If `f(x)={([3+|x-k| ,x<=k ],[a^2-2+(tan(x-k))/(x-k), x>k])`has minimum at x=k,then
(a`)|a|<=2` (b)` |a|<2` (c)` |a|>2` (d)` |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:

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

We are given that this function has a minimum at \( x = k \).

### Step 1: Check Continuity at \( x = k \)

For \( f(x) \) to have a minimum at \( x = k \), it must be continuous at that point. This means we need to check if:

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

**Left-hand limit** as \( x \) approaches \( k \) from the left:

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

**Right-hand limit** as \( x \) approaches \( k \) from the right:

\[ \lim_{x \to k^+} f(x) = \lim_{h \to 0} f(k + h) = a^2 - 2 + \lim_{h \to 0} \frac{\tan(h)}{h} \]

Using the limit \( \lim_{h \to 0} \frac{\tan(h)}{h} = 1 \):

\[ \lim_{x \to k^+} f(x) = a^2 - 2 + 1 = a^2 - 1 \]

Setting the left-hand limit equal to the right-hand limit for continuity:

\[ 3 = a^2 - 1 \]

### Step 2: Solve for \( a^2 \)

Rearranging the equation gives:

\[ a^2 = 3 + 1 = 4 \]

### Step 3: Find the values of \( a \)

Taking the square root of both sides, we find:

\[ |a| = 2 \]

### Step 4: Determine the condition on \( |a| \)

Since we have \( |a| = 2 \), we can analyze the options given in the question:

- (a) \( |a| \leq 2 \) (True, since \( |a| = 2 \))
- (b) \( |a| < 2 \) (False, since \( |a| = 2 \))
- (c) \( |a| > 2 \) (False, since \( |a| = 2 \))
- (d) \( |a| \geq 2 \) (True, since \( |a| = 2 \))

### Conclusion

The correct options are (a) \( |a| \leq 2 \) and (d) \( |a| \geq 2 \). However, since the question asks for the condition that must hold true, the most appropriate answer is:

**Answer: (a) \( |a| \leq 2 \)**