Let `f(x)=cos x and g(x)=[x+1],"where [.]` denotes the greatest integer function, Then `(gof)' (pi//2)` is

Options

A. 0
B. 1
C. `-1`
D. non-existent

Explanation

To solve the problem, we need to find the derivative of the composite function \( (g \circ f)' \left( \frac{\pi}{2} \right) \), where \( f(x) = \cos x \) and \( g(x) = [x + 1] \) (the greatest integer function).

### Step 1: Define the Composite Function
The composite function \( g \circ f \) can be expressed as:
\[ g(f(x)) = g(\cos x) = [\cos x + 1] \]
This means we are taking the cosine of \( x \) and then applying the greatest integer function to \( \cos x + 1 \).

### Step 2: Evaluate \( f \left( \frac{\pi}{2} \right) \)
First, we need to evaluate \( f \left( \frac{\pi}{2} \right) \):
\[ f \left( \frac{\pi}{2} \right) = \cos \left( \frac{\pi}{2} \right) = 0 \]

### Step 3: Evaluate \( g(f(\frac{\pi}{2})) \)
Now, we substitute \( f \left( \frac{\pi}{2} \right) \) into \( g \):
\[ g(f(\frac{\pi}{2})) = g(0) = [0 + 1] = [1] = 1 \]

### Step 4: Check Continuity of \( g(f(x)) \) at \( x = \frac{\pi}{2} \)
To find \( (g \circ f)' \left( \frac{\pi}{2} \right) \), we need to check the continuity of \( g(f(x)) \) at \( x = \frac{\pi}{2} \). The function \( g(x) = [x + 1] \) is piecewise constant except at integer points.

### Step 5: Evaluate \( \cos x + 1 \) near \( x = \frac{\pi}{2} \)
As \( x \) approaches \( \frac{\pi}{2} \), \( \cos x \) approaches \( 0 \). Therefore, \( \cos x + 1 \) approaches \( 1 \).

### Step 6: Determine the Behavior of \( g(f(x)) \)
Since \( g(x) \) is constant around \( x = 1 \) (except at integer points), we find:
\[ g(f(x)) = g(\cos x) = [\cos x + 1] = 1 \text{ for } x \text{ near } \frac{\pi}{2} \]

### Step 7: Find the Derivative
Since \( g(f(x)) \) is constant (equal to 1) in a neighborhood around \( x = \frac{\pi}{2} \), its derivative is:
\[ (g \circ f)' \left( \frac{\pi}{2} \right) = 0 \]

### Final Answer
Thus, the value of \( (g \circ f)' \left( \frac{\pi}{2} \right) \) is:
\[ \boxed{0} \]