A composite slab consists of two materials having coefficients of thermal conductivity K and 2 K , thickness x and 4 x respectively. The temperatures of two outer surfaces of a composite slab are $T_2$ and $T_1$ respectively ( $\mathrm{T}_2>\mathrm{T}_1$ ). The rate of heat transfer through the slab in a steady state is $\left[\frac{A\left(T_2-T_1\right) K}{x}\right] f$, where $f$ is equal to

The thermal resistance for a material is given by $R = \frac{L}{KA}$, where $L$ is thickness, $K$ is thermal conductivity, and $A$ is cross-sectional area.

For material 1 with thickness $x$ and conductivity $K$, $R_1 = \frac{x}{KA}$.

For material 2 with thickness $4x$ and conductivity $2K$, $R_2 = \frac{4x}{2KA} = \frac{2x}{KA}$.

The total resistance for materials in series is $R_{total} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}$.

Heat transfer rate is $\frac{Q}{t} = \frac{\Delta T}{R_{total}} = \frac{T_2 - T_1}{\frac{3x}{KA}} = \frac{A(T_2 - T_1)K}{3x}$.

Comparing with the given form $\left[\frac{A(T_2-T_1) K}{x}\right] f$, we find $f = \frac{1}{3}$.

Final answer: $\boxed{\text{D}}$