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

Options

A. 5
B. 6
C. `11//2`
D. `15//2`

Answer

To solve the problem, we need to analyze the piecewise function given by:

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

We need to determine the value of \( a \) such that \( f(x) \) has a local minimum at \( x = 1 \) and \( a \geq 5 \).

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

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

\[ |x - 1| = 1 - x \]

Thus,

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

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

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

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

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

For \( f(x) \) to have a local minimum at \( x = 1 \), it must be continuous at that point. Therefore, we need to check the values of \( f(1) \) from both sides:

1. From the left (as \( x \) approaches 1 from the left):

\[ f(1^-) = -1 + (1 + a) = a \]

2. From the right (as \( x \) approaches 1 from the right):

\[ f(1^+) = 2(1) + 3 = 5 \]

Setting these equal for continuity:

\[ a = 5 \]

### Step 4: Check the condition for local minimum

Next, we need to check if \( f(x) \) has a local minimum at \( x = 1 \). We can do this by checking the derivative of \( f(x) \):

- For \( x < 1 \):

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

- For \( x > 1 \):

\[ f'(x) = 2 \]

At \( x = 1 \):

- \( f'(x) \) changes from \(-1\) (negative) to \(2\) (positive), indicating that \( x = 1 \) is indeed a local minimum.

### Conclusion

Since we found that \( a = 5 \) satisfies the condition for continuity and the local minimum, we conclude:

The value of \( a \) is \( 5 \).