Statement 2 is the false one — negative feedback widens the bandwidth — so the correct set is 1, 3 and 4, option 2.

Everything follows from one factor. With loop gain \(A\beta\), the desensitivity factor \(1+A\beta\) divides some quantities and multiplies others:

Why bandwidth increases. This is the classic result and the point of the question. The gain-bandwidth product is a property of the amplifier and is preserved:

\(A_{f}\times f_{H,f}=\dfrac{A}{1+A\beta}\times f_{H}\left(1+A\beta\right)=A\,f_{H}\)

Feedback trades gain for bandwidth at a fixed product. An amplifier of gain 1000 flat to 10 kHz becomes, with \(1+A\beta=10\), an amplifier of gain 100 flat to 100 kHz. Physically, at frequencies where the open-loop gain has begun to fall the loop gain falls with it, so less feedback is applied there — and less feedback means less reduction of gain. The response is flattened by the feedback withdrawing exactly where the amplifier is weakest.

Why distortion falls. Distortion is generated inside the amplifier, so it appears at the output but not at the input. The feedback path returns it to the input in opposition, and the amplifier's own gain then works to cancel it — reduced by the same \(1+A\beta\).

Why statement 3 needs one qualification. Noise generated within the amplifier is reduced exactly as distortion is, so the signal-to-noise ratio improves. But noise already present on the input signal is amplified along with the signal and cannot be improved by any amount of feedback — feedback cannot distinguish wanted from unwanted at the input. The statement is true for internally generated noise, which is what it is normally taken to mean.

Hence, statements 1, 3 and 4 are correct.