(a) is correct, and it is what distinguishes a band-pass filter from a low-pass or high-pass one. A band-pass response has two half-power (−3 dB) frequencies: a lower critical frequency f₁ and an upper one f₂, between which signals pass and outside which they are attenuated. Low-pass and high-pass responses have only one such frequency each.

(b) is correct, and it is the definition of Q for a band-pass response :

\(Q=\dfrac{f_{0}}{BW}\qquad\text{where}\qquad BW=f_{2}-f_{1}\ \text{and}\ f_{0}=\sqrt{f_{1}f_{2}}\)

Note that the centre frequency is the geometric mean of the two critical frequencies, not the arithmetic mean — a detail that matters for wide-band filters, though for a narrow band the two nearly coincide.

So the answer is option 1.

Why (c) is the near miss. Q certainly can be computed from f₁ and f₂, since both f₀ and BW derive from them. But the statement as written is incomplete: knowing the two critical frequencies individually is not the same as knowing the ratio of centre frequency to bandwidth, and the standard definition — the one the question is testing — is stated in terms of f₀ and BW. Statement (b) is the precise formulation and (c) the loose one, so where both are offered, (b) is the intended choice.

Statement (d), "a voltage gain", is true of an active filter but not of a filter response as such: a passive RLC band-pass network has a maximum gain of one (or less), and gain is not part of what defines a band-pass characteristic. It is an attribute of the implementation, not of the response.

The physical meaning of Q is worth carrying: it is \(2\pi\) times the ratio of energy stored to energy lost per cycle, so a high-Q circuit rings for many cycles and rejects everything but a narrow band. In a series RLC band-pass network \(Q=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}\), so lowering the resistance sharpens the response — the same resistance that damps the transient response, tying the frequency-domain and time-domain pictures together.

Hence, the correct statements are (a) and (b).