1. Problem 16 concerns the behavior of the function images of set operations. 2. Statement I: $f(A \cup B) = f(A) \cup f(B)$ - This is always true for functions because the image of the union is the union of the images. 3. Statement II: $f(A \cap B) = f(A) \cap f(B)$ - Generally false. The image of an intersection is a subset of the intersection of images, but equality need not hold because different elements in $A \cap B$ may map to the same point. 4. Statement III: Given $C \subseteq B$, then $f(C) \subseteq f(B)$ - True because if every element of $C$ is in $B$, their images under $f$ must lie within $f(B)$. 5. Conclusion for 16: Only statements I and III must be true. So the answer is (D) I and III only. 6. Problem 17 asks which statement about sequences is FALSE. 7. Option (A): Every unbounded sequence is divergent. - True by definition, since boundedness is necessary for convergence. 8. Option (B): Every bounded sequence is convergent. - False. Bounded sequences may fail to converge (example: $a_n = (-1)^n$). 9. Option (C): If $\lim_{n \to \infty} a_n = a$ and $f$ is continuous at $a$, then $\lim_{n \to \infty} f(a_n) = f(a)$. - True by continuity of $f$. 10. Option (D): The Cauchy criterion interpretation for convergent sequences. - True. 11. Option (E): The equivalence of $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} |a_n - a| = 0$. - True. 12. Conclusion for 17: The false statement is (B). 13. Problem 18 concerns analyzing $g$ from its derivative graph $g'$: 14. Analysis of statement I: Local max at $x=0$ and min at $x=4$. - At $x=0$, we do not have derivative info but $g'(x)$ starts high at 2 at $x=1$ and decreases to $x=3$. - Local maximum requires $g'$ changes from positive to negative. - At $x=4$, $g'$ is $-1$, and goes negative before and after, not consistent with local min. 15. Analysis of statement II: Local max at $x=2$ and local min at $x=5$. - From $x=1$ to 3, $g'$ decreases from 2 to -2. - So at $x=2$, $g'$ passes positive to negative suggests a local max. - At $x=5$, $g'$ at -3 shifts to increase up to 1 at $x=7$, so passes negative to positive, indicating local min. 16. Analysis of statement III: $g(2) = g(5)$. - Since $g'$ is derivative, integral of $g'$ from 2 to 5 gives $g(5) - g(2)$. - Approximate areas under $g'$ between 2 and 5: Negative area from $2$ to $3$, then slight increase but overall negative, so unlikely to be zero. - Cannot guarantee $g(2) = g(5)$. 17. Conclusion for 18: Only statement II must be true; answer (B). Final answers: 16: (D) 17: (B) 18: (B)