If `f(x) = {{:([cos pi x]",",x le 1),(2{x}-1",",x gt 1):}`, where [.] and {.} denotes greatest integer and fractional part of x, then

Options

A. `f'(1^(-)) = 2`
B. `f'(1^(+)) = 2`
C. `f'(1^(-)) = -2`
D. `f'(1^(+)) = 0`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} \lfloor \cos(\pi x) \rfloor & \text{if } x \leq 1 \\ 2\{x\} - 1 & \text{if } x > 1 \end{cases} \]

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

### Step 1: Find \( f(1^-) \)
For \( x \leq 1 \), we have:
\[ f(x) = \lfloor \cos(\pi x) \rfloor \]
As \( x \) approaches \( 1 \) from the left, \( \cos(\pi x) \) approaches \( \cos(\pi) = -1 \). Therefore:
\[ f(1^-) = \lfloor -1 \rfloor = -1 \]

### Step 2: Find \( f'(1^-) \)
To find the derivative from the left, we need to differentiate \( f(x) \) for \( x \leq 1 \):
Since \( f(x) = \lfloor \cos(\pi x) \rfloor \) is a constant function (specifically, -1) in the neighborhood of \( x = 1 \), its derivative is:
\[ f'(1^-) = 0 \]

### Step 3: Find \( f(1^+) \)
For \( x > 1 \), we have:
\[ f(x) = 2\{x\} - 1 \]
The fractional part \( \{x\} \) can be expressed as \( x - \lfloor x \rfloor \). For \( x \) just greater than \( 1 \), \( \lfloor x \rfloor = 1 \). Thus:
\[ f(1^+) = 2(x - 1) - 1 = 2x - 2 - 1 = 2x - 3 \]
As \( x \) approaches \( 1 \) from the right:
\[ f(1^+) = 2(1) - 3 = -1 \]

### Step 4: Find \( f'(1^+) \)
Now we differentiate \( f(x) = 2\{x\} - 1 \) for \( x > 1 \):
\[ f'(x) = 2 \cdot \frac{d}{dx} \{x\} = 2 \cdot 1 = 2 \]
Thus:
\[ f'(1^+) = 2 \]

### Conclusion
We have found:
- \( f'(1^-) = 0 \)
- \( f'(1^+) = 2 \)

### Final Answers
- \( f'(1^-) = 0 \) (not one of the options)
- \( f'(1^+) = 2 \) (this corresponds to option B)