If `f(x)={(x,,,0lexle1),(2-x,,,1ltxle2):}` then point of discontinuity fo `f(f(x))` is [0,2] is,

Options

A. `x=1/2`
B. `x=3/2`
C. `x=1`
D. No value of x

Answer

To find the points of discontinuity for the function \( f(f(x)) \) where

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

we will follow these steps:

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) \) is defined piecewise. We can evaluate its behavior at the boundaries of the intervals:

- At \( x = 0 \), \( f(0) = 0 \).
- At \( x = 1 \), \( f(1) = 1 \).
- At \( x = 2 \), \( f(2) = 0 \).

### Step 2: Determine the range of \( f(x) \)

- For \( 0 \leq x \leq 1 \), \( f(x) = x \) gives us the range \( [0, 1] \).
- For \( 1 < x \leq 2 \), \( f(x) = 2 - x \) gives us the range \( (0, 1] \).

Combining these, the overall range of \( f(x) \) for \( x \in [0, 2] \) is \( [0, 1] \).

### Step 3: Find \( f(f(x)) \)

Now we need to evaluate \( f(f(x)) \):

1. **For \( x \in [0, 1] \)**:
- Here, \( f(x) = x \) which is in the range \( [0, 1] \).
- Thus, \( f(f(x)) = f(x) = x \).

2. **For \( x \in (1, 2] \)**:
- Here, \( f(x) = 2 - x \) which lies in the interval \( (0, 1] \).
- Thus, \( f(f(x)) = f(2 - x) = 2 - x \).

### Step 4: Analyze the continuity of \( f(f(x)) \)

Now we need to check the continuity of \( f(f(x)) \):

- For \( x \in [0, 1] \), \( f(f(x)) = x \) is continuous.
- For \( x \in (1, 2] \), \( f(f(x)) = 2 - x \) is also continuous.

### Step 5: Check the transition at \( x = 1 \)

At \( x = 1 \):
- From the left: \( f(f(1)) = f(1) = 1 \).
- From the right: \( f(f(1)) = f(2 - 1) = f(1) = 1 \).

Since both limits equal \( 1 \) at \( x = 1 \), \( f(f(x)) \) is continuous at \( x = 1 \).

### Conclusion

Since \( f(f(x)) \) is continuous for all \( x \in [0, 2] \), there are no points of discontinuity in the interval \( [0, 2] \).

### Final Answer

The points of discontinuity of \( f(f(x)) \) in the interval \( [0, 2] \) is **none**.

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