An ac source is connected to a capacitor C. Due to decrease in its operating frequency:

${\mathrm{x}}_{\mathrm{c}}=\frac{1}{\mathrm{C\omega }}=\frac{1}{2\mathrm{\pi fC}}\Rightarrow {\mathrm{x}}_{\mathrm{c}}\propto \frac{1}{\mathrm{f}}$
where, f = operating frequency
C = capacitance
As the operating frequency decreases, the capacitive reactance will increase. Hence, options (1) and (2) are incorrect. Displacement current is equal to conduction current.
$\begin{array}{l}{\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{{\mathrm{X}}_{\mathrm{c}}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\left(\frac{1}{2\mathrm{\pi fC}}\right)}\\ {\mathrm{I}}_{\mathrm{rms}}=\left({\mathrm{V}}_{\mathrm{rms}}\right)\left(2\mathrm{\pi fC}\right)\\ \text{ since, }{\mathrm{V}}_{\mathrm{rms}},\mathrm{C}\text{ are constant }\\ {\mathrm{I}}_{\mathrm{rms}}\propto \mathrm{f}\end{array}$
Hence on decreasing the operating frequency, the conduction current decreases.
As displacement current = conduction current. On decreasing the operating frequency the
displacement current will decreases. Hence option (4) is correct and option (3) is incorrect.