c
Given : Inductance, \(L=49\, \mu \mathrm{H}=49 \times 10^{-6}\, \mathrm{H},\)

capacitance \(C=2.5\, \mathrm{nF}=2.5 \times 10^{-9}\, \mathrm{F}\)

Using \(\omega=\frac{1}{\sqrt{L C}}\)

\(=\frac{1}{\sqrt{49 \times 10^{-6} \times \frac{2.5}{10} \times 10^{-9}}}\)

\(=\frac{1}{7 \times 5 \times 10^{-8}}=\frac{10^{8}}{7 \times 5}\)

or, \(\frac{10^{8}}{7 \times 5}=2 \pi \times f=2 \times \frac{22}{7} \times f\)

\((\because \omega=2 \pi f)\)

or, \(f=\frac{10^{7}}{22}=\frac{10^{4}}{22}\, \mathrm{kHz}=454.54\, \mathrm{kHz}\)

Therefore frequency range \(454.54 \pm 12\, \mathrm{kHz}\)

i.e., \(442\,\mathrm{kHz}-466 \,\mathrm{kHz}\)