Consider `f(x)={{:([x]+[-x],xne2),(lambda,x=2):}` where `[.]` denotes the greatest integer function. If `f(x)` is continuous at `x=2` then the value of `lambda` is equal to

Options

A. `-1`
B. 0
C. 1
D. No value of possible

Answer

To determine the value of \(\lambda\) for which the function \(f(x)\) is continuous at \(x = 2\), we need to ensure that the left-hand limit (LHL), right-hand limit (RHL), and the function value at \(x = 2\) are all equal.

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} [x] + [-x] & \text{if } x \neq 2 \\ \lambda & \text{if } x = 2 \end{cases} \]
where \([x]\) denotes the greatest integer function.

### Step 2: Calculate the left-hand limit as \(x\) approaches 2
To find the left-hand limit \(f(2^-)\):
- As \(x\) approaches \(2\) from the left, \(x\) is slightly less than \(2\) (e.g., \(x = 1.999\)).
- Therefore, \([x] = [1.999] = 1\) and \([-x] = [-1.999] = -2\).
- Thus, we have:
\[ f(2^-) = [1.999] + [-1.999] = 1 - 2 = -1 \]

### Step 3: Calculate the right-hand limit as \(x\) approaches 2
To find the right-hand limit \(f(2^+)\):
- As \(x\) approaches \(2\) from the right, \(x\) is slightly greater than \(2\) (e.g., \(x = 2.001\)).
- Therefore, \([x] = [2.001] = 2\) and \([-x] = [-2.001] = -3\).
- Thus, we have:
\[ f(2^+) = [2.001] + [-2.001] = 2 - 3 = -1 \]

### Step 4: Set the limits equal to the function value at \(x = 2\)
For the function \(f(x)\) to be continuous at \(x = 2\), we need:
\[ f(2^-) = f(2) = f(2^+) \]
This gives us:
\[ -1 = \lambda = -1 \]

### Conclusion
Thus, the value of \(\lambda\) for which \(f(x)\) is continuous at \(x = 2\) is:
\[ \lambda = -1 \]