Answer:
For \( f(x) = \cot(\text{sgn}(x - kx + 1)) \), the function will be discontinuous wherever the argument of the cotangent function becomes zero.
The signum function (\(\text{sgn}\)) is zero only when its argument is zero. So, we need to find when \( x - kx + 1 \) is zero.
\[ x - kx + 1 = 0 \]
Now, solve for \( k \):
\[ x(1 - k) + 1 = 0 \]
\[ x = -\frac{1}{1 - k} \]
For the function to have exactly one point of discontinuity, this expression must be true for a single value of \( x \). Therefore, the denominator \( (1 - k) \) cannot be zero.
\[ 1 - k \neq 0 \]
\[ k \neq 1 \]
So, \( k \) should not equal 1 for the function \( f(x) \) to have exactly one point of discontinuity.
