Number of points where the function
`f(x)={(1+["cos"(pix)/2],1ltxle2),(1-{x},0lexlt1),(|sinpix|,-1lexlt0):}`
and `f(1)=0` is continuous but non differentiable :
(where [.] denotes greatest integer function and {.} denotes fractional part function)

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise and determine where it is continuous but not differentiable. The function is given as follows:

\[ f(x) = \begin{cases} 1 + \left\lfloor \cos\left(\pi x / 2\right) \right\rfloor & \text{for } 1 < x \leq 2 \\ 1 - \{x\} & \text{for } 0 \leq x < 1 \\ |\sin(\pi x)| & \text{for } -1 < x < 0 \end{cases} \]

where \( \lfloor \cdot \rfloor \) denotes the greatest integer function and \( \{ \cdot \} \) denotes the fractional part function.

### Step 1: Analyze each piece of the function

1. **For \( 1 < x \leq 2 \)**:
- The expression \( \cos\left(\frac{\pi x}{2}\right) \) varies from \( \cos\left(\frac{\pi}{2}\right) = 0 \) to \( \cos(\pi) = -1 \).
- Thus, \( \lfloor \cos\left(\frac{\pi x}{2}\right) \rfloor \) will be \( -1 \) for \( 1 < x < 2 \) and \( 0 \) at \( x = 2 \).
- Therefore, \( f(x) = 1 - 1 = 0 \) for \( 1 < x < 2 \) and \( f(2) = 1 + 0 = 1 \).

2. **For \( 0 \leq x < 1 \)**:
- The fractional part \( \{x\} = x \) since \( x \) is non-negative and less than 1.
- Thus, \( f(x) = 1 - x \).

3. **For \( -1 < x < 0 \)**:
- The function \( |\sin(\pi x)| \) is continuous and differentiable in this interval.

### Step 2: Check continuity at the boundaries

- **At \( x = 0 \)**:
- From the left, \( f(0^-) = |\sin(0)| = 0 \).
- From the right, \( f(0^+) = 1 - 0 = 1 \).
- Thus, \( f(x) \) is not continuous at \( x = 0 \).

- **At \( x = 1 \)**:
- From the left, \( f(1^-) = 1 - 1 = 0 \).
- From the right, \( f(1^+) = 0 \).
- Thus, \( f(x) \) is continuous at \( x = 1 \).

- **At \( x = 2 \)**:
- From the left, \( f(2^-) = 0 \).
- From the right, \( f(2) = 1 \).
- Thus, \( f(x) \) is not continuous at \( x = 2 \).

### Step 3: Check differentiability at \( x = 1 \)

To check if \( f(x) \) is differentiable at \( x = 1 \):
- The left-hand derivative at \( x = 1 \) is:
\[ \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(1 - (1+h)) - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \]

- The right-hand derivative at \( x = 1 \) is:
\[ \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{0 - 0}{h} = 0 \]

Since the left-hand and right-hand derivatives are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

The function \( f(x) \) is continuous at \( x = 1 \) but not differentiable there. Therefore, the number of points where \( f(x) \) is continuous but non-differentiable is:

\[ \text{Number of points} = 1 \]