Answer is: option1

\( \frac{23}{16} \)

Solution:

The derivative of \( f(x) = x^4 - x \) is:

\[ f'(x) = 4x^3 - 1 \]

Rewriting the equation of the normal line in slope-intercept form:

\[ y = 2x - k \]

From this, the slope of the normal line is:

\[ m_{\text{normal}} = 2 \]

Since the normal line is perpendicular to the tangent line, the relationship between their slopes is:

\[ m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}} = -\frac{1}{2} \]

Thus, we set:

\[ 4x^3 - 1 = -\frac{1}{2} \] \[ 4x^3 = \frac{1}{2} \] \[ x^3 = \frac{1}{8} \] \[ x = \frac{1}{2} \] Find \( y = f(x) \) at \( x = \frac{1}{2} \) \[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 - \frac{1}{2} \] \[ = \frac{1}{16} - \frac{1}{2} \] \[ = \frac{1}{16} - \frac{8}{16} \] \[ = -\frac{7}{16} \]

Since the normal line passes through \( \left(\frac{1}{2}, -\frac{7}{16}\right) \), substituting into \( 2x - y = k \):

\[ 2 \times \frac{1}{2} - \left(-\frac{7}{16}\right) = k \] \[ 1 + \frac{7}{16} = k \] \[ \frac{16}{16} + \frac{7}{16} = k \] \[ k = \frac{23}{16} \] Final Answer: \[ \boxed{\frac{23}{16}} \]