Q and bandwidth are inversely related by definition, so a rise in one is a fall in the other — option 2.

\(Q=\dfrac{f_{0}}{BW}\qquad\Rightarrow\qquad BW=\dfrac{f_{0}}{Q}\)

What Q measures. It is the ratio of energy stored to energy dissipated per cycle:

\(Q=2\pi\dfrac{\text{energy stored}}{\text{energy lost per cycle}}\)

A high-Q antenna stores a great deal of reactive energy in the near field and radiates comparatively little of it each cycle. Because so much energy is tied up in the resonance, the structure responds strongly at one frequency and falls away sharply on either side — a narrow bandwidth. A low-Q antenna dissipates its stored energy quickly into radiation, so it is far less selective and works over a wide band.

Why physical size decides the matter. This is the deep reason behind the question. The Chu-Harrington limit shows that for an antenna enclosed in a sphere of radius a,

\(Q\approx\dfrac{1}{(ka)^{3}}+\dfrac{1}{ka},\qquad k=\dfrac{2\pi}{\lambda}\)

so Q rises steeply as the antenna is made small compared with the wavelength. An electrically small antenna is necessarily high-Q and therefore necessarily narrowband — a limit of physics, not of engineering skill. This is why a pocket radio's ferrite rod must be retuned across the band while a full-size dipole needs no such adjustment.

How wideband antennas are made follows directly: increase the effective volume and thicken the conductors, since a fatter element stores less energy per unit of radiation. A thick cylindrical dipole is broader than a thin-wire one, a biconical broader still, and frequency-independent forms such as the log-periodic and the spiral achieve very low Q by making their geometry scale with frequency.

Options 3 and 4 are not statements about a relationship at all — a bandwidth of zero would mean an antenna that works at exactly one frequency and nowhere else, and "equal to unity" gives no units.

Hence, when Q increases the bandwidth decreases.