We test each implication for a continuous, strictly positive f on [1,∞). Exactly one is forced; the rest fall to counterexamples.
The true statement (option 3). On the whole domain x≥1, so x≥1>0 and therefore
0<f(x)≤xf(x)(x≥1).
If ∫1∞xf(x)dx converges, the direct comparison test gives
∫1∞f(x)dx≤∫1∞xf(x)dx<∞,
so ∫1∞f(x)dx converges. This implication always holds.
Option 1 is false (∫f conv. ⇒ ∫f2 conv.). Build continuous triangular spikes centered at each integer n with height hn=n and base width wn=2/n3. Each spike contributes area 21hnwn=1/n2 to ∫f, so ∫1∞f=∑1/n2<∞. But a triangle of height h and base w has ∫(tri)2=31h2w, giving 31n2⋅2/n3=3n2 per spike, so ∫1∞f2=32∑n1=∞. Convergence of ∫f does not transfer to ∫f2.
Option 2 is false (∫f2 conv. ⇒ ∫f conv.). Take f(x)=1/x. Then ∫1∞f2=∫1∞x−2dx=1 converges, yet ∫1∞f=∫1∞x−1dx diverges.
Option 4 is false (∫1/f conv. ⇒ ∫f conv.). Take f(x)=1/x2. Then 1/f=x2 and ∫1∞1/f=∫1∞x2dx diverges, so the hypothesis already fails — but even reversing roles, with f(x)=x2 we get ∫1/f=∫x−2<∞ while ∫f=∫x2=∞. No valid implication.
Hence the only universally valid statement is option 3.
Pitfall: Squaring or reciprocating an integrand changes its tail behavior unpredictably — there is no general implication between ∫f and ∫f2 (in either direction). The one safe move here is monotone domination: on [1,∞), f≤xf, so a convergent ∫xf squeezes ∫f.
Note on the answer key: the official preliminary key (کلید اولیه) marks option 1, but for a positive continuous f on [1,∞) that statement is false — take f(x)=1/x: ∫1∞f2=∫1∞x−2 converges while ∫1∞f=∫1∞x−1 diverges. The only universally true implication is option 3: since 0<f(x)≤xf(x) for x≥1, ∫1∞f≤∫1∞xf<∞. Correct answer: option 3.