If `f(x)=cosx,0lexle(pi)/2` then the real number c of the mean value theorem is

Options

A. `(pi)/6`
B. `(pi)/4`
C. `sin^(-1)(2/(pi))`
D. `cos^(-1)(2/(pi))`

Explanation

To find the real number \( c \) in the context of the Mean Value Theorem (MVT) for the function \( f(x) = \cos x \) on the interval \( [0, \frac{\pi}{2}] \), we will follow these steps:

### Step 1: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
In our case, \( f(x) = \cos x \) is continuous and differentiable on the interval \([0, \frac{\pi}{2}]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = 0 \) and \( b = \frac{\pi}{2} \):
\[ f(0) = \cos(0) = 1 \]
\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

### Step 3: Compute \( \frac{f(b) - f(a)}{b - a} \)
Now, we can calculate:
\[ \frac{f\left(\frac{\pi}{2}\right) - f(0)}{\frac{\pi}{2} - 0} = \frac{0 - 1}{\frac{\pi}{2} - 0} = \frac{-1}{\frac{\pi}{2}} = -\frac{2}{\pi} \]

### Step 4: Find the derivative \( f'(x) \)
Next, we find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\cos x) = -\sin x \]

### Step 5: Set up the equation from the Mean Value Theorem
According to the MVT, we need to find \( c \) such that:
\[ f'(c) = -\frac{2}{\pi} \]
Substituting the derivative we found:
\[ -\sin c = -\frac{2}{\pi} \]
This simplifies to:
\[ \sin c = \frac{2}{\pi} \]

### Step 6: Solve for \( c \)
To find \( c \), we take the inverse sine:
\[ c = \sin^{-1}\left(\frac{2}{\pi}\right) \]

### Conclusion
Thus, the real number \( c \) that satisfies the conditions of the Mean Value Theorem for the function \( f(x) = \cos x \) on the interval \( [0, \frac{\pi}{2}] \) is:
\[ c = \sin^{-1}\left(\frac{2}{\pi}\right) \]