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(D) The inductive reactance is given by ${X}_{L} = 2 \pi fL$. As frequency $f$ becomes very large,${X}_{L} 	o \infty$,which acts as an open circuit.T

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$2$

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(D) The inductive reactance is given by ${X}_{L} = 2 \pi fL$. As frequency $f$ becomes very large,${X}_{L} 	o \infty$,which acts as an open circuit.The capacitive reactance is given by ${X}_{C} = \frac{1}{2 \pi fC}$. As frequency $f$ becomes very large,${X}_{C} 	o 0$,which acts as a short circuit.Applying these conditions to the circuit:$1$. The capacitors in series with resistors act as short circuits (the resistors remain).$2$. The inductor acts as an open circuit.$3$. The middle capacitor acts as a short circuit.Looking at the circuit at very high frequencies,the path through the inductor becomes an open circuit. The remaining part of the circuit simplifies to a $1 \, \Omega$ resistor in series with two parallel branches,each containing a $2 \, \Omega$ resistor (since the capacitors act as short circuits).The equivalent resistance is $R_{eq} = 1 + \frac{2 	imes 2}{2 + 2} = 1 + 1 = 2 \, \Omega$.

At very high frequencies,the effective impedance of the given circuit will be $.... \Omega$.

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