Let f,g : `R to R` be defined by `f(x) = (x-2)|x-2| AA x in R`
`g(x) = sqrt((x-2)^(2)) AA x in R`
`S = {x : x in R " and " f(x) = g(x)}`, number of elements in S is………..

Answer

To solve the problem, we need to find the number of elements in the set \( S = \{ x \in \mathbb{R} : f(x) = g(x) \} \), where \( f(x) = (x-2)|x-2| \) and \( g(x) = \sqrt{(x-2)^2} \).

### Step-by-Step Solution:

1. **Understanding the Functions**:
- The function \( f(x) = (x-2)|x-2| \) can be expressed differently based on the value of \( x \):
- If \( x \geq 2 \): \( f(x) = (x-2)(x-2) = (x-2)^2 \)
- If \( x < 2 \): \( f(x) = (x-2)(-(x-2)) = -(x-2)^2 \)
- The function \( g(x) = \sqrt{(x-2)^2} = |x-2| \) is always non-negative.

2. **Setting Up the Equation**:
- We need to solve the equation \( f(x) = g(x) \).
- This leads to two cases based on the piecewise definition of \( f(x) \).

3. **Case 1: \( x \geq 2 \)**:
- Here, \( f(x) = (x-2)^2 \) and \( g(x) = |x-2| = x-2 \).
- Setting them equal:
\[ (x-2)^2 = x-2 \]
- Rearranging gives:
\[ (x-2)^2 - (x-2) = 0 \]
\[ (x-2)((x-2)-1) = 0 \]
- This gives us two solutions:
\[ x - 2 = 0 \quad \text{or} \quad x - 3 = 0 \]
\[ x = 2 \quad \text{or} \quad x = 3 \]

4. **Case 2: \( x < 2 \)**:
- Here, \( f(x) = -(x-2)^2 \) and \( g(x) = |x-2| = -(x-2) \).
- Setting them equal:
\[ -(x-2)^2 = -(x-2) \]
- Rearranging gives:
\[ (x-2)^2 - (x-2) = 0 \]
- This is the same equation as in Case 1, leading to the same solutions \( x = 2 \) and \( x = 3 \). However, we must check if these solutions are valid in this case:
- For \( x < 2 \), \( x = 2 \) is not valid.

5. **Final Solutions**:
- The valid solutions from both cases are:
- From Case 1: \( x = 2 \) and \( x = 3 \)
- From Case 2: No valid solutions.
- Therefore, the solutions in set \( S \) are \( \{2, 3\} \).

6. **Counting the Elements**:
- The number of elements in set \( S \) is \( 2 \).

### Conclusion:
The number of elements in the set \( S \) is **2**.

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