If `f(x)={3+|x-k|,xlt=k a^2-2+(s n(x-k))/(x-k),x > k` has minimum at `x=k ,` then `a in R` b. `|a|<2` c. `|a|>2` d. `1<|a|<2`

Options

A. `a in R`
B. `|a| lt 2`
C. `|a| gt 2`
D. `1 lt |a| lt 2`

Answer

To solve the problem, we need to analyze the function \( f(x) \) given by:

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

We are tasked with finding the conditions under which \( f(x) \) has a minimum at \( x = k \).

### Step 1: Evaluate \( f(k) \)

First, we find \( f(k) \) using the first case of the function:

\[ f(k) = 3 + |k - k| = 3 + 0 = 3 \]

### Step 2: Evaluate the limit of \( f(x) \) as \( x \) approaches \( k \) from the right

Next, we evaluate the limit of \( f(x) \) as \( x \) approaches \( k \) from the right (i.e., \( x \to k^+ \)):

\[ f(k^+) = a^2 - 2 + \lim_{x \to k} \frac{\sin(x - k)}{x - k} \]

Using the limit property:

\[ \lim_{x \to k} \frac{\sin(x - k)}{x - k} = 1 \]

Thus, we have:

\[ f(k^+) = a^2 - 2 + 1 = a^2 - 1 \]

### Step 3: Set the condition for a minimum at \( x = k \)

For \( f(x) \) to have a minimum at \( x = k \), we need:

\[ f(k^+) \geq f(k) \]

Substituting the values we found:

\[ a^2 - 1 \geq 3 \]

### Step 4: Solve the inequality

Rearranging the inequality gives:

\[ a^2 \geq 4 \]

Taking the square root of both sides, we find:

\[ |a| \geq 2 \]

### Conclusion

The conditions we derived indicate that \( a \) must satisfy \( |a| \geq 2 \). This means the correct answer among the options provided is:

**c. \( |a| > 2 \)**