The behavior where current first increases and then decreases as frequency increases is the characteristic of electrical resonance in a series AC circuit. According to the sources, the current magnitude is $I = V / Z$, where the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ . Inductive reactance ($X_L = \omega L$) increases with frequency, while capacitive reactance ($X_C = 1/\omega C$) decreases . At very low frequencies, $X_C$ is very large, making current small. At the resonant frequency ($f_0 = 1/(2\pi\sqrt{LC})$), $X_L = X_C$, the impedance is at its minimum ($Z=R$), and current is at its maximum . At very high frequencies, $X_L$ becomes very large, causing the current to decrease again. This 'peaking' behavior requires both inductance (L) and capacitance (C) to be present in the circuit. Therefore, combinations (a) and (d) are the correct answers.