Discussion
For that receiver the image frequency lies at
Answer: D
The image sits twice the intermediate frequency above the signal:
\[ f_{image} = 98.1 + 21.4 = 119.5 \ \mathrm{MHz} \]
so 119.5 MHz, comfortably above the broadcast band, which is exactly why a 10.7 MHz intermediate frequency works so well here: the image falls where no broadcast station transmits. 108.8 MHz is the oscillator. The other two are the signal and its low-side alternative. Choosing an intermediate frequency that places the image outside the band is the cleanest solution to image interference.
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