1. State the problem: Find the range of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=\sin x$. 2. Formula and rules: For the sine function we have $-1 \le \sin x \le 1$ for all real $x$ because sine maps to the unit interval and is continuous. 3. Intermediate verification: We check that the endpoints are attained by evaluating $\sin(\pi/2)=1$ and $\sin(3\pi/2)=-1$. 4. Explanation: Because sine is continuous on $\mathbb{R}$ and takes values $-1$ and $1$, by the intermediate value property it attains every value between them, so the range is $[-1,1]$. 5. Final answer for Problem 5: Range$(f)=[-1,1]$. 6. Problem 2.1.1: In a class of 80 students every student studies Economics or Geography or both; 65 study Economics and 50 study Geography; find how many study both. 7. Formula: Use $|E \cup G| = |E| + |G| - |E \cap G|$. 8. Calculation: Here $|E \cup G|=80$, $|E|=65$, $|G|=50$, so $$80 = 65 + 50 - |E \cap G|$$ 9. Solve: $$|E \cap G| = 65 + 50 - 80 = 35$$ 10. Answer: 35 students studied both Economics and Geography. 11. Problem 2.1.2: In a class of 40 each student offers at least one of Mathematics and Physics; 22 offer Physics and 21 offer Mathematics; find how many offer Physics only. 12. Formula: Let $M$ and $P$ be Mathematics and Physics; $|M \cup P| = |M| + |P| - |M \cap P|$. 13. Calculation: Here $|M \cup P|=40$, $|P|=22$, $|M|=21$, so intersection $$|M \cap P| = 21 + 22 - 40 = 3$$ 14. Physics only: $|P| - |M \cap P| = 22 - 3 = 19$. 15. Answer: 19 students offer Physics only. 16. Problem 2.1.3: In a class of 40, 32 offer Mathematics, 24 offer Physics and 4 offer neither; find how many offer both Mathematics and Physics. 17. Calculation: Number offering at least one is $40 - 4 = 36$. 18. Use formula: $$36 = 32 + 24 - |M \cap P|$$ 19. Solve: $$|M \cap P| = 32 + 24 - 36 = 20$$ 20. Answer: 20 students offer both Mathematics and Physics. 21. Problem 2.1.4a: Out of 400 students, 300 offer Biology and 100 offer Chemistry, and 70 offer neither; how many offer both? 22. Calculation: Number offering at least one is $400 - 70 = 330$. 23. Use formula: $$330 = 300 + 100 - |B \cap C|$$ 24. Solve: $$|B \cap C| = 300 + 100 - 330 = 70$$ 25. Answer: 70 students offer both Biology and Chemistry.