Which of the following functions is/are identical to `|x-2|` ?
(a) `f(x)=sqrt(x^(2)-4x+4) " (b) " g(x)=|x|-|2|`
(c ) `h(x)=(|x-2|^(2))/(|x-2|) " (d) "t(x)=|(x^(2)-x-2)/(x+1)|`

Explanation

To determine which of the given functions are identical to \( |x - 2| \), we will analyze each function one by one.

### Step 1: Analyze the function \( p(x) = |x - 2| \)
- The function \( p(x) = |x - 2| \) is defined for all real numbers.
- The range of \( p(x) \) is \( [0, \infty) \) because the absolute value is always non-negative.

### Step 2: Check each option

#### Option (a): \( f(x) = \sqrt{x^2 - 4x + 4} \)
- Simplifying \( f(x) \):
\[ f(x) = \sqrt{(x - 2)^2} \]
- The square root of a square is the absolute value:
\[ f(x) = |x - 2| \]
- **Conclusion**: This function is identical to \( p(x) \).

#### Option (b): \( g(x) = |x| - |2| \)
- Since \( |2| = 2 \), we have:
\[ g(x) = |x| - 2 \]
- This function is not identical to \( |x - 2| \) because it does not have the same behavior. For example, \( g(1) = |1| - 2 = -1 \) while \( p(1) = |1 - 2| = 1 \).
- **Conclusion**: This function is not identical to \( p(x) \).

#### Option (c): \( h(x) = \frac{|x - 2|^2}{|x - 2|} \)
- For \( x \neq 2 \):
\[ h(x) = |x - 2| \]
- However, \( h(x) \) is not defined at \( x = 2 \) because it results in division by zero.
- **Conclusion**: This function is not identical to \( p(x) \) due to the undefined point.

#### Option (d): \( t(x) = \left| \frac{x^2 - x - 2}{x + 1} \right| \)
- Factor the numerator:
\[ x^2 - x - 2 = (x - 2)(x + 1) \]
- Thus,
\[ t(x) = \left| \frac{(x - 2)(x + 1)}{x + 1} \right| \]
- For \( x \neq -1 \):
\[ t(x) = |x - 2| \]
- However, \( t(x) \) is not defined at \( x = -1 \).
- **Conclusion**: This function is not identical to \( p(x) \) due to the undefined point.

### Final Conclusion
The only function that is identical to \( |x - 2| \) is:
- **(a) \( f(x) = \sqrt{x^2 - 4x + 4} \)**