(a) is correct. Whatever quantity is sampled and whatever is fed back, the closed-loop gain of a negative-feedback amplifier has the same form:

\(A_{f}=\dfrac{A}{1+A\beta}\)

Only the meaning of A and β changes between topologies — a voltage ratio in voltage-series feedback, a transresistance in voltage-shunt, and so on — but the expression itself is universal.

(b) is false — negative feedback increases the bandwidth. The gain is reduced by the factor \((1+A\beta)\) and the bandwidth is multiplied by the same factor:

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

— the gain-bandwidth product is preserved, so what is lost in gain is gained in bandwidth. This is the single most useful consequence of feedback.

(c) is false as a general claim. The effect on output impedance depends on what is sampled at the output:

Voltage sampling holds the output voltage steady against load changes, which is exactly what a low output impedance means. Since (c) states the increase unconditionally, it is wrong.

(d) is correct. Feedback desensitises the gain against every source of variation — temperature, ageing, device replacement, supply drift:

\(\dfrac{dA_{f}}{A_{f}}=\dfrac{1}{1+A\beta}\cdot\dfrac{dA}{A}\)

With a loop gain of 1000, a 10 % drift in the device's own gain becomes a 0.01 % drift in the closed-loop gain. Distortion and noise generated inside the loop are reduced by the same factor.

So (a) and (d) — option 3.

The summary of what negative feedback does: it trades gain for everything else — wider bandwidth, lower distortion, gain stability, and input and output impedances shifted in whichever direction the topology dictates. The one thing it can cost is stability in the control sense: if the loop's own phase shift reaches 180° while the loop gain still exceeds unity, the feedback becomes positive and the amplifier oscillates, which is why compensation is designed in.

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