1. **Problem:** Find $f^{100}(x)$ for the function $f(x) = 1 - \frac{1}{x}$ defined for all real $x \neq 0, 1$. 2. **Calculate $f^2(x)$:** $$f^2(x) = f(f(x)) = f\left(1 - \frac{1}{x}\right) = 1 - \frac{1}{1 - \frac{1}{x}} = 1 - \frac{1}{\frac{x-1}{x}} = 1 - \frac{x}{x-1} = \frac{(x-1) - x}{x-1} = \frac{-1}{x-1} = \frac{1}{1-x}.$$ 3. **Calculate $f^3(x)$:** $$f^3(x) = f(f^2(x)) = f\left(\frac{1}{1-x}\right) = 1 - \frac{1}{\frac{1}{1-x}} = 1 - (1-x) = x.$$ This shows that $f^3(x) = x$. 4. **Since $f^3(x) = x$, $f$ has order 3 under composition.** 5. **Calculate $f^{100}(x)$:** Since $100 = 3 \times 33 + 1$, we have $$f^{100}(x) = f^{3 \times 33 + 1}(x) = f(f^{3 \times 33}(x)) = f(x) = 1 - \frac{1}{x}.$$ --- 6. **Problem:** Find the limit of the sequence $$a_n = \sum_{k=1}^n \frac{n}{n^2 + k^2}$$ as $n \to \infty$. 7. Rewrite the term: $$\frac{n}{n^2 + k^2} = \frac{1}{n} \cdot \frac{1}{1 + \left(\frac{k}{n}\right)^2}.$$ 8. Then, $$a_n = \sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{1 + \left(\frac{k}{n}\right)^2}.$$ This is a Riemann sum for the integral $$\int_0^1 \frac{1}{1 + x^2} dx.$$ 9. Compute the integral: $$\int_0^1 \frac{1}{1+x^2} dx = \left[ \arctan x \right]_0^1 = \arctan 1 - \arctan 0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}.$$ So, $$\lim_{n\to\infty} a_n = \frac{\pi}{4}.$$ --- 10. **Problem:** For a bounded, connected set $S \subseteq \mathbb{R}$ with more than one point and a continuous $f$ on $S$, determine which statements must be true: I. $S$ is a closed interval. II. $f(S)$ has a maximum value. III. $f(S)$ has a least upper bound. 11. Since $S$ is connected in $\mathbb{R}$, it must be an interval. However, 'bounded and connected' does not guarantee the interval is closed (e.g., open intervals are connected and bounded). Hence, I is not necessarily true. 12. Since $S$ has more than one point and $f$ is continuous on $S$, $f(S)$ is the continuous image of an interval (possibly open). A continuous function on a closed interval attains its maximum, but on open intervals it might not. Hence II is not necessarily true. 13. However, the image $f(S)$ of a connected set is connected, and since $f$ is continuous on bounded $S$, $f(S)$ is a bounded interval (not necessarily closed), so it has a least upper bound by completeness of $\mathbb{R}$. Thus, III is always true. Answer: (B) III only. --- 14. **Problem:** Find the maximum value of $$f(x,y,z) = x - 3y + 2z$$ subject to $$x^2 + y^2 + z^2 = 9.$$ 15. Use Lagrange multipliers: Set $$\nabla f = \lambda \nabla g,$$ where $g(x,y,z) = x^2 + y^2 + z^2 - 9 = 0.$ 16. Calculate gradients: $$\nabla f = (1, -3, 2), \quad \nabla g = (2x, 2y, 2z).$$ 17. From $\nabla f = \lambda \nabla g$, $$1 = 2 \lambda x, \, -3 = 2 \lambda y, \, 2 = 2 \lambda z.$$ 18. Then, $$x = \frac{1}{2 \lambda}, \, y = -\frac{3}{2 \lambda}, \, z = \frac{1}{\lambda}.$$ 19. Substitute into constraint: $$x^2 + y^2 + z^2 = \left(\frac{1}{2\lambda}\right)^2 + \left(-\frac{3}{2\lambda}\right)^2 + \left(\frac{1}{\lambda}\right)^2 = \frac{1}{4 \lambda^2} + \frac{9}{4 \lambda^2} + \frac{1}{\lambda^2} = \frac{1+9}{4 \lambda^2} + \frac{1}{\lambda^2} = \frac{10}{4 \lambda^2} + \frac{1}{\lambda^2} = \frac{10 + 4}{4 \lambda^2} = \frac{14}{4 \lambda^2} = 9.$$ 20. Solve for $\lambda$: $$\frac{14}{4 \lambda^2} = 9 \implies \frac{14}{4 \times 9} = \lambda^2 \implies \lambda^2 = \frac{14}{36} = \frac{7}{18}.$$ 21. Thus, $$\lambda = \pm \sqrt{\frac{7}{18}} = \pm \frac{\sqrt{14}}{6}.$$ 22. Compute $f(x,y,z)$: $$f = x - 3y + 2z = \frac{1}{2 \lambda} - 3\left(- \frac{3}{2 \lambda} \right) + 2 \cdot \frac{1}{\lambda} = \frac{1}{2 \lambda} + \frac{9}{2 \lambda} + \frac{2}{\lambda} = \left( \frac{1}{2} + \frac{9}{2} + 2 \right) \frac{1}{\lambda} = \left(5 + 2\right) \frac{1}{\lambda} = \frac{7}{\lambda}.$$ 23. For maximum, pick $\lambda > 0$, so $$f_{max} = \frac{7}{\frac{\sqrt{14}}{6}} = 7 \cdot \frac{6}{\sqrt{14}} = \frac{42}{\sqrt{14}}.$$ Answer: (C) $\frac{42}{\sqrt{14}}$. --- 24. **Problem:** Determine possible values for $$d = \dim(\ker(A \circ B)),$$ where $A,B: \mathbb{R}^{12} \to \mathbb{R}^{12}$, $\dim(\ker A) = 3$, $\dim(\ker B) = 5$. 25. Using the Rank-Nullity theorem: $$\dim(\text{Im } B) = 12 - 5 = 7.$$ 26. Since $A$ acts on $\mathbb{R}^{12}$, $$\dim(\ker A) = 3 \implies \text{rank}(A) = 9.$$ 27. The null space of $A \circ B$ contains vectors $v$ with $B(v) \in \ker A$, or $B(v)=0$. So $$\ker(A \circ B) = \{ v : B(v) \in \ker A \} = B^{-1}(\ker A).$$ 28. Then, $$\dim(\ker(A \circ B)) = \dim \ker B + \dim (\text{preimage of } \ker A \text{ in } \text{Im } B).$$ 29. More precisely, since $B : \mathbb{R}^{12} \to \mathbb{R}^{12}$, the dimension of $\ker(A \circ B)$ satisfies $$\dim(\ker B) \leq d \leq \dim(\ker B) + \dim(\ker A)$$ because $\ker B \subseteq \ker(A \circ B)$ and the extra dimension can come from vectors mapped by $B$ into $\ker A$. 30. Here, $$5 \leq d \leq 8.$$ Answer: (D) $5 \leq d \leq 8$.