If `f(x)={{:(3-x^(2)",",xle2),(sqrt(a+14)-|x-48|",",xgt2):}` and if f(x) has a local maxima at x = 2, then greatest value of a is

Options

A. 2013
B. 2012
C. 2011
D. 2010

Explanation

To solve the problem step by step, we need to analyze the function \( f(x) \) and the condition for local maxima at \( x = 2 \).

### Step 1: Define the function
The function is given as:
\[ f(x) = \begin{cases} 3 - x^2 & \text{if } x \leq 2 \\ \sqrt{a + 14} - |x - 48| & \text{if } x > 2 \end{cases} \]

### Step 2: Evaluate \( f(2) \)
First, we need to find \( f(2) \) using the first part of the function:
\[ f(2) = 3 - 2^2 = 3 - 4 = -1 \]

### Step 3: Find the limit as \( x \) approaches 2 from the right
Next, we need to evaluate the limit of \( f(x) \) as \( x \) approaches 2 from the right:
\[ \lim_{h \to 0} f(2 + h) = \lim_{h \to 0} \left( \sqrt{a + 14} - |(2 + h) - 48| \right) \]
Since \( 2 + h \) is greater than 2, we can simplify the absolute value:
\[ |2 + h - 48| = |h - 46| = 46 - h \quad \text{(as } h \text{ approaches } 0\text{)} \]
Thus,
\[ \lim_{h \to 0} f(2 + h) = \sqrt{a + 14} - (46 - h) = \sqrt{a + 14} - 46 + h \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} f(2 + h) = \sqrt{a + 14} - 46 \]

### Step 4: Set up the condition for local maxima
For \( f(x) \) to have a local maximum at \( x = 2 \), we need:
\[ \lim_{h \to 0} f(2 + h) \leq f(2) \]
This gives us:
\[ \sqrt{a + 14} - 46 \leq -1 \]

### Step 5: Solve the inequality
Rearranging the inequality:
\[ \sqrt{a + 14} \leq 45 \]
Squaring both sides:
\[ a + 14 \leq 2025 \]
Thus,
\[ a \leq 2025 - 14 \]
\[ a \leq 2011 \]

### Conclusion
The greatest value of \( a \) such that \( f(x) \) has a local maximum at \( x = 2 \) is:
\[ \boxed{2011} \]