If `f(x)=[x]^({x}) +{x}^([x]) + sin (pix)` , Where [.] and {.} represent the greatest integer function and the fractional part function respectively, then `f'(7/2)` Is equal to

Options

A. `sqrt3In3+pi/4`
B. `sqrt3In3+(3pi)/4`
C. `sqrt3In3+pi+3/4`
D. `sqrt3In3+3/4`

Answer

To solve the problem, we need to find the derivative of the function \( f(x) = [x]^{\{x\}} + \{x\}^{[x]} + \sin(\pi x) \) at \( x = \frac{7}{2} \).

### Step-by-step Solution:

1. **Identify the components of \( f(x) \)**:
- Here, \( [x] \) is the greatest integer function (floor function), and \( \{x\} \) is the fractional part function.
- For \( x = \frac{7}{2} = 3.5 \):
- \( [x] = [3.5] = 3 \)
- \( \{x\} = \{3.5\} = 3.5 - 3 = 0.5 \)

2. **Substitute \( [x] \) and \( \{x\} \) into \( f(x) \)**:
\[ f\left(\frac{7}{2}\right) = 3^{0.5} + 0.5^3 + \sin\left(\pi \cdot \frac{7}{2}\right) \]
- Calculate \( \sin\left(\pi \cdot \frac{7}{2}\right) = \sin\left(\frac{7\pi}{2}\right) \):
- Since \( \frac{7\pi}{2} = 3.5\pi \), we can reduce it:
- \( 3.5\pi \) corresponds to \( \frac{\pi}{2} \) (as \( 3.5\pi - 3\pi = \frac{\pi}{2} \)), thus \( \sin\left(\frac{7\pi}{2}\right) = 1 \).

3. **Calculate \( f\left(\frac{7}{2}\right) \)**:
\[ f\left(\frac{7}{2}\right) = \sqrt{3} + 0.125 + 1 = \sqrt{3} + 1.125 \]

4. **Differentiate \( f(x) \)**:
- We need to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( [x]^{\{x\}} \right) + \frac{d}{dx} \left( \{x\}^{[x]} \right) + \frac{d}{dx} \left( \sin(\pi x) \right) \]

5. **Differentiate each term**:
- For \( [x]^{\{x\}} \):
\[ \frac{d}{dx} \left( [x]^{\{x\}} \right) = [x]^{\{x\}} \ln([x]) \cdot \frac{d}{dx}(\{x\}) = 3^{0.5} \ln(3) \cdot 1 = \sqrt{3} \ln(3) \]
- For \( \{x\}^{[x]} \):
\[ \frac{d}{dx} \left( \{x\}^{[x]} \right) = [x] \cdot \{x\}^{[x]-1} \cdot \frac{d}{dx}(\{x\}) = 3 \cdot (0.5)^2 \cdot 1 = 3 \cdot 0.25 = \frac{3}{4} \]
- For \( \sin(\pi x) \):
\[ \frac{d}{dx} \left( \sin(\pi x) \right) = \pi \cos(\pi x) \]
- At \( x = \frac{7}{2} \), \( \cos\left(\frac{7\pi}{2}\right) = 0 \).

6. **Combine the derivatives**:
\[ f'\left(\frac{7}{2}\right) = \sqrt{3} \ln(3) + \frac{3}{4} + 0 \]

7. **Final result**:
\[ f'\left(\frac{7}{2}\right) = \sqrt{3} \ln(3) + \frac{3}{4} \]