If `f (x)= tan ^(-1) (sgn (n ^(2) -lamda x+1))` has exactly one point of discontinuity, then the value of `lamda ` can be:

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the function \( f(x) = \tan^{-1}(\text{sgn}(n^2 - \lambda x + 1)) \) has exactly one point of discontinuity.

### Step-by-Step Solution:

1. **Understanding the Signum Function**:
The signum function \( \text{sgn}(x) \) is defined as:
\[ \text{sgn}(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \]
The function \( f(x) \) will be discontinuous at the points where the argument of the signum function, \( n^2 - \lambda x + 1 \), is equal to zero.

2. **Setting Up the Equation**:
We need to find when:
\[ n^2 - \lambda x + 1 = 0 \]
Rearranging gives:
\[ \lambda x = n^2 + 1 \quad \Rightarrow \quad x = \frac{n^2 + 1}{\lambda} \]

3. **Condition for Exactly One Point of Discontinuity**:
The quadratic expression \( n^2 - \lambda x + 1 \) will have exactly one point of discontinuity if it has a double root. This occurs when the discriminant of the quadratic equation is zero.

4. **Finding the Discriminant**:
The general form of a quadratic equation is \( ax^2 + bx + c \). Here, we can rewrite our equation as:
\[ x^2 - \lambda x + 1 = 0 \]
The discriminant \( D \) of this quadratic equation is given by:
\[ D = b^2 - 4ac = (-\lambda)^2 - 4 \cdot 1 \cdot 1 = \lambda^2 - 4 \]

5. **Setting the Discriminant to Zero**:
For the quadratic to have exactly one root, we set the discriminant to zero:
\[ \lambda^2 - 4 = 0 \]
Solving this gives:
\[ \lambda^2 = 4 \quad \Rightarrow \quad \lambda = \pm 2 \]

6. **Conclusion**:
Therefore, the values of \( \lambda \) that make \( f(x) \) have exactly one point of discontinuity are:
\[ \lambda = 2 \quad \text{or} \quad \lambda = -2 \]

### Final Answer:
The value of \( \lambda \) can be \( 2 \) or \( -2 \).