}, 1]\).

The function is continuous and differentiable, and since it has one local maximum and one local minimum, we can conclude that there is exactly one distinct real root in this interval. Thus, \(n = 1\).

7. **Relate the root to \(k\)**:
We know that one of the roots is given by \( \cos(k \cos^{-1}(p)) \). We can relate this to the equation:
\[ 4 \cos^3(r) - 3 \cos(r) = p, \]
which can be rewritten using the triple angle formula:
\[ \cos(3r) = p. \]
This implies that \(r = \frac{1}{3} \cos^{-1}(p)\), hence:
\[ \cos(k \cos^{-1}(p)) = \cos\left(\frac{1}{3} \cos^{-1}(p)\right). \]
Thus, we find \(k = \frac{1}{3}\).

8. **Calculate \(n + \frac{1}{k}\)**:
Now we can compute:
\[ n + \frac{1}{k} = 1 + \frac{1}{\frac{1}{3}} = 1 + 3 = 4. \]

### Final Answer:
The value of \(n + \frac{1}{k}\) is \(4\).