Step-by-step Solution:

Since \( f(x) \) is differentiable at \( x = 0 \), it must be continuous at \( x = 0 \). Therefore, \( a = \lim_{x \to 0} f(x) \). Expand the terms in the numerator using the binomial theorem for small \( x \): \[ (1+x)^{\frac{1}{3}} \approx 1 + \frac{1}{3}x + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}x^2 = 1 + \frac{1}{3}x - \frac{1}{9}x^2 \] \[ (1+2x)^{\frac{1}{4}} \approx 1 + \frac{1}{4}(2x) + \frac{\frac{1}{4}(\frac{1}{4}-1)}{2!}(2x)^2 = 1 + \frac{1}{2}x - \frac{3}{8}x^2 \] Substitute these expansions into \( f(x) \) for \( x \neq 0 \): \[ f(x) = \frac{\left(1 + \frac{1}{3}x - \frac{1}{9}x^2\right) - \left(1 + \frac{1}{2}x - \frac{3}{8}x^2\right)}{x} \] \[ f(x) = \frac{\left(\frac{1}{3} - \frac{1}{2}\right)x + \left(-\frac{1}{9} + \frac{3}{8}\right)x^2}{x} \] \[ f(x) = -\frac{1}{6} + \left(\frac{-8 + 27}{72}\right)x = -\frac{1}{6} + \frac{19}{72}x \] Taking the limit as \( x \to 0 \): \[ a = \lim_{x \to 0} f(x) = -\frac{1}{6} \] Since \( f(x) \approx -\frac{1}{6} + \frac{19}{72}x \) for \( x \) near 0, the derivative at \( x = 0 \) is simply the coefficient of \( x \) in this linear approximation: \[ f'(0) = \frac{19}{72} \] We need to find the value of \( a + f'(0) \): \[ a + f'(0) = -\frac{1}{6} + \frac{19}{72} \] \[ = -\frac{12}{72} + \frac{19}{72} = \frac{7}{72} \]