The carrier frequency of a transmitter is provided by a tank circuit of a coil of inductance 49$$\mu$$H and a capacitance of 2.5 nF. It is modulated by an audio signal of 12 kHz. The frequency range occupied by the side bands is:

For an LC tank circuit, the carrier (resonant) frequency is obtained from the well-known resonance formula

$$f_c=\frac{1}{2\pi\sqrt{LC}}.$$

First we convert the given component values into SI units so that all calculations stay consistent:

$$L = 49\,\mu\text{H}=49\times10^{-6}\ \text{H},$$

$$C = 2.5\,\text{nF}=2.5\times10^{-9}\ \text{F}.$$

We now calculate the product $$LC$$:

$$LC = \bigl(49\times10^{-6}\bigr)\bigl(2.5\times10^{-9}\bigr) = 49\times2.5\times10^{-6-9} = 122.5\times10^{-15}.$$

Since $$122.5\times10^{-15}=1.225\times10^{-13}$$, we have

$$LC = 1.225\times10^{-13}.$$

Taking the square root, step by step:

$$\sqrt{LC}=\sqrt{1.225\times10^{-13}} =\sqrt{1.225}\times\sqrt{10^{-13}} \approx1.106\times10^{-6.5}.$$

Because $$10^{-6.5}=3.162\times10^{-7}$$, the numerical value becomes

$$\sqrt{LC}\approx1.106\times3.162\times10^{-7} \approx3.498\times10^{-7}\ \text{s}.$$

Now we substitute this result into the resonance formula. Taking $$2\pi\approx6.283$$, the denominator reads

$$2\pi\sqrt{LC}=6.283\times3.498\times10^{-7} \approx2.197\times10^{-6}.$$

Hence the carrier frequency is

$$f_c=\frac{1}{2\pi\sqrt{LC}} =\frac{1}{2.197\times10^{-6}} \approx4.55\times10^{5}\ \text{Hz} =455\ \text{kHz}.$$

The audio (modulating) signal has a highest frequency of

$$f_m = 12\ \text{kHz}.$$

In amplitude modulation the side bands appear at frequencies $$f_c\pm f_m$$. Therefore

$$\text{Lower side frequency }f_L = f_c - f_m =455\ \text{kHz}-12\ \text{kHz} =443\ \text{kHz},$$

$$\text{Upper side frequency }f_U = f_c + f_m =455\ \text{kHz}+12\ \text{kHz} =467\ \text{kHz}.$$

Thus the side-band spectrum extends from about $$443\ \text{kHz}$$ up to $$467\ \text{kHz}$$. Allowing for the small rounding differences in the listed options, this interval corresponds to

$$442\ \text{kHz}\;-\;466\ \text{kHz}.$$

Hence, the correct answer is Option 3.