(A) is false, but (R) is true — option 4. Everything turns on one word in the assertion: ideal.

Why the assertion is false. An ideal amplifier is by definition perfectly linear — its transfer characteristic is a straight line, so its output is \(v_{o}=A\,v_{i}\) exactly. Feed a sinusoid into it and you get back a scaled sinusoid of the same shape and the same single frequency: an exact replica. The assertion claims the opposite, so it fails.

Class A only strengthens the case. Of the conduction classes it is the one that conducts for the full 360° of the cycle and never cuts off, so it is free of the crossover distortion that afflicts class B. An ideal class A stage is therefore the least distorting arrangement there is.

Why the reason is true. Taken on its own, (R) states a correct fact about real devices. A transistor’s transfer characteristic is not a straight line — for a BJT the collector current is exponential in the base-emitter voltage,

\(I_C=I_Se^{V_{BE}/V_T}\)

and for a FET it is square-law. Expanding such a characteristic about the operating point as a power series,

\(i_c=a_1v_{be}+a_2v_{be}^{2}+a_3v_{be}^{3}+\dots\)

and driving it with \(v_{be}=V_m\cos\omega t\), the squared term produces a DC shift and a component at \(2\omega\), the cubed term one at \(3\omega\), and so on. Frequencies that were never applied appear at the output — harmonic distortion.

So the code is 4. (R) correctly explains why a practical amplifier distorts; it simply has nothing to explain in (A), because the ideal amplifier (A) describes does not distort at all. A true reason attached to a false assertion is exactly the situation option 4 is for.

How the distortion is quantified, for the practical case (R) is really about. Total harmonic distortion collects all the unwanted components:

\(D=\sqrt{D_2^{2}+D_3^{2}+D_4^{2}+\dots}, \qquad D_n=\dfrac{|B_n|}{|B_1|}\)

and the delivered power rises accordingly, \(P=\left(1+D^{2}\right)P_1\). The three-point method estimates the second-harmonic term straight from the waveform extremes:

\(D_2=\dfrac{\tfrac{1}{2}(I_{max}+I_{min})-I_{CQ}}{I_{max}-I_{min}}\)

The ways of reducing it. Keep the signal swing small, so that only the linear term matters — which is why small-signal analysis works at all; apply negative feedback, which divides the distortion generated inside the loop by \(1+A\beta\); or use push-pull operation, since a symmetric pair cancels the even harmonics.

The lesson of the item. In assertion-reason questions read the qualifiers first. “Ideal”, “always”, “never” and “only” are usually load-bearing, and here a single adjective flips the assertion from true to false.

Hence, the answer is (A) is false, but (R) is true.