Let `f(x) = {(|x-1| + a,"if " x le 1),(2x + 3,"if " x gt 1):}` . If f(x) has a local minimum at x=1, then

Options

A. `a gt 5`
B. `0 lt a le 5`
C. `a le 5`
D. `a=5`

Answer

To determine the value of \( a \) such that the function \( f(x) \) has a local minimum at \( x = 1 \), we need to analyze the function defined as:

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

### Step 1: Analyze the function at \( x = 1 \)

First, we need to evaluate \( f(1) \) from both sides.

- For \( x \leq 1 \):
\[ f(1) = |1 - 1| + a = 0 + a = a \]

- For \( x > 1 \):
\[ f(1) = 2(1) + 3 = 2 + 3 = 5 \]

### Step 2: Set the values equal for continuity

For \( f(x) \) to have a local minimum at \( x = 1 \), the function must be continuous at that point. Therefore, we set the two expressions equal to each other:

\[ f(1) = f(1^+) \implies a = 5 \]

### Step 3: Check the conditions for a local minimum

Next, we need to check the behavior of \( f(x) \) around \( x = 1 \):

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

- For \( x > 1 \):
\[ f(x) = 2x + 3 \]

### Step 4: Evaluate the limits as \( x \) approaches 1

As \( x \) approaches 1 from the left:
\[ f(1^-) = a \]

As \( x \) approaches 1 from the right:
\[ f(1^+) = 5 \]

For \( f(x) \) to have a local minimum at \( x = 1 \), we need:
\[ f(1^-) \leq f(1) \leq f(1^+) \]

Substituting the values we found:
\[ a \leq 5 \]

### Step 5: Conclusion

Thus, for \( f(x) \) to have a local minimum at \( x = 1 \), we conclude that:
\[ a \leq 5 \]

### Final Answer

The value of \( a \) must satisfy:
\[ a \leq 5 \]