Step 1: Understanding the Concept:
Continuity of a function at a point \(x = a\) means there is no break, jump, or hole in the graph at that point.
Mathematically, a function \(f(x)\) is continuous at \(x = a\) if and only if:
1. The Left-Hand Limit (\(\text{LHL}\)) as \(x\) approaches \(a\) from the left equals the Right-Hand Limit (\(\text{RHL}\)) as \(x\) approaches \(a\) from the right.
2. These limits must equal the actual value of the function at that point, \(f(a)\).
Condition: \(\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)\).
Step 2: Detailed Explanation:
The function is defined piecewise around the point \(x = 1\).
Step 2.1: Find the Left-Hand Limit (\(\text{LHL}\)) and \(f(1)\):
For values of \(x \leq 1\), the function follows the rule \(f(x) = x^2 + k\).
The \(\text{LHL}\) as \(x\) approaches \(1\) from the left side is:
\[ \text{LHL} = \lim_{x \to 1^-} (x^2 + k) \]
Substituting \(x = 1\):
\[ \text{LHL} = (1)^2 + k = 1 + k \]
Since the inequality includes the equal sign (\(\leq\)), \(f(1)\) is also \(1 + k\).
Step 2.2: Find the Right-Hand Limit (\(\text{RHL}\)):
For values of \(x>1\), the function follows the rule \(f(x) = 2x + 1\).
The \(\text{RHL}\) as \(x\) approaches \(1\) from the right side is:
\[ \text{RHL} = \lim_{x \to 1^+} (2x + 1) \]
Substituting \(x = 1\):
\[ \text{RHL} = 2(1) + 1 = 2 + 1 = 3 \]
Step 2.3: Equate for Continuity:
For the function to be continuous at \(x = 1\), we must have \(\text{LHL} = \text{RHL}\).
\[ 1 + k = 3 \]
Step 2.4: Solve for \(k\):
Subtract \(1\) from both sides:
\[ k = 3 - 1 \]
\[ k = 2 \]
Step 3: Final Answer:
The value of the constant \(k\) that ensures the function is continuous at \(x = 1\) is \(2\).