Let `f(x)={|x-2|+a^2-6a+9, x < 2 and 5-2x, x >= 2` If `lim_(x->2) [f(x)]` exists the possible values a can take is/are (where [.] represents the grestest integer function)

Options

A. 2
B. `5/2`
C. 3
D. `7/2`

Answer

To solve the problem, we need to find the values of \( a \) such that the limit \( \lim_{x \to 2} f(x) \) exists. This requires that the left-hand limit (LHL) and the right-hand limit (RHL) at \( x = 2 \) are equal.

### Step 1: Define the function \( f(x) \)
The function is defined as:
\[ f(x) = \begin{cases} |x - 2| + a^2 - 6a + 9 & \text{if } x < 2 \\ 5 - 2x & \text{if } x \geq 2 \end{cases} \]

### Step 2: Calculate the left-hand limit (LHL) as \( x \to 2 \)
For \( x < 2 \):
\[ f(x) = |x - 2| + a^2 - 6a + 9 \]
As \( x \) approaches 2 from the left, \( |x - 2| = 2 - x \). Therefore:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} ((2 - x) + a^2 - 6a + 9) \]
Substituting \( x = 2 - h \) where \( h \to 0 \):
\[ = \lim_{h \to 0} (2 - (2 - h) + a^2 - 6a + 9) = \lim_{h \to 0} (h + a^2 - 6a + 9) = a^2 - 6a + 9 \]

### Step 3: Calculate the right-hand limit (RHL) as \( x \to 2 \)
For \( x \geq 2 \):
\[ f(x) = 5 - 2x \]
Thus:
\[ \lim_{x \to 2^+} f(x) = 5 - 2(2) = 5 - 4 = 1 \]

### Step 4: Set the limits equal for continuity
For the limit to exist, we need:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \]
This gives us the equation:
\[ a^2 - 6a + 9 = 1 \]

### Step 5: Solve the equation
Rearranging the equation:
\[ a^2 - 6a + 9 - 1 = 0 \implies a^2 - 6a + 8 = 0 \]
Factoring the quadratic:
\[ (a - 4)(a - 2) = 0 \]
Thus, the solutions are:
\[ a = 4 \quad \text{or} \quad a = 2 \]

### Conclusion
The possible values of \( a \) for which the limit exists are \( a = 2 \) and \( a = 4 \).