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 will analyze the function \( f(x) \) and the conditions given in the question.

### Step 1: Define the function
The function is defined 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 \( f(2) \)
Since \( x = 2 \) falls under the first case of the function, we calculate:
\[ f(2) = 3 - 2^2 = 3 - 4 = -1 \]

### Step 3: Analyze the behavior of \( f(x) \) around \( x = 2 \)
To have a local maximum at \( x = 2 \), we need:
\[ \lim_{h \to 0} f(2 + h) \leq f(2) \]
This means that for values of \( x \) just greater than 2, \( f(2 + h) \) must be less than or equal to \( f(2) \).

### Step 4: Calculate \( f(2 + h) \)
For \( x > 2 \):
\[ f(2 + h) = \sqrt{a + 14} - |(2 + h) - 48| = \sqrt{a + 14} - |h - 46| \]
As \( h \to 0 \), \( |h - 46| \) approaches \( 46 \):
\[ f(2 + h) = \sqrt{a + 14} - 46 \]

### Step 5: Set up the inequality
For \( f(2 + h) \) to be less than or equal to \( f(2) \):
\[ \sqrt{a + 14} - 46 \leq -1 \]
Rearranging gives:
\[ \sqrt{a + 14} \leq 45 \]

### Step 6: Square both sides
Squaring both sides of the inequality:
\[ a + 14 \leq 2025 \]
This simplifies to:
\[ a \leq 2025 - 14 \]
\[ a \leq 2011 \]

### Step 7: Conclusion
The greatest value of \( a \) that satisfies the condition is:
\[ \boxed{2011} \]