If `f(x)={(1-x",", 0 le x le 2),(x-1",", 2 le x le 4),(1",", 4 le x le 6):}` then find,
`f(0) +f((1)/(2)) +f(1)+ f((45)/(48))`

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the problem, we need to evaluate the function \( f(x) \) at specific points and then sum the results. The function \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} 1 - x & \text{for } 0 \leq x \leq 2 \\ x - 1 & \text{for } 2 < x \leq 4 \\ 1 & \text{for } 4 < x \leq 6 \end{cases} \]

We need to find \( f(0) + f\left(\frac{1}{2}\right) + f(1) + f\left(\frac{45}{48}\right) \).

### Step-by-step Solution:

1. **Calculate \( f(0) \)**:
- Since \( 0 \) is in the interval \( [0, 2] \), we use the first piece of the function:
\[ f(0) = 1 - 0 = 1 \]

2. **Calculate \( f\left(\frac{1}{2}\right) \)**:
- \( \frac{1}{2} \) is also in the interval \( [0, 2] \):
\[ f\left(\frac{1}{2}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]

3. **Calculate \( f(1) \)**:
- \( 1 \) is still in the interval \( [0, 2] \):
\[ f(1) = 1 - 1 = 0 \]

4. **Calculate \( f\left(\frac{45}{48}\right) \)**:
- First, we need to check where \( \frac{45}{48} \) lies. Since \( \frac{45}{48} \approx 0.9375 \), it is also in the interval \( [0, 2] \):
\[ f\left(\frac{45}{48}\right) = 1 - \frac{45}{48} = \frac{48 - 45}{48} = \frac{3}{48} = \frac{1}{16} \]

5. **Sum the results**:
- Now we can sum all the values we calculated:
\[ f(0) + f\left(\frac{1}{2}\right) + f(1) + f\left(\frac{45}{48}\right) = 1 + \frac{1}{2} + 0 + \frac{1}{16} \]

- To add these fractions, we need a common denominator. The least common multiple of \( 1, 2, \) and \( 16 \) is \( 16 \):
\[ 1 = \frac{16}{16}, \quad \frac{1}{2} = \frac{8}{16}, \quad 0 = \frac{0}{16}, \quad \frac{1}{16} = \frac{1}{16} \]

- Now we can add them:
\[ \frac{16}{16} + \frac{8}{16} + \frac{0}{16} + \frac{1}{16} = \frac{16 + 8 + 0 + 1}{16} = \frac{25}{16} \]

### Final Result:
\[ f(0) + f\left(\frac{1}{2}\right) + f(1) + f\left(\frac{45}{48}\right) = \frac{25}{16} \]