If the function defined by :
`f(x)={{:(2x-1", "xlt2),(a", "x=2),(x+1", "xgt2):}`
is continuous at x = 2, find the value of 'a'.
Also discuss the continuity of f(x) at x = 3.

Answer

To determine the value of 'a' that makes the function \( f(x) \) continuous at \( x = 2 \), we need to ensure that the left-hand limit, right-hand limit, and the function value at that point are all equal.

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

### Step 2: Calculate the left-hand limit as \( x \) approaches 2
The left-hand limit \( f(2^-) \) is calculated using the expression for \( x < 2 \):
\[ f(2^-) = 2(2) - 1 = 4 - 1 = 3 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 2
The right-hand limit \( f(2^+) \) is calculated using the expression for \( x > 2 \):
\[ f(2^+) = 2 + 1 = 3 \]

### Step 4: Set the limits equal to the function value at \( x = 2 \)
For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ f(2^-) = f(2) = f(2^+) \]
Substituting the values we calculated:
\[ 3 = a = 3 \]

### Conclusion for Step 1
Thus, the value of \( a \) that makes the function continuous at \( x = 2 \) is:
\[ a = 3 \]

### Step 5: Discuss continuity at \( x = 3 \)
Now, we need to check the continuity of \( f(x) \) at \( x = 3 \):
- For \( x > 2 \), the function is defined as \( f(x) = x + 1 \).
- Since \( x + 1 \) is a polynomial function, it is continuous everywhere.

### Conclusion for Step 2
Thus, \( f(x) \) is continuous at \( x = 3 \).

### Final Answer
The value of \( a \) is \( 3 \) and the function \( f(x) \) is continuous at \( x = 3 \).

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