Let `f(x) = {(|x-1| + a",",x le 1),(2x + 3",",x gt 1):}`. If f(x) has local minimum x=1 and `a ge 5` then a is equal to

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 1 \) and has a local minimum at that point. The function is defined as:

\[ f(x) = \begin{cases} |x - 1| + a & \text{if } x \leq 1 \\ 2x + 3 & \text{if } x > 1 \end{cases} \]

### Step 1: Determine the expression for \( f(x) \) when \( x \leq 1 \)

For \( x \leq 1 \), we have:

\[ f(x) = |x - 1| + a = (1 - x) + a = 1 - x + a \]

### Step 2: Determine the expression for \( f(x) \) when \( x > 1 \)

For \( x > 1 \), we have:

\[ f(x) = 2x + 3 \]

### Step 3: Check continuity at \( x = 1 \)

To ensure continuity at \( x = 1 \), we need:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

Calculating \( f(1) \):

\[ f(1) = |1 - 1| + a = 0 + a = a \]

Now, calculate \( \lim_{x \to 1^-} f(x) \):

\[ \lim_{x \to 1^-} f(x) = 1 - 1 + a = a \]

Next, calculate \( \lim_{x \to 1^+} f(x) \):

\[ \lim_{x \to 1^+} f(x) = 2(1) + 3 = 5 \]

Setting these equal for continuity:

\[ a = 5 \]

### Step 4: Local minimum condition at \( x = 1 \)

To confirm that \( x = 1 \) is a local minimum, we need to check the derivative of \( f(x) \) around \( x = 1 \).

1. For \( x < 1 \):

\[ f'(x) = -1 \]

2. For \( x > 1 \):

\[ f'(x) = 2 \]

At \( x = 1 \):

- The derivative from the left is \( -1 \) (decreasing).
- The derivative from the right is \( 2 \) (increasing).

Since the derivative changes from negative to positive at \( x = 1 \), this confirms that \( x = 1 \) is indeed a local minimum.

### Conclusion

Given that \( a \geq 5 \) and we found \( a = 5 \) for continuity, the value of \( a \) is:

\[ \boxed{5} \]